Showing that B has no discontinuities at the surface

[Mike400]

Consider a magnetic dipole distribution in space having magnetization $\mathbf{M}$. The potential at any point is given by:

$\displaystyle\psi=\dfrac{\mu_0}{4 \pi} \int_{V'} \dfrac{ \rho}{|\mathbf{r}-\mathbf{r'}|} dV' + \dfrac{\mu_0}{4 \pi} \oint_{S'} \dfrac{\sigma}{|\mathbf{r}-\mathbf{r'}|} dS'=\psi^{V}+\psi^{S}$

The $\mathbf{H}$ field is:

$\displaystyle\mathbf{H}=\dfrac{\mu_0}{4 \pi} \int_{V'} \rho \dfrac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dV' + \dfrac{\mu_0}{4 \pi} \oint_{S'} \sigma \dfrac{ \mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dS'=\mathbf{H}^{V}+\mathbf{H}^{S}$

The $\mathbf{B}$ field is:

$\mathbf{B}=\mathbf{H} + \mu_0 \mathbf{M}=\mathbf{H}^{V} + \mathbf{H}^{S} + \mu_0 \mathbf{M}$

$\mathbf{H}^{V}$ has no discontinuity.

$\mathbf{H}^{S}$ has discontinuity of $\mu_0 \mathbf{M} \cdot \hat{n}$ at the surface $S'$

$\mu_0 \mathbf{M}$ has discontinuity of $\mu_0 \mathbf{M}$ at the surface $S'$

From these knowledge, how shall one deduce that $\mathbf{B}$ is continuous at the surface?

My try: (I am getting a contradiction)

We need to show that $\mu_0 \mathbf{M} \cdot \hat{n}+\mu_0 \mathbf{M}=0$, i.e. $\mathbf{M} \cdot \hat{n}= -\mathbf{M}$
Since the surface could be oriented at any angle w.r.t. $\mathbf{M}$ at the surface, this is a contradiction. Where am I going wrong?

 

[vanhees71]

I guess what you mean is that due to
$$\vec{\nabla} \cdot \vec{B}=0$$
$\vec{B}$'s normal component at the surface must be continuous.

I'm also a bit lost what $\rho$ and $\sigma$ have to do with $\vec{M}$. Without a surface magnetization (which is a bit unusual; I'm not sure, where one would find such a thing in nature) the correct solution of magnetostatics of a (hard) ferro magnet is
$$\vec{H}=-\vec{\nabla} \phi$$
with
$$\phi(\vec{x})=-\vec{\nabla}_x \cdot \int_{\mathbb{R}} \mathrm{d}^3 x' \frac{\vec{M}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$

 

[Mike400]

$\vec{B}$'s normal component at the surface must be continuous.

I know that and can be simply deduced from the equation $\mathbf{B}=\mathbf{H} + \mu_0 \mathbf{M}=\mathbf{H}^{V} + \mathbf{H}^{S} + \mu_0 \mathbf{M}$. But should the tangential component of $\vec{B}$ must be continuous too? The equation shows a discontinuity in the tangential component.

I'm also a bit lost what $ρ$ and $σ$ have to do with $\vec{M}$

$ρ=-\nabla \cdot \mathbf{M}$ and $σ=\mathbf{M} \cdot \hat{n}$

 

[vanhees71]

The tangential component of $\vec{H}$ can have a singularity due to a surface current, and usually also $\vec{B}$ has one there too.

The standard example is a homogeneously magnetized body. The magnetization is equivalent to a current density
$$\vec{j}_{\text{m}}=\vec{\nabla} \times \vec{M},$$
and in this approximation that's a surface current density.

E.g., take a homogeneously magnetized sphere of radius $a$ around the origin of the coordinate system. Then we can write (with $r=|\vec{x}|$)
$$\vec{M}(\vec{x})=M \vec{e}_3 \Theta(a-r).$$
The curl is
$$\vec{\nabla} \times \vec{M}=-M \vec{e}_3 \times \vec{\nabla} \Theta(a-r).$$
Now
$$\vec{\nabla} \Theta(a-r)=\frac{\mathrm{d}}{\mathrm{d} r} \Theta(a-r) \vec{\nabla r} = -\frac{\vec{x}}{r} \delta(a-r)=-\frac{\vec{x}}{a} \delta(a-r)$$
and thus
$$\vec{j}_{\text{m}}=\frac{M}{a} \vec{e}_3 \times \vec{x} \delta(r-a),$$
i.e., you have $\delta$-function like singularity across the surface of the sphere which means that there's a surface-current density.

It's a good example to calculate the magnetic displacement $\vec{H}$ and the magnetic field $\vec{B}$ for this example, which is analytically solvable. Then the concepts of charge and current densities as well as their singular cases, i.e., surface charge and current densities become clear.