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Jean-Sébastien
2022-02-15 10:32:42 +01:00
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commit 6874e66024
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@@ -1,7 +1,7 @@
<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-02-14 Mon 20:35 -->
<!-- 2022-02-15 Tue 10:14 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1614,14 +1614,14 @@ In one dimension, the potential is a single-variable
function \(\phi (x)\) and the Laplace equation reads
</p>
<div class="eqlabel" id="org937348d">
<div class="eqlabel" id="orgdfca409">
<p>
<a id="Lap_1d"></a><a href="./ems_ca_fe_L.html#Lap_1d"><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="orgf7c0132">
<div class="alteqlabels" id="org36d5863">
</div>
@@ -1634,16 +1634,16 @@ function \(\phi (x)\) and the Laplace equation reads
</p>
<p>
The solution to this
The solution to this is
</p>
<div class="eqlabel" id="org5437a41">
<div class="eqlabel" id="orga37bc11">
<p>
<a id="Lap_1d_sol"></a><a href="./ems_ca_fe_L.html#Lap_1d_sol"><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="org777e948">
<div class="alteqlabels" id="orgc5114c3">
<ul class="org-ul">
<li>Gr (3.6)</li>
</ul>
@@ -1702,14 +1702,14 @@ In two dimensions, the potential becomes a function
of two variables (here: \(x\) and \(y\)), so Laplace's
equation now reads
</p>
<div class="eqlabel" id="orgf26c98b">
<div class="eqlabel" id="orgc587278">
<p>
<a id="Lap_2d"></a><a href="./ems_ca_fe_L.html#Lap_2d"><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="org59e7c1d">
<div class="alteqlabels" id="orgd96eb38">
</div>
@@ -1758,15 +1758,31 @@ The Laplace equation is (we repeat)
<p>
<b>Theorem</b>: if \(\phi\) satisfies Laplace, then its value at
a point equals its value averaged over any sphere
\(S_{\bf r}\) centered on this point,
a point equals its value averaged over a sphere
\(S_R({\bf r})\) of any radius \(R\) centered on this point
(and of course not containing any charges),
</p>
<div class="eqlabel" id="orgf7426f4">
<p>
<a id="p_ball_avg"></a><a href="./ems_ca_fe_L.html#p_ball_avg"><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="org1c3162d">
</div>
</div>
<p>
\[
\phi({\bf r}) = \frac{1}{4\pi R^2} \oint_{S_{\bf r}} da' ~\phi ({\bf r}')
\phi({\bf r}) = \frac{1}{4\pi R^2} \oint_{S_R({\bf r})} da' ~\phi ({\bf r}')
\tag{p_ball_avg}\label{p_ball_avg}
\]
</p>
<details id="orgf3cada4">
<summary id="org840c868">
<details id="orgbc3026b">
<summary id="orge4e657e">
<strong>Physicist's proof</strong>
</summary>
<p>
@@ -1779,11 +1795,8 @@ of the sphere.
<p>
We know that the field created by the sphere coincides
with that of a point charge \(q\) at the origin.
Since the potential at \({\bf r = 0}\) created by the charge
\(q'\) at \({\bf r'}\) is simply \(\phi_{q', {\bf r}'} (0) = \frac{q'}{4\pi \varepsilon_0 R'}\),
the work
required to bring the \(q'\) charge into position is thus
simply \(W = q \times \phi_{q', {\bf r}'} (0) = \frac{q q'}{4 \pi \varepsilon_0 R'}\) by <a href="./ems_es_efo_e.html#Wab">Wab</a>.
The work required to bring the \(q'\) charge into position is thus
simply \(W = \frac{q q'}{4 \pi \varepsilon_0 R'}\) by <a href="./ems_es_efo_e.html#Wab">Wab</a>.
</p>
<p>
@@ -1814,14 +1827,12 @@ Equating this with the previous results shows that
<p>
\[
\phi_{q', {\bf r'}} (0) = \frac{1}{4\pi R^2} \oint_{S_R} da ~\phi_{q', {\bf r}'} ({\bf r})
\frac{q'}{4\pi \varepsilon_0 R'} = \frac{1}{4\pi R^2} \oint_{S_R} da ~\phi_{q', {\bf r}'} ({\bf r})
\]
</p>
<p>
namely that for the potential created by a single point
charge \(q'\) at \(R'\),
the value at a point (here the origin)
but this also equals the potential at \({\bf r} = 0\) created by the charge
\(q'\) at \({\bf r'}\), <i>i.e.</i> \(\phi_{q', {\bf r}'} (0) = \frac{q'}{4\pi \varepsilon_0 R'}\).
In other words, we have thus shown that for the potential created by a single point
charge \(q'\) at \(R'\), the value at a point (here the origin)
coincides with the value averaged over a sphere
or an arbitrary radius \(R\) centered on the same point.
</p>
@@ -1833,8 +1844,8 @@ proving the theorem.
</p>
</details>
<details id="org02ee258">
<summary id="orgf24411c">
<details id="orgf289197">
<summary id="orge3fd0c7">
<strong>Formal proof</strong>
</summary>
@@ -1845,13 +1856,13 @@ a ball of radius \(R\) centered on \({\bf r}\):
<p>
\[
f_{S_R} ({\bf r}) \equiv \frac{1}{4\pi R^2}\oint_{S_R} da' ~ f ({\bf r} + {\bf r}')
f_{S_R} ({\bf r}) \equiv \frac{1}{4\pi R^2}\oint_{S_R ({\bf r})} da' ~ f ({\bf r}')
\]
</p>
<p>
For convenience we will hereafter put \({\bf r} = 0\).
In spherical coordinates, we have \(da' = R^2 sin \theta d\theta d\varphi \equiv R^2 d\Omega\).
In spherical coordinates defined around the point \({\bf r}\),
we have \(da' = R^2 sin \theta d\theta d\varphi \equiv R^2 d\Omega\).
Differentiating with respect to \(R\),
</p>
@@ -1863,7 +1874,7 @@ Differentiating with respect to \(R\),
<p>
with \(f\) differentiated with respect to the radial coordiate.
We can rewrite this by noting that \(R^2 d\Omega \hat{\bf r}\)
We can rewrite this by noting that \(R^2 d\Omega ~\hat{\bf r}\)
is the normal differential surface area \(d{\bf a}\), while
\(\left.\frac{\partial f}{\partial r}\right|_{r=R}\) is the radial component of the gradient
of \(f\) in spherical coordinates. Thus,
@@ -1884,10 +1895,23 @@ we get the following general
<p>
<b>Theorem</b>:
</p>
<div class="eqlabel" id="orgd8bf4d9">
<p>
<a id="dfdR_intLap"></a><a href="./ems_ca_fe_L.html#dfdR_intLap"><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="org49a36be">
</div>
</div>
<p>
\[
\frac{d}{dR} f_{S_R} = \frac{1}{4\pi R^2} \int_{V_R} d\tau ~\nabla^2 f
\tag{dfdR_intLap}\label{dfdR_intLap}
\]
</p>
@@ -1904,7 +1928,7 @@ an infinitesimally small ball, we get the result announced above.
</details>
<p>
<b>Theorem (Earnshaw, mathematical versoin)</b>: \(\phi\) has no local extrema except at the boundaries.
<b>Theorem (Earnshaw, mathematical version)</b>: \(\phi\) has no local extrema except at the boundaries.
</p>
<p>
@@ -1931,9 +1955,24 @@ are necessarily positive, we thus require \(f_x &gt; 0\), \(f_y &gt; 0\) and \(f
of the \(f_x + f_y + f_z = 0\) condition above.
</p>
<div class="info div" id="orgfec03b0">
<div class="eqlabel" id="orgfeb6aae">
<p>
<a id="Earnshaw"></a><a href="./ems_ca_fe_L.html#Earnshaw"><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="org7f4d8f6">
</div>
</div>
<div class="info div" id="orgf970a43">
<p>
<b>Earnshaw's theorem (physical version)</b> <br>
</p>
<p>
It is impossible to find a static distribution of charges which generates an electrostatic field
displaying a stable equilibrium position in empty space.
</p>
@@ -1967,7 +2006,7 @@ We start by mentioning some cases, and interpreting them thereafter.
the electrostatic potential is uniquely determined
by Poisson's equation, which
is explicitly solved by <a href="./ems_es_ep_d.html#p_vcd">p_vcd</a>.
One can eplicitly verify this:
One can explicitly verify this:
</p>
<p>
@@ -1980,23 +2019,22 @@ One can eplicitly verify this:
<p>
where we have used <a href="./c_m_dd_3d.html#Lap1or">Lap1or</a>, and the fact that the delta function is always resolved since we
integrate over all space. Note: it is implicitly assumed that the integral in <a href="./ems_es_ep_d.html#p_vcd">p_vcd</a>
converges, <i>i.e.</i> that the charge density \(\rho({\bf r})\) is sufficiently well-behaved (does not
become singular).
converges, <i>i.e.</i> that the charge density \(\rho({\bf r})\) is sufficiently well-behaved.
</p>
<p>
<b>Charge density in closed volume, and boundary surface charge density are known</b>: the electrostatic potential is uniquely determined
<b>"Known boundary charge" case: charge density in closed volume and boundary surface charge density are both known</b>: the electrostatic potential is uniquely determined
in a certain volume \({\cal V}\) bounded by boundary \({\cal S}\), provided the charge density
\(\rho ({\bf x})\) is given everywhere within \({\cal V}\), vanishes outside of \({\cal V}\),
\(\rho ({\bf r})\) is given everywhere within \({\cal V}\), vanishes outside of \({\cal V}\),
and the value of the surface charge density \(\sigma\) is given everywhere on the boundary \({\cal S}\).
Of course, \({\cal S}\) need not be a connected surface.
</p>
<p>
<b>Charge density in closed volume, and potential at boundary are known</b>: the electrostatic potential is uniquely determined
<b>"Known boundary potential" case: charge density in closed volume and potential at boundary are both known</b>: the electrostatic potential is uniquely determined
in a certain volume \({\cal V}\) bounded by boundary \({\cal S}\), provided the charge density
\(\rho ({\bf x})\) is given everywhere within \({\cal V}\), and the value of \(V\) is given everywhere on the
\(\rho ({\bf r})\) is given everywhere within \({\cal V}\), and the value of \(V\) is given everywhere on the
boundary \({\cal S}\). Of course, \({\cal S}\) need not be a connected surface.
</p>
@@ -2048,7 +2086,7 @@ 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: 2022-02-14 Mon 20:35</p>
<p class="date">Created: 2022-02-15 Tue 10:14</p>
<p class="validation"></p>
</div>