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.
