Pre-Quantum Electrodynamics
Spherical Coordinatesems.ca.sv.sph
In spherical coordinates, the Laplace equation takes the following form:
\begin{equation} \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial V}{\partial r}\right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial V}{\partial \theta}\right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 V}{\partial \phi^2} = 0 \label{Gr(3.53)} \end{equation}
If you are dealing with a problem having azimuthal symmetry, \(V\) is independent of \(\phi\) and the equation simplifies to:
\begin{equation} \frac{\partial}{\partial r} \left(r^2 \frac{\partial V}{\partial r}\right) + \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial V}{\partial \theta}\right) = 0. \label{Gr(3.54)} \end{equation}Look for solution in form
\[ V(r, \theta) = R(r) \Theta (\theta). \label{Gr(3.55)} \]
Put in (\ref{Gr(3.54)}), divide by \(V\):
\[ \frac{1}{R} \frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) + \frac{1}{\Theta \sin \theta} \frac{d}{d\theta} \left(\sin \theta \frac{d\Theta}{d\theta} \right) = 0. \label{Gr(3.56)} \]
Separation of variables logic: each term must be a constant,
\[ \frac{1}{R} \frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) = l(l+1), \hspace{1cm} \frac{1}{\Theta \sin \theta} \frac{d}{d\theta} \left(\sin \theta \frac{d\Theta}{d\theta} \right) = -l(l+1) \label{Gr(3.57)} \]
so the problem, originally involving {\it partial} differentials, now is given by {\it ordinary} differential equations. Radial equation:
\[ \frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) = l(l+1) R \label{Gr(3.58)} \]
Search for solution of form \(r^\alpha\): \(\frac{d}{dr} (r^2 \alpha r^{\alpha - 1}) = \alpha (\alpha + 1) r^{\alpha} = l(l+1) r^{\alpha}\) so we get \(\alpha = l\) or \(-(l+1)\). (\ref{Gr(3.58)}) thus has the general solution
\[ R(r) = A r^l + \frac{B}{r^{l+1}} \label{Gr(3.59)} \]
Angular equation:
\[ \frac{d}{d\theta} \left(\sin \theta \frac{d\Theta}{d\theta} \right) = -l(l+1) \sin \theta \Theta \label{Gr(3.60)} \]
has solutions in terms of {\bf Legendre polynomials} of the variable \(\cos \theta\): \[ \Theta(\theta) = P_l (\cos \theta) \label{Gr(3.61)} \] \(P_l(x)\): convenient formula is the {\bf Rodrigues formula}:
\[ P_l(x) = \frac{1}{2^l l!} \left( \frac{d}{dx} \right)^l (x^2 - 1)^l \label{Gr(3.62)} \]
Actually, a more practical formula is Bonnet's recursion relation
\[ (l + 1) P_{l+1} (x) = (2l + 1) x P_l (x) - l P_{l-1} (x) \]
First few examples:
\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). \end{align}Prefactor: chosen such that
\[ P_l(1) = 1 \label{Gr(3.63)} \]
Angular equation (\ref{Gr(3.60)}) is second order: should be 2 solutions ! These other solutions blow up at \(\theta = 0\) and/or \(\theta = \pi\) (unacceptable on physical grounds). Ex.: second solution for \(l = 0\) is
\[ \Theta (\theta) = \ln \left( \tan \frac{\theta}{2} \right) \label{Gr(3.64)} \]
General solution to problem with azimuthal symmetry:
\[ V(r,\theta) = \sum_{l=0}^{\infty} \left( A_l r^l + \frac{B_l}{r^{l+1}} \right) P_l (\cos \theta) \label{Gr(3.65)} \]
Example: separation of variables (spherical coordinates)
The potential \(V_0 (\theta)\) is specified on the surface of a hollow sphere of radius \(R\). Find potential inside sphere. \paragraph{Solution:} (this is a case of Dirichlet boundary conditions) here, \(B_l = 0\) \(\forall l\) since potential cannot diverge at origin. Formal solution:
\[ V(r,\theta) = \sum_{l=0}^\infty A_l r^l P_l (\cos \theta) \label{Gr(3.66)} \]
Boundary condition:
\[ V(R,\theta) = \sum_{l=0}^\infty A_l R^l P_l (\cos \theta) = V_0 (\theta) \label{Gr(3.67)} \]
Use fact that Legendre polynomials are orthogonal functions:
\[ \int_{-1}^1 dx P_l (x) P_{l'} (x) = \frac{2}{2l + 1} \delta_{l l'}, \label{LegendreOrthogonality} \]
or in other words
\[ \int_0^\pi d\theta \sin \theta P_l (\cos \theta) P_{l'} (\cos \theta) = \frac{2}{2l + 1} \delta_{l l'} \label{Gr(3.68)} \]
We thus get
\[ A_l = \frac{2l + 1}{2R^l} \int_0^\pi d\theta \sin \theta P_l (\cos \theta) V_0 (\theta). \label{Gr(3.69)} \]
Specific example: choose
\[ V_0 (\theta) = k \sin^2 (\theta/2) \label{Gr(3.70)} \]
This is \(V_0 (\theta) = \frac{k}{2} (1 - \cos \theta) = \frac{k}{2} (P_0 (\cos \theta) - P_1 (\cos \theta))\).
Thus, \(A_0 = k/2\), \(A_1 = -k/2\), and all others are zero, so
\[ V(r, \theta) = \frac{k}{2} (1 - \frac{r}{R} \cos \theta). \label{Gr(3.71)} \]
Example: surface charge density on sphere
A surface charge density \(\sigma_0 (\theta)\) is glued over the surface of a spherical shell of radius \(R\). Find \(V\) inside and outside sphere. \paragraph{Solution:} (this is a case of Neumann boundary conditions) Could use direct integration. Try separation of variables. In interior:
\[ V^
(other terms blow up as \(r \rightarrow 0\), so need \(B_l^< = 0\) here). Exterior:
\[ V^>(r, \theta) = \sum_{l=0}^{\infty} \frac{B_l^>}{r^{l+1}} P_l (\cos \theta), \hspace{1cm} r \geq R \label{Gr(3.79)} \]
(other terms blow up as \(r \rightarrow \infty\), so need \(A_l^> = 0\) here). Since the potential is continuous at \(r = R\):
\[ \sum_{l=0}^{\infty} A_l^< R^l P_l (\cos \theta) = \sum_{l=0}^{\infty} \frac{B_l^>}{R^{l+1}} P_l (\cos \theta) \label{Gr(3.80)} \]
Since Legendre polynomials are orthogonal, \[ B_l^> = A_l^< R^{2l + 1}. \label{Gr(3.81)} \] Surface charge induces discontinuity in derivative of \(V\) from (\ref{Gr(2.36)}):
\[ \left( \frac{\partial V^>}{\partial r} - \frac{\partial V^
so
\[ -\sum_{l=0}^\infty (l+1) \left(\frac{B_l^>}{R^{l+2}} + l A_l^< R^{l-1} \right) P_l (\cos \theta) = -\frac{\sigma_0 (\theta)}{\varepsilon_0}, \rightarrow \sum_{l=0}^\infty (2l+1) A_l^< R^{l-1} P_l (\cos \theta) = \frac{\sigma_0 (\theta)}{\varepsilon_0} \label{Gr(3.83)} \]
Coefficients: from orthogonality relation (\ref{Gr(3.68)}),
\[ A_l^< = \frac{1}{2\varepsilon_0 R^{l-1}} \int_0^\pi d\theta \sin \theta \sigma_0 (\theta) P_l (\cos \theta). \label{Gr(3.84)} \]
Specific case: choose
\[ \sigma_0 (\theta) = k \cos \theta = k P_1 (\cos \theta) \label{Gr(3.85)} \]
All \(A_l^< = 0\) except for \(l = 1\), in which case
\[ A_1^< = \frac{k}{2\varepsilon_0} \int_0^\pi d\theta \sin \theta [P_l(\cos \theta)]^2 = \frac{k}{3\varepsilon_0}. \]
Potential inside/outside the sphere are then:
\[ V^< (r,\theta) = \frac{k}{3\varepsilon_0} r\cos \theta \hspace{3mm}\mbox{for}~ r \leq R, \hspace{10mm} V^> (r, \theta) = \frac{k R^3}{3\varepsilon_0} \frac{\cos \theta}{r^2} \hspace{3mm}\mbox{for}~ r \geq R. \label{eq:PotentialUniformlyPolarizedSphere} \]
Created: 2022-02-08 Tue 06:55