# Find the field inside and outside a spherical geometry

• Nitacii
Nitacii
Homework Statement
Given the current time dependent density ##\mathbf{J}## find the EM field inside and outside a sphere.
Relevant Equations
##\mathbf{J} = k(t) \delta(r-a) \sin \theta \hat{\boldsymbol{\phi}}##
So I tried to solve this using the Hertz potentials. I choose the magnetic one since this one corresponds to the magnetisation.
Before I start let me note that I denote a unit vector with a hat, while ##{x,y,z}## are the Cartesian coordinates and ##{r,\theta,\phi}## are the spherical coordinates.

First we need to relate the current to the magnetisation. Since for a surface current ##\mathbf{K}##, the magnetization satisfies the equation ##\mathbf{K} =\mathbf{M} \times \hat{\mathbf{n}}## we get
$$\mathbf{M}|_{r =a} = k(t) \hat{\mathbf{z}}.$$

The Hertz potential ##\boldsymbol{\Pi}_m## satisfies the equation

$$\nabla^2 \boldsymbol{\Pi}_m - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \boldsymbol{\Pi}_m = -\mu_0 \mathbf{M}$$
since the magnetiazion is non-zero only in the ##\hat{\mathbf{z}}## direction and utilizing the spherical symmetry of the problem I assumed that ##\boldsymbol{\Pi}_m = \Pi_z(r,t) \hat{\mathbf{z}}##, then the equation for the potential is

$$\frac{1}{r} \frac{\partial^2}{\partial r^2} \left(r \Pi_z(r,t) \right) - \frac{1}{c^2} \frac{\partial^2}{\partial t^2 }\Pi_z(r,t) = -\mu_0 H(t) \delta (r-a).$$

Now this is basically where I get stuck, I'm unable to solve this equations, I've tried combining the spherical Bessel functions so that the field is regular at the origin and zero at infinity (an outgoing wave), but I am not able to solve the equations above.

I assume when I would get the homogeneous equation for the potential I would then assume that the potential is continuous and impose the Maxwell equations by checking the boundary, note that ##\mathbf{A} = \nabla \times \boldsymbol{\Pi}## and ##\mathbf{B} = \nabla \times \nabla \times \boldsymbol{\Pi}_m##.

So the equations is basically what form of solutions satisfy the equation for the potential?

Edit: I've found an article which solves exactly the same problem avaiable on Arxiv, but they use Laplace transform and for me the derivation is quite unclear.

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Note that $\mathbf{J}$ is zero everywhere except on $r = a$ where it is not defined, but causes a jump in $\mathbf{B}$ across $r = a$. Therefore you need to take the general solutions in a vacuum which are proportional to $\sin \theta$ or $\cos \theta$, which are different for $r < a$ and $r > a$, and adjust the arbitrary constants to satisfy the boundary conditions at $r = a$.

hutchphd and Nitacii
That's why I utilised the Hertz vectors, this automatically sets the correct dependence the on ##\theta##. If we reconstruct the vector potential ##\mathbf{A}## we get

$$\mathbf{A} = \nabla \times \boldsymbol{\Pi} = - \sin \theta \frac{\partial}{\partial r} \Pi_z(r,t) \hat{\boldsymbol{\phi}} = A_\phi \hat{\boldsymbol{\phi}}.$$

This seemed the correct reasoning because if we take then the equation for the vector potential ##\mathbf{A}## we get

$$\frac{1}{r} \frac{\partial^2}{\partial r^2}\left(r A_\phi\right)+\frac{1}{r^2} \frac{\partial}{\partial \theta}\left[\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta A_\phi\right)\right]-\frac{1}{c^2} \frac{\partial^2}{\partial t^2} A_\phi =-\mu_0 J_\phi$$
inputing the sine dependence on this we get

$$\frac{\partial^2}{\partial r^2}(r A_\phi)-\frac{2}{r^2}(r A)-\frac{1}{c^2} \frac{\partial^2}{\partial t^2}(r A_\phi)= - \delta(r-a) \mu_0 k(t).$$

this is the same dependence on the ##\theta## variable in the article I've just mentioned in the edit of the original question. Perhaps this is wrong...

Anyway I'm not sure what the general solution for ##\mathbf{\Pi}_m## is.

Perhpas I've didn't phrase the question clearly. I'm just asking what form of $\Pi_z$ do I have to use.

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