Update 2022-03-02 15:47
This commit is contained in:
Jean-Sébastien
+157
-63
@@ -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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<!-- 2022-03-01 Tue 08:14 -->
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<!-- 2022-03-02 Wed 15:45 -->
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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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@@ -1638,38 +1638,45 @@ Total energy should be sum of these two. Derivation from scratch.
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<p>
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Suppose that at time \(t\), we have fields \({\bf E}\) and \({\bf B}\) produced by some charge
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and current distributions \(\rho\) and \({\bf J}\). In an interval \(dt\), how much work is
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done by EM forces ? From Lorentz force law:
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done by EM forces? From Lorentz force law:
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\[
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{\bf F} \cdot d{\bf l} = q({\bf E} + {\bf v} \times {\bf B}) \cdot {\bf v} dt = q ~{\bf E} \cdot {\bf v} dt
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\]
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Really, we're looking at a small volume element \(d\tau\) carrying charge \(\rho d\tau\), moving
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at velocity \({\bf v}\) such that \({\bf J} = \rho {\bf v}\). Thus,
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</p>
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<div class="eqlabel" id="org0197499">
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<p>
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<a id="dWdt_intEJ"></a><a href="./emd_ce_poy.html#dWdt_intEJ"><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="org9aacea3">
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<ul class="org-ul">
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<li>Gr (8.6)</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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\[
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\frac{dW}{dt} = \int_{\cal V} d\tau ~ {\bf E} \cdot {\bf J}
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\label{Gr(8.6)}
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\tag{dWdt_intEJ}\label{dWdt_intEJ}
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\]
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The integrand is the work done per unit time, per unit volume, {\it i.e.} the power delivered per unit volume.
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In terms of fields alone: use Ampère-Maxwell:
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\[
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{\bf E} \cdot {\bf J} = \frac{1}{\mu_0} {\bf E} \cdot ({\boldsymbol \nabla} \times {\bf B}) - \varepsilon_0 {\bf E} \cdot \frac{\partial {\bf E}}{\partial t}
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\]
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Using product rule 6,
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Using <a href="./c_m_dc_pr.html#div_xprod">div_xprod</a>,
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\[
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{\boldsymbol ∇} ⋅ ({\bf E} × {\bf B}) = {\bf B} ⋅ ({\boldsymbol ∇} × {\bf E})
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</p>
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<ul class="org-ul">
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<li>{\bf E} ⋅ ({\boldsymbol ∇} × {\bf B}),</li>
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</ul>
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<p>
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{\boldsymbol \nabla} \cdot ({\bf E} \times {\bf B}) = {\bf B} \cdot ({\boldsymbol \nabla} \times {\bf E}) - {\bf E} \cdot ({\boldsymbol \nabla} \times {\bf B}),
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\]
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Invoking Faraday \({\boldsymbol \nabla} \times {\bf E} = - \partial {\bf B}/\partial t\),
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\[
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{\bf E} ⋅ ({\boldsymbol ∇} × {\bf B}) = - {\bf B} ⋅ \frac{∂ {\bf B}}{∂ t}
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</p>
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<ul class="org-ul">
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<li>{\boldsymbol ∇} ⋅ ({\bf E} × {\bf B}).</li>
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</ul>
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<p>
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{\bf E} \cdot ({\boldsymbol \nabla} \times {\bf B}) = - {\bf B} \cdot \frac{\partial {\bf B}}{\partial t} - {\boldsymbol \nabla} \cdot ({\bf E} \times {\bf B}).
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\]
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But obviously,
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\[
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@@ -1679,28 +1686,35 @@ But obviously,
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\]
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so we get
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\[
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{\bf E} ⋅ {\bf J} = -\frac{1}{2} \frac{\partial}{\partial t} \left( ε_0 E^2 + \frac{1}{\mu_0} B^2 \right)
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</p>
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<ul class="org-ul">
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<li>\frac{1}{\mu_0} {\boldsymbol ∇} ⋅ ({\bf E} × {\bf B}).</li>
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</ul>
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<p>
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{\bf E} \cdot {\bf J} = -\frac{1}{2} \frac{\partial}{\partial t} \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) - \frac{1}{\mu_0} {\boldsymbol \nabla} \cdot ({\bf E} \times {\bf B}).
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\label{Gr(8.8)}
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\]
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Substituting this in \ref{Gr(8.6)} and using the divergence theorem,
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Substituting this in <a href="./emd_ce_poy.html#dWdt_intEJ">dWdt_intEJ</a> and using the divergence theorem,
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we obtain
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</p>
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<div class="main div" id="orgda8af3a">
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<div class="main div" id="orgf118f4f">
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<p>
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{\bf Poynting's theorem}
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\[
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\frac{dW}{dt} = -\frac{d}{d t} ∫_{\cal V} dτ \frac{1}{2} \left( ε_0 E^2 + \frac{1}{\mu_0} B^2 \right)
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<b>Poynting's theorem</b>
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</p>
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<ul class="org-ul">
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<li>\frac{1}{\mu_0} \oint_{\cal S} d{\bf a} ⋅ ({\bf E} × {\bf B})</li>
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</ul>
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<div class="eqlabel" id="org1b7cdac">
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<p>
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\label{Gr(8.9)}
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<a id="👉Thm"></a><a href="./emd_ce_poy.html#👉Thm"><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="orgd7b6cac">
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<ul class="org-ul">
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<li>Gr (8.9)</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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\[
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\frac{dW}{dt} = -\frac{d}{d t} \int_{\cal V} d\tau \frac{1}{2} \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) - \frac{1}{\mu_0} \oint_{\cal S} d{\bf a} \cdot ({\bf E} \times {\bf B})
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\tag{👉Thm}\label{👉Thm}
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\]
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</p>
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@@ -1713,41 +1727,92 @@ energy is carried by EM fields out of \({\cal V}\) across its boundary surface.
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<p>
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Energy per unit time, per unit area carried by EM fields:
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Energy per unit time, per unit area carried by EM fields: given by the
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</p>
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<div class="core div" id="org8a1c10e">
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<div class="core div" id="orgf3198a5">
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<p>
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<b>Poynting vector</b>
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</p>
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<div class="eqlabel" id="org8725431">
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<p>
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<a id="PoyntingVec"></a><a href="./emd_ce_poy.html#PoyntingVec"><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="org6f2879d">
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<ul class="org-ul">
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<li>Gr (8.10)</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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{\bf Poynting vector}
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\[
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{\bf S} \equiv \frac{1}{\mu_0} ({\bf E} \times {\bf B})
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\label{Gr(8.10)}
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\]
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{\bf S} \equiv \frac{1}{\mu_0} ({\bf E} \times {\bf B})
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\tag{PoyntingVec}\label{PoyntingVec}
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\]
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</p>
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</div>
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<p>
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We can thus express Poynting's theorem more compactly:
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</p>
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<div class="core div" id="org57576be">
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<div class="core div" id="org3a4bb91">
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<p>
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<b>Poynting's theorem</b> (integral form)
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</p>
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<div class="eqlabel" id="orgbf2cc63">
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<p>
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<a id="PoyntingThm_int"></a><a href="./emd_ce_poy.html#PoyntingThm_int"><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="org5f484e3">
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<ul class="org-ul">
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<li>Gr (8.11)</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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{\bf Poynting's theorem}
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\[
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\frac{dW}{dt} = - \frac{dU_{em}}{dt} - \oint_{\cal S} d{\bf a} \cdot {\bf S}.
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\label{Gr(8.11)}
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\]
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\frac{dW}{dt} = - \frac{dU_{em}}{dt} - \oint_{\cal S} d{\bf a} \cdot {\bf S}.
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\tag{PoyntingThm_int}\label{PoyntingThm_int}
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\]
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</p>
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</div>
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<p>
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where we have defined the total
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</p>
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<div class="core div" id="org43eb64b">
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<div class="core div" id="orgbc7eb16">
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<p>
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<b>Energy in electromagnetic fields</b>
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</p>
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<div class="eqlabel" id="org89872c3">
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<p>
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<a id="Uem"></a><a href="./emd_ce_poy.html#Uem"><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="org67613ca">
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<ul class="org-ul">
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<li>Gr (8.5)</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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{\bf Energy in electromagnetic fields}
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\[
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U_{em} \equiv \frac{1}{2} \int d\tau \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right)
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\label{Gr(8.5)}
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\]
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U_{em} \equiv \frac{1}{2} \int d\tau \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right)
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\tag{Uem}\label{Uem}
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\]
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</p>
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</div>
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@@ -1764,13 +1829,30 @@ Then,
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\]
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so we get the
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</p>
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<div class="core div" id="org29b53c2">
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<div class="core div" id="org487db23">
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<p>
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<b>Poynting theorem</b> (differential form)
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</p>
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<div class="eqlabel" id="org9593699">
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<p>
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<a id="PoyntingThm"></a><a href="./emd_ce_poy.html#PoyntingThm"><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="org129becb">
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<ul class="org-ul">
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<li>Gr (8.14)</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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{\bf Poynting theorem (differential form)}
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\[
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\frac{\partial}{\partial t} u_{em} + {\boldsymbol \nabla} \cdot {\bf S} = 0
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\label{Gr(8.14)}
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\]
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\frac{\partial}{\partial t} u_{em} + {\boldsymbol \nabla} \cdot {\bf S} = 0
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\tag{PoyntingThm}\label{PoyntingThm}
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\]
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</p>
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</div>
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@@ -1781,26 +1863,38 @@ and has a similar for to the continuity equation
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<div class="example div" id="orgbb7ac4b">
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<div class="example div" id="orgd9e0ab5">
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<p>
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<b>Example: Joule heating</b>
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</p>
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<p>
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<b>Task</b>: characterize the energy flow for a current-carrying wire.
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</p>
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<p>
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<b>Solution</b>: the energy per unit time delivered to wire the wire can
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be obtained from Poynting's theorem.
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</p>
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<p>
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\paragraph{Example 8.1} Current in a wire: Joule heating. Energy per unit time delivered to wire: from Poynting.
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Assuming that the field is uniform, the electric field parallel to the wire is
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\[
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{\boldsymbol E} = \frac{V}{L} \hat{\boldsymbol x},
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\]
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{\boldsymbol E} = \frac{V}{L} \hat{\boldsymbol x},
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\]
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where \(V\) is the potential difference between the ends ald \(L\) is the length. Magnetic field is circumferential:
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wire of radius \(a\),
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\[
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{\boldsymbol B} = \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol \varphi}
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\]
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{\boldsymbol B} = \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol \varphi}
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\]
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Poynting:
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\[
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{\boldsymbol S} = \frac{1}{\mu_0} \frac{V}{L} \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol x} \times \hat{\boldsymbol \varphi} = -\frac{VI}{2\pi a L} \hat{\boldsymbol s}
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\]
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{\boldsymbol S} = \frac{1}{\mu_0} \frac{V}{L} \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol x} \times \hat{\boldsymbol \varphi} = -\frac{VI}{2\pi a L} \hat{\boldsymbol s}
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\]
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and points radially inwards. Energy per unit time passing surface of wire:
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\[
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\int d{\bf a} \cdot {\bf S} = S (2\pi a L) = -V I
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\]
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\int d{\bf a} \cdot {\bf S} = S (2\pi a L) = -V I
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\]
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where the minus sign means energy is flowing {\it in} (the wire heats up),
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and the value is as expected.
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</p>
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@@ -1826,7 +1920,7 @@ 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: 2022-03-01 Tue 08:14</p>
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<p class="date">Created: 2022-03-02 Wed 15:45</p>
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<p class="validation"></p>
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</div>
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