Heat Capacity of One Einstein Mode

Teach me

One Einstein mode in a heat bath — the ladder, the k_BT ruler, and where the step stands

Guided Inquiry

1 · The ladder and the budget

Let’s get to know the ladder. Every system in this large ensemble climbs the same one, rungs ε = 10 meV apart. The bars give the fraction of systems on each rung. The bracket beside them is the thermal budget k_BT, drawn to the same energy scale. How does the bracket compare with the first gap?

2 · Predict: cool to θ/10

Time for a prediction. Suppose you cool from 40 K to 11.6 K — one tenth of θ. What do you think this mode’s heat capacity C/k_B would do?

3 · Cool it and watch

Let’s try it. A new chart beside the ladder now plots C/k_B against temperature, with a readout to match. Drag T down toward 12 K and watch three things: the bars spill onto the ground rung, the k_BT bracket shrink under the first gap, and the marker walk down the step.

4 · Frozen out, not gone

How did your prediction do? Check it against the C/k_B and P₀ readouts: almost every system now sits on the ground rung and C has collapsed far below k_B. The mode hasn’t vanished — it’s frozen out, waiting for a budget that can buy a rung.

5 · Predict: double the spacing

Let’s move somewhere more interesting. Warm T until the C/k_B readout reads 0.500 — the marker then sits exactly halfway up the step. Now suppose you double ε from 10 to 20 meV. What do you think happens to the step?

6 · Drag the spacing

Let’s find out. A second chart has appeared above C: average energy ⟨E⟩ per system versus T — its knee marks where the budget starts affording rungs. Now drag ε slowly from 10 to 20 meV. Watch the rungs spread against the unchanged bracket while the step below and the knee above slide together — and keep an eye on the parked marker as it gets left behind.

7 · Same shape, new address

So what moved? The half-step marker on the lower chart sits at T = 0.335 θ — read where it moved to. The parked marker fell off the step into freeze-out. Nothing about the shape changed — only where it stands, because C depends on T/θ alone.

8 · Why diamond broke Dulong–Petit

Now for a famous real-world puzzle. At 300 K the ruler reads k_BT ≈ 25.8 meV, yet this widest ladder still delivers C/k_B ≈ 0.82 — check the readout. Diamond’s stiffest vibrations sit near 165 meV, about four times this slider’s widest gap. What do you think that does to C at room temperature?

Listen

Level spacing ε10.0 meV

2 meV40 meV

Temperature T (log scale)40.0 K

5 K1500 K

Readouts

θ = ε/k_B

116.0 K

the temperature the spacing sets

k_BT

3.44 meV

thermal budget — compare it with ε

C/k_B

0.517

heat capacity of this one mode

P₀

0.945

fraction of systems on the ground rung

Fraction of systems on rung n (canonical ensemble, E_n = nε, no zero-point term).

Average energy per system; in the hot limit it approaches k_BT − ε/2, never plain k_BT.

Heat capacity of ONE mode: it climbs monotonically to k_B (equipartition of one mode), never 3k_B.

The scaling collapse: C/k_B depends on T/θ alone, so (T, ε) → (2T, 2ε) leaves it untouched.

The Physics

Heat capacity steps · what this sim shows, the physics behind it, and every control

At a glance

A vibration in a solid can take in energy only in whole steps of size ε. This sim sets that step against the thermal energy k_BT and shows what the comparison does to the mode's heat capacity, the heat it can absorb per kelvin.

What you'll learn

How a ladder of equally spaced levels fills at a given temperature.
How one mode's heat capacity depends on k_BT compared with ε.
How the heat-capacity curve moves when ε changes.
How the same comparison applies to real solids and gases.

One mode: rungs at E_n = nε, bars for the fraction of systems on each rung, and the thermal energy k_BT on the same scale.

What's on screen

Rungs (blue): levels n = 0 to 9 on a fixed 0–120 meV scale.
Bars (orange): the fraction of systems on each rung. The wedge gives the percentage above rung 9.
k_BT bracket (amber): the thermal energy, on the same scale as the rungs.
⟨E⟩ chart (teal): average energy against T, on a log axis.
C/k_B chart (orange): heat capacity against T, with an amber half-step marker. Dots mark the current T.
Readouts: θ = ε/k_B, k_BT, C/k_B and P₀.

Nothing changes with time: each picture is the equilibrium state at the current ε and T.

See itSet ε = 10 meV and warm the bath to 300 K. Watch for: bars on every rung up to n = 9, and the dot on the C/k_B chart close to the dashed line at 1.

The controls

Press ⌖ on a row to see where that control sits in the sim.

Control	What it changes	Range
Level spacing ε	The gap between neighbouring rungs. It also sets θ = ε/k_B.	2–40 meV opens at 10 meV
Temperature T	The temperature of the bath. The slider is logarithmic, so equal steps are equal ratios of T.	5–1500 K opens at 40 K
Readouts	θ = ε/k_B (the spacing written as a temperature), k_BT, C/k_B for this one mode, and P₀, the fraction of systems on the ground rung.	panel
↻ Reset	Returns ε and T to their opening values.	top bar

Fixed: the ladder shows rungs n = 0 to 9 on a 0–120 meV scale, and the chart axes never rescale, so a curve can slide but not stretch. There is no zero-point energy: E₀ = 0.

The ladder in a heat bath

The bars are not one system jiggling. They show how a large ensemble of identical modes is spread over the rungs when it is in equilibrium with a bath at temperature T. One sum over the whole ladder, and its derivatives, give everything the sim shows:

Z is the partition function, β = 1/k_BT. The sum runs over every rung, not just the ten drawn.

The rung populations. P₀ is the sim's P₀ readout.

x compares the gap with the thermal energy. Each step up the ladder multiplies the population by e^−x.
θ = ε/k_B is the spacing written as a temperature, 11.60 K for each meV. For ε = 10 meV, θ = 116.0 K.
At T = θ, k_BT = ε. Each rung then holds 1/e ≈ 0.37 of the one below, and P₀ = 0.632.

See itSet the temperature to θ, so that k_BT = ε. Watch for: k_BT equal to ε in the card, and P₀ reading 0.632.

Why a cold mode stops absorbing heat

Heat capacity is the energy the mode takes in per kelvin of warming, C = d⟨E⟩/dT. Energy goes in only by lifting systems up the ladder.

When k_BT is much smaller than ε (x ≫ 1), almost every system is on the ground rung. A small rise in T lifts only the few in the Boltzmann tail, a fraction e^−x of the ensemble, so the mode takes in almost nothing. It is frozen out: the rungs are still there, but the budget cannot pay for the first one.

At T = θ/10 (x = 10) this gives 0.00454, and so does the exact expression.

The fall is exponential in x, not a power of T. That is why the C/k_B curve sits near zero and then climbs steeply.

The same mode, ε = 10 meV, at three temperatures: θ/10, 40 K and 1500 K.

See itCool the mode to T = θ/10, keeping your ε (held at 5 meV or more so that θ/10 stays on the slider). Watch for: C/k_B dropping below 0.01 while P₀ climbs to 1.

The plateau and the half-step

Differentiating ⟨E⟩ gives the whole curve in one variable:

ε T

These sliders move the sim too. T is on a log scale, 5–1500 K.

The plateau is k_B per mode: ½k_B from kinetic and ½k_B from potential energy. Dulong–Petit's 3R per mole counts three such modes for each atom.
The half-step, C = ½k_B, is at x = 2.983, so T½ = 0.3352 θ. For ε = 10 meV that is 38.9 K, the amber marker on the C chart.
The approach is quick. C/k_B = 0.92 at T = θ and 0.999 at T = 10θ.
⟨E⟩ approaches k_BT − ε/2, not k_BT, because the lowest rung is at E = 0.

See itPark T on the half-step for your ε. Watch for: C/k_B reading 0.500, with the dot on the amber half-step marker.

See itWarm the mode to ten times θ (ε capped at 12 meV so that 10θ fits on the slider). Watch for: the C/k_B dot resting on the dashed line at 1.

What changing ε does

C/k_B depends on T and ε only through x = θ/T, so doubling both leaves it unchanged:

The step slides; its shape and height stay. Doubling ε doubles θ, and the whole curve moves to twice the temperature. The plateau is still k_B. The half-step goes from 38.9 K to 77.8 K.
⟨E⟩ scales differently. It is ε times a function of x, so doubling T and ε doubles ⟨E⟩. At T = θ it is 5.82 meV for ε = 10 meV and 11.64 meV for ε = 20 meV.

C/k_B for ε = 10 and 20 meV on the sim's axes. The curves have the same shape.

See itFrom the half-step at ε = 10 meV, double ε and T together. Watch for: C/k_B reading the same before and after, while the half-step marker moves to twice the temperature.

Where it matters

One curve serves every ladder. The dots mark where T = 300 K falls for four of them.

Many simple solids hold about 3R ≈ 24.9 J mol⁻¹ K⁻¹ at room temperature (Dulong–Petit, 1819): three modes per atom, each on its k_B plateau.
Diamond has stiff bonds, so ε is large. An Einstein fit gives θ_E ≈ 1320 K (ε ≈ 114 meV), and C/k_B ≈ 0.24 per mode at 300 K, about a quarter of the classical value.
Hydrogen gas. The H₂ bond vibration has θ ≈ 6330 K, so at 300 K it is frozen out and C_V ≈ 5/2 R, not 7/2 R.

This sim models one mode. Real solids and molecules combine many.

Go deeper: where one ladder stops being enough

Low T in solids. Real solids follow C ∝ T³ because their lowest-frequency modes have arbitrarily small ε. One fixed ladder falls off exponentially instead. Debye's model repairs this.
One ε is a stand-in for diamond. Its vibrations form a band that reaches about 165 meV, where the same formula gives C/k_B ≈ 0.07 at 300 K, against 0.24 for the fitted 114 meV.
Zero-point energy. With levels at (n + ½)ε, ⟨E⟩ rises by ε/2 and C does not change.

Summary

In one line. A mode can absorb heat only if k_BT can pay for a rung, so its heat capacity depends on T/θ alone: near zero when k_BT ≪ ε, and k_B when k_BT ≫ ε.

Key points

Rung populations fall as e^−nx, with x = θ/T.
C/k_B = x²e^x/(e^x − 1)²: about x²e^−x for x ≫ 1, 1 for x ≪ 1, and ½ at T = 0.3352 θ.
ε sets where the step stands, not its height or shape.
One mode plateaus at k_B. A solid's 3k_B per atom is three modes.

Common mix-ups

"Heat capacity is a fixed property of a material." The same ladder gives C/k_B = 0.005 at θ/10 and 0.999 at 10θ.
"A frozen-out mode has gone away." The rungs are still there. Warm the bath and the full k_B returns.
"The plateau is 3k_B." That is per atom, which has three modes. One mode plateaus at k_B.

Things to try on your own

Predict T½ for ε = 25 meV, then find where C/k_B reads 0.500 and compare.
At ε = 10 meV, raise T until P₀ reads 0.632 and compare that T with θ.
Set ε = 40 meV and T = 300 K and read C/k_B. Then work out which ε would put the half-step at 300 K, and whether the slider reaches it.

Predict before reveal	The C/k_B chart and readout stay hidden until the freeze-out prediction is committed; the star control is predicted before it is dragged.
Misconception as a choice	“Heat capacity is a fixed property” is a selectable option; its rebuttal appears only to the student who picks it.
Invariant made visible	C/k_B depends on T/θ alone — on a log-T axis doubling ε translates the step rigidly, shape preserved to the last digit.
Fixed axes, honest scale	Energy axis 0–120 meV, both temperature axes log 5–1500 K, C axis 0–1.1: nothing autoscales, so the slide is real and readable.
Operational quantities	The k_BT bracket is drawn to the rungs' own energy scale, and every bar is a counted fraction of the ensemble.
Honest tail	Probability above the drawn rungs is shown as a fading wedge with its percentage, never renormalized away.
Readout discipline	Two readouts at load; C/k_B and P₀ appear only at the step that asks the student to read them.
Toggleable formalism	Boltzmann factors, ⟨E⟩, C and the scaling identity live behind ⚛︎ The Physics, so the sim works before or after the algebra.