# Energy of Fermi Gas $T>0$

I'm trying to plot $$\frac{E(T)}{N\epsilon_F}$$ vs $$\frac{T}{T_F}$$

I know that the total energy comes from $$E(T) = \int_{0}^{\inf} \frac{3}{2}\frac{N}{\epsilon_F}(\frac{\epsilon}{\epsilon_F})^{1/2} \frac{\epsilon}{e^{-\beta\mu+\beta x}+1} d\epsilon$$

I already have the values for $$\frac{\mu}{\epsilon_F}$$ vs $$\frac{T}{T_F}$$

The question is how to leave the integral in terms of $$\frac{T}{T_F}$$ to plot.

The plot should look like this.

• What is the issue? Can't you just replace $\beta \mu = \mu / (kT)$ by $\mu / \epsilon_F \times T_F/T$ and $\beta \epsilon = \epsilon/(k T)$ by $\epsilon/\epsilon_F \times T_F/T$? – QuantumApple Apr 27 at 13:11
• @QuantumApple I don't have the value for $\epsilon_F$, that's why I'm plotting $\frac{E(T)}{N\epsilon_F}$, so it'd work in the first substitution you propose, but not in the second one – phy_research Apr 27 at 13:21

If you do the change of variable $$x = \epsilon/\epsilon_F$$, everything should nicely come adimensioned in the end:

$$\frac{E(T)}{N\epsilon_F} = \frac{3}{2} \int_{0}^{+\infty} (\frac{\epsilon}{\epsilon_F})^{1/2} \frac{\epsilon/\epsilon_F}{e^{-\frac{\mu}{\epsilon_F}\frac{T_F}{T}+\frac{\epsilon}{\epsilon_F}\frac{T_F}{T}}+1} d\left(\frac{\epsilon}{\epsilon_F} \right) = \frac{3}{2} \int_{0}^{+\infty} \frac{x^{3/2}}{e^{-\frac{\mu}{\epsilon_F}\frac{T_F}{T}+x\frac{T_F}{T}}+1} dx$$

At $$T = 0$$, $$\mu = \epsilon_F$$ so that the Fermi-Dirac function is $$1$$ for $$x < 1$$ and $$0$$ for $$x > 1$$, such that the integral reduces to the integral of $$x^{3/2}$$ from $$0$$ to $$1$$ which is $$2/5$$, yielding the classical result $$E(T=0) = \frac{3}{5} N \epsilon_F$$.

• That's a clever variable change, worked perfectly, thanks! – phy_research Apr 27 at 14:51
• Also, I've just noticed but your plot seems off. To my knowledge, the red curve should go to $0.6$ when $T \to 0$. Is this normal? – QuantumApple Apr 27 at 17:10