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<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-02-13 Sun 21:20 -->
<!-- 2022-02-14 Mon 20:35 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1603,68 +1603,151 @@ Let's consider the spatial function in the potential for a single point source c
\frac{1}{|{\bf r} - {\bf r}_s|}
\]
How does this look when we're at large distances \(|{\bf r}| \gg |{\bf r}_s|\) ?
We can formally expand this in powers of \(|{\bf r}_s|/|{\bf r}|\). For simplicity, let's start by
putting \({\bf r} = r \hat{\bf z}, r &gt; 0\) and \({\bf r}_s = r_s \hat{\bf z}\), with \(|r_s| &lt; r\).
Then, by Taylor expanding, we get
We can formally expand this in powers of \(|{\bf r}_s|/|{\bf r}|\). For simplicity, let's start by
putting \({\bf r} = r ~\hat{\bf z}, r &gt; 0\) and \({\bf r}_s = r_s \hat{\bf z}\), with \(|r_s| &lt; r\).
By Taylor expanding, we get
\[
\frac{1}{|{\bf r} - {\bf r}_s|} = \frac{1}{r - r_s} = \frac{1}{r} \sum_{l=0}^{\infty} \left(\frac{r_s}{r}\right)^l.
\frac{1}{|{\bf r} - {\bf r}_s|} = \frac{1}{r - r_s} = \frac{1}{r} \sum_{l=0}^{\infty} \left(\frac{r_s}{r}\right)^l
\]
Formally, we could do this for any vector \({\bf r}_s\) such that \(|{\bf r}_s| &lt; |{\bf r}|\) by
Taylor expanding with the \({\boldsymbol \nabla}\) operator,
</p>
<div class="eqlabel" id="org6e3ad46">
<p>
<a id="1or_grad"></a><a href="./ems_ca_me_a.html#1or_grad"><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="orgf4dd0b8">
</div>
</div>
<p>
\[
\frac{1}{|{\bf r} - {\bf r}_s|} = \sum_{l=0}^{\infty} \frac{1}{l!} \left( - {\bf r}_s \cdot {\boldsymbol \nabla} \right)^l \frac{1}{r}
= \frac{1}{r} - {\bf r}_s \cdot {\boldsymbol \nabla} \frac{1}{r} + \frac{1}{2} \left({\bf r}_s \cdot {\boldsymbol \nabla} \right)^2 \frac{1}{r} + ...
\tag{1or_grad}\label{1or_grad}
\]
However, it is more practical to exploit the fact that in the configuration above,
the problem has azimuthal symmetry (everything on the \(\hat{\bf z}\) axis), and therefore
the potential takes the form of the general solution of Laplace's equation (\ref{Gr(3.65)}) with \(\theta = 0\).
Reading the parameters, we get \(A_l = 0\), \(B_l = r_s^l\). Putting back a generic angle,
we thus get
\[
the problem has azimuthal symmetry (since everything is on the \(\hat{\bf z}\) axis),
and therefore
the potential takes the form of the general solution of Laplace's equation <a href="./ems_ca_sv_sph.html#Lap_sph_az_sol">Lap_sph_az_sol</a> with \(\theta = 0\).
Reading the parameters, we get \(A_l = 0\), \(B_l = r_s^l\). Putting back a generic angle
(the coefficients remain the same), we thus get
</p>
<div class="eqlabel" id="org92f9788">
<p>
<a id="1or_Leg"></a><a href="./ems_ca_me_a.html#1or_Leg"><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="org86f4738">
<ul class="org-ul">
<li>Gr(3.94)</li>
</ul>
</div>
</div>
\begin{equation}
\frac{1}{|{\bf r} - {\bf r}_s|} = \sum_{l=0}^{\infty} \frac{r_s^l}{r^{l+1}} P_l (\cos \theta),
\hspace{1cm} \cos \theta \equiv \hat{\bf r} \cdot \hat{\bf r}_s, \hspace{1cm} r_s &lt; r.
\label{Gr(3.94)}
\]
\tag{1or_Leg\label{1or_Leg}
\end{equation}
<p>
We thus see that our beloved Legendre polynomials are quite handy beasts indeed.
Considering an arbitrary charge distribution over a volume \({\cal V}\),
we can expand the potential at a point \({\bf r}\) outside \({\cal V}\) according to
(here, we put the origin of our coordinate system closer to all points in \({\cal V}\) than to \({\bf r}\)
to ensure convergence)
</p>
<div class="eqlabel" id="orgf817fc0">
<p>
<a id="p_Leg"></a><a href="./ems_ca_me_a.html#p_Leg"><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="org25fb513">
<ul class="org-ul">
<li>Gr (3.95)</li>
</ul>
</div>
</div>
<p>
\[
V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \sum_{l=0}^{\infty} \frac{1}{|{\bf r}|^{l+1}}
\int_{\cal V} d\tau_s |{\bf r}_s|^l P_l (\hat{\bf r} \cdot \hat{\bf r}_s) \rho({\bf r}_s),
\phi({\bf r}) = \frac{1}{4\pi \varepsilon_0} \sum_{l=0}^{\infty} \frac{1}{|{\bf r}|^{l+1}}
\int_{\cal V} d\tau_s |{\bf r}_s|^l P_l (\hat{\bf r} \cdot \hat{\bf r}_s) ~\rho({\bf r}_s),
\hspace{1cm} |{\bf r}| &gt; |{\bf r}_s| ~\forall~ {\bf r}_s \in {\cal V}.
\label{Gr(3.95)}
\tag{p_Leg}\label{p_Leg}
\]
This is an exact expansion of our potential if we keep all terms. However, the power of
this comes from the fact that truncating the series gives a very good approximation to the
Keeping all the terms up to infinite order, this is an exact expansion of our potential.
The real power of this however
comes from the fact that truncating the series to only its first (few) terms
gives a very good approximation to the
original potential, as long as we are sufficiently far away from the source charges.
</p>
<p>
\paragraph{%DO NOT DO IT LIKE THIS !
Ex. 3.10:} an {\bf electric dipole} consists of two equal and opposite charges \(\pm q\) separated
by a distance \(d\). For definiteness: we put \(q\) at position \({\bf d}/2\) and \(-q\) at \(-{\bf d}/2\).
The potential is
\[
V({\bf r}) = \frac{q}{4\pi \varepsilon_0} \left( \frac{1}{|{\bf r} - {\bf d}/2|} - \frac{1}{|{\bf r} + {\bf d}/2|} \right).
\]
From the law of cosines, \(|{\bf r} \pm {\bf d}/2|^2 = r^2 + (d/2)^2 \mp rd \cos \theta
= r^2 \left( 1 \mp \frac{d}{r} \cos \theta + \frac{d^2}{4r^2}\right)\) where \(\theta\) is the angle
between \(\hat{\bf r}\) and \(\hat{\bf d}\).
As a first very basic example, let us revisit the configuration we
had in equation <a href="./ems_ca_mi.html#p_dip_z">p_dip_z</a> which represented a (physical) electric dipole, namely
two equal and opposite charges \(\pm q\) separated by a distance \(d\).
</p>
<p>
For \(r \gg d\), we can expand,
To be more general, we here put \(q\) at position \({\bf d}/2\) and \(-q\) at \(-{\bf d}/2\).
The potential is then
\[
\frac{1}{|{\bf r} \mp {\bf d}/2|} \simeq \frac{1}{r} \left( 1 \pm \frac{d}{r} \cos \theta \right)^{1/2}
\simeq \frac{1}{r} \left( 1 \pm \frac{d}{2r} \cos \theta \right).
\phi({\bf r}) = \frac{q}{4\pi \varepsilon_0} \left( \frac{1}{|{\bf r} - {\bf d}/2|} - \frac{1}{|{\bf r} + {\bf d}/2|} \right).
\]
Putting things together,
We can write \(|{\bf r} \pm {\bf d}/2|^2 = r^2 \mp {\bf r} \cdot {\bf d} + (d/2)^2
= r^2 \left( 1 \mp \frac{{\bf d} \cdot \hat{\bf r}}{r} + \frac{d^2}{4r^2}\right)\).
For \(r \gg d\), we can expand (immediately dropping terms of order \(d^2/r^2\))
\[
V({\bf r}) \simeq \frac{1}{4\pi \varepsilon_0} \frac{q d \cos \theta}{r^2}
\label{Gr(3.90)}
\frac{1}{|{\bf r} \mp {\bf d}/2|} \simeq \frac{1}{r} \left( 1 \pm \frac{{\bf d} \cdot \hat{\bf r}}{r} \right)^{1/2}
\simeq \frac{1}{r} \left( 1 \pm \frac{{\bf d} \cdot \hat{\bf r}}{2r} \right).
\]
Putting things together, the leading term in the expansion <a href="./ems_ca_me_a.html#p_Leg">p_Leg</a> for the
potential of the physical dipole is given by
</p>
<div class="eqlabel" id="org65131af">
<p>
<a id="p_physdi"></a><a href="./ems_ca_me_a.html#p_physdi"><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="org09e00db">
<ul class="org-ul">
<li>Gr (3.90)</li>
</ul>
</div>
</div>
<p>
\[
\phi({\bf r}) \simeq \frac{1}{4\pi \varepsilon_0} \frac{q~ {\bf d} \cdot \hat{\bf r}}{r^2}
\tag{p_physdi}\label{p_physdi}
\]
</p>
<p>
Thus, although the dipole is electrically neutral overall, its influence does
not vanish. Compared to a point charge, it displays two important differences:
i) it falls off with one more (inverse) power of distance, and ii) it carries
a vector direction, so its influence is not isotropic.
</p>
<p>
In the next section, we will revisit such dipoles, but now coming from more
generic charge configurations.
</p>
</div>
</div>
@@ -1687,7 +1770,7 @@ 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: 2022-02-13 Sun 21:20</p>
<p class="date">Created: 2022-02-14 Mon 20:35</p>
<p class="validation"></p>
</div>