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Maxwell's relations are a set of equations in thermodynamics which are derivable from the symmetry of second derivatives and from the definitions of the thermodynamic potentials. These relations are named for the nineteenth-century physicist James Clerk Maxwell.

Equations

The structure of Maxwell relations is a statement of equality among the second derivatives for continuous functions. It follows directly from the fact that the order of differentiation of an analytic function of two variables is irrelevant (Schwarz theorem). In the case of Maxwell relations the function considered is a thermodynamic potential and $x_{i}$ and $x_{j}$ are two different natural variables for that potential, we have

Schwarz's theorem (general)

${\frac {\partial }{\partial x_{j}}}\left({\frac {\partial \Phi }{\partial x_{i}}}\right)={\frac {\partial }{\partial x_{i}}}\left({\frac {\partial \Phi }{\partial x_{j}}}\right)$

where the partial derivatives are taken with all other natural variables held constant. For every thermodynamic potential there are ${\textstyle {\frac {1}{2}}n(n-1)}$ possible Maxwell relations where $n$ is the number of natural variables for that potential.

The four most common Maxwell relations

The four most common Maxwell relations are the equalities of the second derivatives of each of the four thermodynamic potentials, with respect to their thermal natural variable (temperature $T$ , or entropy $S$ ) and their mechanical natural variable (pressure $P$ , or volume $V$ ):

Maxwell's relations (common)

${\begin{aligned}+\left({\frac {\partial T}{\partial V}}\right)_{S}&=&-\left({\frac {\partial P}{\partial S}}\right)_{V}&=&{\frac {\partial ^{2}U}{\partial S\partial V}}\\+\left({\frac {\partial T}{\partial P}}\right)_{S}&=&+\left({\frac {\partial V}{\partial S}}\right)_{P}&=&{\frac {\partial ^{2}H}{\partial S\partial P}}\\+\left({\frac {\partial S}{\partial V}}\right)_{T}&=&+\left({\frac {\partial P}{\partial T}}\right)_{V}&=&-{\frac {\partial ^{2}F}{\partial T\partial V}}\\-\left({\frac {\partial S}{\partial P}}\right)_{T}&=&+\left({\frac {\partial V}{\partial T}}\right)_{P}&=&{\frac {\partial ^{2}G}{\partial T\partial P}}\end{aligned}}\,\!$

where the potentials as functions of their natural thermal and mechanical variables are the internal energy $U(S,V)$ , enthalpy $H(S,P)$ , Helmholtz free energy $F(T,V)$ , and Gibbs free energy $G(T,P)$ . The thermodynamic square can be used as a mnemonic to recall and derive these relations. The usefulness of these relations lies in their quantifying entropy changes, which are not directly measurable, in terms of measurable quantities like temperature, volume, and pressure.

Each equation can be re-expressed using the relationship $\left({\frac {\partial y}{\partial x}}\right)_{z}=1\left/\left({\frac {\partial x}{\partial y}}\right)_{z}\right.$ which are sometimes also known as Maxwell relations.

Derivations

Short derivation

This section is based on chapter 5 of.^[1]

Suppose we are given four real variables $(x,y,z,w)$ , restricted to move on a 2-dimensional $C^{2}$ surface in $\mathbb {R} ^{4}$ . Then, if we know two of them, we can determine the other two uniquely (generically).

In particular, we may take any two variables as the independent variables, and let the other two be the dependent variables, then we can take all these partial derivatives.

Proposition: $\left({\frac {\partial w}{\partial y}}\right)_{z}=\left({\frac {\partial w}{\partial x}}\right)_{z}\left({\frac {\partial x}{\partial y}}\right)_{z}$

Proof: This is just the chain rule.

Proposition: $\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}\left({\frac {\partial z}{\partial x}}\right)_{y}=-1$

Proof. We can ignore $w$ . Then locally the surface is just $ax+by+cz+d=0$ . Then $\left({\frac {\partial x}{\partial y}}\right)_{z}=-{\frac {b}{a}}$ , etc. Now multiply them.

Proof of Maxwell's relations:

There are four real variables $(T,S,p,V)$ , restricted on the 2-dimensional surface of possible thermodynamic states. This allows us to use the previous two propositions.

It suffices to prove the first of the four relations, as the other three can be obtained by transforming the first relation using the previous two propositions. Pick $V,S$ as the independent variables, and $E$ as the dependent variable. We have $dE=-pdV+TdS$ .

Now, $\partial _{V,S}E=\partial _{S,V}E$ since the surface is $C^{2}$ , that is, $\left({\frac {\partial \left({\frac {\partial E}{\partial S}}\right)_{V}}{\partial V}}\right)_{S}=\left({\frac {\partial \left({\frac {\partial E}{\partial V}}\right)_{S}}{\partial S}}\right)_{V}$ which yields the result.

Another derivation

Based on.^[2]

Since $dU=TdS-PdV$ , around any cycle, we have $0=\oint dU=\oint TdS-\oint PdV$ Take the cycle infinitesimal, we find that ${\frac {\partial (P,V)}{\partial (T,S)}}=1$ . That is, the map is area-preserving. By the chain rule for Jacobians, for any coordinate transform $(x,y)$ , we have ${\frac {\partial (P,V)}{\partial (x,y)}}={\frac {\partial (T,S)}{\partial (x,y)}}$ Now setting $(x,y)$ to various values gives us the four Maxwell relations. For example, setting $(x,y)=(P,S)$ gives us $\left({\frac {\partial T}{\partial P}}\right)_{S}=\left({\frac {\partial V}{\partial S}}\right)_{P}$

Extended derivations

Maxwell relations are based on simple partial differentiation rules, in particular the total differential of a function and the symmetry of evaluating second order partial derivatives.

Derivation

Derivation of the Maxwell relation can be deduced from the differential forms of the thermodynamic potentials:
The differential form of internal energy $U$ is $dU=T\,dS-P\,dV$ This equation resembles total differentials of the form $dz=\left({\frac {\partial z}{\partial x}}\right)_{y}\!dx+\left({\frac {\partial z}{\partial y}}\right)_{x}\!dy$ It can be shown, for any equation of the form, $dz=M\,dx+N\,dy$ that $M=\left({\frac {\partial z}{\partial x}}\right)_{y},\quad N=\left({\frac {\partial z}{\partial y}}\right)_{x}$ Consider, the equation $dU=T\,dS-P\,dV$ . We can now immediately see that $T=\left({\frac {\partial U}{\partial S}}\right)_{V},\quad -P=\left({\frac {\partial U}{\partial V}}\right)_{S}$ Since we also know that for functions with continuous second derivatives, the mixed partial derivatives are identical (Symmetry of second derivatives), that is, that ${\frac {\partial }{\partial y}}\left({\frac {\partial z}{\partial x}}\right)_{y}={\frac {\partial }{\partial x}}\left({\frac {\partial z}{\partial y}}\right)_{x}={\frac {\partial ^{2}z}{\partial y\partial x}}={\frac {\partial ^{2}z}{\partial x\partial y}}$ we therefore can see that ${\frac {\partial }{\partial V}}\left({\frac {\partial U}{\partial S}}\right)_{V}={\frac {\partial }{\partial S}}\left({\frac {\partial U}{\partial V}}\right)_{S}$ and therefore that $\left({\frac {\partial T}{\partial V}}\right)_{S}=-\left({\frac {\partial P}{\partial S}}\right)_{V}$

Derivation of Maxwell Relation from Helmholtz Free energy

The differential form of Helmholtz free energy is $dF=-S\,dT-P\,dV$ $-S=\left({\frac {\partial F}{\partial T}}\right)_{V},\quad -P=\left({\frac {\partial F}{\partial V}}\right)_{T}$ From symmetry of second derivatives ${\frac {\partial }{\partial V}}\left({\frac {\partial F}{\partial T}}\right)_{V}={\frac {\partial }{\partial T}}\left({\frac {\partial F}{\partial V}}\right)_{T}$ and therefore that $\left({\frac {\partial S}{\partial V}}\right)_{T}=\left({\frac {\partial P}{\partial T}}\right)_{V}$ The other two Maxwell relations can be derived from differential form of enthalpy $dH=T\,dS+V\,dP$ and the differential form of Gibbs free energy $dG=V\,dP-S\,dT$ in a similar way. So all Maxwell Relationships above follow from one of the Gibbs equations.

Extended derivation

Combined form first and second law of thermodynamics,

T\,dS=dU+P\,dV

(Eq.1)

$U$ , $S$ , and $V$ are state functions. Let,

$U=U(x,y)$
$S=S(x,y)$
$V=V(x,y)$
$dU=\left({\frac {\partial U}{\partial x}}\right)_{y}\!dx+\left({\frac {\partial U}{\partial y}}\right)_{x}\!dy$
$dS=\left({\frac {\partial S}{\partial x}}\right)_{y}\!dx+\left({\frac {\partial S}{\partial y}}\right)_{x}\!dy$
$dV=\left({\frac {\partial V}{\partial x}}\right)_{y}\!dx+\left({\frac {\partial V}{\partial y}}\right)_{x}\!dy$

Substitute them in Eq.1 and one gets, $T\left({\frac {\partial S}{\partial x}}\right)_{y}\!dx+T\left({\frac {\partial S}{\partial y}}\right)_{x}\!dy=\left({\frac {\partial U}{\partial x}}\right)_{y}\!dx+\left({\frac {\partial U}{\partial y}}\right)_{x}\!dy+P\left({\frac {\partial V}{\partial x}}\right)_{y}\!dx+P\left({\frac {\partial V}{\partial y}}\right)_{x}\!dy$ And also written as, $\left({\frac {\partial U}{\partial x}}\right)_{y}\!dx+\left({\frac {\partial U}{\partial y}}\right)_{x}\!dy=T\left({\frac {\partial S}{\partial x}}\right)_{y}\!dx+T\left({\frac {\partial S}{\partial y}}\right)_{x}\!dy-P\left({\frac {\partial V}{\partial x}}\right)_{y}\!dx-P\left({\frac {\partial V}{\partial y}}\right)_{x}\!dy$ comparing the coefficient of dx and dy, one gets $\left({\frac {\partial U}{\partial x}}\right)_{y}=T\left({\frac {\partial S}{\partial x}}\right)_{y}-P\left({\frac {\partial V}{\partial x}}\right)_{y}$ $\left({\frac {\partial U}{\partial y}}\right)_{x}=T\left({\frac {\partial S}{\partial y}}\right)_{x}-P\left({\frac {\partial V}{\partial y}}\right)_{x}$ Differentiating above equations by $y$ , $x$ respectively

\left({\frac {\partial ^{2}U}{\partial y\partial x}}\right)=\left({\frac {\partial T}{\partial y}}\right)_{x}\left({\frac {\partial S}{\partial x}}\right)_{y}+T\left({\frac {\partial ^{2}S}{\partial y\partial x}}\right)-\left({\frac {\partial P}{\partial y}}\right)_{x}\left({\frac {\partial V}{\partial x}}\right)_{y}-P\left({\frac {\partial ^{2}V}{\partial y\partial x}}\right)

(Eq.2)

and

\left({\frac {\partial ^{2}U}{\partial x\partial y}}\right)=\left({\frac {\partial T}{\partial x}}\right)_{y}\left({\frac {\partial S}{\partial y}}\right)_{x}+T\left({\frac {\partial ^{2}S}{\partial x\partial y}}\right)-\left({\frac {\partial P}{\partial x}}\right)_{y}\left({\frac {\partial V}{\partial y}}\right)_{x}-P\left({\frac {\partial ^{2}V}{\partial x\partial y}}\right)

(Eq.3)

$U$ , $S$ , and $V$ are exact differentials, therefore, $\left({\frac {\partial ^{2}U}{\partial y\partial x}}\right)=\left({\frac {\partial ^{2}U}{\partial x\partial y}}\right)$ $\left({\frac {\partial ^{2}S}{\partial y\partial x}}\right)=\left({\frac {\partial ^{2}S}{\partial x\partial y}}\right)$ $\left({\frac {\partial ^{2}V}{\partial y\partial x}}\right)=\left({\frac {\partial ^{2}V}{\partial x\partial y}}\right)$ Subtract Eq.2 and Eq.3 and one gets $\left({\frac {\partial T}{\partial y}}\right)_{x}\left({\frac {\partial S}{\partial x}}\right)_{y}-\left({\frac {\partial P}{\partial y}}\right)_{x}\left({\frac {\partial V}{\partial x}}\right)_{y}=\left({\frac {\partial T}{\partial x}}\right)_{y}\left({\frac {\partial S}{\partial y}}\right)_{x}-\left({\frac {\partial P}{\partial x}}\right)_{y}\left({\frac {\partial V}{\partial y}}\right)_{x}$ Note: The above is called the general expression for Maxwell's thermodynamical relation.

Maxwell's first relation: Allow $x = S$ and $y = V$ and one gets; $\left({\frac {\partial T}{\partial V}}\right)_{S}=-\left({\frac {\partial P}{\partial S}}\right)_{V}$
Maxwell's second relation: Allow $x = T$ and $y = V$ and one gets; $\left({\frac {\partial S}{\partial V}}\right)_{T}=\left({\frac {\partial P}{\partial T}}\right)_{V}$
Maxwell's third relation: Allow $x = S$ and $y = P$ and one gets; $\left({\frac {\partial T}{\partial P}}\right)_{S}=\left({\frac {\partial V}{\partial S}}\right)_{P}$
Maxwell's fourth relation: Allow $x = T$ and $y = P$ and one gets; $\left({\frac {\partial S}{\partial P}}\right)_{T}=-\left({\frac {\partial V}{\partial T}}\right)_{P}$
Maxwell's fifth relation: Allow $x = P$ and $y = V$ and one gets; $\left({\frac {\partial T}{\partial P}}\right)_{V}\left({\frac {\partial S}{\partial V}}\right)_{P}-\left({\frac {\partial T}{\partial V}}\right)_{P}\left({\frac {\partial S}{\partial P}}\right)_{V}=1$
Maxwell's sixth relation: Allow $x = T$ and $y = S$ and one gets; $\left({\frac {\partial P}{\partial T}}\right)_{S}\left({\frac {\partial V}{\partial S}}\right)_{T}-\left({\frac {\partial P}{\partial S}}\right)_{T}\left({\frac {\partial V}{\partial T}}\right)_{S}=1$

Derivation based on Jacobians

If we view the first law of thermodynamics, $dU=T\,dS-P\,dV$ as a statement about differential forms, and take the exterior derivative of this equation, we get $0=dT\,dS-dP\,dV$ since $d(dU)=0$ . This leads to the fundamental identity $dP\,dV=dT\,dS.$

The physical meaning of this identity can be seen by noting that the two sides are the equivalent ways of writing the work done in an infinitesimal Carnot cycle. An equivalent way of writing the identity is ${\frac {\partial (T,S)}{\partial (P,V)}}=1.$

The Maxwell relations now follow directly. For example, $\left({\frac {\partial S}{\partial V}}\right)_{T}={\frac {\partial (T,S)}{\partial (T,V)}}={\frac {\partial (P,V)}{\partial (T,V)}}=\left({\frac {\partial P}{\partial T}}\right)_{V},$ The critical step is the penultimate one. The other Maxwell relations follow in similar fashion. For example, $\left({\frac {\partial T}{\partial V}}\right)_{S}={\frac {\partial (T,S)}{\partial (V,S)}}={\frac {\partial (P,V)}{\partial (V,S)}}=-\left({\frac {\partial P}{\partial S}}\right)_{V}.$

General Maxwell relationships

The above are not the only Maxwell relationships. When other work terms involving other natural variables besides the volume work are considered or when the number of particles is included as a natural variable, other Maxwell relations become apparent. For example, if we have a single-component gas, then the number of particles N is also a natural variable of the above four thermodynamic potentials. The Maxwell relationship for the enthalpy with respect to pressure and particle number would then be:

$\left({\frac {\partial \mu }{\partial P}}\right)_{S,N}=\left({\frac {\partial V}{\partial N}}\right)_{S,P}\qquad ={\frac {\partial ^{2}H}{\partial P\partial N}}$

where $μ$ is the chemical potential. In addition, there are other thermodynamic potentials besides the four that are commonly used, and each of these potentials will yield a set of Maxwell relations. For example, the grand potential $\Omega (\mu ,V,T)$ yields:^[3] ${\begin{aligned}\left({\frac {\partial N}{\partial V}}\right)_{\mu ,T}&=&\left({\frac {\partial P}{\partial \mu }}\right)_{V,T}&=&-{\frac {\partial ^{2}\Omega }{\partial \mu \partial V}}\\\left({\frac {\partial N}{\partial T}}\right)_{\mu ,V}&=&\left({\frac {\partial S}{\partial \mu }}\right)_{V,T}&=&-{\frac {\partial ^{2}\Omega }{\partial \mu \partial T}}\\\left({\frac {\partial P}{\partial T}}\right)_{\mu ,V}&=&\left({\frac {\partial S}{\partial V}}\right)_{\mu ,T}&=&-{\frac {\partial ^{2}\Omega }{\partial V\partial T}}\end{aligned}}$

References

^ Pippard, A. B. (1957-01-01). Elements of Classical Thermodynamics:For Advanced Students of Physics (1st ed.). Cambridge: Cambridge University Press. ISBN 978-0-521-09101-5.
^ Ritchie, David J. (2002-02-01). "Answer to Question #78. A question about the Maxwell relations in thermodynamics". American Journal of Physics. 70 (2): 104–104. doi:10.1119/1.1410956. ISSN 0002-9505.
^ "Thermodynamic Potentials" (PDF). University of Oulu. Archived (PDF) from the original on 19 December 2022.

[1] Pippard, A. B. (1957-01-01). Elements of Classical Thermodynamics:For Advanced Students of Physics (1st ed.). Cambridge: Cambridge University Press. ISBN 978-0-521-09101-5.

[2] Ritchie, David J. (2002-02-01). "Answer to Question #78. A question about the Maxwell relations in thermodynamics". American Journal of Physics. 70 (2): 104–104. doi:10.1119/1.1410956. ISSN 0002-9505.

[3] "Thermodynamic Potentials" (PDF). University of Oulu. Archived (PDF) from the original on 19 December 2022.

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Maxwell relations