Categories: Physics, Thermodynamics.

# Thermodynamic potential

Thermodynamic potentials are state functions whose minima or maxima represent equilibrium states of a system. Such functions are either energies (hence potential) or entropies.

Which potential (of many) decides the equilibrium states for a given system? That depends which variables are assumed to already be in automatic equilibrium. Such variables are known as the natural variables of that potential. For example, if a system can freely exchange heat with its surroundings, and is consequently assumed to be at the same temperature $$T = T_{\mathrm{sur}}$$, then $$T$$ must be a natural variable.

The link from natural variables to potentials is established by thermodynamic ensembles.

Once enough natural variables have been found, the appropriate potential can be selected from the list below. All non-natural variables can then be calculated by taking partial derivatives of the potential with respect to the natural variables.

Mathematically, the potentials are related to each other by Legendre transformation.

## Internal energy

The internal energy $$U$$ represents the capacity to do both mechanical and non-mechanical work, and to release heat. It is simply the integral of the fundamental thermodynamic relation:

\begin{aligned} \boxed{ U(S, V, N) \equiv T S - P V + \mu N } \end{aligned}

It is a function of the entropy $$S$$, volume $$V$$, and particle count $$N$$: these are its natural variables. An infinitesimal change $$\dd{U}$$ is as follows:

\begin{aligned} \boxed{ \dd{U} = T \dd{S} - P \dd{V} + \mu \dd{N} } \end{aligned}

The non-natural variables are temperature $$T$$, pressure $$P$$, and chemical potential $$\mu$$. They can be recovered by differentiating $$U$$ with respect to the natural variables $$S$$, $$V$$, and $$N$$:

\begin{aligned} \boxed{ T = \Big( \pdv{U}{S} \Big)_{V,N} \qquad P = - \Big( \pdv{U}{V} \Big)_{S,N} \qquad \mu = \Big( \pdv{U}{N} \Big)_{S,V} } \end{aligned}

It is convention to write those subscripts, to help keep track of which function depends on which variables. They are meaningless; these are normal partial derivatives.

## Enthalpy

The enthalpy $$H$$ of a system, in units of energy, represents its capacity to do non-mechanical work, plus its capacity to release heat. It is given by:

\begin{aligned} \boxed{ H(S, P, N) \equiv U + P V } \end{aligned}

It is a function of the entropy $$S$$, pressure $$P$$, and particle count $$N$$: these are its natural variables. An infinitesimal change $$\dd{H}$$ is as follows:

\begin{aligned} \boxed{ \dd{H} = T \dd{S} + V \dd{P} + \mu \dd{N} } \end{aligned}

The non-natural variables are temperature $$T$$, volume $$V$$, and chemical potential $$\mu$$. They can be recovered by differentiating $$H$$ with respect to the natural variables $$S$$, $$P$$, and $$N$$:

\begin{aligned} \boxed{ T = \Big( \pdv{H}{S} \Big)_{P,N} \qquad V = \Big( \pdv{H}{P} \Big)_{S,N} \qquad \mu = \Big( \pdv{H}{N} \Big)_{S,P} } \end{aligned}

## Helmholtz free energy

The Helmholtz free energy $$F$$ represents the capacity of a system to do both mechanical and non-mechanical work, and is given by:

\begin{aligned} \boxed{ F(T, V, N) \equiv U - T S } \end{aligned}

It depends on the temperature $$T$$, volume $$V$$, and particle count $$N$$: these are natural variables. An infinitesimal change $$\dd{H}$$ is as follows:

\begin{aligned} \boxed{ \dd{F} = - P \dd{V} - S \dd{T} + \mu \dd{N} } \end{aligned}

The non-natural variables are entropy $$S$$, pressure $$P$$, and chemical potential $$\mu$$. They can be recovered by differentiating $$F$$ with respect to the natural variables $$T$$, $$V$$, and $$N$$:

\begin{aligned} \boxed{ S = - \Big( \pdv{F}{T} \Big)_{V,N} \qquad P = - \Big( \pdv{F}{V} \Big)_{T,N} \qquad \mu = \Big( \pdv{F}{N} \Big)_{T,V} } \end{aligned}

## Gibbs free energy

The Gibbs free energy $$G$$ represents the capacity of a system to do non-mechanical work:

\begin{aligned} \boxed{ G(T, P, N) \equiv U + P V - T S } \end{aligned}

It depends on the temperature $$T$$, pressure $$P$$, and particle count $$N$$: they are natural variables. An infinitesimal change $$\dd{G}$$ is as follows:

\begin{aligned} \boxed{ \dd{G} = V \dd{P} - S \dd{T} + \mu \dd{N} } \end{aligned}

The non-natural variables are entropy $$S$$, volume $$V$$, and chemical potential $$\mu$$. These can be recovered by differentiating $$G$$ with respect to the natural variables $$T$$, $$P$$, and $$N$$:

\begin{aligned} \boxed{ S = - \Big( \pdv{G}{T} \Big)_{P,N} \qquad V = \Big( \pdv{G}{P} \Big)_{T,N} \qquad \mu = \Big( \pdv{G}{N} \Big)_{T,P} } \end{aligned}

## Landau potential

The Landau potential or grand potential $$\Omega$$, in units of energy, represents the capacity of a system to do mechanical work, and is given by:

\begin{aligned} \boxed{ \Omega(T, V, \mu) \equiv U - T S - \mu N } \end{aligned}

It depends on temperature $$T$$, volume $$V$$, and chemical potential $$\mu$$: these are natural variables. An infinitesimal change $$\dd{\Omega}$$ is as follows:

\begin{aligned} \boxed{ \dd{\Omega} = - P \dd{V} - S \dd{T} - N \dd{\mu} } \end{aligned}

The non-natural variables are entropy $$S$$, pressure $$P$$, and particle count $$N$$. These can be recovered by differentiating $$\Omega$$ with respect to the natural variables $$T$$, $$V$$, and $$\mu$$:

\begin{aligned} \boxed{ S = - \Big( \pdv{\Omega}{T} \Big)_{V,\mu} \qquad P = - \Big( \pdv{\Omega}{V} \Big)_{T,\mu} \qquad N = - \Big( \pdv{\Omega}{\mu} \Big)_{T,V} } \end{aligned}

## Entropy

The entropy $$S$$, in units of energy over temperature, is an odd duck, but nevertheless used as a thermodynamic potential. It is given by:

\begin{aligned} \boxed{ S(U, V, N) \equiv \frac{1}{T} U + \frac{P}{T} V - \frac{\mu}{T} N } \end{aligned}

It depends on the internal energy $$U$$, volume $$V$$, and particle count $$N$$: they are natural variables. An infinitesimal change $$\dd{S}$$ is as follows:

\begin{aligned} \boxed{ \dd{S} = \frac{1}{T} \dd{U} + \frac{P}{T} \dd{V} - \frac{\mu}{T} \dd{N} } \end{aligned}

The non-natural variables are $$1/T$$, $$P/T$$, and $$\mu/T$$. These can be recovered by differentiating $$S$$ with respect to the natural variables $$U$$, $$V$$, and $$N$$:

\begin{aligned} \boxed{ \frac{1}{T} = \Big( \pdv{S}{U} \Big)_{V,N} \qquad \frac{P}{T} = \Big( \pdv{S}{V} \Big)_{U,N} \qquad \frac{\mu}{T} = - \Big( \pdv{S}{N} \Big)_{U,V} } \end{aligned}

## References

1. H. Gould, J. Tobochnik, Statistical and thermal physics, 2nd edition, Princeton.