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The Einstein solid

A solid as $N$ quantum oscillators sharing $q$ units of energy, the multiplicity $\binom{q + N - 1}{q}$ from dots and lines, and its high-temperature limit.

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.

1. What you will learn

By the end of this lesson you will be able to count the oscillators of a solid, turn an energy into a number of units, find the multiplicity of an Einstein solid with the dots-and-lines argument, and use the high-temperature approximation.

2. What you already have

You can find the multiplicity of a two-state system as a binomial coefficient, and you know from the equipartition lesson that each atom in a solid vibrates in three directions. From quantum physics you know that an oscillator of frequency $f$ has evenly spaced energy levels, $hf$ apart. This lesson combines the three to count the microstates of a solid.

3. Words for this lesson

4. Counting ways to share energy among oscillators

In 1907 Einstein modeled a solid as a collection of identical quantum oscillators, one for each direction of vibration of each atom, so a solid of $N_{\text{atoms}}$ atoms has $N = 3N_{\text{atoms}}$ oscillators. Each oscillator's energy levels are evenly spaced by $\epsilon = hf$, so the energy of the whole solid above its ground state is a whole number $q$ of units:

$$U = q\epsilon.$$

A microstate says how many units each oscillator holds: with $N = 3$ and $q = 2$, the microstates are $(2,0,0)$, $(0,2,0)$, $(0,0,2)$, $(1,1,0)$, $(1,0,1)$ and $(0,1,1)$ — six of them. The units are not labeled; there is no "first quantum" and "second quantum", only how many sit on each oscillator.

To count microstates in general, draw each one as a row of $q$ dots split by $N - 1$ vertical lines into $N$ groups. The microstate $(1, 0, 3)$ is $\bullet \mid\ \mid \bullet\bullet\bullet$. Every arrangement of $q$ dots and $N - 1$ lines in a row is a different microstate, and every microstate is exactly one arrangement. So the multiplicity is the number of ways to choose which of the $q + N - 1$ positions hold dots:

$$\Omega(N, q) = \binom{q + N - 1}{q} = \frac{(q + N - 1)!}{q!\,(N - 1)!}.$$

The multiplicity climbs steeply with energy. For three oscillators, one to four units give $3$, $6$, $10$ and $15$ microstates; for a real solid, with $N$ near $10^{23}$, adding a single unit multiplies $\Omega$ by an enormous factor. That steep climb is what will make energy flow from one solid to another.

5. The method, step by step, and how to check it

1. Find $N$. An oscillator is one direction of vibration of one atom. A cluster of $5$ atoms has $15$ oscillators; a mole of atoms has $1.8 \times 10^{24}$.
2. Find $q$. Divide the energy above the ground state by the level spacing. A non-integer result means the energy given is not allowed.
3. Build the binomial coefficient $\binom{q + N - 1}{q}$. It is also $\binom{q + N - 1}{N - 1}$ — choosing the positions of the lines instead of the dots — and the smaller of $q$ and $N - 1$ makes the arithmetic shorter.
4. Evaluate by cancelling: $\binom{8}{3} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.

Checks. With one oscillator, $\Omega = 1$: all the energy must sit on it. With no energy, $\Omega = 1$: every oscillator is in its ground state. With one unit, $\Omega = N$: the unit can be on any oscillator. For small cases, list the microstates and count them directly, as with $N = 3$, $q = 2$ above. The commonest wrong answer, $N^q$, fails the first check: with one oscillator it gives $1$, but with two units and two oscillators it gives $4$ instead of the true $3$ — $(2,0)$, $(1,1)$, $(0,2)$ — because it counts $(1,1)$ twice.

6. Very large numbers, and the high-temperature limit

For a macroscopic solid the multiplicity is too large to evaluate directly, and a useful approximation comes from Stirling's formula, $\ln n! \approx n\ln n - n$, which the next lesson develops. When there are many more units than oscillators, $q \gg N$ — the high-temperature limit, where every oscillator holds many units on average — it gives

$$\Omega(N, q) \approx \left(\frac{eq}{N}\right)^N.$$

Even with six oscillators and six units, far from the limit, this gives $e^6 \approx 403$ against the exact $462$. For a solid with $N = 10^{23}$ oscillators averaging $10$ units each, $\Omega \approx (27)^{10^{23}}$, a number whose logarithm is about $3.3 \times 10^{23}$. Numbers like this are never written out; physicists work with their logarithms, which is exactly what entropy will be.

The formula also shows the key behavior: $\ln\Omega \approx N\ln q + \text{constant}$ grows as the logarithm of the energy. So each extra unit of energy raises $\ln \Omega$ by about $N/q$ — a lot when the solid is cold and holds few units, and less when it is hot. That diminishing return is the reason heat flows from hot to cold.

7. What the model gets right and what it leaves out

Einstein's model was the first explanation of why solids' heat capacities fall at low temperature. When $kT$ is much smaller than $\epsilon$, the solid has too little energy to put even one unit on most oscillators, the vibrations freeze, and the heat capacity drops toward zero — the behavior diamond shows at room temperature. At high temperature, when each oscillator holds many units, the model recovers the Dulong-Petit value of $3R$ per mole.

Its simplification is that every oscillator has the same frequency. In a real crystal the atoms vibrate together in waves with a range of frequencies, down to very low ones for long wavelengths. Debye's 1912 refinement accounts for that spread and predicts a heat capacity proportional to $T^3$ at low temperature, which matches experiment better than Einstein's exponential fall. But for counting microstates and seeing why energy spreads, the Einstein solid is the cleanest model available, and it is the one the next lessons use.

8. A recursion that checks any count

The multiplicity of an Einstein solid obeys a simple recursion that gives an independent check on any value. Look at the last oscillator: it holds some number $j$ of units, from $0$ to $q$, and the other $N - 1$ oscillators share the remaining $q - j$. Summing over $j$,

$$\Omega(N, q) = \sum_{j = 0}^{q} \Omega(N - 1, q - j).$$

For three oscillators and two units that is $\Omega(2, 2) + \Omega(2, 1) + \Omega(2, 0) = 3 + 2 + 1 = 6$, as the direct list found. Building a small table this way, column by column, reproduces every binomial coefficient in this lesson without a single factorial, and it is how a computer counts the states of a solid far too large to list. It also makes plain why each extra oscillator multiplies the possibilities: every way of sharing the energy among the others can be combined with every amount the new oscillator might take.

9. In the world: why diamond feels cold and conducts heat so well

Diamond's carbon atoms are light and joined by very stiff bonds, so its vibrations have high frequencies. In the Einstein model its level spacing is about $\epsilon = 0.11$ eV, set by its Einstein temperature of about $1300$ K ($\epsilon = kT_E$). At room temperature, $kT = 0.026$ eV, only about a quarter of a unit is available per oscillator.

The counting shows what that means. With $N$ oscillators and only $q \approx 0.25N$ units, most oscillators hold nothing: the multiplicity is dominated by arrangements in which a scattering of oscillators hold one unit each. Adding energy to such a solid raises its temperature quickly, and the heat capacity is far below $3R$ — the measured $509$ J/(kg K) against a Dulong-Petit value of $2080$. Lead, whose heavy atoms and soft bonds give an Einstein temperature near $90$ K, has every oscillator well into the many-unit regime at room temperature and obeys Dulong-Petit closely.

The same stiff lattice carries vibrational waves at high speed, which is why diamond conducts heat about five times better than copper at room temperature, and why a real diamond held to the lips feels colder than a glass imitation. Jewelers' "thermal testers" use exactly that property to tell them apart in seconds.

10. In the world: counting states in a nanoparticle

The Einstein solid's formula is exact for any number of oscillators, which makes it a working tool for nanoscale physics, where a particle may contain only a few hundred atoms. A gold nanoparticle of $300$ atoms has $900$ oscillators. With gold's Einstein temperature near $130$ K, an energy unit is about $0.011$ eV, and at room temperature the particle holds roughly $q \approx N kT/\epsilon \approx 900 \times 2.3 \approx 2000$ units.

The multiplicity $\binom{2899}{2000}$ is about $10^{778}$, so large that a calculator overflows, yet small enough, compared with a macroscopic solid, that adding or removing a single unit changes it by a factor of about $(q + N)/q \approx 1.45$. That makes the temperature of a nanoparticle a slightly fuzzy quantity: its energy fluctuates by a measurable fraction as it exchanges units with its surroundings, an effect that matters in nanoparticle catalysts and in the tiny thermometers used in quantum devices. For a mole of gold the same fractional fluctuations are about one part in $10^{12}$, and temperature is as sharp as it ever needs to be.

11. Units of energy are not labeled

The most natural way to count is to give each unit of energy a choice of oscillator: $N$ choices for each of $q$ units, $N^q$ in all. It is wrong, because it treats the units as distinguishable. With two oscillators and two units it counts "unit A on the first and unit B on the second" and "unit B on the first and unit A on the second" as different, when both describe the same physical state — each oscillator holding one unit. Energy has no identity to swap.

The opposite error is to borrow the two-state formula $\binom{N}{q}$, which allows each oscillator at most one unit. That describes a two-state paramagnet, not a solid: an oscillator's ladder of levels goes on up, and the heavily occupied oscillators are exactly the ones that matter at high temperature. The dots-and-lines picture gets both right, and small cases listed by hand are the check.

12. Three oscillators, two units, listed and counted

1. List the microstates with both units on one oscillator.

   $(2, 0, 0), \quad (0, 2, 0), \quad (0, 0, 2)$

   The pair of units can sit on any of the three oscillators.

2. List the microstates with one unit on each of two oscillators.

   $(1, 1, 0), \quad (1, 0, 1), \quad (0, 1, 1)$

   Choose which oscillator is left empty.

3. Count the microstates on both lists.

   $\Omega = 3 + 3 = 6$

   Every way of sharing two units among three oscillators is on one of the two lists.

4. Count the symbols for the formula.

   $q + N - 1 = 2 + 3 - 1 = 4$

   Two dots and two lines.

5. Check with the formula.

   $\Omega = \binom{4}{2} = \dfrac{4 \times 3}{2 \times 1} = 6$

   The formula agrees with the direct count.

13. Adding energy to a small solid

1. Four oscillators hold three units. Count the symbols.

   $q + N - 1 = 3 + 4 - 1 = 6$

   Three dots and three lines.

2. Evaluate the multiplicity.

   $\Omega(4, 3) = \binom{6}{3} = \dfrac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

   Choose which three of six positions hold dots.

3. Add one unit and count the symbols again.

   $q + N - 1 = 4 + 4 - 1 = 7$

   Four dots and three lines.

4. Evaluate the new multiplicity.

   $\Omega(4, 4) = \binom{7}{4} = \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

   Here it is quicker to choose the three positions of the lines.

5. Find the factor the extra unit multiplied by.

   $\dfrac{35}{20} = 1.75 = \dfrac{q + N}{q + 1} = \dfrac{3 + 4}{3 + 1} = \dfrac{7}{4}$

   In general $\Omega(N, q + 1)/\Omega(N, q) = (q + N)/(q + 1)$.

6. Compare with the wrong count.

   $4^3 = 64 \ne 20$

   Labeling the units would triple-count many microstates.

14. A six-oscillator solid and the high-temperature formula

1. Two atoms vibrating in three directions each hold six units. Count the oscillators.

   $N = 2 \times 3 = 6$

   Each direction of each atom is one oscillator.

2. Count the symbols.

   $q + N - 1 = 6 + 6 - 1 = 11$

   Six dots and five lines.

3. Write the multiplicity.

   $\Omega = \binom{11}{6} = \dfrac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1}$

   Choose the five positions of the lines; $\binom{11}{6} = \binom{11}{5}$.

4. Evaluate the binomial coefficient.

   $\Omega = \dfrac{55\,440}{120} = 462$

   Cancel before multiplying if doing it by hand.

5. Evaluate the high-temperature approximation.

   $\left(\dfrac{eq}{N}\right)^N = \left(\dfrac{e \times 6}{6}\right)^6 = e^6 \approx 403$

   The formula assumes $q \gg N$, which is not true here.

6. Compare the logarithms.

   $\ln 462 \approx 6.14, \qquad \ln 403 = 6$

   The logarithms agree to about two percent, and for large $N$ the agreement becomes excellent.

7. Find the multiplicity for one more unit.

   $\Omega(6, 7) = 462 \times \dfrac{6 + 6}{6 + 1} = 462 \times \dfrac{12}{7} = 792$

   Using the ratio $(q + N)/(q + 1)$ avoids recomputing the binomial coefficient.

15. Your turn: three oscillators share four units. How many microstates are there?

1. Count the symbols.

   $q + N - 1 = 4 + 3 - 1 = 6$

   Four dots and two lines.

2. Write the binomial coefficient.

   $\Omega = \binom{6}{4} = \binom{6}{2}$

   Choosing the lines is quicker than choosing the dots.

3. Your turn: work this step out. Its working is at the end of the packet.

   Evaluate the binomial coefficient.

16. Guided practice

An Einstein solid has $6$ oscillators sharing $2$ units of energy. How many microstates does it have?

• $21$
• $36$
• $15$
• $64$

17. Guided practice

Complete the worked solution: an Einstein solid of $7$ oscillators with level spacing $6$ meV holds $18$ meV. How many microstates does it have?

1. Divide the energy by the level spacing.

   $q = \dfrac{18}{6} =$ _q

   Energy comes in whole units of $\epsilon$.

2. Add the number of units to the number of oscillators and subtract one.

   $q + 7 - 1 =$ _m

   That is the number of dots and lines together.

3. Evaluate the binomial coefficient.

   $\Omega = \binom{q + N - 1}{q} =$ _w

   Choose which positions in the row hold dots.

18. Guided practice

Oscillators with levels $29$ meV apart share units of energy. Match each small solid to its multiplicity.

19. Practice

Three oscillators have levels spaced $\epsilon = 19$ meV apart. Fill in the energy above the ground state and the multiplicity for each number of units.

20. Practice

How many microstates does an Einstein solid of $8$ oscillators holding $3$ units of energy have?

Answer:

21. Practice

A tiny Einstein solid has $4$ oscillators, each with energy levels spaced $\epsilon = 3$ meV apart. It holds $15$ meV of energy above its ground state. How many microstates does it have?

Answer:

22. Somewhere new

In a model of a two-atom cluster, each atom vibrates in three directions and every vibration is an oscillator with the same level spacing. The cluster holds $2$ units of vibrational energy. How many microstates does it have?

Answer:

23. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

24. Test question

Three oscillators have levels spaced $\epsilon = 16$ meV apart. Fill in the energy above the ground state and the multiplicity for each number of units.

25. What you can do now

You can count the microstates of an Einstein solid. Explain to someone why the count is not $N^q$, using two oscillators and two units.

Working for the steps left to you

15. Your turn: three oscillators share four units. How many microstates are there?, step 3