# Kinectic energy of a three-dimensional fermi gas of N free electrons at absolute zero

Show that the kinectic energy of a three-dimensional fermi gas of N free electrons at absolute zero is (Mathematica code used)
u = 3/5 N Subscript[\[Epsilon], F]

Now I know total energy of N particles is this integral

u = \!$$\*SubsuperscriptBox[\(\[Integral]$$, $$0$$, $$\[Infinity]$$]$$\ \[Epsilon]\ P[\[Epsilon]] \[DifferentialD]\[Epsilon]$$\)

which is made up of the density of the states and probability of the electron to occupy level with energy \[Epsilon] at temp T.

So P[\[Epsilon]] is this big horrible looking thing. My guess is that there must be a be an easy way to integrate it that comes about from absolute zero tempature because the final answer seems so nice.

Any help, would be nice. thanks!

You need to express the Fermi momentum of the system through its number of particles:

$$N = \frac{V}{(2\pi)^{3}} \int_{0}^{k_{F}}{4\pi k^{2} dk}$$

The Fermi energy $E_{F}$ is defined as the kinetic energy of the particles with Fermi momentum

Once you had done that, the total kinetic energy of the particles is:
$$(E_{\mathrm{kin}})_{\mathrm{tot}} = \frac{V}{(2\pi)^{3}} \, \int_{0}^{k_{F}}{\frac{\hbar^{2} k^{2}}{2 m} \, {4 \pi k^{2} \, dk}$$

In the final result you need to eliminate $k_{F}$.

ahhk, yeah that works. you could have explained it more but I got it in the end.
I'll fill in the gaps for anyone else who might look at this in the future.

use Ef=((khbar)^2)/2m to get the it to look right. and the big integral, you need to know that f(E) = 0 for E>Ef and f(E) = 1 for 0<E<Ef

So put the upper limit to Ef (because anything above this range in the integral equals zero) and also anything inside this range "E<Ef" aka 0 to Ef the f(E)=1, so this makes the integral really easy.

Thanks again Dickfore.