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Temperature defined as $1/T = \partial S/\partial U$, why heat flows from hot to cold, equipartition recovered from counting, Einstein's relation for a solid, and negative temperatures.
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By the end of this lesson you will be able to define temperature from entropy, predict the direction of heat flow from entropy slopes, derive the energy of a solid and a gas as a function of temperature, find the temperature of a small solid by finite differences, and say what a negative temperature means.
You know that two systems in contact settle into the split of energy with the largest total entropy, and you can find the entropy of an Einstein solid and of an ideal gas. From calculus you know that at the maximum of a function of one variable its derivative is zero. This lesson uses that fact to give temperature its exact meaning.
| Term | What it means |
|---|---|
| Temperature | $\dfrac{1}{T} = \left(\dfrac{\partial S}{\partial U}\right)_{N,V}$: the reciprocal of how fast entropy rises with energy. |
| Thermal equilibrium | The state in which two systems in contact have equal $\partial S/\partial U$, and so equal temperatures. |
| Zeroth law | Two systems each in equilibrium with a third are in equilibrium with each other. |
| Finite-difference temperature | $T \approx \Delta U/\Delta S$, from one small step in energy. |
| Einstein temperature | $\epsilon/k$, the temperature at which $kT$ equals an oscillator's energy unit. |
| Negative temperature | A state whose entropy falls as its energy rises, so $\partial S/\partial U < 0$. |
| Partial derivative | A derivative with the other variables, here $N$ and $V$, held fixed. |
Two systems A and B in thermal contact share a fixed total energy $U_A + U_B$. They settle where the total entropy $S_A(U_A) + S_B(U_B)$ is largest. At that maximum, moving a little energy $dU$ from B to A changes nothing to first order:
$$\frac{\partial S_A}{\partial U_A} - \frac{\partial S_B}{\partial U_B} = 0.$$
So in equilibrium the two slopes are equal. That is also the one thing a thermometer can report, and it gives temperature its definition:
$$\frac{1}{T} = \left(\frac{\partial S}{\partial U}\right)_{N,V}.$$
The reciprocal is chosen so that the units come out as kelvin and so that heat flows from high $T$ to low $T$. Away from equilibrium, a small transfer $dU$ from B to A changes the total entropy by $(1/T_A - 1/T_B)\,dU$; the second law requires this to be positive, so $dU > 0$ exactly when $T_A < T_B$. Energy flows into the system whose entropy rises faster per joule — the colder one — because that raises the total.
The definition reproduces what equipartition gave. An Einstein solid in the high-temperature limit has $S = Nk\ln(eq/N)$ with $U = q\epsilon$, so $S = Nk\ln U + \text{constant}$ and
$$\frac{1}{T} = \frac{Nk}{U} \quad\Rightarrow\quad U = NkT,$$
one $kT$ per oscillator: two quadratic terms each at $\tfrac{1}{2}kT$. A monatomic ideal gas has $S = \tfrac{3}{2}Nk\ln U + \ldots$, which gives $U = \tfrac{3}{2}NkT$. Nothing about temperature has been assumed; it has been derived from counting.
Another way: picture
Picture two businesses sharing a fixed budget, each turning money into satisfaction at a rate that falls as it gets richer. Money flows to whichever adds more satisfaction per dollar, until the rates match. Entropy is the satisfaction and energy the money. A cold system is the poor business, for which each joule buys a lot of entropy; a hot system is the rich one. Temperature measures how little entropy one more joule buys.
Another way: steps
Checks. The result must have units of kelvin: joules over joules per kelvin. For a large solid it should reduce to $U = NkT$ at high temperature, and for a monatomic gas to $U = \tfrac{3}{2}NkT$. The temperature should rise as energy is added to any ordinary system, because each extra joule buys less entropy — the entropy-energy curve is concave. When a finite-difference estimate falls as energy rises, the system is not ordinary: it is a paramagnet past the point where adding energy reduces the number of microstates, and its temperature is negative.
Using Stirling's approximation on the full multiplicity of a large Einstein solid, not only its high-temperature limit, gives
$$\frac{S}{k} = (q + N)\ln(q + N) - q\ln q - N\ln N.$$
Differentiating with respect to $U = q\epsilon$:
$$\frac{1}{T} = \frac{k}{\epsilon}\ln\frac{q + N}{q} \quad\Rightarrow\quad \frac{q}{N} = \frac{1}{e^{\epsilon/kT} - 1}.$$
This is Einstein's result. At high temperature, $\epsilon/kT \ll 1$, the exponential is about $1 + \epsilon/kT$ and $q/N \approx kT/\epsilon$: the equipartition value. At low temperature the exponential is huge and $q/N$ falls as $e^{-\epsilon/kT}$: the oscillators freeze. The crossover is at the Einstein temperature, $\epsilon/k$. For lead it is about $90$ K, so at room temperature each oscillator holds almost three units and lead obeys Dulong-Petit; for diamond it is about $1300$ K, and at room temperature an oscillator holds on average only about one hundredth of a unit.
The finite difference for a small solid is the same formula in miniature: adding one unit multiplies $\Omega$ by $(q + N)/(q + 1)$, so $kT \approx \epsilon/\ln\left(\frac{q + N}{q + 1}\right)$, which approaches Einstein's expression as $q$ grows.
For most systems entropy rises without limit as energy is added. A two-state paramagnet is different: its energy has a maximum, reached when every dipole points against the field, and that state has multiplicity one. Between zero energy (half the dipoles each way, maximum multiplicity) and the maximum energy, adding energy reduces the entropy. There $\partial S/\partial U < 0$ and the temperature is negative.
Negative temperatures are not colder than absolute zero; they are hotter than any positive temperature. A system at negative temperature gives energy to anything at positive temperature, because losing energy raises its entropy while the other system's entropy rises too. Such states were first made in 1951 by Purcell and Pound, using the nuclear spins of a lithium fluoride crystal, which are well isolated from the crystal's vibrations for a few minutes. They are also the physics of a laser, whose population inversion — more atoms in the upper state than the lower — is a negative temperature of the two levels involved.
A mercury thermometer, a thermocouple and an infrared sensor measure quite different things — a volume, a voltage, a flux of radiation — yet they agree on the temperature of a cup of tea. The entropy definition explains why. Each instrument is a small system brought into equilibrium with the tea, and equilibrium means equal $\partial S/\partial U$. Whatever property the instrument displays, it is calibrated as a function of its own slope, which is now the tea's slope. This is the content of the zeroth law: two systems each in equilibrium with a third are in equilibrium with each other, so a thermometer can stand between them.
Einstein's relation $q/N = 1/(e^{\epsilon/kT} - 1)$ turns a single number for each solid, its Einstein temperature $\epsilon/k$, into a prediction of how active its vibrations are at room temperature, $300$ K.
For lead, $\epsilon/k \approx 90$ K, so $\epsilon/kT = 0.3$ and $q/N = 1/(e^{0.3} - 1) = 2.86$ units per oscillator. Nearly three units each puts lead deep in the high-temperature regime, and its heat capacity, $129$ J/(kg K) or $26.7$ J/(mol K), is within a few percent of $3R$. For copper, $\epsilon/k \approx 240$ K, $q/N = 0.82$, and the heat capacity is still close to $3R$. For diamond, $\epsilon/k \approx 1300$ K, $\epsilon/kT = 4.33$ and $q/N = 1/(e^{4.33} - 1) = 0.013$: barely one oscillator in eighty holds a unit, and the molar heat capacity is $6.1$ J/(mol K), a quarter of $3R$.
The Einstein temperatures themselves track the physics of the atoms: heavy atoms on soft bonds vibrate slowly and have small energy units, light atoms on stiff bonds vibrate fast. Materials engineers use the related Debye temperature — $105$ K for lead, $343$ K for copper, $2230$ K for diamond — to predict how a material's heat capacity and thermal conductivity will behave in cryogenic equipment, where most solids are well below it.
A helium-neon laser, the red beam in a supermarket scanner of the 1980s, works because a pair of energy levels in its neon atoms is driven to a negative temperature. Collisions with excited helium atoms pump neon into an upper level $1.96$ eV above a lower one, and keep it there faster than it can decay, so the upper level holds more atoms than the lower.
In thermal equilibrium at any positive temperature the ratio of populations would be $e^{-\Delta E/kT}$, less than one. Treat the inverted populations as if they came from a temperature and solve: a ratio $N_{\text{upper}}/N_{\text{lower}} = 1.1$ gives $-\Delta E/kT = \ln 1.1 = 0.095$, so $kT = -1.96/0.095 = -20.6$ eV and $T \approx -2.4 \times 10^{5}$ K. That negative temperature is hotter than any positive one: the inverted atoms give their energy to the light passing through, and a photon of the right energy stimulates the emission of another identical to it. The amplification that makes the beam is the flow of energy out of a negative-temperature system, exactly as the entropy argument predicts.
It is natural to think of temperature as a measure of how much energy something holds. It is not, and the definition shows why. A swimming pool at $25$ °C holds vastly more thermal energy than a kettle of boiling water, yet heat flows from the kettle into the pool. What decides the direction is the slope $\partial S/\partial U$: the pool, with its enormous number of molecules, gains a lot of entropy per joule, and the hot kettle very little. Temperature is about how much entropy one more joule buys, which depends on energy per degree of freedom, not on energy in total.
A related slip is to invert the relation: to think that the system whose entropy rises faster is the hotter one. The reciprocal matters. A steep entropy curve means each joule is precious — that system is cold, and it is the one that receives energy.
System A gains $0.0025$ J/K of entropy per joule; system B gains $0.0040$ J/K. Find A's temperature.
$T_A = \dfrac{1}{0.0025} = 400\ \text{K}$
$1/T = \partial S/\partial U$.
Find B's temperature.
$T_B = \dfrac{1}{0.0040} = 250\ \text{K}$
The steeper entropy curve belongs to the colder system.
Find the change in total entropy if one joule moves from A to B.
$\Delta S = -0.0025 + 0.0040 = +0.0015\ \text{J/K}$
A loses what it would have gained; B gains more.
Apply the second law.
$\Delta S > 0 \quad\Rightarrow\quad \text{energy flows from A to B}$
The total entropy rises, so the transfer happens spontaneously.
Say when the flow stops.
$\dfrac{\partial S_A}{\partial U_A} = \dfrac{\partial S_B}{\partial U_B} \quad\Rightarrow\quad T_A = T_B$
As A cools its slope rises and as B warms its slope falls, until they meet.
Write the high-temperature entropy of $N$ oscillators holding $q$ units.
$\dfrac{S}{k} = N\ln\dfrac{eq}{N}$
From $\Omega \approx (eq/N)^N$ when $q \gg N$.
Replace $q$ by the energy.
$q = \dfrac{U}{\epsilon} \quad\Rightarrow\quad S = Nk\ln U + Nk\ln\dfrac{e}{N\epsilon}$
The logarithm of a product splits into a sum.
Differentiate with respect to $U$.
$\dfrac{\partial S}{\partial U} = \dfrac{Nk}{U}$
The second term does not depend on $U$.
Set the slope equal to $1/T$.
$\dfrac{1}{T} = \dfrac{Nk}{U}$
The definition of temperature.
Solve for the energy.
$U = NkT$
Multiply both sides by $TU$.
Compare with equipartition.
$U = N \times 2 \times \tfrac{1}{2}kT = NkT$
Each oscillator has two quadratic terms, kinetic and potential: the counting and equipartition agree.
A solid of $50$ oscillators holds $50$ units of $\epsilon = 0.020$ eV. Write the temperature for one step of energy.
$T \approx \dfrac{\Delta U}{\Delta S} = \dfrac{\epsilon}{k\,\Delta(S/k)}$
One unit added, $\Delta U = \epsilon$.
Find how much one unit multiplies the multiplicity.
$\dfrac{\Omega(51)}{\Omega(50)} = \dfrac{q + N}{q + 1} = \dfrac{100}{51} = 1.961$
All but one factor of each factorial cancels.
Take the logarithm.
$\Delta\left(\dfrac{S}{k}\right) = \ln 1.961 = 0.673$
The entropy gained, in units of $k$.
Find $kT$ in electronvolts.
$kT = \dfrac{0.020}{0.673} = 0.0297\ \text{eV}$
Energy added over entropy gained.
Convert to kelvin.
$T = \dfrac{0.0297}{8.617 \times 10^{-5}} \approx 345\ \text{K}$
Boltzmann's constant is $8.617 \times 10^{-5}$ eV/K.
Compare with Einstein's large-solid formula.
$kT = \dfrac{\epsilon}{\ln(1 + N/q)} = \dfrac{0.020}{\ln 2} = 0.0289\ \text{eV}, \quad T \approx 335\ \text{K}$
Within three percent even for fifty oscillators.
Say what adding energy does.
$q = 100: \quad \Delta\left(\dfrac{S}{k}\right) = \ln\dfrac{150}{101} = 0.396, \quad T \approx 587\ \text{K}$
Each unit now buys less entropy, so the temperature is higher.
Find the two multiplicities.
$\Omega(2) = \binom{4}{2} = 6, \qquad \Omega(3) = \binom{5}{3} = 10$
$q$ dots and two lines.
Find the rise in $S/k$.
$\Delta\left(\dfrac{S}{k}\right) = \ln\dfrac{10}{6} = 0.511$
The logarithm of the ratio.
Invert for the temperature.
For system A, $\partial S/\partial U = 4 \times 10^{-3}$ K$^{-1}$; for system B, $\partial S/\partial U = 5 \times 10^{-3}$ K$^{-1}$. They are put in thermal contact. Which way does energy flow, and which system is hotter?
Complete the worked solution: an Einstein solid of 4 oscillators goes from $3$ to $4$ units. By what factor does its multiplicity grow, and what does that say about its entropy?
Evaluate the multiplicity before.
$\Omega(3) = \binom{6}{3} =$ a
$q$ dots and $N - 1$ lines.
Evaluate the multiplicity after.
$\Omega(4) = \binom{7}{4} =$ b
One more dot in the row.
Divide the second by the first.
$\dfrac{\Omega(q + 1)}{\Omega(q)} =$ r
It should equal $(q + N)/(q + 1)$, which checks both counts.
Turn the ratio into a rise in entropy.
$\Delta S = k\ln\dfrac{\Omega(q + 1)}{\Omega(q)}$
The rise in entropy per unit of energy is what sets the temperature.
Each system's entropy depends on its energy as given, with constants not shown ($N = 4 \times 10^{22}$ in the first two). Match each to what $1/T = \partial S/\partial U$ implies.
| $U = NkT$ | $U = \tfrac{3}{2}NkT$ | $U \propto T^4$ | a negative temperature | |
|---|---|---|---|---|
| Einstein solid, $S = Nk\ln U + \text{const}$ | ||||
| monatomic gas, $S = \tfrac{3}{2}Nk\ln U + \text{const}$ | ||||
| radiation in a box, $S \propto U^{3/4}$ | ||||
| paramagnet where entropy falls as energy rises |
An Einstein solid of three oscillators, with units of $27$ meV, has multiplicities $1, 3, 6, 10, 15$ for $q = 0$ to $4$. For each step up in energy, fill in the rise in $S/k$ and the temperature as $kT/\epsilon$, both to four decimal places.
| rise in $S/k$ | $kT/\epsilon$ | |
|---|---|---|
| from 0 to 1 unit | ||
| from 1 to 2 units | ||
| from 2 to 3 units | ||
| from 3 to 4 units |
The entropy of $2$ mol of a monatomic ideal gas depends on its energy as $S = \tfrac{3}{2}nR\ln U + \text{constant}$. Its energy is $U = 6481.8$ J. What is its temperature? Use $R = 8.31$ J/(mol K).
Answer: K
A small Einstein solid of 30 oscillators holds $60$ units of energy, each $\epsilon = 0.020$ eV. Estimate its temperature, in kelvin, from the change in entropy when one more unit is added. Use $k = 8.617 \times 10^{-5}$ eV/K.
Answer: K
Heat capacity data give silver an Einstein temperature $\epsilon/k \approx 170$ K. At a room temperature of $300$ K, what is the average number of energy units per oscillator?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An Einstein solid of three oscillators, with units of $16$ meV, has multiplicities $1, 3, 6, 10, 15$ for $q = 0$ to $4$. For each step up in energy, fill in the rise in $S/k$ and the temperature as $kT/\epsilon$, both to four decimal places.
| rise in $S/k$ | $kT/\epsilon$ | |
|---|---|---|
| from 0 to 1 unit | ||
| from 1 to 2 units | ||
| from 2 to 3 units | ||
| from 3 to 4 units |
You can find temperatures from how entropy depends on energy. Explain to someone why a swimming pool, holding far more energy than a kettle, still takes heat from it.
15. Your turn: three oscillators go from $2$ to $3$ units of $\epsilon$. Find $kT/\epsilon$., step 3
$\dfrac{kT}{\epsilon} = \dfrac{1}{0.511} \approx 1.96$
Energy added, one unit, over entropy gained.