Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here by "low temperature" I meant it in the scale of the characteristic $\hbar \omega$ of the system.

One can calculate and show that in the low temperature regime $C_V$ of phonons goes like $T^3$ for free fermions it goes as $T$ and for free bosons its like $T^{\frac{3}{2}}$ In most of the above cases the power comes from some complicated calculation.

  • I was wondering if there is a heuristic argument for "understanding" these powers.

  • Further is the power of $T$ characteristic of the system? Like if in a substance one sees the low temperature behaviour of $C_V$ to be going as $T$ then does it says something like that the effective degree of freedom of the system is that of free fermions?

  • Are there other examples of such calculation? Like typical systems where one can calculate the low temperature dependance of $C_V$?

share|cite|improve this question
I think your $T$-dependencies are wrong: You cannot say that free bosons have such-and-such temperature dependence of $C_V$ if you don't know what the dispersion relation of them is. Photons are bosons, but so are hydrogen atoms, and they have very different heat capacities – Lagerbaer Aug 15 '11 at 0:35
up vote 2 down vote accepted

First, since $C_V = \left(\frac{\partial U}{\partial T}\right)_V$, the problem of looking for the power of $T$ in $C_V$ is equivalent to the problem of looking for the power of $T$ in $U$, the internal energy of your system.

This energy is given by an integral over the density of states $g(E) dE$ and the appropriate distribution function (Fermi-Dirac or Bose-Einstein):

$$U = \int dE E \cdot g(E) n(E)$$

The distribution function includes an exponential of the form $E/kT$, which we replace by a new variable $x$, so that $E = x kT$ and $dE/dx = kT$.

Hence, if $g(E) \sim E^a$, you end up with a factor $T^{(a + 2)}$. All other components in your integral don't depend on $T$, so you have $U \sim T^{a+2}$ and $$C_V \sim T^{a+1}$.

So the only thing you have to figure out is how the density of states scales with energy. For a three dimensional photon gas, for example, we have a density of states that is proportional to $E^2$, so the specific heat is proportional to $T^{2 + 1} = T^3$.

For a two-dimensional free electron gas, we have $g(E) = const$, so the heat capacity should be proportional to $T^{0+1} = T$. For a one-dimensional electron gas, we have $g(E) \sim E^{-1/2}$, so $C_V \sim T^{-1/2 + 1} = T^{1/2}$.

Conversely, if you measure the low temperature behavior, this effectively means you measure $a$. This tells you what the density of states approximately scales like, at least in the lower energy regime. This does not tell you, however, what sort of elementary excitations you are looking at, as a peculiar band structure of the electrons could in principle give rise to a density of states that you'd normally expect in a photon gas. (Graphene would be an example, with its linear dispersion at the Dirac point).

To your last question: Yes. For every system where you can come up with a density of states, you can then carry out the calculation and determine the specific heat. Another examples would be a ferromagnet at low temperature, whose elementary excitations are spin-waves (magnons).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.