Physical interpretation of changing variables $S(T,V)$ to $S(T,p)$
I'm confused in this situation. In an example of this book (Example 16.5. Thermal Physics, by Blundell) they ask to prove that
$$C_P - C_V = VT\frac{\alpha^2}{\beta_T}\tag{1}$$
but using the fact that $S=S(T,V)$. Then they use the fact that
$$C_V = T\biggl(\frac{\partial S}{\partial T}\biggr)_V\tag{2}$$
and that
$$C_P = T\biggl(\frac{\partial S}{\partial T}\biggr)_P\tag{3}$$
My question is, the relation (3) is assuming that $S=S(T,p)$, so what happened with the initial assumption that $S=S(T,V)$? Or is there a way to obtain (3) without using the fact that $S=S(T,p)$?
Finally, if the only way of obtain (3) is assuming that $S=S(T,p)$, then what is the physical meaning of the change $S=S(T,V)$ to $S=S(T,p)$? I mean, how can I justify that change in the variables?
David Z♦
Ana S. H.Ana S. H.
There is no such thing as $S(T,p)$. $T$ and $p$ are both intensive variables, while $S$ is extensive. Just knowing $T$ and $p$ tells you nothing about how large the system is, and therefore cannot tell you the entropy.
However, if you fix some extensive quantity, like the particle number $N$, you can have $S(T,p,N)$. This is implicitly what the book is doing, so $\left(\frac{\partial S}{\partial T}\right)_p$ could be written more explicitly as $\left(\frac{\partial S}{\partial T}\right)_{p,N}$
As to why you can change what variables you use to write the entropy, this is because entropy is really a function of the physical state, not of some particular variables. It's then a question of how you specify the state of the system. If I tell you "I have a such-and-such system with pressure $p$ and temperature $T$ and $N$ molecules", you can go and use the equation of state to figure out $V$ for yourself. Therefore, if someone originally gives you entropy as a function of $V$, you can eliminate it by solving for $V$ and write entropy in terms of $p$ instead.
Mark EichenlaubMark Eichenlaub
1 - From the equation of state you can express p as a function of the volume and the temperature $$ p=p(T,V) $$ then $$ S(T,p)=S(T,p(T,V))=S'(T,V) $$ defining a new function $S'$ that depends only on temperature and volume, now as is common in physics by abuse of notation we will note this new function as $S(T,V)$, but you are right that mathematically is not exacly the same $S$
2 - With this change of variable you use only intensive quantities, for example the intensive entropy $s(T,p)=S(T,p)/N$ has a well defined limit when $N,V\to \infty$
IkiperuIkiperu
A good question, here my attempt at an answer.
To describe a thermodynamic system, you can ask for the values of certain thermodynamic quantities: Pressure $p$, Volume $V$, particle Number $N$, chemical potential $\mu$, temperature $T$, entropy $S$, internal energy $E$.
As it turns out, however, these quantities aren't entirely independent. For an ideal mono-atomic gas, for example, $E = 3/2 N k_B T$, so if you fix particle number $N$ and temperature $T$, the internal energy is already determined.
For a thermodynamic system, one can show that specifying three quantities is enough to determine all other quantities, and you have some freedom in which of those three quantities you specify: You could specify temperature, volume, and particle number, and that would give you every other quantity, including entropy, $S(T,V,N)$. Or you could specify temperature, pressure, and particle number, which gives you $S(T,p,N)$.
Mathematically, you go from $S(T,V,N)$ to $S(T,p,N)$ by expressing volume in terms of temperature, pressure and particle number, $V(T,p,N)$ and substituting that into $S$.
(In your example it looks like there are only two variables. If it is understood that we are looking at systems where $N$ cannot change, it is usually omitted from the function's arguments)
There is one thing to be careful about, though: For your three variables to specify the system, you can't use "conjugate" variables: You can't describe a (general) system by specifying pressure, Volume and temperature, for example, because pressure and volume are conjugate. (Note: It does work for the special case of the ideal gas because of $pV = NkT$.
LagerbaerLagerbaer
There is a systematic approach to verify such identities. Usually, the equations of state of a thermodynamical substance are given by expressing $T$ and $S$ as functions of $p$ and $V$, symbolically, $T=f(p,V)$, $S=g(p,V)$ for suitable functions $f$ and $g$ of two variables---this is presumably because these variables can be measured directly without reference to thermodynamics. Thus for the dieal gas $f(p,V)=p V$ and $g(p,V)= \frac 1 {\gamma-1} (\ln p + \gamma \ln V)$ (we are omitting physical constants to simplify the exposition). One can then express all of the relevant quantities in terms of $p$, $V$ and $f$ and $g$, together with their partials (which we denote by $f_1$ and $f_2$, etc.). Thus for the ideal gas $f_1=V$, $f_2=p$, $g_1= \frac 1{p(\gamma-1)}$ and $g_2= \frac {\gamma}{V(\gamma-1)}$. If one carries this out with both sides of the identity to be proved, then it reduces to simple algebra.
In the example under discussion we have $\beta_T=-\frac 1 V\frac{\partial V}{\partial p}|_T$ which reduces to $\frac 1 V \frac {f_1}{f_2}$. Similarly $\alpha$ reduces to $\frac 1V \frac 1 {f_2}$, $C_p$ to $f \frac {g_2}{f_2}$ and $C_V$ to $f \frac {g_1}{f_1}$. Substituting, we see that both sides reduce to the same expression $\frac f {f_1f_2}$.
We remark that the method of reducing the quantities is a simple exercise in the chain rule and inverse function theorem. It is explicated in the arXiv article "On the mathematics of thermodynamics" (1102.1540). The whole procedure can be codified in a simple Mathematica programme and this is done in the follow-up article "Thermodynamical identities---a systematic approach" (1108.4760).
This might seem rather elaborate for just one identity but it has several advantages. Firstly, it is a systematic approach which works for all such identities. Secondly, it makes the basic mechanism much more transparent. In this case it reveals the interesting fact that both sides depend only on $f$ and its derivatives, i.e., are independent of the form of the dependence of entropy on $p$ and $V$ in contrast to the constituent parts $C_p$ and $C_V$---at first sight rather surprising considering the definitions of the latter which are directly in terms of $S$.
mathuser4891mathuser4891
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