Is intrinsic energy of magnetic material given by $H^2/8\pi$, not $B^2/8\pi$?

Question

In most textbook about electromagnetism, it is taught that the field energy of magnetic field B is given by $B^2/8\pi$, which is indeed consistent with continuity equation of energy, particularly in vaccum. However, in many magnetism book, theorist use H-field instead of B-field due to historical convention, and they also quote magnetic energy as $H^2/8\pi$, not $B^2/8\pi$.

I found that there is more than just convention in a formulation, when I considered interaction between two magnetic dipole. When one tries to calculate interaction energy between two magnetic dipole $\vec{m}_1$ and $\vec{m}_2$, the interaction energy gives different result: First with $B^2/8\pi$,

$$\int\frac{1}{8\pi}\left((B_1+B_2)^2-B_1^2-B_2^2\right)=\int \frac{1}{4\pi}\vec{B}_{1}\cdot\vec{B}_{2}d^3r$$$$=\frac{3(\vec{m}_1\cdot\hat{r})(\vec{m}_2\cdot\hat{r})-\vec{m}_1\cdot\vec{m}_2}{r^3}=+\vec{m}_1\cdot\vec{B}_2=+\vec{m}_2\cdot\vec{B}_1$$ On the other hand, H-field can be written as $$\vec{H}=\vec{B}-4\pi\vec{m}\delta^3(\vec{r})$$ So the interaction energy from $H^2/8\pi$ $$\int \frac{1}{4\pi}\vec{H}_{1}\cdot\vec{H}_{2}d^3r=\int \frac{1}{4\pi}\vec{B}_{1}\cdot\vec{B}_{2}d^3r-\vec{m}_1\cdot\vec{B}_2-\vec{m}_2\cdot\vec{B}_1=-\vec{m}_1\cdot\vec{B}_2=-\vec{m}_2\cdot\vec{B}_1$$ (If one bothers about delta function, using sphere with uniform magnetization gives same result.) (Both magnetic dipole has infinite self-energy due to zero - size, which must be excluded.)

In fact, if one interpret $\vec{m}_2$ as a current source (or magnetic pole source) creating $\vec{B}_{ext}$ on dipole $\vec{m}=\vec{m}_1$, the energy in form of $B^2/8\pi$ suggests $U_{int}=+\vec{m}\cdot\vec{B}_{ext}$, while energy in form of $H^2/8\pi$ suggests $U_{int}=-\vec{m}\cdot\vec{B}_{ext}$.

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The difference between these two formula originate in the way one constitute the system. When one uses magnetic energy density as $B^2/8\pi$, he will constitute the system by moving charge (with sufficient minus charge density fixed in space, compensating the charge density of current therefore not including any electric field energy), or current loop. Both the magnetic dipole and source of external field will be regarded as these current loop.

However, when one uses magnetic energy density as $H^2/8\pi$, he will constitute the system by moving magnetic monopole from infinity, and placing these monopole in current position without any motion. (since H has direct correspondence with E, when there is no free current $J_f$) In this case, both dipole and source of external field will be regarded as set of magnetic monopole.

Now, I will interpret the term $U_B =+\vec{m}\cdot\vec{B}_{ext}$ and $U_H=-\vec{m}\cdot\vec{B}_{ext}$ by specifying particular procedure of reproducing the system. In order to simplify the situation, the source of field $\vec{B}_{ext}$ is regarded as a point dipole $\vec{m}_{ext}$. The procedures are:

1. The magnetic dipole $\vec{m}_{ext}$ is created in $\vec{r}_{2}$ with direction $\hat{n}_2$ as a field source.
2. The magnetic dipole $\vec{m}$ is created far away from $\vec{r}_{2}$ with direction $\hat{n}_1$. (infinite distance)
3. The magnetic dipole $\vec{m}$ translates from infinity to $\vec{r}_{1}$ without change of its direction, preserving both $\vec{m}$ and $\vec{m}_{ext}$.

When one excludes infinite energy of creating such point entity, the energy necessary for each procedure is given as the following.

In the case of current loop, corresponding with energy term $B^2/8\pi$:

1. Zero.
2. Zero. Since $\vec{m}$ and $\vec{m}_{ext}$ are separated, there is no change of magnetic flux $\Phi_{B}$ in $\vec{m}_{ext}$, while $\vec{m}$ increases from 0 to |m|=IS.
3. ⓐ One must exert force against $F=\triangledown(\vec{m}\cdot\vec{B}_{ext})$, thus $W_{mech}=-\vec{m}\cdot\vec{B}_{ext}$.
   ⓑ The flux in $\vec{m}_{ext}$ changes from 0 to $\Phi_B = MI$. Thus emf $\epsilon=-\frac{\partial \Phi_B}{\partial t}$ is induced, and one must supply energy $\int(-\epsilon)I_{ext}dt=MI I_{ext}$ to the loop, maintaining $I_{ext}$ against flux change. ⓒ The moving current loop of $\vec{m}$ induces motional emf, which changes flux from 0 to $\Phi_B = MI_{ext}$. There must be energy $\int(-\epsilon)Idt=MI I_{ext}$ supplied to the current loop.

Since mutual inductance M is given by $M=\left(3(\vec{S}\cdot\hat{r})(\vec{S}_{ext}\cdot\hat{r})-\vec{S}\cdot\vec{S}_{ext}\right)/r^3$, it is easily shown that $MI I_{ext}=+\vec{m}\cdot\vec{B}_{ext}$, which is consistent with calculation using $B^2/8\pi$. This concludes that total energy used to construct the system is $+\vec{m}\cdot\vec{B}_{ext}$, which consists of 3 part:

a) $-\vec{m}\cdot\vec{B}$ from work against mechanical force $\triangledown(\vec{m}\cdot\vec{B})$,

b) $+\vec{m}\cdot\vec{B}$ from maintaining current $\it{I}$ of dipole $\vec{m}=\it{I}\vec{S}$ against emf from change in $\vec{B}_{ext}$,

c) $+\vec{m}\cdot\vec{B}$ from maintaining external source of current producing $\vec{B}_{ext}$.

There might be different interpretation in dividing energy $+\vec{m}\cdot\vec{B}_{ext}$ depending on choice of procedure. Nonetheless, the total energy necessary for constituting such system is independent to choice of procedure, so I will use this interpretation in order to clarify the distinction between current loop model and monopole model.

If one repeats same thing in magnetic monopole model, then one gets:

1. Zero energy needed.
2. Zero energy needed.
3. ⓐ Mechanical work is same as $W_{mech}=-\vec{m}\cdot\vec{B}_{ext}$. ⓑⓒ No energy supply needed.

Thus total energy needed is $-\vec{m}\cdot\vec{B}_{ext}$, which is purely mechanic work. This result coincides with calculation on energy $H^2/8\pi$. Since this scheme calculates pure mechanical work, it will be appropriate for calculation of magnetostatic energy: The energy of dipoles given as $U=\sum U_{ij}$ with $U_{ij}=-\vec{m}_i\cdot\vec{B}_j$ will be acquired by using $H^2/8\pi$ rather than $B^2/8\pi$.

In conclusion, I arrived at the following statement:

The energy in form of $H^2/8\pi$ correctly calculates magnetostatic energy, while $B^2/8\pi$ includes additional energy to maintain current in the loop against external flux change.

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Now I concentrate on actual magnetic moment in magnetic material, and calculate 'free energy' of material determining its equilibrium state under external field $\vec{B}_{ext}$. Since the energy supplied in current source won't effect equilibrium property of material, it won't be included in such 'free energy': ⓑ part of energy won't be included. In addition, no such energy is needed to maintain 'current' of magnetic moment, since these moments are quantum mechanical phenomenon: they are neither current loop nor magnetic monopole. Thus ⓒ part of energy won't be included in 'free energy'. Nonetheless, experiments prove that the force exerted at magnetic material is given as $-\int \triangledown(\vec{M}\cdot\vec{B}_{ext})d^3r$, so there must be mechanical force exerting on individual magnetic moment, which provides a mechanism material reaching its equilibrium state. Thus, ⓐ part of energy won't be included in 'free energy', which is a solely contribution of it.

I admit that the real procedure magnetic material acquiring its equilibrium state will differ much from procedure given in preceding section, so these arguments are just my intuition...

Nevertheless, the situation argued below corresponds with magnetic monopole scheme with energy $H^2/8\pi$. The source of external field is practically current loop, which differs from construction in magnetic monopole scheme. (There the source was also magnetic monopole) However, since contribution of source is excluded, this difference doesn't change the following conclusion:

The 'free energy' describing equilibrium property of magnetic material must be described with $H^2/8\pi$ rather than $B^2/8\pi$.

Domain formation in ferromagnetic material seems to support this conclusion. The direction of each domain is largely determined by 'Pole Avoidance Principle': these domains arrange so that $\delta M_{\perp}=0$ in its boundary, leading to zero $\sigma_b$ and zero H-field. If one accept magnetostatic energy as $H^2/8\pi$ rather than $B^2/8\pi$, then this phenomenon can be explained, since pole avoidance minimizes $H^2/8\pi$ to zero, minimizing free energy. Indeed, there are no such 'Surface Current Avoidance Principle' in ferromagnetic material.

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My question is the following: Is the argument given below correct?

The 'free energy' describing equilibrium property of magnetic material must be described with $H^2/8\pi$ rather than $B^2/8\pi$ (at least a contribution with correct sign).

I know real free energy will depend on constitutional relation between B and H, and I'm just talking about classical electromagnetic contribution. Regarding this constitutional relation, all the argument below is nonsense in practical calculation. I just want to clarify whether $B^2/8\pi$ in classical electrodynamics is fundamentally wrong or not, since it's formal derivation is based on current loop model, while real magnetic dipole isn't a current loop.

Answer 1

thus there are no such energy necessary to maintain such 'current' of magnetic dipole.

Energy is not necessary to maintain magnetic moment or its associated amperian current in general, not just in the spin case. If the magnetic moment is due to current running in circles, adding energy (doing work against the induced electric field) is needed to increase this current, but zero energy (work) is needed to maintain it.

I arrived at the conclusion that magnetic energy term $H^2/8\pi$ is appropriate at situation with magnetic material, rather than $B^2/8\pi$ used in electromagnetic textbooks. (I'm not referring to energy density $\delta u=\vec{E}\cdot\delta\vec{D}+\vec{B}\cdot\delta\vec{H}$ in material.) However, this result is not quite satisfactory, since it seems to contradict with Poynting theorem and Gauge invariance. What point is missing in this whole argument?

Specification of what kind of energy you're looking for or what the question really is. "There are many different formulae for magnetic energy giving different results" is not a real problem, because there are many different concepts of energy.

One such concept is work needed to arrange already existing constant magnetic moments into a configuration; another is work needed to create such configuration from moving charged particles (currents). This difference is due to a different model of the magnetic moment.

There are still more concepts of magnetic energy, e.g. we have total EM energy of a space region, and we have potential energy of magnetic moment in an external field (which is not primarily associated with a specific space region, but more with the magnetic moment itself).

The Poynting theorem is a mathematical relation that electric and magnetic field obey; neither this theorem nor its energetic interpretation cover or exhaust all knowledge about the concepts of energy in EM theory. It is quite OK for some energy concept to not be directly related to the Poynting theorem.

Answer 2

Magnetic ''material''... ''dipole''... ''dipole density''... the force exerted on a dipole... force exerted on dipole density... These are VERY wild creatures when you look at them from a macroscopic view.

I have not read your OP thoroughly, but I guess you have some problems regarding the definition of energy of a magnetic ''material'' immersed in an external field and wonder what type of force is exerted on it.

First of first: the energy density of a magnetic field in vacuum is $\mu_0 |H|^2/2$. When the magnetic matter is present, it is true only when you presume a linear constitutive relation. For a more general statement, you can only determine its energy after you've decided ''how you want to talk to the system''. This is via doing work on the system, and you know better than I do that there are multiple ways to do that. For example, you can manipulate external currents $J_f$ and you do the following work on all space, including the matter: \begin{equation} \delta W_H = - \int_{all~space} A\cdot \delta J_f ~dv = - \int_{all~space} B\cdot\delta H~dv \end{equation} Or suppose you have an experimental setting in which you manipulate the vector potential $A$ as the independent thermodynamic variable. In this case, the work done is \begin{equation} \delta W_B = \int_{all~space} \delta A\cdot J_f ~dv = \int_{all~space} H\cdot\delta B~dv \end{equation} These two works are stored as the energies $E_H$ and $E_B$ in all space, respectively. These two can be totally different, BUT I also say that they are equivalent in the sense that they differ by a change of experimental ensemble: In the first one you are describing the system ''effectively'' in terms of magnetic charges, but with the latter you are describing the system ''effectively'' in terms of Amperian currents. To transit from one description to the other you only need to perform a Legendre transformation (that is to say, change your experimental apparatus you are using to talk to the system): \begin{equation} E_B = E_H + \int_{all~space} H\cdot B~dv \end{equation} You can extract any information you need about the system by either of these relations, and these are of course the easiest experimental procedures to talk to the system, yet theoretically they are not much interesting, to me at least, for they bring the state of all space into describing the state of the very system, and for also they are dependent on total fields $B$, or $H$. I am more interested in a description in which the independent variables are magnetic moments, and their interactions with each other and with the external (applied) fields are taken into account. One of the main reasons, which also has relevance when I will talk about the forces, is that upon coarse-graining, although effective magnetic moment (magnetization) is well-defined, there is no way we can agree upon a meaning to the total fields inside matter. You can show that by a clever Legendre transform, you can transform either of potential energies $E_B$ or $E_H$ into one that expands only over the system alone and is a functional of $M(r)$ and $B_{ext}$.

And, about forces. There is no misunderstanding about the force acting on a dipole. But on a magnetized matter, there is no unique representation of the force whatsoever. This is due to the fact that when you coarse-grain, you lose all information on lower scales. What I mean is that although you can meaningfully calculate the action of a macroscopic magnetic field (either $H$ or $B$) on a responsive constituent (like charges or currents) at a large distance, you should also not forget about the forces that neighboring spins exert on one another. This force is a local one, and you fail to describe it with the usual macroscopic electromagnetism. Since this force is local, you can say that in lieu of acting at a distance, it propagates as a stress tensor. Now the un-uniqueness regarding the force acting on a macroscopic body is that you cannot separate the short-range and long-range forces uniquely. But you can do the following trick: guess a proper form of long-range force, admissible in terms of macroscopic electromagnetism, and pour all the missing information in a tensor field! So, what is this admissible form of force?

It depends on your experimental apparatus again: $B$ acts on Amperian currents, $H$ acts on magnetic charges. If you discard the self-fields for a moment, the force acting on a magnetization density is \begin{align} F_i &= \mu_0\int_V M_j~ H^{ext}_{j,i}~dv = \mu_0\int_V (-\nabla\cdot M)H^{ext}_i~dv + \mu_0\oint_S (n\cdot M) H^{ext}_i~da\\ F_i &= \int_V M_j ~B^{ext}_{j,i}~dv = \int_V [(\nabla\times M)\times B^{ext}]_i~dv + \oint_S [(-n\times M)\times B^{ext}]_i~da \end{align} But when you consider the self-fields as well, you still get meaningful relations in terms of the total fields, but you have to know that since the total fields $H$ or $B$ do not have meanings inside matter, the terms you obtain for forces do not mean they are actual forces; you have to take into account short-range forces as well for a complete treatment of the problem.

Brown's classic Magnetoelastic interactions helped me A LOT when I was struggling with the same problems you are facing at the moment. I hope you enjoy it too. See sections 5.3 and 5.4, in particular.