Update 2022-03-15 10:07
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Jean-Sébastien
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@@ -1,7 +1,7 @@
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<!DOCTYPE html>
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<html lang="en">
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<head>
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<!-- 2022-03-07 Mon 20:38 -->
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<!-- 2022-03-15 Tue 08:10 -->
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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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@@ -1310,10 +1310,6 @@ Table of contents
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</summary>
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<ul>
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<li>
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<a href="./d_m.html#d_m">Diagnostics: Mathematical Preliminaries</a><span class="headline-id">d.m</span>
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</li>
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<li>
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<a href="./d_ems.html#d_ems">Diagnostics: Electromagnetostatics</a><span class="headline-id">d.ems</span>
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</li>
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@@ -1352,6 +1348,10 @@ Table of contents
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<li>
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<a href="./d_red.html#d_red">Diagnostics: Relativistic Electrodynamics</a><span class="headline-id">d.red</span>
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</li>
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<li>
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<a href="./d_m.html#d_m">Diagnostics: Compendium - Mathematics</a><span class="headline-id">d.m</span>
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</li>
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</ul>
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@@ -1628,63 +1628,96 @@ Useful strategy: represent fields in terms of potentials.
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</p>
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<p>
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Easiest:
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Easiest: as we already saw (<a href="./ems_ms_vp_A.html#BcurlA">BcurlA</a>), we can write the magnetic
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field as a pure curl (since its divergence always vanishes):
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</p>
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<div class="core div" id="org5bf4193">
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<div class="core div" id="orgb63eb69">
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<p>
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\[
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{\boldsymbol B} = {\boldsymbol \nabla} \times {\boldsymbol A}
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\]
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{\boldsymbol B} = {\boldsymbol \nabla} \times {\boldsymbol A}
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\]
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</p>
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</div>
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<p>
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Putting this into Faraday's law gives
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Putting this into Faraday's law <a href="./emd_Fl_Fl.html#Fl">Fl</a> gives
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\[
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{\boldsymbol \nabla} \times \left({\boldsymbol E} + \frac{\partial {\boldsymbol A}}{\partial t} \right) = 0
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\]
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so this can be written as the gradient of a scalar (by choice: \(-{\boldsymbol \nabla} V\)) so we get
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so this can be written as the gradient of a scalar.
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Making the choice \(-{\boldsymbol \nabla} \phi\) for this, we get
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</p>
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<div class="core div" id="orgffce8dc">
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<div class="core div" id="org23c61e0">
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<div class="eqlabel" id="org51cbd1b">
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<p>
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\[
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{\boldsymbol E} = -{\boldsymbol \nabla} V - \frac{\partial {\boldsymbol A}}{\partial t}
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\label{eq:E_from_Potentials}
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\]
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<a id="E_phiA"></a><a href="./emf_svp.html#E_phiA"><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="org25412c6">
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</div>
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<p>
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Using this potential representation for \({\boldsymbol E}\) and \({\boldsymbol B}\) automatically fulfills the two homogeneous Maxwell equations. For the inhomogeneous equations, substituting (\ref{eq:E_from_Potentials}) into Gauss's law gives
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</p>
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<div class="main div" id="org26743e8">
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<p>
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\[
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{\boldsymbol \nabla}^2 V + \frac{\partial}{\partial t} {\boldsymbol \nabla} \cdot {\boldsymbol A} = -\frac{\rho}{\varepsilon_0}
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\label{eq:LaplacianV}
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\]
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</p>
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</div>
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<p>
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whereas Amp{\`ere}-Maxwell becomes
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\[
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{\boldsymbol \nabla} \times \left({\boldsymbol \nabla} {\boldsymbol A}\right) = \mu_0 {\boldsymbol J} - \mu_0 \varepsilon_0 {\boldsymbol \nabla} \left(\frac{\partial V}{\partial t}\right) - \mu_0 \varepsilon_0 \frac{\partial^2 {\boldsymbol A}}{\partial t^2}
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{\boldsymbol E} = -{\boldsymbol \nabla} \phi - \frac{\partial {\boldsymbol A}}{\partial t}
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\tag{E_phiA}\label{E_phiA}
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\]
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which becomes after simple rearrangement and use of the identity \({\boldsymbol \nabla} \times \left({\boldsymbol \nabla} \times {\boldsymbol A}\right) = {\boldsymbol \nabla} ({\boldsymbol \nabla} \cdot {\boldsymbol A}) - {\boldsymbol \nabla}^2 {\boldsymbol A}\),
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</p>
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<div class="main div" id="orga86c8c2">
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</div>
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<p>
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Using this potential representation for \({\boldsymbol E}\) and \({\boldsymbol B}\) automatically fulfills the two homogeneous Maxwell equations. For the inhomogeneous equations, substituting <a href="./emf_svp.html#E_phiA">E_phiA</a> into Gauss's law gives
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</p>
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<div class="main div" id="orgf26abb4">
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<div class="eqlabel" id="org986b9b3">
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<p>
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<a id="Lapphi"></a><a href="./emf_svp.html#Lapphi"><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="orgcd04ad9">
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</div>
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</div>
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<p>
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\[
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\left( {\boldsymbol ∇}^2 {\boldsymbol A} - μ_0 ε_0 \frac{∂^2 {\boldsymbol A}}{∂ t^2} \right)
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{\boldsymbol \nabla}^2 \phi + \frac{\partial}{\partial t} {\boldsymbol \nabla} \cdot {\boldsymbol A} = -\frac{\rho}{\varepsilon_0}
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\tag{Lapphi}\label{Lapphi}
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\]
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</p>
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<ul class="org-ul">
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<li>{\boldsymbol ∇} \left({\boldsymbol ∇} ⋅ {\boldsymbol A} + μ_0 ε_0 \frac{\partial V}{\partial t} \right) = -μ_0 {\boldsymbol J}</li>
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</ul>
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</div>
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<p>
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\label{eq:LaplacianA}
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whereas Ampère-Maxwell becomes
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\[
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{\boldsymbol \nabla} \times \left({\boldsymbol \nabla} \times {\boldsymbol A}\right) = \mu_0 {\boldsymbol J} - \mu_0 \varepsilon_0 {\boldsymbol \nabla} \left(\frac{\partial \phi}{\partial t}\right) - \mu_0 \varepsilon_0 \frac{\partial^2 {\boldsymbol A}}{\partial t^2}
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\]
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which becomes after simple rearrangement and use of the <a href="./c_m_dc_d2.html#curlcurl">curlcurl</a> identity \({\boldsymbol \nabla} \times \left({\boldsymbol \nabla} \times {\boldsymbol A}\right) = {\boldsymbol \nabla} ({\boldsymbol \nabla} \cdot {\boldsymbol A}) - {\boldsymbol \nabla}^2 {\boldsymbol A}\),
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</p>
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<div class="main div" id="org135124e">
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<div class="eqlabel" id="org73667ac">
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<p>
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<a id="LapA"></a><a href="./emf_svp.html#LapA"><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="org3f09a58">
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</div>
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</div>
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<p>
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\[
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\left( {\boldsymbol \nabla}^2 {\boldsymbol A} - \mu_0 \varepsilon_0 \frac{\partial^2 {\boldsymbol A}}{\partial t^2} \right) - {\boldsymbol \nabla} \left({\boldsymbol \nabla} \cdot {\boldsymbol A} + \mu_0 \varepsilon_0 \frac{\partial \phi}{\partial t} \right) = -\mu_0 {\boldsymbol J}
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\tag{LapA}\label{LapA}
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\]
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</p>
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@@ -1711,7 +1744,7 @@ target="_blank">Creative Commons Attribution 4.0 International License</a>.
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</div>
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<div id="postamble" class="status">
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<p class="author">Author: Jean-Sébastien Caux</p>
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<p class="date">Created: 2022-03-07 Mon 20:38</p>
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<p class="date">Created: 2022-03-15 Tue 08:10</p>
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<p class="validation"></p>
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</div>
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