I have seen case studies of the 3D Debye model where the vibrational modes of a solid is taken to be harmonic with dispersion relation $\omega = c_sk$. It is said that for temperatures much less than the Debye temperature, the heat capacity at constant volume $$C_V\sim T^3$$.
Now I want to show that for bosons with dispersion relation $\omega\sim A\sqrt k$ has heat capacity $$C_V\sim T^4$$ for $T\ll T_{Debye}$.
In the case studies I have read, I can't find where the dispersion relation comes into play. I have no idea how to see this. Please help!
Perhaps I could get a response if I change the question to: How does the dispersion relation come into the calculations for the Debye model? Links to notes or an answer would be very much appreciated!!
@MarkMitchison kindly pointed out that the density of state depends on the dispersion relation. What are the general definitions of $$E, p$$ in terms of $$\omega, k$$? E.g. for the first case $E=\hbar \omega$ and $p=\hbar k$. So the question is: what are the respective values for $E,p$ in general?
Best Answer
For single quanta with an arbitrary dispersion relation $\omega(\vec{q})$, you can always write $E=\hbar \omega(\vec{q})$ and $\vec{p}=\hbar\vec{q}$. Energies might depend on momenta in some way, but de Broglie's relations hold for all single quanta, whether they're bosons like phonons or photons, fermions like electrons or holes, or your funny looking bosons. It really doesn't matter!
Back to your example:
We have a 2-dimensional* gas of bosons with a dispersion $\omega(\vec{k}) = A \sqrt{|\vec{k}|}$ at a fixed volume $V$ (well, surface area, but "volume" is more general). To find the heat capacity, we first calculate the density of states, which is where the dispersion relation comes in, then from that and Bose-Einstein statistics we calculate the total energy U:
$g(E) = \frac{V}{(2\pi)^2} \frac{d^{2}\vec{k}}{dE}$
Since our dispersion relation is isotropic (angularly symmetric) this simplifies to
$= \frac{V}{(2\pi)^2} 2\pi|\vec{k}| (\frac{dE}{d|\vec{k}|})^{-1}$
$\frac{dE}{d|\vec{k}|} = \frac{A\hbar}{2\sqrt{|\vec{k}|}}$, so
$g(E) = \frac{V}{(2\pi)^2} \frac{4\pi|\vec{k}|\sqrt{|\vec{k}|}}{A\hbar}$
For convenience, call $k = |\vec{k}|$, $E = \frac{k^{1/2}}{\hbar A}$ and this simplifies to
$g(E) = \frac{V}{\pi\hbar A} k^{3/2} = \frac{V(\hbar A)^2}{\pi} E^3$
Now, to calculate our total internal energy, $U$:
$U(T) = \int \frac{E}{exp(E/k_{B}T)-1} g(E)dE = \frac{V(\hbar A)^2}{\pi} \int \frac{E^4}{exp(E/k_{B}T)-1} dE$
$ = \int \frac{z^4}{e^z-1} dz \frac{V(\hbar A)^2}{\pi} (k_{B}T)^{5} $
$U(T) \approx V\frac{120\zeta(5)\hbar^{2}A^{2}k_{B}^5}{\pi} T^5$
Where I've substituted $z = E/k_{B}T$ to simplify the numerical integral, and taken the limits of the integral to be $0$ and $\infty$, which is a fair approximation at low temperatures. This is just a numerical constant, though, and you can usually ignore this when working it through by hand and just throw it into Maple or your favourite computer algebra system later. The important thing is that at low temperatures, the internal energy is proportional to the volume $V$ and $T^5$.
Said more simply, $U(T) \propto V T^5$.
To get the volumetric heat capacity $C_V(T)$ from the internal energy $U(T)$, we simply differentiate by $T$ and then divide by the volume $V$,
$C_{V} = \frac{1}{V}\frac{dU}{dT} \propto \frac{1}{V}\frac{d V T^5}{dT}$
$C_{V} \propto T^4$
Which was what we wanted!
*(I found this after completing the derivations for both 2D and 3D. You didn't specify which in your original question, but 2D is consistent with the $T^4$ dependence of $C_V$.)