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Jean-Sébastien
2022-02-15 10:32:42 +01:00
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build/ems_ca_sv_car.html
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@@ -1,7 +1,7 @@
<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-02-14 Mon 20:35 -->
<!-- 2022-02-15 Tue 10:14 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1597,7 +1597,7 @@ Table of contents
<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><span class="headline-id">ems.ca.sv.car</span></h5>
<div class="outline-text-5" id="text-ems_ca_sv_car">
<div class="example div" id="org5867ca9">
<div class="example div" id="org1c68f32">
<p>
</p>
@@ -1636,20 +1636,21 @@ This thus falls back onto a 2d problem. We need to solve the 2d Laplace equatio
\]
</p>
\begin{align}
&amp;(i) ~\phi(x, y=0) = 0, \hspace{5mm} &amp;(ii) ~\phi(x, y=a) = 0, \nonumber \\
&amp;(iii) ~\phi(x=0, y) = \phi_0 (y), &amp;(iv) ~\phi (x \rightarrow \infty, y) \rightarrow 0.
\end{align}
<p>
\[
(i) ~\phi(x, y = 0) = 0, \hspace{5mm} (ii) ~\phi(x, y = a) = 0, \hspace{5mm} (iii) ~\phi(0, y) = \phi_0 (y), \hspace{5mm} (iv) ~\phi (x \rightarrow \infty) \rightarrow 0.
\]
Let us look for solutions in the form
</p>
<div class="eqlabel" id="org8ec3c2f">
<div class="eqlabel" id="orgbf21d7f">
<p>
<a id="Lap_sv_car"></a><a href="./ems_ca_sv_car.html#Lap_sv_car"><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="orge8be224">
<div class="alteqlabels" id="org42c1fbb">
<ul class="org-ul">
<li>Gr (3.23)</li>
</ul>
@@ -1663,17 +1664,19 @@ Let us look for solutions in the form
\frac{1}{X} \frac{d^2 X}{dx^2} + \frac{1}{Y} \frac{d^2 Y}{dy^2} = 0
\tag{Lap_sv_car}\label{Lap_sv_car}
\]
Since the \(x\) and \(y\) dependecies are separated, the only possibility is that each
piece equals a constant, both adding up to zero. We can thus put
Since the \(x\) and \(y\) dependencies are separated, the only possibility is that each
individual term in the Laplace equation equals a constant,
and that these constants add up to zero. We can thus put (the sign choice anticipates
the solution somewhat)
</p>
<div class="eqlabel" id="org66a021a">
<div class="eqlabel" id="org4a3d5af">
<p>
<a id="Lap_sv_car_sep"></a><a href="./ems_ca_sv_car.html#Lap_sv_car_sep"><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="orgf870493">
<div class="alteqlabels" id="org2c20186">
<ul class="org-ul">
<li>Gr (3.26)</li>
</ul>
@@ -1697,14 +1700,14 @@ Let's look first of all at the solutions of <a href="./ems_ca_sv_car.html#Lap_sv
Since this is a second-order linear differential equation, there are two linearly
independent solutions. The most general solution for \(X\) and \(Y\) can be written
</p>
<div class="eqlabel" id="orgb649426">
<div class="eqlabel" id="org07f6b3c">
<p>
<a id="Lap_sv_car_solXY"></a><a href="./ems_ca_sv_car.html#Lap_sv_car_solXY"><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="orge98b65d">
<div class="alteqlabels" id="org6659867">
<ul class="org-ul">
<li>Gr (3.27)</li>
</ul>
@@ -1747,28 +1750,28 @@ boundary condition \((iii)\). Using the orthogonality relation
<p>
\[
\int_0^a dy \sin(n\pi y/a) \sin (n' \pi y/a) = \delta_{n n'} \frac{a}{2}
\int_0^a dy ~\sin(n\pi y/a) \sin (n' \pi y/a) = \delta_{n n'} \frac{a}{2}
\]
we thus get that the \(C_n\) are Fourien coefficients of the boundary function \(\phi_0(y)\):
we thus get that the \(C_n\) are Fourier coefficients of the boundary function \(\phi_0(y)\):
</p>
<p>
\[
C_n = \frac{2}{a} \int_0^a dy \phi_0(y) \sin(n\pi y/a)
C_n = \frac{2}{a} \int_0^a dy ~\phi_0(y) \sin(n\pi y/a)
\]
</p>
<p>
<b>Specific example</b>: say that \(\phi_0(y) = \phi_0\), <i>i.e.</i> just a constant. Then,
</p>
<div class="eqlabel" id="orgc7da523">
<div class="eqlabel" id="orgbe0e2ff">
<p>
<a id="Cn"></a><a href="./ems_ca_sv_car.html#Cn"><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="orgdc616fb">
<div class="alteqlabels" id="org7c421fe">
<ul class="org-ul">
<li>Gr (3.35)</li>
</ul>
@@ -1779,6 +1782,7 @@ C_n = \frac{2}{a} \int_0^a dy \phi_0(y) \sin(n\pi y/a)
<p>
\[
C_n = \frac{2\phi_0}{a} \int_0^a dy \sin(n\pi y/a) = \frac{2\phi_0}{n\pi} (1 - \cos n\pi) = \frac{4\phi_0}{n\pi} \delta_{n, odd}
\tag{Cn}\label{Cn}
\]
</p>
@@ -1821,7 +1825,7 @@ The solution for the specific case \(\phi_0 (y) = \phi_0\) is thus
</p>
<div class="example div" id="org442561e">
<div class="example div" id="orge84ac93">
<p>
<b>Example: rectangular pipe</b>
</p>
@@ -1844,10 +1848,12 @@ We thus need to solve the 2d Laplace equation <a href="./ems_ca_fe_L.html#Lap_2d
with boundary conditions
</p>
\begin{align}
&amp;(i) ~\phi (x, y = 0) = 0, \hspace{5mm} &amp;(ii) ~\phi (x, y = a) = 0,
\nonumber \\
&amp;(iii) ~\phi(x = b, y) = \phi_0, &amp;(iv) ~\phi(x = -b, y) = \phi_0.
\end{align}
<p>
\[
(i) ~\phi (y = 0) = 0, \hspace{5mm} (ii) ~\phi (y = a) = 0, \hspace{5mm} (iii) ~\phi(x = b) = 0, \hspace{5mm} (iv) ~\phi(x = -b) = 0.
\]
The general solution is obtained from <a href="./ems_ca_sv_car.html#Lap_sv_car_solXY">Lap_sv_car_XY</a>,
\[
\phi(x,y) = (Ae^{kx} + Be^{-kx}) (C\sin ky + D\cos ky).
@@ -1856,7 +1862,7 @@ By symmetry, \(\phi(x,y) = \phi(-x,y)\) so \(A = B\).
Since \(e^{kx} + e^{-kx} = 2\cosh kx\), the generic solution becomes
(redefining \(C\) and \(D\))
\[
\phi(x,y) = \cosh kx (C\sin ky + D\cos ky).
\phi(x,y) = \cosh kx ~(C\sin ky + D\cos ky).
\]
Boundary conditions \((i)\) and \((ii)\) require \(D = 0\), \(k = n\pi/a\) so
</p>
@@ -1876,14 +1882,14 @@ The full solution is then a linear combination of complete set of functions,
The coefficients must be chosen such that \((iii)\) is fulfilled, \(\phi(b,y) = \phi_0\).
This simple case of a constant value \(\phi_0\) gives us the same relation as <a href="./ems_ca_sv_car.html#Cn">Cn</a>, so
</p>
<div class="eqlabel" id="org08d0bc1">
<div class="eqlabel" id="org824a157">
<p>
<a id="p_recpipe"></a><a href="./ems_ca_sv_car.html#p_recpipe"><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="org9c5d186">
<div class="alteqlabels" id="orgb4140a1">
<ul class="org-ul">
<li>Gr (3.42)</li>
</ul>
@@ -1919,7 +1925,7 @@ 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: 2022-02-14 Mon 20:35</p>
<p class="date">Created: 2022-02-15 Tue 10:14</p>
<p class="validation"></p>
</div>