§4 · dilute radiation and the volume law · Light quanta: from entropy to an energy scale

§4 · dilute radiation and the volume law

The constant cannot simply be dropped

Why does the zero-radiation boundary condition matter?

\rho_\nu=A\nu^3\exp(-B\nu/T)

Wien spectral energy density is A times frequency cubed times the exponential of minus B frequency over temperature.

Divide by $A\nu^3$ and take the natural logarithm. Its argument is dimensionless. Solving for inverse temperature turns the entropy derivative from §3 into an explicit function of density.

\frac{1}{T}=-\frac{1}{B\nu}\ln\frac{\rho_\nu}{A\nu^3}

Inverse temperature is minus one over B frequency times the natural logarithm of density over A frequency cubed.

s_\nu=-\frac{\rho_\nu}{B\nu}\left[\ln\frac{\rho_\nu}{A\nu^3}-1\right]+C(\nu)

The integrated entropy density is minus density over B frequency times the bracketed logarithm minus one, plus a frequency-dependent constant.

As positive density approaches zero, $\rho\ln\rho$ tends to zero. The remaining limit is $C(\nu)$ . The boundary condition $s_\nu \to 0$ therefore sets $C(\nu) = 0$ . Without this condition, the volume comparison in the next step has an extra term.

Return to the logarithm, its dimensionless argument, and the zero-density limit. (included in this file)

Letters: the paper writes Wien's law as $\rho=\alpha\nu^3e^{-\beta\nu/T}$ , with the α and β of §2 (A and B here), and writes lg for the natural logarithm (ln here). The paper notes that observations show this law is not exactly valid, but that for large values of ν/T it was confirmed very completely by experiment. It bases its calculation on the law and keeps in mind that the results hold only within certain limits.

\rho_\nu=A\nu^3\exp(-B\nu/T)

Wien spectral energy density is A times frequency cubed times the exponential of minus B frequency over temperature.

Fix ν. Divide both sides by $A\nu^3$ : $\frac{\rho_\nu}{A\nu^3}=\exp(-B\nu/T)$ . The left side has no units, since A carries the units that make $A\nu^3$ a spectral density.
Take the natural logarithm of both sides: $\ln\frac{\rho_\nu}{A\nu^3}=-\frac{B\nu}{T}$ .
Divide both sides by $-B\nu$ to leave 1/T alone. This is the paper's first result in §4, printed with lg.

\frac{1}{T}=-\frac{1}{B\nu}\ln\frac{\rho_\nu}{A\nu^3}

Inverse temperature is minus one over B frequency times the natural logarithm of density over A frequency cubed.

Check the sign. A density that obeys Wien's law is smaller than $A\nu^3$ , so the ratio is below 1 and its logarithm is negative; the minus sign in front makes 1/T positive, as it must be.
§3 found that the entropy derivative is 1/T. Substitute: $\frac{\partial s_\nu}{\partial\rho_\nu}=-\frac{1}{B\nu}\ln\frac{\rho_\nu}{A\nu^3}$ .
Integrate with respect to $\rho_\nu$ at fixed ν. Write $a=A\nu^3$ for short. The antiderivative of $\ln(\rho/a)$ is $\rho\ln(\rho/a)-\rho$ ; differentiating it gives back $\ln(\rho/a)+1-1$ .
Multiply by $-1/(B\nu)$ and add a constant of integration. Because ν was held fixed, the constant may depend on ν, so it is written C(ν).

s_\nu=-\frac{\rho_\nu}{B\nu}\left[\ln\frac{\rho_\nu}{A\nu^3}-1\right]+C(\nu)

The integrated entropy density is minus density over B frequency times the bracketed logarithm minus one, plus a frequency-dependent constant.

Now the boundary condition. As $\rho_\nu$ falls toward zero, $\rho\ln\rho$ also falls toward zero, because the logarithm grows more slowly than ρ shrinks, and so does ρ itself. What remains in the limit is C(ν).
Worked limit, in units of a: at $\rho/a=10^{-3}$ , $\rho\ln(\rho/a)$ is -0.0069; at $\rho/a=10^{-6}$ it is $-1.38\times10^{-5}$ . The density-dependent terms shrink toward zero.
The paper requires the entropy to vanish when there is no radiation (φ = 0 at ρ = 0, end of §3). So C(ν) = 0, and the paper's printed result is the bracket without any constant: $\varphi(\rho,\nu)=-\frac{\rho}{\beta\nu}\left\{\lg\frac{\rho}{\alpha\nu^3}-1\right\}$ .
The constant was fixed by a stated condition; it was not dropped for looking small. Kept, it would add a term proportional to the volume to the comparison in the next step.

Return to the logarithm, its dimensionless argument, and the zero-density limit. (included in this file)

Integrate for the entropy density

Modern teaching equation, not the equation as printed.

s_\nu = -\frac{u_\nu}{B\,\nu}\,\left(\ln\left(\frac{u_\nu}{A\,\nu^{3}}\right) - 1\right) + C\left(\nu\right)

The integrated entropy density is minus density over B frequency times the bracketed logarithm minus one, plus a frequency-dependent constant.

Integrating 1/T with respect to the density gives the entropy density up to a constant C(nu) that may depend on frequency. The constant cannot simply be dropped; the next step fixes it by a boundary condition.

Wien's law solved for the temperature

Modern teaching equation, not the equation as printed.

\frac{1}{T} = -\frac{1}{B\,\nu}\,\ln\left(\frac{u_\nu}{A\,\nu^{3}}\right)

One over the temperature equals minus one over B times the frequency, times the natural logarithm of the density over A times the frequency cubed.

Taking the logarithm of Wien's law and solving for one over the temperature expresses the temperature through the density. With one over T equal to the entropy's slope, this is what gets integrated.

Wien's spectrum

Modern teaching equation, not the equation as printed.

u_\nu = A\,\nu^{3}\,\exp\left(-\frac{B\,\nu}{T}\right)

The spectral energy density equals A times the frequency cubed times the exponential of minus B times the frequency over the temperature.

Wien's law for the energy density on the frequency axis. It fits the measured spectrum where B times the frequency is large compared with the temperature, and fails where it is small.

Assumptions

The narrow spectral region obeys the Wien approximation.
At zero radiation density, radiation entropy density vanishes.

Limits

A positive inferred temperature alone is not a guarantee that the Wien approximation is accurate.
The spectral constants are not initially interpreted as particle energies.

Compare two states, not a compression movie

What is held fixed when volume changes?

Within a narrow band, write $E = V\,\Delta\nu\,\rho_\nu$ and $S = V\,\Delta\nu\,s_\nu$ . Substitute $\rho_\nu = E/(V\,\Delta\nu)$ into the integrated entropy density. Comparing $V$ with $V_0$ at the same $E$ and band removes terms independent of volume.

S(V)-S(V_0)=\frac{E}{B\nu}\ln\frac{V}{V_0}

The entropy difference is E over B frequency times the natural logarithm of the volume ratio.

$E/(B\nu)$ has entropy units. Only after identifying $B = h/k_B$ can it be written $k_B$ times the dimensionless coefficient $E/(h\nu)$ . This distinction matters when interpreting a displayed coefficient as a count.

\Delta S_C=\Delta\nu\,C(\nu)(V-V_0)

An unfixed entropy-density constant would add bandwidth times C of frequency times the volume difference.

Doubling the accessible volume adds $(E/(B\nu))\ln 2$ under the stated constraints. Doubling it again adds the same amount. The radiation-entropy laboratory compares those constrained states and exposes the otherwise hidden constant.

Return to keeping energy and the narrow band fixed. (included in this file)

Letters: the paper's radiation has energy E in the narrow band between ν and ν + dν and fills a volume v; this explanation writes the band width Δν and the volume V. $S_0$ is the entropy when the same radiation fills the volume $v_0$ , written $V_0$ here.

In the band, the energy is volume times band width times density: $E=V\Delta\nu\,\rho_\nu$ , so $\rho_\nu=E/(V\Delta\nu)$ .
The entropy is volume times band width times entropy density: $S=V\Delta\nu\,s_\nu$ .
Insert the entropy density of the previous step, keeping C(ν) for now: $S=-\frac{E}{B\nu}\left[\ln\frac{E}{VA\nu^3\Delta\nu}-1\right]+V\Delta\nu\,C(\nu)$ . The factor $V\Delta\nu$ times $\rho_\nu/(B\nu)$ has become $E/(B\nu)$ .
Write the same expression with $V_0$ in place of V, at the same E, the same ν and the same band.
Subtract. Every term without V cancels. Inside the logarithms, $\ln\frac{E}{VA\nu^3\Delta\nu}-\ln\frac{E}{V_0A\nu^3\Delta\nu}=\ln\frac{V_0}{V}$ , and the minus sign in front turns it into $\ln\frac{V}{V_0}$ .

S(V)-S(V_0)=\frac{E}{B\nu}\ln\frac{V}{V_0}

The entropy difference is E over B frequency times the natural logarithm of the volume ratio.

The C(ν) term does not cancel. It leaves the piece below, proportional to the difference of the volumes. That is why the constant had to be fixed, and the zero-radiation condition fixed it at zero.

\Delta S_C=\Delta\nu\,C(\nu)(V-V_0)

An unfixed entropy-density constant would add bandwidth times C of frequency times the volume difference.

Worked numbers: halve the volume, $V/V_0=1/2$ , and the entropy falls by $(E/B\nu)\ln2$ , about 0.693 times $E/(B\nu)$ . Halve it again and it falls by the same amount again. Only the ratio of the volumes matters, not their size.
Units: B is in seconds times kelvin, so Bν is in kelvin and $E/(B\nu)$ is energy per kelvin, the unit of entropy. Only after reading B as $h/k_B$ , a later identification, does $E/(B\nu)$ become $k_B$ times the pure number $E/(h\nu)$ .
The paper notes that the entropy of monochromatic radiation of low enough density varies with volume by the same law as the entropy of an ideal gas or a dilute solution. §5 asks where that law comes from for a gas.
Check the conditions at both ends: the same energy, the same narrow band, and a density low enough for Wien's law in both volumes. This compares two states. A mirror that moves while the radiation is inside changes frequencies and energy, which is a different comparison.

Return to keeping energy and the narrow band fixed. (included in this file)

What an unfixed constant would add

Modern teaching equation, not the equation as printed.

\Delta S_C = \Delta\nu\,C\left(\nu\right)\,\left(V - V_{0}\right)

An unfixed entropy-density constant would add bandwidth times C of frequency times the volume difference.

If the entropy density's integration constant C(nu) were not set to zero, the comparison of two volumes would pick up this extra term, proportional to the volume difference rather than to the logarithm of the ratio.

The entropy depends on the volume through a logarithm

Modern teaching equation, not the equation as printed.

S\left(V\right) - S\left(V_{0}\right) = \frac{E}{B\,\nu}\,\ln\left(\frac{V}{V_{0}}\right)

The entropy difference is E over B frequency times the natural logarithm of the volume ratio.

At the same energy and band, the radiation's entropy in volume V exceeds that in V₀ by E over Bν times the logarithm of their ratio. Doubling the volume adds (E/Bν) ln 2, and doubling it again adds the same.

Assumptions

Energy E, central frequency ν, and bandwidth Δν are fixed in both states.
The entropy-density constant has been fixed by the zero-radiation condition.
Both states satisfy the dilute Wien-regime and narrow-band assumptions.

Limits

A moving mirror generally changes frequency and energy, so it is not this held-fixed comparison.
The result is not valid merely because one of the two endpoint states is dilute.

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

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

Logarithms: turning products into sums

Why does entropy need a logarithm?

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.

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:

\ln(AB) = \ln A + \ln B, \qquad \ln\left(f^{\,n}\right) = n \ln f

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.

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:

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

S minus S nought equals R over N times the natural log of W.

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.

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.

Worked example: Ten molecules found in half the volume

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:

W = \left(\frac{v}{v_0}\right)^{n}

W equals v over v nought, to the power n.

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.

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.

Where this lesson stops

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.

This lesson builds on

exponentials

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-03

This laboratory is not included in this file. The chapter's authored foundation examples remain available above. Open lq-03 online

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).