Specific Heat

Hey everyone.
I promised to complete a write-up about our findings in class, and here I am.  Earlier, we found the relationship between the total energy (let's call this $U$ for now) and the density of states $D(E)$ :
$$U = \int _{\text{bottom}} ^{\text{top}} D(E) \, f(E) \, E \, dE \quad \text{where} \quad f(E) = \frac{1}{1+\exp{[(E-\mu)/kT]}}$$
For neatness' sake, I'll use $E_b$ and $E_t$ to denote the bottom and the top of the band, respectively.
Q: How do we relate this to specific heat?
From $C_V = \frac{\Delta U}{\Delta T}$ (the specific heat at constant volume), we can write
$$C_V = \frac{\partial}{\partial T} \int_{E_b}^{E_t} D(E)\,f(E)\,E\,dE \\  \int_{E_b}^{E_t} D(E)\, \frac{\partial f(E)}{\partial T}\,E\,dE. \\ \text{Given that} \quad \frac{\partial f(E_b)}{\partial T}D(E_b) \approx 0 \quad \text{and} \quad \frac{\partial f(E_t)}{\partial T}D(E_t) \approx 0 \quad \text{rather strongly,}$$
we can justify $D(E) \approx D(E_0) = constant$, the density of states at the ground energy.  So,
$$C_V \approx D(E_0) \int_{E_b}^{E_t} \frac{\partial f(E)}{\partial T}\,E\,dE = D(E_0) \int_{E_b}^{E_t} \frac{\frac{E-\mu}{kT^2}\exp{[(E-\mu)/kT]}}{(1+\exp{[(E-\mu)/kT])^2}}\,E\,dE$$
where $\mu$ is something we didn't really get to discuss in depth... anyway,
$$C_V \approx \frac{D(E_0)k}{T} \int_{E_b}^{E_t} \frac{(E-\mu)\exp{[(E-\mu)/kT]}}{(1+\exp{[(E-\mu)/kT])^2}}\,E\,\frac{dE}{kT}.$$
We can make this integration easier by letting our limits go to infinity instead.  Physically, we're only including negligible contributions by doing this, so I'm pretty comfortable with that.  Furthermore, substituting  $x = (E-\mu)/kT$  and  $dx = dE/kT$,  we get
$$C_V \approx \frac{D(E_0)(kT)^2}{T} \int_{-\infty}^{\infty} \frac{\frac{E}{kT}xe^x}{(1+e^x)^2}\,dx  \\   \approx  D(E_0)k^2T \left[ \int_{-\infty}^{\infty} \frac{\frac{E-\mu}{kT}xe^x}{(1+e^x)^2}\,dx + \int_{-\infty}^{\infty} \frac{\frac{\mu}{kT}xe^x}{(1+e^x)^2}\,dx \right] \\ \approx D(E_0)k^2T \left[ \int_{-\infty}^{\infty} \frac{x^2e^x}{(1+e^x)^2}\,dx + \frac{\mu}{kT} \int_{-\infty}^{\infty} \frac{xe^x}{(1+e^x)^2}\,dx \right] \\ \approx D(E_0)k^2T \int_{-\infty}^{\infty} \frac{x^2e^x}{(1+e^x)^2}\,dx$$
since the second integral indeed cancels out by asymmetry.  We are left with an immediately repulsive integral, but closer inspection à la Wolfram reveals true beauty:
$$C_V \approx D(E_0)k^2T \frac{\pi^2}{3}$$
As promised, this constant ($\pi^2/3$) is a small contribution in the scheme of things, since $D(E_0)$, which is proportional to the number of atoms in the crystal, is such a large number by comparison.

Really, the take-away seems to be that heat capacity is linear with respect to temperature... at least to a first order approximation.  At the same time, it's interesting to note that this approach implies that different values of $\mu$ and $E$ don't affect the heat capacity whatsoever, though that could be a result inherent to either the rough handling of $\mu$ or the idealized 1-dimensional nature of the problem.

Have a great 3-day weekend everyone.