2.16: Appendix III- Useful Mathematical Relations

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Consider a set of $n$ independent variables $\{x\ns_1,\ldots,x\ns_n\}$, which can be thought of as a point in $n$-dimensional space. Let $\{y\ns_1,\ldots,y\ns_n\}$ and $\{z\ns_1,\ldots,z\ns_n\}$ be other choices of coordinates. Then $${\pz x_i\over \pz z\ns_k}={\pz x_i\over\pz y\ns_j}\,{\pz y_j\over\pz z\ns_k}\ .$$ Note that this entails a matrix multiplication: $A\ns_{ik}=B\ns_{ij}\,C\ns_{jk}$, where $A\ns_{ik}=\pz x\ns_i/\pz z\ns_k$, $B\ns_{ij}=\pz x\ns_i/\pz y\ns_j$, and $C\ns_{jk}=\pz y\ns_j/\pz z\ns_k$. We define the determinant $${det}\bigg({\pz x\ns_i\over\pz z\ns_k}\bigg)\equiv {\pz (x\ns_1,\ldots,x\ns_n)\over \pz (z\ns_1,\ldots,z\ns_n)}\ .$$ Such a determinant is called a Jacobian. Now if $A=BC$, then ${det}(A)={det}(B)\cdot{det}(C)$. Thus, $${\pz (x\ns_1,\ldots, x\ns_n)\over \pz( z\ns_1,\ldots, z\ns_n)}= {\pz( x\ns_1,\ldots, x\ns_n)\over \pz( y\ns_1,\ldots, y\ns_n)}\cdot {\pz( y\ns_1,\ldots, y\ns_n)\over \pz( z\ns_1,\ldots, z\ns_n)}\ . \label{chain}$$ Recall also that $${\pz x\ns_i\over \pz x\ns_k}=\delta\ns_{ik}\ .$$

Consider the case $n=2$. We have $${\pz( x, y)\over \pz( u, v)}={det}\begin{pmatrix} \pabc{x}{u}{v} & \pabc {x}{v}{u} \\ & \\ \pabc{y}{u}{v} & \pabc{y}{v}{u} \end{pmatrix} = \pabc{x}{u}{v} \pabc{y}{v}{u} - \pabc {x}{v}{u} \pabc{y}{u}{v} \ .$$ We also have $${\pz( x, y)\over \pz( u, v)}\cdot {\pz( u, v)\over \pz( r, s)} = {\pz( x, y)\over \pz( r, s)}\ .$$ From this simple mathematics follows several very useful results.

1) First, write $${\pz(x, y)\over \pz(u, v)}=\Bigg[{\pz(u, v)\over \pz(x, y)}\Bigg]^{-1}\ .$$ Now let $v=y$ : $${\pz(x, y)\over \pz(u, y)}=\pabc{x}{u}{y}={1\over\pabc{u}{x}{y}}\ .$$ Thus, $$\pabc{x}{u}{y}=1\Big/\pabc{u}{x}{y} \label{boxone}$$

2) Second, we have $${\pz(x, y)\over \pz(u, y)}=\pabc{x}{u}{y} \nonumber\\ ={\pz(x, y)\over \pz(x, u)}\cdot {\pz(x, u)\over \pz(u, y)}\\ =-\pabc{y}{u}{x}\pabc{x}{y}{u}\ ,$$ which is to say $$\pabc{x}{y}{u} \pabc{y}{u}{x} = - \pabc{x}{u}{y}\ . \label{boxtwob}$$ Invoking Equation [boxone], we conclude that $$\pabc{x}{y}{u} \pabc{y}{u}{x} \pabc{u}{x}{y} = -1\ . \label{boxtwo}$$

3) Third, we have $${\pz(x, v)\over \pz(u, v)}={\pz(x, v)\over \pz(y, v)}\cdot {\pz(y, v)\over \pz(u, v)}\ ,$$ which says $$\pabc{x}{u}{v}=\pabc{x}{y}{v}\pabc{y}{u}{v} \label{boxthree}$$ This is simply the chain rule of partial differentiation.

4) Fourth, we have $$\begin{split} {\pz(x,y)\over\pz(u,y)}&={\pz(x,y)\over\pz(u,v)}\cdot{\pz(u,v)\over\pz(u,y)}\\ &=\pabc{x}{u}{v}\pabc{y}{v}{u}\pabc{v}{y}{u}-\pabc{x}{v}{u}\pabc{y}{u}{v}\pabc{v}{y}{u}\ , \end{split}$$ which says $$\pabc{x}{u}{y}=\pabc{x}{u}{v}-\pabc{x}{y}{u}\pabc{y}{u}{v} \label{boxfour}$$

5) Fifth, whenever we differentiate one extensive quantity with respect to another, holding only intensive quantities constant, the result is simply the ratio of those extensive quantities. For example, $$\pabc{S}{V}{p,T}={S\over V}\ .$$ The reason should be obvious. In the above example, $S(p,V,T)=V\phi(p,T)$, where $\phi$ is a function of the two intensive quantities $p$ and $T$. Hence differentiating $S$ with respect to $V$ holding $p$ and $T$ constant is the same as dividing $S$ by $V$. Note that this implies $$\pabc{S}{V}{p,T}=\pabc{S}{V}{p,\mu}=\pabc{S}{V}{n,T}={S\over V}\ ,$$ where $n=N/V$ is the particle density.

6) Sixth, suppose we have a function $\Phi(y,v)$ and we write $$d\Phi=x\,dy + u\,dv\ .$$ That is, $$x=\pabc{\Phi}{y}{v}\equiv \Phi\ns_y \qquad,\qquad u=\pabc{\Phi}{v}{y}\equiv \Phi\ns_v\ .$$ Now we may write $$\begin{aligned} dx&=\Phi\ns_{yy}\,dy + \Phi\ns_{yv}\,dv \label{dxe} \\ du&=\Phi\ns_{vy}\,dy + \Phi\ns_{vv}\,dv\ .\label{due}\end{aligned}$$ If we demand $du=0$, this yields $$\pabc{x}{u}{v}={\Phi\ns_{yy}\over \Phi\ns_{vy}}\ . \label{pxuv}$$ Note that $\Phi\ns_{vy}=\Phi\ns_{yv}$ . From the equation $du=0$ we also derive $$\pabc{y}{v}{u}=-{\Phi\ns_{vv}\over \Phi\ns_{vy}}\ . \label{pyvu}$$ Next, we use Equation [due] with $du=0$ to eliminate $dy$ in favor of $dv$, and then substitute into Equation [dxe]. This yields $$\pabc{x}{v}{u}=\Phi\ns_{yv}-{\Phi\ns_{yy}\,\Phi\ns_{vv}\over \Phi\ns_{vy}}\ . \label{pxvu}$$ Finally, Equation [due] with $dv=0$ yields $$\pabc{y}{u}{v}={1\over \Phi\ns_{vy}}\ . \label{pyuv}$$

Combining the results of eqns. [pxuv], [pyvu], [pxvu], and [pyuv], we have $$\begin{split} {\pz(x,y)\over\pz(u,v)}&=\pabc{x}{u}{v}\pabc{y}{v}{u} - \pabc{x}{v}{u}\pabc{y}{u}{v}\\ &=\bigg({\Phi\ns_{yy}\over \Phi\ns_{vy}}\bigg)\bigg(-{\Phi\ns_{vv}\over \Phi\ns_{vy}}\bigg)- \bigg(\Phi\ns_{yv}-{\Phi\ns_{yy}\,\Phi\ns_{vv}\over \Phi\ns_{vy}}\bigg)\bigg({1\over \Phi\ns_{vy}}\bigg)\bvph=-1\ . \label{jacob} \end{split}$$ Thus, if $\Phi=E(S,V)$, then $(x,y)=(T,S)$ and $(u,v)=(-p,V)$, we have $${\pz(T,S)\over\pz(-p,V)}=-1\ . \label{detTSpV}$$

Nota bene: It is important to understand what other quantities are kept constant, otherwise we can run into trouble. For example, it would seem that Equation [jacob] would also yield $${\pz(\mu,N)\over\pz(p,V)}=1\ . \label{detmuNpV}$$ But then we should have $${\pz(T,S)\over\pz(\mu,N)}={\pz(T,S)\over\pz(-p,V)}\cdot {\pz(-p,V)\over\pz(\mu,N)} = +1 \qquad\hbox{(WRONG!)}$$ when according to Equation [jacob] it should be $-1$. What has gone wrong?

The problem is that we have not properly specified what else is being held constant. In Equation [detTSpV] it is $N$ (or $\mu$) which is being held constant, while in Equation [detmuNpV] it is $S$ (or $T$) which is being held constant. Therefore a naive application of the chain rule for determinants yields the wrong result, as we have seen.

Let’s be more careful. Applying the same derivation to $dE=x\,dy + u\,dv + r\,ds$ and holding $s$ constant, we conclude $${\pz(x,y,s)\over\pz(u,v,s)}=\pabc{x}{u}{v,s}\pabc{y}{v}{u,s}-\ \pabc{x}{v}{u,s}\pabc{y}{u}{v,s} = -1\ .$$ Thus, if $$dE= T\,dS + y\,dX + \mu\,dN\quad,$$ where $(y,X)=(-p,V)$ or $(H^\alpha,M^\alpha)$ or $(E^\alpha,P^\alpha)$, the appropriate thermodynamic relations are $$\begin{aligned} {\pz(T,S,N)\over\pz(y,X,N)}&=-1 & {\pz(T,S,\mu)\over\pz(y,X,\mu)}&=-1 \nonumber \\ {\pz(\mu,N,X)\over\pz(T,S,X)}&=-1 & {\pz(\mu,N,y)\over\pz(T,S,y)}&=-1 \bvph \label{TSyXN} \\ {\pz(y,X,S)\over\pz(\mu,N,S)}&=-1 & {\pz(y,X,T)\over\pz(\mu,N,T)}&=-1 \nonumber\end{aligned}$$ For example, $${\pz(T,S,N)\over\pz(-p,V,N)}={\pz(-p,V,S)\over\pz(\mu,N,S)}={\pz(\mu,N,V)\over\pz(T,S,V)}=-1$$ and $${\pz(T,S,\mu)\over\pz(-p,V,\mu)}={\pz(-p,V,T)\over\pz(\mu,N,T)}={\pz(\mu,N,-p)\over\pz(T,S,-p)}=-1\ .$$

If we are careful, then the results in eq. [TSyXN] can be quite handy, especially when used in conjunction with Equation [chain]. For example, we have \[\pabc{S}{V}{T,N}={\pz(T,S,N)\over\pz(T,V,N)}=\stackrel{=\,1}{\overbrace

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} \cdot{\pz(p,V,N)\over\pz(T,V,N)}=\pabc{p}{T}{V,N}\ ,\] which is one of the Maxwell relations derived from the exactness of $dF(T,V,N)$. Some other examples include \[\begin{aligned} \pabc{V}{S}{p,N}&={\pz(V,p,N)\over\pz(S,p,N)}=\stackrel{=\,1}{\overbrace

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}\cdot{\pz(S,T,N)\over\pz(S,p,N)}=\pabc{T}{p}{S,N}\\ \pabc{S}{N}{T,p}&={\pz(S,T,p)\over\pz(N,T,p)}=\stackrel{=\,1}{\overbrace

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} \cdot{\pz(\mu,N,p)\over\pz(N,T,p)}=-\pabc{\mu}{T}{p,N}\ ,\bvph\end{aligned}\] which are Maxwell relations deriving from $d\CH(S,p,N)$ and $dG(T,p,N)$, respectively. Note that due to the alternating nature of the determinant – it is antisymmetric under interchange of any two rows or columns – we have $${\pz(x,y,z)\over\pz(u,v,w)}=-{\pz (y,x,z)\over\pz (u,v,w)} = {\pz (y,x,z)\over\pz (w,v,u)} = \ldots \ .$$

In general, it is usually advisable to eliminate $S$ from a Jacobian. If we have a Jacobian involving $T$, $S$, and $N$, we can write $${\pz(T,S,N)\over\pz(\cds,\cds,N)}={\pz(T,S,N)\over\pz(p,V,N)}\,{\pz(p,V,N)\over\pz(\cds,\cds,N)} ={\pz(p,V,N)\over\pz(\cds,\cds,N)}\ ,$$ where each $\cds$ is a distinct arbitrary state variable other than $N$.

If our Jacobian involves the $S$, $V$, and $N$, we write $${\pz(S,V,N)\over\pz(\cds,\cds,N)}={\pz(S,V,N)\over\pz(T,V,N)}\cdot{\pz(T,V,N)\over\pz(\cds,\cds,N)}={C\ns_V\over T}\cdot {\pz(T,V,N)\over\pz(\cds,\cds,N)}\ .$$

If our Jacobian involves the $S$, $p$, and $N$, we write $${\pz(S,p,N)\over\pz(\cds,\cds,N)}={\pz(S,p,N)\over\pz(T,p,N)}\cdot{\pz(T,p,N)\over\pz(\cds,\cds,N)}={C\ns_p\over T}\cdot {\pz(T,p,N)\over\pz(\cds,\cds,N)}\ .$$

For example, \[\begin{aligned} \pabc{T}{p}{S,N}&={\pz(T,S,N)\over\pz(p,S,N)}=\stackrel{=\,1}{\overbrace

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} \cdot{\pz(p,V,N)\over\pz(p,T,N)}\cdot {\pz(p,T,N)\over\pz(p,S,N)}={T\over C\ns_p}\pabc{V}{T}{p,N}\bvph\\ \pabc{V}{p}{S,N}&={\pz(V,S,N)\over\pz(p,S,N)}={\pz(V,S,N)\over\pz(V,T,N)}\cdot{\pz(V,T,N)\over\pz(p,T,N)}\cdot{\pz(p,T,N)\over\pz(p,S,N)} ={C\ns_V\over C\ns_p}\,\pabc{V}{p}{T,N}\ .\bvph\end{aligned}\] With $\kappa\equiv -{1\over V}\,{\pz V\over\pz p}$ the compressibility, we see that the second of these equations says $\kappa\ns_T\,c\ns_V=\kappa\ns_S\,c\ns_p$ , relating the isothermal and adiabatic compressibilities and the molar heat capacities at constant volume and constant pressure. This relation was previously established in Equation [cpcvktks]