# Electron heat capacity vs phonon heat capacity

For metals, the heat capacity has two contributions. The phonon contribution is proportional to $$T^3$$, while the electron contribution is proportional to $$T$$, as $$T$$ goes to zero. Therefore, for sufficiently low $$T$$, the electron contribution will dominate, while for sufficiently high $$T$$, the phonon contribution will dominate.

Does anyone know any specific values of such transition or crossover temperature? Is it typically on the order of $$K$$ or $$10K$$?

Heat capacity data for elements is well researched. Picking Cu as a good metal where one would expect the electron contribution to be large, Google (Googled 'Cu heat capacity vs temperature') leads quickly to nist.gov with a copy of (from Web Of Science search):

HEAT-CAPACITY OF REFERENCE MATERIALS - CU AND W

By: WHITE, GK; COLLOCOTT, SJ

JOURNAL OF PHYSICAL AND CHEMICAL REFERENCE DATA Volume: ‏ 13 Issue: ‏ 4 Pages: ‏ 1251-1257

As noted in the introduction,

For the reference solids discussed here, the vibrational energy is the major contribution above liquid-helium temperatures.

One does not normally consider W to be a great metal, electron conduction wise, so I think the principle applies broadly, so under 10K, closer to 4K.

I believe you can make an approximate calculation of this temperature for a free electron case. From Kittel (or many other sources), the approximation for heat capacity of phonons as low temperature is: $$C_{ph}=234N_{ph}k_B(\frac{T^3}{\theta^3})$$ Where $$\theta$$ is the Debye Temperature and $$k_B$$ is Boltzmann's constant, and $$N_{ph}$$ is the number or primitive cells in the specimen.

For a metal, in the free electron case: $$C_{el} = 0.5\pi^2N_{el}k_B(\frac{T}{T_F})$$ Where $$T_F$$ is the metals Fermi Temperature and $$N_{el}$$ in this case is the total number of orbitals in the free electron sphere (including spin).

So for the crossover point you equate these two expressions and solve for T and you get: $$T=\sqrt{\frac{\pi^2\theta^3N_{el}}{468N_{ph}T_F}}$$ If you have one electron per unit cell, then I believe that $$N_{ph}$$ and $$N_{el}$$ should cancel out. If you have 2 electrons per unit cell, then this ratio is 2 and etc.

Sodium, for which a free electron approximation should be good, has a Debye Temperature of 158 K, a Fermi Temperature of 37500 K and 1 electron per unit cell. Thus T calculates out to be approximately 1.5 K - if I've done the math on my calculator correctly!

Note: The actual formula for the electron heat capacity is: $$C_{el}=\frac{1}{3}\pi^2D(\epsilon_F)k_B^2T$$ Thus if you know the density of states at the Fermi Level for the metal of interest, $$D(\epsilon_F)$$ you can make a better approximation.