##### Spherical Coordinates
ems.ca.sv.sph

In spherical coordinates, the Laplace equation takes the following form (using sph_Lap):

- Gr (3.53)
- W (11-86)

\begin{equation} \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial \phi}{\partial r}\right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial \phi}{\partial \theta}\right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 \phi}{\partial \varphi^2} = 0 \tag{Lap_sph}\label{Lap_sph} \end{equation}

If you are dealing with a problem having azimuthal symmetry, \(\phi\) is independent of \(\varphi\) and the equation simplifies to:

- Gr (3.54)
- W (11-87)

\begin{equation} \frac{\partial}{\partial r} \left(r^2 \frac{\partial \phi}{\partial r}\right) + \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial \phi}{\partial \theta}\right) = 0. \tag{Lap_sph_az}\label{Lap_sph_az} \end{equation}

Without loss of generality, we can look for a solution in the factorized product form

\[ \phi(r, \theta) = R(r) \Theta (\theta). \label{Gr(3.55)} \]

Substituting this in Lap_sph_az and dividing by \(\phi\) yields

\[ \frac{1}{R} \frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) + \frac{1}{\Theta \sin \theta} \frac{d}{d\theta} \left(\sin \theta \frac{d\Theta}{d\theta} \right) = 0. \label{Gr(3.56)} \]

We can now apply the separation of variables logic: being dependent on separate variables, each term must be constant (the reasons for the convenient choice will become clear later),

\[ \frac{1}{R} \frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) = l(l+1), \hspace{1cm} \frac{1}{\Theta \sin \theta} \frac{d}{d\theta} \left(\sin \theta \frac{d\Theta}{d\theta} \right) = -l(l+1) \]

We thus fall back onto ordinary differential equations, whereas our original problem involved partial differentials.

Let us look at the radial equation first:

\[ \frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) = l(l+1) R \]

Let us search for a solution in the form \(r^\alpha\): since \(\frac{d}{dr} (r^2 \alpha r^{\alpha - 1}) = \alpha (\alpha + 1) r^{\alpha} = l(l+1) r^{\alpha}\), we get \(\alpha = l\) or \(-(l+1)\). The radial equation thus has the general solution

\[ R(r) = A r^l + \frac{B}{r^{l+1}} \]

Separately from this, the angular equation takes the form

\[ \frac{d}{d\theta} \left(\sin \theta \frac{d\Theta}{d\theta} \right) = -l(l+1) \sin \theta ~\Theta \]

This equation is solved by Legendre polynomials of the variable \(\cos \theta\): \[ \Theta(\theta) = P_l (\cos \theta) \]

Legendre polynomials

When using spherical coordinates, one inevitably comes across integrals of the form \[ \int_0^\pi d\theta ~\sin \theta ~f(\theta) \] for generic functions \(f\).

Inspired by the logic of Fourier series, we would like to decompose such generic functions in a basis of "orthonormal" functions under this kind of integral (with the \(\sin \theta\) weight). This idea lead us to the Legendre polynomials, denoted \(P_l\), l = 0, 1, 2, …, and conveniently defined (for trigonometric arguments) to obey the orthogonality relationship (the reason for the normalization on the right-hand side will become clear later)

\[ \int_0^\pi d\theta \sin \theta ~P_l (\cos \theta) P_{l'} (\cos \theta) = \frac{2}{2l + 1} \delta_{l l'} \tag{Leg_orth_trig}\label{Leg_orth_trig} \]

This same relation can be more simply written by using the variable \(x = \cos \theta\),

\[ \int_{-1}^1 dx P_l (x) P_{l'} (x) = \frac{2}{2l + 1} \delta_{l l'}, \tag{Leg_orth}\label{Leg_orth} \]

To get started, we need to define the "seed" polynomial (carrying label \(l=0\)). To make life easy, we set \(P_0 (x) = 1\). Higher polynomials are then sought in the form of power series in \(x\). This leads to the first few Legendre polynomials being:

\begin{align} P_0 (x) &= 1 \nonumber \\ P_1 (x) &= x \nonumber \\ P_2 (x) &= \frac{1}{2} (3x^2 - 1) \nonumber \\ P_3 (x) &= \frac{1}{2} (5x^3 - 3x) \nonumber \\ P_4 (x) &= \frac{1}{8} (35x^4 - 30x^2 + 3) \nonumber \\ P_5 (x) &= \frac{1}{8} (63x^5 - 70x^3 + 15x). \tag{Leg_pols}\label{Leg_pols} \end{align}

The prefactor (and thus the factor in the orthogonality relations Leg_orth_trig and the equivalent Leg_orth is chosen for convenience such that each polynomial takes the value \(1\) when evaluated at argument \(x = 1\),

\[ P_l(1) = 1 \tag{Pl_1_1}\label{Pl_1_1} \]

The Legendre polynomial \(P_l\) obeys the differential equation

\[ \left[\frac{d}{d\theta} \left( \sin \theta \frac{d}{d\theta} \right) + l (l+1) \sin \theta \right] P_l (\cos \theta) = 0. \tag{Leg_de_trig}\label{Leg_de_trig} \] or equivalently

\[ \left[(1 - x^2) \frac{d^2}{dx^2} - 2x \frac{d}{dx} + l(l+1) \right] P_l (x) = 0. \tag{Leg_de}\label{Leg_de} \]

A particularly convenient formula for deriving \(P_l(x)\) is the Rodrigues formula:

\[ P_l(x) = \frac{1}{2^l l!} \left( \frac{d}{dx} \right)^l (x^2 - 1)^l \tag{Rodrigues}\label{Rodrigues} \]

Actually, a more practical formula is Bonnet's recursion relation

\[ (l + 1) P_{l+1} (x) = (2l + 1) x P_l (x) - l P_{l-1} (x) \tag{Bonnet}\label{Bonnet} \]

Going back to the angular equation, let us first remark that this is a second order equation, and should thus have 2 solutions. These other solutions blow up at \(\theta = 0\) and/or \(\theta = \pi\), and we thus exclude them on physical grounds. For example, the (here discarded) second solution for \(l = 0\) is

\[ \Theta (\theta) = \ln \left( \tan \frac{\theta}{2} \right) \]

We therefore come to the culmination of our efforts here, and write the general solution to any problem with azimuthal symmetry (for which the potential takes a finite value for \(\theta = 0, \pi\)) as

- Gr (3.65)

\[ \phi(r,\theta) = \sum_{l=0}^{\infty} \left( A_l r^l + \frac{B_l}{r^{l+1}} \right) P_l (\cos \theta) \tag{Lap_sph_az_sol}\label{Lap_sph_az_sol} \]

Example: potential inside a hollow sphere

Consider a hollow sphere of radius \(R\), with specified potential on the surface equal to a given function \(\phi_0 (\theta)\).

Question: find potential inside the sphere.

Solution: (by the way, this is a case of Dirichlet boundary conditions)

Since the potential cannot diverge at the origin, we set \(B_l = 0\) \(\forall l\).

Our solution must thus take the form

\[ \phi(r,\theta) = \sum_{l=0}^\infty A_l r^l P_l (\cos \theta) \label{Gr(3.66)} \]

The specified boundary condition means that

\[ \phi(R,\theta) = \sum_{l=0}^\infty A_l R^l P_l (\cos \theta) = \phi_0 (\theta) \label{Gr(3.67)} \]

We can now use the fact that Legendre polynomials are orthogonal functions, giving us

\[ A_l = \frac{2l + 1}{2R^l} \int_0^\pi d\theta \sin \theta ~P_l (\cos \theta) \phi_0 (\theta). \]

Specific example: choose

\[ \phi_0 (\theta) = k \sin^2 (\theta/2) \label{Gr(3.70)} \]

This is \(\phi_0 (\theta) = \frac{k}{2} (1 - \cos \theta) = \frac{k}{2} (P_0 (\cos \theta) - P_1 (\cos \theta))\).

Thus, \(A_0 = k/2\), \(A_1 = -k/2\), and all others are zero, so

\[ \phi(r, \theta) = \frac{k}{2} (1 - \frac{r}{R} \cos \theta). \label{Gr(3.71)} \]

Example: surface charge density on sphere

Consider once again a spherical shell of radius \(R\). This time, we affix a surface charge density \(\sigma_0 (\theta)\) over the surface of the shell.

Question: find \(\phi\) inside and outside sphere.

Solution: (by the way, this is a case of Neumann boundary conditions)

We could of course use direct integration of p_scd, but let us save some effort by invoking separation of variables. In the interior of the shell,

\[ \phi^i (r,\theta) = \sum_{l=0}^{\infty} A_l^i r^l P_l (\cos \theta), \hspace{1cm} r \leq R \]

(other terms blow up as \(r \rightarrow 0\), so we need to set \(B_l^i = 0\) here).

In the region exterior to the shell,

\[ \phi^o(r, \theta) = \sum_{l=0}^{\infty} \frac{B_l^o}{r^{l+1}} P_l (\cos \theta), \hspace{1cm} r \geq R \]

(other terms blow up as \(r \rightarrow \infty\), so we need to set \(A_l^o = 0\) here).

Since the potential must be continuous at \(r = R\), we must have

\[ \sum_{l=0}^{\infty} A_l^i R^l P_l (\cos \theta) = \sum_{l=0}^{\infty} \frac{B_l^o}{R^{l+1}} P_l (\cos \theta) \]

Invoking the orthononality of Legendre polynomials thus yields

\[ B_l^o = A_l^i R^{2l + 1}. \]

The surface charge induces a discontinuity in derivative of \(\phi\) according to dpdisc:

\begin{equation*} \left( \frac{\partial \phi^{o}}{\partial r} - \frac{\partial \phi^{i}}{\partial r} \right) = -\frac{\sigma_0 (\theta)}{\varepsilon_0} \end{equation*}

so

\[ -\sum_{l=0}^\infty \left((l+1) \frac{B_l^o}{R^{l+2}} + l A_l^i R^{l-1} \right) P_l (\cos \theta) = -\frac{\sigma_0 (\theta)}{\varepsilon_0}, \] and thus

\[ \sum_{l=0}^\infty (2l+1) A_l^i R^{l-1} P_l (\cos \theta) = \frac{\sigma_0 (\theta)}{\varepsilon_0} \]

The coefficients can be fixed from the orthogonality relation Leg_orth_trig,

\[ A_l^i = \frac{1}{2\varepsilon_0 R^{l-1}} \int_0^\pi d\theta \sin \theta ~\sigma_0 (\theta) P_l (\cos \theta). \]

Specific case: choose

\[ \sigma_0 (\theta) = k \cos \theta = k P_1 (\cos \theta) \label{Gr(3.85)} \]

All \(A_l^i = 0\) except for \(l = 1\), in which case

\[ A_1^i = \frac{k}{2\varepsilon_0} \int_0^\pi d\theta \sin \theta [P_l(\cos \theta)]^2 = \frac{k}{3\varepsilon_0}. \]

The potential inside/outside the sphere is then

\begin{align} \phi^i (r,\theta) &= \frac{k}{3\varepsilon_0} r\cos \theta \hspace{3mm}\mbox{for}~ r \leq R, \nonumber \\ \phi^o (r, \theta) &= \frac{k R^3}{3\varepsilon_0} \frac{\cos \theta}{r^2} \hspace{3mm}\mbox{for}~ r \geq R. \tag{p_uni_ch_sph}\label{p_uni_ch_sph} \end{align}

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42