§5 · velocity composition · Special relativity: from clock operations to electrodynamics

§5 · velocity composition

The denominator changes as well

Why do two ordinary-looking speeds not simply add?

u_x'=\frac{u_x-v}{1-vu_x/c^2},\qquad u_y'=\frac{u_y}{\gamma(1-vu_x/c^2)}

The transformed velocity along x is the velocity along x minus v, divided by one minus v times the velocity along x over c squared; the transverse velocity has the same denominator and an additional gamma.

An object moving at $w$ in the moving frame, in the same direction, moves in the original frame at $U = (v + w)/(1 + vw/c^2)$ . With $v = w = 0.6c$ , $U = 15c/17$ , about $0.882c$ , not the $1.2c$ that plain addition would give.

Put light in, $u_x = c$ , and the formula returns $c$ in every frame, as the second postulate requires. The velocity-composition laboratory shows the relativistic and the Galilean sums side by side, and prints the Galilean one even when it exceeds $c$ .

Return to why two ordinary speeds do not simply add. (included in this file)

Letters: in §5 a point moves in k with constant velocity components $w_ξ$ and $w_η$ , so that $ξ=w_ξ\tau$ , $η=w_η\tau$ , ζ = 0; the resulting speed in K is U, w is the speed in k, α the angle between the velocities v and w, and V the speed of light. The formula below goes the other way, from K to the moving frame, and writes $u_x$ and $u_y$ in K and $u'_x$ and $u'_y$ in the moving frame.

A velocity is a small displacement divided by the small time it takes. Both change between frames, so transform both before dividing.
Along the object's path, take the change of $x'=\gamma(x-vt)$ : $dx'=\gamma(dx-v\,dt)$ .
Take the change of $t'=\gamma(t-vx/c^2)$ along the same path: $dt'=\gamma(dt-v\,dx/c^2)$ .
Divide, and then divide numerator and denominator by dt: $u'_x=\frac{dx'}{dt'}=\frac{u_x-v}{1-vu_x/c^2}$ .
Across the motion $dy'=dy$ , but the time still changes, so $u'_y=\frac{dy}{\gamma(dt-v\,dx/c^2)}=\frac{u_y}{\gamma(1-vu_x/c^2)}$ .

u_x'=\frac{u_x-v}{1-vu_x/c^2},\qquad u_y'=\frac{u_y}{\gamma(1-vu_x/c^2)}

The paper's direction, from k to K, uses the inverse transformation and so carries plus signs. For the point above it prints $x=\frac{w_ξ+v}{1+vw_ξ/V^2}t$ and $y=\frac{\sqrt{1-(v/V)^2}}{1+vw_ξ/V^2}w_ηt$ .
So the law of the parallelogram of velocities holds, in this theory, only as a first approximation. The paper's general result for the resulting speed is symmetric in v and w.
With w along the X axis too, the paper's formula becomes $U=\frac{v+w}{1+vw/V^2}$ .
Worked: v = w = 0.6c. Plain addition would give 1.2c. The formula gives $1.2c/(1+0.36)=1.2c/1.36$ , which is 15c/17, about 0.882c.
Two speeds below light always give a speed below light. The paper shows it by writing v = V − κ and w = V − λ with κ and λ positive and less than V: then $U=V\frac{2V-κ-\lambda}{2V-κ-\lambda+κ\lambda/V}$ , which is less than V because the denominator is larger than the numerator. In the worked case κ = λ = 0.4V and $U=V\times1.2/1.36$ .
Light stays at V: put w = V and $U=\frac{V+v}{1+v/V}=V$ , as the second principle requires.
A worked sideways case in the paper's direction: $w_η=0.5c$ , $w_ξ=0$ , v = 0.6c. Then the sideways speed in K is $\sqrt{1-0.36}\times0.5c=0.4c$ , and the point also moves along x at 0.6c.
The paper remarks that the same formula comes from composing two transformations of §3: a third system moving at w relative to k is related to K by the same equations with v replaced by $(v+w)/(1+vw/V^2)$ .
Speeds in different directions combine by the general formula, not by this aligned one.

Return to why two ordinary speeds do not simply add. (included in this file)

Velocity along the motion

Modern teaching equation, not the equation as printed.

u_x' = \frac{u_x - v}{1 - \frac{v\,u_x}{c^{2}}}

The velocity along x in the moving system is the velocity along x in the stationary system minus the speed of the moving system, divided by one minus that speed times the stationary velocity along x over the speed of light squared.

Speeds do not simply subtract. An object moving at 15/17 of c in the stationary system moves at 0.6c in a system that itself moves at 0.6c; light, moving along x at c in one, moves at c in both.

Velocity across the motion

Modern teaching equation, not the equation as printed.

u_y' = \frac{u_y}{\gamma\,\left(1 - \frac{v\,u_x}{c^{2}}\right)}

Transverse velocity has the same denominator and an additional gamma.

Even a velocity at right angles to the motion changes, because the two systems' clocks run at different rates and are set differently along x.

Assumptions

The velocity comes from two neighbouring events on the object’s path.
The frames move at constant velocity relative to each other, along x.

Limits

The plus-sign formula for combining speeds and the minus-sign change of frame run in opposite directions.
Two boosts in different directions combine into a boost and a rotation.

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

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

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

Exponential change

What kind of change does an exponential describe?

Suppose a quantity changes at a rate proportional to its own size. A radioactive sample loses the same fraction of its atoms every hour; a beam of light loses the same fraction of its intensity in every millimetre of absorbing glass. Take 1,000 atoms that lose half their number each hour:

After one hour 500 remain, after two 250, after three 125.
The losses shrink as the sample does: 500 in the first hour, 250 in the second, 125 in the third.
Each hour multiplies the count by the same factor, one half. Equal steps, equal factors: that is exponential change.

The function e to the power x, with e about 2.718, is the one whose rate of change equals its own value everywhere, which is why it describes this kind of change. In symbols, with N₀ atoms at the start and a constant k that sets how fast they decay (for halving every hour, k is about 0.693 per hour):

\begin{gathered} \frac{dN}{dt} = -kN, \\ \text{so}\quad N(t) = N_0\,e^{-kt} \end{gathered}

If the rate of change of N is minus k times N, then N at time t is N nought times e to the minus k t.

Whatever sits in the exponent must be a pure number with no units. The reason is the series that defines e to the power x:

e^{x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \ldots

e to the x equals one plus x plus x squared over two plus x cubed over six, and so on.

If x were a length, the series would add a length to an area to a volume, which has no meaning. In the decay law, k is a rate per hour and t is in hours, so kt is a pure number.

Two exponentials from the 1905 papers pass that check. Wien's radiation law, which the light-quanta paper uses from §4 on, contains βν/T: β times a frequency is a temperature, so dividing by the temperature T leaves a pure number. The formula below writes e to the power −βν/T as exp(−βν/T). exp(u) is another way of writing e to the power u, used when the power is long.

\rho = \alpha\,\nu^{3} e^{-\beta \nu / T}

Rho equals alpha nu cubed times e to the minus beta nu over T.

The Brownian paper's §4 solution contains x²/4Dt: x² is an area, and a diffusion coefficient D in square metres per second times a time t in seconds is also an area, so their ratio is a pure number.

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

f of x and t equals n over the square root of four pi D, times e to the minus x squared over four D t, divided by the square root of t.

Worked example: The bell curve at its centre and one width out

The Brownian curve's shape comes from its exponential factor, e raised to −x²/4Dt. Follow that factor outward from the centre:

At the centre, x = 0, the exponent is 0 and e⁰ = 1: the curve is at its peak.
At x = √(2Dt), one root-mean-square width out, the exponent is −2Dt/4Dt = −1/2, and e raised to −1/2 is about 0.607 of the peak.
At x = √(4Dt), the exponent is −1, and e raised to −1 is about 0.368 of the peak.
Doubling the time doubles 4Dt, so each of these heights is reached √2 times farther from the centre.

None of these steps depends on the units chosen for x, t and D, as long as they agree with one another. That is what a pure-number exponent guarantees.

Where this lesson stops

Whatever sits in an exponent must be a pure number. Check it by cancelling the units, as with βν/T and x²/4Dt.

Return to the chapter

Events and distant clocks

What does it mean to say that two distant events happen at the same time?

An event is something that happens at one place at one instant: a spark, a collision, a flash leaving a lamp. To give it a time, you read a clock at that place. Reading a distant clock by looking at it would mix in the time the light takes to reach you.

A reference frame is a set of rigid measuring rods and clocks at rest, with a clock at every place, so that each event gets three coordinates of position and one of time from the rods and the clock beside it. Before that works, the clocks must agree, and for clocks far apart agreeing needs a rule.

§1 of the relativity paper gives the rule. Send a flash of light from clock A to clock B and back. Note A's reading when it leaves, B's reading when it arrives, and A's reading when it returns. The clocks are called synchronous if the trip out takes as long, by the clocks, as the trip back:

t_B - t_A = t'_A - t_B

t B minus t A equals t A prime minus t B: the trip out takes as long as the trip back.

The paper calls this a definition. It cannot be checked by measurement, because checking it would need clocks that are already synchronous. §1 adds two assumptions: that the rule is consistent, so that if B agrees with A then A agrees with B, and two clocks that agree with a third agree with each other; and that light's speed measured out and back is a universal constant, which the paper writes V.

Worked example: Setting a distant clock with one light signal

Clock A reads 0 s as a flash leaves it for clock B.
The flash is reflected at B and arrives back at A when A reads 10 s.
By the rule the two trips take equal times, so the reflection happened at 5 s on A's time.
If B's clock read 5 s at the reflection, it is synchronous with A. If it read 7 s, it is 2 s fast, and setting it back 2 s makes the two agree.

t_B = \frac{1}{2}\left(t_A + t'_A\right)

The time at B equals one half of the sum of A's reading when the flash left and A's reading when it came back.

The rule needs no distance. If B is known to be 1.5 × 10⁹ metres away, the same round trip also gives light's speed: 2 × 1.5 × 10⁹ m divided by 10 s is 3 × 10⁸ metres per second.

Where this lesson stops

An event is something that happens at one place and one time. Whether two distant events are simultaneous depends on a rule for setting clocks, which §1 of the relativity paper supplies.

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

Hyperbolic functions and rapidity

Two speeds along one line do not simply add. Is there a number that does?

The hyperbolic cosine and sine are the even and odd halves of the exponential: $\operatorname{cosh}x$ averages $e^x$ and $e^{-x}$ , and $\operatorname{sinh}x$ takes half their difference. Their ratio is the hyperbolic tangent, which stays between −1 and 1. In place of $\cos^2 + \sin^2 = 1$ they satisfy $\operatorname{cosh}^2 x - \operatorname{sinh}^2 x = 1$ .

\begin{gathered} \operatorname{cosh}x = \frac{e^x + e^{-x}}{2}, \\ \operatorname{sinh}x = \frac{e^x - e^{-x}}{2}, \\ \operatorname{tanh}x = \frac{\operatorname{sinh}x}{\operatorname{cosh}x} \end{gathered}

cosh x is e to the x plus e to the minus x, over two; sinh x is e to the x minus e to the minus x, over two; and tanh x is sinh x over cosh x.

For a boost at speed v, define the rapidity φ by $v/c = \operatorname{tanh}\varphi$ . Then the factor γ is $\operatorname{cosh}\varphi$ , and $\gamma v/c$ is $\operatorname{sinh}\varphi$ . Two warnings about letters. Einstein prints γ as β in the relativity and mass-energy papers, and in the light paper β is Wien's constant, so this lesson writes v/c out and never calls it β. And this φ is the modern rapidity, not the φ(v) of §3 of the relativity paper, which is a scale factor that turns out to be 1.

\begin{gathered} \gamma = \operatorname{cosh}\varphi, \\ \frac{\gamma v}{c} = \operatorname{sinh}\varphi, \\ \frac{v}{c} = \operatorname{tanh}\varphi \end{gathered}

gamma is cosh phi, gamma v over c is sinh phi, and v over c is tanh phi.

§5 of the relativity paper composes two speeds along one line: an object moving at w in a frame that itself moves at v moves at $U = (v + w)/(1 + vw/c^2)$ . That is exactly the addition rule for the hyperbolic tangent, so in rapidities the composition is plain addition: $\varphi_U = \varphi_v + \varphi_w$ .

\begin{aligned} &\operatorname{tanh}(\varphi_1 + \varphi_2) \\ &\quad = \frac{\operatorname{tanh}\varphi_1 + \operatorname{tanh}\varphi_2}{1 + \operatorname{tanh}\varphi_1\,\operatorname{tanh}\varphi_2} \end{aligned}

tanh of phi one plus phi two equals tanh phi one plus tanh phi two, over one plus tanh phi one times tanh phi two.

Rapidity is a later aid. Hyperbolic functions were applied to boosts by Varićak in 1910, and the word rapidity is Robb's, from 1911. The 1905 argument composes the speeds directly, and rapidities add only for boosts along one line: two boosts in different directions combine into a boost and a rotation.

Worked example: Adding 0.6c to 0.6c

For $v/c = 0.6$ the rapidity is $\varphi = \frac12\ln\frac{1 + 0.6}{1 - 0.6} = \frac12\ln 4 = \ln 2 = 0.6931$ .
Check: $\operatorname{cosh}(\ln 2) = (2 + \frac12)/2 = 1.25$ , which is γ at 0.6c, and $\operatorname{sinh}(\ln 2) = (2 - \frac12)/2 = 0.75$ , which is $\gamma v/c$ .
Two boosts of 0.6c along one line add their rapidities: $\ln 2 + \ln 2 = \ln 4$ .
Back to a speed: $\operatorname{tanh}(\ln 4) = (4 - \frac14)/(4 + \frac14) = 15/17 = 0.8824$ , so the combined speed is $15c/17$ , the same as $U = 1.2c/1.36$ from §5, and less than c.
$e^{-\ln 2} = 0.5$ and $e^{\ln 2} = 2$ are the stretches of the two light lines under the boost at 0.6c: the Doppler factors for light sent forward and back.

Where this lesson stops

Hyperbolic functions are combinations of e to the x and e to the minus x. This lesson stops at boosts along one line.

This lesson builds on

Exponential change
Logarithms: turning products into sums
matrices-linear-maps

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

Exponential change

Return to the chapter

Static laboratory examples

sr-06

This laboratory is not included in this file. The chapter's authored foundation examples remain available above. Open sr-06 online

Source references

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