# Confusing definition of thermodynamic pressure when calculating the electron degeneracy pressure

When dealing with statistical mechanics, one usually defines pressure as $$p(N,T,V,X):=-\frac{\partial F(N,T,V,X)}{\partial V}.$$ In engineering thermodynamics, I've often seen this definition reformulated in terms of the Legendre transform $$U=F\{S\leftrightarrow T\}$$: $$p(N,T,V,X):=-\frac{\partial }{\partial V}\left(U(N,S(N,T,V,X),V,X)-TS(N,T,V,X))\right)\\=-\frac{\partial U(N,S,V,X) }{\partial S}|_{S=S(N,T,V,X)}\frac{\partial S(N,T,V,X)}{\partial V}-\frac{\partial U(N,S,V,X) }{\partial V}|_{S=S(N,T,V,X)}+T\frac{\partial S(N,T,V,X)}{\partial V}\\=-\frac{\partial U(N,S,V,X) }{\partial V}|_{S=S(N,T,V,X)},$$ which can be rewritten as $$p^*(N,S,V,X):=p(N,T(N,S,V,X),V,X)=-\frac{\partial U(N,S,V,X) }{\partial V},$$ where often the asterix is dropped.

Now, however, I'm reading that the electron degeneracy pressure of a Fermi gas at low temperatures is defined to be $$\tilde{p}(N,T=0,V)=-\frac{\partial \tilde{U}(N,T=0,V) }{\partial V}$$ where $$\tilde{U}$$ is the inner energy in terms of the particle number, temperature and volume. I'm confused, as this definition doesn't match the usual definition of pressure. Is this a different kind of pressure? Why aren't we keeping the entropy constant?

• I'm not sure I understand your question properly, but don't we have $F=\bar{U}$ at $T=0$? Isn't this a natural consequence of that? Jun 21 '20 at 11:51
• @NandagopalManoj This makes sense to me, didn't see this at all! I think you should make this an answer, so that I can accept it. Never really thought about it this way.. but yes at absolute zero inner energy and free energy are the same
– user224659
Jun 21 '20 at 11:58

Since we define the free energy as $$F = U - TS$$, at zero temperature we have $$F = U$$.
The definition for pressure is $$P = -\bigg{(}\frac{\partial F}{\partial V}\bigg{)}_{T,N}$$ So at $$T=0$$ we clearly have $$P = -\bigg{(}\frac{\partial U}{\partial V}\bigg{)}_{T,N}$$ which is what you call $$-\frac{\partial \bar{U}(N,T=0,V)}{\partial V}$$. So this matches our natural definition of pressure.