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Jean-Sébastien
2022-03-15 10:07:27 +01:00
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@@ -1,7 +1,7 @@
<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-03-07 Mon 20:38 -->
<!-- 2022-03-15 Tue 08:10 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1310,10 +1310,6 @@ Table of contents
</summary>
<ul>
<li>
<a href="./d_m.html#d_m">Diagnostics: Mathematical Preliminaries</a><span class="headline-id">d.m</span>
</li>
<li>
<a href="./d_ems.html#d_ems">Diagnostics: Electromagnetostatics</a><span class="headline-id">d.ems</span>
</li>
@@ -1352,6 +1348,10 @@ Table of contents
<li>
<a href="./d_red.html#d_red">Diagnostics: Relativistic Electrodynamics</a><span class="headline-id">d.red</span>
</li>
<li>
<a href="./d_m.html#d_m">Diagnostics: Compendium - Mathematics</a><span class="headline-id">d.m</span>
</li>
</ul>
@@ -1616,37 +1616,44 @@ Table of contents
<p>
Newton's second law remains valid provided we use the relativistic momentum:
</p>
<div class="core div" id="org70ed3d8">
<div class="core div" id="orgf962475">
<p>
{\bf Newton's law (relativistic case)}
<b>Newton's law</b> <i>(relativistic case)</i>
\[
{\boldsymbol F} = \frac{d{\boldsymbol p}}{dt}
\]
{\boldsymbol F} = \frac{d{\boldsymbol p}}{dt}
\]
</p>
</div>
<div class="example div" id="orgaf6f710">
<div class="example div" id="orgac47d6f">
<p>
\paragraph{Example 12.10: Motion under a constant force.} A particle
<b>Example: motion under a constant force</b>
</p>
<p>
A particle
of mass \(m\) is subjected to a constant force \(F\). If it starts at the
origin at \(t=0\), what is it's position as a function of time?
\paragraph{Solution:}
</p>
<p>
<b>Solution</b>:
\[
\frac{dp}{dt} = F, ~~p(0) = 0 ~~\longrightarrow~~
p = Ft
\]
\frac{dp}{dt} = F, ~~p(0) = 0 ~~\longrightarrow~~
p = Ft
\]
and thus
\[
p(t) = \frac{m u(t)}{1 - u(t)^2/c^2} = Ft
~~\longrightarrow~~ u(t) = \frac{(F/m)t}{\sqrt{1 + (Ft/mc)^2}}.
\]
p(t) = \frac{m u(t)}{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,
\[
x(t) = \frac{F}{m}\int_0^t dt' \frac{t'}{\sqrt{1 + (Ft'/mc)^2}}
= \frac{mc^2}{F} \left[ \sqrt{1 + (Ft/mc)^2} - 1 \right].
\]
The particle's world line thus shows {\bf hyperbolic motion}.
x(t) = \frac{F}{m}\int_0^t dt' \frac{t'}{\sqrt{1 + (Ft'/mc)^2}}
= \frac{mc^2}{F} \left[ \sqrt{1 + (Ft/mc)^2} - 1 \right].
\]
The particle's world line thus shows <b>hyperbolic motion</b>.
</p>
</div>
@@ -1654,7 +1661,10 @@ The particle's world line thus shows {\bf hyperbolic motion}.
<p>
\paragraph{Work and energy}
<b>Work and energy</b>
</p>
<p>
In the context of relativity, work is still the line integral of the force:
\[
W \equiv \int {\boldsymbol F} \cdot d{\boldsymbol l}
@@ -1682,7 +1692,10 @@ and we thus get
<p>
\paragraph{Force}
<b>Force</b>
</p>
<p>
Since \({\boldsymbol F}\) involves a derivative with respect to ordinary time,
it does not transform well under Lorentz transformations.
For the example of motion along \(\hat{\boldsymbol x}\),
@@ -1704,11 +1717,25 @@ 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>
<div class="eqlabel" id="orgb7d57c3">
<p>
<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">
<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="org2309428">
</div>
</div>
<p>
\[
\bar{\boldsymbol F}_\perp = \frac{1}{\gamma} {\boldsymbol F}_\perp,
\hspace{10mm}
\bar{F}_\parallel = F_\parallel.
\label{eq:ForceTransfoSimple}
\tag{Ftr0}\label{Ftr0}
\]
</p>
@@ -1717,15 +1744,29 @@ 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>
<div class="main div" id="org9757ffe">
<div class="main div" id="orgb93dc36">
<p>
<b>Minkowski force</b>
</p>
<div class="eqlabel" id="orgca23908">
<p>
<a id="MinkF"></a><a href="./red_rm_Mf.html#MinkF"><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="orgd690932">
</div>
</div>
<p>
{\bf Minkowski force}
\[
{\boldsymbol K} = \frac{d{\boldsymbol p}}{d\tau}
= \left( \frac{dt}{d\tau} \right) \frac{d{\boldsymbol p}}{dt}
= \frac{1}{\sqrt{1 - u^2/c^2}} {\boldsymbol F}
\label{eq:MinkowskiForce}
\]
{\boldsymbol K} = \frac{d{\boldsymbol p}}{d\tau}
= \left( \frac{dt}{d\tau} \right) \frac{d{\boldsymbol p}}{dt}
= \frac{1}{\sqrt{1 - u^2/c^2}} {\boldsymbol F}
\tag{MinkF}\label{MinkF}
\]
</p>
</div>
@@ -1754,7 +1795,7 @@ 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: 2022-03-07 Mon 20:38</p>
<p class="date">Created: 2022-03-15 Tue 08:10</p>
<p class="validation"></p>
</div>