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I have problem solving the following magneto-static problem. I would greatly appreciate help and guidance.

This is how the problem is stated:

A spherical shell of radius $R$, carrying a uniform surface charge $\sigma$, is set spinning at a angular velocity $\mathbf{ω}$, see figure 5.45. Calculate the vector potential by solving the corresponding partial differential $\nabla^2 \mathbf{A} = \mu_0\mathbf{J}$.

I have attempted to solve the problem by first writing down what the surface current is, using spherical coordinate system. $$ \mathbf{J}(\mathbf{r^\prime}) = \sigma \mathbf{v} = \sigma \mathbf{ω} \times \mathbf{ρ} = \sigma\omega r^\prime \sin\theta\:\delta(r^\prime - R)\mathbf{\hat{e}_{\phi}} $$ where $\mathbf{ρ}$ is the distance from the z-axis, if the z-axis is taken to be the axis of rotation and $\theta$ is the angle measured from the z-axis. From this we conclude that $\mathbf{A}$ only depends on the $A_\phi$ component since the surface current only depends on the phi component, and thus we have $\nabla^2 A_\phi = \mu_0 J_\phi$. And we're left with solving a "simple" poission equation. Our problem clearly has azimuthal symmetry. We then proceed to solve Laplace's equation $\nabla^2 A_{\phi}(r,\theta)=0$. We solve this partial differential equation with help of separation of variables. In the end we find that the general solution is a linear combination of separable solutions. $$ A_{\phi}(r,\theta)=\sum_{l=0}^\infty \Big(A_lr^l + \frac{B_l}{r^{l+1}}\Big)P_l(\cos\theta) $$ For the interior region where $(r\leq R), B_l=0$ otherwise the potential would blow up at the origin. $$ A_{in}(r,\theta)=\sum_{l=0}^\infty A_lr^lP_l(\cos\theta) $$ In the exterior region $(r\geq R)$ we get that $A_l=0$, because they don't go to zero at infinity. $$ A_{out}(r,\theta)=\sum_{l=0}^\infty \frac{B_l}{r^{l+1}}P_l(\cos\theta) $$ We need now to join these to functions together at the boundary $r=R$. The vector potential should be continous at the boundary. So we have the following boundary conditions $$ \label{eq:one} A_{out}=A_{in} $$ $$ \Big( \frac{\partial A_{out}}{\partial r}-\frac{\partial A_{in}}{\partial r}\Big)\Big|_{r=R}=-\mu_0J_{\phi}=-\mu_0\sigma\omega R\sin\theta $$ From the first boundary condition we find that $$ B_l=A_lR^{2l+1} $$ Plugging it in the second boundary condition equation we get $$ \sum_{l=0}^\infty(2l+1)A_lR^{l-1}P_l(\cos\theta)=\mu_0\sigma\omega R\sin\theta $$ Using fouriers trick by multiplying both sides with $P_{l^\prime}(\cos\theta)\sin\theta$ and integrating. We can determine the coefficients $A_l$ since the Legendre polynomials are orthogonal functions, $$ \int_0^1 P_{l^\prime}(x)P_l(x)\textrm{d}x = \int_0^\pi P_{l^\prime}(\cos\theta)P_l(\cos\theta)\sin\theta\:\textrm{d}\theta = \frac{2}{2l+1}\delta_{ll^\prime} $$

$$ A_l= \frac{\mu_0\sigma\omega R}{2R^{l-1}} \int_0^\pi \sin\theta\:P_l(\cos\theta) \sin\theta\:\textrm{d}\theta $$ This is where I hit a brick wall. I have attempted to solve the problem by expressing $\sin\theta$ in terms of Legendre polynomials and then use the orthogonal condition to determine the coefficients $A_l$. I know that $\sin^2\theta$ can be expressed using Legendre polynomials in the following way. $$\sin^2\theta=\frac{2}{3}(P_0(\cos\theta)-P_2(\cos\theta))$$

But if I just want to have $\sin\theta$ I need to take the square root of everything and then it just gets really messy. Where did I go wrong in my solution?

Image of the problem

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The $\phi$ component of $\nabla^2\mathbf A$ is not $\nabla^2 A_\phi$, but $$ (\nabla^2 \mathbf A)_\phi = \nabla^2A_\phi - \frac{1}{r^2\sin^2\theta} A_\phi + (\text{terms involving $A_r,A_\theta$}) $$

Thus what we need to solve is not a Laplace equation. The solutions are the associated Legendre polynomials $P_l^1(\cos\theta)$:

$$ A_\phi = \sum_{l=1}^\infty \left( A_l r^l + \frac{B_l}{r^{l+1}} \right)P^1_l(\cos\theta). $$

In particular, $P^1_1(\cos\theta)=-\sin\theta$, which should solve your problem neatly.

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  • $\begingroup$ I'm not sure I follow. The laplacian of a vector is $\nabla ^2 \mathbf{A}=\nabla^2 A_r(r,\phi,\theta)\hat{\mathbf{r}} + \nabla^2 A_\phi(r,\phi,\theta)\hat{\mathbf{\phi}}+\nabla^2 A_r\theta(r,\phi,\theta)\hat{\mathbf{\theta}}$. So the phi component of the laplacian of a vector is thus $\nabla^2 A_\phi(r,\phi,\theta)$. Never mind taking a close look at the wikipedia page I see that you're correct! Thank you very much for your help! $\endgroup$ – Turbotanten Feb 26 '17 at 9:57

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