Why is "method of image current" valid in magnetostatics?

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• euphoricrhino
euphoricrhino
TL;DR Summary
"method of image charge in electrostatics" works because of uniqueness of Laplace equation solutions, what makes the "method of image current" work in magnetostatics?
Hi wise folks,
I am working through Jackson problems, and have just encountered problem 5.17:
A current distribution ##\mathbf{J}(\mathbf{x})## exists in a medium of unit relative permeability adjacent to a semi-infinite slab of material having relative permeability ##\mu_r## and filling the halfspace ##z<0##.
Show that for ##z>0## the magnetic induction can be calculated by replacing the medium of permeability ##\mu_r## by an image current distribution ##\mathbf{J}^*##, with components
\begin{align*}
\left(\frac{\mu_r-1}{\mu_r+1}\right)J_x(x,y,-z) && \left(\frac{\mu_r-1}{\mu_r+1}\right)J_y(x,y,-z) && -\left(\frac{\mu_r-1}{\mu_r+1}\right)J_z(x,y,-z)
\end{align*}
and that for ##z<0## the magnetic induction appears to be due to a current distribution ##2\mu_r\mathbf{J}/(\mu_r+1)## in a medium of unit relative permeability.

It is pretty straightforward to show that the given image current distributions will satisfy the boundary conditions (both tangent and normal) at the ##z=0## plane. But my question is actually: "why is the method of image current even valid"?

Recall that the method of image charge works in electrostatics because of the uniqueness of Laplace equation solutions, in particular the Dirichlet boundary condition - once we have a virtual image charge producing the same potential on the conductor's boundary surface, we know immediately its effect is identical to the situation without image charge but only the conductor.

Even with conductors replaced by dielectrics, as in section 4.4, we can use the method of images since the solution is known to be uniquely expandable into legendre series given symmetry.

But can we say the same thing in magnetostatics? For the problem above, can we claim that once the "ansatz" image currents produce field that meet consistently at the boundary, these image currents will produce identical effect elsewhere (i.e., off the slab boundary)? Here we are not dealing with dirichlet boundary conditions unless ##\mu_r## is infinity or unity. For general ##\mu_r##, the boundary may not even be an equipotential surface (that is, the magnetic scalar potential). Without this condition, how is the method of image current even valid?

Thanks!

Image charges/currents are just a way of expressing boundary conditions. No more, no less. However, you do need to be careful that they are correctly expressing the boundary conditions.

PhDeezNutz, vanhees71 and Dale
The boundary conditions for such situations, i.e., no surface currents at the boundary surface between the media are
$$\vec{n} \cdot (\vec{B}_1-\vec{B}_2)=0, \quad \vec{\nabla} \times (\vec{H}_1-\vec{H}_2)=0,$$
where ##\vec{n}## is the surface normal vector. If you also have ##\vec{J}_{\text{free}}=0##, you can use the "magnetic potential"
$$\vec{H}=-\vec{\nabla} \Phi,$$
which fulfills
$$\vec{\nabla} \cdot (\mu \vec{\nabla} \Phi)=0.$$
That's pretty similar to the analogous boundary-value problem for adjacent dielectrics in electrostatics, and that's why similar ansätze for the solutions work also here.

I can totally see this is the argument behind such rationale. But some dots are still missing for me.

The uniqueness theorem's proof depends on the green identity per Jackson section 1.9. In summary, that proof showed that if there are two solutions of poisson equation ##\Phi_1## and ##\Phi_2## satisfying ##\nabla^2\Phi_1=\nabla^2\Phi_2=-\rho/\epsilon_0##, then for ##U=\Phi_1-\Phi_2##, it's necessary to have
##\begin{align*}
\int_V(U\nabla^2U+\nabla U\cdot\nabla U)d^3x=\oint_SU\frac{\partial U}{\partial n}da \qquad (1)
\end{align*}##

From this, we can conclude that if the RHS of above is zero, then ##\Phi_1=\Phi_2## up to a constant difference, i.e., the solution is unique.
Note how the uniqueness is conditioned upon the RHS above being zero.

Take the semi-infinite dielectric slab for example, the method of images is used there in section 4.4. The hidden argument used there is that "if we find image charge setup that satisfies
then by uniqueness, we can conclude this setup produce the desired potential in the corresponding region"

Now it's not obvious to me that once we find some setup that satisfies (2), there will be no other setup that also satisfy (2) but have different normal or tangential values. If such possibility exists, the RHS of (2) is not zero, then the solution is not unique.

In other words, for RHS of (1) to vanish, we either have to specify the potential value, or its normal derivative everywhere on the boundary. But only satisfying the constraint (2) does not mean the potential or its normal derivative on the boundary takes specific values, therefore we can't say RHS of (1) is satisfied.

Am I missing something obvious?
Thanks

But that's the uniqueness theorem you quote! For given boundary conditions the solution is unique modulo an unimportant additive constant. The idea with the "image charges" is just a heuristical tool to find an ansatz for solving the equations together with the boundary conditions. If you have a solution fulfilling the boundary conditions, that's the only solution there is to the problem.

PhDeezNutz

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