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 $0 = g(S,B_e,I,V,N)$ (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 $B_e$?

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 $B_e$?

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.

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 $0 = g(S,B_e,I,V,N)$ (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 $B_e$?