§3 · temperature and entropy · Light quanta: from entropy to an energy scale

§3 · temperature and entropy

A spectrum can determine an entropy derivative

How does equilibrium temperature constrain radiation entropy?

Let $\rho_\nu$ be energy per volume per frequency interval and $s_\nu$ the corresponding entropy density. At fixed volume, equilibrium maximizes entropy while keeping total energy fixed. Moving a small amount of energy from one spectral interval to another must give no first-order entropy gain at equilibrium.

Consequently the derivatives of entropy with respect to spectral energy density have the same value across the equilibrium spectrum. The thermodynamic identity dS/dE = 1/T, with other constraints fixed, identifies that common value.

\left(\frac{\partial s_\nu}{\partial\rho_\nu}\right)_\nu=\frac{1}{T}

At fixed frequency, the partial derivative of spectral entropy density with respect to spectral energy density is inverse temperature.

Einstein attributes this thermodynamic reasoning to Wien. The next step is to express 1/T as a function of density by using Wien’s spectral law, not by assuming particles of light in advance.

Return to recovering the entropy derivative from temperature. (included in this file)

Letters: the paper puts the radiation in a volume v, writes ρ for its density at each frequency ν, and writes φ(ρ, ν) for its entropy per unit volume and unit frequency. This explanation writes $\rho_\nu$ and $s_\nu$ for the same two quantities. The paper takes it as given that the observable properties of the radiation are fixed once ρ is known for all frequencies; a footnote calls this assumption arbitrary, to be kept until experiment forces it to be dropped.

Radiation of different frequencies can be separated without doing work and without supplying heat, so entropy and energy are sums over frequencies: $S=v\int_0^\infty\varphi(\rho,\nu)\,d\nu$ and $E=v\int_0^\infty\rho\,d\nu$ .
Black radiation is the equilibrium state: for the given energy, its entropy is as large as it can be. Take v = 1 and change the spectrum by a small amount $δρ$ at each frequency.
The energy must not change: $\int_0^\infty δρ\,d\nu=0$ . At a maximum, the entropy must not change to first order either: $\int_0^\infty\frac{\partial\varphi}{\partial\rho}δρ\,d\nu=0$ .
Both statements must hold for every such change. That is possible only if $\int_0^\infty\left(\frac{\partial\varphi}{\partial\rho}-\lambda\right)δρ\,d\nu=0$ with λ a number that does not depend on ν. So $\partial\varphi/\partial\rho$ has the same value at every frequency.
A two-band check of that step: move a small amount of energy q out of band 1 and into band 2, keeping the total fixed. The entropy changes by q times (the derivative in band 2 minus the derivative in band 1). If the two derivatives differed, choosing the sign of q would raise the entropy, so at the maximum they are equal.
Now warm black radiation of volume 1 by dT. Its entropy changes by $dS=\int_0^\infty\frac{\partial\varphi}{\partial\rho}d\rho\,d\nu$ . Because the derivative is the same at every frequency, it comes outside the integral, and what remains is the change of energy: $dS=\frac{\partial\varphi}{\partial\rho}dE$ .
dE is the heat supplied, and the warming is reversible, so thermodynamics also gives $dS=\frac{1}{T}dE$ .
Compare the two expressions for dS. The common derivative is the inverse temperature, which in the explanatory letters is the formula below.

\left(\frac{\partial s_\nu}{\partial\rho_\nu}\right)_\nu=\frac{1}{T}

At fixed frequency, the partial derivative of spectral entropy density with respect to spectral energy density is inverse temperature.

Read it in words: at fixed frequency, adding a little energy density raises the entropy density by that amount divided by the temperature.
So a known radiation law, which gives T as a function of ρ at each ν, gives $\partial\varphi/\partial\rho$ as a function of ρ. Integrating recovers φ, once one more condition fixes the constant of integration: the paper requires φ to vanish when ρ = 0.
The paper credits this reasoning to a celebrated work of W. Wien and includes it for completeness. It uses no picture of how radiation is produced or how it travels.

Return to recovering the entropy derivative from temperature. (included in this file)

Entropy per unit of energy, at one frequency

Modern teaching equation, not the equation as printed.

\left(\frac{\partial s_\nu}{\partial u_\nu}\right)_{\nu} = \frac{1}{T}

At fixed frequency, the partial derivative of the spectral entropy density with respect to the spectral energy density is one over the temperature.

At equilibrium, moving a little energy from one frequency interval to another gains no entropy to first order, so this derivative has one common value across the spectrum. The thermodynamic identity dS/dE = 1/T names that value.

Assumptions

Equilibrium thermodynamics holds, and energy and entropy add up over the spectrum.
Vary the energy distribution at fixed volume and fixed total energy.

Limits

This is a thermodynamic comparison, not a microscopic account of every exchange.
Integrating the derivative requires a boundary condition.

Foundations included in this file

Reading a graph

How does a curve show the relationship between two physical quantities?

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. The first drawing below puts time along the bottom and distance up the side, with a dot for each pair. That is a graph: each dot is one moment, read across for the time and up for the distance.

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.

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.

Worked example: Four particles that average to zero

Four particles end at −3, −1, +1 and +3 micrometres from where they started.
The second drawing puts those positions along the bottom, with 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.

Where this lesson stops

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.

This lesson builds on

A sign records direction

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

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

Rates of change and derivatives

How can a body have a speed at a single instant?

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.

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:

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.

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.

In symbols, with Δt for the short interval, the same recipe reads:

\frac{dx}{dt} = \lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t)}{\Delta t}

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.

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.

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:

\frac{\partial f}{\partial t} = D\,\frac{\partial^2 f}{\partial x^2}

The partial derivative of f with respect to t equals D times the second partial derivative of f with respect to x.

The lesson on partial derivatives, listed at the end of this page, reads that equation one piece at a time.

Worked example: Why the averages settle on 6 m/s

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:

3(t+\Delta t)^2 - 3t^2 = 6t\,\Delta t + 3(\Delta t)^2

Three times t plus delta t, squared, minus three t squared, equals six t delta t plus three delta t squared.

Divide that distance by the time Δt to get the average speed over the interval:

\frac{6t\,\Delta t + 3(\Delta t)^2}{\Delta t} = 6t + 3\,\Delta t

Six t delta t plus three delta t squared, divided by delta t, equals six t plus three delta t.

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.

Where this lesson stops

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.

This lesson builds on

Functions and graphs

Return to the chapter

Entropy and temperature

What does temperature have to do with entropy?

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:

dS=\frac{dE}{T},\qquad\left(\frac{\partial S}{\partial E}\right)_V=\frac{1}{T}

At fixed volume and the stated constraints, entropy change is energy change divided by absolute temperature.

Read the other way round, 1/T is how steeply a body's entropy climbs as its energy grows.

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.

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.

Return to the constraints on the entropy derivative. (included in this file)

Worked example: Three joules into a reservoir held at 300 kelvin

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.

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.

Where this lesson stops

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.

This lesson builds on

Partial derivatives and held-fixed quantities
integration

Return to the chapter

Functions and graphs

How does a graph record the way one quantity depends on another?

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.

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.

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.

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.

Worked example: Mean square displacement plotted against time

\langle x^2\rangle = 2Dt

The mean square displacement equals two times the diffusion coefficient times the time.

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.

Where this lesson stops

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.

This lesson builds on

Reading a graph

Return to the chapter

Partial derivatives and held-fixed quantities

When a quantity depends on two things, what does its rate of change mean?

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.

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:

\left(\frac{\partial f}{\partial x}\right)_t \qquad \left(\frac{\partial f}{\partial t}\right)_x

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.

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:

\left(\frac{\partial V}{\partial p}\right)_T \quad \text{versus} \quad \left(\frac{\partial V}{\partial p}\right)_S

The partial derivative of V with respect to p at fixed temperature T, versus the same derivative at fixed entropy S.

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:

\left(\frac{\partial \varphi}{\partial \rho}\right)_{\nu} = \frac{1}{T}

The partial derivative of phi with respect to rho, at fixed frequency nu, equals one over T.

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.

Worked example: The same diffusion profile, differentiated two ways

Take the spreading profile of the Brownian paper's §4, written for a single particle:

f(x,t) = \frac{1}{\sqrt{4\pi D t}}\,e^{-x^2/4Dt}

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.

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:

\left(\frac{\partial f}{\partial x}\right)_t = -\frac{x}{2Dt}\,f(x,t)

The partial derivative of f with respect to x at fixed t equals minus x over two D t, times f.

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:

\left(\frac{\partial f}{\partial t}\right)_x = \left(-\frac{1}{2t} + \frac{x^2}{4Dt^2}\right) f(x,t)

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.

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.

Where this lesson stops

A partial derivative is not defined until you have said which quantities are held fixed.

This lesson builds on

Rates of change and derivatives

Return to the chapter

Static laboratory examples

lq-04

This laboratory is not included in this file. The chapter's authored foundation examples remain available above. Open lq-04 online

Source references

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