# Pre-Quantum Electrodynamics
Author: Jean-Sébastien Caux

##### Approximate Potential at Large Distance
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Let's consider the spatial function in the potential for a single point source charge: \[ \frac{1}{|{\bf r} - {\bf r}_s|} \] 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 \({\bf r} = r \hat{\bf z}, r > 0\) and \({\bf r}_s = r_s \hat{\bf z}\), with \(|r_s| < r\). Then, by Taylor expanding, we get \[ \frac{1}{|{\bf r} - {\bf r}_s|} = \frac{1}{r - r_s} = \frac{1}{r} \sum_{l=0}^{\infty} \left(\frac{r_s}{r}\right)^l. \] Formally, we could do this for any vector \({\bf r}_s\) such that \(|{\bf r}_s| < |{\bf r}|\) by Taylor expanding with the \({\boldsymbol \nabla}\) operator, \[ \frac{1}{|{\bf r} - {\bf r}_s|} = \sum_{l=0}^{\infty} \frac{1}{l!} \left( - {\bf r}_s \cdot {\boldsymbol \nabla} \right)^l \frac{1}{r} = \frac{1}{r} - {\bf r}_s \cdot {\boldsymbol \nabla} \frac{1}{r} + \frac{1}{2} \left({\bf r}_s \cdot {\boldsymbol \nabla} \right)^2 \frac{1}{r} + ... \] 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 \[ \frac{1}{|{\bf r} - {\bf r}_s|} = \sum_{l=0}^{\infty} \frac{r_s^l}{r^{l+1}} P_l (\cos \theta), \hspace{1cm} \cos \theta \equiv \hat{\bf r} \cdot \hat{\bf r}_s, \hspace{1cm} r_s < r. \label{Gr(3.94)} \] 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) \[ V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \sum_{l=0}^{\infty} \frac{1}{|{\bf r}|^{l+1}} \int_{\cal V} d\tau_s |{\bf r}_s|^l P_l (\hat{\bf r} \cdot \hat{\bf r}_s) \rho({\bf r}_s), \hspace{1cm} |{\bf r}| > |{\bf r}_s| ~\forall~ {\bf r}_s \in {\cal V}. \label{Gr(3.95)} \] 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 original potential, as long as we are sufficiently far away from the source charges.

\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 \(\hat{\bf d}\).

For \(r \gg d\), we can expand, \[ \frac{1}{|{\bf r} \mp {\bf d}/2|} \simeq \frac{1}{r} \left( 1 \pm \frac{d}{r} \cos \theta \right)^{1/2} \simeq \frac{1}{r} \left( 1 \pm \frac{d}{2r} \cos \theta \right). \] Putting things together, \[ V({\bf r}) \simeq \frac{1}{4\pi \varepsilon_0} \frac{q d \cos \theta}{r^2} \label{Gr(3.90)} \]

Author: Jean-Sébastien Caux

Created: 2022-02-08 Tue 06:55

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