§1 · classical energy allocation · Light quanta: from entropy to an energy scale

§1 · classical energy allocation

A finite window cannot cure an infinite total

What happens if every radiation mode receives classical thermal energy?

Counting cavity modes and giving each mode the classical mean energy $k_B T$ produces a spectral energy density $\rho_\nu$ . A density must be multiplied by a frequency width to give energy per unit volume.

\rho_\nu=\frac{8\pi\nu^2}{c^3}k_BT

The classical energy density per unit frequency is eight pi frequency squared times thermal energy, divided by light speed cubed.

U(\nu_c)=\int_0^{\nu_c}\rho_\nu\,d\nu=\frac{8\pi k_BT}{3c^3}\nu_c^3

The energy density below the cutoff frequency is eight pi times Boltzmann's constant times temperature, divided by three times the speed of light cubed, times the cutoff frequency cubed.

At fixed positive temperature, doubling the cutoff multiplies this finite-range total by eight. Extending the model to arbitrarily high frequencies produces an unbounded total. The resonator laboratory keeps that failure distinct from the finite portion displayed on screen.

Return to integrating a spectral density over a stated frequency range. (included in this file)

Return to the energy density per unit of frequency and its units. (included in this file)

Symbols first, in the paper's letters. T is the absolute temperature. R is the gas constant and N the number of real molecules in one gram-equivalent, so R/N is the gas constant for a single molecule; this explanation also writes it $k_B$ , a later name. L in the paper is the speed of light, written c below. ν is a frequency, and $\rho_\nu\,d\nu$ is the energy per unit volume carried by the radiation whose frequency lies between ν and ν + dν.

Start from the kinetic theory of gases. A resonator, an electron bound to a point in space, exchanges energy with the gas molecules by collisions. In equilibrium the mean energy of one of its straight-line vibrations comes out as $\frac{R}{N}T$ , which is $k_BT$ (p. 134).
Planck had related such a resonator to the radiation around it. A resonator of natural frequency ν in equilibrium with radiation has, per vibration component, the mean energy $\frac{L^3}{8\pi\nu^2}\rho_\nu$ (p. 135).
If the radiation at frequency ν is neither steadily gaining nor steadily losing energy, both conditions hold at once. Set the two mean energies equal: $\frac{R}{N}T=\frac{L^3}{8\pi\nu^2}\rho_\nu$ .
Multiply both sides by $8\pi\nu^2/L^3$ to leave the density alone: $\rho_\nu=\frac{R}{N}\frac{8\pi\nu^2}{L^3}T$ (p. 136). In the explanatory letters that is the first formula below.

\rho_\nu=\frac{8\pi\nu^2}{c^3}k_BT

The classical energy density per unit frequency is eight pi frequency squared times thermal energy, divided by light speed cubed.

Check the units. $k_BT$ is an energy. $\nu^2/c^3$ has units of $\mathrm{s}^{-2}$ divided by $\mathrm{cm}^3\mathrm{s}^{-3}$ , which is seconds per cubic centimetre. The product is energy times seconds per volume: energy per unit volume and per unit frequency, which is what a spectral density must be.
A density becomes an energy only after it is multiplied by a frequency width. The energy per unit volume in all frequencies from 0 up to a cutoff $\nu_c$ is the integral of $\rho_\nu$ over that range.
Everything except $\nu^2$ stays constant during the integration, so it comes outside the integral. The integral of $\nu^2$ from 0 to $\nu_c$ is $\nu_c^3/3$ , which gives the second formula.

U(\nu_c)=\int_0^{\nu_c}\rho_\nu\,d\nu=\frac{8\pi k_BT}{3c^3}\nu_c^3

The energy density below the cutoff frequency is eight pi times Boltzmann's constant times temperature, divided by three times the speed of light cubed, times the cutoff frequency cubed.

Worked comparison at one fixed temperature: a cutoff twice as high gives $2^3=8$ times the energy, and a cutoff ten times as high gives $10^3=1000$ times. The total keeps growing as the cutoff grows.
Remove the cutoff, as the paper does at the end of §1: the integral of $\nu^2\,d\nu$ from 0 to infinity is infinite, so the radiation energy in the space would be infinite.
The paper draws two conclusions from this relation: it disagrees with experience, and in this picture there can be no definite distribution of energy between the ether and matter.
Keep names with their dates. A radiation law of this $\nu^2T$ form had been proposed by Rayleigh in June 1900; Jeans's correction of its coefficient came in 1905, parallel to this paper rather than behind it. The phrase 'ultraviolet catastrophe' for the divergence is Ehrenfest's, from 1911, and the paper does not use it.
A plotted curve that stops at some frequency shows where the calculation stopped, not a property of light. The resonator laboratory reports the finite part it draws, and refuses to give the unbounded total as a number, rather than showing it as zero.

Return to integrating a spectral density over a stated frequency range. (included in this file)

The classical allocation of energy to the spectrum

Modern teaching equation, not the equation as printed.

u_\nu = \frac{8\,\pi\,\nu^{2}}{c^{3}}\,k_B\,T

The spectral energy density equals eight pi times the frequency squared over the speed of light cubed, times k B times the temperature.

Give every mode of the radiation the average energy that classical theory assigns, Boltzmann's constant times the temperature, and count the modes per unit volume per unit frequency: the density grows as the frequency squared without limit.

Add up the density below a cutoff

Modern teaching equation, not the equation as printed.

U\left(\nu_{c}\right) = \int_{0}^{\nu_{c}} u_\nu\,\mathrm{d}\nu

The energy density below cutoff frequency nu c is the integral of the spectral density from zero to nu c.

A density per unit frequency becomes an energy per unit volume only when it is added up over a range of frequencies. Here the range runs from zero to a cutoff.

The total grows as the cutoff cubed

Modern teaching equation, not the equation as printed.

U\left(\nu_{c}\right) = \frac{8\,\pi\,k_B\,T}{3\,c^{3}}\,\nu_{c}^{3}

The energy density below the cutoff frequency is eight pi times Boltzmann's constant times the temperature, divided by three times the speed of light cubed, times the cutoff frequency cubed.

Integrating the classical density, which grows as the frequency squared, gives a total that grows as the cutoff cubed. Doubling the cutoff multiplies it by eight, and with no cutoff it has no limit.

Assumptions

Classical equilibrium assigns mean energy k_B T to each radiation mode.
The number of modes per unit volume and per unit frequency grows as the square of the frequency.

Limits

The allocation is a classical model, not the spectrum measured across all frequencies.
A finite cutoff is where the calculation stops, not a property of light.

Two assumptions under the equilibrium of section 1

What does the equilibrium of §1 assume about the electrons and about the radiation?

§1 puts gas molecules, free electrons and bound electrons in a box with mirror walls, and lets the free electrons collide like the molecules of a gas. A footnote says what that amounts to: at equilibrium, gas molecules and electrons have equal mean kinetic energies. Drude, Einstein notes, used the same assumption to derive the ratio of heat conduction to electrical conduction in metals.

That equality is what fixes the mean energy of each straight-line vibration of a resonator. A resonator whose average energy were higher or lower would, through the collisions, keep giving energy to the gas or taking it, so equilibrium needs:

E=\frac{R}{N}T

The mean energy E of one resonator vibration equals R over N times the temperature T.

The second assumption belongs to Planck's relation between a resonator and the radiation around it, which §1 uses. Planck derived it for radiation that is 'as disordered as possible', and a footnote defines the phrase. At one point of the space, expand one component Z of the electric field over a long time T in a Fourier series of sine waves with amplitudes A and phases α. Repeat the expansion from starting moments chosen at random: the amplitudes and phases come out differently each time, with a statistical probability for each combination.

Z=\sum_{\nu=1}^{\infty}A_\nu\sin\left(2\pi\nu\frac{t}{T}+\alpha_\nu\right)

Z equals the sum over nu from one to infinity of A nu times the sine of two pi nu t over T plus alpha nu.

The radiation is as disordered as possible when that probability splits into separate factors, one for each amplitude and one for each phase:

f(A_1,A_2,\ldots,\alpha_1,\alpha_2,\ldots)=F_1(A_1)F_2(A_2)\ldots f_1(\alpha_1)f_2(\alpha_2)\ldots

The joint probability density of all the amplitudes and phases equals a product of separate factors, one for each amplitude and one for each phase.

Then the value of any one amplitude or phase tells nothing about the others. Einstein adds that the radiation in his box comes closer to this the more each pair of an amplitude and its phase depends on emission and absorption by its own particular group of resonators. Neither assumption is derived in §1. The footnotes state them, and the equilibrium, with the unbounded total energy it leads to at the end of §1, rests on both.

Return to the independence the disorder condition asks for. (included in this file)

Letters: E is the mean energy of one straight-line vibration of a resonator (the paper writes it with a bar), R the gas constant, N the number of real molecules in a gram-equivalent and T the absolute temperature. In the footnote, Z is one component of the electric field at a point, T there is a long stretch of time, A the amplitudes and α the phases of the Fourier series.

The first assumption: in the box, free electrons collide with gas molecules the way gas molecules collide with each other.
Its content, from the footnote: at a common temperature the mean kinetic energy of an electron equals the mean kinetic energy of a gas molecule. Drude had made the same assumption to derive the ratio of the thermal to the electrical conductivity of metals.
A resonator's electron shares in that equality. Kinetic theory gives a gas molecule the mean kinetic energy of its forward motion as (3/2)(R/N)T, so the resonator electron has that mean kinetic energy too.
Split the electron's motion into three vibrations along lines at right angles. Each gets a third of the kinetic energy: (1/3) × (3/2)(R/N)T = (1/2)(R/N)T.
In a vibration the time averages of kinetic and potential energy are equal, so each vibration's mean energy is twice its kinetic part: 2 × (1/2)(R/N)T = (R/N)T. That is (R/N)T ÷ (3/2)(R/N)T = 2/3 of the kinetic energy of a free monatomic gas molecule, as the paper says.
If a resonator had more than that on average, collisions would drain the excess into the gas; if less, collisions would feed it. Equilibrium therefore needs exactly:

E=\frac{R}{N}T

The mean energy E of one resonator vibration equals R over N times the temperature T.

Worked value, with today's value of R/N, $1.38\times10^{-23}$ joule per kelvin, which the paper does not use: at 300 kelvin, $\overline{E}=1.38\times10^{-23}\times300\approx4.14\times10^{-21}$ joule. The frequency does not enter, so a resonator of any frequency has the same mean energy.
The second assumption comes with Planck's relation, which §1 borrows to connect a resonator to the radiation. Planck derived it for radiation that is as disordered as possible, and the footnote says what that means.
Pick a point in the box and one component Z of the electric field there. Over a time T much longer than every period of vibration that matters, write Z as a sum of sine waves. (Here T is that stretch of time, not the temperature.)

Z=\sum_{\nu=1}^{\infty}A_\nu\sin\left(2\pi\nu\frac{t}{T}+\alpha_\nu\right)

Z equals the sum over nu from one to infinity of A nu times the sine of two pi nu t over T plus alpha nu.

Each term has an amplitude $A_\nu$ , zero or more, and a phase $\alpha_\nu$ between 0 and 2π. Start the stretch of time at a different moment, chosen at random, and the amplitudes and phases come out different.
Doing that many times gives a statistical probability for each combination of amplitudes and phases, written as a density f of all of them together.
The disorder condition: f is a product of separate factors, one for each amplitude and one for each phase.

f(A_1,A_2,\ldots,\alpha_1,\alpha_2,\ldots)=F_1(A_1)F_2(A_2)\ldots f_1(\alpha_1)f_2(\alpha_2)\ldots

The joint probability density of all the amplitudes and phases equals a product of separate factors, one for each amplitude and one for each phase.

What the product means: learning one amplitude or phase changes nothing about the chances of any other. Worked case: if the first amplitude lies above some value in half of the expansions, and the first phase lies in the first half of its circle in half of them, then disorder gives both at once in 1/2 × 1/2 = 1/4 of the expansions.
When the condition holds only roughly: Einstein says it holds the better, the more each pair of an amplitude and its phase comes from the emission and absorption of its own particular group of resonators.
Neither assumption is proved in §1. They are the ground under the equilibrium from which §1 draws its conclusion: that on this picture the radiation energy grows without limit as higher frequencies are allowed.

Return to the mean energy each vibration shares at equilibrium. (included in this file)

Assumptions

Free electrons collide with the gas molecules as gas molecules collide with each other.
At equilibrium, gas molecules and electrons have equal mean kinetic energies (the footnote to the setup of §1).
Planck's relation between a resonator and the radiation around it holds for radiation that is as disordered as possible (the footnote that defines it).

Limits

The passage explains what the footnotes state; it does not test whether real radiation in a cavity meets the disorder condition.
Einstein cites Drude's theory of metals in support; the passage does not assess that theory.

Foundations included in this file

Fractions and ratios

What does dividing one quantity by another tell you?

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.

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.

Worked example: Fourteen metres in ten seconds

A steady walk covers 14 metres in 10 seconds.
Divide the distance by the time: 14 divided by 10 is 1.4, and the unit divides with it, so the rate is 1.4 metres per second.
After 60 seconds the same walk has covered 84 metres, and 84 divided by 60 is 1.4 again. The walk is steady, so the ratio does not depend on how long it is watched.
Metres and seconds are different units, so neither cancels and what comes out is a new quantity that keeps a unit of its own.
The drawing below is the other case: 2 particles out of 8 has the same unit above and below, they cancel, and 0.25 is left as a plain number.

Where this lesson stops

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.

This lesson builds on

Adding and averaging

Return to the chapter

A sign records direction

How can movement add up to zero?

On the ruler below, the starting point is zero. 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.

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, which is the lower pair of arrows in the drawing.

Worked example: Two walkers, opposite directions, average zero

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.

Where this lesson stops

The sign names a direction relative to a chosen axis; it does not mean a negative distance.

This lesson builds on

Adding and averaging

Return to the chapter

Powers of ten and physical units

How do you write a very small length without counting zeros?

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).

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.

Worked example: Squaring six micrometres

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.

Where this lesson stops

A measurement is a number and a unit together. Squaring it squares both: (6 × 10⁻⁶ m)² is 3.6 × 10⁻¹¹ m².

This lesson builds on

Fractions and ratios

Return to the chapter

Adding and averaging

What does an average keep, and what does it lose?

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.

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.

Worked example: Three mornings with the same average

One morning the four jars hold 3, 1, 1 and 3 marbles. Add them: 8. Divide by the four jars: an average of 2.
Another morning they hold 8, 0, 0 and 0. Add them: 8 again. Divide by four: 2 again, and now three jars are empty.
A third morning every jar holds 2, which is the average itself. Add them: 8. Divide by four: 2.
Write any of those three lists out twice over. The total doubles to 16 and the count doubles to 8 jars, and 16 divided by 8 is still 2.
Four descriptions and one average. It keeps the total and the count, and the differences between the mornings are what it leaves behind.

Where this lesson stops

An average is the total shared equally among the observations, and two very different lists can have the same one.

Return to the chapter

Density is not probability

What does the height of a probability curve mean?

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.

P(a\le X\le b)=\int_a^b p(x)\,dx

The probability of the interval from a to b is the integral of the density over that interval.

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.

Worked example: A flat density across four micrometres

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.

Where this lesson stops

The height of the curve is a density, a probability per unit length. Only the area over an interval is a probability.

This lesson builds on

Adding and averaging

Return to the chapter

Adding continuously

How do you add up something that varies continuously?

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.

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:

\int_a^b p(x)\,dx = \lim_{\Delta x \to 0} \sum_i p(x_i)\,\Delta x

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.

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.

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:

\int_a^b u\frac{dv}{dx}\,dx = [uv]_a^b - \int_a^b v\frac{du}{dx}\,dx

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.

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.

Worked example: A rising density, added up in narrower strips

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.

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.

Where this lesson stops

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.

This lesson builds on

Density is not probability

Return to the chapter

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:

(\Delta_1 + \Delta_2)^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?

Return to the chapter

Quantities and units

Can a number of metres be compared with a number of seconds?

"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.

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.

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.

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.

Worked example: Checking that D comes out in square metres per second

$k_BT$ : 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.

Where this lesson stops

This lesson stops at checking units. Converting the papers' own units into SI units is a separate step, with its own conventions.

This lesson builds on

Powers of ten and physical units
Fractions and ratios

Return to the chapter

Temperature and thermal energy

What does a temperature say about the jostling of molecules and grains?

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.

For any particle in equilibrium with its surroundings, the average energy of motion along each direction is the same:

\left\langle \frac{1}{2} m v_x^2 \right\rangle = \frac{1}{2} k_B T

The average of one half m v x squared equals one half k B T.

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.

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.

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.

Worked example: A nitrogen molecule and a half-micrometre grain at 20 °C

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.

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.

Where this lesson stops

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.

This lesson builds on

Energy of motion and inertia
Powers of ten and physical units
bridge-squaring-square-roots

Return to the chapter

Energy of motion and inertia

If a body moves at the same speed but carries less energy of motion, what has changed?

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.

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.

K=\frac{1}{2}mv^2

In Newtonian mechanics, kinetic energy is one half times inertial mass times speed squared.

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.

Worked example: Two bodies at the same speed, different masses

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.

K_0-K_1=\frac{1}{2}(m_0-m_1)v^2

At the same low speed, the kinetic-energy drop equals one half times the mass decrease times the squared speed.

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.

Where this lesson stops

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.

This lesson builds on

bridge-squaring-square-roots
Fractions and ratios

Return to the chapter

Static laboratory examples

lq-02

This laboratory is not included in this file. The chapter's authored foundation examples remain available above. Open lq-02 online

Source 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, 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.