Total energy of a simple fermi gas

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The total energy density of the gas (in Joules/m$^3$) is given by $$ u = \int_{0}^{\infty} E(p) g(p) F(p)\ dp, $$ where $E(p)$ is the particle energy, $g(p)$ is the density of momentum states at momentum $p$ and $F(p)$ is the occupation index of those states.

For a completely degenerate gas, this is drastically simplified by noting that $F(p)=1$ for $0 < p \leq p_f$, where $p_f$ is the Fermi momentum, and $F(p)=0$ for $p>p_f$.

Thus the integral becomes $$ u = \int^{p_f}_{0} E(p) g(p)\ dp,$$ where if you are using kinetic energy and assuming non-relativistic fermions (as most condensed matter physicists, but not astrophysicists, do) $$ u = \int^{p_f}_{0} \frac{p^2}{2m} g(p)\ dp$$

The density of momentum states function for a gas of spin half fermions is $g(p) = 8\pi p^2/h^3$.

Thus $$ u = \int^{p_f}_{0} \frac{8\pi}{2mh^3} p^4\ dp = \frac{4\pi}{5mh^3} p_{f}^{5} \quad \text{(Fixed typo)}$$

Another way of thinking about this is in terms of the average (kinetic) energy of a particle. The average (kinetic) energy is what we just calculated divided by the number density of particles $$n = \int^{p_f}_{0} g(p)\ dp = \frac{8\pi}{3h^3} p_f^{3}$$ Thus $$ u = \frac{3}{10m}n p_f^2 = \frac{3}{5} nE_f,$$ where $E_f$ is the kinetic energy at $p_f$.

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Adrian Stobbe
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Updated on August 01, 2022

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  • Adrian Stobbe
    Adrian Stobbe over 1 year

    I am a student and working on a fermi gas problem. I already figured out how to calculate the fermi energy of my idealized (no interactions) fermi sphere gas of radius R, but now I want to find out the total energy.

    Is it possible to get it by solving this integral? $ \displaystyle{ E_{total}= \int_0^{p_{f}}}\frac{p^2}{2m}\cdot dp= \frac{p_{f}^3}{6m}$

    Then I would substitute $p_{f}$ by $(\frac{3\cdot n}{\pi})^\frac{1}{3}\cdot \frac{h}{2}$ , where n is the particle density. In this expression I used the fact, that the momentum is described by a fermi sphere. (all momenta lie within that sphere).

    What exactly am I missing? Because the energy clearly must depend on the radius of my fermi gas sphere. I am very new to this topic and very grateful for your help.

  • Adrian Stobbe
    Adrian Stobbe about 7 years
    Thank you very much. I unfortunately don't understand why the number of momentum states divided by $h^3$ is g(p). I thought that the momentum states would be described by a sphere ($\frac{\frac{4\cdot \pi \cdot p{f}^3}{3}}{h^3}$). Why is it geometrically the surface area divided by the volume of each particle in phase space?
  • Adrian Stobbe
    Adrian Stobbe about 7 years
    Problem solved ^^ By integrating g(p) one gets the well known sphere. Anyway why is g(p) )= surface area/ volume? How is this to be interpreted? Thanks in advance
  • Adrian Stobbe
    Adrian Stobbe about 7 years
    Ok with a little bit of pondering I think I have it. For a given p the number of momentum states is the surface area. Multiply by 2 because we can have 2 fermions per state and divide by the unit volume (h^3) and you get the density function g(p). Integrate over p and you get the fermi sphere.
  • Adrian Stobbe
    Adrian Stobbe about 7 years
    Now I just have to find out how to express $p{f}$ in terms of the radius of my fermi gas sphere, because this is what I am asked for.
  • ProfRob
    ProfRob about 7 years
    $g(p)$ is the density of momentum states, i.e. a number per phase-space volume. $p=\hbar k$ of course.
  • Adrian Stobbe
    Adrian Stobbe about 7 years
    That is now clear to me. Is it correct to say that the phase-space volume is $V{phase} \cdot V{sphere}$, where $V{phase}$ is 1/2 of what is n in your calculation. Then I could say:
  • Adrian Stobbe
    Adrian Stobbe about 7 years
    Do you know how I can make this expression depend on the radius of my gas?
  • ProfRob
    ProfRob about 7 years
    @AdrianStobbe Your terminology is not making much sense to me. You asked how to calculate the total energy (density) of the gas and introduced only the variables $p$, $p_f$ and $n$, which is what I have used. If you literally mean a gas of radius $R$, then the total energy is the energy density $\times 4\pi R^3/3$ !