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<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-02-13 Sun 21:20 -->
<!-- 2022-02-14 Mon 20:35 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1598,46 +1598,80 @@ Table of contents
</svg></a><span class="headline-id">ems.ca.sv.sph</span></h5>
<div class="outline-text-5" id="text-ems_ca_sv_sph">
<p>
In spherical coordinates, the Laplace equation takes the following form:
In spherical coordinates, the Laplace equation takes the following form
(using <a href="./c_m_cs_sph.html#sph_Lap">sph_Lap</a>):
</p>
<div class="main div" id="org813844c">
<div class="eqlabel" id="orgfbf5340">
<p>
<a id="Lap_sph"></a><a href="./ems_ca_sv_sph.html#Lap_sph"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="org76f3ef7">
<ul class="org-ul">
<li>Gr (3.53)</li>
<li>W (11-86)</li>
</ul>
</div>
</div>
<div class="main div" id="org8129eda">
<p>
</p>
\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)}
\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}
</div>
<p>
If you are dealing with a problem having <b>azimuthal symmetry</b>,
\(V\) is independent of \(\phi\) and the equation simplifies to:
\(\phi\) is independent of \(\varphi\) and the equation simplifies to:
</p>
<div class="eqlabel" id="org27146e0">
<p>
<a id="Lap_sph_az"></a><a href="./ems_ca_sv_sph.html#Lap_sph_az"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="orgff31561">
<ul class="org-ul">
<li>Gr (3.54)</li>
<li>W (11-87)</li>
</ul>
</div>
</div>
\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)
\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.
\label{Gr(3.54)}
\tag{Lap_sph_az}\label{Lap_sph_az}
\end{equation}
<p>
Look for solution in form
Without loss of generality, we can look for a solution in the
factorized product form
</p>
<p>
\[
V(r, \theta) = R(r) \Theta (\theta).
\phi(r, \theta) = R(r) \Theta (\theta).
\label{Gr(3.55)}
\]
</p>
<p>
Put in (\ref{Gr(3.54)}), divide by \(V\):
Substituting this in <a href="./ems_ca_sv_sph.html#Lap_sph_az">Lap_sph_az</a> and dividing by \(\phi\) yields
</p>
<p>
@@ -1649,81 +1683,127 @@ Put in (\ref{Gr(3.54)}), divide by \(V\):
</p>
<p>
Separation of variables logic: each term must be a constant,
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),
</p>
<p>
\[
\frac{1}{R} \frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) = l(l+1), \hspace{1cm}
\frac{1}{\Theta \sin \theta} \frac{d}{d\theta} \left(\sin \theta \frac{d\Theta}{d\theta} \right) = -l(l+1)
\label{Gr(3.57)}
\]
</p>
<p>
so the problem, originally involving {\it partial} differentials, now is given by
{\it ordinary} differential equations. Radial equation:
We thus fall back onto ordinary differential equations, whereas our original
problem involved partial differentials.
</p>
<p>
Let us look at the radial equation first:
</p>
<p>
\[
\frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) = l(l+1) R
\label{Gr(3.58)}
\]
</p>
<p>
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
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
</p>
<p>
\[
R(r) = A r^l + \frac{B}{r^{l+1}}
\label{Gr(3.59)}
\]
</p>
<p>
Angular equation:
Separately from this, the angular equation takes the form
</p>
<p>
\[
\frac{d}{d\theta} \left(\sin \theta \frac{d\Theta}{d\theta} \right) = -l(l+1) \sin \theta \Theta
\label{Gr(3.60)}
\frac{d}{d\theta} \left(\sin \theta \frac{d\Theta}{d\theta} \right) = -l(l+1) \sin \theta ~\Theta
\]
</p>
<p>
has solutions in terms of {\bf Legendre polynomials} of the variable \(\cos \theta\):
This equation is solved by <b>Legendre polynomials</b> 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>
<p>
A particularly convenient formula for deriving \(P_l(x)\)
is the <b>Rodrigues formula</b>:
</p>
<div class="eqlabel" id="org6cd91e9">
<p>
<a id="Rodrigues"></a><a href="./ems_ca_sv_sph.html#Rodrigues"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="org558db55">
<ul class="org-ul">
<li>Gr (3.62)</li>
</ul>
</div>
</div>
<p>
\[
P_l(x) = \frac{1}{2^l l!} \left( \frac{d}{dx} \right)^l (x^2 - 1)^l
\label{Gr(3.62)}
\tag{Rodrigues}\label{Rodrigues}
\]
</p>
<p>
Actually, a more practical formula is <b>Bonnet's recursion relation</b>
</p>
<div class="eqlabel" id="org78d828a">
<p>
<a id="Bonnet"></a><a href="./ems_ca_sv_sph.html#Bonnet"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="org5115e24">
</div>
</div>
<p>
\[
(l + 1) P_{l+1} (x) = (2l + 1) x P_l (x) - l P_{l-1} (x)
\tag{Bonnet}\label{Bonnet}
\]
</p>
<p>
First few examples:
For future reference, here are the first few Legendre polynomials:
</p>
<div class="eqlabel" id="orga93835c">
<p>
<a id="Leg_pols"></a><a href="./ems_ca_sv_sph.html#Leg_pols"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="orgcd3f06e">
</div>
</div>
\begin{align}
P_0 (x) &amp;= 1 \nonumber \\
@@ -1732,96 +1812,165 @@ P_2 (x) &amp;= \frac{1}{2} (3x^2 - 1) \nonumber \\
P_3 (x) &amp;= \frac{1}{2} (5x^3 - 3x) \nonumber \\
P_4 (x) &amp;= \frac{1}{8} (35x^4 - 30x^2 + 3) \nonumber \\
P_5 (x) &amp;= \frac{1}{8} (63x^5 - 70x^3 + 15x).
\tag{Leg_pols}\label{Leg_pols}
\end{align}
<p>
Prefactor: chosen such that
The prefactor is chosen for convenience such that
</p>
<div class="eqlabel" id="org23dbe5c">
<p>
<a id="Pl_1_1"></a><a href="./ems_ca_sv_sph.html#Pl_1_1"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="org61ef8fc">
</div>
</div>
<p>
\[
P_l(1) = 1
\label{Gr(3.63)}
\tag{Pl_1_1}\label{Pl_1_1}
\]
</p>
<p>
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
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
</p>
<p>
\[
\Theta (\theta) = \ln \left( \tan \frac{\theta}{2} \right)
\label{Gr(3.64)}
\]
</p>
<p>
General solution to problem with azimuthal symmetry:
We therefore come to the culmination of our efforts here, and write
the general solution to <i>any</i> problem with azimuthal symmetry
(for which the potential takes a finite value for \(\theta = 0, \pi\)) as
</p>
<div class="main div" id="org05645c9">
<div class="eqlabel" id="orgdbaeb25">
<p>
<a id="Lap_sph_az_sol"></a><a href="./ems_ca_sv_sph.html#Lap_sph_az_sol"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="orga105225">
<ul class="org-ul">
<li>Gr (3.65)</li>
</ul>
</div>
</div>
<div class="main div" id="orgb454935">
<p>
\[
V(r,\theta) = \sum_{l=0}^{\infty} \left( A_l r^l + \frac{B_l}{r^{l+1}} \right) P_l (\cos \theta)
\label{Gr(3.65)}
\]
\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}
\]
</p>
</div>
<div class="example div" id="orga84dbfd">
<div class="example div" id="orgc60f117">
<p>
<b>Example: separation of variables (spherical coordinates)</b>
<b>Example: potential inside a hollow sphere</b>
</p>
<p>
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:
Consider a hollow sphere of radius \(R\), with specified potential on the surface
equal to a given function \(\phi_0 (\theta)\).
</p>
<p>
<b>Question</b>: find potential inside the sphere.
</p>
<p>
<b>Solution</b>: (by the way, this is a case of Dirichlet boundary conditions)
</p>
<p>
Since the potential cannot diverge at the origin, we set \(B_l = 0\) \(\forall l\).
</p>
<p>
Our solution must thus take the form
</p>
<p>
\[
V(r,\theta) = \sum_{l=0}^\infty A_l r^l P_l (\cos \theta)
\phi(r,\theta) = \sum_{l=0}^\infty A_l r^l P_l (\cos \theta)
\label{Gr(3.66)}
\]
</p>
<p>
Boundary condition:
The specified boundary condition means that
</p>
<p>
\[
V(R,\theta) = \sum_{l=0}^\infty A_l R^l P_l (\cos \theta) = V_0 (\theta)
\phi(R,\theta) = \sum_{l=0}^\infty A_l R^l P_l (\cos \theta) = \phi_0 (\theta)
\label{Gr(3.67)}
\]
</p>
<p>
Use fact that Legendre polynomials are orthogonal functions:
We can now use the fact that Legendre polynomials are orthogonal functions
with orthogonality relation
</p>
<div class="eqlabel" id="org17ec99b">
<p>
<a id="Leg_orth"></a><a href="./ems_ca_sv_sph.html#Leg_orth"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="orga0d0923">
</div>
</div>
<p>
\[
\int_{-1}^1 dx P_l (x) P_{l'} (x) = \frac{2}{2l + 1} \delta_{l l'},
\label{LegendreOrthogonality}
\tag{Leg_orth}\label{Leg_orth}
\]
</p>
<p>
or in other words
or rewritten in terms of trigonometric arguments
</p>
<div class="eqlabel" id="org8d87c5d">
<p>
<a id="Leg_orth_trig"></a><a href="./ems_ca_sv_sph.html#Leg_orth_trig"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="org3a73b0c">
</div>
</div>
<p>
\[
\int_0^\pi d\theta \sin \theta P_l (\cos \theta) P_{l'} (\cos \theta) = \frac{2}{2l + 1} \delta_{l l'}
\label{Gr(3.68)}
\]
</p>
@@ -1831,26 +1980,23 @@ We thus get
<p>
\[
A_l = \frac{2l + 1}{2R^l} \int_0^\pi d\theta \sin \theta P_l (\cos \theta) V_0 (\theta).
\label{Gr(3.69)}
A_l = \frac{2l + 1}{2R^l} \int_0^\pi d\theta \sin \theta ~P_l (\cos \theta) \phi_0 (\theta).
\]
</p>
</div>
<p>
<b>Specific example:</b> choose
</p>
<p>
\[
V_0 (\theta) = k \sin^2 (\theta/2)
\phi_0 (\theta) = k \sin^2 (\theta/2)
\label{Gr(3.70)}
\]
</p>
<p>
This is \(V_0 (\theta) = \frac{k}{2} (1 - \cos \theta) = \frac{k}{2} (P_0 (\cos \theta) - P_1 (\cos \theta))\).
This is \(\phi_0 (\theta) = \frac{k}{2} (1 - \cos \theta) = \frac{k}{2} (P_0 (\cos \theta) - P_1 (\cos \theta))\).
</p>
<p>
@@ -1859,90 +2005,119 @@ Thus, \(A_0 = k/2\), \(A_1 = -k/2\), and all others are zero, so
<p>
\[
V(r, \theta) = \frac{k}{2} (1 - \frac{r}{R} \cos \theta).
\phi(r, \theta) = \frac{k}{2} (1 - \frac{r}{R} \cos \theta).
\label{Gr(3.71)}
\]
</p>
</div>
<div class="example div" id="org6004c2b">
<div class="example div" id="org4350b9a">
<p>
<b>Example: surface charge density on sphere</b>
</p>
<p>
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:
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.
</p>
<p>
<b>Question</b>: find \(\phi\) inside and outside sphere.
</p>
<p>
<b>Solution</b>: (by the way, this is a case of Neumann boundary conditions)
</p>
<p>
We could of course use direct integration of <a href="./ems_es_ep_d.html#p_scd">p_scd</a>, but let us save some
effort by invoking separation of variables. In the interior of the shell,
</p>
<p>
\[
V^
</p>
<p>
(other terms blow up as \(r \rightarrow 0\), so need \(B_l^&lt; = 0\) here). Exterior:
</p>
<p>
\[
V^&gt;(r, \theta) = \sum_{l=0}^{\infty} \frac{B_l^&gt;}{r^{l+1}} P_l (\cos \theta), \hspace{1cm} r \geq R
\label{Gr(3.79)}
\phi^i (r,\theta) = \sum_{l=0}^{\infty} A_l^i r^l P_l (\cos \theta), \hspace{1cm} r \leq R
\]
</p>
<p>
(other terms blow up as \(r \rightarrow \infty\), so need \(A_l^&gt; = 0\) here). Since the potential
is continuous at \(r = R\):
(other terms blow up as \(r \rightarrow 0\), so we need to set \(B_l^i = 0\) here).
</p>
<p>
In the region exterior to the shell,
</p>
<p>
\[
\sum_{l=0}^{\infty} A_l^&lt; R^l P_l (\cos \theta) = \sum_{l=0}^{\infty} \frac{B_l^&gt;}{R^{l+1}} P_l (\cos \theta)
\label{Gr(3.80)}
\phi^o(r, \theta) = \sum_{l=0}^{\infty} \frac{B_l^o}{r^{l+1}} P_l (\cos \theta), \hspace{1cm} r \geq R
\]
</p>
<p>
Since Legendre polynomials are orthogonal,
\[
B_l^&gt; = A_l^&lt; R^{2l + 1}.
\label{Gr(3.81)}
\]
Surface charge induces discontinuity in derivative of \(V\) from <a href="./ems_es_ep_bc.html#dpdisc">dpdisc</a>:
(other terms blow up as \(r \rightarrow \infty\), so we need to set \(A_l^o = 0\) here).
</p>
<p>
Since the potential must be continuous at \(r = R\), we must have
</p>
<p>
\[
\left( \frac{\partial V^&gt;}{\partial r} - \frac{\partial V^
\sum_{l=0}^{\infty} A_l^i R^l P_l (\cos \theta) = \sum_{l=0}^{\infty} \frac{B_l^o}{R^{l+1}} P_l (\cos \theta)
\]
</p>
<p>
Invoking the orthononality of Legendre polynomials thus yields
</p>
<p>
\[
B_l^o = A_l^i R^{2l + 1}.
\]
</p>
<p>
The surface charge induces a discontinuity in derivative of \(\phi\)
according to <a href="./ems_es_ep_bc.html#dpdisc">dpdisc</a>:
</p>
\begin{equation*}
\left( \frac{\partial \phi^{o}}{\partial r} - \frac{\partial \phi^{i}}{\partial r} \right)
= -\frac{\sigma_0 (\theta)}{\varepsilon_0}
\end{equation*}
<p>
so
</p>
<p>
\[
-\sum_{l=0}^\infty (l+1) \left(\frac{B_l^&gt;}{R^{l+2}} + l A_l^&lt; R^{l-1} \right) P_l (\cos \theta) = -\frac{\sigma_0 (\theta)}{\varepsilon_0},
\rightarrow
\sum_{l=0}^\infty (2l+1) A_l^&lt; R^{l-1} P_l (\cos \theta) = \frac{\sigma_0 (\theta)}{\varepsilon_0}
\label{Gr(3.83)}
-\sum_{l=0}^\infty (l+1) \left(\frac{B_l^o}{R^{l+2}} + l A_l^i R^{l-1} \right) P_l (\cos \theta) = -\frac{\sigma_0 (\theta)}{\varepsilon_0},
\]
</p>
<p>
Coefficients: from orthogonality relation (\ref{Gr(3.68)}),
and thus
</p>
<p>
\[
A_l^&lt; = \frac{1}{2\varepsilon_0 R^{l-1}} \int_0^\pi d\theta \sin \theta \sigma_0 (\theta) P_l (\cos \theta).
\label{Gr(3.84)}
\sum_{l=0}^\infty (2l+1) A_l^i R^{l-1} P_l (\cos \theta) = \frac{\sigma_0 (\theta)}{\varepsilon_0}
\]
</p>
<p>
The coefficients can be fixed from the orthogonality relation <a href="./ems_ca_sv_sph.html#Leg_orth_trig">Leg_orth_trig</a>,
</p>
<p>
\[
A_l^i = \frac{1}{2\varepsilon_0 R^{l-1}} \int_0^\pi d\theta \sin \theta ~\sigma_0 (\theta) P_l (\cos \theta).
\]
</p>
</div>
<p>
<b>Specific case</b>: choose
@@ -1956,27 +2131,39 @@ A_l^&lt; = \frac{1}{2\varepsilon_0 R^{l-1}} \int_0^\pi d\theta \sin \theta \sigm
</p>
<p>
All \(A_l^&lt; = 0\) except for \(l = 1\), in which case
All \(A_l^i = 0\) except for \(l = 1\), in which case
</p>
<p>
\[
A_1^&lt; = \frac{k}{2\varepsilon_0} \int_0^\pi d\theta \sin \theta [P_l(\cos \theta)]^2 = \frac{k}{3\varepsilon_0}.
A_1^i = \frac{k}{2\varepsilon_0} \int_0^\pi d\theta \sin \theta [P_l(\cos \theta)]^2 = \frac{k}{3\varepsilon_0}.
\]
</p>
<p>
Potential inside/outside the sphere are then:
The potential inside/outside the sphere is then
</p>
<div class="eqlabel" id="org135e184">
<p>
\[
V^&lt; (r,\theta) = \frac{k}{3\varepsilon_0} r\cos \theta \hspace{3mm}\mbox{for}~ r \leq R,
\hspace{10mm}
V^&gt; (r, \theta) = \frac{k R^3}{3\varepsilon_0} \frac{\cos \theta}{r^2} \hspace{3mm}\mbox{for}~ r \geq R.
\label{eq:PotentialUniformlyPolarizedSphere}
\]
<a id="p_uni_ch_sph"></a><a href="./ems_ca_sv_sph.html#p_uni_ch_sph"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="orgcc2b0d5">
</div>
</div>
\begin{align}
\phi^i (r,\theta) &amp;= \frac{k}{3\varepsilon_0} r\cos \theta \hspace{3mm}\mbox{for}~ r \leq R,
\nonumber \\
\phi^o (r, \theta) &amp;= \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}
</div>
</div>
</div>
@@ -1996,7 +2183,7 @@ target="_blank">Creative Commons Attribution 4.0 International License</a>.
</div>
<div id="postamble" class="status">
<p class="author">Author: Jean-Sébastien Caux</p>
<p class="date">Created: 2022-02-13 Sun 21:20</p>
<p class="date">Created: 2022-02-14 Mon 20:35</p>
<p class="validation"></p>
</div>