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During an exercise session in school we worked on Problem 1.4 in Equilibrium Statistical Physics by Plischke and Bergersen. This chapter is a repetition of thermodynamics, and the problem concerns a magnetic system.

At one point the teacher's assistant wrote the magnetisation $M$ as a function of $S$ and $T$ in the following way, where $H$ is the magnetic field:

$$M = M(T,H) = M(T,H(S,T)) \tag{1}$$

I questioned the validity of this parameterization, and my reasoning is as follows. Consider something more familiar; a non-magnetic one-component system with fixed $N$, having the fundamental equation (in energy form)

$$U = U(S,V) \tag{2}$$

so that

$$dU = T dS - P dV, \tag{3}$$

where

$$T = \left( \frac{\partial U}{\partial S} \right)_V, \quad -P = \left( \frac{\partial U}{\partial V} \right)_S. \tag{4}$$

Now, we all know that we can write $U$ instead as a function of $T$ and $V$. This can be argued from the thermodynamic postulates via the observation that, for a given $V$, the energy $U$ is an increasing and convex function of $S$, and hence the map

$$(S,V) \mapsto (T,V) = \left( \left( \frac{\partial U}{\partial S} \right)_V, V \right) \tag{5}$$

is a bijection, making it a valid reparameterization. This is essentially the argument made in Section 3.3 of Callen's Thermodynamics and an Introduction to Thermostatistics, though he simplifies further by letting $V$ be fixed just like $N$.

We can argue in a similar way to replace $V$ with $P$, but I see no way to argue for replacing $V$ with $T$ or $S$. Hence I don't see how it would be possible to parameterize the thermodynamic state by $S$ and $T$, giving $U = U(S,T)$. Indeed it seems like $S$ and $T$ contain the same information about $U$.

Now, let us return to the original question, about equation (1). Is there something special about magnetic systems that lets us parameterize it in terms of $S$ and $T$? Something particular with magnetization? Or is there something wrong with my argument, even for the simple (non-magnetic) system?

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  • $\begingroup$ I'd agree with you. Did the teacher's assistant give reasons for inserting $H=H(T,S)$? $\endgroup$
    – kricheli
    Commented Sep 2, 2023 at 16:44
  • $\begingroup$ @kricheli He uses it to argue that $(\partial M / \partial H)_T = (\partial M / \partial H)_S + (\partial M / \partial S)_H (\partial S / \partial H)_T$, though I don't really follow his reasoning here either. $\endgroup$
    – ummg
    Commented Sep 2, 2023 at 18:10
  • $\begingroup$ Maybe it's just a mistake. I'm noticing now that we should be able to write $M = M(S(T,H), H)$, and then $(\partial M / \partial H)_T = (\partial M / \partial H)_S + (\partial M / \partial S)_H (\partial S / \partial H)_T$ follows by the chain rule. This must have been the intended argument, right? $\endgroup$
    – ummg
    Commented Sep 2, 2023 at 18:19

2 Answers 2

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You have already pointed out in Eq. (5) that $(S,V) \mapsto (T,V)$ is a bijection, now if you also assume that this bijection is a smooth one then you should be able to substitute in $dU=TdS-pdV$ without any problem and get $$dU^*=TdS-p^*\left(\frac{\partial V}{\partial S}dS+\frac{\partial V}{\partial T}dT\right)\\ =\left(T-p^*\frac{\partial V}{\partial S}\right)dS+p^*\frac{\partial V}{\partial T}dT$$ where $U^*(S,T)=U(S,V(S,T))$ and $p^*(S,T)=p(S,V(S,T)).$ Defined this way you get $$\frac{\partial U^*}{\partial S}=T-p^*\frac{\partial V}{\partial S}\\ \frac{\partial U^*}{\partial T}=p^*\frac{\partial V}{\partial T}$$

In this sense I do not think that there is anything special about magnetic systems. There could be a difference as to when the bijection is smooth because in a 1st order phase transition, such as liquid $\Leftrightarrow $ gas the entropy has a jump while in a 2nd order transition such as occurring at the Curie point, ferrommagnetism $\Leftrightarrow $ paramegnetism, the entropy is constant but not its derivative.

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  • $\begingroup$ "You have already pointed out in Eq. (5) that (S,V)↦(S,T) is a bijection". No: "hence the map (S,V)↦(T,V)=((∂U∂S)V,V)(5) is a bijection" $\endgroup$
    – kricheli
    Commented Sep 2, 2023 at 17:08
  • $\begingroup$ @kricheli Yes, a direct quote from the question: "This can be argued from the thermodynamic postulates via the observation that, for a given $V$, the energy $U$ is an increasing and convex function of $S$, and hence the map $(S,V) \mapsto (T,V) = \left( \left( \frac{\partial U}{\partial S} \right)_V, V \right) \tag{5}$ is a bijection, making it a valid reparameterization. " $\endgroup$
    – hyportnex
    Commented Sep 2, 2023 at 17:14
  • $\begingroup$ Yeah, I'm quoting this in my comment. He is not replacing V with T as you state, but rather S with T! $\endgroup$
    – kricheli
    Commented Sep 2, 2023 at 17:30
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    $\begingroup$ @kricheli Ahh, you are right, it was a typo; I have fixed it with the rest stays, thanks. $\endgroup$
    – hyportnex
    Commented Sep 2, 2023 at 17:35
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    $\begingroup$ @kricheli Correct, and I think examples of singularities such as $\tfrac{\partial T}{\partial V}\vert_{S}=-\tfrac{\partial p}{\partial S}\vert_{V} = \tfrac{\partial^2 U}{\partial S\partial V}=0$ are excludable on grounds of stability. $\endgroup$
    – hyportnex
    Commented Sep 2, 2023 at 18:17
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After thinking about this I have come to the conclusion that the TA was probably just making a mistake. His goal was to argue that

$$\left( \frac{\partial M}{\partial H} \right)_T = \left( \frac{\partial M}{\partial S} \right)_H \left( \frac{\partial S}{\partial H} \right)_T + \left( \frac{\partial M}{\partial H} \right)_S. \tag{I}$$

For the magnetic system in question $M$ and $H$ are conjugate variables; the other conjugate pair being $S$ and $T$. Hence $(S,H)$ and $(T,H)$ are both perfectly valid state space parameterizations (as are $(S,M)$ and $(T,M)$). I therefore think the TA meant to write

$$M = M(S(T,H), H). \tag{II}$$

Then (I) simply follows by the chain rule.

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