0
$\begingroup$

In Callen Thermodynamics and an I second edition p141

Consider a reversible process in a composite system, of which the system of interest is a subsystem. The subsystem is taken from some reference state $(T_0 , P_0)$ to the state of interest $(T_1, P_1)$ by some path in the $T$-$P$ plane. The change in entropy is $${S_1} - {S_2} = \int_{({T_0},{P_0})}^{({T_1},{P_1})} {\left[ {{{\left( {\frac{{\partial S}}{{\partial T}}} \right)}_P}dT + {{\left( {\frac{{\partial S}}{{\partial P}}} \right)}_T}dP} \right]} $$

Why does Callen use intensive variables rather than extensive variables here, i.e the fundamental equation $S=S(U,V,N)$ uses $U,V,N$ as its natural variable. Should there be legendre transformation?

$\endgroup$
5
  • $\begingroup$ I think you should review your question, something is not quite right in there: extensive v. intensive... $\endgroup$
    – hyportnex
    Aug 13, 2023 at 1:25
  • $\begingroup$ You asked about the intensive variables twice (two different times). $\endgroup$ Aug 13, 2023 at 1:38
  • $\begingroup$ See answers and discussion related to this question physics.stackexchange.com/questions/771466/… $\endgroup$ Aug 13, 2023 at 7:04
  • $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 thank you for your recommendation. It's helpful ,thank you $\endgroup$
    – Raffaella
    Aug 14, 2023 at 1:05
  • $\begingroup$ @Matt Hanson Thank you ,I‘ve corrected it ,thank you very much $\endgroup$
    – Raffaella
    Aug 14, 2023 at 1:07

1 Answer 1

2
$\begingroup$

There is no need for a Legendre transformation to substitute new variables for old ones. If you start with a function of the extensives $S=S(U,V,N)$ and you know the dependence of $U$ and $V$ on the intensive parameters $p$ and $T$ for fixed $N$, say $U=f(T,p,N)$ and $V=g(T,p,N)$ then you are free to write $$S=S(U,V,N) = S(f(T,p,N), g(T,p,N), N) = h(t,p,N)$$ and then $$dS=dh= \frac{\partial h}{\partial T}dT + \frac{\partial h}{\partial p}dp$$ for fixed $N$ from which you get the integral as $$S_1-S_0 = \int_{T_0,p_0}^{T_1,p_1} \left[\left(\frac{\partial h}{\partial T}\right)_{p,N}dT + \left(\frac{\partial h}{\partial p}\right)_{T,N}dp\right]$$

Note that while $S=S(U,V,N)=h(T,p,N)$ but $S$ and $h$ are not the same functions, they do have the same values though when $U=f$ and $V=g$. The reason to prefer the intensive parameters is because it is easier to measure with them. For example, it is easier to control external pressure than volume, or bias voltage than load current, etc.

$\endgroup$
2
  • $\begingroup$ Sorry , I still have little doubt here. If you have read about the paragraph, you will find that Callen mentioned nothing about f and g . Certainly we can have f and g though partial derivative and taking it inverse, but this needs fundamental equation in advance(again ,we don't know the fundamental equation) and its good property(otherwise you can't take the inverse).I've struggled the whole yesterday. Anyway ,thank you for your answer!! $\endgroup$
    – Raffaella
    Aug 14, 2023 at 1:01
  • $\begingroup$ There is no mystery here, physicists, even Callen, are lazy and they use the same name/notation for different functions, especially in thermodynamics. It is understandable to some extent because who wants to give a different symbol for entropy every time you change the independent variables. If you do not know the the caloric or thermal equation of state equation $U=f(T, p, N)$, $T=t(p, V, N)$ then you know nothing, There is no instrument that directly measures internal energy or entropy, but we can measure $p,T,N,V$. $\endgroup$
    – hyportnex
    Aug 14, 2023 at 8:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.