How does one "invert" derivatives for intensive variables?

Question

Consider a magnetic system which is characterized by the extensive variables $U,V,N,I$, with $I$ the total magnetic moment of the (homogeneous) system. By the postulates, one can easily swap $S$ for $U$, so that $U = U(S,V,N,I)$.

In Chapter 3.9, Callen gives the following equality. where $B_e$ is the externally applied field: $$\left(\frac{\partial B_e}{\partial S} \right)_{I,V,N} = \left(\frac{\partial T}{\partial I} \right)_{S,V,N}.$$ Since this system is assumed to be a normal magnetic system, this is imply the statement of equality of partial derivatives, because (one can show and/or define that) $$B_e = \left(\frac{\partial U}{\partial I} \right)_{S,V,N}$$ and $$T := \left(\frac{\partial U}{\partial S} \right)_{I,V,N}.$$ But Callen goes on to say that it is possible to "invert" the equality of derivatives above to obtain $$T\left(\frac{\partial S}{\partial B_e} \right)_{I,V,N} = T\left(\frac{\partial I}{\partial T} \right)_{S,V,N}.$$ I'm a bit confused as to how one justifies this. I've previously seen this sort of thing shown with the so-called reciprocity theorem ($(\partial x/\partial y)(\partial y/\partial z)(\partial z/\partial x) = -1$, where all other variables (and potential spectator variables) are held constant) but given that we have four variables involved here and some of them are intensive, I'm not sure how to use this, if at all.

Edit: Is this as simple as observing that $$\left(\frac{\partial B_e}{\partial S} \right)_{I,V,N} = \left(\frac{\partial T}{\partial I} \right)_{S,V,N} \implies \frac{1}{\left(\frac{\partial B_e}{\partial S} \right)_{I,V,N}} = \frac{1}{\left(\frac{\partial T}{\partial I} \right)_{S,V,N}} \stackrel{(1)}{\implies} \left(\frac{\partial S}{\partial B_e} \right)_{I,V,N} = \left(\frac{\partial I}{\partial T} \right)_{S,V,N} \stackrel{(2)}{\implies} T \left(\frac{\partial S}{\partial B_e} \right)_{I,V,N} = T\left(\frac{\partial I}{\partial T} \right)_{S,V,N}$$ where in (1) I am using a relationship for reciprocal derivatives and in (2) I am simply multiplying by $T$ to obtain the desired form. I am skeptical about this argument and expect that I should need a derivative of $U$ with respect to $S$ somewhere to obtain the $T$.

As far as I know, the step (1) requires that, at least "locally", the relation $U = U(S,V,N,I)$ instead be expressible as $0 = g(S,B_e,I,V,N)$ for some function $g$ (and similarly for $T$ and $U$ for $I$ on the RHS); why should I expect this to be true? I've seen conjugate pairs replace each other ($B_e,I$ here) but here we seem to be replacing $U$ with $B_e$?

Answer 1

The fact that the variables are intensive/extensive is irrelevant. These partial derivatives just come from differential calculus.

Slight notation change, I’ll use $M$ for the magnetic moment instead of $I$. By definition, you start with: $$ dU=TdS+BdM $$ I dropped the $V,N$ dependencies since they are irrelevant for your question being always kept constant.

As you noticed, your first equality is merely Schwarz’ theorem, i.e. a Maxwell relation.

For step (1), the reasoning is correct since it is always the same variable that is kept constant, so you are reduced to the case of 1 independent variable: $$ \frac{dy}{dx}=\frac{1}{\frac{dx}{dy}} $$

When in doubt, it is best to think of differentials of independent variables as a basis in the linear algebra sense (this reduction to linear algebra is after all the main purpose of differential calculus). The local inversion of the functions is guaranteed by the implicit function theorem.

The goal is to express the partial derivatives in the $T,S$ basis (and the $B,M$ basis) in terms of the $S,M$ basis where you have Maxwell’s relation.

I did the calculation in a recent answer (replace $p,V$ with $B,M$). I’ll detail the intermediate steps in the following: $$ \begin{align} dS &=\left(\frac{\partial S}{\partial B}\right)_MdB+\left(\frac{\partial S}{\partial M}\right)_BdM\\ dB &= \frac{1}{\left(\frac{\partial S}{\partial B}\right)_M}dS-\frac{\left(\frac{\partial S}{\partial M}\right)_B} {\left(\frac{\partial S}{\partial B}\right)_M}dM\\ \left(\frac{\partial B}{\partial S}\right)_M &= \frac{1}{\left(\frac{\partial S}{\partial B}\right)_M}\\ \left(\frac{\partial B}{\partial M}\right)_S &= -\frac{\left(\frac{\partial S}{\partial M}\right)_B} {\left(\frac{\partial S}{\partial B}\right)_M} \end{align} $$ and similarly: $$ \begin{align} dM &=\left(\frac{\partial M}{\partial T}\right)_SdT+\left(\frac{\partial M}{\partial S}\right)_TdS\\ dT &= \frac{1}{\left(\frac{\partial M}{\partial T}\right)_S}dM-\frac{\left(\frac{\partial M}{\partial S}\right)_T} {\left(\frac{\partial M}{\partial T}\right)_S}dS\\ \left(\frac{\partial T}{\partial M}\right)_S &= \frac{1}{\left(\frac{\partial M}{\partial T}\right)_S}\\ \left(\frac{\partial T}{\partial S}\right)_M &= -\frac{\left(\frac{\partial M}{\partial S}\right)_T} {\left(\frac{\partial M}{\partial T}\right)_S} \end{align} $$

In particular, I recover the formulas used in (1). In fact the expression of the extra partial derivatives I calculated give you the necessary ingredients to derive your reciprocity theorem.

Hope this helps.