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.

enter image description here

  • $\begingroup$ 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 $\epsilon/\epsilon_F \times T_F/T$? $\endgroup$ – QuantumApple Apr 27 at 13:11
  • $\begingroup$ @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 $\endgroup$ – 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 $E(T=0) = \frac{3}{5} N \epsilon_F$.

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  • $\begingroup$ That's a clever variable change, worked perfectly, thanks! $\endgroup$ – phy_research Apr 27 at 14:51
  • $\begingroup$ 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? $\endgroup$ – QuantumApple Apr 27 at 17:10

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