Update 2022-03-22 10:53

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Jean-Sébastien
2022-03-22 10:53:22 +01:00
parent f8fdc5f8b4
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<!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>
@@ -1616,7 +1616,7 @@ Table of contents
<p>
Newton's second law remains valid provided we use the relativistic momentum:
</p>
<div class="core div" id="orgf962475">
<div class="core div" id="org11c93db">
<p>
<b>Newton's law</b> <i>(relativistic case)</i>
\[
@@ -1626,7 +1626,7 @@ Newton's second law remains valid provided we use the relativistic momentum:
</div>
<div class="example div" id="orgac47d6f">
<div class="example div" id="org4050d88">
<p>
<b>Example: motion under a constant force</b>
</p>
@@ -1645,7 +1645,7 @@ p = Ft
\]
and thus
\[
p(t) = \frac{m u(t)}{1 - u(t)^2/c^2} = Ft
p(t) = \frac{m u(t)}{\sqrt{1 - u(t)^2/c^2}} = Ft
~~\longrightarrow~~ u(t) = \frac{(F/m)t}{\sqrt{1 + (Ft/mc)^2}}.
\]
Integrating again to get the displacement,
@@ -1656,6 +1656,14 @@ x(t) = \frac{F}{m}\int_0^t dt' \frac{t'}{\sqrt{1 + (Ft'/mc)^2}}
The particle's world line thus shows <b>hyperbolic motion</b>.
</p>
<figure id="orgd090155">
<object type="image/svg+xml" data="./fig/red/red_motion_under_cst_F.svg" class="org-svg" alt="Relativistic motion under constant force">
Sorry, your browser does not support SVG.</object>
<figcaption><span class="figure-number">Figure 1: </span>A particle at rest, accelerated from \(t=0\) onwards by a constant force (classical versus relativistic motion).</figcaption>
</figure>
</div>
@@ -1673,8 +1681,8 @@ In the context of relativity, work is still the line integral of the force:
The relationship between work done and increased energy also still holds:
\[
W = \int d{\boldsymbol l} \cdot \frac{d{\boldsymbol p}}{dt}
= \int dt \frac{d {\boldsymbol l}}{dt} \cdot \frac{d{\boldsymbol p}}{dt}
= \int dt {\boldsymbol u} \cdot \frac{d{\boldsymbol p}}{dt}
= \int dt ~\frac{d {\boldsymbol l}}{dt} \cdot \frac{d{\boldsymbol p}}{dt}
= \int dt ~{\boldsymbol u} \cdot \frac{d{\boldsymbol p}}{dt}
\]
but
\[
@@ -1718,14 +1726,15 @@ whereas the longitudinal component transforms in a complicated way:
For the specific case where the particle is instantaneously at rest in
the original frame, then
</p>
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<a id="Ftr0"></a><a href="./red_rm_Mf.html#Ftr0"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
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@@ -1739,23 +1748,25 @@ the original frame, then
\]
</p>
</div>
<p>
The way to avoid complicated transformation rules is to define a four-vector
as the derivative of momentum with respect to proper time, which leads to
the definition of the
</p>
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<p>
<b>Minkowski force</b>
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@@ -1795,7 +1806,7 @@ 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>