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Is it possible to create an explicit function of entropy $S(E,V,N)$ starting from Helmholtz free energy given as $F(T,V,N)$?

An example of the other direction is here, but I am struggling to do the inverse.

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Åsmund Hj
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1 Answers1

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The expression can easily be derived from knowing the differential form. Since we know that $$dE = TdS -pdV + \mu dN$$ and we know as well that $$F = E-TS$$ then differentiating that later $$dF = dE - TdS -SdT$$ and using the first expression for $dE$ you obtain the differential expression desired $$dF = -SdT - pdV + \mu dN$$ This expression above tells you that $F$ is a potential function (its integral is independent of path, or its values are uniquely determined by its independent variables) when is expressed in terms of $T,V,N$ as you are asking $F(T,V,N)$. It also tells you that $$S=-\frac{\partial F}{\partial T}$$ as desired, where entropy will be a function of the same variables $S=S(T,V,N)$

rmhleo
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  • I see how I can get entropy from F(T,V,N) by using partial derivatives, but my question is different: Is it possible in general to arrive at a function for S that is explicit in E,V,N? – Åsmund Hj Feb 17 '16 at 19:29
  • I think not in general. Since here the potential function is expressed in $T,N,V$ then the energy is also a function of them $E(T,V,N)$. So in general you could have the system in two different states $F_1 = F(T_1,V,N)$, $F_2 = F(T_2,V,N)$ for which the energy is the same $E_1 = E(T_1,V,N) = E_2 = E(T_2,V,N)$ while the entropy is not $S_1 = S(T_1,V,N) \ne S_2 = S(T_2,V,N)$, in which case the relation between $S$ and $E$ would not be univocal. This relation will be determined by the concrete case, or the additional constraints you impose. – rmhleo Feb 26 '16 at 12:35