# Is there relation between adiabatic process and electron degeneracy pressure?

Electron degeneracy pressure scales with density as:

eq.1.1: $$p \approx \rho^{5/3}$$

By dividing electron degeneracy pressures of 1 mole of electrons at two different densities we obtain:

eq.1.2 $$p_1/p_2 = (\rho_1/\rho_2)^{5/3}$$

eq.2.1: $$p V^\gamma = const.$$

respectively:

eq.2.2: $$p_1 / p_2 = (V_2/V_1)^\gamma$$

for fixed number of particles $$\rho \approx 1/V$$ and we can see that

Ratio of electron degeneracy pressures (eq.1.2) formally resembles ratio of pressures in adiabatic process (eq.2.2) with heat capacity ratio $$\gamma = 5/3 = 1.666$$ which correspond to monoatomic gas (point particles).

=> It seems like a hint that microscopic origin of electron degeneracy pressure can be seen as a result of some internal thermal motion, which cannot be ever cooled (that is $$\delta Q=0$$, which means it is always adiabatic)

Question

• Does it make some sense?
• Is this just a coincidence?
• Did somebody examined this line of thinking?
• Is this something obvious?
• Are there same no. of deg. of freedom in both cases?? – aditya_stack Dec 31 '19 at 19:05

There is a simple relationship. The point is that under an adiabatic compression, the volume occupied in phase space by the particles is conserved, by the adiabatic theorem. If you reduce the volume by a factor of $$8$$, the typical momenta must thus increase by a factor of $$2$$. (This assumes you're working with monatomic gases, as more complicated moleclues have other dimensions of phase space they can expand into, which corresponds to a different value of $$\gamma$$.)
Now consider the forces exerted on the walls by a typical particle. Since the length is reduced by a factor of $$2$$, and the momenta are doubled, the frequency of collisions is increased by a factor of $$4$$. The area is a factor of $$4$$ smaller, so since pressure is force per area, it increases by another factor of $$4$$. And finally, each collision now transfers twice the momentum, doubling the pressure again. So the pressure increases by a factor of $$32 = 8^{5/3}$$.
This logic works identically for both an ordinary monatomic gas, and a degenerate Fermi gas at $$T = 0$$. The difference between the two is just that the degenerate Fermi gas actually occupies all of its phase space, but the basic logic of counting phase space and collisions is the same.