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<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-02-10 Thu 08:32 -->
<!-- 2022-02-13 Sun 21:20 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1598,8 +1598,10 @@ Table of contents
</svg></a><span class="headline-id">ems.ca.fe.L</span></h5>
<div class="outline-text-5" id="text-ems_ca_fe_L">
<p>
Of course, the simplest situation is to start by looking at the region of space
where there is no charge density. The potential then solves Laplace's equation. How can it possibly look ?
In regions of space where there is no charge density,
the potential must solve Laplace's equation.
Let us discuss how solutions to this equation look,
in increasingly complicated situations.
</p>
</div>
@@ -1608,18 +1610,85 @@ where there is no charge density. The potential then solves Laplace's equation.
<h6 id="ems_ca_fe_L_1d"><a href="#ems_ca_fe_L_1d">The Laplace Equation in One Dimension</a></h6>
<div class="outline-text-6" id="text-ems_ca_fe_L_1d">
<p>
In one dimension, the potential is a single-variable
function \(\phi (x)\) and the Laplace equation reads
</p>
<div class="eqlabel" id="org8b9eaa8">
<p>
<a id="Lap_1d"></a><a href="./ems_ca_fe_L.html#Lap_1d"><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="org1a79944">
</div>
</div>
<p>
\[
\frac{d^2 V(x)}{dx^2} = 0 \Longrightarrow V(x) = mx + b
\label{Gr(3.6)}
\frac{d^2 \phi(x)}{dx^2} = 0.
\tag{Lap_1d}\label{Lap_1d}
\]
Properties:
\paragraph{1.} \(V(x)\) is the average of \(V(x + a)\) and \(V(x - a)\) for any \(a\).
\paragraph{2.} Solutions to Laplace's equation have no local maxima or minima.
</p>
<p>
Boundary conditions: always work: two end values, one end value + same end derivative value.
Not always: one end value + derivative value at other end, two end derivative values.
The solution to this
</p>
<div class="eqlabel" id="orgd6d6f8b">
<p>
<a id="Lap_1d_sol"></a><a href="./ems_ca_fe_L.html#Lap_1d_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="org135f906">
<ul class="org-ul">
<li>Gr (3.6)</li>
</ul>
</div>
</div>
<p>
\[
\phi(x) = a x + b
\tag{Lap_1d_sol}\label{Lap_1d_sol}
\]
</p>
<p>
Properties:
</p>
<ul class="org-ul">
<li>
<b>Balance</b>: \(\phi(x)\) is the average of \(\phi(x + dx)\) and \(\phi(x - dx)\) for any \(dx\) (with \(x \pm dx\) still being in
the region where Laplace is satisfied, of course).</li>
<li>
<b>No extrema</b>: \(\phi(x)\) has no local extrema. Max/min
values must occur at boundaries.</li>
</ul>
<p>
In a particular problem, to fix the solution (said
otherwise: to fix the parameters \(a\) and \(b\) in <a href="./ems_ca_fe_L.html#Lap_1d_sol">Lap_1d_sol</a>), we need to appeal to boundary
conditions. Concretely, for a finite segment,
a solution exists and is unique if one is
provided with any of these possibilities:
</p>
<ul class="org-ul">
<li>\(\phi\) at both boundaries</li>
<li>\(\phi\) and \(\frac{d\phi}{dx}\) at one boundary</li>
<li>\(\phi\) at one boundary, \(\frac{d\phi}{dx}\) at the other.</li>
</ul>
<p>
Specifying \(\frac{d\phi}{dx}\) at both boundaries
provides insufficient information, since you get
an inconsistency if the derivatives don't match.
</p>
</div>
</div>
@@ -1629,16 +1698,45 @@ Not always: one end value + derivative value at other end, two end derivative v
<h6 id="ems_ca_fe_L_2d"><a href="#ems_ca_fe_L_2d">The Laplace Equation in Two Dimensions</a></h6>
<div class="outline-text-6" id="text-ems_ca_fe_L_2d">
<p>
\[
\frac{d^2 V}{dx^2} + \frac{d^2 V}{dy^2} = 0.
\]
Properties:
\paragraph{1.} The value of \(V(x,y)\) equals the average value around the point:
\[
V(x,y) = \frac{1}{2\pi R} \oint V dl
\]
\paragraph{2.} \(V\) has no local maxima or minima. All extrema occur at the boundaries.
In two dimensions, the potential becomes a function
of two variables (here: \(x\) and \(y\)), so Laplace's
equation now reads
</p>
<div class="eqlabel" id="orgb8aa28d">
<p>
<a id="Lap_2d"></a><a href="./ems_ca_fe_L.html#Lap_2d"><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="org01054fc">
</div>
</div>
\begin{equation*}
\frac{\partial^2 \phi (x,y)}{\partial x^2}
+ \frac{\partial^2 \phi (x,y)}{\partial y^2} = 0.
\tag{Lap_2d}\label{Lap_2d}
\end{equation*}
<p>
Properties:
</p>
<ul class="org-ul">
<li>
<b>Balance</b>: \(\phi(x,y)\) equals the average value around the point:</li>
</ul>
<p>
\[
\phi(x,y) = \frac{1}{2\pi R} \oint dl ~\phi
\]
</p>
<ul class="org-ul">
<li>
<b>No extrema</b>: \(\phi\) has no local maxima or minima. All extrema occur at the boundaries.</li>
</ul>
</div>
</div>
@@ -1647,42 +1745,197 @@ V(x,y) = \frac{1}{2\pi R} \oint V dl
<h6 id="ems_ca_fe_L_3d"><a href="#ems_ca_fe_L_3d">The Laplace Equation in Three Dimensions</a></h6>
<div class="outline-text-6" id="text-ems_ca_fe_L_3d">
<p>
\[
{\boldsymbol \nabla}^2 V = 0
\]
Properties:
\paragraph{1.} \(V({\bf r})\) is the average value of \(V\) over any spherical surface
centered at \({\bf r}\):
\[
V({\bf r}) = \frac{1}{4\pi R^2} \oint V da
\]
\paragraph{2.} \(V\) can have no local maxima or minima. All extrema occur at the boundaries.
In three dimensions, we will write the potential
as a function of a 3-dimensional vector, \(\phi({\bf r})\).
The Laplace equation is (we repeat)
</p>
<p>
\paragraph{Another way of seeing this} is to write the second derivatives as
\[
\frac{\partial^2 V({\bf r})}{\partial x^2} = f_x ({\bf r}), \hspace{5mm}
\frac{\partial^2 V({\bf r})}{\partial y^2} = f_y ({\bf r}), \hspace{5mm}
\frac{\partial^2 V({\bf r})}{\partial z^2} = f_z ({\bf r}), \hspace{5mm}
{\boldsymbol \nabla}^2 \phi ({\bf r}) = 0
\]
</p>
<p>
<b>Theorem</b>: if \(\phi\) satisfies Laplace, then its value at
a point equals its value averaged over any sphere
\(S_{\bf r}\) centered on this point,
\[
\phi({\bf r}) = \frac{1}{4\pi R^2} \oint_{S_{\bf r}} da' ~\phi ({\bf r}')
\]
</p>
<details id="org8c78d21">
<summary id="org6bdc443">
<strong>Physicist's proof</strong>
</summary>
<p>
Consider a sphere of radius \(R\) centered at the origin
carrying charge \(q\) spread with a uniform surface charge density over its surface. Bring in a point charge \(q'\) from
infinity up to a distance \(R'\) (with \(R' &gt; R\)) from the center
of the sphere.
</p>
<p>
We know that the field created by the sphere coincides
with that of a point charge \(q\) at the origin.
Since the potential at \({\bf r = 0}\) created by the charge
\(q'\) at \({\bf r'}\) is simply \(\phi_{q', {\bf r}'} (0) = \frac{q'}{4\pi \varepsilon_0 R'}\),
the work
required to bring the \(q'\) charge into position is thus
simply \(W = q \times \phi_{q', {\bf r}'} (0) = \frac{q q'}{4 \pi \varepsilon_0 R'}\) by <a href="./ems_es_efo_e.html#Wab">Wab</a>.
</p>
<p>
We can however proceed the other way: fixing \(q'\) in place,
and then bringing the charged sphere into position;
the work (energy) has to coincide with our previous result.
But this energy is now given by the integral of the
potential \(\phi_{q', {\bf r'}}\)
created by \(q'\) (sitting at \({\bf r'}\)) over the sphere
times the surface charge density on the sphere,
namely
</p>
<p>
\[
W = \oint_{S_R} da ~\sigma ~\phi_{q', {\bf r}'} ({\bf r})
\]
</p>
<p>
But \(\sigma = q/4\pi R^2\) and is a constant over the
sphere, so \(W = q \times \frac{1}{4\pi R^2} \oint da ~\phi_{q', {\bf r}'} ({\bf r})\).
</p>
<p>
Equating this with the previous results shows that
</p>
<p>
\[
\phi_{q', {\bf r'}} (0) = \frac{1}{4\pi R^2} \oint_{S_R} da ~\phi_{q', {\bf r}'} ({\bf r})
\]
</p>
<p>
namely that for the potential created by a single point
charge \(q'\) at \(R'\),
the value at a point (here the origin)
coincides with the value averaged over a sphere
or an arbitrary radius \(R\) centered on the same point.
</p>
<p>
By the principle of superposition, this works for an
arbitrary distribution of charges outside the sphere,
proving the theorem.
</p>
</details>
<details id="orgff2611b">
<summary id="org89a3b1b">
<strong>Formal proof</strong>
</summary>
<p>
Consider a function \(f({\bf r})\) and its average over
a ball of radius \(R\) centered on \({\bf r}\):
</p>
<p>
\[
f_{S_R} ({\bf r}) \equiv \frac{1}{4\pi R^2}\oint_{S_R} da' ~ f ({\bf r} + {\bf r}')
\]
</p>
<p>
For convenience we will hereafter put \({\bf r} = 0\).
In spherical coordinates, we have \(da' = R^2 sin \theta d\theta d\phi \equiv R^2 d\Omega\).
Differentiating with respect to \(R\),
</p>
<p>
\[
\frac{d}{dR} f_{S_R} = \frac{1}{4\pi} \oint_{S_R} d\Omega ~\left.\frac{\partial f}{\partial r}\right|_{r=R}
\]
</p>
<p>
with \(f\) differentiated with respect to the radial coordiate.
We can rewrite this by noting that \(R^2 d\Omega \hat{\bf r}\)
is the normal differential surface area \(d{\bf a}\), while
\(\left.\frac{\partial f}{\partial r}\right|_{r=R}\) is the radial component of the gradient
of \(f\) in spherical coordinates. Thus,
</p>
<p>
\[
\frac{d}{dR} f_{S_R} = \frac{1}{4\pi R^2} \oint_{S_R} d{\bf a} \cdot ~\nabla f
\]
</p>
<p>
Invoking the divergence theorem and using the definition
of the Laplacian operator \(\nabla^2 = \nabla \cdot \nabla\),
we get the following general
</p>
<p>
<b>Theorem</b>:
</p>
<p>
\[
\frac{d}{dR} f_{S_R} = \frac{1}{4\pi R^2} \int_{V_R} d\tau ~\nabla^2 f
\]
</p>
<p>
For the electrostatic potential away from charges, we have
\[
\nabla^2 \phi = 0 ~\rightarrow \frac{d}{dR} \phi_{S_R} = 0
\]
namely the ball average is independent of the ball size.
Since the value at the center is simply the average for
an infinitesimally small ball, we get the result announced above.
</p>
</details>
<p>
<b>Theorem (Earnshaw, mathematical versoin)</b>: \(\phi\) has no local extrema except at the boundaries.
</p>
<p>
<b>Proof</b>: write the second derivatives as
</p>
<p>
\[
\frac{\partial^2 \phi({\bf r})}{\partial x^2} = f_x ({\bf r}), \hspace{5mm}
\frac{\partial^2 \phi({\bf r})}{\partial y^2} = f_y ({\bf r}), \hspace{5mm}
\frac{\partial^2 \phi({\bf r})}{\partial z^2} = f_z ({\bf r}), \hspace{5mm}
f_x + f_y + f_z = 0.
\]
The \(f_a ({\bf r})\) represent the three components of the curvature of \(V({\bf r})\).
An extremum of \(V\) at \({\bf r}_e\) would be characterized by \({\boldsymbol \nabla} V |_{{\bf r}_e} \cdot \delta{\bf r} = 0\)
</p>
<p>
The \(f_a ({\bf r})\) represent the three components of the curvature of \(\phi({\bf r})\).
An extremum of \(\phi\) at \({\bf r}_e\) would be characterized by \({\boldsymbol \nabla} \phi |_{{\bf r}_e} \cdot \delta{\bf r} = 0\)
for any infinitesimal displacement \(\delta{\bf r}\) around the extremum point. For a local
minimum, the second derivative form should be greater than zero, \(\sum_{i,j} \frac{\partial^2 V}{\partial r_i \partial r_j} \delta r_i \delta r_j &gt; 0\)
minimum, the second derivative form should be greater than zero, \(\sum_{i,j} \frac{\partial^2 \phi}{\partial r_i \partial r_j} \delta r_i \delta r_j &gt; 0\)
for any displacement vector. Choosing alternately displacements along the three axes,
the form becomes \(f_x (\delta x)^2\), \(f_y (\delta y)^2\) or \(f_z (\delta z)^2\). Since the squared displacements
are necessarily positive, we thus require \(f_x &gt; 0\), \(f_y &gt; 0\) and \(f_z &gt; 0\). This is impossible in view
of the \(f_x + f_y + f_z = 0\) condition above.
</p>
<div class="info div" id="orge499be0">
<div class="info div" id="org950ce9d">
<p>
<b>Earnshaw's theorem</b> <br>
Since solutions to Laplace's equation have no local minimum,
it is impossible to find a static distribution of charges which generates an electrostatic field
with a stable equilibrium position for a test charge.
<b>Earnshaw's theorem (physical version)</b> <br>
It is impossible to find a static distribution of charges which generates an electrostatic field
displaying a stable equilibrium position in empty space.
</p>
</div>
@@ -1692,47 +1945,56 @@ Going back to Poisson's equation, we can make a few comments:
</p>
<ul class="org-ul">
<li>representation (\ref{eq:Poisson}) highlights the 'local' nature of the coupling between electrostatic fields and charges: fields are 'created' where the charges 'sit'. This is also seen by looking at the integrand of (\ref{eq:V_from_rho}). If electrostatics was nonlocal, a modified representation like (\ref{eq:V_from_rho}) would still exist, but not a local differential one like (\ref{eq:Poisson}).</li>
<li>representation <a href="./ems_es_ep_PL.html#Poi">🐟</a> highlights the 'local' nature of the coupling between electrostatic fields and charges: fields are 'created' where the charges 'sit'. This is also seen by looking at the integrand of <a href="./ems_es_ep_d.html#p_vcd">p_vcd</a>. If electrostatics was nonlocal, a modified representation like <a href="./ems_es_ep_d.html#p_vcd">p_vcd</a> would still exist, but not a local differential one like Poisson's equation.</li>
<li>as written, representations (\ref{eq:E_from_rho}) and (\ref{eq:V_from_rho}) require the knowledge of the charge density distribution \(\rho({\bf r})\) throughout space to determine the potential at any given point.</li>
<li>as written, representations <a href="./ems_es_ef_ccd.html#E_vcd">E_vcd</a> and <a href="./ems_es_ep_d.html#p_vcd">p_vcd</a> require the knowledge of the charge density distribution \(\rho({\bf r})\) throughout space to determine the potential at any given point.</li>
<li>(\ref{eq:Poisson}), being purely local, might allow to determine the potential at a specified point, provided we know the charge density distribution around this specified point, and at some set of other reference points (to make the solution unique).</li>
<li>Poisson's equation <a href="./ems_es_ep_PL.html#Poi">🐟</a>, being purely local, might allow to determine the potential at a specified point, provided we know the charge density distribution around this specified point, and at some set of other reference points (to make the solution unique).</li>
</ul>
<p>
We therefore want to ask the question: <b>under what conditions can an electrostatic problem be fully
defined by solving Poisson's equation ?</b> We start by mentioning some cases, and interpreting them thereafter.
We therefore want to ask the question: <i>under what conditions can an electrostatic problem be fully
defined by solving Poisson's equation ?</i>
</p>
<p>
We start by mentioning some cases, and interpreting them thereafter.
</p>
<p>
<b>Charge density is known throughout space</b>: in this case,
the electrostatic potential is uniquely determined
by Poisson's equation, which
is explicitly solved by <a href="./ems_es_ep_d.html#p_vcd">p_vcd</a>.
One can eplicitly verify this:
</p>
<p>
<b>First case</b>: the electrostatic potential is uniquely determined
if \(\rho({\bf r})\) is given throughout all space. In this case, Poisson's equation
is explicitly solved by (\ref{eq:V_from_rho}). Explicit check:
\[
{\boldsymbol \nabla}^2 V ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int_{\mathbb{R}^3} d\tau' \rho({\bf r}') {\boldsymbol \nabla}^2 \frac{1}{|{\bf r} - {\bf r}'|}
= \frac{1}{4\pi \varepsilon_0} \int_{\mathbb{R}^3} d\tau' (-4\pi) \delta ({\bf r} - {\bf r}') = -\frac{\rho ({\bf r})}{\varepsilon_0}.
\]
where we have used (\ref{Gr(1.102)}), and the fact that the delta function is always resolved since we
integrate over all space. Note: it is implicitly assumed that the integral in (\ref{eq:V_from_rho})
</p>
<p>
where we have used <a href="./c_m_dd_3d.html#Lap1or">Lap1or</a>, and the fact that the delta function is always resolved since we
integrate over all space. Note: it is implicitly assumed that the integral in <a href="./ems_es_ep_d.html#p_vcd">p_vcd</a>
converges, <i>i.e.</i> that the charge density \(\rho({\bf r})\) is sufficiently well-behaved (does not
become singular).
</p>
<p>
<b>First case (corollary)</b>: the electrostatic potential is uniquely determined
<b>Charge density in closed volume, and boundary surface charge density are known</b>: the electrostatic potential is uniquely determined
in a certain volume \({\cal V}\) bounded by boundary \({\cal S}\), provided the charge density
\(\rho ({\bf x})\) is given everywhere within \({\cal V}\), vanishes outside of \({\cal V}\),
and the value of the surface charge density \(\sigma\) is given everywhere on the boundary \({\cal S}\).
Of course, \({\cal S}\) need not be a connected surface.
</p>
<p>
This is obvious: we know where all the charges are, so this is really the same as the first case.
</p>
<p>
<b>Second case</b>: the electrostatic potential is uniquely determined
<b>Charge density in closed volume, and potential at boundary are known</b>: the electrostatic potential is uniquely determined
in a certain volume \({\cal V}\) bounded by boundary \({\cal S}\), provided the charge density
\(\rho ({\bf x})\) is given everywhere within \({\cal V}\), and the value of \(V\) is given everywhere on the
boundary \({\cal S}\). Of course, \({\cal S}\) need not be a connected surface.
@@ -1748,14 +2010,20 @@ to obtain \(V\) within \({\cal V}\).
Given a solution \(V_1 ({\bf r})\), we can easily show that it is unique. Suppose there was another solution
\(V_2 ({\bf r})\). Look at the difference, \(U \equiv V_1 - V_2\). In the bulk, \(U\) obeys the Laplace
equation
</p>
<p>
\[
{\boldsymbol \nabla}^2 U = {\boldsymbol \nabla}^2 V_1 - {\boldsymbol \nabla}^2 V_2 = -\frac{\rho}{\varepsilon_0} + \frac{\rho}{\varepsilon_0} = 0.
\]
</p>
<p>
Moreover, \(U ({\bf r}) = 0\) for \({\bf r} \in {\cal S}\). Since solutions to the Laplace equation take
their maximal and minimal value on the boundary, we must have \(U = 0\) \(\forall {\bf r} \in {\cal V}\)
(Griffiths' proof).
</p>
<p>
This all feels a bit amateurish and not very systematic. Can we be more precise and general? What kinds of boundary information do we really need to specify the solution uniquely ?
</p>
@@ -1765,6 +2033,8 @@ This all feels a bit amateurish and not very systematic. Can we be more precise
<br><ul class="navigation-links"><li>Prev:&nbsp;<a href="ems_ca_fe.html">Fundamental Equations for the Electrostatic Potential&emsp;<small>[ems.ca.fe]</small></a></li><li>Next:&nbsp;<a href="ems_ca_fe_g.html">Green's Identities&emsp;<small>[ems.ca.fe.g]</small></a></li><li>Up:&nbsp;<a href="ems_ca_fe.html">Fundamental Equations for the Electrostatic Potential&emsp;<small>[ems.ca.fe]</small></a></li></ul>
<br>
<hr>
<div class="license">
<a rel="license noopener" href="https://creativecommons.org/licenses/by/4.0/"
@@ -1778,7 +2048,7 @@ 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: 2022-02-10 Thu 08:32</p>
<p class="date">Created: 2022-02-13 Sun 21:20</p>
<p class="validation"></p>
</div>