Brian Bi

Brian Bi's answer to What is the physical meaning of the H field? - Quora

To understand the H field, we should read the paper A Dynamical Theory of the Electromagnetic Field, in which Maxwell's equations first appeared. In Part III Maxwell uses [math](\alpha, \beta, \gamma)[/math] to denote the field we (following Heaviside) now call [math]\mathbf{H}[/math].

These quantities are introduced thus (emphasis mine):

Let [math]\alpha, \beta, \gamma[/math] represent the force acting on a unit magnetic pole placed at the given point resolved in the directions of [math]x[/math], [math]y[/math], and [math]z[/math].

That is,

[math]\begin{equation}\mathbf{F} = m\mathbf{H}\end{equation}[/math]

in a manner analogous to

[math]\begin{equation}\mathbf{F} = q\mathbf{E}\end{equation}[/math]

where [math]m[/math] is the magnetic charge of a particle.

Note also that Maxwell uses [math]\mu\alpha, \mu\beta, \mu\gamma[/math] to denote what we now call the B field; he doesn't give the B field its own set of three symbols. This clearly reflects a belief that H was the fundamental field. When Maxwell introduces [math]\mu\alpha, \mu\beta, \mu\gamma[/math], he refers to these quantities as "number of lines of force in unit of area", which roughly corresponds to the modern term "magnetic flux density".

Interlude: That's why the physical meaning of H is lost today: we no longer believe in magnetic charges. H has no physical meaning unless we do.

Now let's return to the twenty-first century and ask ourselves: how could Maxwell and his contemporaries have obtained a sensible theory involving magnetic charge when they never found any, when we now know that none of the magnetic fields we observe are actually due to free magnetic charges at all?

Well, the thing is that [math]\mathbf{H}[/math], unlike [math]\mathbf{B}[/math], is really not source-free. Indeed

[math]\begin{equation}\nabla \cdot \mathbf{H} = -\nabla \cdot \mathbf{M}\end{equation}[/math]

and there's no reason for the right hand side to vanish, in general; we may well have spatially varying magnetization. In fact, even if the magnetization is uniform in the interior of a magnet, it certainly varies at the boundary between the magnet and its surroundings. So what if we call the right hand side the magnetic charge density?

[math]\begin{equation}m = -\nabla \cdot \mathbf{M}\end{equation}[/math]

This bears a suspicious similarity to the equation for what we call the bound electric charge density,

[math]\begin{equation}\rho_b = -\nabla \cdot \mathbf{P}\end{equation}[/math]

But that's all wrong, you say. For bound electric charge density is physically real. It arises from a separation of opposite charges within the material, which causes the spatially averaged charge density in the material to really deviate from zero. When bound charge accumulates on the surface of an object, it is real charge. Yet there are no fundamental magnetic charges, period, so bound magnetic charge cannot be real in the same way. The magnetization [math]\mathbf{M}[/math] comes from the alignment of point magnetic dipoles such as electrons, not from the separation of positive and negative magnetic charges.

And yet... in a static configuration with no free current, we really have

[math]\begin{align}\nabla \cdot \mathbf{H} &= m \\ \nabla \times \mathbf{H} &= 0\end{align}[/math]

implying that under such circumstances, H can be written as

[math]\begin{equation}\mathbf{H} = -\nabla \phi_m\end{equation}[/math]

where the "magnetic potential" [math]\phi_m[/math] satisfies a Poisson's equation:

[math]\begin{equation}\nabla^2 \phi_m = -m\end{equation}[/math]

and therefore we have a "Coulomb's law" for H,

[math]\begin{equation}\mathbf{H} = \frac{1}{4\pi} \int \frac{\mathbf{r} - \mathbf{r}'}{\|\mathbf{r} - \mathbf{r}'\|^3} m \, \mathrm{d}\mathbf{r}'\end{equation}[/math]

So that's the physical meaning of H: H is to magnetic charge as E is to electric charge.

One intriguing question remains. Recall the expressions for the electric field of a point electric dipole at the origin:

[math]\begin{equation}\mathbf{E} = \frac{1}{4\pi\epsilon_0 r^3} (3(\mathbf{p} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{p})\end{equation}[/math]

and the magnetic field of a point magnetic dipole at the origin:

[math]\begin{equation}\mathbf{B} = \frac{\mu_0}{4\pi r^3} (3(\boldsymbol\mu \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \boldsymbol\mu)\end{equation}[/math]

If these expressions are identical, then it implies that the relationship between E and the polarization is exactly analogous to the relationship between B and the magnetization. And yet that isn't the case, because in a static system with no free charges or currents,

[math]\begin{align}\nabla \cdot \mathbf{E} &= \rho_b/\epsilon_0 \\ \nabla \cdot \mathbf{B} &= 0\end{align}[/math]

The divergence of B is not, as one might expect, [math]\mu_0 m[/math], with [math]\rho_b = -\nabla \cdot \mathbf{P}[/math] and [math]m = -\nabla \cdot \mathbf{M}[/math].

How can the E and B fields behave differently when electric dipoles and magnetic dipoles give the field analogously?

And the answer is this: if there really were magnetic charges, and magnetic dipoles were constructed like electric dipoles---from the separation of opposite magnetic charges, as so-called Gilbert dipoles, then the analogy would be precise. But a magnetic dipole constructed from opposite magnetic charges, a Gilbert dipole, differs from a magnetic dipole constructed from an infinitesimal loop of current, an Ampère dipole. They produce the same magnetic field everywhere except the origin, or the very point at which the dipole is located.

For the true field of an electric dipole includes a delta function term, which I omitted above,

[math]\begin{equation}-\frac{1}{3\epsilon_0} \mathbf{p} \delta(\mathbf{r})\end{equation}[/math]

and for the magnetic field, there's a similar delta function term, but not the same one:

[math]\begin{equation}+\frac{2\mu_0}{3} \boldsymbol\mu \delta(\mathbf{r})\end{equation}[/math]

For a Gilbert dipole the delta function term would be like the electric one,

[math]\begin{equation}-\frac{\mu_0}{3} \boldsymbol\mu \delta(\mathbf{r})\end{equation}[/math]

and we would have [math]\mathbf{B} = \mu_0 \mathbf{H}[/math] everywhere since [math]\mathbf{B}[/math] and [math]\mathbf{H}[/math] would obey the same "Coulomb's law" except for a factor of [math]\mu_0[/math]. But the difference in the delta function terms makes the Coulomb's law for B only accurate outside the magnetized material. Inside the material, there's a discrepancy in the amount of the difference between the two delta function terms,

[math]\begin{equation}\mu_0 \boldsymbol \mu \delta(\mathbf{r})\end{equation}[/math]

and when we average over the various dipoles in a tiny region, this becomes

[math]\begin{equation}\mu_0 \mathbf{M}\end{equation}[/math]

which is precisely the extra piece we have to add to get the actual B,

[math]\begin{equation}\mathbf{B} = \mu_0 \mathbf{H} + \mu_0 \mathbf{M}\end{equation}[/math]

So the difference between B and H is the difference between the Ampère model, in which electric currents are the sources of magnetic fields, and the Gilbert model, in which magnetic charges are the sources of magnetic fields. H "misses out" on having a true physical meaning precisely because of this, because all magnetic dipoles are Ampère dipoles and not Gilbert dipoles.