Heat Capacity of a Fermi Gas at Low Temperature

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SUMMARY

The discussion focuses on the calculation of the heat capacity of a Fermi gas at low temperatures, specifically using the formula $$U=\int Z \epsilon D(\epsilon) e^{-\epsilon β}d\epsilon$$. The participants confirm the setup of the integral and the substitution of variables, particularly $$x=\sqrt{\frac{\hbar^2}{2mk_bT}}$$, to simplify the expression for internal energy. The integration limits are confirmed to be from 0 to $$\infty$$, ensuring the accuracy of the calculations involved in determining the heat capacity.

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Diracobama2181
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Homework Statement
Suppose that instead of photons, blackbody radiation were composed
of a single species of neutrinos. The neutrino is a spin-1/2 particle
like an electron, with zero electric charge. Without worrying about
the details of the reactions that neutrinos undergo, suppose that they
can be freely created and destroyed such that they maintain thermal
equilibrium with the walls of a cavity. Treat the neutrinos as a grand
canonical ensemble of free particles of mass m, with chemical potential
µ = −mc2.

a)
Show that the heat capacity per unit volume reduces to the following form at low temperature, where the neutrinos are nonrelativistic and fermion quantum statistics reduce to classical Boltzmann statistics.
$$c_v =\frac{1}{V} \frac{dU}{dT}=\frac{4k_B}{λ^3}e^{βµ} [(βµ)^2-\frac{3}{2}βµ]$$
where
$$λ =(\frac{h^2β}{2πm})^{\frac{1}{2}}$$
Relevant Equations
$$<n_i>=\frac{1}{e^{β(\epsilon-µ)}+1}$$
I find that $$U=\int Z \epsilon D(\epsilon) e^{-\epsilon β}d\epsilon=\frac{gV}{(2\pi)^3}\int Z \frac{(\hbar)^2k^2}{2m}k^2 (4\pi)e^{-β\frac{(\hbar)^2k^2}{2m}}dk$$
where g=2s+1=2, $$Z=e^{βµ}$$ and $$D(\epsilon)=\frac{gV}{(2\pi)^3}k^2 4\pi$$ for the density of states

From here, I can use
$$c_v =\frac{1}{V} \frac{dU}{dT}$$. My question is whether I set this up correctly?

Thank you.
 
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Taking this further, I substitute $$x=\sqrt{\frac{\hbar^2}{2mk_bT}}$$ and get that $$U=\int Z \epsilon D(\epsilon) e^{-\epsilon β}d\epsilon=\frac{gV4\pi}{(2\pi)^3}\int Z \frac{(\hbar)^2}{2m}\sqrt{\frac{\hbar^2}{2mk_bT}}^{\frac{5}{2}}x^4 e^{-x^2}dx=\frac{gV4\pi}{(2\pi)^3}Z \frac{(\hbar)^2}{2m}\sqrt{\frac{\hbar^2}{2mk_bT}}^{\frac{5}{2}}\frac{3\sqrt{\pi}}{8}$$. Is this fine so far?
 
Diracobama2181 said:
Taking this further, I substitute $$x=\sqrt{\frac{\hbar^2}{2mk_bT}}$$ and get that $$U=\int Z \epsilon D(\epsilon) e^{-\epsilon β}d\epsilon=\frac{gV4\pi}{(2\pi)^3}\int Z \frac{(\hbar)^2}{2m}\sqrt{\frac{\hbar^2}{2mk_bT}}^{\frac{5}{2}}x^4 e^{-x^2}dx=\frac{gV4\pi}{(2\pi)^3}Z \frac{(\hbar)^2}{2m}\sqrt{\frac{\hbar^2}{2mk_bT}}^{\frac{5}{2}}\frac{3\sqrt{\pi}}{8}$$. Is this fine so far?
Looks fine to me if calculations are done correctly!
 
Thank you. Just one more question. Would the limits of integration just be from 0 to $$\infty$$?
 
Diracobama2181 said:
Thank you. Just one more question. Would the limits of integration just be from 0 to $$\infty$$?
Yes
 

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