jscaux/pqed_html
1
0
Fork 0
Code Issues Pull Requests Releases Wiki Activity

Update 2022-02-14 06:33

Browse Source
This commit is contained in:
Jean-Sébastien
2022-02-14 06:33:37 +01:00
parent f8446c1405
commit 4cfe8cef59
204 changed files with 1839 additions and 941 deletions
Download Patch File Download Diff File Expand all files Collapse all files
build/ems_ca_fe_uP.html
+78 -56
View File
@@ -1,7 +1,7 @@
<!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>
@@ -1598,44 +1598,86 @@ Table of contents
</svg></a><span class="headline-id">ems.ca.fe.uP</span></h5>
<div class="outline-text-5" id="text-ems_ca_fe_uP">
<p>
Suppose now that we have two solutions to the Poisson equation, \(V_1 ({\bf r})\) and \(V_2 ({\bf r})\).
Defining \(U = V_1 - V_2\), we see that \(U\) manifestly obeys Laplace
within \({\cal V}\), \({\boldsymbol \nabla}^2 U = 0\).
We can now use Green's first identity (\ref{eq:GreensFirstIdentity}) to shed some light on the boundary
problem for the electrostatic potential. Namely, put \(\phi = \psi = U\). This yields
Suppose now that we have two solutions to the Poisson equation, \(\phi_1 ({\bf r})\) and \(\phi_2 ({\bf r})\).
Defining \(\Phi = \phi_1 - \phi_2\), we see that \(\Phi\) manifestly obeys Laplace
within \({\cal V}\), \({\boldsymbol \nabla}^2 \Phi = 0\).
We can now use Green's first identity <a href="./ems_ca_fe_g.html#Green1">Green1</a> to shed some light on the boundary
problem for the electrostatic potential. Namely, put \(\phi = \psi = \Phi\). This yields
\[
\int_{\cal V} d\tau \left( U {\boldsymbol \nabla}^2 U + {\boldsymbol \nabla} U \cdot {\boldsymbol \nabla} U \right)
= \oint_{\cal S} da ~U \frac{\partial U}{\partial n}.
\int_{\cal V} d\tau \left( \Phi {\boldsymbol \nabla}^2 \Phi + {\boldsymbol \nabla} \Phi \cdot {\boldsymbol \nabla} \Phi \right)
= \oint_{\cal S} da ~\Phi \frac{\partial \Phi}{\partial n}.
\]
The first term on the left-hand side vanishes since \(U\) satisfies Laplace.
The right-hand side can be made to vanish if \(U\) obeys either
The first term on the left-hand side vanishes since \(\Phi\) satisfies Laplace.
The right-hand side can be made to vanish if \(\Phi\) obeys either
</p>
\begin{align}
&amp;U|_{\cal S} = 0 &amp;\mbox{({\bf Dirichlet})} \label{eq:Dirichlet}\\
\mbox{or}&amp; &amp; \nonumber \\
&amp;\frac{\partial U}{\partial n}|_{\cal S} = 0 &amp;\mbox{({\bf Neumann})} \label{eq:Newmann}
\end{align}
<div class="eqlabel" id="org25f3c75">
<p>
<a id="Dirichlet"></a><a href="./ems_ca_fe_uP.html#Dirichlet"><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="org6a66dca">
</div>
</div>
\begin{equation}
\left.\Phi\right|_{\cal S} = 0
\tag{Dirichlet}\label{Dirichlet}
\end{equation}
<p>
or
</p>
<div class="eqlabel" id="orgff19131">
<p>
<a id="Neumann"></a><a href="./ems_ca_fe_uP.html#Neumann"><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="org3085021">
</div>
</div>
\begin{equation}
\left.\frac{\partial \Phi}{\partial n}\right|_{\cal S} = 0
\tag{Newmann}\label{Newmann}
\end{equation}
<p>
boundary conditions on each individual boundary surface. In those cases, we are left with
\[
\int_{\cal V} d\tau \left|{\boldsymbol \nabla} U \right|^2 = 0, \longrightarrow {\boldsymbol \nabla} U = 0.
\int_{\cal V} d\tau \left|{\boldsymbol \nabla} \Phi \right|^2 = 0, \longrightarrow {\boldsymbol \nabla} \Phi = 0.
\]
\(U\) is thus constant. For Dirichlet, \(U = 0\) throughout \({\cal V}\), and thus \(V_2 = V_1\) and the solution
\(\Phi\) is thus constant. For Dirichlet, \(\Phi = 0\) throughout \({\cal V}\), and thus \(\phi_2 = \phi_1\) and the solution
is unique. For Neumann, the solution is unique apart from an unimportant constant.
</p>
<p>
We can thus finally state the
</p>
<div class="core div" id="org3a01c26">
<div class="eqlabel" id="org56added">
<p>
<a id="uniq_thm"></a><a href="./ems_ca_fe_uP.html#uniq_thm"><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="orgcd9ef4e">
</div>
</div>
<div class="core div" id="org7d69cab">
<p>
<b>Uniqueness Theorem</b>
</p>
<p>
The solution to Poisson's equation \({\boldsymbol \nabla}^2 V = -\frac{\rho}{\varepsilon_0}\) inside a volume \({\cal V}\)
bounded by a (in general disconnected) surface \({\cal S}\) is uniquely defined provided either Dirichlet \(V |_{{\cal S}_i}\) or Neumann
\(\frac{\partial V}{\partial n} |_{{\cal S}_i}\) boundary conditions are used on each individual surface.
The solution to Poisson's equation \({\boldsymbol \nabla}^2 \phi = -\frac{\rho}{\varepsilon_0}\) inside a volume \({\cal \phi}\)
bounded by a (in general disconnected) surface \({\cal S}\) is uniquely defined provided either Dirichlet \(\phi |_{{\cal S}_i}\) or Neumann
\(\frac{\partial \phi}{\partial n} |_{{\cal S}_i}\) boundary conditions are used on each individual surface.
</p>
</div>
@@ -1651,47 +1693,25 @@ the solution always exists for Dirichlet boundary conditions.
</p>
<p>
<b>Link to earlier cases</b>: the 'second case' above, in which the potential is specified on the
boundaries, is the case of Dirichlet boundary conditions.
The 'first case corollary', where the normal derivative of the potential is given, is a subcase involving
<b>Link to earlier cases</b>:
the previous case in which the potential is specified on the
boundaries, is thus the case of Dirichlet boundary conditions.
The one where the normal derivative of the potential is given,
is a subcase involving
Neumann boundary conditions (subcase, because we could imagine other charges living outside volume \({\cal V}\),
whereas our first case corollary involved only surface charges).
whereas the earlier example involved only surface charges).
</p>
<p>
\paragraph{Note on Griffiths' presentation of uniqueness theorem(s):}
we have used Green's identity to provide a general statement on uniqueness. Griffiths
might mislead you into thinking that there are numerous cases and corollaries.
Here (in italics) is his (confused) way of thinking about it (see Comment/warning below):
<i>Note on Griffiths' presentation of uniqueness theorem(s)</i>:
we have used Green's identity to provide a general statement on uniqueness. Reading Griffiths, you might be misled into thinking that there are numerous cases and corollaries.
</p>
<div class="info div" id="orgfb56c0b">
<p>
\subsubsection*{\it Boundary Conditions and Uniqueness Theorem}
\paragraph{\it First uniqueness theorem:} {\it The solution to Laplace's equation in some volume
\({\cal V}\) is uniquely determined if \(V\) is specified on the boundary surface \({\cal S}\).
{\bf Corollary:} the potential in a volume \({\cal V}\) is uniquely determined if
a) the charge density throughout the region and b) the value of \(V\) on all boundaries
are specified.}
</p>
<p>
\subsubsection*{\it Conductors and the Second Uniqueness Theorem}
\paragraph{\it Second uniqueness theorem:} {\it In a volume \({\cal V}\) surrounded by conductors
and containing a specified charge density \(\rho\), the electric field is uniquely determined
if the total charge on each conductor is given.}
</p>
<p>
{\it Comments: this is the same uniqueness as before, in view of the fact that conductors are
equipotentials, and capacitance relates the charge to the potential.
}
</p>
<div class="info div" id="org7a25aa5">
<p>
{\bf Comment/warning: {\color{blue} uniqueness theorem on uniqueness theorems}}<br>
Do not be misled by Griffiths: there is a {\it unique} uniqueness theorem for the
solution of Poisson's equation, namely the one we have stated starting from Green's first identity.
<b>Comment/warning</b>: <b>uniqueness theorem on uniqueness theorems</b> <br>
Do not be misled: there is a <i>unique</i> uniqueness theorem for the
solution of Poisson's equation, namely the one we have stated starting from Green's first identity.
</p>
</div>
@@ -1699,6 +1719,8 @@ solution of Poisson's equation, namely the one we have stated starting from Gree
</div>
<br><ul class="navigation-links"><li>Prev:&nbsp;<a href="ems_ca_fe_g.html">Green's Identities&emsp;<small>[ems.ca.fe.g]</small></a></li><li>Next:&nbsp;<a href="ems_ca_mi.html">The Method of Images&emsp;<small>[ems.ca.mi]</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/"
@@ -1712,7 +1734,7 @@ 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>