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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?

Chemomechanics
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1 Answers1

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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.

Nandagopal Manoj
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