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Update 2022-03-24 08:43

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Jean-Sébastien
2022-03-24 08:43:21 +01:00
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@@ -1,7 +1,7 @@
<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-03-22 Tue 10:52 -->
<!-- 2022-03-24 Thu 08:42 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1293,7 +1293,7 @@ Table of contents
</summary>
<ul>
<li>
<a href="./qed_t.html#qed_t">QED today</a><span class="headline-id">qed.t</span>
<a href="./qed_L.html#qed_L">Lagrangian</a><span class="headline-id">qed.L</span>
</li>
@@ -1615,7 +1615,7 @@ Table of contents
<div class="outline-text-4" id="text-red_rem_Ltf">
<p>
Now that we know that what one observer sees as an electric field
can be view by another as a magnetic field, we can ask the general
can be viewed by another as a magnetic field, we can ask the general
question of how fields transform upon Lorentz transformations.
</p>
@@ -1674,7 +1674,7 @@ This magnetic field between the plates is thus
<p>
If we now have a further referential frame \(\bar{S}\) moving at velocity
\(\bar{v}\) with respect to the original one, we'd have
\(v\) with respect to \({\cal S}\) (and \(\bar{v}\) with respect to the original one), we'd have
\[
\bar{E}_y = \frac{\bar{\sigma}}{\varepsilon_0}, \hspace{10mm}
\bar{B}_z = - \mu_0 \bar{\sigma} \bar{v}
@@ -1682,8 +1682,8 @@ If we now have a further referential frame \(\bar{S}\) moving at velocity
where
\[
\bar{v} = \frac{v + v_0}{1 + v v_0/c^2}, \hspace{10mm}
\bar{\sigma} = \bar{\gamma} \sigma, \hspace{10mm}
\bar{\gamma} = \frac{1}{1 - \bar{v}^2/c^2}.
\bar{\sigma} = \bar{\gamma} \sigma_0, \hspace{10mm}
\bar{\gamma} = \frac{1}{\sqrt{1 - \bar{v}^2/c^2}}.
\]
</p>
@@ -1691,8 +1691,8 @@ where
<p>
We now want to express \(\bar{\boldsymbol E}, \bar{\boldsymbol B}\) in terms of
\({\boldsymbol E}, {\boldsymbol B}\) and other data in frame \({\cal S}\)
(here: \(v\)). To start, we have
\({\boldsymbol E}, {\boldsymbol B}\) and other data in frame \({\cal S}\).
To start, we have
\[
\bar{E}_y = \frac{\bar{\gamma}}{\gamma_0} \frac{\sigma}{\varepsilon_0},
\hspace{10mm}
@@ -1736,28 +1736,28 @@ These factors cancel so \(\bar{B}_x = B_x\).
<p>
We thus obtain the
</p>
<div class="core div" id="orgf512c1f">
<div class="core div" id="org3cf1e4c">
<p>
<b>EM field transformation laws</b> <i>(motion along \(x\) with velocity \(v\))</i>
</p>
<div class="eqlabel" id="org93bd238">
<div class="eqlabel" id="orgc368b6d">
<p>
<a id="EMtr"></a><a href="./red_rem_Ltf.html#EMtr"><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="orgf5395d1">
<div class="alteqlabels" id="orga9279ba">
</div>
</div>
\begin{align}
\bar{E}_x &amp;= E_x, \hspace{10mm} &amp;
\bar{E}_y &amp;= \gamma (E_y - v B_z), \hspace{10mm} &amp;
\bar{E}_z &amp;= \gamma (E_z + v B_y), \\
\bar{B}_x &amp;= B_x, &amp;
\bar{B}_y &amp;= \gamma \left( B_y + \frac{v}{c^2} E_z \right), &amp;
\bar{B}_x &amp;= B_x, \nonumber \\
\bar{E}_y &amp;= \gamma (E_y - v B_z), &amp;
\bar{B}_y &amp;= \gamma \left( B_y + \frac{v}{c^2} E_z \right), \nonumber \\
\bar{E}_z &amp;= \gamma (E_z + v B_y), &amp;
\bar{B}_z &amp;= \gamma \left( B_z - \frac{v}{c^2} E_y \right)
\tag{EMtr}\label{EMtr}
\end{align}
@@ -1805,7 +1805,7 @@ 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: 2022-03-22 Tue 10:52</p>
<p class="date">Created: 2022-03-24 Thu 08:42</p>
<p class="validation"></p>
</div>