0
$\begingroup$

Let us say we have a system with two states, energy $0$ and $E>0$. At absolute zero all particles are in the lowest state (of energy $0$). To rise one particle to the state $E$ this would require a temperature, $T\sim \frac{E}{k_B}$ (correct me if I am wrong). Only at this temperature will an infinitesimal change in temperature $dT$ cause a change in entropy $dS$ and thus heat. To change the entropy (and therefore the heat) again we must excite another particle, which will thus require a temperature $T\sim \frac{2E}{k_B}$. Nowhere between these two temperutres will a small change in $T$, $dT$ cause a change in entropy. What I am saying (I think) holds in general for $n$ particles, so we in theory should have a heat capacity that looks something like this:

enter image description here

This is clearly wrong, but why?

$\endgroup$
  • $\begingroup$ Intuitively you are sort of getting it, I think, you just have to use the correct model for solids. See e.g. hyperphysics.phy-astr.gsu.edu/hbase/solids/phonon.html $\endgroup$ – CuriousOne Apr 16 '16 at 9:58
  • 1
    $\begingroup$ How this works is dealt with by the Debye model. You need to read up about it then come back to us with any more specific questions. $\endgroup$ – John Rennie Apr 16 '16 at 10:11
  • $\begingroup$ What you're missing is that at any non-zero temperature the population of a level with energy $E$ is given by the Boltzmann distribution $exp(-E/kT)$. So even very near absolute zero there will still be a small number of particles that get excited to the state and absorb energy in doing so. That means the specific heat rises smoothly as the occupancy of the excited levels rises. $\endgroup$ – John Rennie Apr 16 '16 at 10:15
  • 1
    $\begingroup$ Actually, very near absolute zero the excited state population is given by the Bose-Einstein distribution for bosons or the Fermi-Dirac distribution for fermions. Both distributions approach the Boltzmann distribution as the temperature rises. The Debye model describes phonons, which are bosons, so it uses Bose-Einstein statistics. $\endgroup$ – John Rennie Apr 16 '16 at 10:17
  • $\begingroup$ @JohnRennie Thanks for your comments, I understand now why at higher temperatures, the heat capacity should be smooth and continuous, but near absolute zero I am still confused. I think at $T=0$ we must have $C_V=0$ because a infinitesimal change in temperature can't be used to change the occupation numbers at $T=0$, but it can be at higher temperatures. Is this the correct reasoning? And if so is it possible to show this mathematically? $\endgroup$ – Quantum spaghettification Apr 16 '16 at 10:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.