Azimuthal Symmetry

Assume $V (θ) \neq V (θ, ϕ)$ then $Φ (r, θ)$ . Then, $m = 0$ in the general spherical solution. So, $Φ (r, θ) = \sum_{ℓ = 0}^{\infty} (A_{ℓ} r^{ℓ} + \frac{B_{ℓ}}{r^{ℓ + 1}}) P_{ℓ} (\cos θ)$ .

Assume we have a potential $V (θ)$ on a sphere of radius $a$ . Find the potential inside and outside. Then, $Φ_{i n} (r, θ) = \sum_{ℓ = 0}^{\infty} A_{ℓ} r^{ℓ} P_{ℓ} (\cos θ)$ , $Φ_{o u t} (r, θ) = \sum_{ℓ = 0}^{\infty} \frac{B_{ℓ}}{r^{ℓ + 1}} P_{ℓ} (\cos θ)$ .

By orthogonality of the Legendre polynomials, we can specify the inside solution with, $V (θ) = \sum_{ℓ = 0}^{\infty} A_{ℓ} a^{ℓ} P_{ℓ} (\cos θ)$ gives $\int_{0}^{π} V (θ) P_{l} (\cos θ) \sin θ d \cos θ = A_{ℓ^{'}} a^{ℓ^{'}} \frac{2}{2 ℓ^{'} + 1}$ .

Similarly, for the outer potential, $V (θ) = \sum_{ℓ = 0}^{\infty} \frac{B_{ℓ}}{a^{ℓ + 1}} P_{ℓ} (\cos θ)$ gives $\int_{0}^{π} V (θ) P_{l} (\cos θ) \sin θ d \cos θ = \frac{B_{ℓ^{'}}}{a^{ℓ^{'} + 1}} \frac{2}{2 ℓ^{'} + 1}$ .

Or, we can use the continuity of the inner and outer potentials to achieve the same result. $Φ_{i n} (a, θ) = Φ_{o u t} (a, θ)$ . By orthogonality, $A_{ℓ} a^{ℓ} = \frac{B_{ℓ}}{a^{2 ℓ + 1}}$ or $B_{ℓ} = A_{ℓ} a^{2 ℓ + 1}$ .

Example

:CUSTOM_ID: EXAMPLE

Let $V (θ) = + V$ for the top hemisphere and $V (θ) = - V$ for the bottom hemisphere. So, $A_{ℓ} = \frac{2 ℓ + 1}{2 a^{ℓ}} V (\int_{0}^{1} P_{ℓ} (x) d x - \int_{- 1}^{0} P_{ℓ} (x) d x) = \frac{2 ℓ + 1}{2 a^{ℓ}} V (\int_{0}^{1} P_{ℓ} (x) d x - (- 1)^{ℓ} i n t_{0}^{1} P_{ℓ} (x) d x) = V \frac{2 ℓ + 1}{2 a^{ℓ}} (1 - (- 1)^{ℓ}) \int_{0}^{1} P_{ℓ} (x) d x$ Using the recurrence relation of the legendre polynomials, $\int_{0}^{1} P_{ℓ} (x) d x = \frac{P_{ℓ - 1} (0) - P_{ℓ + 1} (0)}{2 ℓ + 1} = (- 1)^{\frac{ℓ - 1}{2}} \frac{ℓ!!}{ℓ (ℓ + 1)!!}$ , $= \frac{V}{a^{ℓ}} (- 1)^{\frac{ℓ - 1}{2}} \frac{2 ℓ + 1}{ℓ} \frac{ℓ!!}{(ℓ + 1)!!}$ .

For $ℓ = 1$ , we get $Φ_{i n, 1} (r, θ) = \frac{3}{2} \frac{V}{a} r P_{1} (\cos θ)$ . For $ℓ = 3$ , we get $Φ_{i n, 3} (r, θ) = - \frac{7}{8} \frac{V}{a^{3}} r^{3} P_{3} (\cos θ)$

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

Recall that in principle, $Φ (\vec{x}) = \frac{- 1}{4 π} \int Φ_{s} \frac{\partial G}{\partial n^{'}} d Ω$ . But this may be untenable.

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

Example

:CUSTOM_ID: 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, $Φ_{σ} \propto \frac{\cos θ}{r^{2}}$ . $E_{σ} - = E_{0} \hat{z}$ . From the external field, the potential in the far field is $Φ_{e x}$ . $Φ = Φ_{e x} + Φ_{σ}$ .

${\vec{E}}_{e x} = E_{0} \hat{z} = - \nabla Φ = - \frac{\partial}{\partial Z} Φ \hat{z}$ . So, $Φ (z) = - E_{-} z = - E_{0} r \cos θ = \sum_{ℓ} A_{ℓ} r^{ℓ} \cos θ = - E_{0} r P_{1} (\cos θ)$ , i.e. $A_{1} = - E_{0}$ . $Φ_{o u t} (z) = Φ_{e x} + Φ_{σ} = - E_{0} r P_{1} (\cos θ) + \sum_{ℓ = 0}^{\infty} \frac{B_{ℓ}}{r^{ℓ + 1}} P_{ℓ} (\cos θ)$ .

At $r = a$ , $Φ (a, θ) = 0 = - E_{0} a \cos θ + \sum_{ℓ} \frac{B_{ℓ}}{a^{ℓ + 1}} P_{ℓ} (\cos θ)$ . Then, $B_{1} = E_{0} a^{3}$ . So, $Φ (r, θ) = - E_{0} r \cos θ + \frac{E_{0} a^{2}}{r^{2}} \cos θ = - E_{0} (r - \frac{a^{3}}{r^{2}}) \cos θ$ .

$σ (θ) = ε_{0} {\frac{\partial Φ}{\partial r} |}_{r = a} = 3 ε_{0} E_{0} \cos θ$ .

Addition Theorem

$\frac{1}{| \vec{x} - {\vec{x}}^{'} |} = \sum_{ℓ = 0}^{\infty} \frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} P_{ℓ} (\cos γ)$ . $r_{<} = m i n (r, r^{'}), r_{>} = m a x (r, r^{'})$ . $\cos γ = \cos θ \cos θ^{'} + \sin θ \sin θ^{'} \cos (ϕ - ϕ^{'})$ .

Proof. Assume we have a point charge at ${\vec{x}}^{'}$ on $z -$ axis. So there is an angle $θ$ between them. The potential at $\vec{x}$ is then, $\frac{1}{| \vec{x} - {\vec{x}}^{'}} = \sum_{ℓ = 0}^{\infty} (A_{ℓ} r^{ℓ} + \frac{B_{ℓ}}{r^{ℓ + 1}}) P_{ℓ} (\cos θ)$ . If $\vec{x}$ is on $z -$ axis. then, $\frac{1}{| \vec{x} - {\vec{x}}^{'} |} = \sum_{ℓ = 0}^{\infty} (A_{ℓ} r^{ℓ} + \frac{B_{ℓ}}{r^{ℓ + 1}}) = \frac{1}{| r - r^{'} |}$ . 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 $r$ . $\frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + \dots = \sum_{ℓ}^{\infty} x^{ℓ}$ . Then, for the second case, $\frac{1}{r - r^{'}} = \frac{r^{' ℓ}}{r^{ℓ + 1}}$ . Or for the first case you can swap the $r^{'}$ and $r$ labels.

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

Note also, $P_{ℓ} (\cos γ) = \frac{4 π}{2 ℓ + 1} \sum_{m = - ℓ}^{ℓ} Y_{ℓ}^{m *} (θ^{'}, ϕ^{'}) Y_{ℓ}^{m} (θ, ϕ)$ .

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 $\tan α = \frac{a}{b}$ . So, we get $\vec{1} | \vec{x} - {\vec{x}}^{'} | = \sum_{ℓ} \frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} P_{ℓ} (\cos α) P_{ℓ} (\cos 0)$ .

Lightning Rod

Conical conducting tip, opening angle $β$ from $z-$axis to the cone, $β < π$ . $\nabla^{2} Φ = 0$ . So, $0 \leq θ \leq β$ . So, Legendre polynomials are not our solution since they are defined on $0 \leq θ \leq π$ . We get Legendre function, $P_{ν} (\cos β) = 0$ . $R_{ν} (r) = A_{ν} r^{ν} + \frac{B}{r^{ν + 1}}$ . In our problem, $R_{ν} (r) = A_{ν} r^{ν}$ . $ν_{k} = ν_{0}, ν_{1}, ν_{2}, \dots$ . So, $Φ (r, θ) = s u m_{k = 0}^{\infty} A_{k} r^{ν_{k}} P_{ν_{k}} (\cos θ)$ .

At the tip, $Φ (r, Θ) = A_{0} r^{ν_{0}} P_{ν_{0}} (\cos θ)$ . Then the electric field, $E_{r} = - \frac{\partial Φ}{\partial r} = ν_{0} A_{0} r^{ν_{0} - 1} P_{ν_{0}}$ and $E_{θ} = \frac{- 1}{r} \frac{\partial ϕ}{\partial θ} A_{0} r^{ν_{0} - 1} \sin θ \frac{d P_{ν_{0}}}{d θ}$ .

$σ = - ε_{0} E_{θ} |_{θ = β}$ . So $| σ |, | E_{r} |, | E_{θ} | \propto \frac{1}{r^{1 - ν_{0}}}$ . As $r \to 0$ , the electric field at the tip blows up. When, $β \to π$ , we get a sharp tip and so we get $ν_{0} \approx {(2 \ln \frac{2}{π - β})}^{- 1} \to 0$ . Then, the electric field and charge density at the tip blows up.