Update 2022-02-14 20:42
This commit is contained in:
Jean-Sébastien
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<!DOCTYPE html>
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<html lang="en">
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<head>
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<!-- 2022-02-13 Sun 21:20 -->
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<!-- 2022-02-14 Mon 20:35 -->
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<meta charset="utf-8">
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<meta name="viewport" content="width=device-width, initial-scale=1">
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<title>Pre-Quantum Electrodynamics</title>
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@@ -1597,140 +1597,303 @@ Table of contents
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<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"/>
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</svg></a><span class="headline-id">ems.ca.sv.car</span></h5>
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<div class="outline-text-5" id="text-ems_ca_sv_car">
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<div class="example div" id="orgde10180">
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<div class="example div" id="org5867ca9">
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<p>
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</p>
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<p>
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<b>Example: infinite grounded metal plates</b>
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</p>
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<p>
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Consider two infinite grounded metal plates positioned parallel to the
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\(xz\) plane, one at \(y = 0\) and the other at \(y = a\).
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</p>
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<p>
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At \(x = 0\), the setup is closed off with an infinite insulated strip maintained
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at potential \(\phi_0 (y)\).
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</p>
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<p>
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<b>Question</b>: find the potential inside the slot.
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</p>
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<p>
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<b>Solution</b>:
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</p>
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<p>
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By translational symmetry along \(z\), the potential must be independent of \(z\).
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This thus falls back onto a 2d problem. We need to solve the 2d Laplace equation
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<a href="./ems_ca_fe_L.html#Lap_2d">Lap_2d</a>
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</p>
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<p>
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{\bf Example: separation of variables (Cartesian coordinates)}<br>
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Two infinite grounded metal plates parallel to the \(xz\) plane, one at \(y = 0\) and the
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other at \(y = a\). End at \(x = 0\) closed off with infinite insulated strip maintained
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at potential \(V_0 (y)\). Find potential inside the slot.
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\paragraph{Solution:} indep of \(z\), so 2d problem. Solve
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\[
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\frac{\partial^2 V}{\partial^2 x} + \frac{\partial^2 V}{\partial^2 y} = 0
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\label{Gr(3.20)}
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\frac{\partial^2 \phi}{\partial^2 x} + \frac{\partial^2 \phi}{\partial^2 y} = 0
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\]
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</p>
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<p>
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\[
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(i) V(x, y = 0) = 0, \hspace{5mm} (ii)V(x, y = a) = 0, \hspace{5mm} (iii) V(0, y) = V_0 (y), \hspace{5mm} (iv) V (x \rightarrow \infty) \rightarrow 0.
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\label{Gr(3.21)}
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(i) ~\phi(x, y = 0) = 0, \hspace{5mm} (ii) ~\phi(x, y = a) = 0, \hspace{5mm} (iii) ~\phi(0, y) = \phi_0 (y), \hspace{5mm} (iv) ~\phi (x \rightarrow \infty) \rightarrow 0.
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\]
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Look for solutions of form
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Let us look for solutions in the form
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</p>
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<div class="eqlabel" id="org8ec3c2f">
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<p>
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<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">
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<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"/>
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||||
<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"/>
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</svg></a>
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</p>
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<div class="alteqlabels" id="orge8be224">
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<ul class="org-ul">
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<li>Gr (3.23)</li>
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</ul>
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</div>
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</div>
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<p>
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\[
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V(x,y) = X(x) Y(y), \hspace{1cm}
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\phi(x,y) = X(x) Y(y), \hspace{1cm}
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\frac{1}{X} \frac{d^2 X}{dx^2} + \frac{1}{Y} \frac{d^2 Y}{dy^2} = 0
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\label{Gr(3.23)}
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\tag{Lap_sv_car}\label{Lap_sv_car}
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\]
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Choose
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Since the \(x\) and \(y\) dependecies are separated, the only possibility is that each
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piece equals a constant, both adding up to zero. We can thus put
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</p>
|
||||
<div class="eqlabel" id="org66a021a">
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<p>
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<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">
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<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"/>
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||||
<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"/>
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||||
</svg></a>
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</p>
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<div class="alteqlabels" id="orgf870493">
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<ul class="org-ul">
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<li>Gr (3.26)</li>
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</ul>
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</div>
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</div>
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<p>
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\[
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\frac{d^2 X_n}{dx^2} = k_n^2 X_n, \hspace{1cm} \frac{d^2Y_n}{dy^2} = -k_n^2 Y_n
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\label{Gr(3.26)}
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\tag{Lap_sv_car_sep}\label{Lap_sv_car_sep}
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\]
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where \(k_n\) is some real number. We can linearly combine solutions of (\ref{Gr(3.26)})
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for different \(k_n\) and still get a solution to (\ref{Gr(3.23)}).
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Let's look first of all at the solutions of (\ref{Gr(3.26)}) for a given \(k_n\).
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where \(k_n\) is some real number.
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Since <a href="./ems_ca_fe_L.html#Lap_2d">Lap_2d</a> is a linear equation for \(\phi\),
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we can linearly combine solutions of <a href="./ems_ca_sv_car.html#Lap_sv_car_sep">Lap_sv_car_sep</a>
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for different \(k_n\) and still get a solution to <a href="./ems_ca_sv_car.html#Lap_sv_car">Lap_sv_car</a>.
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</p>
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<p>
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Let's look first of all at the solutions of <a href="./ems_ca_sv_car.html#Lap_sv_car_sep">Lap_sv_car_sep</a> for a given \(k_n\).
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Since this is a second-order linear differential equation, there are two linearly
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independent solutions. Most general solution:
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independent solutions. The most general solution for \(X\) and \(Y\) can be written
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</p>
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<div class="eqlabel" id="orgb649426">
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<p>
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<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">
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||||
<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"/>
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<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"/>
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</svg></a>
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</p>
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<div class="alteqlabels" id="orge98b65d">
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<ul class="org-ul">
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<li>Gr (3.27)</li>
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</ul>
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</div>
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</div>
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\begin{align}
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X_n(x) &= Ae^{k_nx} + Be^{-k_nx}, \nonumber \\
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Y_n(y) &= C \sin k_ny + D \cos k_ny
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\tag{Lap_sv_car_solXY}\label{Lap_sv_car_solXY}
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\end{align}
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<p>
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The constants can be fixed by fitting the boundary conditions.
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From \((iv)\), we get \(A = 0\). From \((i)\), we get \(D = 0\). We are thus left with
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</p>
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<p>
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\[
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X_n(x) = Ae^{k_nx} + Be^{-k_nx}, \hspace{1cm} Y_n(y) = C \sin k_ny + D \cos k_ny
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\label{Gr(3.27)}
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\phi(x,y) = C_n e^{-k_nx} \sin k_n y
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\]
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Fix constants: from \((iv)\), \(A = 0\). From \((i)\), D = 0. Left with
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\[
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V(x,y) = C_n e^{-k_nx} \sin k_n y
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\label{Gr(3.28)}
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\]
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Then, \((ii)\) requires
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Going further, \((ii)\) requires the momenta \(k_n\) to be quantized according to
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</p>
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<p>
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\[
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k_n = \frac{n\pi}{a}, \hspace{1cm} n = 1, 2, 3, ...
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\label{Gr(3.29)}
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\]
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But we can use as solution any linear combination of the functions defined
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by these momenta. Fix coefficients with Fourier series:
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Since we can use as solution any linear combination of the functions defined
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by these momenta, we obtain the solution in the form of a Fourier series,
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</p>
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<p>
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\[
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V(x,y) = \sum_{n=1}^{\infty} C_n e^{-n\pi x/a} \sin (n\pi x/a)
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\label{Gr(3.30)}
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\phi(x,y) = \sum_{n=1}^{\infty} C_n e^{-n\pi x/a} \sin (n\pi y/a)
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\]
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Needed:
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with yet-to-be-determined coefficients \(C_n\), which have to be chosen to fit
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boundary condition \((iii)\). Using the orthogonality relation
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</p>
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<p>
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\[
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\int_0^a dy \sin(n\pi y/a) \sin (n' \pi y/a) = \delta_{n n'} \frac{a}{2}
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\label{Gr(3.33)}
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\]
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so
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we thus get that the \(C_n\) are Fourien coefficients of the boundary function \(\phi_0(y)\):
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</p>
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<p>
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\[
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C_n = \frac{2}{a} \int_0^a dy V_0(y) \sin(n\pi y/a)
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\label{Gr(3.34)}
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C_n = \frac{2}{a} \int_0^a dy \phi_0(y) \sin(n\pi y/a)
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\]
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</p>
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<p>
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<b>Specific example</b>: say that \(\phi_0(y) = \phi_0\), <i>i.e.</i> just a constant. Then,
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</p>
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<div class="eqlabel" id="orgc7da523">
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<p>
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<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">
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||||
<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"/>
|
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<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"/>
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</svg></a>
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</p>
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<div class="alteqlabels" id="orgdc616fb">
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<ul class="org-ul">
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<li>Gr (3.35)</li>
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</ul>
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</div>
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</div>
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<p>
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\[
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C_n = \frac{2\phi_0}{a} \int_0^a dy \sin(n\pi y/a) = \frac{2\phi_0}{n\pi} (1 - \cos n\pi) = \frac{4\phi_0}{n\pi} \delta_{n, odd}
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\]
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</p>
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</div>
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<p>
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The logic of separation of variables exploited two characteristics of
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the basis functions in the Fourier decomposition, namely
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<b>completeness</b> and <b>orthogonality</b>.
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</p>
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<p>
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Completeness simply means that any differentiable function can be expressed
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in terms of the basis functions:
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</p>
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<p>
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\paragraph{Specific example:} say that \(V_0(y) = V_0\), some constant. Then,
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\[
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C_n = \frac{2V_0}{a} \int_0^a dy \sin(n\pi y/a) = \frac{2V_0}{n\pi} (1 - \cos n\pi) = \frac{4V_0}{n\pi} \delta_{n, odd}
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\label{Gr(3.35)}
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f(x) = \sum_{n=1}^{\infty} C_n f_n (y), \hspace{1cm} \forall ~f \in C^{\infty}
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\]
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</p>
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<p>
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Applicable provided {\bf completeness} and {\bf orthogonality}:
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Orthogonality is a convenient choice made to keep things simple: each function,
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under the chosen integration domain and weighing, has no overlap with others, <i>i.e.</i>
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</p>
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<p>
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Completeness:
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\[
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f(x) = \sum_{n=1}^{\infty} C_n f_n (y), \hspace{1cm} \forall f \in C^{\infty}
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\label{Gr(3.38)}
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\int_0^a dx ~f_n (x) f_{n'} (x) = 0, \hspace{1cm} n \neq n'
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\]
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</p>
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<p>
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Orthogonality:
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The solution for the specific case \(\phi_0 (y) = \phi_0\) is thus
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\[
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\int_0^a dx f_n (x) f_{n'} (x) = 0, \hspace{1cm} n \neq n'
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\label{Gr(3.39)}
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\]
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</p>
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<p>
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The solution for the specific case \(V_0 (y) = V_0\) therefore is
|
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\[
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V(x,y) = \frac{4V_0}{\pi}\sum_{n=1, 3, 5, ...}^{\infty} \frac{1}{n} e^{-n\pi x/a} \sin (n\pi x/a)
|
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\phi(x,y) = \frac{4\phi_0}{\pi}\sum_{n=1, 3, 5, ...}^{\infty} \frac{1}{n} e^{-n\pi x/a} \sin (n\pi x/a)
|
||||
\]
|
||||
</p>
|
||||
|
||||
|
||||
<div class="example div" id="orgc1f6d28">
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||||
<div class="example div" id="org442561e">
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||||
<p>
|
||||
{\bf Example: rectangular pipe}<br>
|
||||
Two infinitely long grounded plates at \(y = 0,a\) are connected at \(x = \pm b\)
|
||||
to metal strips maintained at constant \(V = V_0\). Find the potential in the resulting rectangular pipe.
|
||||
\paragraph{Solution:} indep of \(z\). Laplace: \(\partial^2 V/\partial x^2 + \partial^2 V/\partial y^2 = 0\),
|
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boundary conditions
|
||||
<b>Example: rectangular pipe</b>
|
||||
</p>
|
||||
|
||||
<p>
|
||||
Take two infinitely long grounded plates at \(y = 0\) and \(y=a\),
|
||||
cut at \(x = \pm b\) and joined to metal strips maintained at constant \(\phi = \phi_0\).
|
||||
</p>
|
||||
|
||||
<p>
|
||||
<b>Question</b>: find the potential in the resulting rectangular pipe.
|
||||
</p>
|
||||
|
||||
<p>
|
||||
<b>Solution</b>: by translational invariance along \(z\), the potential must be independent of \(z\).
|
||||
We thus need to solve the 2d Laplace equation <a href="./ems_ca_fe_L.html#Lap_2d">Lap_2d</a>,
|
||||
\[
|
||||
(i) V (y = 0) = 0, \hspace{5mm} (ii) V (y = a) = 0, \hspace{5mm} (iii) V(x = b) = 0, \hspace{5mm} (iv) V(x = -b) = 0.
|
||||
\label{Gr(3.40)}
|
||||
\partial^2 \phi/\partial x^2 + \partial^2 \phi/\partial y^2 = 0,
|
||||
\]
|
||||
Generic solution: as (\ref{Gr(3.27)}),
|
||||
with boundary conditions
|
||||
</p>
|
||||
|
||||
<p>
|
||||
\[
|
||||
V(x,y) = (Ae^{kx} + Be^{-kx}) (C\sin ky + D\cos ky).
|
||||
(i) ~\phi (y = 0) = 0, \hspace{5mm} (ii) ~\phi (y = a) = 0, \hspace{5mm} (iii) ~\phi(x = b) = 0, \hspace{5mm} (iv) ~\phi(x = -b) = 0.
|
||||
\]
|
||||
By symmetry, \(V(x,y) = V(-x,y)\) so \(A = B\). Now \(e^{kx} + e^{-kx} = 2\cosh kx\). Generic solution becomes
|
||||
The general solution is obtained from <a href="./ems_ca_sv_car.html#Lap_sv_car_solXY">Lap_sv_car_XY</a>,
|
||||
\[
|
||||
\phi(x,y) = (Ae^{kx} + Be^{-kx}) (C\sin ky + D\cos ky).
|
||||
\]
|
||||
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 \(D\))
|
||||
\[
|
||||
V(x,y) = \cosh kx (C\sin ky + D\cos ky).
|
||||
\phi(x,y) = \cosh kx (C\sin ky + D\cos ky).
|
||||
\]
|
||||
Boundary conditions \((i)\) and \((ii)\) require \(D = 0\), \(k = n\pi/a\) so
|
||||
</p>
|
||||
|
||||
<p>
|
||||
\[
|
||||
V(x,y) = C \cosh(n\pi x/a) \sin(n\pi y/a)
|
||||
\label{Gr(3.41)}
|
||||
\phi(x,y) = C \cosh(n\pi x/a) \sin(n\pi y/a)
|
||||
\]
|
||||
with \((iv)\) already satisfied if \((iii)\) is. Full solution is linear combination of complete set of functions,
|
||||
with \((iv)\) already satisfied if \((iii)\) is.
|
||||
</p>
|
||||
|
||||
<p>
|
||||
The full solution is then a linear combination of complete set of functions,
|
||||
\[
|
||||
V(x,y) = \sum_{n=1}^{\infty} C_n \cosh(n\pi x/a) \sin(n\pi y/a).
|
||||
\phi(x,y) = \sum_{n=1}^{\infty} C_n \cosh(n\pi x/a) \sin(n\pi y/a).
|
||||
\]
|
||||
Coefficients: chosen such that \((iii)\) is fulfilled, \(V(b,y) = V_0\). From (\ref{Gr(3.35)}):
|
||||
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">
|
||||
<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">
|
||||
<ul class="org-ul">
|
||||
<li>Gr (3.42)</li>
|
||||
</ul>
|
||||
|
||||
</div>
|
||||
|
||||
</div>
|
||||
<p>
|
||||
\[
|
||||
V(x,y) = \frac{4V_0}{\pi} \sum_{n = 1, 3, 5, ...} \frac{1}{n} \frac{\cosh (n\pi x/a)}{\cosh(n\pi b/a)} \sin(n\pi y/a).
|
||||
\label{Gr(3.42)}
|
||||
\phi(x,y) = \frac{4\phi_0}{\pi} \sum_{n = 1, 3, 5, ...} \frac{1}{n} \frac{\cosh (n\pi x/a)}{\cosh(n\pi b/a)} \sin(n\pi y/a).
|
||||
\]
|
||||
</p>
|
||||
|
||||
@@ -1756,7 +1919,7 @@ 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>
|
||||
|
||||
|
||||
Reference in New Issue
Block a user