Specific Heat of Solids

Classical physics predicts that every solid has the same molar heat capacity, $3Nk$, the law of Dulong and Petit, and at room temperature most do. But cool a crystal toward absolute zero and its heat capacity collapses, something classical equipartition cannot explain. Einstein gave the first quantum account by treating every atom as an independent oscillator of a single frequency: the modes freeze out when $kT$ drops below $\hbar\omega_E$, and the heat capacity falls. It was the right idea but the wrong detail. Einstein's curve drops exponentially at low temperature, while real solids follow a gentler power law. Debye fixed it by replacing the single frequency with a whole spectrum of sound waves up to a cutoff, the Debye frequency, and the long-wavelength acoustic modes survive to very low temperature and give the famous $T^3$ law. The scene plots both heat capacities against temperature, sharing the Dulong-Petit plateau at high $T$ but parting ways in the cold, and a sweeping cursor reads each one off. The lower panel puts them on log-log axes, where the Debye curve becomes a straight line of slope three (the $T^3$ law) and the Einstein curve bends sharply away below it: the same data that told physicists in 1912 which model nature actually uses.

Figure 1. Specific heat of solids. Top: the lattice vibrating with thermal amplitude, and the phonon mode spectrum coloured by each mode's activation (warm = active, contributing k to C; cool = frozen). As kT sweeps right, more modes switch on; the warm fraction of the modes is C/3Nk. The Debye spectrum is a continuum g(omega) proportional to omega squared up to omega_Debye; the Einstein model puts every mode at one frequency. Bottom: the resulting heat capacities versus temperature, with the Dulong-Petit plateau and the Debye T^3 asymptote. Source: Ashcroft and Mermin, Solid State Physics, Ch. 23.

temperature T200 K

Debye temperature400 K

Einstein temperature320 K

WHAT TO TRY

Raise the temperature $T$: the $kT$ line sweeps right, more modes turn warm (active), the lattice jiggles harder, and $C/3Nk$ climbs toward the Dulong-Petit value of 1.
Drop $T$ low: the Einstein spike freezes out all at once (its single frequency needs $kT \gtrsim \hbar\omega_E$), while the Debye continuum keeps a few low-frequency modes active, giving the $T^3$ tail.
Compare the warm fraction of the spectrum with the heat capacity below: they are the same number, because $C$ is the activation-weighted area of the density of states.
Raise the Debye temperature (a stiffer, lighter crystal like diamond): every mode needs more heat to activate, so the whole curve shifts to higher $T$.