# Helmholtz free energy

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${\displaystyle S=k_{B}\,\ln \Omega }$
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Helmholtz free energy
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In thermodynamics, the Helmholtz free energy is a thermodynamic potential which measures the “useful” work obtainable from a closed thermodynamic system at a constant temperature and volume. For such a system, the negative of the difference in the Helmholtz energy is equal to the maximum amount of work extractable from a thermodynamic process in which temperature and volume are held constant. Under these conditions, it is minimized at equilibrium. The Helmholtz free energy was developed by Hermann von Helmholtz and is usually denoted by the letter A  (from the German “Arbeit” or work), or the letter F . The IUPAC recommends the letter A  as well as the use of name Helmholtz energy;[1]. In physics, A is called the Helmholtz function or simply “free energy”.

While Gibbs free energy is most commonly used as a measure of thermodynamic potential, especially in the field of chemistry, the isobaric restriction on that quantity is sometimes inconvenient. For example, in explosives research, Helmholtz free energy is often used since explosive reactions by their nature induce pressure changes.

## Definition

The Helmholtz energy is defined as:

${\displaystyle A\equiv U-TS\,}$[2]

where

• A  is the Helmholtz free energy (SI: joules, CGS: ergs),
• U  is the internal energy of the system (SI: joules, CGS: ergs),
• T  is the absolute temperature (kelvins),
• S  is the entropy (SI: joules per kelvin, CGS: ergs per kelvin).

## Mathematical development

From the first law of thermodynamics we have:

${\displaystyle {\rm {d}}U=\delta Q-\delta W\,}$

where ${\displaystyle U}$ is the internal energy, ${\displaystyle \delta Q}$ is the energy added by heating and ${\displaystyle \delta W}$ is the work done by the system. From the second law of thermodynamics, for a reversible process we may say that ${\displaystyle \delta Q=T{\rm {d}}S}$. Also, in case of a reversible change, the work done can be expressed as ${\displaystyle \delta W=pdV}$

Differentiating the expression for ${\displaystyle A}$  we have:

${\displaystyle {\rm {d}}A={\rm {d}}U-(T{\rm {d}}S+S{\rm {d}}T)\,}$
${\displaystyle =(T{\rm {d}}S-p\,{\rm {d}}V)-T{\rm {d}}S-S{\rm {d}}T\,}$
${\displaystyle =-p\,{\rm {d}}V-S{\rm {d}}T\,}$

This relation is also valid for a process which is not reversible because A is a thermodynamic function of state.

## Minimum free energy and maximum work principles

The laws of thermodynamics are only directly applicable to systems in thermal equilibrium. If we wish to describe phenomena like chemical reactions, then the best we can do is to consider suitably chosen initial and final states in which the system is in (metastable) thermal equilibrium. If the system is kept at fixed volume and is in contact with a heat bath at some constant temperature, then we can reason as follows.

Since the thermodynamical variables of the system are well defined in the initial state and the final state, the internal energy increase, ${\displaystyle \Delta U}$, the entropy increase ${\displaystyle \Delta S}$, and the work performed by the system, ${\displaystyle W}$, are well defined quantities. Conservation of energy implies:

${\displaystyle \Delta U_{\text{bath}}+\Delta U+W=0\,}$

The volume of the system is kept constant. This means that the volume of the heat bath does not change either and we can conclude that the heat bath does not perform any work. This implies that the amount of heat that flows into the heat bath is given by:

${\displaystyle Q_{\text{bath}}=\Delta U_{\text{bath}}=-\left(\Delta U+W\right)\,}$

The heat bath remains in thermal equilibrium at temperature T no matter what the system does. Therefore the entropy change of the heat bath is:

${\displaystyle \Delta S_{\text{bath}}={\frac {Q_{\text{bath}}}{T}}=-{\frac {\Delta U+W}{T}}\,}$

The total entropy change is thus given by:

${\displaystyle \Delta S_{\text{bath}}+\Delta S=-{\frac {\Delta U-T\Delta S+W}{T}}\,}$

Since the system is in thermal equilibrium with the heat bath in the initial and the final states, T is also the temperature of the system in these sates. The fact that the system's temperature does not change allows us to express the numerator as the free energy change of the system:

${\displaystyle \Delta S_{\text{bath}}+\Delta S=-{\frac {\Delta A+W}{T}}\,}$

Since the total change in entropy must always be larger or equal to zero, we obtain the inequality:

${\displaystyle W\leq -\Delta A\,}$

If no work is extracted from the system then

${\displaystyle \Delta A\leq 0\,}$

We see that for a system kept at constant temperature and volume, the total free energy during a spontaneous change can only decrease, that the total amount of work that can be extracted is limited by the free energy decrease, and that increasing the free energy requires work to be done on the system.

This result seems to contradict the equation ${\displaystyle dA=-SdT-PdV}$, as keeping T and V constant seems to imply ${\displaystyle dA=0}$ and hence ${\displaystyle F={\text{ constant}}}$. In reality there is no contradiction. After the spontaneous change the system, as described by thermodynamics, is a different system with a different free energy function than it was before the spontaneous change. We can thus say that ${\displaystyle \Delta A=A_{2}-A_{1}\leq 0}$ where the ${\displaystyle F_{i}}$ are different thermodynamic functions of state.

One can imagine that the spontaneous change is carried out in a sequence of infinitesimally small steps. To describe such a system thermodynamically, one needs to enlarge the thermodynamical state space of the system. In case of a chemical reaction one would need to specify the number of particles of each type. The differential of the free energy then generalizes to:

${\displaystyle dA=-SdT-pdV+\sum _{j}\mu _{j}dN_{j}\,}$

where the ${\displaystyle N_{j}}$ are the numbers of particles of type j and the ${\displaystyle \mu _{j}}$ are the corresponding chemical potentials. This equation is then again valid for both reversible and non-reversible changes. In case of a spontaneous change at constant T and V, the last term will thus be negative.

In case there are other external parameters the above equation generalizes to:

${\displaystyle dA=-SdT-\sum _{i}X_{i}dx_{i}+\sum _{j}\mu _{j}dN_{j}\,}$

Here the ${\displaystyle x_{i}}$ are the external variables and the ${\displaystyle X_{i}}$ the corresponding generalized forces.

## Relation to the partition function

A system kept at constant temperature is described by the canonical ensemble. The probability to find the system in some energy eigenstate r is given by:

${\displaystyle P_{r}={\frac {e^{-\beta E_{r}}}{Z}}\,}$

where

${\displaystyle \beta \equiv {\frac {1}{kT}}\,}$
${\displaystyle E_{r}={\text{ energy of eigenstate }}r\,}$
${\displaystyle Z=\sum _{r}e^{-\beta E_{r}}\,}$

Z is called the partition function of the system. The fact that the system does not have a unique energy means that the various thermodynamical quantities must be defined as expectation values. In the thermodynamical limit of infinite system size, the relative fluctuations in these averages will go to zero.

The internal energy of the system is the expectation value of the energy and can be expressed in terms of Z as follows:

${\displaystyle U=\sum _{r}P_{r}E_{r}=-{\frac {\partial \log Z}{\partial \beta }}\,}$

If the system is in state r, then the generalized force corresponding to an external variable x is given by

${\displaystyle X_{r}=-{\frac {\partial E_{r}}{\partial x}}\,}$

The thermal average of this can be written as:

${\displaystyle X=\sum _{r}P_{r}X_{r}={\frac {1}{\beta }}{\frac {\partial \log Z}{\partial x}}\,}$

Suppose the system has one external variable x. Then changing the system's temperature parameter by ${\displaystyle d\beta }$ and the external variable by dx will lead to a change in ${\displaystyle \log Z}$:

${\displaystyle d\left(\log Z\right)={\frac {\partial \log Z}{\partial \beta }}d\beta +{\frac {\partial \log Z}{\partial x}}dx=-U\,d\beta +\beta X\,dx\,}$

If we write ${\displaystyle U\,d\beta }$ as:

${\displaystyle U\,d\beta =d\left(\beta U\right)-\beta \,dU\,}$

we get:

${\displaystyle d\left(\log Z\right)=-d\left(\beta U\right)+\beta \,dU+\beta X\,dx\,}$

This means that the change in the internal energy is given by:

${\displaystyle dU={\frac {1}{\beta }}d\left(\log Z+\beta U\right)-X\,dx\,}$

In the thermodynamic limit, the fundamental thermodynamic relation should hold:

${\displaystyle dU=T\,dS-X\,dx\,}$

This then implies that the entropy of the system is given by:

${\displaystyle S=k\log Z+{\frac {U}{T}}+c\,}$

where c is some constant. The value of c can be determind by considering the limit T → 0. In this limit the entropy becomes ${\displaystyle S=k\log \Omega _{0}}$ where ${\displaystyle \Omega _{0}}$ is the ground state degeneracy. The partition function in this limit is ${\displaystyle \Omega _{0}e^{-\beta U_{0}}}$ where ${\displaystyle U_{0}}$ is the ground state energy. We thus see that ${\displaystyle c=0}$ and that:

${\displaystyle A=-kT\log \left(Z\right)\,}$

## Bogoliubov inequality

Computing the free energy is an intractable problem for all but the simplest models in statistical physics. A powerful approximation method is mean field theory which is a variational method based on the Bogoliubov inequality. This inequality can be formulated as follows.

Suppose we replace the real Hamiltonian ${\displaystyle H}$ of the model by a trial Hamiltonian ${\displaystyle {\tilde {H}}}$ which has different interactions and may depend on extra parameters that are not present in the original model. If we choose this trial Hamiltonian such that

${\displaystyle \left\langle {\tilde {H}}\right\rangle =\left\langle H\right\rangle \,}$

where both averages are taken w.r.t. the canonical distribution defined by the trial Hamiltonian ${\displaystyle {\tilde {H}}}$, then

${\displaystyle A\leq {\tilde {A}}\,}$

where ${\displaystyle A}$ is the free energy of the original Hamiltonian and ${\displaystyle {\tilde {A}}}$ is the free energy of the trial Hamiltonian. By including a large number of parameters in the trial Hamiltonian and minimizing the free energy we can expect to get a close approximation to the exact free energy.

The Bogoliubov inequality is often formulated in a sightly different but equivalent way. If we write the Hamiltonian as:

${\displaystyle H=H_{0}+\Delta H\,}$

where ${\displaystyle H_{0}}$ is exactly solvable, then we can apply the above inequality by defining

${\displaystyle {\tilde {H}}=H_{0}+\left\langle \Delta H\right\rangle _{0}\,}$

Here we have defined ${\displaystyle \left\langle X\right\rangle _{0}}$ to be the average of X over the canonical ensemble defined by ${\displaystyle H_{0}}$. Since ${\displaystyle {\tilde {H}}}$ defined this way differs from ${\displaystyle H_{0}}$ by a constant, we have in general

${\displaystyle \left\langle X\right\rangle _{0}=\left\langle X\right\rangle \,}$

Therefore

${\displaystyle \left\langle {\tilde {H}}\right\rangle =\left\langle H_{0}+\left\langle \Delta H\right\rangle \right\rangle =\left\langle H\right\rangle \,}$

And thus the inequality

${\displaystyle A\leq {\tilde {A}}\,}$

holds. The free energy ${\displaystyle {\tilde {A}}}$ is the free energy of the model defined by ${\displaystyle H_{0}}$ plus ${\displaystyle \left\langle \Delta H\right\rangle }$. This means that

${\displaystyle {\tilde {A}}=\left\langle H_{0}\right\rangle _{0}-TS_{0}+\left\langle \Delta H\right\rangle _{0}=\left\langle H\right\rangle _{0}-TS_{0}\,}$

and thus:

${\displaystyle A\leq \left\langle H\right\rangle _{0}-TS_{0}\,}$

### Proof

For a classical model we can prove the Bogliubov inequality as follows. We denote the canonical probability distributions for the Hamiltonian and the trial Hamiltonian by ${\displaystyle P_{r}}$ and ${\displaystyle {\tilde {P}}_{r}}$, respectively. The inequality:

${\displaystyle \sum _{r}{\tilde {P}}_{r}\log \left({\tilde {P}}_{r}\right)\geq \sum _{r}{\tilde {P}}_{r}\log \left(P_{r}\right)\,}$

then holds. To see this, consider the difference between the l.h.s. and the r.h.s.. We can write this as:

${\displaystyle \sum _{r}{\tilde {P}}_{r}\log \left({\frac {{\tilde {P}}_{r}}{P_{r}}}\right)\,}$

Since

${\displaystyle \log \left(x\right)\geq 1-{\frac {1}{x}}\,}$

it follows that:

${\displaystyle \sum _{r}{\tilde {P}}_{r}\log \left({\frac {{\tilde {P}}_{r}}{P_{r}}}\right)\geq \sum _{r}\left({\tilde {P}}_{r}-P_{r}\right)=0\,}$

where in the last step we have used that both probability distributions are normalized to 1.

We can write the inequality as:

${\displaystyle \left\langle \log \left({\tilde {P}}_{r}\right)\right\rangle \geq \left\langle \log \left(P_{r}\right)\right\rangle \,}$

where the averages are taken w.r.t. ${\displaystyle {\tilde {P}}_{r}}$. If we now substitute in here the expressions for the probability distributions:

${\displaystyle P_{r}={\frac {\exp \left[-\beta H\left(r\right)\right]}{Z}}\,}$

and

${\displaystyle {\tilde {P}}_{r}={\frac {\exp \left[-\beta {\tilde {H}}\left(r\right)\right]}{\tilde {Z}}}\,}$

we get:

${\displaystyle \left\langle -\beta {\tilde {H}}-\log \left({\tilde {Z}}\right)\right\rangle \geq \left\langle -\beta H-\log \left(Z\right)\right\rangle }$

Since the averages of ${\displaystyle H}$ and ${\displaystyle {\tilde {H}}}$ are, by assumption, identical we have:

${\displaystyle A\leq {\tilde {A}}}$

Here we have used that the partition functions are constants w.r.t. taking averages and that the free energy is proportional to minus the logarithm of the partition function.

We can easily generalize this proof to the case of quantum mechanical models. We denote the eigenstates of ${\displaystyle {\tilde {H}}}$ by ${\displaystyle \left|r\right\rangle }$. We denote the diagonal components of the density matrices for the canonical distributions for ${\displaystyle H}$ and ${\displaystyle {\tilde {H}}}$ in this basis as:

${\displaystyle P_{r}=\left\langle r\left|{\frac {\exp \left[-\beta H\right]}{Z}}\right|r\right\rangle \,}$

and

${\displaystyle {\tilde {P}}_{r}=\left\langle r\left|{\frac {\exp \left[-\beta {\tilde {H}}\right]}{\tilde {Z}}}\right|r\right\rangle ={\frac {\exp \left(-\beta {\tilde {E}}_{r}\right)}{\tilde {Z}}}\,}$

where the ${\displaystyle {\tilde {E}}_{r}}$ are the eigenvalues of ${\displaystyle {\tilde {H}}}$

We assume again that the averages of H and ${\displaystyle {\tilde {H}}}$ in the canonical ensemble defined by ${\displaystyle {\tilde {H}}}$ are the same:

${\displaystyle \left\langle {\tilde {H}}\right\rangle =\left\langle H\right\rangle \,}$

where

${\displaystyle \left\langle H\right\rangle =\sum _{r}{\tilde {P}}_{r}\left\langle r\left|H\right|r\right\rangle \,}$

The inequality

${\displaystyle \sum _{r}{\tilde {P}}_{r}\log \left({\tilde {P}}_{r}\right)\geq \sum _{r}{\tilde {P}}_{r}\log \left(P_{r}\right)\,}$

still holds as both the ${\displaystyle P_{r}}$ and the ${\displaystyle {\tilde {P}}_{r}}$ sum to 1. On the l.h.s. we can replace:

${\displaystyle \log \left({\tilde {P}}_{r}\right)=-\beta {\tilde {E}}_{r}-\log \left({\tilde {Z}}\right)\,}$

On the r.h.s. we can use the inequality

${\displaystyle \left\langle \exp \left(X\right)\right\rangle _{r}\geq \exp \left(\left\langle X\right\rangle _{r}\right)\,}$

where we have introduced the notation

${\displaystyle \left\langle Y\right\rangle _{r}\equiv \left\langle r\left|Y\right|r\right\rangle \,}$

for the expectation value of the operator Y in the state r. See here for a proof. Taking the logarithm of this inequality gives:

${\displaystyle \log \left[\left\langle \exp \left(X\right)\right\rangle _{r}\right]\geq \left\langle X\right\rangle _{r}\,}$

This allows us to write:

${\displaystyle \log \left(P_{r}\right)=\log \left[\left\langle \exp \left(-\beta H-\log \left(Z\right)\right)\right\rangle _{r}\right]\geq \left\langle -\beta H-\log \left(Z\right)\right\rangle _{r}\,}$

The fact that the averages of H and ${\displaystyle {\tilde {H}}}$ are the same then leads to the same conclusion as in the classical case:

${\displaystyle A\leq {\tilde {A}}}$

## Generalized Helmholtz energy

In the more general case, the mechanical term (${\displaystyle p{\rm {d}}V}$) must be replaced by the product of the volume times the stress times an infinitesimal strain:[3]

${\displaystyle {\rm {d}}A=V\sum _{ij}\sigma _{ij}\,{\rm {d}}\varepsilon _{ij}-S{\rm {d}}T+\sum _{i}\mu _{i}\,{\rm {d}}N_{i}\,}$

where ${\displaystyle \sigma _{ij}}$ is the stress tensor, and ${\displaystyle \varepsilon _{ij}}$ is the strain tensor. In the case of linear elastic materials which obey Hooke's Law, the stress is related to the strain by:

${\displaystyle \sigma _{ij}=C_{ijkl}\epsilon _{kl}}$

where we are now using Einstein notation for the tensors, in which repeated indices in a product are summed. We may integrate the expression for ${\displaystyle {\rm {d}}A}$ to obtain the Helmholtz energy:

${\displaystyle A={\frac {1}{2}}VC_{ijkl}\epsilon _{kl}^{2}-ST+\sum _{i}\mu _{i}N_{i}\,}$
${\displaystyle ={\frac {1}{2}}V\sigma _{ij}\epsilon _{ij}-ST+\sum _{i}\mu _{i}N_{i}\,}$