# Radiation from a point charge near a conducting plane

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 25.

For another example of radiation, we can return to the example of the point charge ${q}$ at a distance ${z}$ from an infinite conducting plane, which we looked at in electrostatics as an example of the method of images. We saw then that we could replace the conducting plane by an equal and opposite charge ${-q}$ at position ${-z}$. This makes the ${xy}$ plane (the plane of the conductor) an equipotential surface with ${V=0}$, so it satisfies the boundary conditions for the half-space ${z\ge0}$.

Because the potentials of the original and image systems are the same in the physical region (the region outside the conductor), so too are the fields, and since the force on the charge is determined by the fields, the forces in the two systems are also equal, so we have

$\displaystyle \mathbf{F}=-\frac{1}{4\pi\epsilon_{0}}\frac{q^{2}}{\left(2z\right)^{2}}\hat{\mathbf{z}} \ \ \ \ \ (1)$

where the negative sign indicates that the force is towards the plane (or towards the image charge, in the image system).

If the charge is not restrained, this force will cause it to accelerate, so

$\displaystyle \mathbf{a}=\frac{\mathbf{F}}{m}=-\frac{1}{4\pi\epsilon_{0}}\frac{q^{2}}{\left(2z\right)^{2}m}\hat{\mathbf{z}} \ \ \ \ \ (2)$

This acceleration will cause the charge to radiate, and we can calculate the dipole radiation from the formula

$\displaystyle P\cong\frac{\mu_{0}\ddot{p}^{2}}{6\pi c} \ \ \ \ \ (3)$

where ${p}$ is the dipole moment of the charge distribution. The dipole moment of the charge and its image is

$\displaystyle \mathbf{p}=2qz\hat{\mathbf{z}} \ \ \ \ \ (4)$

since the two charges are separated by a distance ${2z}$. Therefore

$\displaystyle \ddot{\mathbf{p}}=2q\ddot{z}\hat{\mathbf{z}}=2q\mathbf{a}=-\frac{1}{2\pi\epsilon_{0}}\frac{q^{3}}{\left(2z\right)^{2}m}\hat{\mathbf{z}} \ \ \ \ \ (5)$

so the power is

 $\displaystyle P$ $\displaystyle \cong$ $\displaystyle \frac{\mu_{0}}{6\pi c}\left(\frac{1}{2\pi\epsilon_{0}}\frac{q^{3}}{\left(2z\right)^{2}m}\right)^{2}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{6\pi c}\left(\frac{c^{2}\mu_{0}}{2\pi}\frac{q^{3}}{\left(2z\right)^{2}m}\right)^{2}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{\mu_{0}cq^{2}}{4\pi}\right)^{3}\frac{1}{6m^{2}z^{4}} \ \ \ \ \ (8)$

Although this is the answer given in Griffiths’s book (in his problem 11.25), I’m not convinced it’s actually correct. The actual charge distribution that we calculated when studying the method of images, that is, using the point charge ${q}$ at position ${z}$ and the surface charge density, given by

$\displaystyle \sigma=\frac{-qz}{2\pi\left(r^{2}+z^{2}\right)^{3/2}} \ \ \ \ \ (9)$

where ${r}$ is the radial distance on the ${xy}$
plane measured from the point directly beneath the point charge. If we take the origin to be the point in the ${xy}$ plane directly beneath the point charge,
the dipole moment of the surface charge comes out to zero (by symmetry; for each charge element at a point ${\mathbf{r}}$ there is an equal charge element at point ${-\mathbf{r}}$ so their dipole moments cancel out), while the dipole moment of the point charge is ${qz\hat{\mathbf{z}}}$, so the total dipole moment of the point charge + surface charge is also ${qz\hat{\mathbf{z}}}$. Since the dipole moment is lower by a factor of 2, the total power radiated is lower by a factor of 4, and would seem to be

$\displaystyle P=\left(\frac{\mu_{0}cq^{2}}{4\pi}\right)^{3}\frac{1}{24m^{2}z^{4}} \ \ \ \ \ (10)$

If we assume that the half space ${z\le0}$ is entirely occupied by the conductor, there can be no free charge anywhere except on the surface of the conductor (and of course the point charge itself). I suspect the problem may have something to do with the fact that the method of images really applies only in electrostatics, though I’m not entirely sure. Comments welcome.

# Fields and radiation from a time-dependent sheet of current

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 24.

Suppose we have an infinite plane of current with surface current density ${K\left(t\right)\hat{\mathbf{z}}}$; that is, the current density at any instant of time is the same everywhere on the ${yz}$ plane and flows in the ${z}$ direction, but ${K}$ can change with time.

To find the fields generated by this surface current, we need first to find the vector potential (we’re assuming that the plane is electrically neutral so that the scalar potential ${V=0}$). From symmetry, ${\mathbf{A}}$ can depend only on ${x}$, the perpendicular distance from the ${yz}$ plane. Suppose we’re at some distance ${x}$ on the ${x}$ axis (it doesn’t matter what ${y}$ and ${z}$ are, but for convenience we’ll take ${y=z=0}$) from the plane. Then the potential is given by

$\displaystyle \mathbf{A}\left(x,t\right)=\frac{\mu_{0}}{4\pi}\hat{\mathbf{z}}\int\frac{K(t-\mathfrak{r}/c)}{\mathfrak{r}}d^{2}\mathbf{r} \ \ \ \ \ (1)$

where the integral is taken over the ${yz}$ plane and for a point on the ${yz}$ plane a distance ${r}$ from the origin, we have

$\displaystyle \mathfrak{r}=\sqrt{x^{2}+r^{2}} \ \ \ \ \ (2)$

It’s easiest to do the integral using polar coordinates, since signals from all points at a given radius ${r}$ on the ${yz}$ plane that arrive at the observation point at time ${t}$ left their points of origin at the same retarded time ${t_{r}=t-\mathfrak{r}/c}$. Therefore

 $\displaystyle \mathbf{A}\left(x,t\right)$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\hat{\mathbf{z}}\int_{0}^{\infty}\int_{0}^{2\pi}\frac{K\left(t-\frac{\sqrt{x^{2}+r^{2}}}{c}\right)}{\sqrt{x^{2}+r^{2}}}r\; d\phi\; dr\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{2}\hat{\mathbf{z}}\int_{0}^{\infty}\frac{K\left(t-\frac{\sqrt{x^{2}+r^{2}}}{c}\right)}{\sqrt{x^{2}+r^{2}}}r\; dr \ \ \ \ \ (4)$

We can simplify the integral using the substitution (remember that ${x}$ is a constant since it is the fixed point of observation):

 $\displaystyle w$ $\displaystyle =$ $\displaystyle \frac{\sqrt{x^{2}+r^{2}}}{c}\ \ \ \ \ (5)$ $\displaystyle dw$ $\displaystyle =$ $\displaystyle \frac{r}{c\sqrt{x^{2}+r^{2}}}dr\ \ \ \ \ (6)$ $\displaystyle r=0$ $\displaystyle \rightarrow$ $\displaystyle w=\frac{x}{c}\ \ \ \ \ (7)$ $\displaystyle r=\infty$ $\displaystyle \rightarrow$ $\displaystyle w=\infty \ \ \ \ \ (8)$

So we get

$\displaystyle \mathbf{A}\left(x,t\right)=\frac{\mu_{0}c}{2}\hat{\mathbf{z}}\int_{x/c}^{\infty}K\left(t-w\right)dw \ \ \ \ \ (9)$

Finally, to get rid of the inconvenient lower bound on the integral, we can use another substitution:

 $\displaystyle u$ $\displaystyle =$ $\displaystyle w-\frac{x}{c}\ \ \ \ \ (10)$ $\displaystyle w=\frac{x}{c}$ $\displaystyle \rightarrow$ $\displaystyle u=0\ \ \ \ \ (11)$ $\displaystyle w=\infty$ $\displaystyle \rightarrow$ $\displaystyle u=\infty\ \ \ \ \ (12)$ $\displaystyle \mathbf{A}\left(x,t\right)$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}c}{2}\hat{\mathbf{z}}\int_{0}^{\infty}K\left(t-\frac{x}{c}-u\right)du \ \ \ \ \ (13)$

From here, we can find the fields:

 $\displaystyle \mathbf{E}\left(x,t\right)$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}c}{2}\hat{\mathbf{z}}\int_{0}^{\infty}\dot{K}\left(t-\frac{x}{c}-u\right)du \ \ \ \ \ (15)$

However, from the chain rule

 $\displaystyle \dot{K}\left(t-\frac{x}{c}-u\right)$ $\displaystyle =$ $\displaystyle -\frac{\partial K\left(t-\frac{x}{c}-u\right)}{\partial u}\ \ \ \ \ (16)$ $\displaystyle \mathbf{E}\left(x,t\right)$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}c}{2}\hat{\mathbf{z}}\int_{0}^{\infty}\frac{\partial K\left(t-\frac{x}{c}-u\right)}{\partial u}du\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}c}{2}\hat{\mathbf{z}}\left[K\left(-\infty\right)-K\left(t-\frac{x}{c}\right)\right] \ \ \ \ \ (18)$

The original form 1 is actually valid only for finite current distributions, so we need to assume that at some finite past time ${t_{0}}$, ${K\left(t\right)=0}$ for all ${t. This means that the portion of the ${yz}$ plane that contributes to ${\mathbf{A}}$ is always finite, since if ${t-\frac{\mathfrak{r}}{c}, ${K=0}$, so ${\mathfrak{r}}$ must be restricted to ${\mathfrak{r}. Therefore, the field is

$\displaystyle \mathbf{E}\left(x,t\right)=-\frac{\mu_{0}c}{2}K\left(t-\frac{x}{c}\right)\hat{\mathbf{z}} \ \ \ \ \ (19)$

The magnetic field is

 $\displaystyle \mathbf{B}\left(x,t\right)$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial A_{z}}{\partial x}\hat{\mathbf{y}}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}c}{2}\hat{\mathbf{y}}\int_{0}^{\infty}\frac{\partial K\left(t-\frac{x}{c}-u\right)}{\partial x}du \ \ \ \ \ (22)$

From the chain rule

$\displaystyle \frac{\partial K\left(t-\frac{x}{c}-u\right)}{\partial x}=\frac{1}{c}\frac{\partial K\left(t-\frac{x}{c}-u\right)}{\partial u} \ \ \ \ \ (23)$

so, again invoking the condition that ${K\left(-\infty\right)=0}$:

 $\displaystyle \mathbf{B}\left(x,t\right)$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}}{2}\hat{\mathbf{y}}\int_{0}^{\infty}\frac{\partial K\left(t-\frac{x}{c}-u\right)}{\partial u}du\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}}{2}\hat{\mathbf{y}}\left[K\left(-\infty\right)-K\left(t-\frac{x}{c}\right)\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{2}K\left(t-\frac{x}{c}\right)\hat{\mathbf{y}} \ \ \ \ \ (26)$

[To be precise, we should note that we’ve been assuming ${x>0}$, so that it represents the distance from the sheet of current rather than a strict ${x}$ coordinate and thus the time ${t_{r}=t-\frac{x}{c}}$ is actually a retarded (earlier) time. If we look on the other side of the sheet where ${x<0}$ then we need to replace ${K\left(t-\frac{x}{c}-u\right)}$ by ${K\left(t+\frac{x}{c}-u\right)}$ to get the correct retarded time. This means that

$\displaystyle \frac{\partial K\left(t+\frac{x}{c}-u\right)}{\partial x}=-\frac{1}{c}\frac{\partial K\left(t+\frac{x}{c}-u\right)}{\partial u} \ \ \ \ \ (27)$

$\displaystyle \mathbf{B}\left(x,t\right)=-\frac{\mu_{0}}{2}K\left(t+\frac{x}{c}\right)\hat{\mathbf{y}} \ \ \ \ \ (28)$

for ${x<0}$.]

Note that the electric field does not reduce to the correct value if we take ${K}$ to be a constant for all time. In that case, since the sheet of current is electrically neutral and nothing is changing with time, ${\mathbf{E}=0}$. This is because we can’t apply the analysis above to states where the current is not localized. [Curiously, though, we do get the correct answer from 18 if ${K}$ is constant, since then ${K\left(-\infty\right)-K\left(t-\frac{x}{c}\right)=0}$. However, this logic doesn’t work with ${\mathbf{B}}$, since for a constant surface current ${\mathbf{B}=\frac{\mu_{0}K}{2}\hat{\mathbf{y}}}$ and taking ${K}$ constant in 25 gives ${\mathbf{B}=0}$. The moral is, we just can’t apply this analysis to infinite current distributions.]

The power radiated can be found from the Poynting vector:

 $\displaystyle \mathbf{S}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}c}{4}K^{2}\left(t-\frac{x}{c}\right)\hat{\mathbf{x}} \ \ \ \ \ (30)$

This is the energy per unit time per unit area radiated away from the sheet of current on each side of the sheet, so the total radiated energy is twice this. Also, the energy that leaves the surface of the plane (rather than the energy detected at some distance ${x}$ from the plane) is found when ${x=0}$, so the total radiated energy is

$\displaystyle E=\frac{\mu_{0}c}{2}K^{2}\left(t\right) \ \ \ \ \ (31)$

Example 1 We can find the fields for a couple of special cases. First we consider

$\displaystyle K\left(t\right)=\begin{cases} 0 & t\le0\\ K_{0} & t>0 \end{cases} \ \ \ \ \ (32)$

From 19 and 26 we have

$\displaystyle \mathbf{E}\left(x,t\right)=\begin{cases} 0 & t<\frac{x}{c}\\ -\frac{\mu_{0}c}{2}K_{0}\hat{\mathbf{z}} & t>\frac{x}{c} \end{cases} \ \ \ \ \ (33)$

$\displaystyle \mathbf{B}\left(x,t\right)=\begin{cases} 0 & t<\frac{x}{c}\\ \frac{\mu_{0}}{2}K_{0}\hat{\mathbf{y}} & t>\frac{x}{c},\; x>0\\ -\frac{\mu_{0}}{2}K_{0}\hat{\mathbf{y}} & t>-\frac{x}{c},\; x<0 \end{cases} \ \ \ \ \ (34)$

It may seem odd that ${\mathbf{E}}$ gets ‘switched on’ and remains constant for ${t>\frac{x}{c}}$, since once the current is on and constant, there is no electric field generated. However, the current gets switched on at one particular time over the entire yz plane, so due to retardation, the observer at point ${x}$ receives news of this switching on from ever increasing circles on the ${yz}$ plane as time progresses, so after the first news arrives (from the point directly below the observer, at a distance of ${x}$) there is a continual stream of news coming in that the current has switched on from circles that are ever further away, which results in a steady electric field.

It is also interesting that the magnetic field takes on the same value as that produced by an infinite sheet of constant current (see Griffiths, Example 5.8). This is because the magnetic field is determined solely by the point on the sheet directly below the observer. Contributions to ${\mathbf{B}}$ from other points on the sheet cancel each other out.

Example 2 This time, the current is linearly increasing

$\displaystyle K\left(t\right)=\begin{cases} 0 & t\le0\\ \alpha t & t>0 \end{cases} \ \ \ \ \ (35)$

From 19 and 26 we have

$\displaystyle \mathbf{E}\left(x,t\right)=\begin{cases} 0 & t<\frac{x}{c}\\ -\frac{\mu_{0}c\alpha}{2}\left(t-\frac{\left|x\right|}{c}\right)\hat{\mathbf{z}} & t>\frac{x}{c} \end{cases} \ \ \ \ \ (36)$

$\displaystyle \mathbf{B}\left(x,t\right)=\begin{cases} 0 & t<\frac{x}{c}\\ \frac{\mu_{0}\alpha}{2}\left(t-\frac{x}{c}\right)\hat{\mathbf{y}} & t>\frac{x}{c},\; x>0\\ -\frac{\mu_{0}\alpha}{2}\left(t+\frac{x}{c}\right)\hat{\mathbf{y}} & t>-\frac{x}{c},\; x<0 \end{cases} \ \ \ \ \ (37)$

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 22.

A radio station’s transmitter consists of a circular current loop acting as a magnetic dipole. The dipole is on a radio tower 200 m above the ground, and is emitting a total power of ${3.5\times10^{4}\mbox{ watts}}$. We can find the position on the ground receiving the maximum power, although unlike the previous problem, we’re concerned only with the actual magnitude of the power and not with the power per unit area of ground. The formula for intensity is

$\displaystyle \left\langle \mathbf{S}\right\rangle =\frac{\mu_{0}m_{0}^{2}\omega^{4}\sin^{2}\theta}{32\pi^{2}c^{3}r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (1)$

For a location on the ground that is a distance ${R}$ from the base of the tower,

 $\displaystyle \sin\theta$ $\displaystyle =$ $\displaystyle \frac{R}{\sqrt{R^{2}+h^{2}}}\ \ \ \ \ (2)$ $\displaystyle r^{2}$ $\displaystyle =$ $\displaystyle R^{2}+h^{2} \ \ \ \ \ (3)$

so the intensity is

$\displaystyle \left\langle \mathbf{S}\right\rangle =\frac{\mu_{0}m_{0}^{2}\omega^{4}}{32\pi^{2}c^{3}}\frac{R^{2}}{\left(R^{2}+h^{2}\right)^{2}}\hat{\mathbf{r}} \ \ \ \ \ (4)$

The position on the ground receiving maximum intensity is determined from

 $\displaystyle \frac{d\left\langle S\right\rangle }{dR}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{4R^{3}}{\left(R^{2}+h^{2}\right)^{3}}+\frac{2R}{\left(R^{2}+h^{2}\right)^{2}}\ \ \ \ \ (6)$ $\displaystyle R_{max}$ $\displaystyle =$ $\displaystyle h \ \ \ \ \ (7)$

and the intensity at this location is

 $\displaystyle \left\langle S\right\rangle _{max}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}^{2}\omega^{4}}{128\pi^{2}c^{3}h^{2}} \ \ \ \ \ (8)$

The total power is obtained by integrating over a sphere of radius ${r}$:

 $\displaystyle \left\langle P\right\rangle$ $\displaystyle =$ $\displaystyle \int\mathbf{S}\cdot d\mathbf{a}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}^{2}\omega^{4}}{32\pi^{2}c^{3}}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{\sin^{2}\theta}{r^{2}}r^{2}\sin\theta d\phi d\theta\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}^{2}\omega^{4}}{12\pi c^{3}} \ \ \ \ \ (11)$

Therefore the maximum intensity can be written in terms of the power:

 $\displaystyle \left\langle S\right\rangle _{max}$ $\displaystyle =$ $\displaystyle \frac{12}{128\pi}\frac{\left\langle P\right\rangle }{h^{2}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{12}{128\pi}\frac{3.5\times10^{4}}{200^{2}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.6\times10^{-2}\mbox{ watts m}^{-2}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.6\times10^{-6}\mbox{ watts cm}^{-2} \ \ \ \ \ (15)$

# Radiation from a charge on a spring

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 21.

Here’s another example of radiation from an accelerated charge. We hang a spring with spring constant ${k}$ from the ceiling and attach a particle of mass ${m}$ and charge ${q}$ to its lower end, which has an equilibrium distance of ${h}$ above the floor. If the particle is pulled down a distance ${d}$ below the equilibrium and released at ${t=0}$, we wish to find the intensity (average power per unit area) that hits the floor.

To solve this problem, we can use the Poynting vector for an arbitrary charge distribution that we found earlier:

$\displaystyle \mathbf{S}=\frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B}\cong\frac{\mu_{0}\ddot{p}^{2}}{16\pi^{2}c}\frac{\sin^{2}\theta}{r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (1)$

If we take the axis of the spring as the ${z}$ axis and the equilibrium position as ${z=0}$ and ${z}$ increasing downwards, the dipole moment of the charge is

 $\displaystyle p\left(t\right)$ $\displaystyle =$ $\displaystyle qd\cos\omega t\ \ \ \ \ (2)$ $\displaystyle \ddot{p}\left(t\right)$ $\displaystyle =$ $\displaystyle -qd\omega^{2}\cos\omega t \ \ \ \ \ (3)$

In this coordinate system, the angle ${\theta}$ is the angle between the ${z}$ axis and a point on the floor on a circle of radius ${R}$ centred on the ${z}$ axis. If the charge is at ${z=0}$ then

 $\displaystyle \sin\theta$ $\displaystyle =$ $\displaystyle \frac{R}{\sqrt{R^{2}+h^{2}}}\ \ \ \ \ (4)$ $\displaystyle \cos\theta$ $\displaystyle =$ $\displaystyle \frac{h}{\sqrt{R^{2}+h^{2}}}\ \ \ \ \ (5)$ $\displaystyle r^{2}$ $\displaystyle =$ $\displaystyle R^{2}+h^{2} \ \ \ \ \ (6)$

Therefore the Poynting vector is

$\displaystyle \mathbf{S}=\frac{\mu_{0}q^{2}d^{2}\omega^{4}R^{2}\cos^{2}\omega t}{16\pi^{2}c\left(R^{2}+h^{2}\right)^{2}}\hat{\mathbf{r}} \ \ \ \ \ (7)$

This gives the power per unit area and unit time radiated along the ${r}$ direction, pointing radially away from the charge. If we want the power received per unit area on the floor, we note that the angle between the normal to ${\hat{\mathbf{r}}}$ and the floor is equal to the angle between ${\hat{\mathbf{r}}}$ and the normal to the floor, the latter of which is the ${z}$ axis. Thus the angle between the normal to ${\hat{\mathbf{r}}}$ and the floor is just ${\theta}$, so that a unit area on the plane normal to ${\hat{\mathbf{r}}}$ spreads out to an area of ${1/\cos\theta}$ when projected onto the floor. Therefore the power per unit time per unit area of floor is

 $\displaystyle S_{floor}$ $\displaystyle =$ $\displaystyle \frac{S}{1/\cos\theta}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}q^{2}d^{2}\omega^{4}R^{2}\cos^{2}\omega t}{16\pi^{2}c\left(R^{2}+h^{2}\right)^{2}}\cos\theta\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}q^{2}d^{2}\omega^{4}R^{2}h\cos^{2}\omega t}{16\pi^{2}c\left(R^{2}+h^{2}\right)^{5/2}} \ \ \ \ \ (10)$

Although we derived this for ${z=0}$, it is a good approximation for all times provided that ${d\ll h}$, so the amplitude of oscillation is very small, since in that case the angle ${\theta}$ changes very little over a cycle.

The intensity of radiation on the floor is the time average of this over one cycle, and since the average of ${\cos^{2}\omega t}$ over a cycle is ${\frac{1}{2}}$, we get

$\displaystyle \left\langle S\right\rangle =\frac{\mu_{0}q^{2}d^{2}\omega^{4}R^{2}h}{32\pi^{2}c\left(R^{2}+h^{2}\right)^{5/2}} \ \ \ \ \ (11)$

The radius ${R_{max}}$ that receives the highest intensity of radiation is found by setting the derivative to zero:

 $\displaystyle \frac{d\left\langle S\right\rangle }{dR}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}q^{2}d^{2}\omega^{4}h}{32\pi^{2}c}\left[\frac{2R}{\left(R^{2}+h^{2}\right)^{5/2}}-\frac{5R^{3}}{\left(R^{2}+h^{2}\right)^{7/2}}\right]\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle R_{max}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{6}h}{3} \ \ \ \ \ (14)$

If the floor is infinitely large, we can integrate 11 over all values of ${R}$ to find the total power received by the floor:

$\displaystyle P_{floor}=\frac{\mu_{0}q^{2}d^{2}\omega^{4}h}{32\pi^{2}c}\int_{0}^{\infty}\frac{R^{2}}{\left(R^{2}+h^{2}\right)^{5/2}}RdR\int_{0}^{2\pi}d\phi \ \ \ \ \ (15)$

The integral can be done by parts (integrate ${\frac{R}{\left(R^{2}+h^{2}\right)^{5/2}}}$ and differentiate ${R^{2}}$) so we get

 $\displaystyle \left\langle P_{floor}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}q^{2}d^{2}\omega^{4}h}{32\pi^{2}c}\frac{4\pi}{3h}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}q^{2}d^{2}\omega^{4}}{24\pi c} \ \ \ \ \ (17)$

The total radiated power should be

$\displaystyle P=\int\mathbf{S}\cdot d\mathbf{a}\cong\frac{\mu_{0}\ddot{p}^{2}}{6\pi c} \ \ \ \ \ (18)$

Averaged over a cycle this comes out to (using 3)

 $\displaystyle \left\langle P\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}\left\langle \ddot{p}^{2}\right\rangle }{6\pi c}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}\frac{1}{2}q^{2}d^{2}\omega^{4}}{6\pi c}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}q^{2}d^{2}\omega^{4}}{12\pi c} \ \ \ \ \ (21)$

Thus half the power gets absorbed by the floor, which is what we’d expect (the other half goes upwards and gets absorbed by the ceiling).

Finally, since the charge is radiating away energy, its total energy is decreasing so the amplitude of oscillation will decrease. The amount of power lost in time ${dt}$ is ${P\; dt}$, which is equal to the negative change in energy, ${-dE}$. We therefore have, defining the amplitude as a function ${A\left(t\right)}$ of time:

 $\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{1}{2}mv^{2}+\frac{1}{2}kz^{2}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(mA^{2}\omega^{2}\sin^{2}\omega t+kA^{2}\cos^{2}\omega t\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(mA^{2}\omega^{2}\sin^{2}\omega t+m\omega^{2}A^{2}\cos^{2}\omega t\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}mA^{2}\omega^{2} \ \ \ \ \ (25)$

Therefore, the energy lost as the amplitude decreases is

 $\displaystyle -dE$ $\displaystyle =$ $\displaystyle -\frac{dE}{dA}dA\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -mA\omega^{2}dA \ \ \ \ \ (27)$

The power for an amplitude ${A}$ is

$\displaystyle \left\langle P\left(A\right)\right\rangle =\frac{\mu_{0}q^{2}A^{2}\omega^{4}}{12\pi c} \ \ \ \ \ (28)$

so

 $\displaystyle \frac{\mu_{0}q^{2}A^{2}\omega^{4}}{12\pi c}dt$ $\displaystyle =$ $\displaystyle -mA\omega^{2}dA\ \ \ \ \ (29)$ $\displaystyle dt$ $\displaystyle =$ $\displaystyle -\frac{12\pi cm}{\mu_{0}q^{2}\omega^{2}}\frac{dA}{A} \ \ \ \ \ (30)$

To find how long it takes for the amplitude to drop from ${A=d}$ to ${A=d/e,}$ we have

 $\displaystyle \int_{0}^{T}dt$ $\displaystyle =$ $\displaystyle -\frac{12\pi cm}{\mu_{0}q^{2}\omega^{2}}\int_{d}^{d/e}\frac{dA}{A}\ \ \ \ \ (31)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{12\pi cm}{\mu_{0}q^{2}\omega^{2}}\left[\ln d-\ln\frac{d}{e}\right]\ \ \ \ \ (32)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{12\pi cm}{\mu_{0}q^{2}\omega^{2}}\ln e\ \ \ \ \ (33)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{12\pi cm}{\mu_{0}q^{2}\omega^{2}} \ \ \ \ \ (34)$

# Radiation from a current loop with time-varying current

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 12.

We’ve looked at the fields produced by a magnetic dipole that oscillates with a regular frequency ${\omega}$. By following the procedure in Griffiths’s section 11.1.4, where he derives the fields due to an electric dipole of arbitrary shape, we can derive the formulas for a magnetic dipole consisting of a circular current loop carrying a time-dependent current ${I\left(t\right)}$ where the time dependence is arbitrary.

We assume the current loop has radius ${b}$ and lies in the ${xy}$ plane with its centre on the ${z}$ axis. Since at any instant, the magnitude of the current is the same everywhere in the loop, we can use the same argument as in the oscillating case to deduce that for some observation point ${\mathbf{r}}$ in the ${xz}$ plane, the vector potential ${\mathbf{A}}$ points in the ${y}$ direction, and thus, since ${\mathbf{A}}$ is always tangential to the loop, its direction in general is in the ${\phi}$ direction. If the loop is electrically neutral, the electric potential is ${V=0}$, so we need to calculate only ${\mathbf{A}}$. The retarded potential is

 $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\int\frac{I\left(t-d/c\right)}{d}d\boldsymbol{\ell}'\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}b}{4\pi}\hat{\boldsymbol{\phi}}\int_{0}^{2\pi}\frac{I\left(t-d/c\right)}{d}\cos\phi'd\phi' \ \ \ \ \ (2)$

where ${\phi'}$ is the azimuthal angle around the loop so that the ${y}$ component of ${d\boldsymbol{\ell}'}$ is ${b\cos\phi'}$ and the retarded time is

$\displaystyle t_{r}\equiv t-\frac{d}{c} \ \ \ \ \ (3)$

and

 $\displaystyle d$ $\displaystyle \equiv$ $\displaystyle \left|\mathbf{r}-\mathbf{r}'\right|\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\left(x-x'\right)^{2}+\left(y-y'\right)^{2}+\left(z-z'\right)^{2}}\ \ \ \ \ (5)$ $\displaystyle \hat{\mathbf{d}}$ $\displaystyle =$ $\displaystyle \frac{\mathbf{r}-\mathbf{r}'}{d} \ \ \ \ \ (6)$

where ${\mathbf{r}'}$ is the position on the loop being integrated over.

For our observation point in the ${xz}$ plane, we have

$\displaystyle \mathbf{r}=r\sin\theta\hat{\mathbf{x}}+r\cos\theta\hat{\mathbf{z}} \ \ \ \ \ (7)$

and for a point on the loop

$\displaystyle \mathbf{r}'=b\cos\phi'\hat{\mathbf{x}}+b\sin\phi'\hat{\mathbf{y}} \ \ \ \ \ (8)$

In what follows, we’ll use the notation ${c_{\theta}\equiv\cos\theta}$, ${s_{\theta}\equiv\sin\theta}$, etc to simplify the notation.

Therefore, assuming ${b\ll r}$ (the loop is very small)

 $\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{r^{2}+b^{2}-2\mathbf{r}\cdot\mathbf{r}'}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle \cong$ $\displaystyle r\left(1-\frac{b}{r}s_{\theta}c_{\phi'}\right)\ \ \ \ \ (10)$ $\displaystyle \frac{1}{d}$ $\displaystyle \cong$ $\displaystyle \frac{1}{r}\left(1+\frac{b}{r}s_{\theta}c_{\phi'}\right) \ \ \ \ \ (11)$

We can expand the current in a Taylor series about ${t_{0}\equiv t-\frac{r}{c}}$:

 $\displaystyle I\left(t-\frac{d}{c}\right)$ $\displaystyle \cong$ $\displaystyle I\left(t-\frac{r}{c}+\frac{b}{c}s_{\theta}c_{\phi'}\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I\left(t_{0}\right)+\dot{I}\left(t_{0}\right)\frac{b}{c}s_{\theta}c_{\phi'}+\frac{1}{2!}\ddot{I}\left(t_{0}\right)\left(\frac{b}{c}s_{\theta}c_{\phi'}\right)^{2}+\dots \ \ \ \ \ (13)$

We are justified in dropping the last term if

 $\displaystyle \frac{1}{2!}\ddot{I}\left(t_{0}\right)\left(\frac{b}{c}s_{\theta}c_{\phi'}\right)^{2}$ $\displaystyle \ll$ $\displaystyle \dot{I}\left(t_{0}\right)\frac{b}{c}s_{\theta}c_{\phi'}\ \ \ \ \ (14)$ $\displaystyle b$ $\displaystyle \ll$ $\displaystyle \frac{c}{\left|\ddot{I}/\dot{I}\right|} \ \ \ \ \ (15)$

If we compare higher derivative terms with the first order term, we get the general condition

$\displaystyle b\ll c\left|\frac{\dot{I}}{d^{n}I/dt^{n}}\right|^{n-1} \ \ \ \ \ (16)$

Assuming this is true, we get from 2:

 $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle \cong$ $\displaystyle \frac{\mu_{0}b}{4\pi r}\hat{\boldsymbol{\phi}}\int_{0}^{2\pi}\left(I\left(t_{0}\right)+\dot{I}\left(t_{0}\right)\frac{b}{c}s_{\theta}c_{\phi'}\right)\left(1+\frac{b}{r}s_{\theta}c_{\phi'}\right)c_{\phi'}d\phi'\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle \cong$ $\displaystyle \frac{\mu_{0}\pi b^{2}}{4\pi r}\left(\frac{I\left(t_{0}\right)}{r}+\frac{\dot{I}\left(t_{0}\right)}{c}\right)s_{\theta}\hat{\boldsymbol{\phi}} \ \ \ \ \ (18)$

where to get the second line, we discarded the term in ${b^{3}}$ and used ${\int_{0}^{2\pi}\cos\phi'd\phi'=0}$ and ${\int_{0}^{2\pi}\cos^{2}\phi'd\phi'=\pi}$. If we’re interested only in the radiation produced by this dipole, we can ignore any terms in the potential that are of order 2 or higher in ${\frac{1}{r}}$, since it is only ${\frac{1}{r^{2}}}$ terms in the Poynting vector that will contribute to radiation that escapes to infinity. Therefore, we can throw away the first term above to get our final approximation:

$\displaystyle \mathbf{A}\left(\mathbf{r},t\right)\cong\frac{\mu_{0}\pi b^{2}}{4\pi rc}\dot{I}\left(t_{0}\right)s_{\theta}\hat{\boldsymbol{\phi}} \ \ \ \ \ (19)$

We can write this in terms of the magnetic moment of the loop, which is

$\displaystyle m\left(t_{0}\right)=\pi b^{2}I\left(t_{0}\right) \ \ \ \ \ (20)$

so

$\displaystyle \mathbf{A}\left(\mathbf{r},t\right)\cong\frac{\mu_{0}}{4\pi rc}\dot{m}\left(t_{0}\right)s_{\theta}\hat{\boldsymbol{\phi}}$

We can now calculate the fields:

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}s_{\theta}}{4\pi rc}\ddot{m}\left(t_{0}\right)\hat{\boldsymbol{\phi}}\ \ \ \ \ (22)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{r}\frac{\partial}{\partial r}\left(rA\right)\hat{\boldsymbol{\theta}}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}s_{\theta}}{4\pi rc}\frac{\partial\dot{m}}{\partial r}\hat{\boldsymbol{\theta}} \ \ \ \ \ (25)$

where in calculating ${\mathbf{B}}$, we ignored the term ${\frac{1}{rs_{\theta}}\frac{\partial}{\partial\theta}\left(s_{\theta}A\right)\hat{\mathbf{r}}}$ since it gives a term containing ${\frac{1}{r^{2}}}$.

Since ${m=m\left(t-\frac{r}{c}\right)}$ we have

 $\displaystyle \frac{\partial\dot{m}}{\partial r}$ $\displaystyle =$ $\displaystyle -\frac{1}{c}\ddot{m}\ \ \ \ \ (26)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}s_{\theta}}{4\pi rc^{2}}\ddot{m}\hat{\boldsymbol{\theta}} \ \ \ \ \ (27)$

The Poynting vector is

 $\displaystyle \mathbf{S}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}s_{\theta}^{2}\ddot{m}^{2}}{16\pi^{2}r^{2}c^{3}}\hat{\mathbf{r}} \ \ \ \ \ (29)$

The power radiated is the integral of ${\mathbf{S}}$ over a large sphere of radius ${r}$:

 $\displaystyle P$ $\displaystyle =$ $\displaystyle \int\mathbf{S}\cdot d\mathbf{a}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi\mu_{0}\ddot{m}^{2}}{16\pi^{2}c^{3}}\int_{0}^{\pi}\frac{\sin^{2}\theta}{r^{2}}r^{2}\sin\theta d\theta\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}\ddot{m}^{2}}{6\pi c^{3}} \ \ \ \ \ (32)$

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 11.

In analyzing radiation from an arbitrary configuration of charge, we made the assumption that the maximum dimension of the source is much smaller than the observation distance, so that we can retain only first order terms in ${r'}$, the variable that is integrated over the source. In some cases, the first order contribution is zero and in that case, we need to look at the next order. This leads to electric quadrupole (and magnetic dipole, but we’ll leave that for now) radiation. A simple model that illustrates this is as follows.

Suppose we have two oscillating electric dipoles situated on the ${z}$ axis, with ${\mathbf{p}_{+}}$ at ${z=+\frac{d}{2}}$ and ${\mathbf{p}_{-}}$ at ${z=-\frac{d}{2}}$. The dipole oscillate exactly ${\pi}$ out of phase, so that the dipole moment of the upper dipole is always the negative of the dipole moment of the lower one. We can work out the fields of this setup by using the same approximations we used in deriving the ordinary oscillating dipole. First, we need to define a few terms. (I’d draw a diagram, but that’s a painful process, so bear with me.)

Let the observation point ${\mathbf{r}}$ make an angle ${\theta}$ with the ${z}$ axis, and let the vector from ${\mathbf{p}_{+}}$ to ${\mathbf{r}}$ be ${\mathbf{r}_{+}}$ and the vector from ${\mathbf{p}_{-}}$ to ${\mathbf{r}}$ be ${\mathbf{r}_{-}}$. The vectors ${\mathbf{r}_{\pm}}$ make angles ${\theta_{\pm}}$ with the ${z}$ axis.

The potential formulas for a dipole at the origin are

 $\displaystyle V\left(r,\theta,t\right)$ $\displaystyle =$ $\displaystyle -\frac{p_{0}\omega\cos\theta}{4\pi\epsilon_{0}rc}\sin\left(\omega\left(t-\frac{r}{c}\right)\right)\ \ \ \ \ (1)$ $\displaystyle \mathbf{A}\left(r,\theta,t\right)$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega}{4\pi r}\hat{\mathbf{z}}\sin\left(\omega\left(t-\frac{r}{c}\right)\right) \ \ \ \ \ (2)$

However, here, the dipoles are not at the origin so we need to adapt these formulas. For ${\mathbf{p}_{+}}$ we must use ${\mathbf{r}_{+}}$ and ${\theta_{+}}$ so we have

$\displaystyle V_{+}=-\frac{p_{0}\omega\cos\theta_{+}}{4\pi\epsilon_{0}r_{+}c}\sin\left(\omega\left(t-\frac{r_{+}}{c}\right)\right) \ \ \ \ \ (3)$

From the law of cosines we have

$\displaystyle r_{+}=\sqrt{r^{2}+\frac{d^{2}}{4}-2r\frac{d}{2}\cos\theta} \ \ \ \ \ (4)$

and from the geometry of the setup

$\displaystyle r\cos\theta=r_{+}\cos\theta_{+}+\frac{d}{2} \ \ \ \ \ (5)$

Now assuming ${d\ll r}$ we have

 $\displaystyle r_{+}$ $\displaystyle \approx$ $\displaystyle r\left(1-\frac{d}{2r}\cos\theta\right)\ \ \ \ \ (6)$ $\displaystyle r\cos\theta$ $\displaystyle \approx$ $\displaystyle r\left(1-\frac{d}{2r}\cos\theta\right)\cos\theta_{+}+\frac{d}{2}\ \ \ \ \ (7)$ $\displaystyle \cos\theta_{+}$ $\displaystyle \approx$ $\displaystyle \frac{r\cos\theta-\frac{d}{2}}{r-\frac{d}{2}\cos\theta}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \frac{1}{r}\left(r\cos\theta-\frac{d}{2}\right)\left(1+\frac{d}{2r}\cos\theta\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \cos\theta+\frac{d}{2r}\left(\cos^{2}\theta-1\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \cos\theta-\frac{d}{2r}\sin^{2}\theta \ \ \ \ \ (11)$

Also,

 $\displaystyle \sin\left(\omega\left(t-\frac{r_{+}}{c}\right)\right)$ $\displaystyle \approx$ $\displaystyle \sin\left[\omega\left(t-\frac{r}{c}\right)+\frac{\omega d}{2c}\cos\theta\right]\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \sin\left[\omega\left(t-\frac{r}{c}\right)\right]+\frac{\omega d}{2c}\cos\theta\cos\left[\omega\left(t-\frac{r}{c}\right)\right] \ \ \ \ \ (13)$

to first order in ${d}$.

Plugging these into 3 we get

 $\displaystyle V_{+}$ $\displaystyle =$ $\displaystyle -\frac{p_{0}\omega}{4\pi\epsilon_{0}cr}\left(\cos\theta-\frac{d}{2r}\sin^{2}\theta\right)\left(1+\frac{d}{2r}\cos\theta\right)\times\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle$ $\displaystyle \left\{ \sin\left[\omega\left(t-\frac{r}{c}\right)\right]+\frac{\omega d}{2c}\cos\theta\cos\left[\omega\left(t-\frac{r}{c}\right)\right]\right\} \nonumber$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle -\frac{p_{0}\omega}{4\pi\epsilon_{0}cr}\left[\cos\theta\sin\left[\omega\left(t-\frac{r}{c}\right)\right]+\frac{d}{2r}\sin\left[\omega\left(t-\frac{r}{c}\right)\right]\cos2\theta+\frac{\omega d}{2c}\cos^{2}\theta\cos\left[\omega\left(t-\frac{r}{c}\right)\right]\right] \ \ \ \ \ (15)$

Under our approximation of ${d\ll r}$ we can drop the middle term to get

$\displaystyle V_{+}\approx-\frac{p_{0}\omega}{4\pi\epsilon_{0}cr}\left[\cos\theta\sin\left[\omega\left(t-\frac{r}{c}\right)\right]+\frac{\omega d}{2c}\cos^{2}\theta\cos\left[\omega\left(t-\frac{r}{c}\right)\right]\right] \ \ \ \ \ (16)$

For ${\mathbf{p}_{-}}$ we can do the same calculation to get (note the opposite sign of ${p_{0}}$ since the dipole is opposite to the top one)

 $\displaystyle V_{-}$ $\displaystyle =$ $\displaystyle \frac{p_{0}\omega\cos\theta_{-}}{4\pi\epsilon_{0}r_{+-}c}\sin\left(\omega\left(t-\frac{r_{-}}{c}\right)\right)\ \ \ \ \ (17)$ $\displaystyle r_{-}$ $\displaystyle \approx$ $\displaystyle r\left(1+\frac{d}{2r}\cos\theta\right)\ \ \ \ \ (18)$ $\displaystyle \cos\theta_{-}$ $\displaystyle \approx$ $\displaystyle \cos\theta-\frac{d}{2r}\sin^{2}\theta\ \ \ \ \ (19)$ $\displaystyle \sin\left(\omega\left(t-\frac{r_{-}}{c}\right)\right)$ $\displaystyle \approx$ $\displaystyle \sin\left[\omega\left(t-\frac{r}{c}\right)\right]-\frac{\omega d}{2c}\cos\theta\cos\left[\omega\left(t-\frac{r}{c}\right)\right] \ \ \ \ \ (20)$

Putting this together, we get

$\displaystyle V_{-}\approx\frac{p_{0}\omega}{4\pi\epsilon_{0}cr}\left[\cos\theta\sin\left[\omega\left(t-\frac{r}{c}\right)\right]-\frac{\omega d}{2c}\cos^{2}\theta\cos\left[\omega\left(t-\frac{r}{c}\right)\right]\right] \ \ \ \ \ (21)$

The total potential is

 $\displaystyle V$ $\displaystyle =$ $\displaystyle V_{+}+V_{-}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{p_{0}\omega^{2}d}{4\pi\epsilon_{0}c^{2}r}\cos^{2}\theta\cos\left[\omega\left(t-\frac{r}{c}\right)\right]\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi r}\cos^{2}\theta\cos\left[\omega\left(t-\frac{r}{c}\right)\right] \ \ \ \ \ (24)$

using ${c^{2}=1/\mu_{0}\epsilon_{0}}$.

For the vector potential, we get

 $\displaystyle \mathbf{A}_{+}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega}{4\pi r_{+}}\hat{\mathbf{z}}\sin\left(\omega\left(t-\frac{r_{+}}{c}\right)\right)\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle -\frac{\mu_{0}p_{0}\omega}{4\pi r}\hat{\mathbf{z}}\left[\sin\left(\omega\left(t-\frac{r}{c}\right)\right)-\frac{d\omega\cos\theta}{2c}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\right]\ \ \ \ \ (26)$ $\displaystyle \mathbf{A}_{-}$ $\displaystyle \approx$ $\displaystyle \frac{\mu_{0}p_{0}\omega}{4\pi r}\hat{\mathbf{z}}\left[\sin\left(\omega\left(t-\frac{r}{c}\right)\right)+\frac{d\omega\cos\theta}{2c}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\right]\ \ \ \ \ (27)$ $\displaystyle \mathbf{A}$ $\displaystyle \approx$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}d}{4\pi rc}\hat{\mathbf{z}}\cos\theta\cos\left(\omega\left(t-\frac{r}{c}\right)\right) \ \ \ \ \ (28)$

With the potentials, we can calculate the fields. To simplify the notation, we’ll use the shorthand

 $\displaystyle c_{\omega}$ $\displaystyle \equiv$ $\displaystyle \cos\left(\omega\left(t-\frac{r}{c}\right)\right)\ \ \ \ \ (29)$ $\displaystyle c_{\theta}$ $\displaystyle \equiv$ $\displaystyle \cos\theta \ \ \ \ \ (30)$

and so on.

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V-\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}d\omega^{2}p_{0}}{4\pi r^{2}}c_{\theta}^{2}c_{\omega}\hat{\mathbf{r}}+\frac{\mu_{0}p_{0}d\omega^{2}}{\pi}c_{\theta}s_{\theta}\left(\frac{\omega s_{\omega}}{4cr}-\frac{c_{\omega}}{2r^{2}}\right)\hat{\boldsymbol{\theta}} \ \ \ \ \ (32)$

Using the approximation ${r\gg c/\omega}$ we can drop all but one term to get

$\displaystyle \mathbf{E}\approx\frac{\mu_{0}p_{0}d\omega^{3}}{4\pi rc}c_{\theta}s_{\theta}s_{\omega}\hat{\boldsymbol{\theta}} \ \ \ \ \ (33)$

For the magnetic field, we get

 $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}p_{0}d\omega^{2}}{\pi rc}c_{\theta}s_{\theta}\left(\frac{\omega s_{\omega}}{4c}-\frac{c_{\omega}}{2r}\right)\hat{\boldsymbol{\phi}} \ \ \ \ \ (35)$

Again, using the approximation ${r\gg c/\omega}$ we drop the second term to get

$\displaystyle \mathbf{B}\approx-\frac{\mu_{0}p_{0}d\omega^{3}}{4\pi rc^{2}}c_{\theta}s_{\theta}s_{\omega}\hat{\boldsymbol{\phi}} \ \ \ \ \ (36)$

The Poynting vector is

 $\displaystyle \mathbf{S}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\mu_{0}}{c}\left(\frac{p_{0}d\omega^{3}}{4\pi rc}\right)^{2}\left(c_{\theta}s_{\theta}s_{\omega}\right)^{2}\hat{\mathbf{r}} \ \ \ \ \ (38)$

The intensity is the time average of ${\mathbf{S}}$:

$\displaystyle \left\langle \mathbf{S}\right\rangle =\frac{\mu_{0}p_{0}^{2}d^{2}\omega^{6}}{32\pi^{2}c^{3}r^{2}}\left(c_{\theta}s_{\theta}\right)^{2}\hat{\mathbf{r}} \ \ \ \ \ (39)$

and the power is the integral of this over a sphere of radius ${r}$:

 $\displaystyle \left\langle P\right\rangle$ $\displaystyle =$ $\displaystyle \int\mathbf{S}\cdot d\mathbf{a}\ \ \ \ \ (40)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\pi\frac{\mu_{0}p_{0}^{2}d^{2}\omega^{6}}{32\pi^{2}c^{3}}\int_{0}^{\pi}\cos^{2}\theta\sin^{3}\theta d\theta\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}p_{0}^{2}d^{2}\omega^{6}}{60\pi c^{3}} \ \ \ \ \ (42)$

# Radiation from a charge falling under gravity

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 10.

If a charge falls under the influence of gravity, it accelerates and therefore radiates. This means that not all of the potential energy lost as the charge falls is converted to kinetic energy, so a charged object falls more slowly than an uncharged one. Will this difference be noticeable?

Suppose we drop a single electron from rest at ${z=0}$. After it has fallen to a position ${z}$, its dipole moment is

$\displaystyle \mathbf{p}=ez\hat{\mathbf{z}} \ \ \ \ \ (1)$

(The dipole moment is in the ${+z}$ direction since the electron’s charge is negative and it falls to a point ${z<0}$.) The power radiated is

$\displaystyle P\cong\frac{\mu_{0}\ddot{p}^{2}}{6\pi c} \ \ \ \ \ (2)$

where

 $\displaystyle \ddot{p}$ $\displaystyle =$ $\displaystyle e\ddot{z}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle eg \ \ \ \ \ (4)$

so

$\displaystyle P=\frac{\mu_{0}e^{2}g^{2}}{6\pi c}=5.7\times10^{-52}\mbox{ J s}^{-1} \ \ \ \ \ (5)$

which is a constant.

To find how much energy is radiated as the electron falls, say, 1 cm, we need to know how long it takes the electron to fall 1 cm. If all its lost potential energy were converted to kinetic, then we get ${v=gt}$ and ${d=\frac{1}{2}gt^{2}}$. Since the power is very small, it’s a safe bet that very little of the energy is radiated, so we can assume that ${d=\frac{1}{2}gt^{2}}$ and then check that our answer is consistent. From this we get

$\displaystyle t=\sqrt{\frac{2\times0.01}{9.8}}=0.045\mbox{ s} \ \ \ \ \ (6)$

so the total energy radiated is

$\displaystyle Pt=2.46\times10^{-53}\mbox{ J} \ \ \ \ \ (7)$

The potential energy lost is

$\displaystyle V=mgh=0.01mg=8.92\times10^{-32}\mbox{ J} \ \ \ \ \ (8)$

so the fraction of potential energy radiated is

$\displaystyle \frac{Pt}{V}=2.76\times10^{-22} \ \ \ \ \ (9)$

So hardly any of the energy is radiated.

# Power radiated by a spinning ring of charge

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 9.

Here’s a generalization of the rotating dipole problem we did earlier. This time we have a circular ring with radius ${b}$ with a linear charge distribution, at ${t=0}$, of

$\displaystyle \lambda=\lambda_{0}\sin\phi \ \ \ \ \ (1)$

where ${\phi}$ is the azimuthal angle. The disk is set spinning with an angular velocity of ${\omega}$.

Because ${\sin\left(\phi+\pi\right)=-\sin\phi}$, this disk is essentially a collection of dipoles with charges ${\pm\lambda b\; d\phi}$ separated by distance ${2b}$. Therefore, at time ${t}$, the dipole moment is

 $\displaystyle \mathbf{p}\left(t\right)$ $\displaystyle =$ $\displaystyle 2b^{2}\lambda_{0}\int_{0}^{\pi}\sin\phi\left[\cos\left(\omega t+\phi\right)\hat{\mathbf{x}}+\sin\left(\omega t+\phi\right)\hat{\mathbf{y}}\right]d\phi\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2b^{2}\lambda_{0}\frac{\pi}{2}\left(-\sin\omega t\hat{\mathbf{x}}+\cos\omega t\hat{\mathbf{y}}\right)\ \ \ \ \ (3)$ $\displaystyle \ddot{\mathbf{p}}\left(t\right)$ $\displaystyle =$ $\displaystyle -\pi b^{2}\lambda_{0}\omega^{2}\left(-\sin\omega t\hat{\mathbf{x}}+\cos\omega t\hat{\mathbf{y}}\right)\ \ \ \ \ (4)$ $\displaystyle \ddot{p}^{2}$ $\displaystyle =$ $\displaystyle \left(\pi b^{2}\lambda_{0}\omega^{2}\right)^{2} \ \ \ \ \ (5)$

The total power radiated is therefore

$\displaystyle P=\int\mathbf{S}\cdot d\mathbf{a}\cong\frac{\mu_{0}\ddot{p}^{2}}{6\pi c}=\frac{\mu_{0}\pi\lambda_{0}^{2}\omega^{4}b^{4}}{6c} \ \ \ \ \ (6)$

Calculating the fields and Poynting vector is more complicated, as they both change with time.

# Electric dipole radiation from an arbitrary source

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 8.

Having examined electromagnetic radiation from an oscillating electric dipole, we can now look at radiation from an arbitrary source of moving charges. The derivation of the results is rather long and Griffiths treats it in detail in his section 11.1.4 so I won’t go over it all again here, except to point out the key assumptions made in the derivation.

 $\displaystyle V\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}\int\frac{\rho\left(\mathbf{r}',t_{r}\right)}{d}d^{3}\mathbf{r}'\ \ \ \ \ (1)$ $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\int\frac{\mathbf{J}\left(\mathbf{r}',t_{r}\right)}{d}d^{3}\mathbf{r}' \ \ \ \ \ (2)$

where

$\displaystyle t_{r}\equiv t-\frac{d}{c} \ \ \ \ \ (3)$

and

 $\displaystyle d$ $\displaystyle \equiv$ $\displaystyle \left|\mathbf{r}-\mathbf{r}'\right|\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\left(x-x'\right)^{2}+\left(y-y'\right)^{2}+\left(z-z'\right)^{2}}\ \ \ \ \ (5)$ $\displaystyle \hat{\mathbf{d}}$ $\displaystyle =$ $\displaystyle \frac{\mathbf{r}-\mathbf{r}'}{d} \ \ \ \ \ (6)$

The first assumption is that the overall size of the charge distribution is much smaller than the distance to the observer, so that

$\displaystyle r_{max}'\ll r \ \ \ \ \ (7)$

This allows us to approximate by saving only up to first order terms in ${r'}$.

The second approximation is that ${r_{max}'}$ is much less than all the terms

$\displaystyle r_{max}'\ll\frac{c}{\left|\left(d^{n}\rho/dt^{n}\right)/\dot{\rho}\right|^{1/\left(n-1\right)}} \ \ \ \ \ (8)$

for ${n\ge2}$. For an oscillating system, ${\rho\left(t\right)=A\cos\omega t}$ so

$\displaystyle \frac{d^{n}\rho}{dt^{n}}=\left(-1\right)^{n}\omega^{n}\rho\left(t\right) \ \ \ \ \ (9)$

so

$\displaystyle \left|\frac{1}{\dot{\rho}}\frac{d^{n}\rho}{dt^{n}}\right|^{1/\left(n-1\right)}=\omega \ \ \ \ \ (10)$

and this assumption is equivalent to ${r_{max}'\ll\lambda}$ that we made in analyzing the oscillating dipole. In practice, it means that we keep up to first order terms in ${r'}$.

After making these two assumptions, we arrive at approximate formulas for the potentials:

 $\displaystyle V\left(\mathbf{r},t\right)$ $\displaystyle \cong$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}\left[\frac{Q}{r}+\frac{\hat{\mathbf{r}}\cdot\mathbf{p}\left(t-r/c\right)}{r^{2}}+\frac{\hat{\mathbf{r}}\cdot\dot{\mathbf{p}}\left(t-r/c\right)}{rc}\right]\ \ \ \ \ (11)$ $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle \cong$ $\displaystyle \frac{\mu_{0}}{4\pi}\frac{\dot{\mathbf{p}}\left(t-r/c\right)}{r} \ \ \ \ \ (12)$

where ${\mathbf{p}}$ is the dipole moment

$\displaystyle \mathbf{p}=\int\mathbf{r}'\rho\left(\mathbf{r}',t-r/c\right)d^{3}\mathbf{r}' \ \ \ \ \ (13)$

and ${Q}$ is the total charge in the system.

By making a further assumption that ${r}$ itself is very large (essentially approaching infinity, since ultimately we are interested only in radiation that makes it to infinity), we arrive at approximate formulas for the fields:

 $\displaystyle \mathbf{E}\left(\mathbf{r},t\right)$ $\displaystyle \cong$ $\displaystyle \frac{\mu_{0}}{4\pi r}\left[\hat{\mathbf{r}}\times\left(\hat{\mathbf{r}}\times\ddot{\mathbf{p}}\right)\right]\ \ \ \ \ (14)$ $\displaystyle \mathbf{B}\left(\mathbf{r},t\right)$ $\displaystyle \cong$ $\displaystyle -\frac{\mu_{0}}{4\pi rc}\left(\hat{\mathbf{r}}\times\ddot{\mathbf{p}}\right) \ \ \ \ \ (15)$

where in both cases ${\ddot{\mathbf{p}}}$ is evaluated at the retarded time ${t-r/c}$. Note that the fields depend on the second time derivative of the dipole moment, which means that no radiation is produced unless the charges are accelerating. The only way to accelerate something is, of course, to apply a force to it so we are doing work on the system, and this work is being converted (at least partly) into radiation.

If we use spherical coordinates with the ${z}$ axis in the direction of ${\ddot{\mathbf{p}}}$ then the fields can be written as

 $\displaystyle \mathbf{E}\left(\mathbf{r},t\right)$ $\displaystyle \cong$ $\displaystyle \frac{\mu_{0}\ddot{p}\left(t-r/c\right)}{4\pi r}\sin\theta\hat{\boldsymbol{\theta}}\ \ \ \ \ (16)$ $\displaystyle \mathbf{B}\left(\mathbf{r},t\right)$ $\displaystyle \cong$ $\displaystyle \frac{\mu_{0}\ddot{p}\left(t-r/c\right)}{4\pi rc}\sin\theta\hat{\boldsymbol{\phi}} \ \ \ \ \ (17)$

The Poynting vector is

$\displaystyle \mathbf{S}=\frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B}\cong\frac{\mu_{0}\ddot{p}^{2}}{16\pi^{2}c}\frac{\sin^{2}\theta}{r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (18)$

and the total radiated power is

$\displaystyle P=\int\mathbf{S}\cdot d\mathbf{a}\cong\frac{\mu_{0}\ddot{p}^{2}}{6\pi c} \ \ \ \ \ (19)$

Example We can apply these formulas to the case of the rotating dipole. In that case, we had a dipole rotating in the ${xy}$ plane, so its dipole moment is given by

$\displaystyle \mathbf{p}\left(t-r/c\right)=p_{0}\left(\cos\omega\left(t-r/c\right)\hat{\mathbf{x}}+\sin\omega\left(t-r/c\right)\hat{\mathbf{y}}\right) \ \ \ \ \ (20)$

Therefore

$\displaystyle \ddot{\mathbf{p}}=-p_{0}\omega^{2}\left(\cos\omega\left(t-r/c\right)\hat{\mathbf{x}}+\sin\omega\left(t-r/c\right)\hat{\mathbf{y}}\right) \ \ \ \ \ (21)$

so from 14 and 15 we get

 $\displaystyle \mathbf{E}\left(\mathbf{r},t\right)$ $\displaystyle \cong$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi r}\left[\hat{\mathbf{r}}\times\left(\hat{\mathbf{r}}\times\left(\cos\omega\left(t-r/c\right)\hat{\mathbf{x}}+\sin\omega\left(t-r/c\right)\hat{\mathbf{y}}\right)\right)\right]\ \ \ \ \ (22)$ $\displaystyle \mathbf{B}\left(\mathbf{r},t\right)$ $\displaystyle \cong$ $\displaystyle \frac{\mu_{0}p_{0}\omega^{2}}{4\pi rc}\left(\hat{\mathbf{r}}\times\left(\cos\omega\left(t-r/c\right)\hat{\mathbf{x}}+\sin\omega\left(t-r/c\right)\hat{\mathbf{y}}\right)\right) \ \ \ \ \ (23)$

which are the same equations we got earlier (after swapping the orders of the cross products):

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi r}\left[\left(\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\hat{\mathbf{x}}+\sin\left(\omega\left(t-\frac{r}{c}\right)\right)\hat{\mathbf{y}}\right)\times\hat{\mathbf{r}}\right]\times\hat{\mathbf{r}}\ \ \ \ \ (24)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi rc}\left[\left(\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\hat{\mathbf{x}}+\sin\left(\omega\left(t-\frac{r}{c}\right)\right)\hat{\mathbf{y}}\right)\times\hat{\mathbf{r}}\right] \ \ \ \ \ (25)$

In this case, it’s not convenient to use the spherical coordinate forms for the fields, since the direction of the dipole moment (and hence its second derivative) is changing with time. However, since the power 19 is obtained by integrating over all angles, it does give the same result, since

 $\displaystyle \ddot{p}^{2}$ $\displaystyle =$ $\displaystyle p_{0}^{2}\omega^{4}\ \ \ \ \ (26)$ $\displaystyle P$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}p_{0}^{2}\omega^{4}}{6\pi c} \ \ \ \ \ (27)$

which is the same as we got earlier.

# Radiation from a magnetic dipole composed of monopoles

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 7

We’ve seen that the fields produced by an oscillating magnetic dipole are

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}\sin\theta}{4\pi r}\left\{ \frac{\omega^{2}}{c}\cos\left[\omega\left(t-r/c\right)\right]+\frac{\omega}{r}\sin\left[\omega\left(t-r/c\right)\right]\right\} \hat{\boldsymbol{\phi}}\ \ \ \ \ (2)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}\cos\theta}{2\pi r^{2}}\left[\frac{1}{r}\cos\left[\omega\left(t-r/c\right)\right]-\frac{\omega}{c}\sin\left[\omega\left(t-r/c\right)\right]\right]\hat{\mathbf{r}}+\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\mu_{0}m_{0}\sin\theta}{4\pi r}\left[\left(\frac{1}{r^{2}}-\frac{\omega^{2}}{c^{2}}\right)\cos\left[\omega\left(t-r/c\right)\right]-\frac{\omega}{rc}\sin\left[\omega\left(t-r/c\right)\right]\right]\hat{\boldsymbol{\theta}}\nonumber$

By making the approximation that the observation distance ${r}$ is much larger than the wavelength of radiation, so that ${r\gg c/\omega}$, these formulas simplify to

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}\omega^{2}\sin\theta}{4\pi cr}\hat{\boldsymbol{\phi}}\cos\left[\omega\left(t-r/c\right)\right]\ \ \ \ \ (5)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}m_{0}\omega^{2}\sin\theta}{4\pi c^{2}r}\hat{\boldsymbol{\theta}}\cos\left[\omega\left(t-r/c\right)\right] \ \ \ \ \ (6)$

Now let’s return to the fantasy world where magnetic monopoles exist, so that another way we can create a magnetic dipole is to connect two magnetic charges by a wire and then drive charge back and forth between the ends of the wire, in the same way that we did for the electric dipole. Earlier, we’ve seen that if we include magnetic charge in Maxwell’s equations, the duality transformation produces fields that still satisfy Maxwell’s equations:

 $\displaystyle \mathbf{E}^{\prime}$ $\displaystyle =$ $\displaystyle \mathbf{E}\cos\alpha+c\mathbf{B}\sin\alpha\ \ \ \ \ (7)$ $\displaystyle c\mathbf{B}^{\prime}$ $\displaystyle =$ $\displaystyle c\mathbf{B}\cos\alpha-\mathbf{E}\sin\alpha\ \ \ \ \ (8)$ $\displaystyle cq_{e}^{\prime}$ $\displaystyle =$ $\displaystyle cq_{e}\cos\alpha+q_{m}\sin\alpha\ \ \ \ \ (9)$ $\displaystyle q_{m}^{\prime}$ $\displaystyle =$ $\displaystyle q_{m}\cos\alpha-cq_{e}\sin\alpha \ \ \ \ \ (10)$

where ${\alpha}$ is a rotation angle in ${\mathbf{E}-\mathbf{B}}$ space. If we start with the fields generated by an oscillating electric dipole and then choose ${\alpha=\frac{\pi}{2}}$ so that we convert all electric charge into magnetic charge, we can generate the fields that would be produced by an oscillating magnetic dipole constructed using magnetic charge as described above. The original fields for the electric dipole are

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V-\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi}\frac{\sin\theta}{r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\hat{\boldsymbol{\theta}}\ \ \ \ \ (12)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi c}\frac{\sin\theta}{r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\hat{\boldsymbol{\phi}} \ \ \ \ \ (14)$

The required transformations with ${\alpha=\frac{\pi}{2}}$ are

 $\displaystyle \mathbf{E}'$ $\displaystyle =$ $\displaystyle c\mathbf{B}\ \ \ \ \ (15)$ $\displaystyle c\mathbf{B}'$ $\displaystyle =$ $\displaystyle -\mathbf{E}\ \ \ \ \ (16)$ $\displaystyle cq_{e}'$ $\displaystyle =$ $\displaystyle q_{m}\ \ \ \ \ (17)$ $\displaystyle q_{m}'$ $\displaystyle =$ $\displaystyle -cq_{e} \ \ \ \ \ (18)$

As the electric dipole moment ${p_{0}=q_{e}l}$ is the product of an electric charge ${q_{e}}$ and the length ${l}$ of the wire joining the two charges, it transforms in the same way as ${q_{e}}$ so we have, if we take the magnetic moment to be ${m_{0}=q_{m}'l}$:

$\displaystyle m_{0}=-cp_{0} \ \ \ \ \ (19)$

Applying these transformations to 12 and 14 we get

 $\displaystyle \mathbf{E}'$ $\displaystyle =$ $\displaystyle -c\frac{\mu_{0}\left(-m_{0}/c\right)\omega^{2}}{4\pi c}\frac{\sin\theta}{r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\hat{\boldsymbol{\phi}}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}\omega^{2}\sin\theta}{4\pi cr}\hat{\boldsymbol{\phi}}\cos\left[\omega\left(t-r/c\right)\right]\ \ \ \ \ (21)$ $\displaystyle \mathbf{B}'$ $\displaystyle =$ $\displaystyle -\left(-\frac{1}{c}\right)\frac{\mu_{0}\left(-m_{0}/c\right)\omega^{2}}{4\pi}\frac{\sin\theta}{r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\hat{\boldsymbol{\theta}}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}m_{0}\omega^{2}\sin\theta}{4\pi c^{2}r}\hat{\boldsymbol{\theta}}\cos\left[\omega\left(t-r/c\right)\right] \ \ \ \ \ (23)$

These fields are the same as 5 and 6 that we got from the current loop. Thus we can’t tell whether magnetic dipole radiation is coming from a current loop or from magnetic monopoles.