{ "arguments": [ { "citations": [ "ap-17-549" ], "experiments": [ "bm-01" ], "help": { "example": "random-walks", "missingStep": "orders-of-magnitude", "why": "temperature-thermal-energy" }, "id": "arg-bm-introduction", "kind": "argument", "limitations": [ "The introduction states what the later sections derive; it derives nothing itself.", "Einstein does not identify his motion with the Brownian motion; he says the reports available to him were too imprecise for a judgment.", "Both consequences are conditional. The passage does not say the motion had been observed with its laws, nor that the existence of molecules was settled in 1905." ], "meaning": { "executionStatus": "static-illustration", "historicalStatus": "pedagogical-reconstruction", "logicalRole": "qualification", "modelStatus": "exact-within-model" }, "paper": "brownian-motion", "premises": [ "The molecular-kinetic theory of heat: the heat of a liquid is the irregular motion of its molecules.", "A body suspended in the liquid, large enough to see in a microscope, is struck by those molecules." ], "prerequisites": [], "question": "What does Einstein say the molecular-kinetic theory requires of suspended bodies, how sure is he that this is the Brownian motion, and what would observing it decide?", "readings": { "full": [ { "kind": "paragraph", "text": "The paper opens with a claim about what a theory requires. The molecular-kinetic theory of heat takes the heat of a liquid to be the irregular motion of its molecules. From that theory, Einstein announces, it will be shown that bodies suspended in a liquid and large enough to see in a microscope must, because of the molecules' thermal motion, move by amounts large enough to be detected easily with a microscope." }, { "id": "temperature-thermal-energy", "kind": "foundation", "returnCaption": "Return to the claim that heat is molecular motion." }, { "kind": "paragraph", "text": "He does not claim to have explained a motion already seen. The motions to be treated, he writes, may be the same as the so-called Brownian molecular motion; but the information he could obtain about it was 'so ungenau, daß ich mir hierüber kein Urteil bilden konnte': so imprecise that he could form no judgment. The identification is left open. The paper predicts a motion and derives its laws; whether the reported motion is that one is a question for observation." }, { "kind": "paragraph", "text": "The second paragraph says what depends on the prediction, in both directions. If the motion, together with the laws the paper expects it to follow, can really be observed, then classical thermodynamics can no longer be regarded as exactly valid even for spaces that can be told apart in a microscope, and an exact determination of the true size of atoms becomes possible. If instead the prediction proves false, that would be a weighty argument against the molecular-kinetic view of heat." }, { "kind": "paragraph", "text": "Why would a visible motion limit thermodynamics? Classical thermodynamics describes a liquid in equilibrium at one temperature by a few quantities, such as its pressure and temperature, which then stay fixed. It has no place for a part of the liquid that keeps moving of its own accord, now one way and now another. On the molecular view such restless departures from the average are always present. For a body of ordinary size they are far too small to notice; the paper argues that for a grain about a thousandth of a millimetre across, the size §5 uses, they are large enough to see." }, { "kind": "paragraph", "text": "Why would it give the size of atoms? §§4 and 5 tie how far such a grain wanders in a given time to N, the number of real molecules in a gram-molecule. With N known, the mass of a single molecule is the mass of a gram-molecule divided by N, and its size can be estimated from the volume a gram-molecule fills. The introduction announces this; the later sections carry it out." }, { "id": "orders-of-magnitude", "kind": "foundation", "returnCaption": "Return to why a grain's motion could give the size of atoms." }, { "kind": "paragraph", "text": "The road there: §1 argues that suspended bodies should exert osmotic pressure just as dissolved molecules do; §2 derives that from the molecular-kinetic theory; §3 turns it into a diffusion coefficient; §§4 and 5 give the spread of a grain over time and the displacement to look for." } ], "margin": [ { "kind": "paragraph", "text": "Robert Brown described the irregular motion of small particles suspended in water in 1828, from observations made in 1827. Later students of the motion included Christian Wiener (1863) and Louis Georges Gouy, who argued in 1888 (Journal de Physique (2) 7, p. 561) that it comes from the thermal agitation of the liquid; Carl Nägeli had argued in 1879 that single molecular impacts are far too weak to move such particles visibly. The 1905 paper cites none of this literature, which fits the introduction's statement that the information available to Einstein was too imprecise for a judgment." }, { "kind": "paragraph", "text": "Einstein's next paper on the subject, 'Zur Theorie der Brownschen Bewegung' (Ann. d. Phys. 19, p. 371, 1906), opens by saying that after this paper appeared, Siedentopf told him that Gouy and other physicists had become convinced by direct observation that the Brownian motion comes from the irregular thermal motion of the liquid's molecules." }, { "kind": "paragraph", "text": "In 1905 the molecular-kinetic theory was in wide use, and it also had critics who treated atoms as a useful hypothesis rather than an established fact. The second paragraph speaks to that disagreement by naming an observation that could count against the theory as well as for it. Jean Perrin's measurements of 1908 and 1909 on suspended grains are later evidence, dated after this paper; they do not appear in it." }, { "kind": "paragraph", "text": "A modern lens: the motion the introduction predicts is what is now called a thermal fluctuation, and the spaces that can be told apart in a microscope are where such fluctuations become visible. The term is later usage; the paper speaks of classical thermodynamics ceasing to hold exactly." } ], "overview": [ { "kind": "paragraph", "text": "On the molecular-kinetic theory of heat, Einstein writes, bodies just visible in a microscope must move enough to be seen, and this may be the Brownian motion, though the reports he could obtain were too imprecise for him to judge. If the motion and its laws are observed, classical thermodynamics is not exactly valid for spaces a microscope can resolve and the true size of atoms can be determined; if the prediction fails, that is a weighty argument against the molecular-kinetic view of heat." } ], "steps": [ { "kind": "paragraph", "text": "Words first. The molecular-kinetic theory of heat is the view that heat is the motion of molecules too small to see. A body suspended in a liquid floats in it without dissolving, like a fine grain of powder stirred into water. The paper's own example, in §5, is a grain a thousandth of a millimetre across: far too small to see by eye, easy to see in a microscope. A gram-molecule is the amount of a substance whose mass in grams equals its molecular weight, what is now called a mole." }, { "items": [ "The claim. If the molecular-kinetic theory is right, the molecules of a liquid are always moving and always striking any grain suspended in it. Einstein announces that the paper will show that, for grains big enough to see in a microscope, these impacts must move the grain by amounts a microscope can detect easily.", "The word 'must' belongs to the theory: the motion is required if the theory holds. The paper does not report having watched it.", "The identification, left open. A jittering of small particles in liquids had been reported and was called the Brownian motion, after the botanist Robert Brown. Einstein writes that his motions may be the same, but that the information he could obtain about it was so imprecise ('so ungenau') that he could form no judgment. So the paper predicts a motion; whether the reported jitter is that motion is for observation to say.", "The stakes, first direction. Suppose the motion, and the laws the paper derives for it, are observed. Then classical thermodynamics, which describes a liquid at one temperature by a few steady quantities such as its pressure and temperature, is not exactly valid for regions as small as a microscope can resolve. And the true size of atoms can be determined exactly." ], "kind": "steps" }, { "id": "temperature-thermal-energy", "kind": "foundation", "returnCaption": "Return to the claim that heat is molecular motion." }, { "items": [ "Why thermodynamics would fail there. A grain that keeps moving, now left, now right, in a liquid at one temperature is an unevenness that those steady quantities do not describe. On the molecular view such unevenness is always present. For an ordinary body it is far too small to notice; for a grain a thousandth of a millimetre across, the paper argues, it is large enough to see.", "Why atoms could be measured. §§4 and 5 connect the grain's wandering over a given time to N, the number of real molecules in a gram-molecule, what is now called Avogadro's number. A gram-molecule of a substance is a known mass of it, so once N is known, one molecule's mass is that mass divided by N, and its size can be estimated from the volume a gram-molecule fills.", "The stakes, second direction. Suppose the predicted motion is looked for carefully and is not there. Einstein says that would be a weighty argument against the molecular-kinetic view of heat. He calls it an argument, and a weighty one: a failed prediction would count heavily against the theory.", "What is not said. The introduction does not claim that the existence of molecules had been settled, that the motion had been measured with these laws, or that Einstein had studied the Brownian reports in detail. Each consequence is conditional: if the motion is observed, then one thing follows; if the prediction fails, then another.", "The plan of the paper. §1 argues that suspended bodies should exert osmotic pressure just as dissolved molecules do. §2 derives that from the molecular-kinetic theory. §3 turns it into a diffusion coefficient. §§4 and 5 give the spread of a grain over time and the displacement a microscope should show." ], "kind": "steps" }, { "id": "orders-of-magnitude", "kind": "foundation", "returnCaption": "Return to why a grain's motion could give the size of atoms." } ] }, "recap": "On the molecular-kinetic theory of heat, bodies just large enough to see in a microscope must move by amounts a microscope can detect. Einstein thinks this may be the Brownian motion, but says the reports he could obtain were too imprecise for him to judge. Observing the motion and its laws would mean classical thermodynamics is not exactly valid for spaces a microscope can resolve and would allow the size of atoms to be determined; a failed prediction would weigh heavily against the molecular view.", "review": "draft", "schemaVersion": 1, "section": "s0", "title": "What the paper sets out to show, and what would decide it" }, { "citations": [ "ap-17-549" ], "experiments": [ "bm-02" ], "help": { "example": "temperature-thermal-energy", "missingStep": "ratios-scaling", "why": "free-energy-osmotic-pressure" }, "id": "arg-bm-osmotic-suspended", "kind": "argument", "limitations": [ "Gravity is set aside, and so are the energy and entropy of the surfaces between bodies and liquid (capillary forces), on the assumption that the moves considered do not change those surfaces.", "§1 states the molecular-kinetic expectation; §2 derives it from the theory. Neither section says which view nature follows; that is for observation.", "The law is for great dilution. Crowded bodies, or bodies that act on one another, depart from it." ], "meaning": { "executionStatus": "static-illustration", "historicalStatus": "pedagogical-reconstruction", "logicalRole": "heuristic-inference", "modelStatus": "approximation" }, "paper": "brownian-motion", "premises": [ "Van 't Hoff's law for a dilute solution of a non-electrolyte held behind a wall that passes the solvent only: pV* = RTz, for large enough V*/z.", "On the molecular-kinetic theory of heat, a dissolved molecule differs from a suspended body only in size.", "The suspended bodies are few enough that neighbouring bodies are far apart." ], "prerequisites": [], "question": "Should small bodies suspended in a liquid press on a wall that holds them back, as dissolved molecules do, and what would tell the two expectations apart?", "readings": { "full": [ { "kind": "paragraph", "text": "Take a liquid of total volume V. In part of it, a volume \\(V^*\\), dissolve z gram-molecules of a non-electrolyte, a substance that does not split into ions, and separate \\(V^*\\) from the pure solvent by a wall that lets the solvent through but not the dissolved substance. The dissolved molecules push on that wall. The push on each unit of its area is the osmotic pressure p, and when \\(V^*/z\\) is large enough, that is, when the solution is dilute, it obeys the law van 't Hoff found, which has the form of the gas law:" }, { "kind": "formula", "latex": "pV^*=RTz", "spoken": "p times V star equals R times T times z." }, { "kind": "paragraph", "text": "R is the gas constant and T the absolute temperature. Pressure is force per unit area: the force on the whole wall is p times the wall's area." }, { "kind": "paragraph", "text": "Now put small bodies suspended in the liquid into \\(V^*\\) in place of the dissolved substance, bodies that also cannot pass through the wall. What does classical thermodynamics expect? Einstein states its answer together with its reason. At a fixed temperature the force on the wall follows from how the free energy of the system changes when the wall is moved. On the usual view the free energy depends on the total masses and kinds of the suspended substance, the liquid and the wall, and on pressure and temperature, but not on where the wall and the suspended bodies are. If moving the wall does not change the free energy, the wall feels no force. So, gravity aside, classical thermodynamics does not expect the suspended bodies to exert any force on the wall." }, { "id": "free-energy-osmotic-pressure", "kind": "foundation", "returnCaption": "Return to the classical expectation for suspended bodies." }, { "kind": "paragraph", "text": "Einstein sets two effects aside so that the comparison is fair. Gravity, which would pull the bodies down, does not concern him here. The energy and entropy of the surfaces where the bodies meet the liquid (capillary forces) would also enter the free energy, but he assumes the moves considered do not change the size or nature of those surfaces, so they drop out." }, { "kind": "paragraph", "text": "The classical expectation is a coherent position, and for ordinary bodies it agrees with experience: a few pebbles held behind a sieve do not push on it measurably. The question is whether it still holds for bodies small enough to be jostled by the molecules of the liquid." }, { "kind": "paragraph", "text": "The molecular-kinetic theory of heat reaches a different view. On it, a dissolved molecule differs from a suspended body only in size (Einstein sets 'lediglich', only, in italics), and there is no reason why a number of suspended bodies should not give the same osmotic pressure as the same number of dissolved molecules. Jostled by the molecular motion of the liquid, the suspended bodies must perform an irregular motion in it, however slow; if the wall keeps them from leaving \\(V^*\\), they exert forces on it, just as dissolved molecules do." }, { "kind": "paragraph", "text": "With n suspended bodies in \\(V^*\\), so that there are \\(\\nu = n/V^*\\) of them in each unit of volume, and with neighbouring bodies far enough apart, the osmotic pressure should be" }, { "kind": "formula", "latex": "p=\\frac{RT}{V^*}\\frac{n}{N}=\\frac{RT}{N}\\,\\nu", "spoken": "p equals R T over V star, times n over N, which equals R T over N, times nu." }, { "kind": "paragraph", "text": "where N is the number of real molecules in a gram-molecule. The first form is van 't Hoff's law with \\(z = n/N\\) gram-molecules; the second says that the pressure depends on the number of bodies per unit volume and on the temperature, and not on their size or mass." }, { "id": "temperature-thermal-energy", "kind": "foundation", "returnCaption": "Return to where the pressure depends on temperature and number, not size." }, { "kind": "paragraph", "text": "The two views disagree about something that can be looked for. If suspended bodies exert osmotic pressure, then wherever their number per unit volume varies, so does the pressure, and it pushes them toward thinner regions; §3 shows that this appears as diffusion, and §§4 and 5 predict how far a grain wanders in a given time. On the classical expectation there is no osmotic pressure to drive such a spread. Observing the predicted wandering, with the predicted size, would count for the molecular-kinetic view, and its absence against it. First, §2 shows that the molecular-kinetic theory really leads to the extended law." } ], "margin": [ { "kind": "paragraph", "text": "Van 't Hoff stated the law for dilute solutions in 1887 (Zeitschrift für physikalische Chemie 1, p. 481), drawing on Wilhelm Pfeffer's measurements of 1877 with membranes that pass water but not sugar (Osmotische Untersuchungen). The paper takes the law as known and cites neither." }, { "kind": "paragraph", "text": "The expectation Einstein attributes to classical thermodynamics is his own characterization of 'the usual view' (der üblichen Auffassung); the paper names no one who held it. The paper does not call that view mistaken. It says the molecular-kinetic theory leads to a different one, and §2 shows how." }, { "kind": "paragraph", "text": "A modern lens: R/N is what is now written \\(k_B\\), Boltzmann's constant, so the extended law reads \\(p = \\nu k_B T\\), the ideal-gas law for particles of any size. The paper's N is what is now called Avogadro's number; since 2019 it has an exact defined value, and in 1905 it was known only roughly. §5 of this paper proposes a way to measure it." }, { "kind": "paragraph", "text": "Later evidence: Jean Perrin's measurements of 1908 and 1909 found that the grains of a suspension at rest thin out with height as a gas of very heavy molecules would, which is what the extended law predicts once gravity is put back. That is later evidence, dated after this paper, and not a premise of it." } ], "overview": [ { "kind": "paragraph", "text": "A dissolved substance held behind a wall that passes only the solvent presses on it with the osmotic pressure. Classical thermodynamics, as Einstein characterizes it, expects small suspended bodies to exert no such force, because the free energy seems not to depend on where they are; on the molecular-kinetic view they differ from dissolved molecules only in size, so equally many of either, far enough apart, should press alike." } ], "steps": [ { "kind": "paragraph", "text": "Letters, as the paper prints them. V is the total volume of the liquid and \\(V^*\\) the part of it behind the wall. z is the number of gram-molecules dissolved in \\(V^*\\); a gram-molecule is what is now called a mole. p is the osmotic pressure, R the gas constant, T the absolute temperature. n is the number of suspended bodies, a count. ν is their number per unit volume, \\(\\nu = n/V^*\\), and not a frequency. N is the number of real molecules in a gram-molecule, now called Avogadro's number. Three letters, three different numbers: n counts the bodies, ν counts them per unit volume, N counts the molecules in a gram-molecule." }, { "items": [ "The wall. It is semipermeable: the solvent, say water, passes through it; the dissolved substance does not. The solution is on one side, pure solvent on the other.", "Osmotic pressure. The dissolved molecules push on the wall. The push on each unit of area is the osmotic pressure p. Pressure is force divided by area, so a wall of area A feels the force \\(pA\\).", "Van 't Hoff's law. For a dilute solution, one in which \\(V^*/z\\) is large, the pressure obeys the same law as an ideal gas of as many molecules in the same volume:" ], "kind": "steps" }, { "kind": "formula", "latex": "pV^*=RTz", "spoken": "p times V star equals R times T times z." }, { "items": [ "Reading it. Doubling the amount dissolved, z, doubles p. Doubling the volume \\(V^*\\) at fixed z halves p. Raising the absolute temperature T raises p in proportion.", "A worked case, with today's value of the gas constant, \\(R \\approx 8.314\\) joule per kelvin per gram-molecule: one gram-molecule in 22.4 litres at 273 K gives \\(p = RTz/V^* \\approx 8.314 \\times 273 / 0.0224 \\approx 101\\,300\\) pascal, about one atmosphere, the pressure one gram-molecule of gas exerts in that volume at 0 °C.", "Classical thermodynamics, for suspended bodies. Replace the dissolved substance by small bodies that float in the liquid and also cannot pass the wall. At a fixed temperature the force on the wall is found from the free energy F of the whole system: if moving the wall a little changes F, there is a force; if not, there is none.", "On the usual view, F depends on the total masses and kinds of the suspended substance, the liquid and the wall, and on pressure and temperature. It does not depend on where the wall is, or where the bodies are. So moving the wall leaves F unchanged, and classical thermodynamics expects no force on the wall from the suspended bodies.", "What is set aside. Gravity, which would make the bodies sink. And the energy and entropy of the surfaces between bodies and liquid, the capillary forces, removed by assuming that the moves considered do not change those surfaces. Without these set-asides the classical expectation would carry extra forces that have nothing to do with the question." ], "kind": "steps" }, { "id": "free-energy-osmotic-pressure", "kind": "foundation", "returnCaption": "Return to the classical expectation for suspended bodies." }, { "items": [ "Why the classical expectation is reasonable. For ordinary bodies it matches experience: a few pebbles held behind a sieve do not press on it measurably. It is a coherent position, and §1 treats it as one: the question it asks is whether the expectation still holds for bodies small enough to be jostled by molecules.", "The molecular-kinetic view. A dissolved molecule and a suspended body differ only in size. Both are struck by the liquid's molecules and move irregularly, the body far more slowly. A body kept in by the wall strikes it now and then, and so pushes on it, as a dissolved molecule does.", "So equally many bodies and molecules should give the same pressure. Take n bodies in \\(V^*\\). They amount to \\(z = n/N\\) gram-molecules' worth of particles, because a gram-molecule holds N particles. Put that z into van 't Hoff's law and divide by \\(V^*\\); since \\(\\nu = n/V^*\\), the last step replaces \\(n/V^*\\) by ν:" ], "kind": "steps" }, { "kind": "formula", "latex": "p=\\frac{RT}{V^*}\\frac{n}{N}=\\frac{RT}{N}\\,\\nu", "spoken": "p equals R T over V star, times n over N, which equals R T over N, times nu." }, { "items": [ "Reading the result. The pressure is RT/N times ν, the number of bodies per unit volume. Their size and mass do not appear. A body a thousand times bigger than a molecule presses no harder, as long as there are equally many per unit volume.", "The condition. Neighbouring bodies must be far enough apart, the same condition as a dilute solution. Crowded bodies, or bodies that act on one another, depart from the law.", "For scale, with today's value of R/N, \\(1.38\\times10^{-23}\\) joule per kelvin, at T = 290 K, the 17 °C of the paper's §5 example, the factor RT/N is about \\(4.0\\times10^{-21}\\) joule. A suspension with \\(10^{16}\\) grains per cubic metre, ten million per cubic millimetre, would press with about \\(4\\times10^{-5}\\) pascal: far too little to measure directly. The paper's test lies in the motion, in §§4 and 5." ], "kind": "steps" }, { "id": "temperature-thermal-energy", "kind": "foundation", "returnCaption": "Return to where the pressure depends on temperature and number, not size." }, { "items": [ "What tells the views apart. On the molecular-kinetic view, wherever the number per unit volume varies, the pressure varies, and it pushes the bodies toward thinner regions: §3 shows that this is diffusion, and §§4 and 5 predict how far a grain wanders in a given time. On the classical expectation there is no osmotic pressure to drive such a spread.", "So observing the predicted wandering, with its predicted size, would count for the molecular-kinetic view, and its absence against it. §1 only states the expectation; §2 shows that the molecular-kinetic theory really leads to it." ], "kind": "steps" } ] }, "recap": "Classical thermodynamics, as Einstein characterizes it, expects no force on the wall from suspended bodies, because the free energy seems not to depend on where they are. On the molecular-kinetic view a dissolved molecule differs from a suspended body only in size, so equally many of either, far enough apart, should exert the same osmotic pressure, RT/N times their number per unit volume.", "review": "draft", "schemaVersion": 1, "section": "s1", "title": "Why a suspended grain should press like a dissolved molecule" }, { "citations": [ "ap-17-549" ], "experiments": [ "bm-03" ], "help": { "example": "probability-independence", "missingStep": "logarithms", "why": "entropy-multiplicity" }, "id": "arg-bm-kinetic-justification", "kind": "argument", "limitations": [ "The footnote to the heading says this section, and Einstein's earlier papers on the foundations of thermodynamics, are not needed to understand the paper's results. The passage explains the section for the reader who wants to know how the law is obtained.", "The argument finds only how the free energy depends on V*; it computes nothing else about the integral B.", "Independence, homogeneity and the absence of forces are assumptions. With particles that act on one another, or crowded ones, J would depend on their positions and the pressure would depart from the dilute law." ], "meaning": { "executionStatus": "static-illustration", "historicalStatus": "pedagogical-reconstruction", "logicalRole": "derivation", "modelStatus": "approximation" }, "paper": "brownian-motion", "premises": [ "Einstein's statistical theory of heat of 1902 and 1903: for a system whose state variables change by equations whose rates satisfy the sum condition of §2, the entropy and the free energy are given by the logarithm of an integral over all its states.", "The particles move independently of one another to a sufficient approximation, the liquid is homogeneous, and no forces act on the particles.", "The total volume of the particles is small compared with the volume V* that holds them." ], "prerequisites": [ { "edge": "cross-reference", "id": "arg-bm-osmotic-suspended" } ], "question": "How can a theory of countless moving molecules give the osmotic pressure of dissolved molecules and suspended bodies without anyone solving their motion?", "readings": { "full": [ { "kind": "paragraph", "text": "A footnote to the heading says what this section assumes and what it is for. It takes as known Einstein's papers on the foundations of thermodynamics (Ann. d. Phys. 9, p. 417, 1902; 11, p. 170, 1903), and says that neither those papers nor this section is needed to understand the results of the present paper. The section is still worth reading, because it answers the question §1 leaves open: how can a theory of countless colliding molecules give a law as simple as van 't Hoff's, for dissolved molecules and suspended bodies alike, without anyone solving their motions?" }, { "kind": "paragraph", "text": "Einstein describes the whole system, liquid, wall and particles, by state variables \\(p_1, \\ldots, p_l\\) that fix its momentary state completely, for example the coordinates and velocity components of all its atoms. How they change in time is given by equations of the form" }, { "kind": "formula", "latex": "\\frac{\\partial p_\\nu}{\\partial t}=\\varphi_\\nu(p_1,\\ldots,p_l)", "spoken": "The rate of change of p nu with time equals phi nu, a function of p 1 through p l." }, { "kind": "paragraph", "text": "with the condition \\(\\sum \\frac{\\partial \\varphi_\\nu}{\\partial p_\\nu} = 0\\). Two of these letters are used elsewhere in the paper for other things: these \\(p_\\nu\\) are not the osmotic pressure p of §1, and these \\(\\varphi_\\nu\\), the rates of change of the state variables, are not the φ of §4." }, { "kind": "paragraph", "text": "For such a system his earlier theory gives the entropy S as an expression containing the logarithm, printed lg and meaning the natural logarithm, of an integral taken over every combination of the state variables that the conditions of the problem allow. In it T is the absolute temperature, Ē (printed with a bar) the energy of the system, and E the energy as a function of the \\(p_\\nu\\). The constant is printed as 2κ, and Einstein ties κ to N by 2κN = R, so 2κ is R/N. For the free energy F he obtains" }, { "kind": "formula", "latex": "F=-\\frac{R}{N}T\\lg\\int e^{-\\frac{EN}{RT}}\\,dp_1\\ldots dp_l=-\\frac{RT}{N}\\lg B", "spoken": "F equals minus R over N times T times the logarithm of the integral of e to the minus E N over R T, over d p 1 through d p l, which equals minus R T over N times the logarithm of B." }, { "kind": "paragraph", "text": "and the integral is named B." }, { "id": "logarithms", "kind": "foundation", "returnCaption": "Return to the free energy as a logarithm of the integral B." }, { "kind": "paragraph", "text": "Even if the molecular picture were fixed in every detail, Einstein says, computing B would be so hard that an exact calculation of F is hardly conceivable. But the pressure needs only how F depends on the volume \\(V^*\\) in which all the particles are held. (Particles, 'Teilchen', is his short word for dissolved molecules and suspended bodies alike.)" }, { "kind": "paragraph", "text": "Put n particles in \\(V^*\\), held there by a semipermeable wall, their total volume small compared with \\(V^*\\). Where the wall stands limits the range of the integral B. Name the coordinates of the particles' centres of gravity \\(x_1, y_1, z_1\\) through \\(x_n, y_n, z_n\\), give each centre a tiny box inside \\(V^*\\), and ask for the part of B that comes from states with every centre in its box. It has the form" }, { "kind": "formula", "latex": "dB=dx_1\\,dy_1\\ldots dz_n\\cdot J", "spoken": "d B equals d x 1, d y 1, and so on up to d z n, times J." }, { "kind": "paragraph", "text": "where the factor J does not depend on the box sizes, nor on \\(V^*\\), that is, on where the wall is. J does not depend on where the boxes are either. Take a second set of boxes, of the same sizes, in other places inside \\(V^*\\); its part of B is \\(dB'\\) with a factor \\(J'\\). Since the sizes are equal," }, { "kind": "formula", "latex": "\\frac{dB}{dB'}=\\frac{J}{J'}", "spoken": "d B over d B prime equals J over J prime." }, { "kind": "paragraph", "text": "Einstein's earlier theory gives these parts a meaning: dB/B is the probability that, at a moment chosen at random, the centres are in the given boxes. If the particles move independently of one another, to a sufficient approximation, the liquid is homogeneous and no forces act on the particles, then equal boxes are equally probable wherever they are, so" }, { "kind": "formula", "latex": "\\frac{dB}{B}=\\frac{dB'}{B}", "spoken": "d B over B equals d B prime over B." }, { "kind": "paragraph", "text": "and with the previous equation, \\(J = J'\\)." }, { "id": "probability-independence", "kind": "foundation", "returnCaption": "Return to why equal boxes are equally probable." }, { "kind": "paragraph", "text": "So J depends neither on \\(V^*\\) nor on where the particles are. Integrating over all positions of the n centres, each ranging over the volume \\(V^*\\), gives" }, { "kind": "formula", "latex": "B=\\int J\\,dx_1\\ldots dz_n=JV^{*n}", "spoken": "B equals the integral of J over d x 1 through d z n, which equals J times V star to the power n." }, { "kind": "paragraph", "text": "and so the free energy is" }, { "kind": "formula", "latex": "F=-\\frac{RT}{N}\\left\\{\\lg J+n\\lg V^*\\right\\}", "spoken": "F equals minus R T over N, times the logarithm of J plus n times the logarithm of V star." }, { "kind": "paragraph", "text": "The pressure on the wall is minus the rate at which F changes as \\(V^*\\) grows:" }, { "kind": "formula", "latex": "p=-\\frac{\\partial F}{\\partial V^*}=\\frac{RT}{V^*}\\frac{n}{N}=\\frac{RT}{N}\\,\\nu", "spoken": "p equals minus the partial derivative of F with respect to V star, which equals R T over V star times n over N, which equals R T over N times nu." }, { "id": "partial-derivatives", "kind": "foundation", "returnCaption": "Return to the pressure as the change of free energy with volume." }, { "kind": "paragraph", "text": "This shows, Einstein concludes, that osmotic pressure is a consequence of the molecular-kinetic theory of heat, and that on this theory equal numbers of dissolved molecules and suspended bodies behave exactly alike as regards osmotic pressure at great dilution. The question of how the theory avoids solving every molecular motion has a plain answer: the hard part of B, the factor J, is never computed. Only its independence of \\(V^*\\) is needed, and that follows from equal boxes being equally probable. The volume enters only through \\(V^{*n}\\), and its logarithm, \\(n\\lg V^*\\), gives the pressure." } ], "margin": [ { "kind": "paragraph", "text": "The footnote cites Einstein's 'Kinetische Theorie des Wärmegleichgewichtes und des zweiten Hauptsatzes der Thermodynamik' (Ann. d. Phys. 9, p. 417, 1902) and 'Eine Theorie der Grundlagen der Thermodynamik' (11, p. 170, 1903); the second footnote cites the 1903 paper for the probability reading of dB/B. A third paper of the series, 'Zur allgemeinen molekularen Theorie der Wärme' (14, p. 354, 1904), is not cited here. J. Willard Gibbs set out a closely related formulation in Elementary Principles in Statistical Mechanics (1902)." }, { "kind": "paragraph", "text": "A modern lens: the condition \\(\\sum \\frac{\\partial \\varphi_\\nu}{\\partial p_\\nu} = 0\\) is the property later texts discuss under the name of Liouville's theorem; Einstein does not use the name. In modern notation 2κ = R/N is \\(k_B\\), B is, up to a constant factor, the classical partition function, and \\(F = -\\frac{RT}{N}\\lg B\\) is \\(F = -k_BT\\ln Z\\)." }, { "kind": "paragraph", "text": "A modern lens on the limits: for particles that act on one another, J depends on their positions, and the osmotic pressure picks up corrections in powers of the number density, the virial expansion of later statistical mechanics. §2 excludes them by its stated assumptions, and §1's condition that neighbours be far apart says the same." } ], "overview": [ { "kind": "paragraph", "text": "Einstein's statistical theory of heat gives the free energy as a logarithm of an integral over all the states of the system. He cannot compute that integral, but he needs only how it changes with the volume the particles are held in, and if they move independently in a uniform liquid with no forces on them, that dependence gives the osmotic pressure (RT/N)ν for dissolved molecules and suspended bodies alike. A footnote says the section is not needed for the paper's results." } ], "steps": [ { "kind": "paragraph", "text": "Letters, as the paper prints them. \\(p_1, \\ldots, p_l\\) are the state variables, l of them, which fix the state of the whole system: for example every atom's coordinates and velocity components. They are not the osmotic pressure p. \\(\\varphi_\\nu\\) is the rate at which \\(p_\\nu\\) changes; it is not the φ of §4. ∂ marks a partial derivative, a rate of change with the other variables held fixed, and Σ a sum. T is the absolute temperature. Ē, printed with a bar, is the energy of the system; E is the energy as a function of the state variables. κ is a constant with 2κN = R, where R is the gas constant and N the number of real molecules in a gram-molecule, so 2κ is R/N. lg is the natural logarithm, today written ln. S is the entropy and F the free energy. B is an integral over all states. \\(V^*\\) is the volume holding the particles, n their number, and \\(x_i, y_i, z_i\\) the coordinates of the centre of the i-th particle." }, { "items": [ "The footnote first. The section takes Einstein's 1902 and 1903 papers as known, and says that neither they nor this section is needed to understand the paper's results. A reader may skip to §3 and lose none of them. What the section adds is an answer to a question: how can a theory of countless colliding molecules give van 't Hoff's simple law, for molecules and suspended bodies alike, without anyone solving their motions?", "The system. Everything, liquid, wall and particles, is described by state variables \\(p_1, \\ldots, p_l\\). How each one changes in time is given by an equation:" ], "kind": "steps" }, { "kind": "formula", "latex": "\\frac{\\partial p_\\nu}{\\partial t}=\\varphi_\\nu(p_1,\\ldots,p_l)", "spoken": "The rate of change of p nu with time equals phi nu, a function of p 1 through p l." }, { "items": [ "The condition \\(\\sum \\frac{\\partial \\varphi_\\nu}{\\partial p_\\nu} = 0\\) says that the motion neither crowds the possible states together nor spreads them apart; the equations of mechanics have this property when the forces do not depend on the velocities. Einstein's earlier theory needs it.", "Entropy and free energy. The earlier theory gives the entropy S through the logarithm of an integral of \\(e^{-EN/RT}\\) over every combination of the state variables that the conditions allow. States of higher energy E count for less, by that exponential factor. Writing 2κ as R/N, the free energy is:" ], "kind": "steps" }, { "kind": "formula", "latex": "F=-\\frac{R}{N}T\\lg\\int e^{-\\frac{EN}{RT}}\\,dp_1\\ldots dp_l=-\\frac{RT}{N}\\lg B", "spoken": "F equals minus R over N times T times the logarithm of the integral of e to the minus E N over R T, over d p 1 through d p l, which equals minus R T over N times the logarithm of B." }, { "items": [ "The integral is named B. So F is minus RT/N times the logarithm of B: whatever makes B larger makes F smaller.", "Hopeless in full, Einstein says: even with the molecular picture fixed in every detail, computing B would be too hard for an exact calculation of F.", "Not needed in full. The pressure on the wall is the change of F when the wall moves, that is, when \\(V^*\\) changes. So only how B depends on \\(V^*\\) is needed.", "The particles. n of them, dissolved molecules or suspended bodies, are held in \\(V^*\\) by a semipermeable wall, and they take up little of \\(V^*\\). The wall's position sets the limits of the integral B.", "Boxes. Give the centre of each particle a tiny box inside \\(V^*\\), of sides \\(dx_1, dy_1, dz_1\\) for the first particle, and so on. The part of B from states with every centre in its box is the product of the box sizes times a factor J:" ], "kind": "steps" }, { "kind": "formula", "latex": "dB=dx_1\\,dy_1\\ldots dz_n\\cdot J", "spoken": "d B equals d x 1, d y 1, and so on up to d z n, times J." }, { "items": [ "J collects everything else: the liquid's molecules, the velocities, the insides of the particles. It does not depend on the box sizes, nor on where the wall is.", "Moving the boxes. Choose a second set of boxes, the same sizes, in other places inside \\(V^*\\). Their part of B is \\(dB'\\), with its own factor \\(J'\\). The box sizes are the same, so dividing one form by the other leaves:" ], "kind": "steps" }, { "kind": "formula", "latex": "\\frac{dB}{dB'}=\\frac{J}{J'}", "spoken": "d B over d B prime equals J over J prime." }, { "items": [ "What the parts mean. dB/B is the probability that, at a moment chosen at random, every centre is in its box. The footnoted 1903 paper supplies this reading.", "The assumptions. The particles move independently of one another, to a sufficient approximation; the liquid is the same everywhere; no forces act on the particles. Then no place inside \\(V^*\\) is preferred, and boxes of equal size are equally probable wherever they are:" ], "kind": "steps" }, { "kind": "formula", "latex": "\\frac{dB}{B}=\\frac{dB'}{B}", "spoken": "d B over B equals d B prime over B." }, { "items": [ "So \\(J = J'\\): J does not depend on where the boxes are, and, from before, not on \\(V^*\\) either.", "Adding up the boxes. To get all of B, let each centre range over the whole of \\(V^*\\). Each centre contributes a factor \\(V^*\\), and there are n centres:" ], "kind": "steps" }, { "kind": "formula", "latex": "B=\\int J\\,dx_1\\ldots dz_n=JV^{*n}", "spoken": "B equals the integral of J over d x 1 through d z n, which equals J times V star to the power n." }, { "items": [ "A worked case. Two particles in a box: doubling the box doubles the room for each centre, so the number of ways to place both grows by 2 × 2 = 4. With 10 particles, doubling the volume multiplies B by \\(2^{10} = 1024\\). In general B grows as \\(V^*\\) to the power n.", "Taking the logarithm. The logarithm turns a product into a sum and a power into a multiple: \\(\\lg(JV^{*n}) = \\lg J + n\\lg V^*\\). So:" ], "kind": "steps" }, { "kind": "formula", "latex": "F=-\\frac{RT}{N}\\left\\{\\lg J+n\\lg V^*\\right\\}", "spoken": "F equals minus R T over N, times the logarithm of J plus n times the logarithm of V star." }, { "id": "logarithms", "kind": "foundation", "returnCaption": "Return to the free energy as a logarithm of the integral B." }, { "items": [ "The pressure. Pressure is minus the rate at which F changes with the volume. lg J does not change with \\(V^*\\). The rate of change of \\(n\\lg V^*\\) with \\(V^*\\) is n divided by \\(V^*\\). The two minus signs cancel, and \\(n/V^*\\) is ν, the number per unit volume:" ], "kind": "steps" }, { "kind": "formula", "latex": "p=-\\frac{\\partial F}{\\partial V^*}=\\frac{RT}{V^*}\\frac{n}{N}=\\frac{RT}{N}\\,\\nu", "spoken": "p equals minus the partial derivative of F with respect to V star, which equals R T over V star times n over N, which equals R T over N times nu." }, { "id": "partial-derivatives", "kind": "foundation", "returnCaption": "Return to the pressure as the change of free energy with volume." }, { "items": [ "The conclusion, as Einstein states it: osmotic pressure is a consequence of the molecular-kinetic theory of heat, and at great dilution equal numbers of dissolved molecules and suspended bodies behave exactly alike as regards osmotic pressure.", "How no molecular motion had to be solved: the factor J, which holds all the hard physics, was never computed. Its independence of \\(V^*\\) was enough, and that came from equal boxes being equally probable. Nothing about a particle's size or mass entered; only its number, the temperature and the volume did.", "Where it would fail: if the particles act on one another, or crowd each other, some placements are more probable than others, J depends on the positions, and the pressure departs from the dilute law. Those are exactly the assumptions Einstein states." ], "kind": "steps" }, { "id": "probability-independence", "kind": "foundation", "returnCaption": "Return to why equal boxes are equally probable." } ] }, "recap": "Einstein writes the free energy as F = -(RT/N) lg B, where B is an integral over every state of the system. He cannot compute B, but he needs only how it depends on the volume V* the particles are held in. If they move independently to a sufficient approximation, in a homogeneous liquid, with no forces on them, B is a factor J that does not depend on V* times V* to the power n, and the change of F with V* gives p = (RT/N)ν for dissolved molecules and suspended bodies alike.", "review": "draft", "schemaVersion": 1, "section": "s2", "title": "How molecular theory gives the osmotic law without solving the motion" }, { "citations": [ "ap-17-549" ], "experiments": [ "bm-01" ], "help": { "example": "random-walks", "missingStep": "bridge-squaring-square-roots", "why": "mean-variance-rms" }, "id": "arg-bm-observable", "kind": "argument", "limitations": [ "The model expectation need not equal the mean of one small sample.", "Net displacement is not total path length." ], "meaning": { "executionStatus": "static-illustration", "historicalStatus": "pedagogical-reconstruction", "logicalRole": "definition", "modelStatus": "exact-within-model" }, "paper": "brownian-motion", "premises": [ "All displacements use the same origin, axis, units and observation interval." ], "prerequisites": [], "question": "What can we measure when left and right cancel?", "readings": { "full": [ { "kind": "paragraph", "text": "A signed mean answers where the ensemble’s centre has moved. It does not answer how far its members have wandered. For a centred distribution, rightward and leftward contributions balance even while the distribution broadens." }, { "kind": "formula", "latex": "\\langle x\\rangle=0,\\qquad\\langle x^2\\rangle>0", "spoken": "The model’s mean displacement can be zero while its mean-square displacement is positive." }, { "kind": "paragraph", "text": "To retain the movement, square each displacement before averaging. Taking the square root of that mean square returns a length: the root-mean-square displacement, or RMS. A mean absolute displacement is a different observable, and neither is the length of a wandering trajectory." }, { "id": "mean-variance-rms", "kind": "foundation", "returnCaption": "Return to the measurement that still shows movement when left and right cancel." } ], "margin": [ { "kind": "paragraph", "text": "The four numbers are this page’s own example. Einstein’s observable is the mean square of the displacement along one axis, \\(\\lambda_x^2\\); its square root, \\(\\lambda_x\\), is the RMS used here." } ], "overview": [ { "kind": "paragraph", "text": "Left and right can cancel in the average while every tracer moves. Squaring before averaging keeps track of the spread." } ], "steps": [ { "id": "bridge-sum-average", "kind": "foundation", "returnCaption": "Return to adding and averaging the four displacements." }, { "id": "bridge-negative-numbers-direction", "kind": "foundation", "returnCaption": "Return to why leftward and rightward displacements can cancel." }, { "items": [ "Put every tracer’s starting point at zero and measure along one line, the \\(x\\)-axis, with right as positive and left as negative. After the same observation time, each tracer’s displacement \\(x\\) is a signed number: where it is now minus where it started.", "The angle brackets stand for an average over the whole ensemble of tracers the model describes: \\(\\langle x\\rangle\\) is the average of the signed displacement, and \\(\\langle x^2\\rangle\\) is the average of its square, the mean square.", "Take four displacements, in micrometres: −3, −1, +1, +3. As many went left as right, and by the same amounts, so they stand in for a centred distribution.", "Their sum is (−3) + (−1) + (+1) + (+3) = 0: the two leftward numbers cancel the two rightward ones. Divide by four: the signed mean is zero.", "The signed mean says where the centre of the group has moved, and here it has not moved at all. It says nothing about how far the members have spread out from that centre.", "Square each displacement first. A negative number times itself is positive, so the squares are 9, 1, 1, 9 square micrometres, and now nothing can cancel.", "The squares add to 20. Divide by four: the mean square is 5 square micrometres. It is positive although the signed mean was zero, which is the statement of the formula:" ], "kind": "steps" }, { "kind": "formula", "latex": "\\langle x\\rangle=0,\\qquad\\langle x^2\\rangle>0", "spoken": "The model’s mean displacement can be zero while its mean-square displacement is positive." }, { "items": [ "In the formula both averages belong to the model, over its whole ensemble. Four numbers drawn at random from such a model would give a signed mean near zero rather than exactly zero; these four were chosen to cancel exactly, so that the arithmetic stays easy to follow.", "The mean square is measured in square micrometres, an area, not a length. To return to a length, take the square root: \\(\\sqrt{5}\\) is approximately 2.236, so the root-mean-square displacement, or RMS, is about 2.236 micrometres.", "The mean distance, ignoring direction, is a third average: (3 + 1 + 1 + 3) / 4 = 2 micrometres, the mean absolute displacement. RMS and mean distance weigh the large displacements differently, so 2.236 and 2 answer different questions. They are not inconsistent answers.", "Neither average is the length of the path a tracer wandered along. A tracer that moves 2 micrometres right and then 1 micrometre left ends 1 micrometre from its start, having travelled 3. The squares record where the tracers ended up, not how far they walked on the way." ], "kind": "steps" }, { "id": "mean-variance-rms", "kind": "foundation", "returnCaption": "Return to the distinction between signed mean, mean square and RMS." } ] }, "recap": "Signed displacements can cancel. Their squares still record movement.", "review": "draft", "schemaVersion": 1, "section": "s4", "title": "Zero average is not no movement" }, { "citations": [ "ap-17-549" ], "experiments": [ "bm-05", "bm-01" ], "help": { "example": "random-walks", "missingStep": "random-walks", "why": "probability-independence" }, "id": "arg-bm-independent-steps", "kind": "argument", "limitations": [ "Correlated steps, bias or an infinite second moment change the argument.", "This pedagogical walk does not describe fixed physical jumps in a liquid." ], "meaning": { "executionStatus": "static-illustration", "historicalStatus": "pedagogical-reconstruction", "logicalRole": "derivation", "modelStatus": "exact-within-model" }, "paper": "brownian-motion", "premises": [ "Increments are independent, have zero mean, and share a finite mean square ℓ².", "One step corresponds to a declared interval τ." ], "prerequisites": [ { "edge": "premise", "id": "arg-bm-observable" } ], "question": "What permits us to add the contributions of many random steps?", "readings": { "full": [ { "kind": "paragraph", "text": "Write the displacement after n steps as the sum of their increments. Expanding its square exposes both the squared increments and their cross terms. Independence factors each expected cross term into the product of two means; centring makes that product zero." }, { "equations": [ "eq-model-bm-sum-of-steps" ], "kind": "formula", "latex": "\\left\\langle\\left(\\sum_{i=1}^n\\Delta_i\\right)^2\\right\\rangle=\\sum_{i=1}^n\\langle\\Delta_i^2\\rangle=nl^2", "spoken": "The mean square of the sum of independent centred increments is the sum of their mean squares." }, { "equations": [ "eq-model-bm-walk-time", "eq-model-bm-walk-diffusivity", "eq-model-bm-mean-square-growth" ], "kind": "formula", "latex": "t=n\\tau,\\qquad D=\\frac{l^2}{2\\tau},\\qquad\\langle x^2\\rangle=2Dt", "spoken": "Elapsed time is n tau; defining D as ell squared over two tau gives mean-square displacement two D t." }, { "id": "probability-independence", "kind": "foundation", "returnCaption": "Return to why the expected cross terms vanish." } ], "margin": [ { "kind": "paragraph", "text": "Einstein’s independence holds over his interval \\(\\tau\\), not at every instant: over very short times a particle’s velocity carries over from one moment to the next, which is why \\(\\tau\\) must not be too small. Paul Langevin’s equation of 1908, which keeps the particle’s inertia, shows where that crossover lies. A simulation that agrees with \\(2Dt\\) checks the arithmetic of the model, not the existence of molecules." } ], "overview": [ { "kind": "paragraph", "text": "If each step is independent of the last and as likely to go left as right, the mean square grows in proportion to time. Four times the time gives twice the RMS displacement, not four times." } ], "steps": [ { "id": "probability-independence", "kind": "foundation", "returnCaption": "Return to the independence and centring assumptions." }, { "items": [ "Name the pieces. A tracer takes \\(n\\) steps, one in each interval of time \\(\\tau\\). Its \\(i\\)-th step is a signed displacement \\(\\Delta_i\\), where \\(i\\) counts the steps 1, 2, 3 and so on up to \\(n\\). Where it ends, measured from where it started, is the sum of its steps: \\(x = \\Delta_1 + \\Delta_2 + \\ldots + \\Delta_n\\).", "The assumptions, stated once. Each step is centred: its average over many tracers is zero, \\(\\langle\\Delta_i\\rangle = 0\\), so on average it goes neither left nor right. Every step has the same mean square, written \\(l^2\\), so \\(l\\) is the root-mean-square length of one step. And the steps are independent: knowing one step tells you nothing about any other.", "Start with two steps and call them \\(A\\) and \\(B\\). The square of their sum is found by multiplying it out, \\((A+B)(A+B)\\), which gives four products, two of which are the same:" ], "kind": "steps" }, { "equations": [ "eq-model-bm-square-of-sum" ], "kind": "formula", "latex": "(A+B)^2=A^2+2AB+B^2", "spoken": "The square of A plus B contains two squares and twice the cross product." }, { "items": [ "Average both sides over many tracers. The average of a sum is the sum of the averages, and a fixed number such as 2 can be taken outside an average, so \\(\\langle(A+B)^2\\rangle = \\langle A^2\\rangle + 2\\langle AB\\rangle + \\langle B^2\\rangle\\).", "The middle term is where independence enters. For independent steps, the average of a product is the product of the averages: \\(\\langle AB\\rangle = \\langle A\\rangle\\langle B\\rangle\\). Both steps are centred, so both averages are zero, and \\(\\langle AB\\rangle = 0\\).", "Check this with the simplest walk, where each step is +1 or −1 with equal chance. The four equally likely pairs (+1, +1), (+1, −1), (−1, +1) and (−1, −1) give the products +1, −1, −1, +1, whose average is zero. The squares \\(A^2\\) and \\(B^2\\) are always 1, so the mean square after two steps is 1 + 0 + 1 = 2.", "Three steps work the same way. Multiplying out \\((\\Delta_1+\\Delta_2+\\Delta_3)^2\\) gives three squares, \\(\\Delta_1^2 + \\Delta_2^2 + \\Delta_3^2\\), and three cross terms, \\(2\\Delta_1\\Delta_2 + 2\\Delta_1\\Delta_3 + 2\\Delta_2\\Delta_3\\), one for each pair of different steps. Each cross term averages to zero, for the reason just given.", "For \\(n\\) steps there are \\(n\\) squares and one cross term for every pair of different steps. Every cross term averages to zero. Each square averages to \\(l^2\\), so the \\(n\\) squares together contribute \\(n\\) times \\(l^2\\). In symbols, where \\(\\sum_{i=1}^n\\) means add up for \\(i\\) from 1 to \\(n\\):" ], "kind": "steps" }, { "equations": [ "eq-model-bm-sum-of-steps" ], "kind": "formula", "latex": "\\left\\langle\\left(\\sum_{i=1}^n\\Delta_i\\right)^2\\right\\rangle=\\sum_{i=1}^n\\langle\\Delta_i^2\\rangle=nl^2", "spoken": "The mean square of the sum of independent centred increments is the sum of their mean squares." }, { "items": [ "Now bring in time. Each step takes the interval \\(\\tau\\), so after \\(n\\) steps the elapsed time is \\(t = n\\tau\\). Dividing both sides by \\(\\tau\\), the number of steps is \\(n = t/\\tau\\).", "Replace \\(n\\) in the mean square: \\(\\langle x^2\\rangle = nl^2 = (t/\\tau)\\,l^2\\). Regroup the same factors as \\((l^2/\\tau)\\,t\\): a constant times the elapsed time.", "Write the constant \\(l^2/\\tau\\) as 2 times \\(l^2/(2\\tau)\\), and give \\(l^2/(2\\tau)\\) its own name, \\(D\\), the diffusivity of the walk. The factor 2 is a choice of convention: with it, this \\(D\\) is the same coefficient that appears in the diffusion equation of the next passage. The three statements together:" ], "kind": "steps" }, { "equations": [ "eq-model-bm-walk-time", "eq-model-bm-walk-diffusivity", "eq-model-bm-mean-square-growth" ], "kind": "formula", "latex": "t=n\\tau,\\qquad D=\\frac{l^2}{2\\tau},\\qquad\\langle x^2\\rangle=2Dt", "spoken": "Elapsed time is n tau; defining D as ell squared over two tau gives mean-square displacement two D t." }, { "items": [ "The mean square grows in proportion to time: four times the time gives four times the mean square. The RMS is its square root, and \\(\\sqrt{4} = 2\\), so four times the time gives twice the RMS, not four times. In the ±1 walk, 4 steps give a mean square of 4 and an RMS of 2; 16 steps give a mean square of 16 and an RMS of 4.", "Each assumption did work. If the steps were not centred, \\(\\langle A\\rangle\\langle B\\rangle\\) would not vanish, and the cross terms would add a part growing like \\(n^2\\), the square of a steady drift. If the steps were not independent, \\(\\langle AB\\rangle\\) need not factor at all. If the mean square of a step were infinite, there would be no \\(l^2\\) to add." ], "kind": "steps" }, { "id": "random-walks", "kind": "foundation", "returnCaption": "Return to the step-count and elapsed-time comparison." } ] }, "recap": "Independent centred steps add mean squares; the root therefore grows as the square root of time.", "review": "draft", "schemaVersion": 1, "section": "s4", "title": "Why the square grows with time" }, { "citations": [ "ap-17-549" ], "experiments": [ "bm-05", "bm-06" ], "help": { "example": "taylor-expansion", "missingStep": "taylor-expansion", "why": "distributions" }, "id": "arg-bm-diffusion-equation", "kind": "argument", "limitations": [ "Cutting the expansion after the second term is an approximation; it becomes exact only in a limit where the jumps shrink with the time step.", "Over very short times a particle’s motion is not independent from one moment to the next, so the diffusion law is not a picture of individual collisions." ], "meaning": { "executionStatus": "static-illustration", "historicalStatus": "pedagogical-reconstruction", "logicalRole": "derivation", "modelStatus": "approximation" }, "paper": "brownian-motion", "premises": [ "The transition density is normalized, symmetric and has finite second moment.", "The density changes little over one jump, so a short expansion describes it on scales larger than a jump." ], "prerequisites": [ { "edge": "premise", "id": "arg-bm-independent-steps" } ], "question": "How can random individual steps produce a deterministic equation?", "readings": { "full": [ { "kind": "paragraph", "text": "Let \\(\\varphi(\\Delta)\\) be the probability density for a displacement \\(\\Delta\\) during \\(\\tau\\). To end at \\(x\\), a tracer must start at \\(x - \\Delta\\) and then make that displacement. Adding over all possible increments gives the transition relation, written here in modern notation." }, { "equations": [ "eq-model-bm-next-density" ], "kind": "formula", "latex": "p(x,t+\\tau)=\\int_{-\\infty}^{\\infty}p(x-\\Delta,t)\\varphi(\\Delta)\\,d\\Delta", "spoken": "The density one interval later is the integral, over every jump, of the density one jump away now, times how likely that jump is." }, { "kind": "paragraph", "text": "Expand to first order in time and second order in displacement. Normalization cancels the zeroth-order term. Symmetry removes the first moment. The second moment remains." }, { "equations": [ "eq-model-bm-jump-moment" ], "kind": "formula", "latex": "D=\\frac{1}{2\\tau}\\int_{-\\infty}^{\\infty}\\Delta^2\\varphi(\\Delta)\\,d\\Delta", "spoken": "D is half the mean-square step divided by the step interval." }, { "equations": [ "eq-model-bm-diffusion-equation" ], "kind": "formula", "latex": "\\frac{\\partial p}{\\partial t}=D\\frac{\\partial^2p}{\\partial x^2}", "spoken": "The retained equation is the diffusion equation." }, { "id": "taylor-expansion", "kind": "foundation", "returnCaption": "Return to the expansion at fixed position or fixed time." } ], "margin": [ { "kind": "paragraph", "text": "Einstein takes an interval \\(\\tau\\) very short compared with the times we observe, yet long enough that a particle’s motions in two successive intervals are independent. That double condition, not a formal limit \\(\\tau \\to 0\\), is what the expansion needs: the jumps must shrink with \\(\\tau\\) so that \\(D\\) stays finite. Einstein writes the density as \\(f(x, t)\\), the number of particles per unit volume, and the jump law as \\(\\varphi(\\Delta)\\), as here." } ], "overview": [ { "kind": "paragraph", "text": "Each tracer’s next position is its present one plus a random jump. Average that rule over every possible jump, take left and right jumps to be equally likely, and on scales larger than one jump the density obeys the diffusion equation." } ], "steps": [ { "id": "taylor-expansion", "kind": "foundation", "returnCaption": "Return to the two expansions and their remainders." }, { "items": [ "Name the pieces. \\(p(x,t)\\) is the density of tracers at position \\(x\\) and time \\(t\\): the chance of finding a tracer between \\(x\\) and a slightly larger \\(x + dx\\) is \\(p(x,t)\\,dx\\). Einstein writes this density as \\(f(x, t)\\) and counts particles rather than chances; the steps below are the same for both.", "\\(\\tau\\) is a short interval of time. During it each tracer moves by a signed displacement \\(\\Delta\\), and \\(\\varphi(\\Delta)\\) is the probability density of that displacement: the chance of a displacement between \\(\\Delta\\) and \\(\\Delta + d\\Delta\\) is \\(\\varphi(\\Delta)\\,d\\Delta\\).", "Two facts about \\(\\varphi\\) are assumed. It is normalized: some displacement certainly happens, so all its chances add up to one, \\(\\int_{-\\infty}^{\\infty}\\varphi(\\Delta)\\,d\\Delta = 1\\). And it is symmetric: a jump to the left is as likely as the same jump to the right, \\(\\varphi(-\\Delta) = \\varphi(\\Delta)\\).", "To be at \\(x\\) at time \\(t + \\tau\\), a tracer must have been at \\(x - \\Delta\\) at time \\(t\\) and then made the displacement \\(\\Delta\\). The chance of that one route is the density at its start times the chance of the jump, \\(p(x - \\Delta, t)\\,\\varphi(\\Delta)\\,d\\Delta\\), because in this model the jump law is the same wherever the tracer starts. Adding over every possible \\(\\Delta\\), from far left to far right, gives the transition relation:" ], "kind": "steps" }, { "equations": [ "eq-model-bm-next-density" ], "kind": "formula", "latex": "p(x,t+\\tau)=\\int_{-\\infty}^{\\infty}p(x-\\Delta,t)\\varphi(\\Delta)\\,d\\Delta", "spoken": "The density one interval later is the integral, over every jump, of the density one jump away now, times how likely that jump is." }, { "items": [ "The integral sign \\(\\int\\) is that adding-up over every value of \\(\\Delta\\). Both sides can now be simplified, using the assumption that \\(p\\) changes little over one interval \\(\\tau\\) and over one jump \\(\\Delta\\).", "The left side, at fixed position. A short time \\(\\tau\\) later, the density is approximately its present value plus its rate of change in time, \\(\\partial p/\\partial t\\), multiplied by \\(\\tau\\). The curly \\(\\partial\\) marks a partial derivative: a rate of change in one variable while the other is held fixed, here time with position held fixed." ], "kind": "steps" }, { "equations": [ "eq-model-bm-taylor-in-time" ], "kind": "formula", "latex": "p(x,t+\\tau)\\approx p(x,t)+\\tau\\frac{\\partial p}{\\partial t}", "spoken": "Keep the density and its first time change on the left." }, { "items": [ "The right side, at fixed time. The density a distance \\(\\Delta\\) to the left of \\(x\\) is approximately its value at \\(x\\), minus the slope \\(\\partial p/\\partial x\\) times \\(\\Delta\\), plus half the curvature \\(\\partial^2p/\\partial x^2\\) times \\(\\Delta^2\\). The slope term carries a minus sign because the step is to the left. The curvature term does not, because \\((-\\Delta)^2 = \\Delta^2\\). Here \\(p\\) and its derivatives are all taken at \\(x\\) and \\(t\\)." ], "kind": "steps" }, { "equations": [ "eq-model-bm-taylor-in-space" ], "kind": "formula", "latex": "p(x-\\Delta,t)\\approx p-\\Delta\\frac{\\partial p}{\\partial x}+\\frac{\\Delta^2}{2}\\frac{\\partial^2p}{\\partial x^2}", "spoken": "Keep value, slope and curvature of the spatial density on the right." }, { "items": [ "Put the spatial expansion into the integral and split it into three integrals, one for each term. The density and its derivatives are taken at \\(x\\), so they do not change as \\(\\Delta\\) runs over its values, and each one comes outside its integral.", "First term: \\(p\\) times \\(\\int\\varphi\\,d\\Delta\\), which is \\(p\\) times one, because \\(\\varphi\\) is normalized.", "Second term: \\(-\\partial p/\\partial x\\) times \\(\\int\\Delta\\varphi\\,d\\Delta\\). For a symmetric \\(\\varphi\\), each positive \\(\\Delta\\) is matched by the negative \\(-\\Delta\\) with the same weight, so this integral is zero. That is why there is no drift.", "Third term: \\(\\frac{1}{2}\\,\\partial^2p/\\partial x^2\\) times \\(\\int\\Delta^2\\varphi\\,d\\Delta\\). This integral is the mean-square jump. It is positive, because \\(\\Delta^2\\) is never negative, so symmetry cannot remove it.", "Set the two sides equal: \\(p + \\tau\\,\\partial p/\\partial t \\approx p + \\frac{1}{2}\\,\\frac{\\partial^2p}{\\partial x^2}\\int\\Delta^2\\varphi\\,d\\Delta\\). Subtract \\(p\\) from both sides; normalization is what makes the two \\(p\\) terms cancel. Then divide both sides by \\(\\tau\\): \\(\\partial p/\\partial t\\) is approximately \\(\\frac{1}{2\\tau}\\int\\Delta^2\\varphi\\,d\\Delta\\) times \\(\\partial^2p/\\partial x^2\\).", "The factor in front of the curvature depends only on the jump law and on \\(\\tau\\), not on \\(x\\) or \\(t\\). Give it a name, \\(D\\):" ], "kind": "steps" }, { "equations": [ "eq-model-bm-jump-moment" ], "kind": "formula", "latex": "D=\\frac{1}{2\\tau}\\int_{-\\infty}^{\\infty}\\Delta^2\\varphi(\\Delta)\\,d\\Delta", "spoken": "D is half the mean-square step divided by the step interval." }, { "items": [ "It is the mean-square jump divided by \\(2\\tau\\), the same form as the \\(l^2/(2\\tau)\\) of the previous passage. With that name, what remains is the diffusion equation:" ], "kind": "steps" }, { "equations": [ "eq-model-bm-diffusion-equation" ], "kind": "formula", "latex": "\\frac{\\partial p}{\\partial t}=D\\frac{\\partial^2p}{\\partial x^2}", "spoken": "The retained equation is the diffusion equation." }, { "items": [ "Read it term by term. Where the density has a peak, its curvature is negative, so \\(\\partial p/\\partial t\\) is negative and the peak falls. Where it has a hollow, the curvature is positive and the hollow fills. \\(D\\) sets how fast.", "What was dropped. The next spatial term carries \\(\\Delta^3\\) and vanishes by symmetry, like the \\(\\Delta\\) term. The first nonzero term left out carries \\(\\Delta^4\\), and on the left the term with \\(\\tau^2\\) was left out. Cutting the expansion there is an approximation. It becomes exact only in a limit where the jumps shrink with the time step so that \\(D\\) stays finite, and the equation describes the density on scales larger than one jump, not individual collisions." ], "kind": "steps" }, { "id": "integration", "kind": "foundation", "returnCaption": "Return to where the jump chances add up to one and the lopsided terms cancel." }, { "id": "partial-derivatives", "kind": "foundation", "returnCaption": "Return to the rates of change taken with position or time held fixed." }, { "id": "diffusion-equation", "kind": "foundation", "returnCaption": "Return to what the resulting density equation predicts." } ] }, "recap": "Normalization preserves probability; symmetry removes drift; the second moment supplies diffusion.", "review": "draft", "schemaVersion": 1, "section": "s4", "title": "From a step law to a density law" }, { "citations": [ "ap-17-549" ], "experiments": [ "bm-06", "bm-01" ], "help": { "example": "distributions", "missingStep": "gaussian-distributions", "why": "distributions" }, "id": "arg-bm-gaussian", "kind": "argument", "limitations": [ "A finite closed box has a different long-time distribution.", "At t = 0 the distribution is a point mass, not an ordinary density.", "Coordinate RMS is not the three-dimensional RMS distance." ], "meaning": { "executionStatus": "static-illustration", "historicalStatus": "pedagogical-reconstruction", "logicalRole": "derivation", "modelStatus": "exact-within-model" }, "paper": "brownian-motion", "premises": [ "D is constant and positive, the line is unbounded, nothing pushes the particles one way, and they all start at one point." ], "prerequisites": [ { "edge": "premise", "id": "arg-bm-diffusion-equation" } ], "question": "How does the density law become a measurable displacement?", "readings": { "full": [ { "kind": "formula", "latex": "p(x,t)=\\frac{e^{-x^2/(4Dt)}}{\\sqrt{4\\pi Dt}}\\quad(t>0)", "spoken": "The point-source solution is the normalized Gaussian density for positive time." }, { "kind": "paragraph", "text": "Its symmetry gives zero mean. Its second moment is 2Dt. These are ensemble statements: the curve assigns probabilities to intervals, not destinations to individual particles." }, { "equations": [ "eq-model-bm-rms-from-mean-square" ], "kind": "formula", "latex": "\\lambda_x=\\sqrt{\\langle x^2\\rangle}=\\sqrt{2Dt}", "spoken": "Coordinate RMS displacement is the square root of two D t." }, { "kind": "paragraph", "text": "The probability of finding a displacement between a and b is the integral of this density over that interval. At time zero, an interval containing the starting point has probability one; no finite bell represents that state." }, { "id": "gaussian-distributions", "kind": "foundation", "returnCaption": "Return to checking the bell curve: its total is one, and its spread is 2Dt." } ], "margin": [ { "kind": "paragraph", "text": "Einstein writes the solution for \\(n\\) particles starting together as \\(f(x, t) = \\frac{n}{\\sqrt{4\\pi D}}\\frac{e^{-x^2/4Dt}}{\\sqrt{t}}\\), then gives the displacement \\(\\lambda_x = \\sqrt{2Dt}\\). Three independent coordinates would give a total mean square of \\(6Dt\\); the paper asks about one coordinate, hence \\(2Dt\\)." } ], "overview": [ { "kind": "paragraph", "text": "Tracers that all start at one point spread into a bell curve whose width grows with time: along one axis the RMS displacement is \\(\\sqrt{2Dt}\\)." } ], "steps": [ { "id": "bridge-letter-for-quantity", "kind": "foundation", "returnCaption": "Return to λx, the displacement after a time t." }, { "id": "gaussian-distributions", "kind": "foundation", "returnCaption": "Return to the bell curve for particles that all start at one point." }, { "id": "exponentials", "kind": "foundation", "returnCaption": "Return to the Gaussian's exponential factor." }, { "items": [ "Set the scene. Every tracer starts at \\(x = 0\\) at time zero, on a line with no walls, with a constant, positive diffusivity \\(D\\) and nothing pushing the tracers one way. Their density \\(p(x,t)\\) then spreads by the diffusion equation of the previous passage. For every positive time \\(t\\), the solution is this bell-shaped curve, a Gaussian:" ], "kind": "steps" }, { "kind": "formula", "latex": "p(x,t)=\\frac{e^{-x^2/(4Dt)}}{\\sqrt{4\\pi Dt}}\\quad(t>0)", "spoken": "The point-source solution is the normalized Gaussian density for positive time." }, { "items": [ "Read its parts. \\(e\\) is the base of natural exponentials, about 2.718, and \\(\\pi\\) is about 3.1416. The exponent \\(-x^2/(4Dt)\\) is zero at \\(x = 0\\) and more negative the farther \\(x\\) is from the start, so the curve is highest at the start and falls away on both sides. The denominator \\(\\sqrt{4\\pi Dt}\\) scales the whole curve so that its total is one, as the next steps check.", "The curve depends on \\(x\\) only through \\(x^2\\), so \\(p(-x,t) = p(x,t)\\): it is symmetric about the start.", "Change variable to make the integrals standard ones. For positive \\(t\\), let \\(u = x/\\sqrt{4Dt}\\). Then \\(x = \\sqrt{4Dt}\\,u\\), \\(x^2 = 4Dt\\,u^2\\) and \\(dx = \\sqrt{4Dt}\\,du\\), and the exponent becomes \\(-u^2\\).", "Total probability. \\(\\int p\\,dx\\) becomes \\(\\frac{1}{\\sqrt{4\\pi Dt}}\\int e^{-u^2}\\sqrt{4Dt}\\,du\\). The two square roots of \\(4Dt\\) cancel, leaving \\(\\frac{1}{\\sqrt\\pi}\\int_{-\\infty}^{\\infty}e^{-u^2}\\,du\\). The Gaussian integral \\(\\int_{-\\infty}^{\\infty}e^{-u^2}\\,du = \\sqrt\\pi\\) is a known result, so the total is one.", "Mean. The mean displacement is \\(\\int x\\,p\\,dx\\). Because \\(p\\) is symmetric, the product \\(x\\,p\\) is odd: its value at \\(-x\\) is minus its value at \\(x\\). The negative side cancels the positive side, and the mean is zero.", "Mean square. \\(\\langle x^2\\rangle = \\int x^2p\\,dx\\). Substitute \\(x^2 = 4Dt\\,u^2\\) and cancel the square roots as before; what remains is \\(4Dt\\) times \\(\\frac{1}{\\sqrt\\pi}\\int u^2e^{-u^2}\\,du\\).", "That last integral is found by integration by parts. Write \\(u^2e^{-u^2}\\) as \\(u\\) times \\(ue^{-u^2}\\), and notice that \\(ue^{-u^2}\\) is the derivative of \\(-\\frac{1}{2}e^{-u^2}\\). Integration by parts moves the derivative across: \\(\\int u^2e^{-u^2}\\,du\\) equals \\(-\\frac{1}{2}ue^{-u^2}\\) evaluated between far left and far right, plus \\(\\frac{1}{2}\\int e^{-u^2}\\,du\\).", "The first part is zero at both ends, because \\(e^{-u^2}\\) falls to zero much faster than \\(u\\) grows. So the integral is \\(\\frac{1}{2}\\sqrt\\pi\\), and divided by \\(\\sqrt\\pi\\) it gives one half:" ], "kind": "steps" }, { "equations": [ "eq-model-bm-gaussian-second-moment" ], "kind": "formula", "latex": "\\langle x^2\\rangle=\\frac{4Dt}{\\sqrt\\pi}\\int_{-\\infty}^{\\infty}u^2e^{-u^2}\\,du=2Dt", "spoken": "The second moment is four D t times the normalized Gaussian second-moment integral, giving two D t." }, { "items": [ "So the mean square is \\(4Dt\\) times one half, which is \\(2Dt\\): the same growth in proportion to time that the walk of independent steps gave. Its square root is a length, the RMS displacement along the axis, which Einstein writes \\(\\lambda_x\\). Taking the nonnegative root:" ], "kind": "steps" }, { "equations": [ "eq-model-bm-rms-from-mean-square" ], "kind": "formula", "latex": "\\lambda_x=\\sqrt{\\langle x^2\\rangle}=\\sqrt{2Dt}", "spoken": "Coordinate RMS displacement is the square root of two D t." }, { "items": [ "These are statements about the ensemble. The curve does not say where one tracer will be; it gives the probability of an interval. The probability of a displacement between two positions \\(x_1\\) and \\(x_2\\) is the integral of \\(p\\) from \\(x_1\\) to \\(x_2\\), the area under the curve there.", "For this curve, the interval from \\(-\\lambda_x\\) to \\(+\\lambda_x\\) holds about 68% of the probability (0.6827), and the interval from \\(-2\\lambda_x\\) to \\(+2\\lambda_x\\) about 95% (0.9545), at every positive time. Only the width changes as the curve spreads.", "At \\(t = 0\\) the formula has no meaning: \\(4Dt\\) is zero, and the curve would be infinitely narrow and infinitely tall. The state itself is simple. Every tracer is at the start, so any interval containing the start has probability one and any interval that misses it has probability zero. No finite bell represents that, which is why the formula carries \\((t>0)\\).", "Last, check that the curve obeys the diffusion equation. Write \\(p\\) as a factor \\((4\\pi Dt)^{-1/2}\\) times \\(e^{-x^2/(4Dt)}\\), and take each rate of change with the other variable held fixed.", "Time, with \\(x\\) fixed. The factor contributes \\(-1/(2t)\\) times \\(p\\), because the rate of change of \\(t^{-1/2}\\) is \\(-\\frac{1}{2}t^{-3/2}\\). The exponential contributes \\(x^2/(4Dt^2)\\) times \\(p\\), because the rate of change of \\(-x^2/(4Dt)\\) with \\(t\\) is \\(+x^2/(4Dt^2)\\). By the product rule, \\(\\partial p/\\partial t = p\\,(-1/(2t) + x^2/(4Dt^2))\\).", "Position, with \\(t\\) fixed. The first derivative is \\(\\partial p/\\partial x = p\\,(-x/(2Dt))\\). Differentiate again with the product rule: \\(\\partial^2p/\\partial x^2 = p\\,(x^2/(4D^2t^2)) + p\\,(-1/(2Dt))\\).", "Multiply that by \\(D\\): \\(p\\,(x^2/(4Dt^2) - 1/(2t))\\), the same as the time derivative. So the curve satisfies \\(\\partial p/\\partial t = D\\,\\partial^2p/\\partial x^2\\)." ], "kind": "steps" }, { "id": "integration", "kind": "foundation", "returnCaption": "Return to the integration by parts that gives the one half." }, { "id": "distributions", "kind": "foundation", "returnCaption": "Return to probabilities of intervals rather than heights." } ] }, "recap": "For an unbounded point source, the Gaussian has mean square 2Dt and coordinate RMS √(2Dt).", "review": "draft", "schemaVersion": 1, "section": "s4", "title": "What the spreading curve predicts" }, { "citations": [ "ap-17-549" ], "experiments": [ "bm-01", "bm-06" ], "help": { "example": "diffusion-equation", "missingStep": "diffusion-equation", "why": "flux-continuity" }, "id": "arg-bm-diffusivity", "kind": "argument", "limitations": [ "Slip at the particle’s surface, its inertia, interactions between particles, unusual liquids, and gases are all left out.", "Stokes’s law and the osmotic-pressure law are brought in from outside; conservation alone does not give them." ], "meaning": { "executionStatus": "static-illustration", "historicalStatus": "pedagogical-reconstruction", "logicalRole": "derivation", "modelStatus": "approximation" }, "paper": "brownian-motion", "premises": [ "Few, small, spherical particles in a uniform liquid, with Stokes’s drag and ideal osmotic pressure.", "R, the gas constant, is known separately; N is the number of molecules in a mole." ], "prerequisites": [ { "edge": "premise", "id": "arg-bm-gaussian" } ], "question": "What fixes D for a small spherical tracer in a liquid?", "readings": { "full": [ { "kind": "paragraph", "text": "The displacement law tells us what a given \\(D\\) predicts. A separate model connects \\(D\\) to a tracer’s physical surroundings. Let \\(b\\) be mobility, so a small force \\(F\\) produces mean drift \\(bF\\). Stokes drag for a sphere gives \\(b = 1/(6\\pi\\eta a)\\)." }, { "id": "viscosity-stokes-drag", "kind": "foundation", "returnCaption": "Return to where the sphere's drag sets its mobility." }, { "kind": "paragraph", "text": "Let \\(c\\) be number density. At isothermal balance the force density \\(cF\\) balances the osmotic-pressure gradient. With ideal osmotic pressure \\(cRT/N\\), the drift flux \\(cbF\\) becomes \\(b\\,(RT/N)\\) times the density gradient. Equating it with the opposite diffusive flux gives \\(D = bRT/N\\)." }, { "id": "free-energy-osmotic-pressure", "kind": "foundation", "returnCaption": "Return to where the osmotic pressure balances the force on each grain." }, { "equations": [ "eq-model-bm-diffusivity-molar", "eq-model-bm-diffusivity" ], "kind": "formula", "latex": "D=\\frac{RT}{6\\pi\\eta aN}=\\frac{k_BT}{6\\pi\\eta a}", "spoken": "Stokes–Einstein diffusivity is the gas constant times temperature, divided by six pi times viscosity times radius times Avogadro's number; equivalently, Boltzmann's constant times temperature, divided by six pi times viscosity times radius." }, { "kind": "paragraph", "text": "The equality kB = R/N relates the two forms. For a prediction using modern constants it is convenient. For an inference of N, using a value of kB derived from that same N would defeat the point." }, { "id": "temperature-thermal-energy", "kind": "foundation", "returnCaption": "Return to where Boltzmann's constant and R/N turn out to be one quantity." }, { "id": "flux-continuity", "kind": "foundation", "returnCaption": "Return to balancing drift and diffusive fluxes." } ], "margin": [ { "kind": "paragraph", "text": "Einstein reaches this relation in §§2–3 through osmotic pressure and Stokes’s law, writing \\(k\\) for the viscosity and \\(P\\) for the radius: \\(D = \\frac{RT}{N}\\frac{1}{6\\pi kP}\\). His \\(k\\) is not Boltzmann’s constant. William Sutherland published the same relation independently in 1905, and Marian Smoluchowski reached the displacement law by another route in 1906. The laboratories use modern SI constants and say so." } ], "overview": [ { "kind": "paragraph", "text": "For small spheres in a liquid, a more viscous liquid gives a smaller \\(D\\): doubling the viscosity halves the mean square at a fixed time and divides the RMS displacement by \\(\\sqrt{2}\\), not by 2." } ], "steps": [ { "id": "ratios-scaling", "kind": "foundation", "returnCaption": "Return to why doubling the viscosity divides the spread by the square root of 2." }, { "items": [ "Name the quantities. \\(T\\) is the absolute temperature, in kelvin. \\(\\eta\\) is the liquid’s dynamic viscosity, its resistance to flow, in pascal seconds. \\(a\\) is the radius of the spherical tracer. \\(R\\) is the molar gas constant, and \\(N\\) is the number of molecules in a mole. Einstein prints the viscosity as \\(k\\) and the radius as \\(P\\); his \\(k\\) is the viscosity, not Boltzmann’s constant.", "\\(D\\) is the diffusivity of the earlier passages: it says how fast the mean square grows, \\(\\langle x^2\\rangle = 2Dt\\). This passage asks what fixes \\(D\\). The route has two halves: how a tracer answers a steady push, and how the crowding of many tracers pushes back.", "First half, the push. Suppose a small steady force \\(F\\) acts on each tracer. In a viscous liquid the tracer soon moves at a steady mean drift speed in proportion to the force. Call the ratio of speed to force the mobility \\(b\\), so the drift speed is \\(bF\\).", "Stokes’s law gives the drag on a sphere moving slowly at speed \\(v\\) through a liquid as \\(6\\pi\\eta av\\). At steady drift the drag balances the push, \\(F = 6\\pi\\eta av\\), so \\(v = F/(6\\pi\\eta a)\\) and the mobility is \\(b = 1/(6\\pi\\eta a)\\). Stokes’s law is brought in from the theory of fluids; the random-walk argument does not supply it." ], "kind": "steps" }, { "id": "viscosity-stokes-drag", "kind": "foundation", "returnCaption": "Return to where the sphere's drag sets its mobility." }, { "items": [ "Second half, the crowding. Let \\(c\\) be the number of tracers per unit volume, which Einstein writes \\(\\nu\\). Few, small tracers in a liquid exert an osmotic pressure, written here Π, of the same form as an ideal gas’s pressure: Π = cRT/N. Dividing \\(R\\) by \\(N\\) turns the gas constant per mole into a constant per molecule.", "Let the density vary along \\(x\\). A gradient is the rate at which a quantity changes with position. With \\(T\\) held fixed, only \\(c\\) varies in cRT/N, so the constant factor comes outside: the gradient of Π is \\(RT/N\\) times the gradient of \\(c\\).", "In a steady balance, the force on the tracers in a thin slab is held by the difference in osmotic pressure across it. Per unit volume that reads: \\(cF\\) equals the gradient of Π, which is \\(RT/N\\) times the gradient of \\(c\\). This osmotic law, like Stokes’s, is brought in from outside; conservation of tracers alone does not give it." ], "kind": "steps" }, { "id": "free-energy-osmotic-pressure", "kind": "foundation", "returnCaption": "Return to where the osmotic pressure balances the force on each grain." }, { "items": [ "Now count flows. A flux is the number of tracers crossing unit area in unit time. Tracers drifting at speed \\(bF\\) with density \\(c\\) carry the drift flux \\(c\\,bF\\). Multiply the balance of the last step by \\(b\\): the drift flux is \\(b\\,(RT/N)\\) times the gradient of \\(c\\).", "Fick’s law gives the diffusive flux: tracers spread from crowded to sparse regions, and the flux is \\(-D\\) times the gradient of \\(c\\), with the minus sign because the flow runs down the gradient.", "In the balance nothing flows overall, so the two fluxes add to zero: \\(b\\,(RT/N)\\) times the gradient, minus \\(D\\) times the gradient, is zero. Divide by the gradient, which is not zero where the density varies: \\(D = bRT/N\\).", "Substitute the Stokes mobility \\(b = 1/(6\\pi\\eta a)\\): \\(D = (RT/N) \\times 1/(6\\pi\\eta a)\\), which is \\(RT\\) divided by \\(6\\pi\\eta aN\\). Einstein found this in §3 and quotes it at the start of §5 as \\(D = \\frac{RT}{N}\\frac{1}{6\\pi kP}\\), with his \\(k\\) for \\(\\eta\\) and his \\(P\\) for \\(a\\).", "Boltzmann’s constant \\(k_B\\) equals \\(R/N\\), the gas constant per molecule. Replacing \\(R/N\\) by \\(k_B\\) gives the second form. Both are one relation:" ], "kind": "steps" }, { "equations": [ "eq-model-bm-diffusivity-molar", "eq-model-bm-diffusivity" ], "kind": "formula", "latex": "D=\\frac{RT}{6\\pi\\eta aN}=\\frac{k_BT}{6\\pi\\eta a}", "spoken": "Stokes–Einstein diffusivity is the gas constant times temperature, divided by six pi times viscosity times radius times Avogadro's number; equivalently, Boltzmann's constant times temperature, divided by six pi times viscosity times radius." }, { "items": [ "Which form to use depends on the question. To predict a spread from modern constants, the \\(k_B\\) form is convenient. To infer \\(N\\) from a spread, it defeats the point: a value of \\(k_B\\) derived from a known \\(N\\) would put the answer into the data. The \\(R\\) form keeps \\(N\\) separate, because \\(R\\) can be measured on gases without counting molecules." ], "kind": "steps" }, { "id": "temperature-thermal-energy", "kind": "foundation", "returnCaption": "Return to where Boltzmann's constant and R/N turn out to be one quantity." }, { "items": [ "Einstein’s own numbers, from §5: water at 17 °C, so \\(T\\) = 290.15 K; the viscosity printed as \\(k\\) = \\(1{,}35\\cdot10^{-2}\\), a bare number that in the centimetre-gram-second units of the time means poise, and is \\(1.35\\times10^{-3}\\) Pa·s; a particle diameter printed as 0,001 mm, so \\(a\\) = 0.0005 mm = \\(5\\times10^{-7}\\) m; and \\(N\\) = \\(6\\times10^{23}\\) from the kinetic theory of gases. The paper does not print \\(R\\); this edition supplies \\(R\\) = 8.31 J/(mol·K).", "The denominator: 6π × (\\(1.35\\times10^{-3}\\) Pa·s) × (\\(5\\times10^{-7}\\) m) = \\(1.272\\times10^{-8}\\) kg/s. The numerator: \\(RT/N\\) = 8.31 × 290.15 / (\\(6\\times10^{23}\\)) = \\(4.019\\times10^{-21}\\) J. Their quotient is \\(D\\) = \\(3.16\\times10^{-13}\\) \\(\\mathrm{m}^2/\\mathrm{s}\\), since a joule divided by a kilogram per second is a square metre per second.", "With \\(\\lambda_x = \\sqrt{2Dt}\\) from §4: at \\(t\\) = 1 s the mean square is \\(2Dt\\) = \\(6.32\\times10^{-13}\\) \\(\\mathrm{m}^2\\), whose square root is \\(7.95\\times10^{-7}\\) m, about 0.8 μm. Einstein prints “\\(8\\cdot10^{-5}\\) cm = 0,8 Mikron”. At \\(t\\) = 60 s the mean square is 60 times larger, so \\(\\lambda_x\\) is \\(\\sqrt{60}\\), about 7.75, times larger: 6.16 μm, which Einstein gives as about 6 micrometres in one minute.", "Now double the viscosity, holding \\(T\\), \\(a\\) and the constants fixed. \\(\\eta\\) sits in the denominator, so \\(D\\) halves, to \\(1.58\\times10^{-13}\\) \\(\\mathrm{m}^2/\\mathrm{s}\\), and the mean square at a fixed time halves with it. The RMS is a square root, so it changes by a factor \\(1/\\sqrt{2}\\), about 0.707: at 1 s it falls from 0.795 μm to 0.562 μm. Doubling the viscosity divides the RMS displacement by \\(\\sqrt{2}\\), not by two." ], "kind": "steps" }, { "id": "flux-continuity", "kind": "foundation", "returnCaption": "Return to the distinction between density and flux." }, { "id": "bridge-squaring-square-roots", "kind": "foundation", "returnCaption": "Return to the square-root response when diffusivity changes." } ] }, "recap": "Combining ideal osmotic pressure with mobility and diffusive balance relates diffusivity to temperature, viscosity and radius.", "review": "draft", "schemaVersion": 1, "section": "s5", "title": "Why viscosity changes the spread" }, { "citations": [ "ap-17-549" ], "experiments": [ "bm-07", "bm-01" ], "help": { "example": "error-and-inference", "missingStep": "diffusion-equation", "why": "error-and-inference" }, "id": "arg-bm-inference", "kind": "argument", "limitations": [ "A synthetic run generated from an assumed N is not independent evidence for N.", "Without the radius, the diffusivity fixes only the product aN.", "Measurement noise, drift, exposure and finite sampling require separate treatment." ], "meaning": { "executionStatus": "static-illustration", "historicalStatus": "pedagogical-reconstruction", "logicalRole": "derivation", "modelStatus": "approximation" }, "paper": "brownian-motion", "premises": [ "A measured mean-square displacement along one axis, or how fast it grows with time.", "Separately measured R, T, η and particle radius a, for few small spheres in a liquid." ], "prerequisites": [ { "edge": "premise", "id": "arg-bm-diffusivity" } ], "question": "Which additional measurements turn displacement into an estimate of N?", "readings": { "full": [ { "equations": [ "eq-model-bm-diffusivity-from-data", "eq-model-bm-avogadro-inference" ], "kind": "formula", "latex": "D=\\frac{\\langle x^2\\rangle}{2t},\\qquad N=\\frac{RTt}{3\\pi\\eta a\\langle x^2\\rangle}", "spoken": "Diffusivity is the mean-square displacement divided by twice the time; the molecular number is the gas constant times temperature times time, divided by three pi times viscosity times radius times the mean-square displacement." }, { "kind": "paragraph", "text": "The first expression refers to the model mean square, or to an estimate obtained from an appropriate sample. The second is an inversion under the stated physical assumptions. An estimate needs uncertainty and checks of those assumptions; rearranging symbols does not remove experimental error." }, { "kind": "paragraph", "text": "Without an independent radius, the same D can result from many pairs of a and N. The data then select a compatible family rather than a unique molecular number. The existing synthetic laboratories explore this relationship but do not supply a historical measurement." }, { "id": "two-measurements-two-unknowns", "kind": "foundation", "returnCaption": "Return to the radius and the molecular number, which a measured D alone cannot separate." }, { "id": "mean-variance-rms", "kind": "foundation", "returnCaption": "Return to the quantity estimated from measured displacements." }, { "id": "error-and-inference", "kind": "foundation", "returnCaption": "Return to what a finite sample and independent inputs can identify." }, { "id": "unit-system-1905", "kind": "foundation", "returnCaption": "Return to the inversion, where R, η and a must be in one system of units." } ], "margin": [ { "kind": "paragraph", "text": "Einstein did not estimate \\(N\\). He took \\(N = 6 \\times 10^{23}\\) from the kinetic theory of gases and predicted a displacement of about 0.8 μm in one second for particles 0.001 mm across in water at 17 °C, closing with the hope that someone would soon test it. Jean Perrin’s measurements from 1908 onward did, and gave values of \\(N\\) near \\(7 \\times 10^{23}\\). This page supplies no dataset of its own." } ], "overview": [ { "kind": "paragraph", "text": "Measured displacements give \\(D\\). Turning \\(D\\) into a count of molecules also needs the temperature, the viscosity and, above all, an independently measured radius. A simulation that assumes the answer cannot measure it." } ], "steps": [ { "items": [ "Start from the displacement law of §4, \\(\\langle x^2\\rangle = 2Dt\\). Here \\(\\langle x^2\\rangle\\) is the mean-square displacement along one axis after an observation time \\(t\\), and \\(D\\) is the tracers’ diffusivity. Divide both sides by \\(2t\\), which is allowed for any positive \\(t\\): \\(D = \\langle x^2\\rangle/(2t)\\).", "Next, the physical relation of the previous passage, \\(D = RT/(6\\pi\\eta aN)\\), where \\(R\\) is the gas constant, \\(T\\) the absolute temperature, \\(\\eta\\) the viscosity (Einstein’s \\(k\\)), \\(a\\) the tracer radius (his \\(P\\)) and \\(N\\) the number of molecules in a mole.", "Solve it for \\(N\\). Multiply both sides by \\(6\\pi\\eta aN\\), giving \\(6\\pi\\eta aND = RT\\). Then divide both sides by \\(6\\pi\\eta aD\\): \\(N = RT/(6\\pi\\eta aD)\\).", "Put the first result in for \\(D\\). Dividing by a fraction is multiplying by its inverse, so dividing by \\(\\langle x^2\\rangle/(2t)\\) multiplies by \\(2t/\\langle x^2\\rangle\\): \\(N = RT \\cdot 2t/(6\\pi\\eta a\\langle x^2\\rangle)\\).", "The 2 above and the 6 below leave a 3 below, since 2/6 = 1/3. Einstein prints this inversion at the end of §5 as \\(N = \\frac{t}{\\lambda_x^2}\\cdot\\frac{RT}{3\\pi kP}\\), with \\(\\lambda_x^2\\) for the mean square, his \\(k\\) for \\(\\eta\\) and his \\(P\\) for \\(a\\). The two results together:" ], "kind": "steps" }, { "equations": [ "eq-model-bm-diffusivity-from-data", "eq-model-bm-avogadro-inference" ], "kind": "formula", "latex": "D=\\frac{\\langle x^2\\rangle}{2t},\\qquad N=\\frac{RTt}{3\\pi\\eta a\\langle x^2\\rangle}", "spoken": "Diffusivity is the mean-square displacement divided by twice the time; the molecular number is the gas constant times temperature times time, divided by three pi times viscosity times radius times the mean-square displacement." }, { "items": [ "Read what each symbol needs. In the formula \\(\\langle x^2\\rangle\\) is the model’s mean square. From a finite set of tracked particles you get an estimate of it, and that estimate has an uncertainty the formula does not show. Rearranging the symbols carries that uncertainty into \\(N\\); it does not remove it.", "\\(R\\), \\(T\\), \\(\\eta\\) and \\(a\\) must each be measured separately, and all in one system of units. The inversion also holds only under the assumptions that gave \\(D = RT/(6\\pi\\eta aN)\\): few, small spheres in a liquid, Stokes’s drag and ideal osmotic pressure. An estimate needs checks of those assumptions as well as its uncertainty.", "The radius matters most. In \\(D = RT/(6\\pi\\eta aN)\\), \\(a\\) and \\(N\\) appear only as the product \\(aN\\). A measured \\(D\\) fixes that product, \\(aN = RT/(6\\pi\\eta D)\\), and nothing more: halve \\(a\\) and double \\(N\\), and \\(D\\) is unchanged. Without an independent radius, the data select a family of compatible pairs rather than one molecular number.", "Run Einstein’s §5 case through the formulas. His inputs are \\(T\\) = 290.15 K (water at 17 °C), \\(\\eta\\) = \\(1.35\\times10^{-3}\\) Pa·s, \\(a\\) = \\(5\\times10^{-7}\\) m and \\(N\\) = \\(6\\times10^{23}\\), with \\(R\\) = 8.31 J/(mol·K) supplied by this edition. The previous passage found \\(D\\) = \\(3.16\\times10^{-13}\\) \\(\\mathrm{m}^2/\\mathrm{s}\\) from them, so at \\(t\\) = 1 s the mean square is \\(6.32\\times10^{-13}\\) \\(\\mathrm{m}^2\\), an RMS of 0.795 μm.", "Now invert. \\(RTt\\) = 8.31 × 290.15 × 1 = 2411 J·s/mol. \\(3\\pi\\eta a\\langle x^2\\rangle\\) = 3π × (\\(1.35\\times10^{-3}\\)) × (\\(5\\times10^{-7}\\)) × (\\(6.32\\times10^{-13}\\)) = \\(4.02\\times10^{-21}\\) J·s. Their quotient is \\(N\\) = \\(6.0\\times10^{23}\\) per mole: the \\(N\\) that went in.", "That checks the arithmetic of the inversion and nothing else. It cannot count molecules, because the mean square fed into it was computed from an assumed \\(N\\). In §5 Einstein ran the relation forward in this way, as a prediction, and printed the inversion for a future measurement to use; he did not estimate \\(N\\) himself.", "The same numbers show the family. With this \\(D\\), the product \\(aN\\) is \\(3.0\\times10^{17}\\) metres per mole. A radius of 0.5 μm gives \\(N\\) = \\(6\\times10^{23}\\); had the radius been 1 μm, the same \\(D\\) would give \\(N\\) = \\(3\\times10^{23}\\). Only an independent measurement of \\(a\\) chooses between them.", "A simulation that uses an assumed molecular number can test this inversion’s arithmetic in the same way, but cannot independently establish that number in nature." ], "kind": "steps" }, { "id": "two-measurements-two-unknowns", "kind": "foundation", "returnCaption": "Return to the radius and the molecular number, which a measured D alone cannot separate." }, { "id": "mean-variance-rms", "kind": "foundation", "returnCaption": "Return to estimating a mean square rather than a squared mean." }, { "id": "error-and-inference", "kind": "foundation", "returnCaption": "Return to what a finite sample and independent inputs can identify." }, { "id": "unit-system-1905", "kind": "foundation", "returnCaption": "Return to the inversion, where R, η and a must be in one system of units." } ] }, "recap": "Displacement constrains diffusivity; independent radius, viscosity, temperature and gas-constant information are needed to infer molecular number.", "review": "draft", "schemaVersion": 1, "section": "s5", "title": "What would let us count molecules?" } ], "citations": [ { "id": "ap-17-132", "kind": "citation", "locator": "Annalen der Physik (4), 17, 132–148 (1905).", "schemaVersion": 1, "title": "A. Einstein, Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt", "url": "https://doi.org/10.1002/andp.19053220607" }, { "id": "ap-17-549", "kind": "citation", "locator": "Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.", "schemaVersion": 1, "title": "A. Einstein, On the motion of particles suspended in liquids at rest required by the molecular-kinetic theory of heat", "url": "https://doi.org/10.1002/andp.19053220806" }, { "id": "ap-17-891", "kind": "citation", "locator": "Annalen der Physik (4), 17, 891–921 (1905).", "schemaVersion": 1, "title": "A. Einstein, Zur Elektrodynamik bewegter Körper", "url": "https://doi.org/10.1002/andp.19053221004" }, { "id": "ap-18-639", "kind": "citation", "locator": "Annalen der Physik (4), 18, 639–641 (1905). External 1923 Perrett–Jeffery translation, electronically transcribed by John Walker; its notation was modernized. A reference for this explanatory preview, not this edition’s reviewed translation or pinned facsimile.", "schemaVersion": 1, "title": "A. Einstein, Does the inertia of a body depend upon its energy content?", "url": "https://www.fourmilab.ch/etexts/einstein/E_mc2/www/" }, { "id": "bipm-si-definitions", "kind": "citation", "locator": "Modern SI definitions; not historical evidence", "schemaVersion": 1, "title": "BIPM: SI measurement units and defining constants", "url": "https://www.bipm.org/en/measurement-units" }, { "id": "nist-normal-variance", "kind": "citation", "locator": "Modern statistical reference: chi-square confidence limits", "schemaVersion": 1, "title": "NIST/SEMATECH e-Handbook: confidence limits for a standard deviation", "url": "https://www.itl.nist.gov/div898/handbook/eda/section3/eda358.htm" } ], "equations": [ { "argument": "arg-bm-observable", "assumptions": [ "The same one-coordinate ideal Brownian model as the RMS relation.", "A positive observation interval is required for the quotient.", "Lines joining recorded points are a rendering convention, not a velocity measurement." ], "bindings": [ { "experimentId": "bm-01", "instanceSlot": "primary", "outputId": "modelApparentSpeed", "quantityId": "modelApparentSpeed", "termId": "eq-model-bm-apparent-speed.t.apparentSpeed" }, { "experimentId": "bm-01", "instanceSlot": "primary", "outputId": "rmsDisplacement1d", "quantityId": "rmsDisplacement1d", "termId": "eq-model-bm-apparent-speed.t.rms" }, { "experimentId": "bm-01", "instanceSlot": "primary", "outputId": "observationInterval", "quantityId": "observationInterval", "termId": "eq-model-bm-apparent-speed.t.time" } ], "explanation": "This quotient depends on the observation interval. It is not instantaneous physical velocity. At zero interval the quotient is undefined, even though the displacement is zero.", "id": "eq-model-bm-apparent-speed", "kind": "equation", "notation": "modern-pedagogical", "notes": [ { "explanation": "This defines an interval-dependent comparison. It does not introduce a physical instantaneous Brownian velocity.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-apparent-speed.op.definition", "title": "Define the observable" }, { "explanation": "A distance-per-interval statistic. It is not a molecular collision speed.", "foundation": "random-walks", "nodeId": "eq-model-bm-apparent-speed.t.apparentSpeed", "title": "Apparent coordinate speed" }, { "explanation": "The distance grows as the square root of time, while the denominator grows linearly. The quotient therefore decreases as the interval grows.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-apparent-speed.op.division", "title": "Divide by the same interval" }, { "explanation": "The same accepted one-coordinate model displacement used by the neighboring RMS equation.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-apparent-speed.t.rms", "title": "Model coordinate RMS" }, { "explanation": "A zero interval yields no apparent-speed value. The interface keeps the explanation instead of fabricating zero or infinity.", "foundation": "random-walks", "nodeId": "eq-model-bm-apparent-speed.t.time", "title": "Positive observation interval" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-apparent-speed.t.apparentSpeed", "text": "The apparent coordinate speed" }, { "nodeId": "eq-model-bm-apparent-speed.op.division", "text": " divides" }, { "nodeId": "eq-model-bm-apparent-speed.t.rms", "text": " the typical coordinate distance" }, { "nodeId": "eq-model-bm-apparent-speed.t.time", "text": " by the chosen positive interval." } ], "spoken": "Apparent coordinate speed equals the model coordinate root mean square displacement divided by the observation interval.", "title": "Why the apparent speed depends on how you watch", "tree": { "kind": "relation", "left": { "kind": "symbol", "quantityId": "modelApparentSpeed", "termId": "eq-model-bm-apparent-speed.t.apparentSpeed" }, "opId": "eq-model-bm-apparent-speed.op.definition", "operator": "define", "right": { "denominator": { "kind": "symbol", "quantityId": "observationInterval", "termId": "eq-model-bm-apparent-speed.t.time" }, "kind": "quotient", "numerator": { "kind": "symbol", "quantityId": "rmsDisplacement1d", "termId": "eq-model-bm-apparent-speed.t.rms" }, "opId": "eq-model-bm-apparent-speed.op.division" } }, "unitSystem": "si" }, { "argument": "arg-bm-inference", "assumptions": [ "Radius, viscosity and temperature are measured independently of the displacement data.", "The same dilute-sphere model as the diffusivity.", "These are modern teaching equations in SI notation, not a transcription of the printed paper. The paper writes k for the viscosity." ], "bindings": [], "explanation": "With temperature, viscosity and radius measured independently, the observed mean square gives Avogadro's number. The radius has to come from somewhere else: the spread alone cannot separate it from N.", "id": "eq-model-bm-avogadro-inference", "kind": "equation", "live": false, "notation": "modern-pedagogical", "notes": [ { "explanation": "Drag and the observed spread.", "foundation": "bridge-fractions-ratios", "nodeId": "eq-model-bm-avogadro-inference.op.denominator", "title": "Three pi eta a times the mean square" }, { "explanation": "From many displacements.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-avogadro-inference.op.mean", "title": "The observed mean square" }, { "explanation": "Thermal energy per mole, times the interval.", "foundation": "entropy-temperature", "nodeId": "eq-model-bm-avogadro-inference.op.numerator", "title": "R T t" }, { "explanation": "Per mole.", "foundation": "bridge-fractions-ratios", "nodeId": "eq-model-bm-avogadro-inference.op.quotient", "title": "The ratio" }, { "explanation": "Counting molecules becomes a measurement of spreading.", "foundation": "error-and-inference", "nodeId": "eq-model-bm-avogadro-inference.op.relation", "title": "What it asserts" }, { "explanation": "Each observed displacement, squared.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-avogadro-inference.op.square", "title": "x squared" }, { "explanation": "How far one particle has moved along x since observation began.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-avogadro-inference.t.displacement", "title": "Displacement since the start" }, { "explanation": "The number of molecules per mole this measurement infers. An estimate, not the defined constant.", "foundation": "error-and-inference", "nodeId": "eq-model-bm-avogadro-inference.t.estimate", "title": "Estimated Avogadro number" }, { "explanation": "Energy per mole per kelvin.", "foundation": "entropy-temperature", "nodeId": "eq-model-bm-avogadro-inference.t.gas", "title": "Gas constant" }, { "explanation": "The radius of the suspended sphere, not its diameter.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-avogadro-inference.t.radius", "title": "Particle radius" }, { "explanation": "Temperature in kelvin.", "foundation": "entropy-temperature", "nodeId": "eq-model-bm-avogadro-inference.t.temperature", "title": "Absolute temperature" }, { "explanation": "The time over which displacement is observed.", "foundation": "random-walks", "nodeId": "eq-model-bm-avogadro-inference.t.time", "title": "Observation interval" }, { "explanation": "The liquid's resistance to shear. The paper writes it k, which is not Boltzmann's constant.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-avogadro-inference.t.viscosity", "title": "Viscosity" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-avogadro-inference.t.estimate", "text": "N" }, { "text": " is " }, { "nodeId": "eq-model-bm-avogadro-inference.t.gas", "text": "R" }, { "text": " times " }, { "nodeId": "eq-model-bm-avogadro-inference.t.temperature", "text": "T" }, { "text": " times " }, { "nodeId": "eq-model-bm-avogadro-inference.t.time", "text": "t" }, { "text": " over three pi times " }, { "nodeId": "eq-model-bm-avogadro-inference.t.viscosity", "text": "the viscosity" }, { "text": ", " }, { "nodeId": "eq-model-bm-avogadro-inference.t.radius", "text": "the radius" }, { "text": " and the mean of " }, { "nodeId": "eq-model-bm-avogadro-inference.t.displacement", "text": "x" }, { "text": " squared." } ], "spoken": "Avogadro's number N equals the gas constant times the temperature times the observation time, divided by the product of three pi, the viscosity, the particle radius and the mean square displacement.", "title": "From a measured spread to the number of molecules", "tree": { "kind": "relation", "left": { "kind": "symbol", "quantityId": "avogadroNumberEstimate", "termId": "eq-model-bm-avogadro-inference.t.estimate" }, "opId": "eq-model-bm-avogadro-inference.op.relation", "operator": "=", "right": { "denominator": { "args": [ { "kind": "number", "value": "3" }, { "kind": "constant", "name": "pi" }, { "kind": "symbol", "quantityId": "viscosity", "termId": "eq-model-bm-avogadro-inference.t.viscosity" }, { "kind": "symbol", "quantityId": "particleRadius", "termId": "eq-model-bm-avogadro-inference.t.radius" }, { "argument": { "base": { "kind": "symbol", "quantityId": "displacement1d", "termId": "eq-model-bm-avogadro-inference.t.displacement" }, "exponent": { "den": 1, "num": 2 }, "kind": "power", "opId": "eq-model-bm-avogadro-inference.op.square" }, "kind": "average", "opId": "eq-model-bm-avogadro-inference.op.mean" } ], "kind": "product", "opId": "eq-model-bm-avogadro-inference.op.denominator" }, "kind": "quotient", "numerator": { "args": [ { "kind": "symbol", "quantityId": "molarGasConstant", "termId": "eq-model-bm-avogadro-inference.t.gas" }, { "kind": "symbol", "quantityId": "temperature", "termId": "eq-model-bm-avogadro-inference.t.temperature" }, { "kind": "symbol", "quantityId": "observationInterval", "termId": "eq-model-bm-avogadro-inference.t.time" } ], "kind": "product", "opId": "eq-model-bm-avogadro-inference.op.numerator" }, "opId": "eq-model-bm-avogadro-inference.op.quotient" } }, "unitSystem": "si" }, { "argument": "arg-bm-diffusion-equation", "assumptions": [ "Independent steps with a symmetric step law and a finite mean square step.", "Times long compared with one step, distances large compared with one step.", "These are modern teaching equations in SI notation, not a transcription of the printed paper." ], "bindings": [], "explanation": "Where the density curves upward it grows, and where it curves downward it falls, at a rate set by D. A peak therefore flattens and spreads.", "id": "eq-model-bm-diffusion-equation", "kind": "equation", "live": false, "notation": "modern-pedagogical", "notes": [ { "explanation": "The second derivative: how the slope of the density changes along x.", "foundation": "derivatives", "nodeId": "eq-model-bm-diffusion-equation.op.curvature", "title": "Curvature in space" }, { "explanation": "The spreading rate: larger D, faster flattening.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-diffusion-equation.op.product", "title": "D times the curvature" }, { "explanation": "How fast the density at one x changes.", "foundation": "partial-derivatives", "nodeId": "eq-model-bm-diffusion-equation.op.rate", "title": "Rate of change in time" }, { "explanation": "The density changes in time exactly as fast as D times its curvature in space.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-diffusion-equation.op.relation", "title": "What it asserts" }, { "explanation": "Probability per metre of finding the particle at x after time t.", "foundation": "distributions", "nodeId": "eq-model-bm-diffusion-equation.t.density", "title": "Probability density" }, { "explanation": "Probability per metre of finding the particle at x after time t.", "foundation": "distributions", "nodeId": "eq-model-bm-diffusion-equation.t.density2", "title": "Probability density" }, { "explanation": "How fast the mean square displacement grows: half its rate of growth. Not a speed.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-diffusion-equation.t.diffusion", "title": "Diffusion coefficient" }, { "explanation": "The coordinate x along which the particles spread.", "foundation": "functions-graphs", "nodeId": "eq-model-bm-diffusion-equation.t.position", "title": "Position" }, { "explanation": "The rate is per unit of time, at one fixed place x.", "foundation": "random-walks", "nodeId": "eq-model-bm-diffusion-equation.t.time", "title": "Time" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-diffusion-equation.t.density", "text": "The rate of change of the density" }, { "text": " in " }, { "nodeId": "eq-model-bm-diffusion-equation.t.time", "text": "time" }, { "text": " is " }, { "nodeId": "eq-model-bm-diffusion-equation.t.diffusion", "text": "D" }, { "text": " times its second derivative along " }, { "nodeId": "eq-model-bm-diffusion-equation.t.position", "text": "x" }, { "text": "." } ], "spoken": "The rate of change of the probability density with time equals D times its second derivative with respect to position.", "title": "The diffusion equation", "tree": { "kind": "relation", "left": { "expression": { "kind": "symbol", "quantityId": "probabilityDensity", "termId": "eq-model-bm-diffusion-equation.t.density" }, "kind": "derivative", "opId": "eq-model-bm-diffusion-equation.op.rate", "order": 1, "partial": true, "variable": { "kind": "symbol", "quantityId": "fieldTimeCoordinate", "termId": "eq-model-bm-diffusion-equation.t.time" } }, "opId": "eq-model-bm-diffusion-equation.op.relation", "operator": "=", "right": { "args": [ { "kind": "symbol", "quantityId": "diffusionCoefficient", "termId": "eq-model-bm-diffusion-equation.t.diffusion" }, { "expression": { "kind": "symbol", "quantityId": "probabilityDensity", "termId": "eq-model-bm-diffusion-equation.t.density2" }, "kind": "derivative", "opId": "eq-model-bm-diffusion-equation.op.curvature", "order": 2, "partial": true, "variable": { "kind": "symbol", "quantityId": "positionCoordinate1d", "termId": "eq-model-bm-diffusion-equation.t.position" } } ], "kind": "product", "opId": "eq-model-bm-diffusion-equation.op.product" } }, "unitSystem": "si" }, { "argument": "arg-bm-inference", "assumptions": [ "The mean square is estimated from many observed displacements over one interval t.", "These are modern teaching equations in SI notation, not a transcription of the printed paper." ], "bindings": [], "explanation": "Turn the relation around: a measured mean square over a known interval gives D.", "id": "eq-model-bm-diffusivity-from-data", "kind": "equation", "notation": "modern-pedagogical", "notes": [ { "explanation": "Twice the interval.", "foundation": "bridge-fractions-ratios", "nodeId": "eq-model-bm-diffusivity-from-data.op.denominator", "title": "Two t" }, { "explanation": "An estimate from a finite sample.", "foundation": "error-and-inference", "nodeId": "eq-model-bm-diffusivity-from-data.op.mean", "title": "Their mean" }, { "explanation": "Metres squared per second.", "foundation": "bridge-fractions-ratios", "nodeId": "eq-model-bm-diffusivity-from-data.op.quotient", "title": "Mean square over two t" }, { "explanation": "D is measurable from displacements alone.", "foundation": "error-and-inference", "nodeId": "eq-model-bm-diffusivity-from-data.op.relation", "title": "What it asserts" }, { "explanation": "Each observed displacement, squared.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-diffusivity-from-data.op.square", "title": "x squared" }, { "explanation": "How fast the mean square displacement grows: half its rate of growth. Not a speed.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-diffusivity-from-data.t.diffusion", "title": "Diffusion coefficient" }, { "explanation": "How far one particle has moved along x since observation began.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-diffusivity-from-data.t.displacement", "title": "Displacement since the start" }, { "explanation": "The time over which displacement is observed.", "foundation": "random-walks", "nodeId": "eq-model-bm-diffusivity-from-data.t.time", "title": "Observation interval" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-diffusivity-from-data.t.diffusion", "text": "D" }, { "text": " is the mean of " }, { "nodeId": "eq-model-bm-diffusivity-from-data.t.displacement", "text": "x" }, { "text": " squared over two " }, { "nodeId": "eq-model-bm-diffusivity-from-data.t.time", "text": "t" }, { "text": "." } ], "spoken": "D equals the mean square displacement over two t.", "title": "Read D off the observations", "tree": { "kind": "relation", "left": { "kind": "symbol", "quantityId": "diffusionCoefficient", "termId": "eq-model-bm-diffusivity-from-data.t.diffusion" }, "opId": "eq-model-bm-diffusivity-from-data.op.relation", "operator": "=", "right": { "denominator": { "args": [ { "kind": "number", "value": "2" }, { "kind": "symbol", "quantityId": "observationInterval", "termId": "eq-model-bm-diffusivity-from-data.t.time" } ], "kind": "product", "opId": "eq-model-bm-diffusivity-from-data.op.denominator" }, "kind": "quotient", "numerator": { "argument": { "base": { "kind": "symbol", "quantityId": "displacement1d", "termId": "eq-model-bm-diffusivity-from-data.t.displacement" }, "exponent": { "den": 1, "num": 2 }, "kind": "power", "opId": "eq-model-bm-diffusivity-from-data.op.square" }, "kind": "average", "opId": "eq-model-bm-diffusivity-from-data.op.mean" }, "opId": "eq-model-bm-diffusivity-from-data.op.quotient" } }, "unitSystem": "si" }, { "argument": "arg-bm-diffusivity", "assumptions": [ "A dilute suspension of spheres, each moving slowly through a Newtonian liquid (Stokes drag).", "Avogadro's number is the quantity the paper sets out to measure.", "These are modern teaching equations in SI notation, not a transcription of the printed paper. The paper writes k for the viscosity." ], "bindings": [], "explanation": "Einstein's form: the gas constant per molecule, R over N, stands where the modern form writes Boltzmann's constant. A larger N would make each molecule's kick smaller and the spreading slower.", "id": "eq-model-bm-diffusivity-molar", "kind": "equation", "live": false, "notation": "modern-pedagogical", "notes": [ { "explanation": "Stokes's drag coefficient for one sphere, times Avogadro's number.", "foundation": "bridge-fractions-ratios", "nodeId": "eq-model-bm-diffusivity-molar.op.denominator", "title": "Six pi eta a N sub A" }, { "explanation": "Thermal energy per mole.", "foundation": "entropy-temperature", "nodeId": "eq-model-bm-diffusivity-molar.op.numerator", "title": "R times T" }, { "explanation": "Kicks drive spreading; drag resists it.", "foundation": "bridge-fractions-ratios", "nodeId": "eq-model-bm-diffusivity-molar.op.quotient", "title": "Thermal energy over drag" }, { "explanation": "The spreading rate is fixed by temperature, viscosity, size and the number of molecules in a mole.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-diffusivity-molar.op.relation", "title": "What it asserts" }, { "explanation": "Molecules per mole: what the measurement is meant to find.", "foundation": "bridge-scientific-notation-units", "nodeId": "eq-model-bm-diffusivity-molar.t.avogadro", "title": "Avogadro's number" }, { "explanation": "How fast the mean square displacement grows: half its rate of growth. Not a speed.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-diffusivity-molar.t.diffusion", "title": "Diffusion coefficient" }, { "explanation": "Energy per mole per kelvin.", "foundation": "entropy-temperature", "nodeId": "eq-model-bm-diffusivity-molar.t.gas", "title": "Gas constant" }, { "explanation": "The radius of the suspended sphere, not its diameter.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-diffusivity-molar.t.radius", "title": "Particle radius" }, { "explanation": "Temperature in kelvin.", "foundation": "entropy-temperature", "nodeId": "eq-model-bm-diffusivity-molar.t.temperature", "title": "Absolute temperature" }, { "explanation": "The liquid's resistance to shear. The paper writes it k, which is not Boltzmann's constant.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-diffusivity-molar.t.viscosity", "title": "Viscosity" } ], "paper": "brownian-motion", "printedGlyphs": { "avogadroConstant": "N_A" }, "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-diffusivity-molar.t.diffusion", "text": "D" }, { "text": " is " }, { "nodeId": "eq-model-bm-diffusivity-molar.t.gas", "text": "R" }, { "text": " times " }, { "nodeId": "eq-model-bm-diffusivity-molar.t.temperature", "text": "T" }, { "text": " over six pi times " }, { "nodeId": "eq-model-bm-diffusivity-molar.t.viscosity", "text": "the viscosity" }, { "text": ", " }, { "nodeId": "eq-model-bm-diffusivity-molar.t.radius", "text": "the radius" }, { "text": " and " }, { "nodeId": "eq-model-bm-diffusivity-molar.t.avogadro", "text": "Avogadro's number" }, { "text": "." } ], "spoken": "D equals R T over six pi eta a N sub A.", "title": "The same diffusivity, written with the gas constant", "tree": { "kind": "relation", "left": { "kind": "symbol", "quantityId": "diffusionCoefficient", "termId": "eq-model-bm-diffusivity-molar.t.diffusion" }, "opId": "eq-model-bm-diffusivity-molar.op.relation", "operator": "=", "right": { "denominator": { "args": [ { "kind": "number", "value": "6" }, { "kind": "constant", "name": "pi" }, { "kind": "symbol", "quantityId": "viscosity", "termId": "eq-model-bm-diffusivity-molar.t.viscosity" }, { "kind": "symbol", "quantityId": "particleRadius", "termId": "eq-model-bm-diffusivity-molar.t.radius" }, { "kind": "symbol", "quantityId": "avogadroConstant", "termId": "eq-model-bm-diffusivity-molar.t.avogadro" } ], "kind": "product", "opId": "eq-model-bm-diffusivity-molar.op.denominator" }, "kind": "quotient", "numerator": { "args": [ { "kind": "symbol", "quantityId": "molarGasConstant", "termId": "eq-model-bm-diffusivity-molar.t.gas" }, { "kind": "symbol", "quantityId": "temperature", "termId": "eq-model-bm-diffusivity-molar.t.temperature" } ], "kind": "product", "opId": "eq-model-bm-diffusivity-molar.op.numerator" }, "opId": "eq-model-bm-diffusivity-molar.op.quotient" } }, "unitSystem": "si" }, { "argument": "arg-bm-diffusivity", "assumptions": [ "Dilute, approximately spherical tracers with no-slip Stokes drag in a homogeneous Newtonian liquid.", "Wall corrections, interactions, inertia, and observation noise are not included.", "The modern SI 2019 constant set is used; this is not a historical inversion exercise." ], "bindings": [ { "experimentId": "bm-01", "instanceSlot": "primary", "outputId": "diffusionCoefficient", "quantityId": "diffusionCoefficient", "termId": "eq-model-bm-diffusivity.t.diffusion" }, { "experimentId": "bm-01", "instanceSlot": "primary", "outputId": "boltzmannConstant", "quantityId": "boltzmannConstant", "termId": "eq-model-bm-diffusivity.t.boltzmann" }, { "experimentId": "bm-01", "instanceSlot": "primary", "outputId": "temperature", "quantityId": "temperature", "termId": "eq-model-bm-diffusivity.t.temperature" }, { "experimentId": "bm-01", "instanceSlot": "primary", "outputId": "viscosity", "quantityId": "viscosity", "termId": "eq-model-bm-diffusivity.t.viscosity" }, { "experimentId": "bm-01", "instanceSlot": "primary", "outputId": "particleRadius", "quantityId": "particleRadius", "termId": "eq-model-bm-diffusivity.t.radius" } ], "explanation": "Thermal energy competes with viscous drag. Doubling viscosity halves the diffusion coefficient while reducing RMS displacement only by the square root of two. Radius means radius, not diameter.", "id": "eq-model-bm-diffusivity", "kind": "equation", "notation": "modern-pedagogical", "notes": [ { "explanation": "This equation assumes the dilute spherical tracer and Stokes-drag model; dimensional consistency alone does not establish those assumptions.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-diffusivity.op.equality", "title": "An ideal model" }, { "explanation": "A squared-distance-per-time coefficient. This value comes from the accepted model calculation.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-diffusivity.t.diffusion", "title": "Diffusion coefficient" }, { "explanation": "Increasing the denominator reduces diffusivity at fixed thermal energy. This is a model dependence, not a rule that every larger quantity must reduce motion.", "foundation": "flux-continuity", "nodeId": "eq-model-bm-diffusivity.op.division", "title": "Why divide by drag?" }, { "explanation": "Boltzmann constant times absolute temperature supplies energy per particle.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-diffusivity.op.thermalEnergy", "title": "Thermal energy scale" }, { "explanation": "The SI 2019 constant set supplies this value. It must not be treated as independent historical evidence when trying to infer molecular number.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-diffusivity.t.boltzmann", "title": "A known modern constant" }, { "explanation": "Use kelvin. The laboratory holds viscosity as an independently supplied parameter.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-diffusivity.t.temperature", "title": "Absolute temperature" }, { "explanation": "Six pi times viscosity times radius is Stokes drag per speed for an isolated sphere in the assumed low-Reynolds regime.", "foundation": "flux-continuity", "nodeId": "eq-model-bm-diffusivity.op.drag", "title": "The drag coefficient" }, { "explanation": "Viscosity is in the denominator. Doubling it at fixed temperature and radius halves D, not the RMS displacement.", "foundation": "flux-continuity", "nodeId": "eq-model-bm-diffusivity.t.viscosity", "title": "Dynamic viscosity" }, { "explanation": "The input is a radius, not a diameter. Confusing them changes the predicted coefficient by a factor of two.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-diffusivity.t.radius", "title": "Particle radius" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-diffusivity.t.diffusion", "text": "The diffusion coefficient" }, { "nodeId": "eq-model-bm-diffusivity.op.thermalEnergy", "text": " is thermal energy" }, { "nodeId": "eq-model-bm-diffusivity.op.division", "text": " divided by" }, { "nodeId": "eq-model-bm-diffusivity.op.drag", "text": " the viscous drag coefficient." } ], "spoken": "The diffusion coefficient equals the Boltzmann constant times the absolute temperature, divided by six times pi times the dynamic viscosity times the particle radius.", "title": "Resistance to motion controls spreading", "tree": { "kind": "relation", "left": { "kind": "symbol", "quantityId": "diffusionCoefficient", "termId": "eq-model-bm-diffusivity.t.diffusion" }, "opId": "eq-model-bm-diffusivity.op.equality", "operator": "=", "right": { "denominator": { "args": [ { "kind": "number", "value": "6" }, { "kind": "constant", "name": "pi" }, { "kind": "symbol", "quantityId": "viscosity", "termId": "eq-model-bm-diffusivity.t.viscosity" }, { "kind": "symbol", "quantityId": "particleRadius", "termId": "eq-model-bm-diffusivity.t.radius" } ], "kind": "product", "opId": "eq-model-bm-diffusivity.op.drag" }, "kind": "quotient", "numerator": { "args": [ { "kind": "symbol", "quantityId": "boltzmannConstant", "termId": "eq-model-bm-diffusivity.t.boltzmann" }, { "kind": "symbol", "quantityId": "temperature", "termId": "eq-model-bm-diffusivity.t.temperature" } ], "kind": "product", "opId": "eq-model-bm-diffusivity.op.thermalEnergy" }, "opId": "eq-model-bm-diffusivity.op.division" } }, "unitSystem": "si" }, { "argument": "arg-bm-gaussian", "assumptions": [ "The Gaussian density that solves the diffusion equation from a point start, one coordinate.", "These are modern teaching equations in SI notation, not a transcription of the printed paper." ], "bindings": [], "explanation": "Changing to the unitless u moves every unit into the factor in front, and leaves an integral worth the square root of pi over two. The mean square is 2Dt.", "id": "eq-model-bm-gaussian-second-moment", "kind": "equation", "layout": "rows", "live": false, "notation": "modern-pedagogical", "notes": [ { "explanation": "Averaging x squared over the Gaussian density, rewritten in the unitless variable u.", "foundation": "gaussian-distributions", "nodeId": "eq-model-bm-gaussian-second-moment.op.definition", "title": "The mean square as an integral" }, { "explanation": "The width of the Gaussian, squared, up to a factor two.", "foundation": "gaussian-distributions", "nodeId": "eq-model-bm-gaussian-second-moment.op.fourDt", "title": "Four D t" }, { "explanation": "e to the minus u squared: the bell curve without its units.", "foundation": "exponentials", "nodeId": "eq-model-bm-gaussian-second-moment.op.gaussian", "title": "The Gaussian shape" }, { "explanation": "Every value of u, from minus infinity to infinity.", "foundation": "integration", "nodeId": "eq-model-bm-gaussian-second-moment.op.integral", "title": "Over the whole line" }, { "explanation": "Each u squared counts as often as the Gaussian says it occurs.", "foundation": "integration", "nodeId": "eq-model-bm-gaussian-second-moment.op.integrand", "title": "u squared, weighted" }, { "explanation": "Averaged over many particles.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-gaussian-second-moment.op.mean", "title": "The mean square" }, { "explanation": "The exponent is negative, so large u are rare.", "foundation": "bridge-negative-numbers-direction", "nodeId": "eq-model-bm-gaussian-second-moment.op.negate", "title": "Minus" }, { "explanation": "4Dt over the square root of pi: what is left after changing to u.", "foundation": "gaussian-distributions", "nodeId": "eq-model-bm-gaussian-second-moment.op.prefactor", "title": "The factor in front" }, { "explanation": "The integral is a pure number, and working it out gives the mean square 2Dt.", "foundation": "gaussian-distributions", "nodeId": "eq-model-bm-gaussian-second-moment.op.relation", "title": "What it asserts" }, { "explanation": "The integral is the square root of pi over two, so the mean square is 2Dt.", "foundation": "gaussian-distributions", "nodeId": "eq-model-bm-gaussian-second-moment.op.result", "title": "Two D t" }, { "explanation": "The area under e to the minus u squared, which normalizes the density.", "foundation": "gaussian-distributions", "nodeId": "eq-model-bm-gaussian-second-moment.op.rootPi", "title": "Root pi" }, { "explanation": "All the metres and seconds sit in the factor in front; the integral is a pure number.", "foundation": "integration", "nodeId": "eq-model-bm-gaussian-second-moment.op.scaled", "title": "Units times a number" }, { "explanation": "Squared, so leftward and rightward moves both count.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-gaussian-second-moment.op.square", "title": "x squared" }, { "explanation": "The square of the scaled displacement.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-gaussian-second-moment.op.uSquare", "title": "u squared" }, { "explanation": "The same scaled displacement, squared.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-gaussian-second-moment.op.uSquareExp", "title": "u squared in the exponent" }, { "explanation": "How fast the spread grows. Not a speed.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-gaussian-second-moment.t.diffusion", "title": "Diffusion coefficient" }, { "explanation": "The same D.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-gaussian-second-moment.t.diffusion2", "title": "Diffusion coefficient" }, { "explanation": "How far one particle has moved along x since observation began.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-gaussian-second-moment.t.displacement", "title": "Displacement since the start" }, { "explanation": "x divided by the square root of 4Dt, a pure number.", "foundation": "bridge-scientific-notation-units", "nodeId": "eq-model-bm-gaussian-second-moment.t.scaled", "title": "Scaled displacement" }, { "explanation": "The same u, inside the exponent.", "foundation": "exponentials", "nodeId": "eq-model-bm-gaussian-second-moment.t.scaledExp", "title": "Scaled displacement" }, { "explanation": "The variable the integral runs over.", "foundation": "integration", "nodeId": "eq-model-bm-gaussian-second-moment.t.scaledVar", "title": "Summed over u" }, { "explanation": "The time over which displacement is observed.", "foundation": "random-walks", "nodeId": "eq-model-bm-gaussian-second-moment.t.time", "title": "Observation interval" }, { "explanation": "The same t.", "foundation": "random-walks", "nodeId": "eq-model-bm-gaussian-second-moment.t.time2", "title": "Observation interval" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "text": "The mean of " }, { "nodeId": "eq-model-bm-gaussian-second-moment.t.displacement", "text": "x" }, { "text": " squared is four " }, { "nodeId": "eq-model-bm-gaussian-second-moment.t.diffusion", "text": "D" }, { "text": " times " }, { "nodeId": "eq-model-bm-gaussian-second-moment.t.time", "text": "t" }, { "text": " over root pi times the integral of " }, { "nodeId": "eq-model-bm-gaussian-second-moment.t.scaled", "text": "u" }, { "text": " squared times e to the minus u squared, which is two D t." } ], "spoken": "The mean of x squared equals four D t over root pi, times the integral from minus infinity to infinity of u squared e to the minus u squared, d u, which equals two D t.", "title": "The mean square from the Gaussian", "tree": { "kind": "relation", "left": { "kind": "relation", "left": { "argument": { "base": { "kind": "symbol", "quantityId": "displacement1d", "termId": "eq-model-bm-gaussian-second-moment.t.displacement" }, "exponent": { "den": 1, "num": 2 }, "kind": "power", "opId": "eq-model-bm-gaussian-second-moment.op.square" }, "kind": "average", "opId": "eq-model-bm-gaussian-second-moment.op.mean" }, "opId": "eq-model-bm-gaussian-second-moment.op.definition", "operator": "=", "right": { "args": [ { "denominator": { "degree": 2, "kind": "root", "opId": "eq-model-bm-gaussian-second-moment.op.rootPi", "radicand": { "kind": "constant", "name": "pi" } }, "kind": "quotient", "numerator": { "args": [ { "kind": "number", "value": "4" }, { "kind": "symbol", "quantityId": "diffusionCoefficient", "termId": "eq-model-bm-gaussian-second-moment.t.diffusion" }, { "kind": "symbol", "quantityId": "observationInterval", "termId": "eq-model-bm-gaussian-second-moment.t.time" } ], "kind": "product", "opId": "eq-model-bm-gaussian-second-moment.op.fourDt" }, "opId": "eq-model-bm-gaussian-second-moment.op.prefactor" }, { "expression": { "args": [ { "base": { "kind": "symbol", "quantityId": "scaledDisplacement", "termId": "eq-model-bm-gaussian-second-moment.t.scaled" }, "exponent": { "den": 1, "num": 2 }, "kind": "power", "opId": "eq-model-bm-gaussian-second-moment.op.uSquare" }, { "argument": { "argument": { "base": { "kind": "symbol", "quantityId": "scaledDisplacement", "termId": "eq-model-bm-gaussian-second-moment.t.scaledExp" }, "exponent": { "den": 1, "num": 2 }, "kind": "power", "opId": "eq-model-bm-gaussian-second-moment.op.uSquareExp" }, "kind": "negate", "opId": "eq-model-bm-gaussian-second-moment.op.negate" }, "kind": "function", "name": "exp", "opId": "eq-model-bm-gaussian-second-moment.op.gaussian" } ], "kind": "product", "opId": "eq-model-bm-gaussian-second-moment.op.integrand" }, "kind": "integral", "lower": { "argument": { "kind": "constant", "name": "infinity" }, "kind": "negate" }, "opId": "eq-model-bm-gaussian-second-moment.op.integral", "upper": { "kind": "constant", "name": "infinity" }, "variable": { "kind": "symbol", "quantityId": "scaledDisplacement", "termId": "eq-model-bm-gaussian-second-moment.t.scaledVar" } } ], "kind": "product", "opId": "eq-model-bm-gaussian-second-moment.op.scaled" } }, "opId": "eq-model-bm-gaussian-second-moment.op.relation", "operator": "=", "right": { "args": [ { "kind": "number", "value": "2" }, { "kind": "symbol", "quantityId": "diffusionCoefficient", "termId": "eq-model-bm-gaussian-second-moment.t.diffusion2" }, { "kind": "symbol", "quantityId": "observationInterval", "termId": "eq-model-bm-gaussian-second-moment.t.time2" } ], "kind": "product", "opId": "eq-model-bm-gaussian-second-moment.op.result" } }, "unitSystem": "si" }, { "argument": "arg-bm-diffusion-equation", "assumptions": [ "Jumps in successive intervals are independent and symmetric, one coordinate.", "These are modern teaching equations in SI notation, not a transcription of the printed paper." ], "bindings": [], "explanation": "D is the mean squared jump in one short interval, divided by twice that interval. Nothing about the fluid appears here yet; that comes with Stokes' law.", "id": "eq-model-bm-jump-moment", "kind": "equation", "live": false, "notation": "modern-pedagogical", "notes": [ { "explanation": "Adds up the contribution of every possible jump, from any leftward to any rightward one.", "foundation": "integration", "nodeId": "eq-model-bm-jump-moment.op.integral", "title": "Over every jump" }, { "explanation": "One over two tau turns a squared distance per jump into a rate.", "foundation": "bridge-fractions-ratios", "nodeId": "eq-model-bm-jump-moment.op.perInterval", "title": "Per two intervals" }, { "explanation": "The rate factor times the averaged squared jump.", "foundation": "bridge-fractions-ratios", "nodeId": "eq-model-bm-jump-moment.op.product", "title": "Times" }, { "explanation": "D is set by the jumps themselves: their mean square, per unit of time.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-jump-moment.op.relation", "title": "What it asserts" }, { "explanation": "Squared, so leftward and rightward jumps both add.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-jump-moment.op.square", "title": "Jump squared" }, { "explanation": "Twice the interval; the two comes from the second derivative of the density.", "foundation": "taylor-expansion", "nodeId": "eq-model-bm-jump-moment.op.twoTau", "title": "Two tau" }, { "explanation": "Each squared jump counts in proportion to how often it happens.", "foundation": "distributions", "nodeId": "eq-model-bm-jump-moment.op.weighted", "title": "Weighted by how likely" }, { "explanation": "Paper 2's phi: probability per metre of a jump of this size. A density, not a probability.", "foundation": "distributions", "nodeId": "eq-model-bm-jump-moment.t.density", "title": "How likely a jump is" }, { "explanation": "How fast the spread grows, in square metres per second. Not a speed.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-jump-moment.t.diffusion", "title": "Diffusion coefficient" }, { "explanation": "Paper 2's tau: short enough for the density to change little, long enough for jumps to be independent.", "foundation": "random-walks", "nodeId": "eq-model-bm-jump-moment.t.interval", "title": "The short interval" }, { "explanation": "How far a particle moves in one interval tau.", "foundation": "random-walks", "nodeId": "eq-model-bm-jump-moment.t.jump", "title": "A jump" }, { "explanation": "phi is read at this jump.", "foundation": "functions-graphs", "nodeId": "eq-model-bm-jump-moment.t.jumpArg", "title": "The jump it describes" }, { "explanation": "The variable the integral runs over.", "foundation": "integration", "nodeId": "eq-model-bm-jump-moment.t.jumpVar", "title": "Summed over jumps" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-jump-moment.t.diffusion", "text": "The diffusion coefficient" }, { "text": " is one over two " }, { "nodeId": "eq-model-bm-jump-moment.t.interval", "text": "tau" }, { "text": " times the mean of the squared jump " }, { "nodeId": "eq-model-bm-jump-moment.t.jump", "text": "Delta" }, { "text": ", each jump weighted by its density " }, { "nodeId": "eq-model-bm-jump-moment.t.density", "text": "phi" }, { "text": "." } ], "spoken": "D equals one over two tau times the integral, from minus infinity to infinity, of Delta squared times phi of Delta, d Delta.", "title": "The diffusion coefficient from the jumps", "tree": { "kind": "relation", "left": { "kind": "symbol", "quantityId": "diffusionCoefficient", "termId": "eq-model-bm-jump-moment.t.diffusion" }, "opId": "eq-model-bm-jump-moment.op.relation", "operator": "=", "right": { "args": [ { "denominator": { "args": [ { "kind": "number", "value": "2" }, { "kind": "symbol", "quantityId": "stepInterval", "termId": "eq-model-bm-jump-moment.t.interval" } ], "kind": "product", "opId": "eq-model-bm-jump-moment.op.twoTau" }, "kind": "quotient", "numerator": { "kind": "number", "value": "1" }, "opId": "eq-model-bm-jump-moment.op.perInterval" }, { "expression": { "args": [ { "base": { "kind": "symbol", "quantityId": "displacementIncrement", "termId": "eq-model-bm-jump-moment.t.jump" }, "exponent": { "den": 1, "num": 2 }, "kind": "power", "opId": "eq-model-bm-jump-moment.op.square" }, { "at": { "kind": "symbol", "quantityId": "displacementIncrement", "termId": "eq-model-bm-jump-moment.t.jumpArg" }, "kind": "symbol", "quantityId": "transitionKernel", "termId": "eq-model-bm-jump-moment.t.density" } ], "kind": "product", "opId": "eq-model-bm-jump-moment.op.weighted" }, "kind": "integral", "lower": { "argument": { "kind": "constant", "name": "infinity" }, "kind": "negate" }, "opId": "eq-model-bm-jump-moment.op.integral", "upper": { "kind": "constant", "name": "infinity" }, "variable": { "kind": "symbol", "quantityId": "displacementIncrement", "termId": "eq-model-bm-jump-moment.t.jumpVar" } } ], "kind": "product", "opId": "eq-model-bm-jump-moment.op.product" } }, "unitSystem": "si" }, { "argument": "arg-bm-independent-steps", "assumptions": [ "Independent, zero-mean steps.", "These are modern teaching equations in SI notation, not a transcription of the printed paper." ], "bindings": [], "explanation": "The squares of independent steps add, so the mean square grows in proportion to time, not to its square.", "id": "eq-model-bm-mean-square-growth", "kind": "equation", "live": false, "notation": "modern-pedagogical", "notes": [ { "explanation": "Averaged over many particles.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-mean-square-growth.op.mean", "title": "The mean square" }, { "explanation": "Linear in time.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-mean-square-growth.op.product", "title": "Two D t" }, { "explanation": "Four times as long gives four times the mean square.", "foundation": "random-walks", "nodeId": "eq-model-bm-mean-square-growth.op.relation", "title": "What it asserts" }, { "explanation": "Squared, so both directions count.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-mean-square-growth.op.square", "title": "x squared" }, { "explanation": "How fast the mean square displacement grows: half its rate of growth. Not a speed.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-mean-square-growth.t.diffusion", "title": "Diffusion coefficient" }, { "explanation": "How far one particle has moved along x since observation began.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-mean-square-growth.t.displacement", "title": "Displacement since the start" }, { "explanation": "The time over which displacement is observed.", "foundation": "random-walks", "nodeId": "eq-model-bm-mean-square-growth.t.time", "title": "Observation interval" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-mean-square-growth.t.displacement", "text": "The mean of x" }, { "text": " squared is two times " }, { "nodeId": "eq-model-bm-mean-square-growth.t.diffusion", "text": "D" }, { "text": " times " }, { "nodeId": "eq-model-bm-mean-square-growth.t.time", "text": "t" }, { "text": "." } ], "spoken": "The mean square displacement equals two D t.", "title": "The mean square grows in proportion to time", "tree": { "kind": "relation", "left": { "argument": { "base": { "kind": "symbol", "quantityId": "displacement1d", "termId": "eq-model-bm-mean-square-growth.t.displacement" }, "exponent": { "den": 1, "num": 2 }, "kind": "power", "opId": "eq-model-bm-mean-square-growth.op.square" }, "kind": "average", "opId": "eq-model-bm-mean-square-growth.op.mean" }, "opId": "eq-model-bm-mean-square-growth.op.relation", "operator": "=", "right": { "args": [ { "kind": "number", "value": "2" }, { "kind": "symbol", "quantityId": "diffusionCoefficient", "termId": "eq-model-bm-mean-square-growth.t.diffusion" }, { "kind": "symbol", "quantityId": "observationInterval", "termId": "eq-model-bm-mean-square-growth.t.time" } ], "kind": "product", "opId": "eq-model-bm-mean-square-growth.op.product" } }, "unitSystem": "si" }, { "argument": "arg-bm-diffusion-equation", "assumptions": [ "Jumps in successive intervals are independent, and the jump density phi is the same at every place and every time.", "These are modern teaching equations in SI notation, not a transcription of the printed paper." ], "bindings": [], "explanation": "A particle found at x after one more interval tau was somewhere else a jump earlier. Add up every jump Delta that could have brought it: the density at x minus Delta now, times how likely a jump of that size is.", "id": "eq-model-bm-next-density", "kind": "equation", "layout": "break", "live": false, "notation": "modern-pedagogical", "notes": [ { "explanation": "The density one interval later is fixed by the density now and the jump density alone.", "foundation": "probability-independence", "nodeId": "eq-model-bm-next-density.op.relation", "title": "What it asserts" }, { "explanation": "The time t plus one step interval tau: a later time, not a longer interval.", "foundation": "random-walks", "nodeId": "eq-model-bm-next-density.op.later", "title": "One interval later" }, { "explanation": "Adds up the arrivals from every possible jump, leftward and rightward.", "foundation": "integration", "nodeId": "eq-model-bm-next-density.op.integral", "title": "Over every jump" }, { "explanation": "The density a jump away, weighted by the chance of that jump.", "foundation": "probability-independence", "nodeId": "eq-model-bm-next-density.op.weighted", "title": "Where it came from, times how likely" }, { "explanation": "x minus Delta: the place a jump of Delta starts from if it ends at x.", "foundation": "bridge-negative-numbers-direction", "nodeId": "eq-model-bm-next-density.op.origin", "title": "Where it came from" }, { "explanation": "Probability per metre of finding the particle at x, one interval after t.", "foundation": "distributions", "nodeId": "eq-model-bm-next-density.t.density", "title": "The density later" }, { "explanation": "The place where the density is read.", "foundation": "functions-graphs", "nodeId": "eq-model-bm-next-density.t.x", "title": "Position" }, { "explanation": "The time now. Every particle was at x = 0 when it read zero.", "foundation": "functions-graphs", "nodeId": "eq-model-bm-next-density.t.time", "title": "Time" }, { "explanation": "Paper 2's tau: short, yet long enough for successive jumps to be independent.", "foundation": "random-walks", "nodeId": "eq-model-bm-next-density.t.interval", "title": "The step interval" }, { "explanation": "The density at the starting place x minus Delta, at time t.", "foundation": "distributions", "nodeId": "eq-model-bm-next-density.t.densityNow", "title": "The density now" }, { "explanation": "The same x.", "foundation": "functions-graphs", "nodeId": "eq-model-bm-next-density.t.x2", "title": "Position" }, { "explanation": "How far a particle moves in one interval tau.", "foundation": "random-walks", "nodeId": "eq-model-bm-next-density.t.jump", "title": "A jump" }, { "explanation": "The same t: the density now.", "foundation": "functions-graphs", "nodeId": "eq-model-bm-next-density.t.time2", "title": "Time" }, { "explanation": "Paper 2's phi: probability per metre of a jump of this size.", "foundation": "distributions", "nodeId": "eq-model-bm-next-density.t.kernel", "title": "How likely a jump is" }, { "explanation": "phi is read at this jump.", "foundation": "functions-graphs", "nodeId": "eq-model-bm-next-density.t.jumpArg", "title": "The jump it describes" }, { "explanation": "The variable the integral runs over.", "foundation": "integration", "nodeId": "eq-model-bm-next-density.t.jumpVar", "title": "Summed over jumps" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-next-density.t.density", "text": "The density" }, { "text": " one " }, { "nodeId": "eq-model-bm-next-density.t.interval", "text": "interval" }, { "text": " later is the " }, { "nodeId": "eq-model-bm-next-density.t.densityNow", "text": "density now" }, { "text": " one " }, { "nodeId": "eq-model-bm-next-density.t.jump", "text": "jump" }, { "text": " away, weighted by " }, { "nodeId": "eq-model-bm-next-density.t.kernel", "text": "how likely that jump is" }, { "text": ", summed over every jump." } ], "spoken": "p of x and t plus tau equals the integral, over every jump Delta from minus infinity to infinity, of p of x minus Delta and t, times phi of Delta.", "title": "The density one interval later", "tree": { "kind": "relation", "left": { "args": [ { "kind": "symbol", "quantityId": "positionCoordinate1d", "termId": "eq-model-bm-next-density.t.x" }, { "args": [ { "kind": "symbol", "quantityId": "fieldTimeCoordinate", "termId": "eq-model-bm-next-density.t.time" }, { "kind": "symbol", "quantityId": "stepInterval", "termId": "eq-model-bm-next-density.t.interval" } ], "kind": "sum", "opId": "eq-model-bm-next-density.op.later" } ], "kind": "symbol", "quantityId": "probabilityDensity", "termId": "eq-model-bm-next-density.t.density" }, "opId": "eq-model-bm-next-density.op.relation", "operator": "=", "right": { "expression": { "args": [ { "args": [ { "args": [ { "kind": "symbol", "quantityId": "positionCoordinate1d", "termId": "eq-model-bm-next-density.t.x2" }, { "argument": { "kind": "symbol", "quantityId": "displacementIncrement", "termId": "eq-model-bm-next-density.t.jump" }, "kind": "negate" } ], "kind": "sum", "opId": "eq-model-bm-next-density.op.origin" }, { "kind": "symbol", "quantityId": "fieldTimeCoordinate", "termId": "eq-model-bm-next-density.t.time2" } ], "kind": "symbol", "quantityId": "probabilityDensity", "termId": "eq-model-bm-next-density.t.densityNow" }, { "at": { "kind": "symbol", "quantityId": "displacementIncrement", "termId": "eq-model-bm-next-density.t.jumpArg" }, "kind": "symbol", "quantityId": "transitionKernel", "termId": "eq-model-bm-next-density.t.kernel" } ], "kind": "product", "opId": "eq-model-bm-next-density.op.weighted" }, "kind": "integral", "lower": { "argument": { "kind": "constant", "name": "infinity" }, "kind": "negate" }, "opId": "eq-model-bm-next-density.op.integral", "upper": { "kind": "constant", "name": "infinity" }, "variable": { "kind": "symbol", "quantityId": "displacementIncrement", "termId": "eq-model-bm-next-density.t.jumpVar" } } }, "unitSystem": "si" }, { "argument": "arg-bm-gaussian", "assumptions": [ "The Gaussian spreading of independent steps, one coordinate.", "These are modern teaching equations in SI notation, not a transcription of the printed paper." ], "bindings": [], "explanation": "The root of the mean square is a distance in metres. It grows as the square root of time: four times as long, twice as far.", "id": "eq-model-bm-rms-from-mean-square", "kind": "equation", "live": false, "notation": "modern-pedagogical", "notes": [ { "explanation": "Lambda x is the root mean square.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-rms-from-mean-square.op.definition", "title": "The definition" }, { "explanation": "Averaged over many particles.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-rms-from-mean-square.op.mean", "title": "The mean square" }, { "explanation": "The model's mean square after time t.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-rms-from-mean-square.op.product", "title": "Two D t" }, { "explanation": "The measurable distance follows from D and t alone.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-rms-from-mean-square.op.relation", "title": "What it asserts" }, { "explanation": "Back to metres.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-rms-from-mean-square.op.rootMean", "title": "Its square root" }, { "explanation": "Grows as the square root of t.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-rms-from-mean-square.op.rootModel", "title": "Square root of 2 D t" }, { "explanation": "Squared, so leftward and rightward moves both count.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-rms-from-mean-square.op.square", "title": "x squared" }, { "explanation": "How fast the mean square displacement grows: half its rate of growth. Not a speed.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-rms-from-mean-square.t.diffusion", "title": "Diffusion coefficient" }, { "explanation": "How far one particle has moved along x since observation began.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-rms-from-mean-square.t.displacement", "title": "Displacement since the start" }, { "explanation": "The root mean square displacement along one coordinate.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-rms-from-mean-square.t.rms", "title": "Typical distance along x" }, { "explanation": "The time over which displacement is observed.", "foundation": "random-walks", "nodeId": "eq-model-bm-rms-from-mean-square.t.time", "title": "Observation interval" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-rms-from-mean-square.t.rms", "text": "The typical distance" }, { "text": " is the square root of the mean of " }, { "nodeId": "eq-model-bm-rms-from-mean-square.t.displacement", "text": "x" }, { "text": " squared, which is the square root of two times " }, { "nodeId": "eq-model-bm-rms-from-mean-square.t.diffusion", "text": "D" }, { "text": " times " }, { "nodeId": "eq-model-bm-rms-from-mean-square.t.time", "text": "t" }, { "text": "." } ], "spoken": "Lambda x is the square root of the mean square displacement, which is the square root of two D t.", "title": "The typical distance is the root of the mean square", "tree": { "kind": "relation", "left": { "kind": "relation", "left": { "kind": "symbol", "quantityId": "rmsDisplacement1d", "termId": "eq-model-bm-rms-from-mean-square.t.rms" }, "opId": "eq-model-bm-rms-from-mean-square.op.definition", "operator": "=", "right": { "degree": 2, "kind": "root", "opId": "eq-model-bm-rms-from-mean-square.op.rootMean", "radicand": { "argument": { "base": { "kind": "symbol", "quantityId": "displacement1d", "termId": "eq-model-bm-rms-from-mean-square.t.displacement" }, "exponent": { "den": 1, "num": 2 }, "kind": "power", "opId": "eq-model-bm-rms-from-mean-square.op.square" }, "kind": "average", "opId": "eq-model-bm-rms-from-mean-square.op.mean" } } }, "opId": "eq-model-bm-rms-from-mean-square.op.relation", "operator": "=", "right": { "degree": 2, "kind": "root", "opId": "eq-model-bm-rms-from-mean-square.op.rootModel", "radicand": { "args": [ { "kind": "number", "value": "2" }, { "kind": "symbol", "quantityId": "diffusionCoefficient", "termId": "eq-model-bm-rms-from-mean-square.t.diffusion" }, { "kind": "symbol", "quantityId": "observationInterval", "termId": "eq-model-bm-rms-from-mean-square.t.time" } ], "kind": "product", "opId": "eq-model-bm-rms-from-mean-square.op.product" } } }, "unitSystem": "si" }, { "argument": "arg-bm-observable", "assumptions": [ "Independent, zero-mean Gaussian displacement increments in a homogeneous liquid.", "One-coordinate model statistic, not a measured speed or a sample estimate.", "The overdamped regime is assumed rather than established from additional particle and fluid measurements." ], "bindings": [ { "experimentId": "bm-01", "instanceSlot": "primary", "outputId": "rmsDisplacement1d", "quantityId": "rmsDisplacement1d", "termId": "eq-model-bm-rms.t.rms" }, { "experimentId": "bm-01", "instanceSlot": "primary", "outputId": "diffusionCoefficient", "quantityId": "diffusionCoefficient", "termId": "eq-model-bm-rms.t.diffusion" }, { "experimentId": "bm-01", "instanceSlot": "primary", "outputId": "observationInterval", "quantityId": "observationInterval", "termId": "eq-model-bm-rms.t.time" } ], "explanation": "Squaring measures spread without cancellation between directions. Taking the positive square root turns squared distance back into a distance. Four times the observation interval gives twice the model RMS, not four times.", "id": "eq-model-bm-rms", "kind": "equation", "notation": "modern-pedagogical", "notes": [ { "explanation": "This equality concerns the ideal model. A finite synthetic ensemble fluctuates around it; its sample RMS is displayed separately.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-rms.op.equality", "title": "A model relation" }, { "explanation": "One coordinate, not the total three-dimensional distance. Model RMS and sample RMS have different meanings.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-rms.t.rms", "title": "Coordinate RMS" }, { "explanation": "Two D t has units of squared length. Its positive square root has units of length and defines the typical displacement.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-rms.op.squareRoot", "title": "Why a square root?" }, { "explanation": "Independent zero-mean increments add their variances. The definition of D makes the coordinate mean square equal to 2 D t.", "foundation": "probability-independence", "nodeId": "eq-model-bm-rms.op.meanSquare", "title": "Build the mean square" }, { "explanation": "This is the accepted model diffusivity, not a rate inferred from the synthetic data.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-rms.t.diffusion", "title": "Diffusion coefficient" }, { "explanation": "This is the interval used for the displayed displacement, not the simulation frame rate. Re-observing the trial preserves its paths.", "foundation": "random-walks", "nodeId": "eq-model-bm-rms.t.time", "title": "Observation interval" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-rms.t.rms", "text": "The typical coordinate distance" }, { "nodeId": "eq-model-bm-rms.op.squareRoot", "text": " is the positive square root" }, { "nodeId": "eq-model-bm-rms.t.diffusion", "text": " of twice the diffusion coefficient" }, { "nodeId": "eq-model-bm-rms.t.time", "text": " times the observation interval." } ], "spoken": "The model coordinate root mean square displacement equals the square root of two times the diffusion coefficient times the observation interval.", "title": "From spreading to a measurable distance", "tree": { "kind": "relation", "left": { "kind": "symbol", "quantityId": "rmsDisplacement1d", "termId": "eq-model-bm-rms.t.rms" }, "opId": "eq-model-bm-rms.op.equality", "operator": "=", "right": { "degree": 2, "kind": "root", "opId": "eq-model-bm-rms.op.squareRoot", "radicand": { "args": [ { "kind": "number", "value": "2" }, { "kind": "symbol", "quantityId": "diffusionCoefficient", "termId": "eq-model-bm-rms.t.diffusion" }, { "kind": "symbol", "quantityId": "observationInterval", "termId": "eq-model-bm-rms.t.time" } ], "kind": "product", "opId": "eq-model-bm-rms.op.meanSquare" } } }, "unitSystem": "si" }, { "argument": "arg-bm-independent-steps", "assumptions": [ "Ordinary algebra; A and B are any numbers.", "These are modern teaching equations in SI notation, not a transcription of the printed paper." ], "bindings": [], "explanation": "The cross term 2AB is what averages to zero when the next step is independent of the ones before it, which is why the mean squares simply add.", "id": "eq-model-bm-square-of-sum", "kind": "equation", "live": false, "notation": "modern-pedagogical", "notes": [ { "explanation": "A times A.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-square-of-sum.op.aSquared", "title": "A squared" }, { "explanation": "B times B.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-square-of-sum.op.bSquared", "title": "B squared" }, { "explanation": "Twice A times B: it vanishes on average for an independent, zero-mean B.", "foundation": "probability-independence", "nodeId": "eq-model-bm-square-of-sum.op.cross", "title": "The cross term" }, { "explanation": "Each term of the bracket times each term.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-square-of-sum.op.expanded", "title": "Three terms" }, { "explanation": "An identity: true for every A and B.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-square-of-sum.op.relation", "title": "What it asserts" }, { "explanation": "The whole bracket times itself, not each term squared.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-square-of-sum.op.squareSum", "title": "The sum, squared" }, { "explanation": "The sum to be squared.", "foundation": "bridge-sum-average", "nodeId": "eq-model-bm-square-of-sum.op.sum", "title": "A plus B" }, { "explanation": "A stands for any number; here, the sum of the earlier steps.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-square-of-sum.t.a", "title": "Any number" }, { "explanation": "A stands for any number; here, the sum of the earlier steps.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-square-of-sum.t.a2", "title": "Any number" }, { "explanation": "A stands for any number; here, the sum of the earlier steps.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-square-of-sum.t.a3", "title": "Any number" }, { "explanation": "B stands for any number; here, the next step.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-square-of-sum.t.b", "title": "Any other number" }, { "explanation": "B stands for any number; here, the next step.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-square-of-sum.t.b2", "title": "Any other number" }, { "explanation": "B stands for any number; here, the next step.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-square-of-sum.t.b3", "title": "Any other number" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-square-of-sum.t.a", "text": "A" }, { "text": " plus " }, { "nodeId": "eq-model-bm-square-of-sum.t.b", "text": "B" }, { "text": ", squared, is " }, { "nodeId": "eq-model-bm-square-of-sum.t.a2", "text": "A" }, { "text": " squared plus two " }, { "nodeId": "eq-model-bm-square-of-sum.t.a3", "text": "A" }, { "nodeId": "eq-model-bm-square-of-sum.t.b3", "text": "B" }, { "text": " plus " }, { "nodeId": "eq-model-bm-square-of-sum.t.b2", "text": "B" }, { "text": " squared." } ], "spoken": "A plus B, squared, equals A squared plus two A B plus B squared.", "title": "The square of a sum", "tree": { "kind": "relation", "left": { "base": { "args": [ { "kind": "symbol", "quantityId": "genericNumberA", "termId": "eq-model-bm-square-of-sum.t.a" }, { "kind": "symbol", "quantityId": "genericNumberB", "termId": "eq-model-bm-square-of-sum.t.b" } ], "kind": "sum", "opId": "eq-model-bm-square-of-sum.op.sum" }, "exponent": { "den": 1, "num": 2 }, "kind": "power", "opId": "eq-model-bm-square-of-sum.op.squareSum" }, "opId": "eq-model-bm-square-of-sum.op.relation", "operator": "=", "right": { "args": [ { "base": { "kind": "symbol", "quantityId": "genericNumberA", "termId": "eq-model-bm-square-of-sum.t.a2" }, "exponent": { "den": 1, "num": 2 }, "kind": "power", "opId": "eq-model-bm-square-of-sum.op.aSquared" }, { "args": [ { "kind": "number", "value": "2" }, { "kind": "symbol", "quantityId": "genericNumberA", "termId": "eq-model-bm-square-of-sum.t.a3" }, { "kind": "symbol", "quantityId": "genericNumberB", "termId": "eq-model-bm-square-of-sum.t.b3" } ], "kind": "product", "opId": "eq-model-bm-square-of-sum.op.cross" }, { "base": { "kind": "symbol", "quantityId": "genericNumberB", "termId": "eq-model-bm-square-of-sum.t.b2" }, "exponent": { "den": 1, "num": 2 }, "kind": "power", "opId": "eq-model-bm-square-of-sum.op.bSquared" } ], "kind": "sum", "opId": "eq-model-bm-square-of-sum.op.expanded" } }, "unitSystem": "si" }, { "argument": "arg-bm-independent-steps", "assumptions": [ "The n steps are independent, each with mean zero and the same mean square l squared.", "These are modern teaching equations in SI notation, not a transcription of the printed paper." ], "bindings": [], "explanation": "Square the sum of n steps and average. Every cross term between two different steps averages to zero, because the steps are independent and centred. Only the n squared steps remain, each contributing l squared.", "id": "eq-model-bm-sum-of-steps", "kind": "equation", "live": false, "notation": "modern-pedagogical", "notes": [ { "explanation": "Each step has the same mean square l squared, so n of them give n l squared.", "foundation": "random-walks", "nodeId": "eq-model-bm-sum-of-steps.op.relation", "title": "Equal step sizes" }, { "explanation": "Independent, centred steps: every cross term averages to zero, leaving only the squares.", "foundation": "probability-independence", "nodeId": "eq-model-bm-sum-of-steps.op.crossTermsVanish", "title": "The cross terms drop out" }, { "explanation": "The average, over many walkers, of the square of where each ends up.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-sum-of-steps.op.meanSquareOfSum", "title": "Mean square of the net displacement" }, { "explanation": "The net displacement squared, so left and right count alike.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-sum-of-steps.op.squareOfSum", "title": "Squared" }, { "explanation": "The n steps added, each with its sign.", "foundation": "random-walks", "nodeId": "eq-model-bm-sum-of-steps.op.sumOfSteps", "title": "The net displacement" }, { "explanation": "The sum, over the n steps, of each step's mean square.", "foundation": "bridge-sum-average", "nodeId": "eq-model-bm-sum-of-steps.op.sumOfMeanSquares", "title": "Add the mean squares" }, { "explanation": "The average of one step's square over many walkers.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-sum-of-steps.op.meanStepSquare", "title": "One step's mean square" }, { "explanation": "Step i squared.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-sum-of-steps.op.stepSquare", "title": "A step squared" }, { "explanation": "n equal contributions.", "foundation": "bridge-sum-average", "nodeId": "eq-model-bm-sum-of-steps.op.nTimes", "title": "n times" }, { "explanation": "The mean square of one step.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-sum-of-steps.op.lSquare", "title": "l squared" }, { "explanation": "Delta sub i: the i-th signed step of the walk.", "foundation": "random-walks", "nodeId": "eq-model-bm-sum-of-steps.t.step", "title": "Step i" }, { "explanation": "The same Delta sub i.", "foundation": "random-walks", "nodeId": "eq-model-bm-sum-of-steps.t.step2", "title": "Step i" }, { "explanation": "n: how many independent steps the walk takes.", "foundation": "random-walks", "nodeId": "eq-model-bm-sum-of-steps.t.n", "title": "Number of steps" }, { "explanation": "The same n.", "foundation": "random-walks", "nodeId": "eq-model-bm-sum-of-steps.t.n2", "title": "Number of steps" }, { "explanation": "The same n.", "foundation": "random-walks", "nodeId": "eq-model-bm-sum-of-steps.t.n3", "title": "Number of steps" }, { "explanation": "l: the root mean square length of one step.", "foundation": "mean-variance-rms", "nodeId": "eq-model-bm-sum-of-steps.t.l", "title": "Step length" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "text": "The mean square of the " }, { "nodeId": "eq-model-bm-sum-of-steps.op.sumOfSteps", "text": "net displacement" }, { "text": " is the " }, { "nodeId": "eq-model-bm-sum-of-steps.op.sumOfMeanSquares", "text": "sum of the steps' mean squares" }, { "text": ": " }, { "nodeId": "eq-model-bm-sum-of-steps.t.n3", "text": "n" }, { "text": " times " }, { "nodeId": "eq-model-bm-sum-of-steps.t.l", "text": "l" }, { "text": " squared." } ], "spoken": "The mean of the square of the sum, for i from 1 to n, of Delta sub i, equals the sum, for i from 1 to n, of the mean of Delta sub i squared, which equals n l squared.", "title": "The mean square of n independent steps", "tree": { "kind": "relation", "left": { "kind": "relation", "left": { "argument": { "base": { "expression": { "index": "i", "kind": "symbol", "quantityId": "displacementIncrement", "termId": 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"eq-model-bm-sum-of-steps.op.sumOfMeanSquares", "to": { "kind": "symbol", "quantityId": "walkStepCount", "termId": "eq-model-bm-sum-of-steps.t.n2" } } }, "opId": "eq-model-bm-sum-of-steps.op.relation", "operator": "=", "right": { "args": [ { "kind": "symbol", "quantityId": "walkStepCount", "termId": "eq-model-bm-sum-of-steps.t.n3" }, { "base": { "kind": "symbol", "quantityId": "stepRms", "termId": "eq-model-bm-sum-of-steps.t.l" }, "exponent": { "den": 1, "num": 2 }, "kind": "power", "opId": "eq-model-bm-sum-of-steps.op.lSquare" } ], "kind": "product", "opId": "eq-model-bm-sum-of-steps.op.nTimes" } }, "unitSystem": "si" }, { "argument": "arg-bm-diffusion-equation", "assumptions": [ "Jumps small compared with the distance over which the density changes, so terms beyond Delta squared are dropped.", "These are modern teaching equations in SI notation, not a transcription of the printed paper." ], "bindings": [], "explanation": "Expanding the density about x in the jump Delta keeps its value, its slope and its curvature. Averaged over the jump density, the slope term cancels and the curvature term leaves the diffusion coefficient.", "id": "eq-model-bm-taylor-in-space", "kind": "equation", "live": false, "notation": "modern-pedagogical", "notes": [ { "explanation": "The second derivative of p along x.", "foundation": "partial-derivatives", "nodeId": "eq-model-bm-taylor-in-space.op.curve", "title": "Curvature along x" }, { "explanation": "Half Delta squared times the curvature.", "foundation": "taylor-expansion", "nodeId": "eq-model-bm-taylor-in-space.op.curveTerm", "title": "The curvature term" }, { "explanation": "The Taylor coefficient of the second derivative.", "foundation": "taylor-expansion", "nodeId": "eq-model-bm-taylor-in-space.op.half", "title": "Delta squared over two" }, { "explanation": "The jump times itself; it will be averaged over phi.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-taylor-in-space.op.jumpSquare", "title": "Delta squared" }, { "explanation": "Where the particle came from: a position minus a displacement is a position.", "foundation": "bridge-negative-numbers-direction", "nodeId": "eq-model-bm-taylor-in-space.op.origin", "title": "x minus Delta" }, { "explanation": "The density one jump away is, approximately, its value here corrected by its slope and its curvature.", "foundation": "taylor-expansion", "nodeId": "eq-model-bm-taylor-in-space.op.relation", "title": "What it asserts" }, { "explanation": "The first three terms of a Taylor series in Delta.", "foundation": "taylor-expansion", "nodeId": "eq-model-bm-taylor-in-space.op.series", "title": "Value, slope and curvature" }, { "explanation": "How fast p changes along x.", "foundation": "partial-derivatives", "nodeId": "eq-model-bm-taylor-in-space.op.slope", "title": "Slope along x" }, { "explanation": "Delta times the slope, subtracted because the jump is backwards.", "foundation": "taylor-expansion", "nodeId": "eq-model-bm-taylor-in-space.op.slopeTerm", "title": "The slope term" }, { "explanation": "p.", "foundation": "distributions", "nodeId": "eq-model-bm-taylor-in-space.t.curveP", "title": "Density" }, { "explanation": "x.", "foundation": "derivatives", "nodeId": "eq-model-bm-taylor-in-space.t.curveX", "title": "Position" }, { "explanation": "p where a particle stood before a jump of Delta, at the same time t.", "foundation": "distributions", "nodeId": "eq-model-bm-taylor-in-space.t.density", "title": "Density one jump back" }, { "explanation": "Delta, one interval's displacement.", "foundation": "random-walks", "nodeId": "eq-model-bm-taylor-in-space.t.jump", "title": "The jump" }, { "explanation": "The same Delta.", "foundation": "random-walks", "nodeId": "eq-model-bm-taylor-in-space.t.jump2", "title": "The jump" }, { "explanation": "The same Delta.", "foundation": "random-walks", "nodeId": "eq-model-bm-taylor-in-space.t.jump3", "title": "The jump" }, { "explanation": "p.", "foundation": "distributions", "nodeId": "eq-model-bm-taylor-in-space.t.slopeP", "title": "Density" }, { "explanation": "x.", "foundation": "derivatives", "nodeId": "eq-model-bm-taylor-in-space.t.slopeX", "title": "Position" }, { "explanation": "The time at which the density is read, held fixed while x varies.", "foundation": "random-walks", "nodeId": "eq-model-bm-taylor-in-space.t.time", "title": "Time" }, { "explanation": "p at x, the value the series starts from.", "foundation": "distributions", "nodeId": "eq-model-bm-taylor-in-space.t.value", "title": "The density here" }, { "explanation": "x, where the density is read.", "foundation": "derivatives", "nodeId": "eq-model-bm-taylor-in-space.t.x", "title": "Position" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-taylor-in-space.t.density", "text": "The density" }, { "text": " one jump back is its value here, minus " }, { "nodeId": "eq-model-bm-taylor-in-space.t.jump2", "text": "Delta" }, { "text": " times the slope, plus half Delta squared times the curvature." } ], "spoken": "p at x minus Delta and t is approximately p minus Delta times the partial derivative of p with respect to x, plus Delta squared over two times the second partial derivative of p with respect to x.", "title": "The density one jump away", "tree": { "kind": "relation", "left": { "args": [ { "args": [ { "kind": "symbol", "quantityId": "positionCoordinate1d", "termId": "eq-model-bm-taylor-in-space.t.x" }, { "argument": { "kind": "symbol", "quantityId": "displacementIncrement", "termId": "eq-model-bm-taylor-in-space.t.jump" }, "kind": "negate" } ], "kind": "sum", "opId": "eq-model-bm-taylor-in-space.op.origin" }, { "kind": "symbol", "quantityId": "fieldTimeCoordinate", "termId": "eq-model-bm-taylor-in-space.t.time" } ], "kind": "symbol", "quantityId": "probabilityDensity", "termId": "eq-model-bm-taylor-in-space.t.density" }, "opId": "eq-model-bm-taylor-in-space.op.relation", "operator": "approx", "right": { "args": [ { "kind": "symbol", "quantityId": "probabilityDensity", "termId": "eq-model-bm-taylor-in-space.t.value" }, { "argument": { "args": [ { "kind": "symbol", "quantityId": "displacementIncrement", "termId": "eq-model-bm-taylor-in-space.t.jump2" }, { "expression": { "kind": "symbol", "quantityId": "probabilityDensity", "termId": "eq-model-bm-taylor-in-space.t.slopeP" }, "kind": "derivative", "opId": "eq-model-bm-taylor-in-space.op.slope", "order": 1, "partial": true, "variable": { "kind": "symbol", "quantityId": "positionCoordinate1d", "termId": "eq-model-bm-taylor-in-space.t.slopeX" } } ], "kind": "product", "opId": "eq-model-bm-taylor-in-space.op.slopeTerm" }, "kind": "negate" }, { "args": [ { "denominator": { "kind": "number", "value": "2" }, "kind": "quotient", "numerator": { "base": { "kind": "symbol", "quantityId": "displacementIncrement", "termId": "eq-model-bm-taylor-in-space.t.jump3" }, "exponent": { "den": 1, "num": 2 }, "kind": "power", "opId": "eq-model-bm-taylor-in-space.op.jumpSquare" }, "opId": "eq-model-bm-taylor-in-space.op.half" }, { "expression": { "kind": "symbol", "quantityId": "probabilityDensity", "termId": "eq-model-bm-taylor-in-space.t.curveP" }, "kind": "derivative", "opId": "eq-model-bm-taylor-in-space.op.curve", "order": 2, "partial": true, "variable": { "kind": "symbol", "quantityId": "positionCoordinate1d", "termId": "eq-model-bm-taylor-in-space.t.curveX" } } ], "kind": "product", "opId": "eq-model-bm-taylor-in-space.op.curveTerm" } ], "kind": "sum", "opId": "eq-model-bm-taylor-in-space.op.series" } }, "unitSystem": "si" }, { "argument": "arg-bm-diffusion-equation", "assumptions": [ "tau short compared with the time over which the density changes, so terms in tau squared are dropped.", "These are modern teaching equations in SI notation, not a transcription of the printed paper." ], "bindings": [], "explanation": "Over a short interval tau, the density at a fixed place changes by about tau times its rate of change there. Terms in tau squared are dropped.", "id": "eq-model-bm-taylor-in-time", "kind": "equation", "live": false, "notation": "modern-pedagogical", "notes": [ { "explanation": "Later density is about the density now plus the change over tau.", "foundation": "taylor-expansion", "nodeId": "eq-model-bm-taylor-in-time.op.relation", "title": "What it asserts" }, { "explanation": "The time t plus one step interval tau: a later time, not a longer interval.", "foundation": "random-walks", "nodeId": "eq-model-bm-taylor-in-time.op.later", "title": "One interval later" }, { "explanation": "The value now plus the first correction; the next term, in tau squared, is dropped.", "foundation": "taylor-expansion", "nodeId": "eq-model-bm-taylor-in-time.op.series", "title": "Now, plus the change" }, { "explanation": "The rate of change times the interval.", "foundation": "derivatives", "nodeId": "eq-model-bm-taylor-in-time.op.change", "title": "The change over tau" }, { "explanation": "How fast the density changes with time, with x held fixed.", "foundation": "partial-derivatives", "nodeId": "eq-model-bm-taylor-in-time.op.rate", "title": "Rate at one place" }, { "explanation": "Probability per metre of finding the particle at x, one interval after t.", "foundation": "distributions", "nodeId": "eq-model-bm-taylor-in-time.t.density", "title": "The density later" }, { "explanation": "The place where the density is read, held fixed.", "foundation": "functions-graphs", "nodeId": "eq-model-bm-taylor-in-time.t.x", "title": "Position" }, { "explanation": "The time now. Every particle was at x = 0 when it read zero.", "foundation": "functions-graphs", "nodeId": "eq-model-bm-taylor-in-time.t.time", "title": "Time" }, { "explanation": "Paper 2's tau: short compared with the time over which the density changes.", "foundation": "random-walks", "nodeId": "eq-model-bm-taylor-in-time.t.interval", "title": "The step interval" }, { "explanation": "The density at x at time t.", "foundation": "distributions", "nodeId": "eq-model-bm-taylor-in-time.t.value", "title": "The density now" }, { "explanation": "The same x.", "foundation": "functions-graphs", "nodeId": "eq-model-bm-taylor-in-time.t.x2", "title": "Position" }, { "explanation": "The same t.", "foundation": "functions-graphs", "nodeId": "eq-model-bm-taylor-in-time.t.time2", "title": "Time" }, { "explanation": "The same tau.", "foundation": "random-walks", "nodeId": "eq-model-bm-taylor-in-time.t.interval2", "title": "The step interval" }, { "explanation": "The density whose rate is taken.", "foundation": "distributions", "nodeId": "eq-model-bm-taylor-in-time.t.rateP", "title": "Density" }, { "explanation": "The rate is per unit of time.", "foundation": "partial-derivatives", "nodeId": "eq-model-bm-taylor-in-time.t.rateT", "title": "Time" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-taylor-in-time.t.density", "text": "The density" }, { "text": " one " }, { "nodeId": "eq-model-bm-taylor-in-time.t.interval", "text": "interval" }, { "text": " later is about " }, { "nodeId": "eq-model-bm-taylor-in-time.t.value", "text": "the density now" }, { "text": " plus the interval times " }, { "nodeId": "eq-model-bm-taylor-in-time.op.rate", "text": "its rate of change" }, { "text": " at that place." } ], "spoken": "p of x and t plus tau is approximately p of x and t, plus tau times the partial derivative of p with respect to t.", "title": "The density a short interval later", "tree": { "kind": "relation", "left": { "args": [ { "kind": "symbol", "quantityId": "positionCoordinate1d", "termId": "eq-model-bm-taylor-in-time.t.x" }, { "args": [ { "kind": "symbol", "quantityId": "fieldTimeCoordinate", "termId": "eq-model-bm-taylor-in-time.t.time" }, { "kind": "symbol", "quantityId": "stepInterval", "termId": "eq-model-bm-taylor-in-time.t.interval" } ], "kind": "sum", "opId": "eq-model-bm-taylor-in-time.op.later" } ], "kind": "symbol", "quantityId": "probabilityDensity", "termId": "eq-model-bm-taylor-in-time.t.density" }, "opId": "eq-model-bm-taylor-in-time.op.relation", "operator": "approx", "right": { "args": [ { "args": [ { "kind": "symbol", "quantityId": "positionCoordinate1d", "termId": "eq-model-bm-taylor-in-time.t.x2" }, { "kind": "symbol", "quantityId": "fieldTimeCoordinate", "termId": "eq-model-bm-taylor-in-time.t.time2" } ], "kind": "symbol", "quantityId": "probabilityDensity", "termId": "eq-model-bm-taylor-in-time.t.value" }, { "args": [ { "kind": "symbol", "quantityId": "stepInterval", "termId": "eq-model-bm-taylor-in-time.t.interval2" }, { "expression": { "kind": "symbol", "quantityId": "probabilityDensity", "termId": "eq-model-bm-taylor-in-time.t.rateP" }, "kind": "derivative", "opId": "eq-model-bm-taylor-in-time.op.rate", "order": 1, "partial": true, "variable": { "kind": "symbol", "quantityId": "fieldTimeCoordinate", "termId": "eq-model-bm-taylor-in-time.t.rateT" } } ], "kind": "product", "opId": "eq-model-bm-taylor-in-time.op.change" } ], "kind": "sum", "opId": "eq-model-bm-taylor-in-time.op.series" } }, "unitSystem": "si" }, { "argument": "arg-bm-independent-steps", "assumptions": [ "Independent steps of mean square length l squared.", "These are modern teaching equations in SI notation, not a transcription of the printed paper." ], "bindings": [], "explanation": "A walk of steps of length l every tau spreads like diffusion with this D.", "id": "eq-model-bm-walk-diffusivity", "kind": "equation", "live": false, "notation": "modern-pedagogical", "notes": [ { "explanation": "The two is one dimension's convention.", "foundation": "bridge-fractions-ratios", "nodeId": "eq-model-bm-walk-diffusivity.op.denominator", "title": "Two tau" }, { "explanation": "Squared length per second.", "foundation": "bridge-fractions-ratios", "nodeId": "eq-model-bm-walk-diffusivity.op.quotient", "title": "Step square per unit time" }, { "explanation": "The microscopic walk sets the macroscopic D.", "foundation": "random-walks", "nodeId": "eq-model-bm-walk-diffusivity.op.relation", "title": "What it asserts" }, { "explanation": "The mean square of one step.", "foundation": "bridge-squaring-square-roots", "nodeId": "eq-model-bm-walk-diffusivity.op.square", "title": "l squared" }, { "explanation": "How fast the mean square displacement grows: half its rate of growth. Not a speed.", "foundation": "diffusion-equation", "nodeId": "eq-model-bm-walk-diffusivity.t.diffusion", "title": "Diffusion coefficient" }, { "explanation": "The short time between successive steps of the walk.", "foundation": "random-walks", "nodeId": "eq-model-bm-walk-diffusivity.t.interval", "title": "Step interval" }, { "explanation": "The root mean square length of one step.", "foundation": "random-walks", "nodeId": "eq-model-bm-walk-diffusivity.t.step", "title": "Step length" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-walk-diffusivity.t.diffusion", "text": "D" }, { "text": " is " }, { "nodeId": "eq-model-bm-walk-diffusivity.t.step", "text": "l" }, { "text": " squared over two " }, { "nodeId": "eq-model-bm-walk-diffusivity.t.interval", "text": "tau" }, { "text": "." } ], "spoken": "D equals l squared over two tau.", "title": "The diffusivity of a walk", "tree": { "kind": "relation", "left": { "kind": "symbol", "quantityId": "diffusionCoefficient", "termId": "eq-model-bm-walk-diffusivity.t.diffusion" }, "opId": "eq-model-bm-walk-diffusivity.op.relation", "operator": "=", "right": { "denominator": { "args": [ { "kind": "number", "value": "2" }, { "kind": "symbol", "quantityId": "stepInterval", "termId": "eq-model-bm-walk-diffusivity.t.interval" } ], "kind": "product", "opId": "eq-model-bm-walk-diffusivity.op.denominator" }, "kind": "quotient", "numerator": { "base": { "kind": "symbol", "quantityId": "stepRms", "termId": "eq-model-bm-walk-diffusivity.t.step" }, "exponent": { "den": 1, "num": 2 }, "kind": "power", "opId": "eq-model-bm-walk-diffusivity.op.square" }, "opId": "eq-model-bm-walk-diffusivity.op.quotient" } }, "unitSystem": "si" }, { "argument": "arg-bm-independent-steps", "assumptions": [ "Steps of equal duration tau.", "These are modern teaching equations in SI notation, not a transcription of the printed paper." ], "bindings": [], "explanation": "n steps of duration tau take time t.", "id": "eq-model-bm-walk-time", "kind": "equation", "live": false, "notation": "modern-pedagogical", "notes": [ { "explanation": "The total time of n steps.", "foundation": "random-walks", "nodeId": "eq-model-bm-walk-time.op.product", "title": "n times tau" }, { "explanation": "Counting steps and measuring time are the same thing here.", "foundation": "random-walks", "nodeId": "eq-model-bm-walk-time.op.relation", "title": "What it asserts" }, { "explanation": "How many independent steps fit in time t.", "foundation": "random-walks", "nodeId": "eq-model-bm-walk-time.t.count", "title": "Number of steps" }, { "explanation": "The short time between successive steps of the walk.", "foundation": "random-walks", "nodeId": "eq-model-bm-walk-time.t.interval", "title": "Step interval" }, { "explanation": "The time over which displacement is observed.", "foundation": "random-walks", "nodeId": "eq-model-bm-walk-time.t.time", "title": "Observation interval" } ], "paper": "brownian-motion", "review": "draft", "schemaVersion": 1, "sentence": [ { "nodeId": "eq-model-bm-walk-time.t.time", "text": "t" }, { "text": " is " }, { "nodeId": "eq-model-bm-walk-time.t.count", "text": "n" }, { "text": " times " }, { "nodeId": "eq-model-bm-walk-time.t.interval", "text": "tau" }, { "text": "." } ], "spoken": "t equals n tau.", "title": "Time counts steps", "tree": { "kind": "relation", "left": { "kind": "symbol", "quantityId": "observationInterval", "termId": "eq-model-bm-walk-time.t.time" }, "opId": "eq-model-bm-walk-time.op.relation", "operator": "=", "right": { "args": [ { "kind": "symbol", "quantityId": "walkStepCount", "termId": "eq-model-bm-walk-time.t.count" }, { "kind": "symbol", "quantityId": "stepInterval", "termId": "eq-model-bm-walk-time.t.interval" } ], "kind": "product", "opId": "eq-model-bm-walk-time.op.product" } }, "unitSystem": "si" } ], "foundations": [ { "citations": [], "example": [ { "items": [ "Four particles end at −3, −1, +1 and +3 micrometres from where they started.", "Mark those positions along the bottom and draw a bar one particle tall above each.", "The bars stand two on each side of zero, at the same distances out. The picture is symmetric, so the average position is 0.", "Yet no bar stands at zero: every particle moved. The graph shows at a glance what the average hides." ], "kind": "steps" } ], "exampleTitle": "Four particles that average to zero", "explanation": [ { "kind": "paragraph", "text": "Walk for a minute and note how far you have gone every ten seconds: 0 m at the start, 14 m after ten seconds, 28 m after twenty, and so on. Draw a line along the bottom of a page for time and one up the side for distance, and put a dot for each pair. That is a graph: each dot is one moment, read across for the time and up for the distance." }, { "kind": "paragraph", "text": "The dots of a steady walk fall on a straight line, and the steeper the line, the faster the walk. This one climbs 14 metres for every 10 seconds: 1.4 metres per second. A flat line would mean standing still, and a line that bends upward would mean speeding up." }, { "kind": "paragraph", "text": "The papers draw other pairs the same way. Along the bottom goes what you choose or wait for, such as a position or a time; up the side goes what you then find, such as how many particles sit there. A bell-shaped curve, high in the middle and falling away on both sides, says that most particles are near the centre and fewer are farther out." } ], "id": "bridge-a-graph", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-negative-numbers-direction", "kind": "proof-edge" } ], "question": "How does a curve show the relationship between two physical quantities?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "Before reading a curve, read its two labels and their units, and ask whether its height is a count, a density or a running total.", "summary": "A graph puts one quantity along the bottom and another up the side, one point for each pair. How steeply the curve climbs is a rate; the area under it is a total.", "title": "Reading a graph" }, { "citations": [], "example": [ { "items": [ "In a group of 4 particles, 1 moves to the right: the fraction is 1/4 = 0.25.", "In a group of 8, 2 move to the right: the fraction is 2/8 = 0.25.", "Doubling both the number that moved and the size of the group leaves the fraction at 0.25." ], "kind": "steps" } ], "exampleTitle": "One in four, two in eight", "explanation": [ { "kind": "paragraph", "text": "The fraction 6/2 is 3, because 2 fits into 6 three times; it also says that 6 is three times 2. Every fraction a/b works the same way. The bottom number can never be zero, since no count of zeros adds up to 6." }, { "kind": "paragraph", "text": "When the two quantities have different units, their ratio is a new quantity with a unit of its own: 12 micrometres travelled in 4 seconds is 3 micrometres per second. When they have the same unit, the units cancel and the ratio is a plain number: 2 particles out of 8 is 0.25 of the group." } ], "id": "bridge-fractions-ratios", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-sum-average", "kind": "proof-edge" } ], "question": "What does dividing one quantity by another tell you?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "A ratio of two quantities in the same unit is a plain number. A ratio of different units, such as micrometres per second, is a new quantity and keeps its unit.", "summary": "A fraction a/b says how many times b fits into a. Multiplying the top and the bottom by the same number leaves it unchanged.", "title": "Fractions and ratios" }, { "citations": [ "ap-17-549", "ap-17-132" ], "example": [ { "items": [ "§5 of the Brownian paper prints λx = 8 × 10⁻⁵ cm = 0.8 micrometres for t = 1 second, and about 6 micrometres for one minute, for a particle 0.001 mm across in water at 17 °C.", "Worked out again with the paper's own constants, R = 8.31 and N = 6 × 10²³ (the constant set einstein-1905-brownian-printed), the values are 0.7948 μm at t = 1 s and 6.1564 μm at t = 60 s.", "With the modern Boltzmann constant in place of R/N (the exact value of the set modern-si-2019), and the same particle and water, they are 0.7935 μm and 6.1467 μm.", "In every line t means the same thing, the time elapsed. Only its value changes, and λx changes with it." ], "kind": "steps" } ], "exampleTitle": "The paper's two estimates, and where they come from", "explanation": [ { "kind": "paragraph", "text": "Fill in the blanks: \"After ___ seconds the particle had wandered about ___ micrometres.\" The Brownian paper fills them twice. After 1 second, about 0.8 micrometres; after 60 seconds, about 6." }, { "kind": "paragraph", "text": "Write t for the first blank and λx for the second, and the sentence becomes a relation between two quantities. The letter t keeps meaning the time elapsed, in seconds, while its value goes from 1 to 60. The letter λx keeps meaning how far the particle has wandered along one direction while its value grows from about 0.8 to about 6 micrometres." }, { "kind": "paragraph", "text": "A letter's meaning belongs to the place where it is used, not to the letter. In the light-quanta paper L is the speed of light in §§1 and 2, and in §9 it is the light energy absorbed. Each passage says what its letters mean, and so does every formula on this site, in its legend." } ], "id": "bridge-letter-for-quantity", "kind": "foundation", "prerequisites": [], "question": "What does a letter such as t mean in a formula?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "A letter names a quantity: its value can change while its meaning stays the same. The rule that turns t into λx is the subject of the Brownian paper's §4.", "summary": "A letter holds the place of a quantity whose value can change. The letter t means the time elapsed, in seconds, whatever number it takes; the sentence around the formula says what each letter means.", "title": "A letter stands for a quantity" }, { "citations": [], "example": [ { "kind": "paragraph", "text": "Two walkers both finish one unit from their start, one at −1 and the other at +1. The signed average is zero. The average distance is one." } ], "exampleTitle": "Two walkers, opposite directions, average zero", "explanation": [ { "kind": "paragraph", "text": "Put the starting point at zero on a ruler. A final mark three units to the right has displacement +3; one three units to the left has displacement −3. Adding the signed displacements gives zero. Adding the distances from the start gives six." }, { "kind": "paragraph", "text": "Displacement compares where something ends with where it started, whatever it did in between. A walker who goes three units right and comes back has displacement zero after walking six." } ], "id": "bridge-negative-numbers-direction", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-sum-average", "kind": "proof-edge" } ], "question": "How can movement add up to zero?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "The sign names a direction relative to a chosen axis; it does not mean a negative distance.", "summary": "Choose right as positive and left as negative. Opposite displacements can cancel without either particle staying still.", "title": "A sign records direction" }, { "citations": [ "ap-17-132" ], "example": [ { "items": [ "Two tokens: 4 equally likely outcomes, 1 of them both left, so the probability is 1/4.", "Four tokens: 2⁴ = 16 outcomes, 1 of them all left, so the probability is 1/16, which is (½)⁴.", "A run of 100 drops of four tokens might show all left 5 times: a frequency of 5/100, near 1/16 (about 6 in 100) but not equal to it.", "Any one drop shows all left or it does not. The 1/16 is about the long run." ], "kind": "steps" } ], "exampleTitle": "Counting outcomes for two tokens and for four", "explanation": [ { "kind": "paragraph", "text": "Drop two tokens into a box, each landing in the left or the right half with equal chance. There are four outcomes: left and left, left and right, right and left, right and right. One of the four is both left, so the model probability of both left is 1/4." }, { "kind": "paragraph", "text": "Now drop them 100 times and count. Both might land left 23 times: a frequency of 23/100. Over more and more trials the frequency tends to settle near 1/4, but any particular count can differ from 25." }, { "kind": "paragraph", "text": "A single drop either lands both left or it does not. A probability of 1/4 is not a quarter of an outcome. It says how often, over many drops." }, { "kind": "paragraph", "text": "The light-quanta paper counts in exactly this way in §5. The chance that n independent points are all in a part v of a volume v₀ is (v/v₀)ⁿ, and the paper calls it a statistical probability: a frequency over many moments." } ], "id": "bridge-probability-notation", "kind": "foundation", "prerequisites": [], "question": "What does a probability of 1/4 say about one try, and about a hundred?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "A probability counts favoured outcomes among equally likely ones; a frequency counts what happened. How a probability becomes an entropy is the lesson on entropy and the number of ways.", "summary": "A model probability counts the favoured outcomes among equally likely ones. A frequency counts what happened over many trials. A single trial either happens or does not. The three answer different questions.", "title": "A probability, a frequency and a single trial" }, { "citations": [], "example": [ { "items": [ "1 micrometre (1 μm) is 10⁻⁶ metres, or 0.000001 m.", "The paper's displacement after one minute, about 6 μm, is 6 × 10⁻⁶ m.", "Squaring it squares the number and the unit: (6 × 10⁻⁶ m)² = 36 × 10⁻¹² m² = 3.6 × 10⁻¹¹ m²." ], "kind": "steps" } ], "exampleTitle": "Squaring six micrometres", "explanation": [ { "kind": "paragraph", "text": "The Brownian paper works its example, in §5, for particles 0.001 mm across. In metres that is 0.000001 m: easy to miscount by one zero, which is a factor of ten. Written as 10⁻⁶ m, the −6 says the decimal point has moved six places to the left, and that length has a name of its own, the micrometre (μm)." }, { "kind": "paragraph", "text": "The unit goes through the arithmetic with the number. A micrometre is 10⁻⁶ m, so a square micrometre is 10⁻¹² m², not 10⁻⁶ m²: squaring a length squares its unit, power of ten included." } ], "id": "bridge-scientific-notation-units", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-fractions-ratios", "kind": "proof-edge" } ], "question": "How do you write a very small length without counting zeros?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "A measurement is a number and a unit together. Squaring it squares both: (6 × 10⁻⁶ m)² is 3.6 × 10⁻¹¹ m².", "summary": "A power of ten moves the decimal point. A measurement is a number together with its unit, and the unit goes through every step of a calculation.", "title": "Powers of ten and physical units" }, { "citations": [], "example": [ { "items": [ "The displacements −3, −1, +1, +3 have squares 9, 1, 1, 9. Their average is 5, so the RMS is √5, about 2.236.", "Double each displacement: −6, −2, +2, +6. The squares are 36, 4, 4, 36, and their average is 20.", "The mean square went from 5 to 20, four times as large. The RMS went from about 2.236 to √20, about 4.472: twice as large." ], "kind": "steps" } ], "exampleTitle": "Doubling every displacement, and what the RMS does", "explanation": [ { "kind": "paragraph", "text": "Both +3 and −3 have square 9, so squaring removes the sign, and it weights a large distance more than a small one. A square root returns to the original unit: the square root of a square micrometre is a micrometre." }, { "kind": "paragraph", "text": "The root mean square, or RMS, uses both steps: square each displacement, average the squares, then take the square root. The result is a typical distance from the start, in the original unit." }, { "kind": "formula", "latex": "\\sqrt{4q}=2\\sqrt q\\quad(q\\ge 0)", "spoken": "The square root of four q is twice the square root of q, for nonnegative q." } ], "id": "bridge-squaring-square-roots", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-sum-average", "kind": "proof-edge" } ], "question": "If the mean square grows four times, why does the typical distance only double?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "Four times the mean square is twice the RMS. That is why, in the Brownian paper, waiting four times as long doubles the typical distance.", "summary": "Squaring multiplies a number by itself, and the square root undoes it. Four times a square is the square of twice the number, so four times a mean square is twice a root mean square.", "title": "Squares and square roots" }, { "citations": [], "example": [ { "items": [ "Add 3 + 1 + 1 + 3 to get 8.", "Count the values: there are four.", "Divide 8 by 4 to get 2.", "List the values twice: 16 divided by 8 is still 2." ], "kind": "steps" } ], "exampleTitle": "Adding four numbers and dividing by four", "explanation": [ { "kind": "paragraph", "text": "Four jars hold 3, 1, 1 and 3 marbles. Pour them together and there are 8. Share the 8 equally between the four jars and each holds 2. Two is the average, although no jar held 2 to begin with." }, { "kind": "paragraph", "text": "The average keeps the total and the count and nothing else, so 3, 1, 1, 3 and 2, 2, 2, 2 have the same average. Listing every value twice doubles both the total and the count, and the average stays at 2." } ], "id": "bridge-sum-average", "kind": "foundation", "prerequisites": [], "question": "What does an average keep, and what does it lose?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "An average is the total shared equally among the observations, and two very different lists can have the same one.", "summary": "Add the values, then divide by how many there are. The average keeps the total and the count, and loses the individual values.", "title": "Adding and averaging" }, { "citations": [], "example": [ { "kind": "paragraph", "text": "The list of averages can be explained, not only watched. Start at any time t and let the interval be Δt. In that interval the ball moves 3(t + Δt)² − 3t² metres. Since (t + Δt)² is t² + 2tΔt + (Δt)², that distance is:" }, { "equations": [ "eq-model-fd-distance-in-interval" ], "kind": "formula", "latex": "3(t+\\Delta t)^2 - 3t^2 = 6t\\,\\Delta t + 3(\\Delta t)^2", "spoken": "Three times t plus delta t, squared, minus three t squared, equals six t delta t plus three delta t squared." }, { "kind": "paragraph", "text": "Divide that distance by the time Δt to get the average speed over the interval:" }, { "equations": [ "eq-model-fd-average-speed-in-interval" ], "kind": "formula", "latex": "\\frac{6t\\,\\Delta t + 3(\\Delta t)^2}{\\Delta t} = 6t + 3\\,\\Delta t", "spoken": "Six t delta t plus three delta t squared, divided by delta t, equals six t plus three delta t." }, { "kind": "paragraph", "text": "The 3Δt term shrinks with the interval and leaves 6t metres per second. At t = 1 s that is 6 m/s. The extra 3, 0.3, 0.03 and 0.003 in the list above are that 3Δt term, for Δt of 1, 0.1, 0.01 and 0.001 seconds." } ], "exampleTitle": "Why the averages settle on 6 m/s", "explanation": [ { "kind": "paragraph", "text": "A speed needs two readings of position and the time between them: divide the distance covered by the time taken, and you have the average speed over that interval. An instant has no duration, so that recipe cannot be used at an instant as it stands." }, { "kind": "paragraph", "text": "So shorten the interval and watch what happens. A ball rolls so that after t seconds it has gone 3t² metres: 3 m after one second, 12 m after two. Start at the one-second mark and average over shorter and shorter stretches:" }, { "items": [ "Over the next second it goes from 3 m to 12 m: 9 m in 1 s, an average of 9 m/s.", "Over the next tenth of a second it reaches 3.63 m: 0.63 m in 0.1 s, an average of 6.3 m/s.", "Over the next hundredth it reaches 3.0603 m: 0.0603 m in 0.01 s, an average of 6.03 m/s.", "Over the next thousandth it reaches 3.006003 m: 0.006003 m in 0.001 s, an average of 6.003 m/s." ], "kind": "steps" }, { "kind": "paragraph", "text": "The averages close in on 6 m/s. That number is the derivative: the ball's speed at the instant t = 1 s. The interval is never set to zero, which would mean dividing zero distance by zero time. It only shrinks, and the averages settle." }, { "kind": "paragraph", "text": "In symbols, with Δt for the short interval, the same recipe reads:" }, { "equations": [ "eq-model-fd-derivative-limit" ], "kind": "formula", "latex": "\\frac{dx}{dt} = \\lim_{\\Delta t \\to 0} \\frac{x(t+\\Delta t) - x(t)}{\\Delta t}", "spoken": "d x by d t is the value that the change in position divided by the change in time settles on as delta t shrinks towards zero." }, { "kind": "paragraph", "text": "A derivative carries units: those of the quantity that changes, divided by those of the quantity you vary. Position in metres, varied over time in seconds, gives metres per second. On a graph of position against time, the derivative at a point is the steepness of the curve there." }, { "kind": "paragraph", "text": "Einstein's Brownian paper needs rates like this for a concentration of particles, which changes along a tube and in time at once. With two things changing, a rate has to say which one it follows, so the paper writes a curly ∂ and holds the other fixed:" }, { "equations": [ "eq-model-fd-diffusion-concentration" ], "kind": "formula", "latex": "\\frac{\\partial f}{\\partial t} = D\\,\\frac{\\partial^2 f}{\\partial x^2}", "spoken": "The partial derivative of f with respect to t equals D times the second partial derivative of f with respect to x." }, { "kind": "paragraph", "text": "The lesson on partial derivatives, listed at the end of this page, reads that equation one piece at a time." } ], "id": "derivatives", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:functions-graphs", "kind": "proof-edge" } ], "question": "How can a body have a speed at a single instant?", "returnCaptions": [ { "callingAnchor": "brownian-motion:s4", "caption": "Return to Brownian §4 diffusion rate derivation" } ], "review": "draft", "schemaVersion": 1, "stoppingPoint": "A derivative is the value an average rate settles on as the interval shrinks, in units of one quantity per unit of the other. The interval shrinks but is never set to zero, so nothing is ever divided by zero.", "summary": "Work out the average rate over shorter and shorter intervals. The value those averages settle on is the rate at the instant, and it is called the derivative.", "title": "Rates of change and derivatives" }, { "citations": [], "example": [ { "kind": "paragraph", "text": "Suppose the dye's typical distance from where it started is 1 mm after one minute. Double D, keeping everything else the same: the mean square doubles, so the typical distance becomes about 1.4 mm, the square root of 2 times as far, not twice. Keep the original D and wait four minutes instead of one: the typical distance doubles, to 2 mm. Neither comparison says where any single particle goes." } ], "exampleTitle": "Doubling the diffusion coefficient, then quadrupling the time", "explanation": [ { "kind": "paragraph", "text": "Put a drop of dye into a long tube of still water. At first the dye sits in one spot; an hour later it has spread along the tube, thinner and wider. The diffusion coefficient D says how fast that spreading goes." }, { "kind": "paragraph", "text": "The rule it follows is local. Look at one spot. If the dye is thicker on both sides than at the spot itself, a dip, dye drifts in from both sides and the spot fills. If the spot is a peak, thicker than its neighbours, dye drifts out and it thins. The more sharply the dip or peak is curved, the faster it changes, and D sets the pace. With p for how much dye sits per metre of tube at place x and time t:" }, { "equations": [ "eq-model-fd-diffusion-density" ], "kind": "formula", "latex": "\\frac{\\partial p}{\\partial t}=D\\frac{\\partial^2p}{\\partial x^2}", "spoken": "The time rate of change of density is D times its spatial curvature." }, { "kind": "paragraph", "text": "The right-hand side is D times the curvature, which is positive at a dip and negative at a peak. For one particle instead of a drop of dye, p is the probability per metre of finding it at x. Where the tube ends matters too: a closed tube keeps all its dye, and after a long time spreads it evenly, not in the ever-widening bell of an endless tube." }, { "kind": "paragraph", "text": "The same rule can be read as a flow. The amount crossing a point each second runs from thick to thin, faster where the density falls more steeply:" }, { "equations": [ "eq-model-fd-flux-law" ], "kind": "formula", "latex": "J=-D\\frac{\\partial p}{\\partial x}", "spoken": "Fick’s constitutive law sends flux down the density gradient." }, { "kind": "paragraph", "text": "Combining the flux law with conservation gives the diffusion equation for constant D. The flux law itself is a model premise in this route, not something conservation alone proves." } ], "id": "diffusion-equation", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:flux-continuity", "kind": "proof-edge" }, { "foundationId": "foundation:bridge-squaring-square-roots", "kind": "proof-edge" } ], "question": "What changes when D changes?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "D is measured in square metres per second. Before using it, say where the tube ends and what the model assumes, since both change the answer.", "summary": "D sets how fast a spreading cloud widens. Along one line, the mean square distance of its particles from where they started grows by 2D every second.", "title": "What the diffusion coefficient means" }, { "citations": [], "example": [ { "items": [ "Spread the probability evenly over 4 μm. The density is 1 divided by 4 μm, or 0.25 per micrometre.", "The area over the whole 4 μm is 0.25 per μm × 4 μm = 1.", "A 1 μm piece has probability 0.25 × 1 = 0.25, and a 2 μm piece has 0.5.", "Squeeze the same probability into 2 μm and the height doubles to 0.5 per micrometre, while the total stays 1. The height was never a probability." ], "kind": "steps" } ], "exampleTitle": "A flat density across four micrometres", "explanation": [ { "kind": "paragraph", "text": "Mark where 100 particles end up on a ruler divided into 2 μm bins. If 30 land in one bin, that bin holds a probability of about 0.3, and the density there is 0.3 divided by 2 μm, or 0.15 per micrometre. Measured in metres the same density is 150,000 per metre, while the bin still holds 0.3: the height depends on the unit of length, and the probability does not." }, { "kind": "formula", "latex": "P(a\\le X\\le b)=\\int_a^b p(x)\\,dx", "spoken": "The probability of the interval from a to b is the integral of the density over that interval." }, { "kind": "paragraph", "text": "Under a smooth curve, any single exact position has probability zero, since an interval of zero width has no area. At the starting instant of the Brownian problem every particle sits at one point. No curve can draw that: the whole probability, 1, sits on the single point, and a curve appears only once the particles have begun to spread." } ], "extensionSections": [ { "body": [ { "kind": "paragraph", "text": "A distribution can be fixed without its causes being fixed. The spreading of Brownian particles has one curve for each value of D, and in the Brownian paper D = RT/(N · 6πkP), with k the liquid's viscosity and P the particle's radius. At a fixed temperature and viscosity, a radius twice as large with an N half as large gives the same D, so it gives the same curve. No number of observed displacements can tell the two apart: the samples depend only on the product N·P, so only that product can be read from them." }, { "id": "two-measurements-two-unknowns", "kind": "foundation", "returnCaption": "Return to one curve from different causes" }, { "kind": "paragraph", "text": "A change of variable reshapes a distribution. If an estimate of D is turned into an estimate of N by dividing a constant by it, the distribution of N is not the distribution of D relabelled. Equal steps in D become unequal steps in N, so the density is stretched in some places and squeezed in others, while each pair of matching intervals keeps the same probability." }, { "items": [ "Suppose an estimate of D comes out as 4 or 6 with equal chance. Its average is 5, the true value, so it has no bias.", "Estimate N as 60 ÷ D. The two outcomes give 15 and 10, whose average is 12.5, but 60 ÷ 5 is 12. The turned estimate is too large by about 4 per cent on average, though nothing about the measurement changed.", "An interval survives the turn with its ends swapped: D between 4 and 6 is the same event as N between 10 and 15, so the two have the same probability.", "For an estimate built from q = 100 squared Gaussian displacements the same effect is smaller: the turned estimate is too large on average by the factor q/(q − 2) = 1.0204, about 2 per cent." ], "kind": "steps" }, { "id": "error-and-inference", "kind": "foundation", "returnCaption": "Return to a curve turned around" } ], "citations": [], "id": "same-curve-different-causes", "kind": "foundation-extension", "ownerBead": "am-found-statistics-inference-pzqv", "schemaVersion": 1, "targetFoundation": "foundation:distributions", "title": "One curve from different causes, and a curve turned around" } ], "id": "distributions", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-sum-average", "kind": "proof-edge" } ], "question": "What does the height of a probability curve mean?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "The height of the curve is a density, a probability per unit length. Only the area over an interval is a probability.", "summary": "A density is probability per unit length. A probability belongs to an interval, and it is the area under the curve over that interval.", "title": "Density is not probability" }, { "citations": [ "ap-17-132" ], "example": [ { "items": [ "Take one gram-molecule of gas, n = N = 6.02 × 10²³ particles, and ask for all of them in the left half: v/v₀ = ½.", "The probability is (½) to the power 6.02 × 10²³, far too small to write out. Its logarithm is N ln ½, about −4.2 × 10²³.", "The entropy change is \\(k_B\\) × N × ln ½ = R ln ½ = 8.314 × (−0.693), about −5.76 joules per kelvin.", "Thermodynamics gives the same number for compressing an ideal gas to half its volume at fixed temperature, R ln ½. Counting ways and measuring heat agree." ], "kind": "steps" }, { "kind": "paragraph", "text": "§6 of the paper finds the same form for dilute radiation, with E/(βν) standing where the gas has Rn/N, and reads it as radiation behaving, in this respect, like independent quanta." } ], "exampleTitle": "Squeezing a gas into half its volume", "explanation": [ { "kind": "paragraph", "text": "Four particles move independently in a box. The chance that all four happen to be in the left half at a given moment is ½ × ½ × ½ × ½ = 1/16. For a hundred particles it is (½)¹⁰⁰, about 8 × 10⁻³¹: possible, but never seen. A state is more probable when more of the ways the particles can be arranged produce it." }, { "kind": "paragraph", "text": "Entropy measures this probability on a logarithmic scale. Boltzmann's principle, as §5 of the light-quanta paper states it, makes the entropy of a system a function of the probability W of its state, and that function must be a logarithm:" }, { "equations": [ "eq-model-fd-entropy-boltzmann" ], "kind": "formula", "latex": "S - S_0 = k_B \\ln W", "spoken": "S minus S nought equals k B times the natural logarithm of W." }, { "kind": "paragraph", "text": "Why a logarithm: two independent systems have a combined probability W = W₁ × W₂, while their entropies add, S = S₁ + S₂. Only a logarithm turns the product into the sum, and that is the argument §5 gives. The constant \\(k_B\\) is R/N, which is how the paper writes it, and the paper prints the natural logarithm as lg." }, { "kind": "paragraph", "text": "Apply it to the box. Squeezing n independent particles from a volume v₀ into a part v of it has probability W = (v/v₀)ⁿ, so the entropy changes by S − S₀ = n \\(k_B\\) ln(v/v₀), a decrease, since v/v₀ is less than 1. §5 derives exactly this, and remarks that it needs no assumption about the law by which the molecules move." }, { "kind": "paragraph", "text": "The probability here is a frequency: the fraction of moments at which the state is found. §5 insists on this, and criticizes calculations in which the equally probable cases are simply postulated." } ], "id": "entropy-multiplicity", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:logarithms", "kind": "proof-edge" }, { "foundationId": "foundation:probability-independence", "kind": "proof-edge" }, { "foundationId": "foundation:entropy-temperature", "kind": "cross-link" } ], "question": "Why is entropy the logarithm of a probability?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "This lesson stops at entropy as the logarithm of a probability. What the comparison with radiation shows, and what it does not, is §6 of the light-quanta paper.", "summary": "Entropy measures, on a logarithmic scale, how probable a state is: the entropy difference is Boltzmann's constant times the natural logarithm of the probability W. Two independent systems multiply their probabilities and add their entropies, and only a logarithm does both. §5 of the light-quanta paper uses this rule, which it calls Boltzmann's principle.", "title": "Entropy and the number of ways" }, { "citations": [ "ap-17-132" ], "example": [ { "items": [ "A reservoir so large that its temperature stays at 300 kelvin receives 3 joules of heat, slowly.", "Its entropy rises by 3 J ÷ 300 K = 0.01 joule per kelvin.", "The same 3 joules given to a reservoir at 150 kelvin would raise its entropy by 0.02 joule per kelvin, twice as much.", "A small body warms as it takes in heat, so its temperature changes on the way. Its entropy change is found by adding dE/T over the warming, not by dividing by one temperature." ], "kind": "steps" }, { "kind": "paragraph", "text": "The relation fixes how entropy changes, not its absolute value: two entropy formulas that differ by a constant have the same slope everywhere, and a separate condition is needed to fix the constant. For an entropy per unit volume, that constant multiplied by two different volumes can change the difference between two states." } ], "exampleTitle": "Three joules into a reservoir held at 300 kelvin", "explanation": [ { "kind": "paragraph", "text": "Add a small amount of heat to a body, slowly and reversibly, and keep its volume fixed so that it does no work. Its energy rises by that amount, and its entropy rises by the energy divided by its absolute temperature. So one joule raises the entropy of a body at 100 kelvin by 0.01 joule per kelvin, and of a body at 1000 kelvin by only 0.001: the same energy counts for more in a cold body. In symbols, with dE a small gain of energy and T the temperature:" }, { "kind": "formula", "latex": "dS=\\frac{dE}{T},\\qquad\\left(\\frac{\\partial S}{\\partial E}\\right)_V=\\frac{1}{T}", "spoken": "At fixed volume and the stated constraints, entropy change is energy change divided by absolute temperature." }, { "kind": "paragraph", "text": "Read the other way round, 1/T is how steeply a body's entropy climbs as its energy grows." }, { "kind": "paragraph", "text": "The subscript V matters. If the volume can change, the body can do work while it takes in energy, and the relation gains another term. The derivative describes one kind of change, at fixed volume, not every process." }, { "kind": "paragraph", "text": "The relation also says why bodies in contact end at one temperature. Share a fixed total of energy between two bodies, and the total entropy is greatest when moving a little energy from one to the other no longer changes it: when their entropy slopes, and so their temperatures, are equal. §3 of the light-quanta paper uses the same relation for radiation, setting its entropy slope equal to 1/T." }, { "id": "partial-derivatives", "kind": "foundation", "returnCaption": "Return to the constraints on the entropy derivative." } ], "id": "entropy-temperature", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:partial-derivatives", "kind": "proof-edge" }, { "foundationId": "foundation:integration", "kind": "cross-link" } ], "question": "What does temperature have to do with entropy?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "Before using 1/T as an entropy slope, say what is held fixed. Before comparing two entropies, say whether they are totals or per unit volume, and what fixes the constant.", "summary": "Add a little energy to a body at fixed volume and its entropy rises by that energy divided by its absolute temperature. The same energy raises the entropy of a cold body more than that of a hot one.", "title": "Entropy and temperature" }, { "citations": [ "nist-normal-variance", "bipm-si-definitions" ], "example": [ { "items": [ "Take a made-up relation, N = 12/D, with D measured as 4. The estimate is N = 12/4 = 3.", "Suppose the method gives D between 3 and 6. A larger D means a smaller N, so N runs from 12/6 = 2 to 12/3 = 4: the ends swap over.", "Now suppose N = 12/(aD), with a an unknown scale. The same D fits many pairs of a and N, and no interval for N follows until a is measured separately.", "Declaring a = 2 halves both the estimate and its interval without changing a single measurement. It changes an assumption, not the data.", "If three inputs each come from a method that misses at most once in 100 uses, and D's method at most twice, the combination misses at most 5 times in 100, even if the misses are related. Without the stated miss rates, this arithmetic guarantees nothing." ], "kind": "steps" } ], "exampleTitle": "Inverting an interval, and watching its endpoints swap", "explanation": [ { "kind": "paragraph", "text": "Some questions the data cannot answer, however carefully you measure. If all you know of a rectangle is that its area is 12 square centimetres, its sides could be 3 and 4, 2 and 6, or 1 and 12. The area fixes their product, not each side; to get one side you must measure the other." }, { "kind": "paragraph", "text": "The Brownian paper meets the same limit. In its §3 the diffusion coefficient is D = RT/(N · 6πkP), with k the liquid's viscosity and P the particle's radius. So watching particles spread gives D, and with the temperature and viscosity known, D fixes the product of N and P, not each of them: a smaller particle and a larger N give the same spreading. To get N, the radius has to be measured some other way." }, { "kind": "paragraph", "text": "An estimate comes from a finite sample, and a different sample gives a different number. Two questions follow: whether the method lands on the right value on average, which is its bias, and how widely its answers scatter from sample to sample. Fitting a drift from the same displacements uses up part of the information and leaves less to estimate the scatter." }, { "kind": "paragraph", "text": "Uncertainty comes in four kinds, and each needs its own statement. Sampling uncertainty is the scatter from one sample to the next. Measurement uncertainty is the error in each reading, such as where the microscope places a particle. Model uncertainty is the chance that the formula itself does not fit these particles. Numerical uncertainty is the error of the computation. A confidence interval is not an interval-arithmetic enclosure, which is guaranteed to contain the exact result of a calculation, and a numerical error bar, which bounds the arithmetic, is not a measurement uncertainty." }, { "kind": "paragraph", "text": "A 95 per cent confidence interval comes from a method that, used on many samples, catches the true value in 95 of every 100 of them, provided its assumptions hold. It is not a 95 per cent chance that this one interval contains the value. A plot of many repeated intervals, with the misses left in, shows what the promise means." }, { "kind": "paragraph", "text": "For a diffusion coefficient the promise has an exact form. With M independent displacements in d coordinates there are q = dM squared displacements, and their sum divided by 2Dt has a known spread from sample to sample: the chi-square distribution with q degrees of freedom. Its 2.5 and 97.5 per cent points give the 95 per cent interval. In a synthetic example with 50 displacements in two coordinates, so q = 100, and an estimate of 0.43 μm²/s, those points are 74.2219 and 129.5612:" }, { "kind": "formula", "latex": "\\begin{aligned} &\\left[\\frac{100 \\times 0.43}{129.5612},\\ \\frac{100 \\times 0.43}{74.2219}\\right] \\\\ &\\quad = [0.3319,\\ 0.5793]\\ \\mu\\mathrm{m}^2/\\mathrm{s} \\end{aligned}", "spoken": "The interval runs from 43 divided by 129.5612 to 43 divided by 74.2219, that is from 0.3319 to 0.5793 square micrometres per second." }, { "kind": "paragraph", "text": "Inverting an estimate adds a bias of its own. The molecular number is found by dividing by the estimated D, and although that estimate is right on average, its reciprocal is not: with q = 100 degrees of freedom the average of the inverted estimate is too large by the factor q/(q − 2) = 1.020408, about two per cent. The interval itself survives inversion, with its ends swapped, as the worked example shows." }, { "kind": "paragraph", "text": "Reading positions adds noise of its own. If each position is read with an error of spread σ in each coordinate, each measured step gains 2σ² in its mean square per coordinate, and two neighbouring steps share one reading, so their errors have a covariance of −σ². Neighbouring steps are then no longer independent, and cutting one path into overlapping windows does not make independent samples." }, { "kind": "paragraph", "text": "An interval that treats the radius, viscosity and temperature as exact holds only on that condition. Allowing for their uncertainty needs to know how each was measured and how often its own interval misses. Without that, a plus-or-minus has no stated coverage and is not a confidence statement." }, { "kind": "paragraph", "text": "A path simulated from a chosen N is useful for checking that a method recovers the N it was given. It is not evidence about N in nature. Since 2019 the SI fixes Avogadro's number by definition, so recovering it from modern data checks consistency. Measuring N, as Perrin did from 1908, needs inputs measured independently of it." } ], "id": "error-and-inference", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:distributions", "kind": "proof-edge" }, { "foundationId": "foundation:mean-variance-rms", "kind": "proof-edge" }, { "foundationId": "foundation:gaussian-distributions", "kind": "proof-edge" }, { "foundationId": "foundation:probability-independence", "kind": "proof-edge" }, { "foundationId": "foundation:bridge-squaring-square-roots", "kind": "proof-edge" }, { "foundationId": "foundation:bridge-probability-notation", "kind": "proof-edge" } ], "question": "What can a set of measurements tell you, and what can it not?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "Say what the data can tell apart, which method and model you used, what you held exact, and whether the result recovers a value you put in, estimates it independently, or checks consistency.", "summary": "An estimate is worth as much as the model and the inputs behind it. Before trusting a number, ask whether the data can tell the unknowns apart, how much the answer would move on a new sample, and which inputs were measured independently.", "title": "Uncertainty, evidence and inference" }, { "citations": [], "example": [ { "kind": "paragraph", "text": "The Brownian curve's shape comes from its exponential factor, e raised to −x²/4Dt. Follow that factor outward from the centre:" }, { "items": [ "At the centre, x = 0, the exponent is 0 and e⁰ = 1: the curve is at its peak.", "At x = √(2Dt), one root-mean-square width out, the exponent is −2Dt/4Dt = −1/2, and e raised to −1/2 is about 0.607 of the peak.", "At x = √(4Dt), the exponent is −1, and e raised to −1 is about 0.368 of the peak.", "Doubling the time doubles 4Dt, so each of these heights is reached √2 times farther from the centre." ], "kind": "steps" }, { "kind": "paragraph", "text": "None of these steps depends on the units chosen for x, t and D, as long as they agree with one another. That is what a pure-number exponent guarantees." } ], "exampleTitle": "The bell curve at its centre and one width out", "explanation": [ { "kind": "paragraph", "text": "Suppose a quantity changes at a rate proportional to its own size. A radioactive sample loses the same fraction of its atoms every hour; a beam of light loses the same fraction of its intensity in every millimetre of absorbing glass. Take 1,000 atoms that lose half their number each hour:" }, { "items": [ "After one hour 500 remain, after two 250, after three 125.", "The losses shrink as the sample does: 500 in the first hour, 250 in the second, 125 in the third.", "Each hour multiplies the count by the same factor, one half. Equal steps, equal factors: that is exponential change." ], "kind": "steps" }, { "kind": "paragraph", "text": "The function e to the power x, with e about 2.718, is the one whose rate of change equals its own value everywhere, which is why it describes this kind of change. In symbols, with N₀ atoms at the start and a constant k that sets how fast they decay (for halving every hour, k is about 0.693 per hour):" }, { "kind": "formula", "latex": "\\begin{gathered} \\frac{dN}{dt} = -kN, \\\\ \\text{so}\\quad N(t) = N_0\\,e^{-kt} \\end{gathered}", "spoken": "If the rate of change of N is minus k times N, then N at time t is N nought times e to the minus k t." }, { "kind": "paragraph", "text": "Whatever sits in the exponent must be a pure number with no units. The reason is the series that defines e to the power x:" }, { "kind": "formula", "latex": "e^{x} = 1 + x + \\frac{x^{2}}{2} + \\frac{x^{3}}{6} + \\ldots", "spoken": "e to the x equals one plus x plus x squared over two plus x cubed over six, and so on." }, { "kind": "paragraph", "text": "If x were a length, the series would add a length to an area to a volume, which has no meaning. In the decay law, k is a rate per hour and t is in hours, so kt is a pure number." }, { "kind": "paragraph", "text": "Two exponentials from the 1905 papers pass that check. Wien's radiation law, which the light-quanta paper uses from §4 on, contains βν/T: β times a frequency is a temperature, so dividing by the temperature T leaves a pure number. The formula below writes e to the power −βν/T as exp(−βν/T). exp(u) is another way of writing e to the power u, used when the power is long." }, { "equations": [ "eq-model-fd-wien-law" ], "kind": "formula", "latex": "\\rho = \\alpha\\,\\nu^{3} e^{-\\beta \\nu / T}", "spoken": "Rho equals alpha nu cubed times e to the minus beta nu over T." }, { "kind": "paragraph", "text": "The Brownian paper's §4 solution contains x²/4Dt: x² is an area, and a diffusion coefficient D in square metres per second times a time t in seconds is also an area, so their ratio is a pure number." }, { "kind": "formula", "latex": "f(x,t) = \\frac{n}{\\sqrt{4\\pi D}}\\,\\frac{e^{-x^{2}/4Dt}}{\\sqrt{t}}", "spoken": "f of x and t equals n over the square root of four pi D, times e to the minus x squared over four D t, divided by the square root of t." } ], "id": "exponentials", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:derivatives", "kind": "proof-edge" }, { "foundationId": "foundation:logarithms", "kind": "cross-link" } ], "question": "What kind of change does an exponential describe?", "returnCaptions": [ { "callingAnchor": "brownian-motion:s4", "caption": "Return to Brownian §4 Gaussian distribution and exponential solution" } ], "review": "draft", "schemaVersion": 1, "stoppingPoint": "Whatever sits in an exponent must be a pure number. Check it by cancelling the units, as with βν/T and x²/4Dt.", "summary": "An exponential describes change by equal factors: a quantity whose rate of change is proportional to its own size. Whatever sits in the exponent must be a pure number, with no units.", "title": "Exponential change" }, { "citations": [], "example": [ { "items": [ "In one second, 7 particles cross into the stretch and 5 cross out. The count inside rises by 2.", "Next second, 5 cross in and 5 cross out. The count is unchanged, although 10 particles crossed." ], "kind": "steps" } ], "exampleTitle": "Seven particles in, five out", "explanation": [ { "kind": "paragraph", "text": "Take a short stretch of a tube. If more particles cross into it than cross out, the number inside goes up. The flux J counts the net number crossing a surface per unit area and per second, with rightward crossings positive; the concentration c counts particles per unit volume. They measure different things, and the equation below connects them." }, { "equations": [ "eq-model-fd-continuity" ], "kind": "formula", "latex": "\\frac{\\partial c}{\\partial t}=-\\frac{\\partial J}{\\partial x}", "spoken": "The rate at which the concentration changes equals minus the rate at which the flux changes along x." }, { "kind": "paragraph", "text": "The minus sign is the bookkeeping. If the rightward flux is larger at the right end of the stretch than at the left, more particles leave on the right than arrive on the left, and the concentration falls." } ], "id": "flux-continuity", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-sum-average", "kind": "proof-edge" } ], "question": "If particles keep crossing the edges of a region, what decides whether it fills up or empties?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "The contents change by the net crossing. A steady count does not mean the particles have stopped, only that as many leave as arrive.", "summary": "A region gains what enters and loses what leaves. If nothing is made or destroyed inside, the change in its contents is the difference.", "title": "Counting what crosses a boundary" }, { "citations": [ "ap-17-549" ], "example": [ { "items": [ "Dissolve 0.1 gram-molecule of sugar in each litre: 100 in each cubic metre, so ν = 100 × 6.02 × 10²³ = 6.02 × 10²⁵ molecules per cubic metre.", "At T = 293 K, RT/N = 8.314 × 293 ÷ 6.02 × 10²³ = 4.05 × 10⁻²¹ joules. Multiply by ν: p ≈ 2.4 × 10⁵ Pa, about 2.4 atmospheres, enough to hold up a column of water nearly 25 metres tall.", "Now a suspension of grains, a million in each cubic millimetre: ν = 10¹⁵ per cubic metre. The same law gives p = 4.05 × 10⁻²¹ × 10¹⁵ ≈ 4 × 10⁻⁶ Pa.", "Each grain counts exactly as much as one sugar molecule. The pressure is tiny only because there are fewer grains, by a factor of about 6 × 10¹⁰." ], "kind": "steps" }, { "kind": "paragraph", "text": "While the particles are few and far apart, the pressure depends on their number and the temperature, not on their size or mass. That is what lets §3 treat a grain like a molecule and weigh its jostling against a force." } ], "exampleTitle": "A sugar solution and a suspension of grains", "explanation": [ { "kind": "paragraph", "text": "Put sugar water on one side of a membrane that lets water through but not sugar, and pure water on the other. Water flows toward the sugar until the sugar side stands higher: the sugar pushes outward on the membrane. That push on each square metre is the osmotic pressure." }, { "kind": "paragraph", "text": "For a dilute solution van 't Hoff found in 1887 that the pressure is the one an ideal gas of the same particles would exert in the same volume. With ν particles in each cubic metre at absolute temperature T, R the gas constant and N the number of molecules in a gram-molecule:" }, { "equations": [ "eq-model-fd-osmotic-pressure" ], "kind": "formula", "latex": "p = \\frac{RT}{N}\\,\\nu", "spoken": "The osmotic pressure p equals R T over N, times the number of particles per unit volume, nu." }, { "kind": "paragraph", "text": "The paper's §1 asks why a grain floating in water, large enough to see under a microscope, should be any different. Classical thermodynamics expected no force at all from suspended grains: the free energy of the system did not seem to depend on where the wall or the grains were. On the kinetic theory of heat, the paper answers, a dissolved molecule and a suspended grain differ only in size. The grains are jostled into a slow, irregular motion, and a wall that keeps them in must be pushed as a wall that keeps molecules in is pushed. So n grains in a volume V*, ν = n/V* in each unit of volume, press on it with p = (RT/N)ν." }, { "kind": "paragraph", "text": "§2 checks this inside the kinetic theory with free energy, F = E − TS: the energy minus the temperature times the entropy. At a fixed temperature a system settles where F can no longer decrease. Moving the wall that confines the grains by a small step changes F, and requiring that change to vanish gives the same pressure, (RT/N)ν." }, { "kind": "paragraph", "text": "The law holds for dilute suspensions and solutions, where the particles are far apart and do not act on one another. The paper states it for a large volume per gram-molecule; crowded particles depart from it." } ], "id": "free-energy-osmotic-pressure", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:entropy-temperature", "kind": "proof-edge" }, { "foundationId": "foundation:quantities-units", "kind": "proof-edge" }, { "foundationId": "foundation:bridge-scientific-notation-units", "kind": "cross-link" }, { "foundationId": "foundation:bridge-fractions-ratios", "kind": "cross-link" } ], "question": "Why should grains floating in water push on a wall the way dissolved molecules do?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "This lesson stops at the pressure of a dilute suspension. How that pressure, balanced against a force on each grain, fixes how fast the grains spread is §3 of the Brownian paper.", "summary": "In a dilute solution, dissolved molecules held behind a wall that lets water through push on it with the pressure an ideal gas of the same particles would exert: p = (RT/N)ν. §1 of the Brownian paper argues that suspended grains must push in the same way, and §2 checks it by the bookkeeping of free energy.", "title": "Osmotic pressure and free energy" }, { "citations": [], "example": [ { "equations": [ "eq-model-fd-mean-square-line" ], "kind": "formula", "latex": "\\langle x^2\\rangle = 2Dt", "spoken": "The mean square displacement equals two times the diffusion coefficient times the time." }, { "items": [ "Take D = 0.5 square micrometres per second, so 2D = 1 square micrometre per second.", "At t = 1 s the mean square displacement is 1 µm²; at t = 4 s it is 4 µm²; at t = 9 s, 9 µm².", "Plotted against time, these points lie on a straight line through the origin whose slope is 2D, 1 µm² per second.", "Their square roots, the root-mean-square displacements, are 1, 2 and 3 µm: four times the time gives twice the distance." ], "kind": "steps" } ], "exampleTitle": "Mean square displacement plotted against time", "explanation": [ { "kind": "paragraph", "text": "A bath fills from a tap. After one minute it holds 10 litres, after two 20, after three 30. For each time there is exactly one amount of water, and that pairing, one output for each input, is a function. Plot the pairs with time along the bottom and litres up the side and they fall on a straight line that climbs 10 litres for every minute. The slope, in litres per minute, is how fast the bath fills." }, { "kind": "paragraph", "text": "In the papers the pairs are physical: the mean square displacement of a Brownian particle for each elapsed time, or the concentration for each position along a tube. A function says how one thing responds when another changes." }, { "kind": "paragraph", "text": "A graph puts the input along the horizontal axis and the output up the vertical one, so each point is one pair. Both axes carry units, and so does everything read off the graph: the slope of mean square displacement against time is in square micrometres per second, and the area under a probability density is a plain probability." }, { "kind": "paragraph", "text": "A formula such as ⟨x²⟩ = 2Dt describes an average over many particles. One particle's squared displacement scatters widely around the line; the average over many lies close to it. The Brownian paper predicts the average, and ends by hoping that a researcher will soon decide the question by observation." } ], "id": "functions-graphs", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-a-graph", "kind": "proof-edge" } ], "question": "How does a graph record the way one quantity depends on another?", "returnCaptions": [ { "callingAnchor": "brownian-motion:s5", "caption": "Return to Brownian §5 displacement distribution plot" } ], "review": "draft", "schemaVersion": 1, "stoppingPoint": "A graph is a record of paired quantities, each with its units. A slope read off it has units too, and a slope stated without them explains nothing.", "summary": "A function gives exactly one output for each input. Its graph lays those pairs out on two axes, and every slope or area read off it carries units made from the two axes' units.", "title": "Functions and graphs" }, { "citations": [], "example": [ { "kind": "paragraph", "text": "In the formula above, x is the displacement, t the time since the start and D the diffusion coefficient. Two things need checking: that the total probability under the curve is 1, and that the mean square displacement is 2Dt. One change of variable turns both into standard integrals." }, { "kind": "formula", "latex": "u=\\frac{x}{\\sqrt{4Dt}},\\qquad dx=\\sqrt{4Dt}\\,du", "spoken": "Let u equal x over the square root of four D t. Then d x equals the square root of four D t, times d u." }, { "kind": "formula", "latex": "\\begin{aligned}\\int_{-\\infty}^{\\infty}p(x,t)\\,dx&=\\frac{1}{\\sqrt{\\pi}}\\int_{-\\infty}^{\\infty}e^{-u^2}\\,du\\\\&=1\\end{aligned}", "spoken": "The total probability becomes one over the square root of pi, times the integral of e to the minus u squared, and that equals one." }, { "kind": "paragraph", "text": "The last step needs the integral of e to the minus u squared to be √π. One way to see it is to square the integral, which makes it an integral over a whole plane, and then add the plane up in rings around the origin: a ring of radius r and width dr has area 2πr dr. This check is a modern one; the paper does not show it." }, { "kind": "formula", "latex": "\\begin{aligned}I^2&=\\int_{-\\infty}^{\\infty}\\int_{-\\infty}^{\\infty}e^{-(u^2+v^2)}\\,du\\,dv\\\\&=\\int_{0}^{\\infty}e^{-r^2}\\,2\\pi r\\,dr=\\pi\\end{aligned}", "spoken": "I squared is the integral of e to the minus u squared plus v squared over the whole plane, which equals the integral from zero to infinity of e to the minus r squared times two pi r d r, which is pi." }, { "kind": "paragraph", "text": "For the mean square, the same substitution leaves the integral of u² times e to the minus u squared. Integrating by parts once, the end terms vanish at both infinities, and half the integral above remains:" }, { "equations": [ "eq-model-fd-gaussian-moment-integral" ], "kind": "formula", "latex": "\\frac{1}{\\sqrt\\pi}\\int_{-\\infty}^{\\infty}u^2e^{-u^2}\\,du=\\frac12", "spoken": "The normalized integral of u squared times e to the minus u squared is one half." }, { "kind": "paragraph", "text": "Substituting back, the mean square is 4Dt times one half, which is 2Dt: the value the Brownian paper's §4 gives for the mean of the squared displacement, λx² = 2Dt." } ], "exampleTitle": "Checking the curve: total probability one, and mean square 2Dt", "explanation": [ { "kind": "paragraph", "text": "Release many particles at one point in still water and look again after a time t. They have spread out: most are still near the start, fewer are farther away, and as many went left as right. Count how many ended up at each distance and the counts make a bell-shaped curve, the Gaussian. One number sets its width, √(2Dt), the root-mean-square distance. For D = 0.5 µm² per second and t = 1 s the width is 1 µm, and:" }, { "items": [ "Within one width of the start, between −1 µm and +1 µm, end about 68 per cent of the particles.", "Within two widths, ±2 µm, end about 95 per cent; within three, about 99.7 per cent.", "So about one particle in three ends more than one width from where it started, and about one in twenty more than two widths away." ], "kind": "steps" }, { "kind": "paragraph", "text": "Wait four times as long and the width doubles, because it grows as the square root of the time. The curve itself, with p the probability per micrometre of finding a particle at displacement x after time t, is:" }, { "equations": [ "eq-model-fd-gaussian-density" ], "kind": "formula", "latex": "p(x,t)=\\frac{1}{\\sqrt{4\\pi Dt}}\\exp\\left(-\\frac{x^2}{4Dt}\\right)", "spoken": "p of x and t equals one over the square root of four pi D t, times e to the minus x squared over four D t." }, { "kind": "paragraph", "text": "The factor in front keeps the total probability at one as the curve spreads: a wider bell has to be a lower one. The exponential makes the curve fall away on both sides, the same way to the left as to the right, because x appears squared." } ], "extensionSections": [ { "body": [ { "kind": "paragraph", "text": "The bell curve says where one particle is likely to end. It also says how far a whole sample's width can stray from the model's. Divide each measured displacement by √(2Dt) and it becomes a draw from the standard bell curve, whose mean square is 1. The sum of q such squares follows a curve of its own, the chi-square distribution with q degrees of freedom, whose mean is q." }, { "items": [ "With q = 100 squared displacements, 95 samples in 100 have a mean square between 0.742 and 1.296 times 2Dt. Their widths, the square roots, lie between 0.86 and 1.14 times the true width.", "With q = 4, the same 95 samples in 100 run from 0.121 to 2.79 times 2Dt: a mean square about an eighth of the true one, or nearly three times it.", "The spread narrows as q grows, but it is never zero for a finite sample." ], "kind": "steps" }, { "kind": "paragraph", "text": "Turned around, the same two points of the chi-square curve make an interval for D from a single sample. With q = 100, and \\(D_{\\mathrm{est}}\\) for the estimate:" }, { "kind": "formula", "latex": "\\left[\\frac{100\\,D_{\\mathrm{est}}}{129.56},\\ \\frac{100\\,D_{\\mathrm{est}}}{74.22}\\right]", "spoken": "From one hundred D est over one hundred twenty-nine point five six, to one hundred D est over seventy-four point two two." }, { "kind": "paragraph", "text": "Used on sample after sample, this recipe catches the true D in 95 of every 100 of them. The promise is about the recipe, repeated. It says nothing more about any one interval once it has been drawn." }, { "kind": "paragraph", "text": "All of this assumes the displacements follow the bell curve and are independent of one another. Reading errors and overlapping windows break the second assumption, and then q is not the number of independent squares, so the interval's promise no longer holds as stated." }, { "id": "error-and-inference", "kind": "foundation", "returnCaption": "Return to how far a sample's width can stray" } ], "citations": [], "id": "sample-width-and-its-interval", "kind": "foundation-extension", "ownerBead": "am-found-statistics-inference-pzqv", "schemaVersion": 1, "targetFoundation": "foundation:gaussian-distributions", "title": "How far a sample's width can stray" } ], "id": "gaussian-distributions", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:integration", "kind": "proof-edge" }, { "foundationId": "foundation:mean-variance-rms", "kind": "proof-edge" } ], "question": "After a time t, where is a spreading particle likely to be?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "The bell curve says where many particles end up. Any one particle can end anywhere, and about one in three ends more than one width from its start.", "summary": "The chance of finding a diffusing particle at a displacement x follows a bell curve whose mean square is 2Dt. About 68 per cent of the chance lies within one root-mean-square width, √(2Dt), of the start.", "title": "The Gaussian and its width" }, { "citations": [], "example": [ { "items": [ "A particle's density rises steadily across a 4 µm stretch, from 0 at the left end to 0.5 per micrometre at the right. The whole area is a triangle, ½ × 4 × 0.5 = 1.", "Four strips 1 µm wide, each taking the height at its left edge: 0 + 0.125 + 0.25 + 0.375 = 0.75.", "Eight strips 0.5 µm wide: the heights add to 1.75, and times 0.5 that is 0.875.", "Sixteen strips 0.25 µm wide: 0.9375. The shortfall halves each time the strips do: 0.25, then 0.125, then 0.0625.", "The totals settle on 1, the area of the triangle. That settled value is the integral." ], "kind": "steps" }, { "kind": "paragraph", "text": "A flat density gives the same total for every cut, which is why the uniform stretch above needed no strips. A curved one, such as the Brownian bell curve, does, and the integral is the value the totals settle on." } ], "exampleTitle": "A rising density, added up in narrower strips", "explanation": [ { "kind": "paragraph", "text": "Chance spread along a line is measured per unit length: a density. Where the density is the same all along a stretch, the chance of the stretch is the density times its length. A particle equally likely to be anywhere in a 4 µm stretch has density 0.25 per micrometre, and the chance of finding it in the first 1.5 µm is 0.25 × 1.5 = 0.375." }, { "kind": "paragraph", "text": "Where the density changes from place to place, cut the line into narrow strips. Each strip contributes its height times its width, as if the density were flat across it. Add the strips, then cut them narrower and add again. For a smooth curve the totals settle on one value, and that value is the integral:" }, { "kind": "formula", "latex": "\\int_a^b p(x)\\,dx = \\lim_{\\Delta x \\to 0} \\sum_i p(x_i)\\,\\Delta x", "spoken": "The integral of p of x from a to b is the value that the sum of p of x i times delta x settles on as delta x shrinks." }, { "kind": "paragraph", "text": "The units multiply. A probability density is a probability per metre, so a strip of it, times a width in metres, is a plain probability. For a density the whole area must be 1, because the particle has to be somewhere. The Brownian paper's §4 writes exactly that for the chances of a jump Δ, ∫φ(Δ) dΔ = 1, and the light-quanta paper's §3 adds entropy over every frequency, S = v∫φ dν." }, { "kind": "paragraph", "text": "One more tool recurs in the papers: integration by parts. It is the product rule for derivatives added up over an interval, and it moves a derivative from one factor to the other at the price of an endpoint term:" }, { "kind": "formula", "latex": "\\int_a^b u\\frac{dv}{dx}\\,dx = [uv]_a^b - \\int_a^b v\\frac{du}{dx}\\,dx", "spoken": "The integral of u times the derivative of v equals u v at b minus u v at a, minus the integral of v times the derivative of u." }, { "kind": "paragraph", "text": "The endpoint term, uv at b minus uv at a, is part of the answer. It disappears only when uv is zero at both ends, as it is for a density that falls to zero far out on both sides." } ], "id": "integration", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:distributions", "kind": "proof-edge" } ], "question": "How do you add up something that varies continuously?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "An integral is a total of height times width over a range, in the units of both. The widths and any endpoint term are part of the calculation, not decoration on the integral sign.", "summary": "An integral adds up many thin strips, each a height times a width, and narrows the strips until the total stops changing. Its units are the height's units times the width's.", "title": "Adding continuously" }, { "citations": [], "example": [ { "kind": "paragraph", "text": "In §5 Einstein asks how probable it is that n molecules, each moving independently through a volume v₀, are all found at one moment in a part v of it. Each molecule is in that part with probability v/v₀, and independent probabilities multiply:" }, { "equations": [ "eq-model-fd-molecules-in-part" ], "kind": "formula", "latex": "W = \\left(\\frac{v}{v_0}\\right)^{n}", "spoken": "W equals v over v nought, to the power n." }, { "items": [ "Take n = 10 molecules and v = v₀/2. Then W = (1/2)¹⁰ = 1/1024, about 0.000977.", "The logarithm turns the tenth power into ten times: ln W = 10 × ln(1/2) = 10 × (−0.693) = −6.93.", "So the entropy of the state with every molecule in one half is lower by 6.93 × R/N: 6.93 times Boltzmann's constant." ], "kind": "steps" }, { "kind": "paragraph", "text": "For a mole of gas, n = N, and the same formula gives S − S₀ = R ln(v/v₀). That is the entropy change thermodynamics gives for an ideal gas compressed from v₀ to v at constant temperature." } ], "exampleTitle": "Ten molecules found in half the volume", "explanation": [ { "kind": "paragraph", "text": "Adding is easier than multiplying, and a logarithm turns one into the other. The natural logarithm of 2 is about 0.693, and of 4 about 1.386. Add them and you get 2.079, which is the natural logarithm of 8, the product of 2 and 4." }, { "kind": "paragraph", "text": "The natural logarithm ln x answers one question: to what power must e, about 2.718, be raised to give x? Raised to 0.693, e gives 2, and raised to 1.386 it gives 4. Multiplying two powers of e adds the powers, so raised to 2.079 it gives 8. Written for any numbers, the logarithm of a product is the sum of the logarithms, and a power becomes a multiple:" }, { "equations": [ "eq-model-fd-log-product", "eq-model-fd-log-power" ], "kind": "formula", "latex": "\\ln(AB) = \\ln A + \\ln B, \\qquad \\ln\\left(f^{\\,n}\\right) = n \\ln f", "spoken": "The log of A times B is log A plus log B, and the log of f to the power n is n times log f." }, { "kind": "paragraph", "text": "Entropy needs exactly this. In §5 of the light-quanta paper, Einstein takes two independent systems. Their entropies add, S = S₁ + S₂, while the probabilities of their states multiply, W = W₁W₂. If entropy is some function φ of probability, then φ(W₁W₂) = φ(W₁) + φ(W₂), and he concludes that φ is a logarithm:" }, { "equations": [ "eq-model-fd-entropy-probability" ], "kind": "formula", "latex": "S - S_0 = \\frac{R}{N} \\ln W", "spoken": "S minus S nought equals R over N times the natural log of W." }, { "kind": "paragraph", "text": "R is the gas constant and N the number of molecules in a mole, so R/N is what is now called Boltzmann's constant. The step from φ(W₁W₂) = φ(W₁) + φ(W₂) to the logarithm needs φ to be a smooth function, which Einstein takes for granted." }, { "kind": "paragraph", "text": "Einstein prints lg for the natural logarithm, as German physics did in 1905. Today lg usually means the base-10 logarithm and ln the natural one. So a printed lg 2 means ln 2, about 0.693, not log₁₀ 2, about 0.301." } ], "id": "logarithms", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:exponentials", "kind": "cross-link" } ], "question": "Why does entropy need a logarithm?", "returnCaptions": [ { "callingAnchor": "brownian-motion:s2", "caption": "Return to Brownian §2 osmotic pressure and free energy" } ], "review": "draft", "schemaVersion": 1, "stoppingPoint": "A logarithm turns products into sums, which is why it connects probabilities that multiply to entropies that add. In the 1905 papers, lg means the natural logarithm.", "summary": "A logarithm turns multiplying into adding. When two systems are independent, the probabilities of their states multiply while their entropies add, so entropy has to be a logarithm of probability.", "title": "Logarithms: turning products into sums" }, { "citations": [], "example": [ { "items": [ "For −3, −1, +1, +3 the signed sum is 0, so the mean is 0.", "The distances, ignoring sign, are 3, 1, 1, 3; their mean is 2.", "The squares are 9, 1, 1, 9; their mean is 5.", "The RMS is the square root of 5, about 2.236. With a mean of 0, the variance is also 5 and the standard deviation is also about 2.236.", "A small random sample need not have a signed mean of exactly zero, even when the model's mean is zero." ], "kind": "steps" } ], "exampleTitle": "Four displacements, four different averages", "explanation": [ { "kind": "paragraph", "text": "Four particles move −3, −1, +1 and +3 micrometres. Their average position has not changed, yet they have spread out. Three averages answer three different questions." }, { "equations": [ "eq-model-fd-sample-mean" ], "kind": "formula", "latex": "\\langle x\\rangle=\\frac{1}{M}\\sum_{i=1}^{M}x_i", "spoken": "Add the M signed displacements and divide by M." }, { "equations": [ "eq-model-fd-sample-mean-square" ], "kind": "formula", "latex": "\\langle x^2\\rangle=\\frac{1}{M}\\sum_{i=1}^{M}x_i^2", "spoken": "Square each displacement before adding and dividing by M." }, { "equations": [ "eq-model-fd-variance" ], "kind": "formula", "latex": "\\operatorname{Var}(x)=\\langle x^2\\rangle-\\langle x\\rangle^2", "spoken": "Variance is the mean square minus the square of the mean." }, { "kind": "paragraph", "text": "RMS means root mean square: average the squares, then take the square root. The standard deviation is the square root of the variance, so it measures spread around the mean, where the RMS measures distance from zero. The two are equal only when the mean is zero. The average of the distances, ignoring sign, is a third statistic and gives a different number." } ], "extensionSections": [ { "body": [ { "kind": "paragraph", "text": "In the Brownian problem the model's mean square displacement along one coordinate is 2Dt. A sample's mean square is an estimate of it: average the squares of the displacements actually measured, then divide by 2t to estimate D. If the four displacements above, −3, −1, +1 and +3 μm, were measured after t = 1 s along one coordinate, their mean square of 5 μm² would give D = 5 ÷ 2 = 2.5 μm² per second." }, { "kind": "paragraph", "text": "Averaged over many samples, this estimate lands on the true value: it has no bias. One sample scatters, and the scatter shrinks only with the square root of the number q of independent squared displacements. For displacements drawn from the bell curve, the spread of the estimate, as a fraction of the true value, is:" }, { "kind": "formula", "latex": "\\frac{\\mathrm{spread}}{D}=\\sqrt{\\frac{2}{q}}", "spoken": "The spread of the estimate, divided by D, equals the square root of two over q." }, { "items": [ "With q = 4, as above, the spread is about 71 per cent of D, so four particles say little.", "With q = 100 it is about 14 per cent.", "To halve the spread, measure four times as many independent displacements." ], "kind": "steps" }, { "kind": "paragraph", "text": "If the particles might also drift, the mean has to be estimated from the same numbers. The average squared deviation from that estimated mean then comes out too small, by the factor (M − 1)/M on average, because the estimated mean sits closer to the sample than the true one does. Dividing the sum by M − 1 instead of M removes that bias." }, { "kind": "paragraph", "text": "The microscope adds an error of its own. A measured step is the difference of two position readings, so if each reading is off by an error of spread σ in each coordinate, the step's mean square gains 2σ² per coordinate. For D = 0.5 μm² per second and t = 1 s the true mean square is 2Dt = 1 μm². A reading error of σ = 0.1 μm adds 0.02 μm², 2 per cent; σ = 0.3 μm adds 0.18 μm², 18 per cent. Unless it is subtracted, all of it is read as extra diffusion." }, { "kind": "paragraph", "text": "Two neighbouring steps share the reading between them, so their errors are related, with a covariance of −σ². The steps are then not independent, and cutting one path into overlapping windows does not help: the windows reuse the same readings, so they do not add independent squares to q." }, { "id": "error-and-inference", "kind": "foundation", "returnCaption": "Return to one sample's mean square as an estimate of 2Dt" } ], "citations": [], "id": "mean-square-as-an-estimate", "kind": "foundation-extension", "ownerBead": "am-found-statistics-inference-pzqv", "schemaVersion": 1, "targetFoundation": "foundation:mean-variance-rms", "title": "One sample's mean square, and what the microscope adds" } ], "id": "mean-variance-rms", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-negative-numbers-direction", "kind": "proof-edge" }, { "foundationId": "foundation:bridge-squaring-square-roots", "kind": "proof-edge" } ], "question": "If particles wander both ways, which average shows how far they got?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "Say which average you mean. Squaring then averaging, averaging then squaring, and averaging the distances give three different numbers.", "summary": "The mean says where the centre is. The variance says how widely the values spread around it. The RMS says how far, typically, the values lie from zero.", "title": "Mean, variance and RMS" }, { "citations": [ "ap-17-132", "ap-17-549", "ap-17-891" ], "example": [ { "items": [ "The Earth: v/c = 29,800 ÷ 299,792,458 ≈ 9.9 × 10⁻⁵. Squared, that is about 9.9 × 10⁻⁹, near 10⁻⁸.", "Water's molecules: 1,000 kg per cubic metre divided by 0.018015 kg per gram-molecule, times 6.02 × 10²³, is 3.34 × 10²⁸ molecules per cubic metre.", "Molecules near 600 m/s crossing a surface at a quarter of n times their mean speed give about 5 × 10³⁰ strikes per square metre each second.", "A sphere 1 μm across has a surface of 3.1 × 10⁻¹² m², so it takes about 1.6 × 10¹⁹ strikes a second. The inputs are rough, so the result is an order of magnitude, 10¹⁹ to 10²⁰." ], "kind": "steps" } ], "exampleTitle": "Two estimates, done with powers of ten", "explanation": [ { "kind": "paragraph", "text": "Two numbers of the same order differ by less than a factor of ten. Numbers several orders apart belong to different worlds, and comparing them usually tells you which effect can be ignored." }, { "kind": "paragraph", "text": "Sizes. A water molecule is about 0.3 nanometres across and a Brownian grain about 1 micrometre: the grain is roughly 3,000 times larger. The light-quanta paper prints the mass of a hydrogen atom as 1/N gram, 1.62 × 10⁻²⁴ g." }, { "kind": "paragraph", "text": "Speeds against light. The ratio v/c is about 3.3 × 10⁻⁶ for a rifle bullet at 1,000 m/s, 8.3 × 10⁻⁷ for a jet at 250 m/s, and 9.9 × 10⁻⁵ for the Earth in its orbit at 29.8 km/s. The corrections of relativity go as the square of v/c, so for the Earth they are about 10⁻⁸. Only Kaufmann's fast electrons, a large fraction of the speed of light, reached speeds where they are not small." }, { "kind": "paragraph", "text": "Energies of light. One quantum hν of red light (650 nm) carries about 1.91 electron volts, green (530 nm) 2.34 eV, and ultraviolet (250 nm) 4.96 eV. The light paper's photoelectric estimate, about 4.3 volts, is of the same order." }, { "kind": "paragraph", "text": "Molecular kicks. An estimate of how often water molecules strike a grain 1 μm across gives about 1.6 × 10¹⁹ blows a second, and about 6 × 10¹⁹ for a grain of 1 μm radius. The assumptions are rough, so the honest statement is 10¹⁹ to 10²⁰ blows a second: far too many for any single blow to be seen." } ], "id": "orders-of-magnitude", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-scientific-notation-units", "kind": "proof-edge" }, { "foundationId": "foundation:bridge-fractions-ratios", "kind": "proof-edge" } ], "question": "How big, how fast and how energetic are the things in the 1905 papers, measured against each other?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "This lesson places numbers on a scale of powers of ten. Converting the papers' printed units into SI units is a separate step, with its own conventions.", "summary": "An order of magnitude is a power of ten. Placing the papers' numbers on that scale shows which effects are tiny, which are comparable, and which can safely be ignored.", "title": "Orders of magnitude" }, { "citations": [], "example": [ { "kind": "paragraph", "text": "Take the spreading profile of the Brownian paper's §4, written for a single particle:" }, { "equations": [ "eq-model-fd-partial-profile" ], "kind": "formula", "latex": "f(x,t) = \\frac{1}{\\sqrt{4\\pi D t}}\\,e^{-x^2/4Dt}", "spoken": "f of x and t equals one over the square root of four pi D t, times e to the minus x squared over four D t." }, { "kind": "paragraph", "text": "Hold t fixed and differentiate with respect to x. Only the exponent depends on x, and the derivative of −x²/4Dt with respect to x is −x/2Dt:" }, { "equations": [ "eq-model-fd-partial-slope-x" ], "kind": "formula", "latex": "\\left(\\frac{\\partial f}{\\partial x}\\right)_t = -\\frac{x}{2Dt}\\,f(x,t)", "spoken": "The partial derivative of f with respect to x at fixed t equals minus x over two D t, times f." }, { "kind": "paragraph", "text": "That is the slope of the profile in one snapshot: downhill to the right of the centre, uphill to the left. Now hold x fixed and differentiate with respect to t. Both the factor in front and the exponent depend on t:" }, { "equations": [ "eq-model-fd-partial-rate-t" ], "kind": "formula", "latex": "\\left(\\frac{\\partial f}{\\partial t}\\right)_x = \\left(-\\frac{1}{2t} + \\frac{x^2}{4Dt^2}\\right) f(x,t)", "spoken": "The partial derivative of f with respect to t at fixed x equals minus one over two t, plus x squared over four D t squared, all times f." }, { "kind": "paragraph", "text": "That is what a probe fixed at one place records. Near the centre, where x² is less than 2Dt, the bracket is negative: the concentration falls as the cloud spreads away. Farther out it is positive: the concentration rises as the cloud arrives. Differentiate the first result once more with respect to x and multiply by D, and you get the second. That is the §4 equation ∂f/∂t = D ∂²f/∂x², a rate at one place equal to a curvature at one time." } ], "exampleTitle": "The same diffusion profile, differentiated two ways", "explanation": [ { "kind": "paragraph", "text": "The concentration of particles in a tube depends on where you look, x, and when you look, t. Ask how fast the concentration changes and there are two different answers: how it changes as you move along the tube at one instant, and how it changes at one place as time passes." }, { "kind": "paragraph", "text": "A partial derivative picks one of them. ∂f/∂x is the rate of change along the tube with the time t held fixed, like comparing points in one photograph. ∂f/∂t is the rate of change in time at a fixed position x, like watching one spot under the microscope. The curly ∂ says that something else is being held fixed, and a subscript names it:" }, { "kind": "formula", "latex": "\\left(\\frac{\\partial f}{\\partial x}\\right)_t \\qquad \\left(\\frac{\\partial f}{\\partial t}\\right)_x", "spoken": "The partial derivative of f with respect to x at fixed t, and the partial derivative of f with respect to t at fixed x." }, { "kind": "paragraph", "text": "Thermodynamics depends on the same care. How much a gas's volume changes as the pressure changes depends on whether its temperature T is held fixed, with heat flowing in and out, or its entropy S is held fixed, with the gas insulated. The two derivatives have different values, and ∂V/∂p written alone does not say which is meant:" }, { "kind": "formula", "latex": "\\left(\\frac{\\partial V}{\\partial p}\\right)_T \\quad \\text{versus} \\quad \\left(\\frac{\\partial V}{\\partial p}\\right)_S", "spoken": "The partial derivative of V with respect to p at fixed temperature T, versus the same derivative at fixed entropy S." }, { "kind": "paragraph", "text": "In §3 of the light-quanta paper, radiation fills a fixed volume and φ is its entropy per unit volume and per unit interval of frequency, a function of the energy density ρ and of ν. Varying ρ with the frequency ν held fixed, Einstein finds that the derivative is 1/T:" }, { "equations": [ "eq-model-fd-entropy-slope" ], "kind": "formula", "latex": "\\left(\\frac{\\partial \\varphi}{\\partial \\rho}\\right)_{\\nu} = \\frac{1}{T}", "spoken": "The partial derivative of phi with respect to rho, at fixed frequency nu, equals one over T." }, { "kind": "paragraph", "text": "In §6 of the relativity paper, Maxwell's equations are rewritten in the coordinates ξ, η, ζ, τ of a moving system. A derivative with respect to x taken with y, z and t held fixed then becomes a combination of a derivative with respect to ξ and one with respect to τ, because holding the resting time t fixed does not hold the moving time τ fixed." } ], "id": "partial-derivatives", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:derivatives", "kind": "proof-edge" } ], "question": "When a quantity depends on two things, what does its rate of change mean?", "returnCaptions": [ { "callingAnchor": "brownian-motion:s3", "caption": "Return to Brownian §3 osmotic equilibrium" }, { "callingAnchor": "brownian-motion:s4", "caption": "Return to Brownian §4 diffusion equation derivation" } ], "review": "draft", "schemaVersion": 1, "stoppingPoint": "A partial derivative is not defined until you have said which quantities are held fixed.", "summary": "A quantity that depends on more than one variable has more than one rate of change. A partial derivative varies one variable and holds the others fixed, and it means nothing until you say which ones are held fixed.", "title": "Partial derivatives and held-fixed quantities" }, { "citations": [], "example": [ { "items": [ "Now let the second flip always copy the first. Only right-right and left-left can happen, one chance in two each.", "The product of the two steps is +1 every time, so the mixed term no longer averages zero.", "You always end 2 m from the start, so the squared distance is 4 every time: the average is 4, not 2. Independence removed that extra 2." ], "kind": "steps" } ], "exampleTitle": "What changes when the second step copies the first", "explanation": [ { "kind": "paragraph", "text": "Flip a coin twice and take one step for each flip: a metre to the right for heads, a metre to the left for tails. How far from the start do you end up, on average?" }, { "items": [ "The four outcomes, right-right, right-left, left-right and left-left, are equally likely: one chance in four each.", "They leave you 2 m to the right, back at the start, back at the start, or 2 m to the left.", "Square those distances so that left and right count alike: 4, 0, 0 and 4. Their average is 2, one square metre for each step." ], "kind": "steps" }, { "kind": "paragraph", "text": "Combining the two steps added nothing beyond one square metre each. In symbols, with Δ₁ and Δ₂ for the two steps, squaring the sum gives each step squared and a mixed term, twice their product:" }, { "equations": [ "eq-model-fd-square-of-two-steps" ], "kind": "formula", "latex": "(\\Delta_1 + \\Delta_2)^2 = \\Delta_1^2 + 2\\Delta_1\\Delta_2 + \\Delta_2^2", "spoken": "Delta one plus delta two, squared, equals delta one squared plus two delta one delta two plus delta two squared." }, { "kind": "paragraph", "text": "In the coin walk the product of the two steps is +1 for right-right and left-left and −1 for the other two, so on average it is zero and the mixed term adds nothing. Two conditions make that happen. Independence: learning the first step does not change the chances for the second, so the chance of a pair is the product of the two chances. Centring: each step averages zero, left as likely as right. Then the average of the product is the product of the averages, zero times zero:" }, { "kind": "formula", "latex": "\\langle \\Delta_1 \\Delta_2 \\rangle = \\langle \\Delta_1 \\rangle \\langle \\Delta_2 \\rangle = 0", "spoken": "The average of delta one times delta two equals the average of delta one times the average of delta two, which is zero." }, { "kind": "paragraph", "text": "Both conditions matter. If each step drifts, averaging +1, the product of the averages is 1, not 0. If the second step tends to copy the first, the product is positive more often than negative. The Brownian paper's §4 assumes that a particle's motions in successive intervals are independent, as long as the intervals are not chosen too small." } ], "id": "probability-independence", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-sum-average", "kind": "proof-edge" }, { "foundationId": "foundation:bridge-negative-numbers-direction", "kind": "proof-edge" } ], "question": "When you square a sum of random steps, why do the mixed terms average to zero?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "Before dropping a mixed term, ask two things: does learning one step change the chances for the next, and does each step average zero?", "summary": "Squaring a sum of two steps gives each step squared plus twice their product. If the steps are independent and each averages zero, the product averages zero too, and only the squares are left to add up.", "title": "Probability and independence" }, { "citations": [ "ap-17-549", "ap-17-132" ], "example": [ { "items": [ "\\(k_BT\\): joules per kelvin times kelvins is joules, which are newton-metres.", "\\(6\\pi\\eta a\\): 6π has no units, and pascal-seconds times metres is (newtons per square metre) × seconds × metres, which is newton-seconds per metre.", "Divide: newton-metres divided by newton-seconds per metre is square metres per second.", "With numbers: 1.380649 × 10⁻²³ × 290.15 ÷ (6π × 1.35 × 10⁻³ × 5 × 10⁻⁷) = 3.15 × 10⁻¹³ m²/s. That is the modern Boltzmann constant (set modern-si-2019) with the paper's water at 17 °C and its particle." ], "kind": "steps" } ], "exampleTitle": "Checking that D comes out in square metres per second", "explanation": [ { "kind": "paragraph", "text": "\"The particle moved 6\" says nothing until the unit is added: 6 micrometres, or 6 seconds? A physical quantity is a number times a unit, and the number alone changes when the unit does. Six micrometres is 0.006 millimetres: the same length, written with another unit." }, { "kind": "paragraph", "text": "Quantities of different kinds cannot be added or compared: 3 metres plus 2 seconds has no meaning. So in a correct equation every term, on both sides, carries the same units. Checking that is the quickest test a formula can pass or fail." }, { "kind": "paragraph", "text": "Take the Brownian diffusion coefficient, \\(D = k_BT/(6\\pi\\eta a)\\). Boltzmann's constant is in joules per kelvin, T in kelvins, the viscosity η in pascal-seconds and the radius a in metres. The units work out to square metres per second, which is what a diffusion coefficient must be. The worked example below goes through it." }, { "kind": "paragraph", "text": "Units also tell apart quantities that share a letter. The light paper's \\(\\rho_\\nu\\) is an energy per unit volume per unit of frequency, joules per cubic metre per hertz, not per unit of wavelength: two different quantities with different units. And its L is the speed of light in §§1 and 2 but an absorbed energy in §9. A unit check starts from what the quantity is, not from its letter." } ], "id": "quantities-units", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-scientific-notation-units", "kind": "proof-edge" }, { "foundationId": "foundation:bridge-fractions-ratios", "kind": "proof-edge" } ], "question": "Can a number of metres be compared with a number of seconds?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "This lesson stops at checking units. Converting the papers' own units into SI units is a separate step, with its own conventions.", "summary": "A physical quantity is a number times a unit. Only quantities of the same kind can be added or compared, so every term of a correct equation carries the same units, and checking that catches many slips.", "title": "Quantities and units" }, { "citations": [], "example": [ { "kind": "paragraph", "text": "A hundred steps of typical size 1 m leave a typical distance of 10 m from the start, because the square root of 100 is 10. Four hundred steps leave 20 m: four times the steps, twice the distance. No single walker has to end exactly 10 or 20 m away; these are typical sizes over many walkers. Nor is the distance from the start the length of the path: to end 10 m from home, the walker has walked 100 m." } ], "exampleTitle": "A hundred steps, then four hundred", "explanation": [ { "kind": "paragraph", "text": "A walker flips a coin before every step: heads, one metre to the right; tails, one metre to the left. After four steps, how far from the start is the walker likely to be? Not four metres: the steps partly cancel. Count the walks, and square each final distance so that left and right count alike:" }, { "items": [ "After one step the walker is 1 m away, every time. The squared distance is 1.", "After two steps there are four equally likely walks, ending 2 m right, at the start, at the start, or 2 m left. The squared distances 4, 0, 0 and 4 average 2.", "After four steps there are sixteen equally likely walks. Two end 4 m out, eight end 2 m out and six end at the start. The squared distances average (2 × 16 + 8 × 4) ÷ 16 = 4." ], "kind": "steps" }, { "kind": "paragraph", "text": "The average squared distance grows by one square metre per step: after n steps it is n. The typical distance, its square root, grows only as the square root of n: 1 m after one step, about 1.4 m after two, 2 m after four, 10 m after a hundred. To get twice as far takes four times as many steps." }, { "kind": "paragraph", "text": "In symbols, call the steps Δ₁, Δ₂ and so on, and the position after n steps X, their sum:" }, { "kind": "formula", "latex": "X_n=\\Delta_1+\\ldots+\\Delta_n", "spoken": "Position after n steps is the sum of the n signed increments." }, { "kind": "paragraph", "text": "Suppose each step averages zero and has the same mean square l², where l is the typical size of one step (1 m in the coin walk). Squaring the sum gives the n squared steps and many products of two different steps. When the steps are independent and each averages zero, every such product averages zero, as the lesson on independence shows for two coin steps. Only the squares are left to add up:" }, { "equations": [ "eq-model-fd-walk-mean-square", "eq-model-fd-walk-rms" ], "kind": "formula", "latex": "\\langle X_n^2\\rangle=nl^2,\\qquad \\sqrt{\\langle X_n^2\\rangle}=l\\sqrt n", "spoken": "The mean square is n times ell squared, and RMS displacement is ell times the square root of n." } ], "id": "random-walks", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:probability-independence", "kind": "proof-edge" }, { "foundationId": "foundation:mean-variance-rms", "kind": "proof-edge" } ], "question": "Why does typical displacement grow as the square root of the step count?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "The coin walk is a picture, not a claim that molecules push a particle in fixed jumps. The Brownian paper needs only that the particle's moves in successive intervals are independent of one another.", "summary": "Random steps partly cancel, but their squares add up. After n steps of one metre the average squared distance from the start is n, so the typical distance grows only as the square root of n.", "title": "From steps to spread" }, { "citations": [ "ap-17-549", "ap-17-132" ], "example": [ { "items": [ "A sphere of radius 0.5 μm in the paper's water at 17 °C spreads 0.7948 μm in 1 s (the paper's constants, set einstein-1905-brownian-printed).", "Double the radius to 1 μm. \\(D = k_BT/(6\\pi\\eta a)\\) halves.", "λx = √(2Dt) is multiplied by √½ = 0.7071, so it becomes 0.7948 × 0.7071 = 0.5620 μm, not half of 0.7948.", "Doubling the viscosity instead does exactly the same: D halves and λx is multiplied by 0.7071." ], "kind": "steps" } ], "exampleTitle": "Double the radius: how far does the particle get?", "explanation": [ { "kind": "paragraph", "text": "Double the side of a square and its area grows four times. Double the side of a cube and its volume grows eight times. Length, area and volume scale as the first, second and third power of the side." }, { "kind": "paragraph", "text": "Physical laws scale in the same ways. Stokes's drag on a sphere is in proportion to its radius: double the radius, double the drag. So a particle's diffusion coefficient D is inversely proportional to its radius: double the radius, and D halves." }, { "equations": [ "eq-model-fd-spread-root-time", "eq-model-fd-stokes-einstein" ], "kind": "formula", "latex": "\\lambda_x=\\sqrt{2Dt},\\qquad D=\\frac{k_BT}{6\\pi\\eta a}", "spoken": "lambda x is the square root of 2 D t, and D is k B T over six pi eta a." }, { "kind": "paragraph", "text": "The spread of a diffusing particle grows as the square root of D times the time. So halving D does not halve the spread: it divides it by √2, about 1.41. The tempting first thought, that twice the radius means half as far, is out by that factor, and so is the same thought about twice the viscosity." }, { "kind": "paragraph", "text": "Scaling can also run away. Before 1905, giving every mode of radiation in a box its classical share of heat energy made the total grow as the cube of the highest frequency allowed: widen the range tenfold and the energy grows a thousandfold, without limit. That is the difficulty §1 of the light-quanta paper sets out." } ], "id": "ratios-scaling", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-fractions-ratios", "kind": "proof-edge" }, { "foundationId": "foundation:bridge-squaring-square-roots", "kind": "proof-edge" } ], "question": "If one quantity doubles, what happens to the others?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "This lesson stops at how a change scales. Why D depends on the radius and the viscosity in the first place is the lesson on viscosity and Stokes drag.", "summary": "Many laws say how one quantity scales with another: in proportion, inversely, as a square, or as a square root. Knowing which lets you predict the effect of a change without working out any absolute number.", "title": "Ratios and scaling" }, { "citations": [], "example": [ { "equations": [ "eq-model-fd-taylor-square" ], "kind": "formula", "latex": "(x+\\Delta)^2=x^2+2x\\Delta+\\Delta^2", "spoken": "For the square function, the local expansion is exact through the quadratic term." }, { "kind": "paragraph", "text": "Keep only the value and the slope term, 4 + 2 × 2 × Δ, and see what dropping Δ² costs as the step grows:" }, { "items": [ "Δ = 0.1: 4.4 instead of 4.41. The error, 0.01, is about 0.2 per cent.", "Δ = 0.5: 6 instead of 6.25. The error, 0.25, is 4 per cent.", "Δ = 2: 12 instead of 16. The error, 4, is a quarter of the answer.", "Halving the step quarters the error, because the dropped term is Δ²." ], "kind": "steps" } ], "exampleTitle": "What the dropped term costs", "explanation": [ { "kind": "paragraph", "text": "What is 2.1 squared, without a calculator? Start from what you know: 2 squared is 4. Picture a square 2 wide growing to 2.1 wide. It gains two strips, each 2 long and 0.1 thick, which add 0.4, and a small corner, 0.1 by 0.1, which adds 0.01. So 2.1 squared is 4.41. The first correction, 0.4, did almost all the work; the second, 0.01, is forty times smaller." }, { "kind": "paragraph", "text": "The same recipe works for any smooth curve, close to a point where you know it. Start with its value there. Add its slope times the step Δ: the slope, or first derivative, is the change in output per unit of input. Then add half its curvature times Δ²: the curvature, or second derivative, says how fast the slope itself changes. Further terms carry higher powers of Δ, so they are small only when Δ is:" }, { "kind": "formula", "latex": "\\begin{aligned}f(x+\\Delta)&=f(x)+\\Delta f'(x)\\\\&\\quad+\\frac{\\Delta^2}{2}f''(x)+\\ldots\\end{aligned}", "spoken": "The value a step delta away equals the value, plus delta times the slope, plus half of delta squared times the curvature, plus further terms." }, { "kind": "paragraph", "text": "The Brownian paper does exactly this in §4. It expands the number of particles at x + Δ in powers of the jump Δ and keeps the terms up to Δ², because only very small jumps contribute and each further surviving term is much smaller than the one before. The Δ term averages away, since jumps to the left and to the right are equally likely, and the Δ² term becomes the diffusion coefficient. A real particle does not make independent jumps over arbitrarily short times, so the expansion holds for times long enough for successive jumps to be independent and short enough for each jump to be small." } ], "id": "taylor-expansion", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-squaring-square-roots", "kind": "proof-edge" } ], "question": "Why can a curve be replaced, close to one point, by a few simple terms?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "An expansion is a claim about size. Say which terms you keep, and why the ones you drop are smaller: here, because they carry higher powers of a small step.", "summary": "Close to a point, a smooth curve is described by its value there, its slope and its curvature. Dropping the rest needs a reason: how small the step is, and how fast the dropped terms shrink with it.", "title": "Approximating a curve near a point" }, { "citations": [ "ap-17-549" ], "example": [ { "items": [ "At T = 293 K, \\(k_BT\\) = 1.381 × 10⁻²³ × 293 = 4.05 × 10⁻²¹ J, so the average energy of motion along one direction is half of that, 2.02 × 10⁻²¹ J.", "A nitrogen molecule has a mass of 4.65 × 10⁻²⁶ kg. Setting \\(\\frac{1}{2} m v_x^2\\) equal to \\(\\frac{1}{2} k_BT\\) gives a typical speed along one direction of √(4.05 × 10⁻²¹ ÷ 4.65 × 10⁻²⁶), about 295 metres a second.", "A grain of radius 0.5 μm and density 1200 kg/m³ has a mass of 6.3 × 10⁻¹⁶ kg, about 1.4 × 10¹⁰ times the molecule's. The same average energy gives it about 2.5 millimetres a second.", "In water that motion is turned about within some 70 nanoseconds: the grain's mass divided by its drag coefficient 6πηa. That is far too brief to follow, so what a microscope sees is the net result of the zigzag." ], "kind": "steps" }, { "kind": "paragraph", "text": "This is why the Brownian paper asks for a displacement over a time rather than a speed: the thermal speed is real, but it turns about far faster than any observer can watch." } ], "exampleTitle": "A nitrogen molecule and a half-micrometre grain at 20 °C", "explanation": [ { "kind": "paragraph", "text": "The air in a room at 20 °C is a crowd of molecules moving in every direction and colliding billions of times a second. Warm the room and, on average, they move faster. The kinetic theory of heat reads temperature this way: through the average energy of motion of the particles, taken over very many of them." }, { "kind": "paragraph", "text": "For any particle in equilibrium with its surroundings, the average energy of motion along each direction is the same:" }, { "equations": [ "eq-model-fd-equipartition" ], "kind": "formula", "latex": "\\left\\langle \\frac{1}{2} m v_x^2 \\right\\rangle = \\frac{1}{2} k_B T", "spoken": "The average of one half m v x squared equals one half k B T." }, { "kind": "paragraph", "text": "Here m is the particle's mass, \\(v_x\\) its speed along x, T the absolute temperature, and \\(k_B\\) = 1.38 × 10⁻²³ joules per kelvin is Boltzmann's constant. It equals R/N, the gas constant shared out per molecule. The Brownian paper writes RT/N where a modern text writes \\(k_BT\\); the paper's own letter k means the viscosity, not this constant." }, { "kind": "paragraph", "text": "The average is the point. One molecule's energy of motion changes at every collision, from almost nothing to several times the average. The temperature fixes how energy is shared out over many particles. It is not the energy of any one of them." }, { "kind": "paragraph", "text": "Nothing in the rule mentions size. A grain ten billion times heavier than a molecule has the same average energy of motion, so it moves more slowly, by the square root of the mass ratio. That is why a suspended grain belongs to the same thermal story as a dissolved molecule, as §1 of the paper insists." } ], "id": "temperature-thermal-energy", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:work-energy", "kind": "proof-edge" }, { "foundationId": "foundation:bridge-scientific-notation-units", "kind": "cross-link" }, { "foundationId": "foundation:bridge-squaring-square-roots", "kind": "cross-link" } ], "question": "What does a temperature say about the jostling of molecules and grains?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "This lesson stops at the average energy of motion. How the jostling of many molecules makes a grain wander, and how far, is §4 of the Brownian paper.", "summary": "In equilibrium at a given temperature, every kind of particle, molecule or grain, has the same average energy of motion along each direction it can move: half of Boltzmann's constant times the temperature. The temperature fixes that average over many particles. It is not the energy of any one of them.", "title": "Temperature and thermal energy" }, { "citations": [ "ap-17-549" ], "example": [ { "items": [ "The question: can a measured spreading rate tell you how big the particle is?", "What is given: the measured D, the temperature T, the viscosity η, and the relation \\(D = RT/(6\\pi\\eta a N)\\).", "The first thought, and a reasonable one: solve the relation for a. The algebra looks as if it allows that.", "The decisive step: solving gives \\(a = RT/(6\\pi\\eta N D)\\), the radius in terms of N. Every choice of N gives its own a. Double a and halve N, and \\(6\\pi\\eta a N\\), and with it D, is exactly what it was. The answer is a curve, not a number.", "The limitation: a second relation that depends on a and N in another way is needed. How well it fixes the pair depends on the angle at which the two curves cross and on the two measurements being independent, and what comes out is a region, not a point." ], "kind": "steps" } ], "exampleTitle": "Why solving for the radius gives a curve", "explanation": [ { "kind": "paragraph", "text": "In §5 of the Brownian paper the diffusion coefficient of a suspended sphere is \\(D = \\frac{RT}{N}\\cdot\\frac{1}{6\\pi\\eta a}\\). Einstein writes k for the viscosity and P for the radius; this lesson uses η and a. At a fixed temperature and viscosity, a measured D fixes one combination of the two unknowns, the product of the molecular number N and the radius a." }, { "equations": [ "eq-model-fd-product-na" ], "kind": "formula", "latex": "N a = \\frac{RT}{6\\pi\\eta D}", "spoken": "N times a equals R T divided by six pi eta D." }, { "kind": "paragraph", "text": "Every pair with that product fits the measurement equally well. Drawn in a plane with the radius along one axis and the molecular number along the other, the pairs form a curve. Double the radius and halve N, and the product, and with it D, is exactly what it was. Displacements alone cannot say which point on the curve is the true one, and measuring D more carefully does not change that." }, { "kind": "paragraph", "text": "Einstein's dissertation measured something else: how much a dissolved substance thickens water. The volume taken up by the dissolved particles, as a fraction of the whole, is \\(\\varphi = \\frac{4}{3}\\pi a^3 N n\\), where n is the number of moles dissolved per unit volume. The viscosity relation therefore fixes a different combination, \\(a^3 N\\). Its curve in the same plane has another shape, and the two curves cross." }, { "equations": [ "eq-model-fd-cube-na" ], "kind": "formula", "latex": "a^3 N = \\frac{3(\\eta^*/\\eta - 1)}{4\\pi c\\, n}", "spoken": "a cubed times N equals three times the quantity eta star over eta minus one, divided by four pi c n." }, { "kind": "paragraph", "text": "Here \\(\\eta^*\\) is the viscosity of the solution, η that of the water, and c the coefficient of φ in the viscosity law: 1 as the dissertation printed it in 1906, and 5/2 after Einstein's correction of 1911. This lesson takes the relation as given: deriving it, and the history of its correction, belong to the dissertation. Dividing the second combination by the first leaves \\(a^2\\) alone, so the crossing gives the radius, and then either relation gives N." }, { "equations": [ "eq-model-fd-radius-squared" ], "kind": "formula", "latex": "a^2 = \\frac{3(\\eta^*/\\eta - 1)}{4\\pi c\\, n}\\cdot\\frac{6\\pi\\eta D}{RT}", "spoken": "a squared equals three times the quantity eta star over eta minus one, over four pi c n, times six pi eta D over R T." }, { "kind": "paragraph", "text": "A crossing is not yet an answer. Two curves that cross at a shallow angle fix their crossing poorly, so a small error in either measurement moves it a long way. Each measurement has an uncertainty, which widens its curve into a band, and the honest result is the region where the two bands overlap, not a point. Two measurements that rest on the same calibration can cross and still add nothing new. And each relation holds only within its own domain: dilute suspensions, spheres much larger than the molecules of the liquid." } ], "id": "two-measurements-two-unknowns", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:ratios-scaling", "kind": "proof-edge" }, { "foundationId": "foundation:error-and-inference", "kind": "proof-edge" }, { "foundationId": "foundation:functions-graphs", "kind": "cross-link" } ], "question": "Can a measured spreading rate tell you how big the particle is?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "One measurement of one combination of two quantities leaves a curve, not a point.", "summary": "The Brownian relation fixes only the product of the molecular number and the particle's radius, so one measured diffusion coefficient leaves a curve of possible pairs. A second relation that depends on the two in another way crosses that curve, and the crossing fixes both, within a region set by the uncertainties of the two measurements.", "title": "Two measurements, two unknowns" }, { "citations": [ "ap-17-132", "ap-17-549", "ap-17-891" ], "example": [ { "items": [ "The inputs: β = 4,866 · 10⁻¹¹ and ν = 1,03 · 10¹⁵ per second are printed in §8, and so is E = 9,6 · 10³. R = 8.31 × 10⁷ erg per mole per kelvin is the edition's editorial input, because the paper does not print it.", "With P′ = 0, Π = Rβν/E = 8.31 × 10⁷ × 4.866 × 10⁻¹¹ × 1.03 × 10¹⁵ ÷ 9.6 × 10³ = 4.34 × 10⁸ abvolts.", "Multiply by the ratio 10⁻⁸ volts per abvolt: 4.34 volts. The paper prints about 4,3 volts.", "In §9 the conversion runs the other way. An energy per gram-equivalent divided by E is a potential: the printed Rβν = 6,4 · 10¹² erg gives 6.4 × 10¹² ÷ 9.6 × 10³ = 6.67 × 10⁸ abvolts, or 6.67 volts.", "Stark's ionization voltage of 10 volts is 10⁹ abvolts, and 10⁹ × 9.6 × 10³ = 9.6 × 10¹² erg per gram-equivalent, the upper bound for J that §9 prints.", "E itself, 9.6 × 10³ abcoulombs, is 9.6 × 10⁴ coulombs per gram-equivalent. Beside it, the modern Faraday constant is 96 485 coulombs per mole (set modern-si-2019); Einstein's value is 0.50 percent below it." ], "kind": "steps" } ], "exampleTitle": "The photoelectric check, converted to volts", "explanation": [ { "kind": "paragraph", "text": "Einstein's papers measure length in centimetres, mass in grams and time in seconds, the CGS system. Energy is in ergs, and one erg is 10⁻⁷ joules. Viscosity is in grams per centimetre per second, the unit now called the poise, and one poise is 0.1 pascal-seconds. These mechanical factors are exact, because they follow from 1 cm = 0.01 m and 1 g = 0.001 kg." }, { "kind": "paragraph", "text": "To convert, multiply by a ratio equal to one. Since 1 poise is 0.1 pascal-seconds, the ratio (0.1 Pa s)/(1 P) equals one, and multiplying by it changes the unit without changing the viscosity. The Brownian paper gives water's viscosity as 1,35 · 10⁻², with no unit name; it is in poise, so the viscosity is 1.35 × 10⁻³ Pa s. The paper calls this viscosity k, which is not Boltzmann's constant." }, { "kind": "formula", "latex": "\\begin{aligned} &1.35\\times10^{-2}\\,\\mathrm{P}\\times\\frac{0.1\\,\\mathrm{Pa\\,s}}{1\\,\\mathrm{P}} \\\\ &\\quad = 1.35\\times10^{-3}\\,\\mathrm{Pa\\,s} \\end{aligned}", "spoken": "1.35 times ten to the minus two poise, times 0.1 pascal-seconds per poise, equals 1.35 times ten to the minus three pascal-seconds." }, { "kind": "paragraph", "text": "The same ratio works for compound units. The gas constant R = 8.31 × 10⁷ erg per mole per kelvin is 8.31 joules per mole per kelvin. Neither the light paper nor the Brownian paper prints this number; the edition supplies the standard value of 1905 as an editorial input and labels it that way." }, { "kind": "paragraph", "text": "Electricity had two CGS systems. The electrostatic system builds its charge unit, the statcoulomb, from the force between charges. The electromagnetic system builds its current unit from the force between currents; its units are the abcoulomb and the abvolt. One abcoulomb is 10 coulombs, one abvolt is 10⁻⁸ volts, one statvolt is 299.792458 volts, and one gauss is 10⁻⁴ tesla. Planck's 1901 elementary charge, 4,69 · 10⁻¹⁰ statcoulombs, is 1.56 × 10⁻¹⁹ coulombs." }, { "kind": "paragraph", "text": "These electrical correspondences are conventional, not exact. They assume the magnetic constant \\(\\mu_0\\) is exactly 4π × 10⁻⁷ henry per metre. Since the 2019 revision of the SI it is a measured quantity, equal to that value within about one part in a billion. The site marks every electrical conversion conventional." }, { "kind": "paragraph", "text": "The light paper's photoelectric check in §8 uses the electromagnetic system. It takes E = 9,6 · 10³ for the charge of one gram-equivalent of a monovalent ion, so a potential Π comes out in abvolts, and the paper multiplies by 10⁻⁸ to get volts. It prints the result as about 4,3 volts. On the result line the factor is set as 10⁷; the edition records this as a misprint, because the sentence four lines earlier gives 10⁻⁸. The check can also be run per electron, with Planck's charge in statcoulombs and the result in statvolts, and it gives 4.31 volts; the edition keeps that as a documented alternative and follows the page." }, { "kind": "paragraph", "text": "The paper's R, E and J are amounts per gram-equivalent, a mole of monovalent ions. Dividing by N gives the amount per molecule, R/N, or per electron, ε. With the paper's E and the N it prints in §2, 6,17 · 10²³, the charge on one ion is 9.6 × 10³ ÷ 6.17 × 10²³ = 1.56 × 10⁻²⁰ abcoulombs, which is 1.56 × 10⁻¹⁹ coulombs. The paper does not print this quotient." }, { "kind": "paragraph", "text": "The relativity paper writes its fields in Gaussian units, where electric and magnetic fields share one unit. That is why its transformation reads β(Y − (v/V)N), with the dimensionless ratio v/V; here N is a magnetic field component, not Avogadro's number. SI measures the electric field in volts per metre and the magnetic field in tesla, so the same law reads \\(\\gamma(E_y - vB_z)\\), with the speed itself multiplying the magnetic field. One statvolt per centimetre is 2.998 × 10⁴ volts per metre." }, { "kind": "formula", "latex": "\\begin{gathered} 1\\,\\mathrm{abV} = 10^{-8}\\,\\mathrm{V}, \\\\ 1\\,\\mathrm{abC} = 10\\,\\mathrm{C}, \\\\ 1\\,\\mathrm{statV} = 299.792458\\,\\mathrm{V}, \\\\ 1\\,\\mathrm{G} = 10^{-4}\\,\\mathrm{T} \\end{gathered}", "spoken": "One abvolt corresponds to ten to the minus eight volts, one abcoulomb to ten coulombs, one statvolt to 299.792458 volts, and one gauss to ten to the minus four tesla, all by convention." } ], "id": "unit-system-1905", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-scientific-notation-units", "kind": "proof-edge" }, { "foundationId": "foundation:bridge-fractions-ratios", "kind": "proof-edge" }, { "foundationId": "foundation:quantities-units", "kind": "cross-link" } ], "question": "How do Einstein's numbers, printed in centimetres, grams, ergs and the old electrical units, become SI units?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "A conversion factor is a ratio equal to one. This lesson converts units only: it never turns a historical constant's value into a modern one, and it treats every electrical correspondence as a convention.", "summary": "The papers use the centimetre-gram-second system and its two electrical variants. A printed number becomes an SI number when it is multiplied by a ratio equal to one, such as 0.1 pascal-seconds per poise. The electrical ratios are conventional correspondences, not exact identities.", "title": "The units of 1905" }, { "citations": [ "ap-17-549" ], "example": [ { "items": [ "Take a sphere of radius a = 0.5 μm = 5 × 10⁻⁷ m, moving at v = 1 μm/s = 10⁻⁶ m/s through water, with η = 10⁻³ Pa·s.", "Multiply: 6π × 10⁻³ × 5 × 10⁻⁷ × 10⁻⁶ = 6π × 5 × 10⁻¹⁶, about 9.4 × 10⁻¹⁵ newtons.", "Twice the speed gives twice the drag, 1.9 × 10⁻¹⁴ N. A sphere of twice the radius at the same speed feels twice the drag too.", "Read backwards, as §3 does: a steady force of 9.4 × 10⁻¹⁵ N keeps this sphere drifting at 1 μm/s. The speed per unit force, 1/(6πηa), is what the paper needs next." ], "kind": "steps" }, { "kind": "paragraph", "text": "The flow here is very slow in the sense that matters: the liquid's inertia is less than a millionth of its viscous resistance (the ratio is 1000 kg/m³ × 10⁻⁶ m/s × 5 × 10⁻⁷ m ÷ 10⁻³ Pa·s = 5 × 10⁻⁷), so Stokes's law applies with room to spare." } ], "exampleTitle": "The drag on a half-micrometre sphere in water", "explanation": [ { "kind": "paragraph", "text": "Stir honey and then water with the same spoon at the same speed. The honey pushes back far harder. Viscosity is the number that measures this. Water at room temperature has a viscosity of about 1 millipascal-second (0.001 Pa·s); honey's is thousands of times larger." }, { "kind": "paragraph", "text": "Now move a small sphere slowly through the liquid. The liquid pushes back with a force that grows in step with the speed: twice as fast, twice the drag. It also grows with the sphere's radius and with the viscosity. George Gabriel Stokes worked out the whole law in 1851:" }, { "equations": [ "eq-model-fd-stokes-drag" ], "kind": "formula", "latex": "F = 6\\pi\\eta a v", "spoken": "The drag F equals six pi times the viscosity eta, times the radius a, times the speed v." }, { "kind": "paragraph", "text": "F is the drag in newtons, η the viscosity, a the sphere's radius and v its speed. The factor 6π comes from the way the liquid flows around a sphere; another shape would have another factor." }, { "kind": "paragraph", "text": "§3 of the Brownian paper runs the law backwards. A steady force K on a sphere of radius P, in a liquid whose viscosity the paper writes k, makes the sphere drift at the speed K/(6πkP). There k is the viscosity, not Boltzmann's constant, and P is the radius, not a pressure. For the law itself the paper refers to Kirchhoff's lectures on mechanics." }, { "kind": "paragraph", "text": "The law holds only in a regime. The flow must be slow and smooth, so that the liquid's own inertia plays no part; that is true for a micrometre sphere moving a few micrometres a second. The sphere must also be much larger than the liquid's molecules, which are about 0.3 nanometres across, so that the liquid acts as a smooth continuum. In a gas, where a molecule travels about 70 nanometres between collisions, a small particle partly slips through, and the drag is smaller than Stokes's law says." } ], "id": "viscosity-stokes-drag", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:quantities-units", "kind": "proof-edge" }, { "foundationId": "foundation:bridge-scientific-notation-units", "kind": "cross-link" }, { "foundationId": "foundation:bridge-fractions-ratios", "kind": "cross-link" }, { "foundationId": "foundation:flux-continuity", "kind": "cross-link" } ], "question": "How hard does a liquid hold back a small sphere drifting through it?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "This lesson stops at the drag on one slow sphere. Why the same 6πηa also sets how fast a crowd of such spheres spreads by diffusion is the subject of §§3 and 5 of the Brownian paper.", "summary": "A slow sphere in a liquid feels a drag that grows in step with its speed, its radius and the liquid's viscosity: F = 6πηav. The Brownian paper runs this law backwards, to turn a force on a particle into the speed it drifts at.", "title": "Viscosity and Stokes drag" }, { "citations": [ "ap-18-639" ], "example": [ { "items": [ "Two bodies move at 2 metres per second.", "The 3 kg body has ½ × 3 × 2² = 6 joules of energy of motion; the 2 kg body has ½ × 2 × 2² = 4 joules.", "The speed is the same, and the energies differ by 2 joules because the masses differ by 1 kg." ], "kind": "steps" }, { "equations": [ "eq-model-fd-kinetic-difference" ], "kind": "formula", "latex": "K_0-K_1=\\frac{1}{2}(m_0-m_1)v^2", "spoken": "At the same low speed, the kinetic-energy drop equals one half times the mass decrease times the squared speed." }, { "kind": "paragraph", "text": "Here the two masses are given, so the example proves nothing about mass and energy. The 1905 argument runs the other way: it computes the energy difference from the light the body gives off, finds that it has this form, and reads the mass difference from the coefficient of ½v²." } ], "exampleTitle": "Two bodies at the same speed, different masses", "explanation": [ { "kind": "paragraph", "text": "A force that pushes a body through a distance does work on it, and the work becomes the body's energy of motion, its kinetic energy. That is separate from energy stored inside the body, as heat or in its chemistry." }, { "kind": "paragraph", "text": "For one body at everyday speeds, doubling the speed multiplies the energy of motion by four. At a fixed speed, doubling the mass doubles it. The mass-energy paper's title asks about a body's Trägheit, its inertia: its resistance to a change in its motion, which is measured by pushing it, not by weighing it." }, { "equations": [ "eq-model-fd-kinetic-energy" ], "kind": "formula", "latex": "K=\\frac{1}{2}mv^2", "spoken": "In Newtonian mechanics, kinetic energy is one half times inertial mass times speed squared." }, { "kind": "paragraph", "text": "This formula holds at speeds small compared with the speed of light, and it is not exact close to that speed. So the mass-energy argument works with the exact expression and reads the mass from its low-speed limit, where it takes this form." } ], "id": "work-energy", "kind": "foundation", "prerequisites": [ { "foundationId": "foundation:bridge-squaring-square-roots", "kind": "cross-link" }, { "foundationId": "foundation:bridge-fractions-ratios", "kind": "cross-link" } ], "question": "If a body moves at the same speed but carries less energy of motion, what has changed?", "review": "draft", "schemaVersion": 1, "stoppingPoint": "At the same low speed, less energy of motion means less mass. The comparison gives only a difference in mass, never the total energy stored inside the body.", "summary": "At everyday speeds, the energy of motion depends on speed and on mass. Compare two states at the same speed, and any difference in that energy is a difference in mass.", "title": "Energy of motion and inertia" } ], "paper": { "citation": "ap-17-549", "description": "Particles just large enough to see under a microscope must wander, Einstein argues, and how far they wander in a given time would let us count molecules. Read the argument, and open any step it leaves out.", "germanTitle": "Über die von der molekularkinetischen Theorie der Wärme geforderte Bewegung von in ruhenden Flüssigkeiten suspendierten Teilchen", "id": "brownian-motion", "kind": "paper", "schemaVersion": 1, "sections": [ { "arguments": [ "arg-bm-introduction" ], "id": "s0", "title": "Introduction · A motion the theory requires" }, { "arguments": [ "arg-bm-osmotic-suspended" ], "id": "s1", "title": "§1 · Osmotic pressure from suspended particles" }, { "arguments": [ "arg-bm-kinetic-justification" ], "id": "s2", "title": "§2 · Osmotic pressure from the molecular-kinetic theory" }, { "arguments": [ "arg-bm-observable", "arg-bm-independent-steps", "arg-bm-diffusion-equation", "arg-bm-gaussian" ], "id": "s4", "title": "§4 · From random displacement to diffusion" }, { "arguments": [ "arg-bm-diffusivity", "arg-bm-inference" ], "id": "s5", "title": "§5 · From displacement to molecular scale" } ], "sourceNotice": "This is newly written explanation in modern notation, and its editorial review is pending. It is not the German source, an English translation, or a complete edition of the paper. The German source face holds a machine-drafted transcription with hand correction that no one has reviewed yet, and the facsimile face shows the pinned journal pages; an English translation is not ready. The headings name the part of the argument each passage discusses; they are not a list of the paper’s paragraphs.", "sourceStatus": "in-preparation", "status": "explanation-preview", "title": "Brownian motion: from wandering to a measurable law" }, "schemaVersion": 1 }