Consider a magnetic system subject to a magnetic field. Here we work with the fields $H,M,B$. Now, how does a change in the Helmholz free energy depend on $H,M,B$? I have three sources that seem to give contradicting claims for this.

The first source is Wikipedia. Hete it states:

Accordingly the change during a quasi-static process in the Helmholtz free energy, $F$, […] will be $$\Delta F = -SdT -pdV + \frac{1}{4\pi} \int_V H \cdot \Delta B dV.$$

When we approximate the magnetic fields to be constant over the volume, and also assume that the volume is constant, and ignore constants, we get $$dF = -SdT + H \cdot dB.\tag{1}$$

However, now consider these lecture notes. Here it says

$$dF= -SdT -MdB.\tag{2}$$

To add to the confusion, I am following a course in statistical physics and here the lecture notes say that $$dF =-SdT +HdM.\tag{3}$$

Am I making any mistakes in the approximations? And which of the formulas is correct? Or are they somehow equivalent?


2 Answers 2


Magnetic work in equation ($1$) is correct. It can be derived from analyzing the work required to vary the magnetic $B$ field. To be precise, because, in general, the fields are not uniform, it should be written as an integral over the volume (see, for instance, Landau & Lifshitz Electrodynamics of continuous media ): $$ dF = -S dT+\int dV {\bf H}\cdot \delta {\bf B}.\tag{1} $$ Thus, in general, $F$ is a function of the temperature $T$ and a functional of the magnetic field ${\bf B}({\bf r})$. The field ${\bf H}({\bf r})$ should be related to ${\bf B}$, in general, by a non-linear functional relation: $$ {\bf H}({\bf r})= \tilde {\bf H}({\bf r},[{\bf B}],T).\tag{2} $$

Equation $(2)$ depends on the materials' presence, geometry, and characteristics in the region where fields are present. It allows us to close the Maxwell equations for the magnetic fields. Notice that in general, even if the fields in the absence of material, ${\bf B_0}$ and ${\bf H_0}$ are uniform and aligned (${\bf B_0}=\mu_0 {\bf H_0}$), the fields in the presence of a material sample (${\bf B}$ and ${\bf H}$) are point dependent and not proportional.

If we want to introduce the magnetization density ${\bf M}({\bf r})$, defined via $$ {\bf B}= \mu_0\left( {\bf H}+{\bf M} \right), $$ we can do it, but the correct expression for the work is $$ \mu_0 \int dV {\bf H}\cdot \delta {\bf H} + \mu_0 \int dV {\bf H}\cdot \delta {\bf M}. \tag{3} $$ This last equation differs from the expression of OP's equation ($3$) for the presence of the first integral. Such a term is not the work for creating the field in the absence of the material because, as noted above, the ${\bf H}$ field does depend on the presence of the sample. Therefore, it cannot be ignored when we need the thermodynamic work.

A more convenient reformulation of eq. $(1)$, is by rewriting the term $\int dV {\bf H}\cdot \delta {\bf B}$ in terms of ${\bf M}$ and ${\bf H_0}$, the field in the absence of the sample. It can be shown (see Stratton's textbook on Electromagnetism or the cited Landau & Lifshitz) that we get $$ \mu_0 \int dV {\bf H_0}\cdot \delta {\bf H_0} + \mu_0 \int dV {\bf H_0}\cdot \delta {\bf M}. \tag{3'} $$ The huge advantage of equation $(3')$ over equation ($3$) is that now the field ${\bf H_0}$ is entirely under our control, whatever sample we introduce in it. It can be chosen as uniform, and the first term in equation $(3')$ is the work to build the field without the sample. Therefore, most thermodynamic manipulations can ignore its contribution to free energy.

Summarizing, the differential of the free energy without the work to build the ${\bf H_0}$ field, $F'(T,[{\bf M}])$ can be written as $$ dF' = -S dT+ \mu_0 \int dV {\bf H_0}\cdot \delta {\bf M}.\tag{4} $$ and, by choosing a uniform field ${\bf H_0}$ and an ellipsoidal sample (in such a case the magnetization is uniform) aligned with the field, we can finally write $$ dF' = -SdT + \mu_0 H_0 d\tilde M \tag{5} $$ where $\tilde M$ is the total dipole moment of the sample.

Finally, if instead of $F'(T,\tilde M)$ we have better working with a function of the external field, we can introduce the Legendre transform of $F'$, a new function. $ \tilde F(T,H_0) = F' - \mu_0 H_0 \tilde M $ whose differential is $$ d \tilde F = -S dT - \mu_0 \tilde M dH_0. \tag{6} $$

Summarizing, OP's equations $(2)$ and $(3)$ do not follow directly, in general from equation $(1)$, because the contribution to the free energy from the term $\frac{\mu_0}{2} \int dV H^2$ is sample-dependent.

There are only a few cases where equations $(5)$ and $(6)$ coincide with OP's equations $(2)$ and $(3)$: the cases where the demagnetizing field, i.e. the difference ${\bf H}-{\bf H_0}$ vanishes. Such cases require a special shape of the sample.

  • $\begingroup$ wouldn't the last expression (6) be $d \tilde F = -S dT - \mu_0 \tilde M dH_0$ ? $\endgroup$
    – Botond
    Dec 13, 2023 at 12:25
  • 1
    $\begingroup$ @Botond Of course. Thanks for signaling the typo. I'll correct it immediately. $\endgroup$ Dec 13, 2023 at 13:32

The 'real' internal energy is $dU=TdS+HdM$, and hence the Helmholtz free energy is given by $dF=-SdT+HdM$. Therefore, Equation 2 is incorrect [when $F$ is the Helmholtz free energy] and Equation 3 is correct. In the source of Equation 2, they call $F$ the 'free energy' and not the 'Helmholtz free energy'. On page 15, they say:

Sometimes, one sees other definitions of the free energy of a magnetic system from that in Eq. (32). To clarify this, in the first edition of the book, Kittel uses the term $F_A(B, T)$ for the free energy we call $F$ here, and adopts a different notation, $F_B$, for another free energy which one often sees in the literature, which is defined by $$F_B=F+MB.$$ ... The difference between $F_B$ and $F$ is rather like that between the Helmholtz and Gibbs free energies, and some authors use that terminology. However, I prefer to use the terms “Helmholtz” and “Gibbs” just for free energies which differ by a factor of $PV$ .

So they compare their $F$ with the Gibbs free energy, not with the Helmholz free energy.

Equation 1 is also correct, and can be reconciled with Equation 3. For this, we can first rewrite Equation 1:


Now we can ignore the $HdH$ term to obtain Equation 3. We can do this because it is the work of the external field on itself, which is not relevant for the system.

  • $\begingroup$ Unfortunately, if $H$ is the external $ H$ field, it is not the same as appearing in equation (1) of the OP. I'll write an answer to clarify this often-neglected point. $\endgroup$ Nov 11, 2023 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.