# Azimuthal Symmetry

Assume $$V(\theta)\neq V(\theta,\phi)$$ then $$\Phi(r,\theta)$$. Then, $$m=0$$ in the general spherical solution. So, $$\Phi(r,\theta) = \sum_{\ell=0}^\infty \left(A_\ell r^\ell+\frac{B_\ell}{r^{\ell+1}}\right)P_\ell(\cos\theta)$$.

Assume we have a potential $$V(\theta)$$ on a sphere of radius $$a$$. Find the potential inside and outside. Then, $$\Phi_{in}(r,\theta)=\sum_{\ell=0}^\infty A_\ell r^\ell P_\ell(\cos\theta)$$, $$\Phi_{out}(r,\theta)=\sum_{\ell=0}^\infty \frac{B_\ell}{r^{\ell+1}} P_\ell(\cos\theta)$$.

By orthogonality of the Legendre polynomials, we can specify the inside solution with, $$V(\theta) = \sum_{\ell=0}^\infty A_\ell a^\ell P_\ell(\cos\theta)$$ gives $$\int_0^\pi V(\theta)P_l(\cos\theta)\:\sin\theta d\cos\theta = A_{\ell'}a^{\ell'}\frac{2}{2\ell'+1}$$.

Similarly, for the outer potential, $$V(\theta) = \sum_{\ell=0}^\infty \frac{B_\ell}{a^{\ell+1}} P_\ell(\cos\theta)$$ gives $$\int_0^\pi V(\theta)P_l(\cos\theta)\:\sin\theta d\cos\theta = \frac{B_{\ell'}}{a^{\ell'+1}}\frac{2}{2\ell'+1}$$.

Or, we can use the continuity of the inner and outer potentials to achieve the same result. $$\Phi_{in}(a,\theta) = \Phi_{out}(a,\theta)$$. By orthogonality, $$A_\ell a^\ell = \frac{B_\ell}{a^{2\ell+1}}$$ or $$B_\ell = A_\ell a^{2\ell+1}$$.

## Example

:CUSTOMID: EXAMPLE

Let $$V(\theta) = +V$$ for the top hemisphere and $$V(\theta) = -V$$ for the bottom hemisphere. So, $$A_\ell = \frac{2\ell+1}{2a^\ell}V\left(\int_0^1P_\ell(x)\:dx -\int_{-1}^0P_\ell(x)\:dx\right) = \frac{2\ell+1}{2a^\ell}V\left(\int_0^1P_\ell(x)\:dx - (-1)^\ell int_0^1P_\ell(x)\:dx\right) = V\frac{2\ell+1}{2a^\ell}\left(1 - (-1)^\ell\right)\int_0^1P_\ell(x)\:dx$$ Using the recurrence relation of the legendre polynomials, $$\int_0^1 P_\ell(x)dx = \frac{P_{\ell-1}(0)-P_{\ell+1}(0)}{2\ell+1}=(-1)^{\frac{\ell-1}{2}}\frac{\ell!!}{\ell(\ell+1)!!}$$, $$= \frac{V}{a^\ell}(-1)^{\frac{\ell-1}{2}}\frac{2\ell+1}{\ell}\frac{\ell!!}{(\ell+1)!!}$$.

For $$\ell=1$$, we get $$\Phi_{in,1}(r,\theta) = \frac{3}{2}\frac{V}{a}rP_1(\cos\theta)$$. For $$\ell=3$$, we get $$\Phi_{in,3}(r,\theta) = -\frac{7}{8}\frac{V}{a^3}r^3P_3(\cos\theta)$$

For the outside, $$B_\ell = a^{2\ell+1}A_\ell$$.

Recall that in principle, $$\Phi(\vec{x}) = \frac{-1}{4\pi}\int\Phi_s\frac{\partial G}{\partial n'}\:d\Omega$$. But this may be untenable.

Note that the general form of a dipole is $$\frac{\cos\theta}{r^2}$$.

## Example

:CUSTOMID: EXAMPLE

Put a sphere with no charge in an electric field $$\vec{E} = E_0\hat{z}$$. This polarizes the sphere. Note for the conductor, there is no field inside so the charges are arranged to counteract the applied field. Assume the potential on the sphere is zero. In the far field, this looks like an electric dipole. Hence, $$\Phi_{\sigma} \propto \frac{\cos\theta}{r^2}$$. $$E_\sigma - = E_{0}\hat{z}$$. From the external field, the potential in the far field is $$\Phi_{ex}$$. $$\Phi = \Phi_{ex} + \Phi_{\sigma}$$.

$$\vec{E}_{ex} = E_0\hat{z} = -\nabla\Phi = -\frac{\partial}{\partial Z}\Phi\hat{z}$$. So, $$\Phi(z) = -E_-z = -E_0r\cos\theta = \sum_\ell A_\ell r^\ell\cos\theta = -E_0rP_1(\cos\theta)$$, i.e. $$A_1=-E_0$$. $$\Phi_{out}(z) = \Phi_{ex} + \Phi_{\sigma} = -E_0 rP_1(\cos\theta) + \sum_{\ell=0}^\infty \frac{B_\ell}{r^{\ell+1}}P_\ell(\cos\theta)$$.

At $$r=a$$, $$\Phi(a,\theta)=0=-E_0a\cos\theta + \sum_\ell \frac{B_\ell}{a^{\ell+1}}P_\ell(\cos\theta)$$. Then, $$B_1 = E_0a^3$$. So, $$\Phi(r,\theta) = -E_0r\cos\theta + \frac{E_0a^2}{r^2}\cos\theta = -E_0\left(r-\frac{a^3}{r^2}\right)\cos\theta$$.

$$\sigma(\theta) = \varepsilon_0\left.\frac{\partial\Phi}{\partial r}\right|_{r=a} = 3\varepsilon_0 E_0\cos\theta$$.

$$\frac{1}{|\vec{x}-\vec{x}'|} = \sum_{\ell=0}^\infty\frac{r_{<}^\ell}{r_{>}^{\ell+1}}P_\ell(\cos\gamma)$$. $$r_{<} = min(r,r'), r_{>} = max(r,r')$$. $$\cos\gamma = \cos\theta\cos\theta' + \sin\theta\sin\theta'\cos(\phi-\phi')$$.

Proof. Assume we have a point charge at $$\vec{x}'$$ on $$z-$$ axis. So there is an angle $$\theta$$ between them. The potential at $$\vec{x}$$ is then, $$\frac{1}{|\vec{x}-\vec{x}'} = \sum_{\ell=0}^\infty (A_\ell r^\ell + \frac{B_\ell}{r^{\ell+1}})P_\ell(\cos\theta)$$. If $$\vec{x}$$ is on $$z-$$ axis. then, $$\frac{1}{|\vec{x}-\vec{x}'|} = \sum_{\ell=0}^\infty\left(A_\ell r^\ell + \frac{B_\ell}{r^{\ell+1}}\right) = \frac{1}{|r-r'|}$$. If $$rr'$$ then $$\frac{1}{r-r'} = \frac{1}{r}\frac{1}{1-\frac{r'}{r}}$$. So we are either inside or outside the sphere of radius $$r$$. $$\frac{1}{1-x} = 1+x+x^2+x^3+\cdots = \sum_\ell^\infty x^\ell$$. Then, for the second case, $$\frac{1}{r-r'}=\frac{r'^{\ell}}{r^{\ell+1}}$$. Or for the first case you can swap the $$r'$$ and $$r$$ labels.

So, on $$z-$$ axis, $$\frac{1}{|\vec{x}-\vec{x}'|} = \sum_{\ell}\frac{r_{<}^\ell}{r_{>}^{\ell+1}}$$.

Note also, $$P_\ell(\cos\gamma) = \frac{4\pi}{2\ell+1}\sum_{m=-\ell}^\ell Y_\ell^{m*}(\theta',\phi')Y_\ell^m(\theta,\phi)$$.

## In Class Problem

If the ring is a distance $$b$$ up from the x-y plane. Then, we get $$\vec{x}'$$ on the ring to be $$|\vec{x}'|^2 = a^2 + b^2$$. Let $$\tan\alpha = \frac{a}{b}$$. So, we get $$\vec{1}{|\vec{x}-\vec{x}'|} = \sum_\ell \frac{r_{<}^\ell}{r_{>}^{\ell+1}}P_\ell(\cos\alpha)P_\ell(\cos0)$$.

## Lightning Rod

Conical conducting tip, opening angle $$\beta$$ from $z-$axis to the cone, $$\beta<\pi$$. $$\nabla^2\Phi=0$$. So, $$0\leq \theta\leq \beta$$. So, Legendre polynomials are not our solution since they are defined on $$0\leq\theta\leq\pi$$. We get Legendre function, $$P_\nu(\cos\beta) = 0$$. $$R_\nu(r) = A_\nu r^\nu + \frac{B}{r^{\nu + 1}}$$. In our problem, $$R_\nu(r) = A_\nu r^\nu$$. $$\nu_k=\nu_0,\nu_1,\nu_2,\cdots$$. So, $$\Phi(r,\theta)=sum_{k=0}^\infty A_k r^{\nu_k} P_{\nu_k}(\cos\theta)$$.

At the tip, $$\Phi(r,\Theta) = A_0r^{\nu_0}P_{\nu_0}(\cos\theta)$$. Then the electric field, $$E_r = -\frac{\partial\Phi}{\partial r}=\nu_0 A_0r^{\nu_0-1}P_{\nu_0}$$ and $$E_\theta = \frac{-1}{r}\frac{\partial\phi}{\partial \theta}A_0r^{\nu_0-1}\sin\theta\frac{d P_{\nu_0}}{d\theta}$$.

$$\sigma=-\varepsilon_0 E_\theta|_{\theta=\beta}$$. So $$|\sigma|,|E_r|,|E_\theta|\propto \frac{1}{r^{1-\nu_0}}$$. As $$r\to 0$$, the electric field at the tip blows up. When, $$\beta\to \pi$$, we get a sharp tip and so we get $$\nu_0\approx \left(2\ln\frac{2}{\pi-\beta}\right)^{-1}\to0$$. Then, the electric field and charge density at the tip blows up.

Created: 2023-06-25 Sun 02:29

Validate