# The Equation of State for a Degenerate Fermi gas

I have read in Chandrasekhar's paper The highly collapsed configurations of a stellar mass Appendix I the equation of a degenerate Fermi gas as follows:

$$n=\frac{8\pi}{h^3}\int^{p_0}_0 p^2dp$$

and

$$P=\frac{8\pi}{3h^3}\int^{p_0}_0 p^3\frac{dE}{dp}dp$$

I have derived myself the proper energy density:

$$\rho=\frac{8\pi}{h^3}\int^{p_0}_0\sqrt{(pc)^2+(mc^2)^2} p^2dp$$

where $n$ is the number density of electrons and $P$ is the pressure. $E$ is the kinetic energy of a free electron.

I have no problem understanding the first equation for $n$ but I am confused with the second one. Could anyone please explain to how to obtain the equation for $P$ and tell me if my energy density expression is correct?

## 1 Answer

The formula for the pressure $$P=\frac{8\pi}{3h^3}\int^{p_0}_0 p^3\frac{dE}{dp}dp$$ is valid for a simple reason: $dE/dp$ is nothing else than the expression for the speed $v$. Check it for $E=p^2/2m$; the $p$-derivative is $p/m=v$. So the integrand only differs from the integrand for the electron number density $n$ in the first formula by the extra factor of $pv/3$ (I included the new prefactor $1/3$ into the integral). You said that you agree with the first formula so I assume it is correct.

But the addition of $pv/3$ to the formula for the number density (per volume) is exactly right to calculate the pressure. If you divide the identity described in the previous sentence by $V$, the "number density" becomes simply "number of electrons" while $pv/3$ will be divided by $V$. So the claim is equivalent to the claim that one electron contributes $pv/(3 V)$ to the pressure.

But that's true for the usual reason why radiation has $p=\rho/3$. Put the electron in a box of dimensions $V=L_xL_yL_z$. I will assume $L_x=L_y=L_z=L$ for simplicity, a cube. It takes the time $L_x/v_x$ for the electron to hit one of the $x={\rm const}$ walls. Each $x$-collision leads to the sign flip of $p_x$, so it deposits the momentum $2p_x$ to the walls. When converted to force, i.e. momentum change per unit time, the force on these $x$-walls is $2p_x / (L_x/v_x) = 2p_x v_x/L$. The total force on $x,y,z$ six walls of the cube is simply the sum over $x,y,z$ i.e. the inner product $2\vec p\cdot \vec v/L$, and the pressure is the force per area. The area is $6L^2$ so the force per area is $2\vec p\cdot \vec v / L / 6L^2 = \vec p\cdot \vec v / 3V$ as I wanted to prove.

Yes, the energy density differs from the formula for $n$, the number density, by an extra factor in the integrand which is the square root formula for the energy of an electron. One must understand that this doesn't include any potential energy but it does include the latent $E=m_0 c^2$ energy of the electrons.

• Hello Luboš Motl!Thank you so much for your reply. I have also read in Pathria's statistics book about how this can be formally derived. You did give me a very intuitive picture about this equation. Thanks again for your help. – HanaKaze Aug 24 '14 at 16:13