Heat Capacity of a Fermi Gas at Low Temperature

In summary, the conversation discusses the use of the formula $$U=\int Z \epsilon D(\epsilon) e^{-\epsilon β}d\epsilon=\frac{gV}{(2\pi)^3}\int Z \frac{(\hbar)^2k^2}{2m}k^2 (4\pi)e^{-β\frac{(\hbar)^2k^2}{2m}}dk$$ to calculate the internal energy of a system, where g=2 and Z=e^{βµ}. It also mentions the substitution of $$x=\sqrt{\frac{\hbar^2}{2mk_bT}}$$ and the use of the formula $$c_v =\frac{
  • #1
Diracobama2181
75
2
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.
 
Last edited:
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  • #2
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?
 
  • #3
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!
 
  • #4
Thank you. Just one more question. Would the limits of integration just be from 0 to $$\infty$$?
 
  • #5
Diracobama2181 said:
Thank you. Just one more question. Would the limits of integration just be from 0 to $$\infty$$?
Yes
 

1. What is the definition of heat capacity of a Fermi gas at low temperature?

The heat capacity of a Fermi gas at low temperature is a measure of the amount of heat energy needed to raise the temperature of the gas by a certain amount. It is often denoted as C and is measured in units of energy per degree Celsius (J/K).

2. How is the heat capacity of a Fermi gas at low temperature calculated?

The heat capacity of a Fermi gas at low temperature can be calculated using the formula C = (π2/3) * kB2 * T * N, where kB is the Boltzmann constant, T is the temperature in Kelvin, and N is the number of particles in the gas.

3. What is the significance of studying the heat capacity of a Fermi gas at low temperature?

Studying the heat capacity of a Fermi gas at low temperature is important for understanding the behavior of matter at extremely low temperatures, which can have applications in fields such as condensed matter physics, quantum computing, and superconductivity.

4. What factors can affect the heat capacity of a Fermi gas at low temperature?

The heat capacity of a Fermi gas at low temperature can be affected by factors such as the number of particles in the gas, the temperature, and the density of the gas. Additionally, the type of particles in the gas can also play a role in determining the heat capacity.

5. Are there any real-world applications of the heat capacity of a Fermi gas at low temperature?

Yes, the heat capacity of a Fermi gas at low temperature has real-world applications in fields such as nuclear physics, astrophysics, and material science. It is also important in the development of technologies such as cryogenics and superconductors.

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