Pre-Quantum Electrodynamics
Spherical Coordinatesems.ca.sv.sph
In spherical coordinates, the Laplace equation takes the following form (using sph_Lap):
\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:
\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 no 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 \(\cos \theta\): \[ \Theta(\theta) = P_l (\cos \theta) \]
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
\[ (l + 1) P_{l+1} (x) = (2l + 1) x P_l (x) - l P_{l-1} (x) \tag{Bonnet}\label{Bonnet} \]
For future reference, here are the first few Legendre polynomials:
\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 is chosen for convenience such that
\[ P_l(1) = 1 \tag{Pl_1_1}\label{Pl_1_1} \]
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
\[ \phi(r,\theta) = \sum_{l=0}^{\infty} \left( A_l r^l + \frac{B_l}{r^{l+1}} \right) P_l (\cos \theta) \tag{Lap_sph_az_sol}\label{Lap_sph_az_sol} \]
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 with orthogonality relation
\[ \int_{-1}^1 dx P_l (x) P_{l'} (x) = \frac{2}{2l + 1} \delta_{l l'}, \tag{Leg_orth}\label{Leg_orth} \]
or rewritten in terms of trigonometric arguments
\[ \int_0^\pi d\theta \sin \theta P_l (\cos \theta) P_{l'} (\cos \theta) = \frac{2}{2l + 1} \delta_{l l'} \]
We thus get
\[ A_l = \frac{2l + 1}{2R^l} \int_0^\pi d\theta \sin \theta ~P_l (\cos \theta) \phi_0 (\theta). \]
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
\[ -\sum_{l=0}^\infty (l+1) \left(\frac{B_l^o}{R^{l+2}} + l A_l^i R^{l-1} \right) P_l (\cos \theta) = -\frac{\sigma_0 (\theta)}{\varepsilon_0}, \] 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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Created: 2022-02-14 Mon 20:35