CND-13

Einstein solid heat capacity

C_V = 3 N k_B (θ_E/T)² e^{θ_E/T} / (e^{θ_E/T}−1)². Independent oscillators.

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PhononsEinstein solid

Governing equation

C_V=3Nk_B\left(\dfrac{\theta_E}{T}\right)^2\dfrac{e^{\theta_E/T}}{(e^{\theta_E/T}-1)^2}

$T \in \left[5, 600\right]$

where

\theta_E: Einstein temperature (K)
T: Temperature (K)
n: Amount (mol)
C_V: Heat capacity (J/K)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

C_V=3Nk_B\left(\dfrac{\theta_E}{T}\right)^2\dfrac{e^{\theta_E/T}}{(e^{\theta_E/T}-1)^2}

$T \in \left[5, 600\right]$

$T = 5$ $C_V = 1.7\times 10^{-13}$

Along the curve

②

Domain of the variable — blue

T_{\min} = 5,\quad T_{\max} = 600

$T \in \left[5, 600\right]$

$T = 5$ $T = 5$

Along the curve

③

Response curve — green

C_V=3Nk_B\left(\dfrac{\theta_E}{T}\right)^2\dfrac{e^{\theta_E/T}}{(e^{\theta_E/T}-1)^2}

$T \in \left[5, 600\right]$

$T = 5$ $C_V = 1.7\times 10^{-13}$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}C_V}{\mathrm{d}T}

$T \in \left[5, 600\right]$

$T = 8.742$ $C_V^{\prime} = 9.43\times 10^{-5}$

Along the curve

Lecture brief

Historical brief

Drude electrons, Bloch waves, BCS pairing (1957) and Wiedemann–Franz heat are the first solids-and-metals laws. The lab is conductivity, gap and phonon heat in closed form. This sheet (CND-13 — Einstein solid heat capacity) is the form associated with Einstein solid. Working symbols: $\theta_E$ , $T$ , $n$ $\rightarrow$ $C_V$ . Einstein (1907) explained why C_V vanishes as T → 0, before Debye's T³.

Purpose

Purpose: compute

C_V

from

\theta_E

T

n

in Condensed matter via

C_V=3Nk_B\left(\dfrac{\theta_E}{T}\right)^2\dfrac{e^{\theta_E/T}}{(e^{\theta_E/T}-1)^2}

C_V = 3 N k_B (θ_E/T)² e^{θ_E/T} / (e^{θ_E/T}−1)². Independent oscillators. Use it when a real condensed matter question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

\theta_E = 200.000\,\mathrm{K}

T = 80.000\,\mathrm{K}

n = 1.000\,\mathrm{mol}

, the governing relation

C_V=3Nk_B\left(\dfrac{\theta_E}{T}\right)^2\dfrac{e^{\theta_E/T}}{(e^{\theta_E/T}-1)^2}

yields

C_V = 15.187\,\mathrm{J/K}

. A grid of oscillators, a saturating C. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Einstein temperature \theta_E200 KTemperature T80 KAmount n1 mol

Outputs

Heat capacity C_V15.187 J/K

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Narration of this film

A grid of oscillators, a saturating C.

Einstein (1907) explained why C_V vanishes as T → 0, before Debye's T³.

Reading speed

Watch on YouTube