Read · Light quanta: from entropy to an energy scale

§5 · independent configurations

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Detail

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A definition

Which probability the entropy uses

What does Einstein mean by the probability of a state, and why does he insist on it?

The printed paragraph this passage explains

§5, paragraph 1: Einstein notes that molecular theory often uses the word probability in a sense that differs from its mathematical definition, and says he will show in another paper that statistical probability is enough; here he gives only the general principle and some special cases.

§5 opens with a warning about a word. When entropy is calculated from molecular theory, Einstein writes, 'probability' is often used in a sense that does not match its definition in the theory of probability. In particular, which cases count as equally probable is often fixed by hypothesis, even where the theoretical picture is definite enough to deduce them instead.

He announces that he will show in a separate paper that the statistical probability is all that thermal questions need, and hopes this removes a logical difficulty that still stands in the way of applying Boltzmann's principle. Here he states the principle only in general and applies it to special cases. The first special case shows what statistical probability means: the probability that, at a moment chosen at random, the system is found in the state considered.

Open the foundation: A probability, a frequency and a single trial

S-S_0=\frac{R}{N}\lg W

The entropy difference S minus S zero equals R over N times the logarithm of W.

Here W is the probability of the state with entropy S relative to a starting state with entropy $S_0$ , R is the gas constant, N the number of real molecules in a gram-equivalent, and lg is the natural logarithm as the paper writes it. Read with a statistical W, the formula needs no list of equally likely cases: a state the system is found in at 1 moment in 16, compared with a state it is always in, has W = 1/16, and its entropy is lower by R/N times ln 16.

The reading matters for everything after it. The probability that n independent points are all in part of a volume, and in §6 the probability that all the energy of monochromatic radiation is in part of its volume, are statistical probabilities in exactly this sense. The step from entropy to energy quanta is only as good as that reading of W.

Show every step here: Which probability the entropy uses

Letters: the paper writes W for the probability of a state, S for its entropy, R for the gas constant, N for the number of real molecules in a gram-equivalent, and lg for the natural logarithm, which this explanation also writes ln.

The warning: in molecular calculations of entropy, the word 'probability' is often used in a sense that differs from its definition in the theory of probability.
The shortcut Einstein objects to: decide by hypothesis which cases are equally probable, then count cases. He objects where the model is definite enough to deduce those probabilities instead of assuming them.
His alternative, the statistical probability: watch the system over a long time and ask what fraction of the moments it spends in the state. Equivalently, pick a moment at random and ask how likely it is that the system is in the state then.
Worked example: four points that move independently in a box, each equally likely to be anywhere in it. At a random moment one point is in the left half with probability 1/2, so all four are there with probability 1/2 × 1/2 × 1/2 × 1/2 = 1/16. Watched for a long time, the box spends about 1 moment in 16 with all four points on the left.
Boltzmann's principle, as Einstein uses it: the entropy of a system is a function of the probability of its state. The next paragraphs of §5 show that the function has to be a logarithm, because entropies of separate systems add while their probabilities multiply.

S-S_0=\frac{R}{N}\lg W

The entropy difference S minus S zero equals R over N times the logarithm of W.

Reading the formula: S − $S_0$ is the entropy of the state minus the entropy of the starting state, and W is the probability of the state relative to the starting state.
In the worked example, relative to the box with its points anywhere, W = 1/16. So S − $S_0$ = (R/N) ln(1/16) = −(R/N) ln 16. Since ln 16 = 4 ln 2 ≈ 4 × 0.693 ≈ 2.77, the state with all four points on the left has an entropy lower by about 2.77 × R/N.
Nothing in this used a list of equally likely cases decided in advance: the 1/16 came from the independence of the points and the fraction of the volume, both properties of the model.
Einstein promises a separate paper to show that the statistical probability suffices for thermal questions; this paper does not contain that proof, so here it remains his announced position.

A tool for this step

Probability and independence

When you square a sum of random steps, why do the mixed terms average to zero?

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?

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.

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:

\left(\Delta_{1} + \Delta_{2}\right)^{2} = \Delta_{1}^{2} + 2\,\Delta_{1}\,\Delta_{2} + \Delta_{2}^{2}

Delta one plus delta two, squared, equals delta one squared plus two delta one delta two plus delta two squared.

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:

\langle \Delta_1 \Delta_2 \rangle = \langle \Delta_1 \rangle \langle \Delta_2 \rangle = 0

The average of delta one times delta two equals the average of delta one times the average of delta two, which is zero.

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.

Worked example: What changes when the second step copies the first

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.

Where this lesson stops

Before dropping a mixed term, ask two things: does learning one step change the chances for the next, and does each step average zero?

Modern qualifications

Boltzmann had tied entropy to the probability of a state in the kinetic theory of gases; Einstein calls the relation Boltzmann's principle and writes its constant as R/N. Planck's 1901 paper, which §2 cites (Ann. d. Phys. 4, p. 561), writes that constant as k, the letter later written $k_B$ . Before 1905 Einstein had published three papers on the molecular foundations of thermodynamics in the Annalen der Physik (1902, 1903 and 1904). The separate paper promised in §5 is not named, and this passage does not identify it; nor does §5 say whose use of probability it criticizes.

Assumptions and limits: Which probability the entropy uses

Assumed here

The probability of a state is its statistical probability: the chance of finding the system in that state at a moment chosen at random.
Boltzmann's principle, as Einstein uses it: the entropy of a system is a function of the probability of its state.
The argument compares entropy differences between states; it needs no absolute entropy.

What this does not establish

Einstein says he will show in a separate paper that statistical probability is enough; this paper does not give that proof, and this passage does not supply it.
A moment chosen at random presupposes a system in a steady condition, watched over a long time; the passage does not treat systems that are not.

Source context: German source · English · Interlinear gloss · Facsimile

A derivation

Independence supplies the exponent

Why is the probability f to the power n, rather than just f?

The 8 printed paragraphs this passage explains

§5, paragraph 2: If entropy is a function of the probability of a state, two systems that do not act on each other have entropies that add while their probabilities multiply, because their states are independent events.
§5, paragraph 3: It follows that entropy is a universal constant times the logarithm of the probability, plus a constant. Kinetic theory gives that constant as R/N, so the entropy difference between two states is R/N times the logarithm of their relative probability.
§5, paragraph 4: A special case: n moving points, such as molecules, in a volume v₀, with nothing assumed about how they move except that no part of the space and no direction is preferred, and so few of them that they do not act on each other.
§5, paragraph 5: This system, an ideal gas or a dilute solution for example, has some entropy. Imagine all n points moved into a part v of the volume with nothing else changed: the new state has a different entropy, to be found from Boltzmann's principle.
§5, paragraph 6: The question: how probable is it that, at a moment chosen at random, all n independently moving points happen to be in the part v?
§5, paragraph 7: That statistical probability is v/v₀ raised to the power n, and Boltzmann's principle turns it into an entropy difference of R(n/N) times the logarithm of v/v₀.
§5, paragraph 8: Remarkably, this equation, from which Boyle and Gay-Lussac's law and the matching law of osmotic pressure follow, needs no assumption about the law by which the molecules move.
§5, footnote 1: The footnote derives the gas law from that entropy difference: the work p dv equals T dS, which gives pv = R(n/N)T.

Each independent point has probability f of being inside the chosen fraction. For all n to be inside, multiply n independent probabilities. Taking the logarithm makes the exponent n a coefficient.

Open the foundation: Entropy and the number of ways

W = f^{n}

\Delta S_{\mathrm{gas}} = k_B\,\ln\left(W\right) = n\,k_B\,\ln\left(f\right)

The probability is f to the n; the entropy difference is Boltzmann's constant times its logarithm, or n times Boltzmann's constant times the logarithm of f.

With four independent points and half the volume, the probability is 1/16. If all four positions are exactly the same uniformly distributed position, the probability is 1/2 instead. Having four labels is not enough to justify four independent factors.

The concentration from V₀ to V smaller than V₀ has a negative entropy difference. Reversing the constrained comparison reverses the sign. The independent-configurations laboratory makes the probability and entropy readouts separate.

Open the foundation: Probability and independence

Show every step here: Independence supplies the exponent

A tool for this step

A probability, a frequency and a single trial

What does a probability of 1/4 say about one try, and about a hundred?

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.

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.

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.

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.

Worked example: Counting outcomes for two tokens and for four

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.

Every way two tokens can land

Each box is one way the drop can come out, with the first token above the second. There are 4 boxes because a box divides in two and there are 2 tokens, and the shaded one is the single outcome with both tokens on the left.

What it shows, in words

The 4 boxes, from the left, are: both tokens left; the first left and the second right; the first right and the second left; both right. They are equally likely, and counting them is the whole of where one in 4 comes from. The middle two are the ones a reader is most likely to collapse into one, and the drawing keeps them apart: the tokens are in a different order, and either order is a way the drop can happen.

The drawing is about one drop, and says nothing about how a run of drops will come out. How often both land left over a hundred drops is a different number, counted rather than reasoned, and the paragraph above it says so.

Where this lesson stops

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.

Letters: the paper writes W for the probability of a state and $S-S_0=\frac{R}{N}\lg W$ for Boltzmann's principle, and for the case below $W=(v/v_0)^n$ . This explanation writes f for $v/v_0$ and $k_B$ for R/N.

Boltzmann's principle, as the paper states it: the entropy of a system is a function of the probability of its state. For two systems that do not act on each other, the entropies add and the probabilities multiply. A function that turns products into sums is a logarithm, so the entropy is C times ln W plus a constant, and the paper takes C = R/N from the kinetic theory of gases.
The setting: n movable points in a volume $v_0$ , so few that they do not act on each other, with no part of the space and no direction preferred. They could be the molecules of a dilute gas or of a dissolved substance.
The question: at a moment chosen at random, what is the probability that all n points are in a chosen part v of $v_0$ ?
One point is equally likely to be anywhere, so the chance that it is in v is the fraction of the volume, $f=v/v_0$ .
Two independent points are both inside with probability f × f. For n independent points, multiply n times: $W=f^n$ .
Worked case: n = 4 and f = 1/2. Mark each point as in or out: there are 2 × 2 × 2 × 2 = 16 equally likely patterns, and exactly one has all four in, so W = 1/16, which is $(1/2)^4$ .
Take the logarithm: $\ln W=n\ln f$ . In the worked case, $\ln(1/16)=4\ln(1/2)$ , about -2.77.
Multiply by $k_B$ , the paper's R/N, to get the entropy difference.

A tool for this step

Entropy and the number of ways

Why is entropy the logarithm of a probability?

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.

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:

S - S_{0} = k_B\,\ln\left(W\right)

S minus S nought equals k B times the natural logarithm of W.

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.

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.

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.

Worked example: Squeezing a gas into half its volume

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.

§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.

JavaScript is off, so the controls below cannot change anything. The construction shows its first setting, and its closing paragraph, “What it shows, in words”, describes the rest.

Try it: counting arrangements

Cut the box into equal parts and let each particle land in any part with the same chance, whatever the others do. Every arrangement is then as likely as every other, and the probability of a state is the share of arrangements that produce it. Choose how many particles, and how much of the box they must all be found in.

How many particles

All of them in

Of 16 equally likely arrangements, 1 put all 4 in the left half. So W = (1/2)⁴ = 0.0625.

ln W = 4 × ln(1/2) = −2.77, so the entropy changes by S − S₀ = k_B ln W = −3.83 × 10⁻²³ joules per kelvin.

Under each box, the marks say where particle 1, particle 2 and so on are, in that order. The same picture can come up more than once, because the particles are told apart: particle 1 alone outside and particle 2 alone outside are two different arrangements.

How many arrangements give each count in the left half
Particles in the left half	Arrangements
all 4	1
3	4
2	6
1	4
0	1

L L L Lall inside
L L L R
L L R L
L L R R
L R L L
L R L R
L R R L
L R R R
R L L L
R L L R
R L R L
R L R R
R R L L
R R L R
R R R L
R R R R

What it shows, in words

With four particles and the box cut in half there are 16 arrangements, and one of them has all four on the left, so W = 1/16. Each particle adds a factor of ½ to W and adds ln ½, about −0.693, to ln W: the chances multiply and their logarithms add. Two on each side comes up in 6 of the 16 arrangements, more than any other split, which is why particles left to themselves are found spread out. At a hundred particles W is about 8 × 10⁻³¹. For a gram-molecule W cannot be written as a decimal at all, yet the entropy change is an ordinary number, R ln ½, about −5.76 joules per kelvin. Squeezing into nine tenths of the box lowers the entropy less, but for a gram-molecule it is still a state that is possible but never seen.

Where this lesson stops

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.

W = f^{n}

\Delta S_{\mathrm{gas}} = k_B\,\ln\left(W\right) = n\,k_B\,\ln\left(f\right)

The probability is f to the n; the entropy difference is Boltzmann's constant times its logarithm, or n times Boltzmann's constant times the logarithm of f.

In the paper's letters this is $S-S_0=R\frac{n}{N}\lg\frac{v}{v_0}$ . Here n/N is the number of gram-molecules of points, so R times it is the gas constant for that amount.
A footnote draws the gas law from it: the work $p\,dv$ equals $T\,dS=R\frac{n}{N}\frac{dv}{v}$ , so $pv=R\frac{n}{N}T$ , the law of Boyle and Gay-Lussac and the same law for osmotic pressure. The paper remarks that no assumption about how the molecules move was needed.
Independence is doing the work. If the four points were locked to one shared position, the chance of finding all of them in half the volume would be 1/2, not 1/16. Four labels do not by themselves give four independent factors.
The sign: concentrating the points into a smaller part (f less than 1) makes ln f negative, so the concentrated state has the lower entropy. Reversing the comparison reverses the sign.

A tool for this step

Probability and independence

When you square a sum of random steps, why do the mixed terms average to zero?

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?

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.

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:

\left(\Delta_{1} + \Delta_{2}\right)^{2} = \Delta_{1}^{2} + 2\,\Delta_{1}\,\Delta_{2} + \Delta_{2}^{2}

Delta one plus delta two, squared, equals delta one squared plus two delta one delta two plus delta two squared.

\langle \Delta_1 \Delta_2 \rangle = \langle \Delta_1 \rangle \langle \Delta_2 \rangle = 0

The average of delta one times delta two equals the average of delta one times the average of delta two, which is zero.

Worked example: What changes when the second step copies the first

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.

Where this lesson stops

Before dropping a mixed term, ask two things: does learning one step change the chances for the next, and does each step average zero?

Explore the equations in this step

Read each operation, check its units and assumptions, or open the mathematical step behind it. These are modern teaching equations, not the equations as printed.

Explore the equation · Modern model notation

The count comes down as a coefficient

\Delta S_{\mathrm{gas}} = k_B\,\ln\left(W\right) = n\,k_B\,\ln\left(f\right)

is times the logarithm of , which is times times the logarithm of .

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Explain this equation

Each term and operation, in words

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Read the equation aloud in words

The entropy change is k B times the logarithm of W, which is n times k B times the logarithm of f.

Boltzmann's principle turns the probability into an entropy difference. The logarithm of f to the n is n times the logarithm of f, so the number of independent points becomes the coefficient, and that coefficient is what section 6 compares with the radiation's entropy.

Model assumptions and every term’s meaning

n points move independently of one another, each equally likely to be anywhere in the volume V₀.
The chosen part V is a fraction f = V/V₀ of that volume.
Boltzmann's principle, as the paper states it: an entropy difference is Boltzmann's constant times the logarithm of a relative probability.
These are modern teaching equations in SI notation, not a transcription of the printed paper. The paper writes R/N for Boltzmann's constant.

Boltzmann's constant times ln W: Boltzmann's principle, the name the paper gives it. Read the prerequisite
Logarithm of the fraction: Negative when f is less than one, so confining the points lowers their entropy. Read the prerequisite
Logarithm of the probability: Entropy differences go as the logarithm of relative probability. Read the prerequisite
Boltzmann's principle: The entropy difference from the probability of the configuration. Read the prerequisite
n times Boltzmann's constant times ln f: The logarithm of f to the n is n times the logarithm of f. Read the prerequisite
What it asserts: The entropy change is proportional to the number of independent points. Read the prerequisite
Boltzmann's constant: Turns the logarithm of a probability into an entropy; R over N in the paper. Read the prerequisite
Entropy change of the points: The entropy with the points confined minus the entropy with them free: a difference, not an absolute entropy. Read the prerequisite
Number of independent points: How many points move independently of one another. Read the prerequisite
Fraction of the volume: The share of the volume the points must all be in, V over V₀; between 0 and 1. Read the prerequisite
Probability of the configuration: The chance that, at one instant, all n points are in the chosen part of the volume. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

Multiply one chance per independent point

W = f^{n}

that all points are inside of the volume is f multiplied by itself n times.

Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.

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Each term and operation, in words

No live values

This equation states a relation. No instrument on this page computes its terms.

Read the equation aloud in words

The probability W is f to the power n.

Each of n independent points is inside the fraction f of the volume with probability f, so the chance that all of them are inside at once is f multiplied by itself n times. Independence does the work: four labels on one shared position would give f, not f to the fourth.

Model assumptions and every term’s meaning

n points move independently of one another, each equally likely to be anywhere in the volume V₀.
The chosen part V is a fraction f = V/V₀ of that volume.
These are modern teaching equations in SI notation, not a transcription of the printed paper.

f multiplied by itself n times: One factor of f for each point. Only independent chances multiply like this. Read the prerequisite
What it asserts: The chance that every point is inside at once is the product of their separate chances. Read the prerequisite
Number of independent points: How many points move independently of one another. Read the prerequisite
Fraction of the volume: The share of the volume the points must all be in, V over V₀; between 0 and 1. Read the prerequisite
Probability of the configuration: The chance that, at one instant, all n points are in the chosen part of the volume. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Modern qualifications

This is Boltzmann’s principle read backwards. Einstein computes how the entropy of dilute radiation changes with volume, compares it with the entropy of $n$ independent points, $k\ln W$ with $W = f^n$ , and concludes that such radiation behaves thermodynamically as if it consisted of $n$ independent energy quanta. For large $n$ , $f^n$ is astronomically small, so the laboratory reports its logarithm.

Assumptions and limits: Independence supplies the exponent

Assumed here

Each point is uniformly distributed over the initial volume.
The point positions are statistically independent for the independent model.
Entropy differences are related to logarithms of relative configuration probabilities.

What this does not establish

The locked-position example is an authored mathematical counterexample, not an established alternative theory of radiation.
Spontaneous concentration probability and constrained-state entropy are related but are not the same observable.

What is getting in the way?

An unfamiliar word or symbol
f is a fraction of the volume: the chosen part divided by the whole, one half when the part is half of it. n is the number of points. W is a probability, the chance that at one moment all n points are inside the chosen part, and ln is the natural logarithm.
Open the explanation that addresses this
An algebraic move
Multiplying n equal factors f is written fⁿ, and the logarithm turns that power into a product: ln fⁿ = n ln f. So k_B ln W becomes n k_B ln f, the number of points times the change for one. Because f is less than 1, ln f is negative, and so is the entropy change for crowding the points into the smaller part.
Open the explanation that addresses this
The physical reason for a step
In the model each point moves over the whole volume without regard to the others, as the molecules of a dilute gas do. So the chance that any one point is in the chosen part is f, whatever the other points are doing. That independence is a premise of the model; the counting does not prove it.
Open the explanation that addresses this
The connection to the picture
Split the volume into the chosen part and the rest, and take one snapshot. One point is in the part a fraction f of the time. For two, the part must catch the first and also the second: f of the time, and then f of those times, which is f × f. Each further point multiplies by f again.
Open the explanation that addresses this
The purpose of the calculation
Section 6 uses the result in reverse. The entropy of dilute monochromatic radiation changes with volume the way n k_B ln f does for n independent points, so in that respect such radiation behaves as if it consisted of n independent energy quanta. The exponent is what carries the comparison, which is why the passage says it comes from independence.
Open the explanation that addresses this
Too much at once
One example: four points, half the volume. One point is in the left half with chance 1/2. Two are, with chance 1/2 × 1/2 = 1/4. All four are, with chance 1/16. If the four were locked together and moved as one, the chance would stay 1/2. That difference is the whole passage.
Open the explanation that addresses this

Source context: German source · English · Interlinear gloss · Facsimile

References

A. Einstein, Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt. Annalen der Physik (4), 17, 132–148 (1905).

A. Einstein, On the motion of particles suspended in liquids at rest required by the molecular-kinetic theory of heat. Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

A. Einstein, Zur Elektrodynamik bewegter Körper. Annalen der Physik (4), 17, 891–921 (1905).

A. Einstein, Does the inertia of a body depend upon its energy content? 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 translation or its pinned facsimile.

R. A. Millikan, A Direct Photoelectric Determination of Planck's "h". Physical Review (2), 7 (3), 355–388 (1916).

26 foundation readings sit behind this argument.

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