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Update 2022-03-22 10:53

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Jean-Sébastien
2022-03-22 10:53:22 +01:00
parent f8fdc5f8b4
commit 2d1d8d39a4
209 changed files with 1240 additions and 1226 deletions
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build/red_rem_mre.html
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@@ -1,7 +1,7 @@
<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-03-15 Tue 08:10 -->
<!-- 2022-03-22 Tue 10:52 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1622,7 +1622,7 @@ implies the existence of magnetism, and vice-versa.
<p>
To illustrate this, we take a simple example
</p>
<aside id="orgbe19a87">
<aside id="org964fb67">
<p>
See Purcell \&amp; Morin, section 5.9.
</p>
@@ -1663,7 +1663,7 @@ The dilation factors in the test particle's frame are thus
\gamma_\pm &amp;= \frac{1}{\sqrt{1 - (v \mp u)^2/(c^2 \mp uv)^2}}
= \frac{c^2 \mp uv}{\sqrt{(c^2 \mp uv)^2 - c^2 (v \mp u)^2}} \nonumber \\
&amp;= \frac{c^2 \mp uv}{\sqrt{(c^2 - v^2)(c^2 - u^2)}}
= \gamma \frac{1 \mp uv/c^2}{\sqrt{q - u^2/c^2}}
= \gamma \frac{1 \mp uv/c^2}{\sqrt{1 - u^2/c^2}}
\end{align}
<p>
so the resultant line charge in the test frame is
@@ -1680,25 +1680,25 @@ neutral in one reference frame, can appear to be charged in another.
In the frame of the test particle, there is an electric field
equal to that of a uniformly charged wire:
\[
E = \frac{\lambda_{\mbox{test}}}{2\pi \epsilon_0 s}
E = \frac{\lambda_{\mbox{test}}}{2\pi \epsilon_0 r}
\]
so the force (in the test frame) is
\[
F_{\mbox{test}} = q E = -\frac{\lambda v}{\pi \epsilon_0 c^2 s} \frac{q u}{\sqrt{1 - u^2/c^2}}.
F_{\mbox{test}} = q E = -\frac{\lambda v}{\pi \epsilon_0 c^2 r} \frac{q u}{\sqrt{1 - u^2/c^2}}.
\]
If there is a force on our test charge in this test frame, there must
also be one in the lab frame. Using equation <a href="./red_rm_Mf.html#Ftr0">Ftr0</a>,
\[
F = \sqrt{1 - u^2/c^2} ~F_{\mbox{test}}
= -\frac{\lambda v}{\pi \varepsilon_0 c^2} \frac{qu}{s}
= -\frac{\lambda v}{\pi \varepsilon_0 c^2} \frac{qu}{r}
\]
which upon recognizing \(c^2 = \frac{1}{\varepsilon_0 \mu_0}\)
and the current \(I = 2 \lambda v\) becomes
\[
F = - q u ~\frac{\mu_0 I}{2\pi s}
F = - q u ~\frac{\mu_0 I}{2\pi r}
\]
which you will recognize as the magnetic part of the Lorentz force
for a charge \(q\) moving at velocity \(v\) in the presence of the
for a charge \(q\) moving at velocity \(u\) in the presence of the
magnetic field of a long straight wire carrying current \(I\).
</p>
</div>
@@ -1722,7 +1722,7 @@ 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: 2022-03-15 Tue 08:10</p>
<p class="date">Created: 2022-03-22 Tue 10:52</p>
<p class="validation"></p>
</div>