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Two Einstein solids sharing energy: the multiplicity of each split as a product, the most probable split, Stirling's approximation, and why the peak is so sharp that energy flow is irreversible.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find the multiplicity and probability of each way two solids can share energy, find the most probable split, estimate large factorials with Stirling's approximation, and explain why energy flow between large systems is irreversible.
You can count the microstates of an Einstein solid, $\binom{q + N - 1}{q}$, and you know that in an isolated system every microstate is equally likely. This lesson puts two solids side by side and lets them trade energy, which is the simplest model of two objects at different temperatures touching.
| Term | What it means |
|---|---|
| Thermal contact | Two systems able to exchange energy, while the pair as a whole is isolated. |
| Macrostate of the pair | A split of the total energy: $q_A$ units in A and $q_B = q - q_A$ in B. |
| Multiplicity of a split | $\Omega_A(q_A)\,\Omega_B(q_B)$, the product of the two solids' counts. |
| Most probable macrostate | The split with the largest multiplicity, where the pair spends most of its time. |
| Stirling's approximation | $\ln N! \approx N\ln N - N$, accurate to a fraction of a percent for large $N$. |
| Irreversible | A change whose reverse is possible in principle but so improbable that it never happens. |
| Thermal equilibrium | The macrostate the pair settles into: energy shared so that each oscillator holds the same average. |
Put an Einstein solid A of $N_A$ oscillators next to a solid B of $N_B$ oscillators, so that they can trade units of energy but the pair as a whole is isolated with $q$ units in total. A macrostate of the pair is a split: $q_A$ units in A and $q_B = q - q_A$ in B. Each microstate of A can be combined with each microstate of B, so the multiplicity of a split is a product:
$$\Omega_{\text{pair}}(q_A) = \Omega_A(q_A)\,\Omega_B(q - q_A).$$
For two solids of three oscillators sharing six units the splits have multiplicities $28, 63, 90, 100, 90, 63, 28$ for $q_A = 0$ to $6$, out of $462$ in all. By the fundamental assumption, the pair is found in the even split $100$ times out of $462$ and in the most lopsided split only $28$ times.
With a few oscillators the peak is broad, and the energy wanders widely. With many it is astonishingly sharp. For two solids of $N = 10^{20}$ oscillators each, the multiplicity falls to a negligible fraction of its peak value when $q_A$ is off the peak by one part in ten billion. So once the solids are in contact, the energy moves toward the most probable split and stays there — not because anything pushes it, but because almost all the microstates are there.
For large solids the peak is where each oscillator holds the same average energy,
$$\frac{q_A}{N_A} = \frac{q_B}{N_B},$$
which the next lessons will identify as equal temperatures. Energy flowing from a hot solid to a cold one is the pair climbing to that peak.
Another way: picture
Picture the splits as the rungs of a ladder and the pair as someone stepping from rung to rung at random, one unit of energy at a time. Each rung's width is its number of microstates. With small solids the rungs near the middle are a little wider, and the walker drifts around them. With solids of everyday size the middle rung is wider than all the others put together by a factor with more digits than there are atoms in the universe, and a walker who reaches it never, in practice, finds the way off.
Another way: steps
Checks. The counts of all the splits must add up to the total from step 4, which catches almost any slip. For two identical solids the table of products is symmetric about the even split. The most probable split should put roughly equal energy per oscillator in each solid, more so the larger the solids. And the product for the most lopsided split, all the energy in one solid, is always the count of that solid alone, since the empty one has a single microstate.
Factorials of large numbers cannot be evaluated directly, but their logarithms can be estimated. Since $\ln N! = \ln 1 + \ln 2 + \cdots + \ln N$, and the sum is close to the integral of $\ln x$,
$$\ln N! \approx \int_1^N \ln x\,dx \approx N\ln N - N.$$
For $N = 100$ this gives $360.5$ against the exact $363.7$, an error under one percent, and the relative error shrinks as $N$ grows.
Applied to two identical solids with $q \gg N$, it shows that near the peak the multiplicity of the pair has the shape of a Gaussian,
$$\Omega(q_A) \approx \Omega_{\max}\,e^{-N\left(\frac{2x}{q}\right)^2}, \qquad x = q_A - \frac{q}{2},$$
whose width is about $q/(2\sqrt{N})$. The relative width, width over position, goes as $1/\sqrt{N}$. For $N = 100$ that is ten percent: small solids fluctuate visibly. For $N = 10^{20}$ it is one part in ten billion. No instrument could see a fluctuation that size, and one ten times larger has a probability suppressed by a factor of $e^{-100}$. That is why the macroscopic world looks deterministic even though every step underneath is random.
Suppose solid A starts with all the energy and B with none. The pair begins in the most lopsided split, the one with the fewest microstates of any with energy in A. As units hop at random between neighboring oscillators, some cross into B, and because there are vastly more microstates with energy shared than with energy concentrated, the pair drifts toward the even split. Run the process backward — every unit returning to A — and nothing in the rules of hopping forbids it. It is simply outnumbered.
This is the statistical meaning of the second law, and it answers a puzzle that troubled physicists in the nineteenth century: the laws of mechanics are the same forward and backward in time, yet heat always flows from hot to cold. The laws do not pick a direction; the counting does. A process is irreversible when its reverse requires a system to move from many microstates to few. The next lesson attaches a name and a unit to that count, and the second law becomes a statement about entropy.
A computer can play the random hopping out. Start two solids of $300$ and $200$ oscillators with all $100$ units in the smaller one, and at each step move one unit from a randomly chosen oscillator to another chosen at random. Within a few thousand steps the energy of the smaller solid has fallen to about $40$ units, where the counting puts the peak ($q_A/N_A = q_B/N_B$ gives $q_A = 100 \times 200/500 = 40$), and it then wanders by a few units either side of it, never returning to $100$.
The path is noisy and never exactly repeats, but its destination is fixed by the multiplicities alone. That is the sense in which thermal equilibrium is a statistical state rather than a static one: the microstate keeps changing every instant, while the macrostate, the split of energy, stays within a narrow band around the most probable value.
A $0.30$ kg mug of coffee at $80$ °C sits in a room holding about $60$ kg of air at $20$ °C, with furniture and walls holding far more. Energy leaves the coffee until the temperatures match, and the counting says where it goes.
Treat each as a collection of oscillators with the same energy per oscillator in equilibrium. Water's heat capacity is about $4186$ J/(kg K), so the coffee has a heat capacity of about $1260$ J/K; the air alone has $60 \times 718 \approx 43\,000$ J/K at constant volume. Heat capacity counts active oscillators, so the air has about $34$ times as many. The final temperature is the heat-capacity-weighted average: $(1260 \times 80 + 43\,000 \times 20)/44\,260 \approx 21.7$ °C — and with the walls and furniture included the room hardly changes at all.
The coffee has given up about $1260 \times 58 \approx 73\,000$ J, and the room's air has risen by under two degrees. No law says the energy must leave the coffee. There are simply overwhelmingly more microstates with the energy spread across the room's $10^{27}$ molecules than concentrated in the coffee's $10^{25}$, which is why a cooling drink is the most ordinary irreversible process there is.
The sharpness of the peak depends on $1/\sqrt{N}$, and in small enough systems the fluctuations it predicts are measured routinely. A standard example is the thermal noise in an electrical resistor. The electrons in a resistor share energy with its lattice, and the fluctuations in how that energy is shared produce a random voltage whose average square is $4kTR\,\Delta f$ — the Johnson-Nyquist formula.
For a $1$ MΩ resistor at room temperature, measured over a bandwidth of $10$ kHz, the rms noise voltage is $\sqrt{4 \times 1.38 \times 10^{-23} \times 300 \times 10^{6} \times 10^{4}} \approx 13$ μV. Radio astronomers cool their receivers to a few kelvin to shrink it, and audio engineers choose resistor values to keep it below hearing.
Johnson measured it at Bell Telephone Laboratories in 1926 and Nyquist explained it with exactly the reasoning of this lesson: energy is constantly sloshing between parts of a system in contact, and the smaller the parts, the larger the slosh. Measuring the noise voltage is even one of the most precise ways to measure temperature.
Asked how energy divides between two objects in contact, most people say "equally". It divides equally per oscillator — per degree of freedom — so a solid with three times as many oscillators ends with three times the energy. A cup of hot tea set in a bathtub of cool water does not split its energy with the bath half and half; the bath, with thousands of times as many molecules, takes nearly all of it, and the tea ends barely warmer than the bath started.
A second misconception is that the most probable split is the only possible one. For small systems it is not even likely: two six-oscillator solids spend under a quarter of their time in the even split. What makes large systems settle is not that other splits are forbidden but that they are outnumbered by factors that grow exponentially with $N$. Every fluctuation the counting allows really happens; in a mole of material they are simply too small to see.
Solids A and B have two oscillators each and share three units. Write the multiplicity of a split.
$\Omega(q_A) = \binom{q_A + 1}{q_A}\binom{q_B + 1}{q_B} = (q_A + 1)(q_B + 1)$
With two oscillators, $\binom{q + 1}{q} = q + 1$.
Evaluate the lopsided splits.
$q_A = 0: \ 1 \times 4 = 4, \qquad q_A = 3: \ 4 \times 1 = 4$
One solid empty, the other holding everything.
Evaluate the middle splits.
$q_A = 1: \ 2 \times 3 = 6, \qquad q_A = 2: \ 3 \times 2 = 6$
Sharing gives more microstates.
Add them and check.
$4 + 6 + 6 + 4 = 20 = \binom{3 + 4 - 1}{3}$
The pair is one solid of four oscillators holding three units.
Find the probability of an even-as-possible split.
$P(q_A = 1 \text{ or } 2) = \dfrac{12}{20} = 0.6$
With so few oscillators the lopsided splits still turn up two times in five.
Solid A has two oscillators, solid B four; they share four units. Write each solid's count.
$\Omega_A = q_A + 1, \qquad \Omega_B = \binom{q_B + 3}{q_B}$
Two oscillators need one line; four need three.
Evaluate B's counts for $q_B = 4, 3, 2, 1, 0$.
$35, \quad 20, \quad 10, \quad 4, \quad 1$
These are $\binom{7}{4}, \binom{6}{3}, \binom{5}{2}, \binom{4}{1}, \binom{3}{0}$.
Multiply for each split, $q_A = 0$ to $4$.
$1 \times 35 = 35, \quad 2 \times 20 = 40, \quad 3 \times 10 = 30, \quad 4 \times 4 = 16, \quad 5 \times 1 = 5$
A's count rises as B's falls, and the product peaks between.
Add them and check.
$35 + 40 + 30 + 16 + 5 = 126 = \binom{4 + 6 - 1}{4}$
The total for six oscillators holding four units.
Find the most probable split.
$q_A = 1, \ q_B = 3: \quad P = \dfrac{40}{126} \approx 0.32$
The larger solid holds most of the energy.
Compare the energy per oscillator.
$\dfrac{q_A}{N_A} = \dfrac{1}{2} = 0.5, \qquad \dfrac{q_B}{N_B} = \dfrac{3}{4} = 0.75$
Roughly equal per oscillator; the match becomes exact as the solids grow.
Write Stirling's approximation for $\ln 100!$.
$\ln 100! \approx 100\ln 100 - 100$
$\ln N! \approx N\ln N - N$ for large $N$.
Evaluate the logarithm.
$\ln 100 = 4.6052$
A calculator, or $2\ln 10 = 2 \times 2.3026$.
Evaluate the approximation.
$100 \times 4.6052 - 100 = 360.52$
The leading terms only.
Compare with the exact value.
$\ln 100! = 363.74$
Found by summing $\ln 1 + \ln 2 + \cdots + \ln 100$.
Find the relative error.
$\dfrac{363.74 - 360.52}{363.74} \approx 0.9\%$
Under one percent even for a number as small as $100$.
Add the next term and compare again.
$\tfrac{1}{2}\ln(2\pi \times 100) = \tfrac{1}{2}\ln 628.3 \approx 3.22, \qquad 360.52 + 3.22 = 363.74$
With the correction the agreement is essentially exact.
Judge the error for a macroscopic number.
$N = 10^{23}: \quad \dfrac{\tfrac{1}{2}\ln(2\pi N)}{N\ln N} \approx \dfrac{27}{5.3 \times 10^{24}} \approx 5 \times 10^{-24}$
For real systems the leading terms are all that ever matter.
Write the counts for $q_A = 0$ to $4$.
$(q_A + 1)(5 - q_A): \quad 5, \ 8, \ 9, \ 8, \ 5$
With two oscillators each solid's count is its units plus one.
Add the five counts together.
$5 + 8 + 9 + 8 + 5 = 35 = \binom{7}{4}$
Four oscillators holding four units.
Find the most probable split and its probability.
Two large Einstein solids of the same material, with $2 \times 10^{22}$ and $7 \times 10^{22}$ oscillators, are put in contact and share $360 \times 10^{22}$ units of energy. In the most probable macrostate, how many units does the smaller solid hold?
Complete the worked solution: solid A of $4$ oscillators holds $4$ units and solid B of $4$ oscillators holds $4$. How many microstates does the pair have in this split?
Evaluate solid A's multiplicity.
$\Omega_A = \binom{7}{4} =$ x
$q_A$ dots and $N_A - 1$ lines.
Evaluate solid B's multiplicity.
$\Omega_B = \binom{7}{4} =$ y
The same formula for the other solid.
Multiply them.
$\Omega_A\Omega_B =$ p
Each microstate of A pairs with each microstate of B.
Two Einstein solids of three oscillators each, with levels $27$ meV apart, share six units. Match each quantity to its value.
| $462$ | $100$ | $28$ | $3$ | |
|---|---|---|---|---|
| microstates of the pair, over all splits | ||||
| microstates with three units in each solid | ||||
| microstates with all six units in solid A | ||||
| units in A in the most probable split |
Solid A has $3$ oscillators and solid B has $4$; they share three units. Fill in the multiplicity of the pair for each split, and the total.
| microstates of the pair | |
|---|---|
| no units in A | |
| one unit in A | |
| two units in A | |
| three units in A | |
| total |
Use Stirling's approximation to estimate $\ln(50!)$. Take $\ln 50 = 3.912$.
Answer:
Einstein solid A has $4$ oscillators and solid B has $4$. In thermal contact they share $8$ units of energy. What is the probability of finding $4$ units in A and $4$ in B? Give it to three decimal places.
Answer:
A hot silver nanocrystal of $500$ atoms touches a cold silver nanocrystal of $800$ atoms. Between them they hold $35100$ units of vibrational energy. Treating both as Einstein solids with the same level spacing, how many units does the cold crystal hold once they reach the most probable split?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Solid A has $2$ oscillators and solid B has $2$; they share three units. Fill in the multiplicity of the pair for each split, and the total.
| microstates of the pair | |
|---|---|
| no units in A | |
| one unit in A | |
| two units in A | |
| three units in A | |
| total |
You can count the microstates of two solids sharing energy and find how they most probably share it. Explain to someone why a hot cup of tea in a bathtub gives nearly all its energy to the bath.
15. Your turn: two solids of two oscillators each share four units. Which split is most probable, and how probable is it?, step 3
$q_A = 2: \quad P = \dfrac{9}{35} \approx 0.26$
The even split wins, but only just, with so few oscillators.