--- title: Thermal Properties — Einstein and Debye Models module: Lattice Dynamics moduleNumber: 4 lessonNumber: 3 order: 403 summary: > The lattice heat capacity follows the classical Dulong-Petit value at high temperature but collapses toward zero as T approaches zero, a purely quantum effect. This lesson derives that behavior from the Einstein model of a single frequency, then the Debye model of a linear phonon spectrum with a cutoff, obtaining the Debye T-cubed law at low temperature and the Debye interpolation across all temperatures, and closes with thermal expansion and the Gruneisen parameter. topics: [Lattice Dynamics] draft: false sources: - book: Kittel ref: "Ch. 5 — Phonons II: Thermal Properties" - book: Ashcroft & Mermin ref: "Ch. 23 — Quantum Theory of the Harmonic Crystal" - book: Simon ref: "Ch. 2–3 — Specific Heat of Solids: Einstein and Debye" --- The heat capacity of an insulating solid is the clearest test of lattice dynamics. Classical statistical mechanics predicts a constant value independent of temperature; experiment finds that value only at high temperature and sees the heat capacity fall to zero as $T\to 0$, proportional to $T^3$. The resolution is the quantization of the [phonon modes](/condensed-matter/lattice-dynamics/phonons-quantization-and-dos): modes with $\hbar\omega \gg k_B T$ are frozen out and stop absorbing heat. This lesson works through two models of increasing fidelity — Einstein's single frequency and Debye's linear spectrum — and extracts the universal low- temperature law. ## The Dulong–Petit law and its failure The lattice energy is a sum over modes of the mean phonon energy. Using the Bose occupation and including the density of states $g(\omega)$, $$ U = \int_0^\infty \d\omega\;g(\omega)\,\hbar\omega \left(\frac{1}{e^{\hbar\omega/k_B T} - 1} + \frac{1}{2}\right). $$ The zero-point term is temperature-independent and drops out of the heat capacity $C = (\partial U/\partial T)_V$. Classically, equipartition assigns $k_B T$ to each of the $3N$ vibrational modes (kinetic plus potential), giving $U = 3N k_B T$ and the constant heat capacity known as the **Dulong–Petit law**: $$ C_{\text{DP}} = 3N k_B = 3n R, $$ with $n$ the number of moles and $R = N_A k_B = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$. The molar value $3R \approx 24.9\ \text{J mol}^{-1}\text{K}^{-1}$ matches most solids at room temperature. Below a material-specific temperature it fails: $C$ drops steadily toward zero. Equipartition assumes every mode carries $k_B T$, but a mode of frequency $\omega$ holds that energy only when $k_B T \gtrsim \hbar\omega$. As $T$ falls, the high-frequency modes freeze first, then progressively lower ones, and the heat capacity follows them down. $$ % caption: Molar heat capacity versus temperature. At high temperature C % approaches the classical Dulong-Petit plateau 3R; below a characteristic % temperature the modes freeze out and C falls smoothly to zero, vanishing as T % cubed at the lowest temperatures. \begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black!70] (0,0) -- (6.4,0) node[below right] {$T$}; \draw[->, black!70] (0,0) -- (0,3.4) node[left] {$C$}; % plateau line 3R \draw[black, dashed] (0,2.8) -- (6.2,2.8); \node[anchor=east] at (-0.05,2.8) {$3R$}; % C(T) curve: saturating rise, T^3 at low T \draw[acc, very thick, domain=0:6.1, samples=160, variable=\t] plot ({\t},{2.8*\t*\t*\t/(\t*\t*\t + 9)}); \node[acc, anchor=west] at (3.6,2.35) {$C(T)$}; \node[black, anchor=west] at (0.35,0.55) {$T^3$}; \end{tikzpicture} $$ ## The Einstein model Einstein modelled every atom as an independent three-dimensional oscillator of a single frequency $\omega_E$, so $g(\omega) = 3N\,\delta(\omega - \omega_E)$. The energy and heat capacity follow from one oscillator times $3N$: $$ U = 3N\hbar\omega_E\left(\frac{1}{e^{\hbar\omega_E/k_B T}-1}+\frac12\right), \qquad C = 3N k_B\left(\frac{\Theta_E}{T}\right)^2 \frac{e^{\Theta_E/T}}{\bigl(e^{\Theta_E/T}-1\bigr)^2}, $$ where the **Einstein temperature** $\Theta_E = \hbar\omega_E/k_B$ sets the scale. Two limits check the result. For $T \gg \Theta_E$, expanding the exponentials gives $C \to 3N k_B$, recovering Dulong–Petit. For $T \ll \Theta_E$, $$ C \approx 3N k_B\left(\frac{\Theta_E}{T}\right)^2 e^{-\Theta_E/T}, $$ which does fall to zero, capturing the qualitative freeze-out. The exponential is too abrupt, though: real solids follow $T^3$, not $e^{-\Theta_E/T}$. The Einstein model fails at low $T$ because it gives every mode the same high frequency, so all modes freeze together. In a real crystal the acoustic branches extend to arbitrarily low frequency, and those low-frequency modes stay thermally active down to much lower temperature. Einstein's picture works well for the nearly dispersionless **optical** branches, which do cluster near one frequency. ## The Debye model Debye kept the low-frequency acoustic modes that Einstein discarded. He replaced the true dispersion with a linear one, $\omega = v_s\,k$, valid for the long- wavelength acoustic modes that dominate at low $T$, and used it for all modes. The density of states of a linear, isotropic spectrum in three dimensions is $$ g(\omega) = \frac{V}{2\pi^2}\,\frac{3\omega^2}{v_s^3} = \frac{9N}{\omega_D^3}\,\omega^2, \qquad \omega \le \omega_D, $$ quadratic in $\omega$. Because a crystal of $N$ atoms has only $3N$ modes, the spectrum must be cut off at a maximum frequency, the **Debye frequency** $\omega_D$, fixed by $\int_0^{\omega_D} g(\omega)\,\d\omega = 3N$. The corresponding **Debye temperature** is $\Theta_D = \hbar\omega_D/k_B$. > **Definition (Debye frequency and temperature).** The **Debye frequency** > $\omega_D$ is the cutoff imposed on the linear acoustic spectrum so that the > total number of modes equals $3N$: $\omega_D = v_s\,(6\pi^2 N/V)^{1/3}$. The > **Debye temperature** $\Theta_D = \hbar\omega_D/k_B$ marks the crossover between > classical ($T \gg \Theta_D$) and quantum ($T \ll \Theta_D$) behavior of the > lattice. It ranges from about $105\ \text{K}$ for soft, heavy metals such as > lead to $2200\ \text{K}$ for diamond, tracking the stiffness and light atomic > mass that raise the sound speed. $$ % caption: Debye density of states. The quadratic curve g proportional to omega % squared is truncated abruptly at the Debye frequency omega_D, the cutoff that % keeps the total mode count at 3N. The shaded area equals 3N. \begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black!70] (0,0) -- (5.0,0) node[below right] {frequency}; \draw[->, black!70] (0,0) -- (0,3.2) node[left] {$g$}; % quadratic up to cutoff at x=3.2 \fill[acc!12] (0,0) -- plot[domain=0:3.2, samples=80, variable=\x] ({\x},{0.28*\x*\x}) -- (3.2,0) -- cycle; \draw[very thick, domain=0:3.2, samples=80, variable=\x] plot ({\x},{0.28*\x*\x}); % cutoff vertical \draw[acc, very thick] (3.2,0) -- (3.2,{0.28*3.2*3.2}); \draw[black, dashed] (3.2,0) -- (3.2,2.9); \node[anchor=north] at (3.2,-0.05) {limit}; \node[anchor=east] at (2.6,1.4) {quadratic}; \end{tikzpicture} $$ The energy integral becomes, with $x = \hbar\omega/k_B T$, $$ U = \frac{9N}{\omega_D^3}\int_0^{\omega_D}\frac{\hbar\omega^3}{e^{\hbar\omega/k_B T}-1} \,\d\omega = 9N k_B T\left(\frac{T}{\Theta_D}\right)^3 \int_0^{\Theta_D/T}\frac{x^3}{e^x - 1}\,\d x. $$ Differentiating gives the Debye heat capacity $$ C = 9N k_B\left(\frac{T}{\Theta_D}\right)^3 \int_0^{\Theta_D/T}\frac{x^4 e^x}{(e^x-1)^2}\,\d x, $$ a universal function of the single ratio $T/\Theta_D$. All monatomic solids collapse onto one curve when $C$ is plotted against $T/\Theta_D$. ## The T³ law The low-temperature limit is the model's triumph. As $T\to 0$ the upper limit $\Theta_D/T\to\infty$ and the integral becomes a constant, $$ \int_0^\infty \frac{x^3}{e^x-1}\,\d x = \frac{\pi^4}{15}. $$ The energy is then $U = \tfrac{3}{5}\pi^4 N k_B T\,(T/\Theta_D)^3$, and the heat capacity is the **Debye $T^3$ law** $$ C = \frac{12\pi^4}{5}\,N k_B\left(\frac{T}{\Theta_D}\right)^3 \approx 234\,N k_B\left(\frac{T}{\Theta_D}\right)^3. $$ The physical origin is a mode count. At temperature $T$ only modes with $\hbar\omega \lesssim k_B T$ are thermally active, i.e. those inside a sphere of radius $k_{\text{th}} \sim k_B T/\hbar v_s$ in $\vec k$-space. The number of such modes scales as the sphere volume, $k_{\text{th}}^3 \propto T^3$, and each contributes about $k_B$, so $C \propto T^3$. The exponent counts the three dimensions of $\vec k$-space that the acoustic modes fill. > **Worked example.** Estimate the Debye temperature of copper from its low- > temperature heat capacity. The molar lattice heat capacity obeys $C_{\text{lat}} > = 234\,R\,(T/\Theta_D)^3$. At $T = 10\ \text{K}$ the measured heat capacity of > copper is $C = 0.055\ \text{J mol}^{-1}\text{K}^{-1}$, of which the electronic > term $\gamma T = (0.695\ \text{mJ mol}^{-1}\text{K}^{-2})(10\ \text{K}) = 0.007\ > \text{J mol}^{-1}\text{K}^{-1}$ must be subtracted, leaving a lattice > contribution $C_{\text{lat}} = 0.048\ \text{J mol}^{-1}\text{K}^{-1}$. Then > $C_{\text{lat}}/R = 0.048/8.314 = 5.77\times10^{-3}$, so > $$ > \left(\frac{T}{\Theta_D}\right)^3 = \frac{5.77\times10^{-3}}{234} > = 2.47\times10^{-5}, > \qquad \frac{T}{\Theta_D} = 0.0292, > $$ > giving $\Theta_D = 10/0.0292 \approx 343\ \text{K}$, matching the tabulated value > for copper. Subtracting the electronic term first is essential: at $10\ \text{K}$ > it is more than a tenth of the total heat capacity. ## The full interpolation and comparison Between the two limits the Debye integral must be evaluated numerically, but it interpolates smoothly from $C = 3N k_B$ at high $T$ to $C \propto T^3$ at low $T$. The contrast with Einstein is sharpest at low temperature, where the Einstein curve drops exponentially and the Debye curve as a power law, matching data. $$ % caption: Heat capacity of the Einstein and Debye models against reduced % temperature. Both reach the Dulong-Petit plateau at high T. At low T the % Einstein curve (dashed) falls exponentially and undershoots the data, while the % Debye curve (solid) follows the correct T-cubed power law. \begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black!70] (0,0) -- (6.2,0) node[below right] {reduced temperature}; \draw[->, black!70] (0,0) -- (0,3.4) node[left] {$\frac{C}{3Nk_B}$}; \draw[black, dashed] (0,2.8) -- (6.0,2.8); \node[anchor=east] at (-0.05,2.8) {$1$}; % Debye (solid): T^3 rise then saturate \draw[acc, very thick, domain=0:6.0, samples=160, variable=\t] plot ({\t},{2.8*\t*\t*\t/(\t*\t*\t + 2.2)}); % Einstein (dashed): later, sharper rise \draw[densely dashed, thick, domain=0.5:6.0, samples=160, variable=\t] plot ({\t},{2.8*exp(-1.4/\t)}); \node[acc, anchor=west] at (2.6,2.55) {Debye}; \node[anchor=north west] at (3.0,1.3) {Einstein}; \end{tikzpicture} $$ A clean way to separate the lattice from the electronic term in a metal plots $C/T$ against $T^2$. The total low-temperature heat capacity is $$ C = \gamma T + \beta T^3, \qquad C/T = \gamma + \beta T^2, $$ with $\gamma T$ the electronic term (linear in $T$) and $\beta T^3$ the Debye lattice term. The plot is a straight line whose intercept is $\gamma$ and whose slope $\beta$ gives $\Theta_D$ through $\beta = (12\pi^4/5)N k_B/\Theta_D^3$. $$ % caption: Separating the electronic and lattice terms. Plotting C over T % against T squared linearizes C = gamma T + beta T cubed: the intercept is the % electronic coefficient gamma, and the slope beta fixes the Debye temperature. \begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black!70] (0,0) -- (5.6,0) node[below right] {$T^2$}; \draw[->, black!70] (0,0) -- (0,3.2) node[left] {$\frac{C}{T}$}; % straight line with positive intercept \draw[acc, very thick] (0,0.8) -- (5.2,2.9); % intercept marker \fill[black] (0,0.8) circle (2pt); \draw[black, dashed] (0,0.8) -- (-0.02,0.8); \node[anchor=east] at (-0.1,0.8) {intercept}; % slope annotation \node[black, anchor=north west] at (2.6,2.2) {slope}; \draw[black] (3.2,1.85) -- (4.2,1.85) -- (4.2,2.25); \end{tikzpicture} $$ ## Thermal expansion and the Grüneisen parameter A strictly harmonic crystal would not expand: the mean position of each atom sits at the symmetric minimum of a parabola regardless of amplitude. Real solids expand because the mode frequencies depend on volume — a mild anharmonic effect captured by keeping harmonic modes but letting each $\omega_s(\vec k)$ shift as the lattice is compressed or dilated (the quasiharmonic approximation). The sensitivity is the **Grüneisen parameter** $$ \gamma_G = -\frac{\d\ln\omega}{\d\ln V}, $$ the fractional drop in a mode frequency per fractional increase in volume, of order unity for most solids. Thermodynamics then ties the linear thermal- expansion coefficient $\alpha$ to the heat capacity: $$ \alpha = \frac{\gamma_G\,C_V}{3 B V}, $$ with $B$ the bulk modulus and $C_V$ the heat capacity per unit volume; the volume coefficient is $3\alpha$. The proportionality to $C_V$ predicts that thermal expansion, like heat capacity, falls as $T^3$ at low temperature and saturates at high temperature — a link confirmed across many materials. The [anharmonic mechanism](/condensed-matter/lattice-dynamics/anharmonicity-and-thermal-transport) behind this expansion, and behind the finite thermal conductivity that pure harmonic theory cannot produce, is the subject of the next lesson.