---
title: "Thermodynamic potential"
date: 2021-07-07
categories:
- Physics
- Thermodynamics
layout: "concept"
---

**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](/know/category/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](/know/concept/legendre-transform/).


## 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](/know/concept/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.