## Electric quadrupole radiation

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

$\displaystyle P=\frac{\mu_{0}e^{2}g^{2}}{6\pi c}=5.7\times10^{-54}\mbox{ J s}^{-1} \ \ \ \ \ (5)$

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

$\displaystyle t=\sqrt{\frac{2\times0.01}{9.8}}=0.319\mbox{ s} \ \ \ \ \ (6)$

so the total energy radiated is

$\displaystyle Pt=1.82\times10^{-54}\mbox{ J} \ \ \ \ \ (7)$

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

$\displaystyle \frac{Pt}{V}=2\times10^{-23} \ \ \ \ \ (9)$

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.

To calculate the retarded potentials we start with

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

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.

## Radiation resistance of an oscillating magnetic dipole

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 6

We can find the radiation resistance of an oscillating magnetic dipole produced by an AC current in a circular wire loop of radius ${b}$ in the same way as for an oscillating electric dipole. The average power radiated by the magnetic dipole is the integral of the intensity over a sphere, so we have

 $\displaystyle \left\langle P\right\rangle$ $\displaystyle =$ $\displaystyle \int\left\langle \mathbf{S}\right\rangle \cdot d\mathbf{a}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\frac{\mu_{0}m_{0}^{2}\omega^{4}\sin^{2}\theta}{32\pi^{2}c^{3}r^{2}}\hat{\mathbf{r}}\cdot d\mathbf{a}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}^{2}\omega^{4}}{32\pi^{2}c^{3}}2\pi\int_{0}^{\pi}\frac{\sin^{2}\theta}{r^{2}}r^{2}\sin\theta d\theta\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}^{2}\omega^{4}}{12\pi c^{3}} \ \ \ \ \ (4)$

The resistance ${R}$ required in the wire loop to generate the same power loss through heat is given by ${P=\left\langle I^{2}\right\rangle R}$ where the current is

$\displaystyle I\left(t\right)=I_{0}\cos\omega t \ \ \ \ \ (5)$

with the maximum current ${I_{0}}$ given in terms of the maximum magnetic moment ${m_{0}}$:

$\displaystyle I_{0}=\frac{m_{0}}{\pi b^{2}} \ \ \ \ \ (6)$

Since the average of ${\cos^{2}x}$ over a complete cycle is ${\frac{1}{2}}$ we get

 $\displaystyle \left\langle I^{2}\right\rangle R$ $\displaystyle =$ $\displaystyle \frac{m_{0}^{2}}{2\pi^{2}b^{4}}R\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}^{2}\omega^{4}}{12\pi c^{3}}\ \ \ \ \ (8)$ $\displaystyle R$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}\pi\omega^{4}b^{4}}{6c^{3}}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{8\mu_{0}\pi^{5}c}{3}\frac{b^{4}}{\lambda^{4}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3.08\times10^{5}\frac{b^{4}}{\lambda^{4}}\;\Omega \ \ \ \ \ (11)$

where in the fourth line we used ${\omega/c=2\pi/\lambda}$.

To compare this to the electric dipole’s radiation resistance, which is given in terms of the length ${l}$ of the wire joining the two charges as:

$\displaystyle R_{e}=787\frac{l^{2}}{\lambda^{2}}\;\Omega \ \ \ \ \ (12)$

we can take ${l=2\pi b}$ so the lengths of wire in the two cases are the same. Then

 $\displaystyle \frac{R_{e}}{R_{m}}$ $\displaystyle =$ $\displaystyle \frac{787\left(2\pi\right)^{2}}{3.08\times10^{5}}\frac{\lambda^{2}}{b^{2}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.1\frac{\lambda^{2}}{b^{2}} \ \ \ \ \ (14)$

Since we’re assuming that ${b\ll\lambda}$, ${R_{e}\gg R_{m}}$.

## Fields of an oscillating magnetic dipole

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 5

We can analyze an oscillating magnetic dipole in a similar way to the electric dipole. We begin with a small circular current loop of radius ${b}$ in the ${xy}$ plane, centred at the origin. The current is driven to be alternating, so that

$\displaystyle I\left(t\right)=I_{0}\cos\omega t \ \ \ \ \ (1)$

The magnetic dipole moment of a current loop is

$\displaystyle \mathbf{m}=I\mathbf{a} \ \ \ \ \ (2)$

where ${\mathbf{a}}$ is the vector area of the loop, which for a planar circular loop is just ${\pi b^{2}\hat{\mathbf{z}}}$. Thus

 $\displaystyle \mathbf{m}\left(t\right)$ $\displaystyle =$ $\displaystyle \pi b^{2}I_{0}\cos\left(\omega t\right)\hat{\mathbf{z}}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle m_{0}\cos\left(\omega t\right)\hat{\mathbf{z}} \ \ \ \ \ (4)$

If the loop is electrically neutral, the electric potential is ${V=0}$, so we need to calculate only ${\mathbf{A}}$. The retarded potential is

$\displaystyle \mathbf{A}\left(\mathbf{r},t\right)=\frac{\mu_{0}}{4\pi}\int\frac{I_{0}\cos\omega\left(t-d/c\right)}{d}d\boldsymbol{\ell}' \ \ \ \ \ (5)$

where the integral is taken around the loop and the retarded time is

$\displaystyle t_{r}\equiv t-\frac{d}{c} \ \ \ \ \ (6)$

and

 $\displaystyle d$ $\displaystyle \equiv$ $\displaystyle \left|\mathbf{r}-\mathbf{r}'\right|\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\left(x-x'\right)^{2}+\left(y-y'\right)^{2}+\left(z-z'\right)^{2}}\ \ \ \ \ (8)$ $\displaystyle \hat{\mathbf{d}}$ $\displaystyle =$ $\displaystyle \frac{\mathbf{r}-\mathbf{r}'}{d} \ \ \ \ \ (9)$

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

To work out the integral, we can start by considering ${\mathbf{r}}$ to be some point in the ${xz}$ plane. The line integral can be broken down into pairs of increments on the circle located at ${\left(x,\pm y,0\right)}$, that is, each pair of points is symmetric about the ${x}$ axis. The increment ${d\boldsymbol{\ell}_{-}}$ at ${\left(x,-y,0\right)}$ has components ${\left(dx,dy,0\right)}$ while the increment ${d\boldsymbol{\ell}_{+}}$ at ${\left(x,y,0\right)}$ has components ${\left(-dx,dy,0\right)}$. Thus the ${x}$ components cancel in pairs in the integral, while the ${y}$ components add in pairs, so the net result of the integral around the entire circle is a vector pointing in the ${y}$ direction. Since the circle is symmetric about the ${z}$ axis, we can generalize this result to deduce that ${\mathbf{A}}$ has a direction that is always tangential to the circle, which means that, in spherical coordinates, it is in the ${\phi}$ direction and has the same magnitude at all points around the circle.

Griffiths goes through the calculation of ${\mathbf{A}}$ in detail in his section 11.1.3 so I won’t repeat that here, other than to note that he uses the same approximations as were used with the electric dipole, namely that the radius of the loop ${b}$ is much less than the observation distance ${r}$, and that ${b}$ is also much smaller than the wavelength of the radiation, represented by the condition ${b\ll c/\omega}$. The result is

$\displaystyle \mathbf{A}\left(r,\theta,t\right)=\frac{\mu_{0}m_{0}}{4\pi}\frac{\sin\theta}{r}\left[\frac{1}{r}\cos\left[\omega\left(t-r/c\right)\right]-\frac{\omega}{c}\sin\left[\omega\left(t-r/c\right)\right]\right]\hat{\boldsymbol{\phi}} \ \ \ \ \ (10)$

At this stage, Griffiths invokes a further approximation by assuming that ${r\gg c/\omega}$ (observer is much further away than the wavelength of the radiation). However, we can calculate the fields without making that approximation to see how much of an effect that approximation has. Since ${V=0}$ we have

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}\sin\theta}{4\pi r}\left\{ \frac{\omega^{2}}{c}\cos\left[\omega\left(t-r/c\right)\right]+\frac{\omega}{r}\sin\left[\omega\left(t-r/c\right)\right]\right\} \hat{\boldsymbol{\phi}}\ \ \ \ \ (12)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}\cos\theta}{2\pi r^{2}}\left[\frac{1}{r}\cos\left[\omega\left(t-r/c\right)\right]-\frac{\omega}{c}\sin\left[\omega\left(t-r/c\right)\right]\right]\hat{\mathbf{r}}+\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\mu_{0}m_{0}\sin\theta}{4\pi r}\left[\left(\frac{1}{r^{2}}-\frac{\omega^{2}}{c^{2}}\right)\cos\left[\omega\left(t-r/c\right)\right]-\frac{\omega}{rc}\sin\left[\omega\left(t-r/c\right)\right]\right]\hat{\boldsymbol{\theta}}\nonumber$

These fields have the same form as those we worked out earlier for a spherical wave in vacuum.

The Poynting vector can be worked out and simplified using Maple to combine the trig products using double angle formulas:

 $\displaystyle \mathbf{S}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}m_{0}^{2}\sin^{2}\theta}{32\pi^{2}c^{3}r^{5}}\left[\left(2c^{2}\omega^{2}r-\omega^{4}r^{3}\right)\cos\left[2\omega\left(t-r/c\right)\right]+\left(\omega c^{3}-2\omega^{3}cr^{2}\right)\sin\left[2\omega\left(t-r/c\right)\right]-\omega^{4}r^{3}\right]\hat{\mathbf{r}}+\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\mu_{0}m_{0}^{2}\sin\left(2\theta\right)}{32\pi^{2}c^{3}r^{5}}\left[2c^{2}\omega^{2}r\cos\left[2\omega\left(t-r/c\right)\right]+\left(\omega c^{3}-\omega^{3}cr^{2}\right)\sin\left[2\omega\left(t-r/c\right)\right]\right]\hat{\boldsymbol{\theta}}\nonumber$

To get the average intensity, we integrate ${\mathbf{S}}$ over a complete cycle, that is, for ${t=0}$ to ${t=2\pi/\omega}$ and then multiply by ${\omega/2\pi}$ to get the average. Integrating over one cycle causes each of the double angle trig functions to go through two complete cycles, so they all integrate to zero and the only term that is left is the ${-\omega^{4}r^{3}}$ term in the radial component, so we get

$\displaystyle \left\langle \mathbf{S}\right\rangle =\frac{\mu_{0}m_{0}^{2}\omega^{4}\sin^{2}\theta}{32\pi^{2}c^{3}r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (17)$

This value is the same as that obtained by assuming ${r\gg c/\omega}$ from the start (equation 11.39 in Griffiths).

## Radiation from a rotating dipole

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 4

We can simulate a rotating dipole by superimposing two perpendicular oscillating dipoles. If our rotating dipole is located at the origin and rotates about the ${z}$ axis (so the axis of the dipole lies in the ${xy}$ plane), then we get

$\displaystyle \mathbf{p}=p_{0}\left(\cos\omega t\hat{\mathbf{x}}+\sin\omega t\hat{\mathbf{y}}\right) \ \ \ \ \ (1)$

Since the fields obey the superposition principle (fields from 2 sources just add), we can work through the formulas we found earlier to get the fields and thus the radiated power. To simplify the notation, we’ll use the shorthand

 $\displaystyle c_{\omega}$ $\displaystyle \equiv$ $\displaystyle \cos\left(\omega\left(t-\frac{r}{c}\right)\right)\ \ \ \ \ (2)$ $\displaystyle c_{\theta}$ $\displaystyle \equiv$ $\displaystyle \cos\theta\ \ \ \ \ (3)$ $\displaystyle c_{\phi}$ $\displaystyle \equiv$ $\displaystyle \cos\phi \ \ \ \ \ (4)$

with analogous notation for the sines of these quantities.

The fields for a dipole pointing in an arbitrary direction are

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}\omega^{2}}{4\pi r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\left(\hat{\mathbf{p}}_{0}\times\hat{\mathbf{r}}\right)\times\hat{\mathbf{r}}\ \ \ \ \ (5)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}\omega^{2}}{4\pi rc}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\left(\hat{\mathbf{p}}_{0}\times\hat{\mathbf{r}}\right) \ \ \ \ \ (6)$

Superposing the two perpendicular dipoles we get

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi r}\left[\left(c_{\omega}\hat{\mathbf{x}}+s_{\omega}\hat{\mathbf{y}}\right)\times\hat{\mathbf{r}}\right]\times\hat{\mathbf{r}}\ \ \ \ \ (7)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi rc}\left[\left(c_{\omega}\hat{\mathbf{x}}+s_{\omega}\hat{\mathbf{y}}\right)\times\hat{\mathbf{r}}\right] \ \ \ \ \ (8)$

To do the cross products we convert the rectangular unit vectors to spherical unit vectors:

 $\displaystyle \hat{\mathbf{x}}$ $\displaystyle =$ $\displaystyle s_{\theta}c_{\phi}\hat{\mathbf{r}}+c_{\theta}c_{\phi}\hat{\boldsymbol{\theta}}-s_{\phi}\hat{\boldsymbol{\phi}}\ \ \ \ \ (9)$ $\displaystyle \hat{\mathbf{y}}$ $\displaystyle =$ $\displaystyle s_{\theta}s_{\phi}\hat{\mathbf{r}}+c_{\theta}s_{\phi}\hat{\boldsymbol{\theta}}+c_{\phi}\hat{\boldsymbol{\phi}} \ \ \ \ \ (10)$

Then

 $\displaystyle \hat{\mathbf{x}}\times\hat{\mathbf{r}}$ $\displaystyle =$ $\displaystyle -c_{\theta}c_{\phi}\hat{\boldsymbol{\phi}}-s_{\phi}\hat{\boldsymbol{\theta}}\ \ \ \ \ (11)$ $\displaystyle \left(\hat{\mathbf{x}}\times\hat{\mathbf{r}}\right)\times\hat{\mathbf{r}}$ $\displaystyle =$ $\displaystyle -c_{\theta}c_{\phi}\hat{\boldsymbol{\theta}}+s_{\phi}\hat{\boldsymbol{\phi}}\ \ \ \ \ (12)$ $\displaystyle \hat{\mathbf{y}}\times\hat{\mathbf{r}}$ $\displaystyle =$ $\displaystyle -c_{\theta}s_{\phi}\hat{\boldsymbol{\phi}}+c_{\phi}\hat{\boldsymbol{\theta}}\ \ \ \ \ (13)$ $\displaystyle \left(\hat{\mathbf{y}}\times\hat{\mathbf{r}}\right)\times\hat{\mathbf{r}}$ $\displaystyle =$ $\displaystyle -c_{\theta}s_{\phi}\hat{\boldsymbol{\theta}}-c_{\phi}\hat{\boldsymbol{\phi}} \ \ \ \ \ (14)$

Plugging everything in and collecting terms we get

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi r}\left[\hat{\boldsymbol{\theta}}c_{\theta}\left(-c_{\omega}c_{\phi}-s_{\omega}s_{\phi}\right)+\hat{\boldsymbol{\phi}}\left(c_{\omega}s_{\phi}-s_{\omega}c_{\phi}\right)\right]\ \ \ \ \ (15)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi rc}\left[-\hat{\boldsymbol{\theta}}\left(c_{\omega}s_{\phi}-s_{\omega}c_{\phi}\right)+\hat{\boldsymbol{\phi}}c_{\theta}\left(-c_{\omega}c_{\phi}-s_{\omega}s_{\phi}\right)\right] \ \ \ \ \ (16)$

Defining

 $\displaystyle E_{\theta}$ $\displaystyle \equiv$ $\displaystyle -c_{\theta}\left(c_{\omega}c_{\phi}+s_{\omega}s_{\phi}\right)\ \ \ \ \ (17)$ $\displaystyle E_{\phi}$ $\displaystyle \equiv$ $\displaystyle \left(c_{\omega}s_{\phi}-s_{\omega}c_{\phi}\right) \ \ \ \ \ (18)$

we can write the fields as

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi r}\left[\hat{\boldsymbol{\theta}}E_{\theta}+\hat{\boldsymbol{\phi}}E_{\phi}\right]\ \ \ \ \ (19)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi rc}\left[-\hat{\boldsymbol{\theta}}E_{\phi}+\hat{\boldsymbol{\phi}}E_{\theta}\right] \ \ \ \ \ (20)$

In this form, it’s obvious that ${\mathbf{E}\cdot\mathbf{B}=0}$, so ${\mathbf{E}}$ and ${\mathbf{B}}$ are perpendicular.

The Poynting vector is

 $\displaystyle \mathbf{S}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{c}\left(\frac{p_{0}\omega^{2}}{4\pi r}\right)^{2}\left(E_{\theta}^{2}+E_{\phi}^{2}\right)\hat{\mathbf{r}}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{c}\left(\frac{p_{0}\omega^{2}}{4\pi r}\right)^{2}\left(c_{\theta}^{2}\left(c_{\omega}c_{\phi}+s_{\omega}s_{\phi}\right)^{2}+\left(c_{\omega}s_{\phi}-s_{\omega}c_{\phi}\right)^{2}\right)\hat{\mathbf{r}} \ \ \ \ \ (23)$

The average energy radiated is the average of ${\mathbf{S}}$ over a single time cycle, so it’s the average over the terms involving ${c_{\omega}}$ and ${s_{\omega}}$. These are of two types: terms involving ${c_{\omega}^{2}}$ or ${s_{\omega}^{2}}$ and the cross terms involving ${s_{\omega}c_{\omega}}$. The average of ${s_{\omega}c_{\omega}}$ over a cycle is zero and the average of ${c_{\omega}^{2}}$ or ${s_{\omega}^{2}}$ is ${\frac{1}{2}}$, so the cross terms contribute nothing and we get

 $\displaystyle \left\langle \mathbf{S}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{c}\left(\frac{p_{0}\omega^{2}}{4\pi r}\right)^{2}\left[\frac{1}{2}c_{\theta}^{2}\left(c_{\phi}^{2}+s_{\phi}^{2}\right)+\frac{1}{2}\left(c_{\phi}^{2}+s_{\phi}^{2}\right)\right]\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{2c}\left(\frac{p_{0}\omega^{2}}{4\pi r}\right)^{2}\left(1+\cos^{2}\theta\right) \ \ \ \ \ (25)$

The average radiated power is maximum in the ${\pm z}$ directions where ${\theta=0,\pi}$ and minimum (though not zero) in the ${xy}$ plane, where ${\theta=\frac{\pi}{2}}$. There is no dependence on ${\phi}$ which is what we’d expect on average since the dipole rotates uniformly through all values of ${\phi}$. [There is a dependence on ${\phi}$ within each cycle, since the radiated power in a given azimuthal direction depends on where the dipole is in its rotation.]

The total average radiated power is

 $\displaystyle \left\langle P\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{2c}\left(\frac{p_{0}\omega^{2}}{4\pi}\right)^{2}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{1+\cos^{2}\theta}{r^{2}}r^{2}\sin\theta d\phi d\theta\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}p_{0}^{2}\omega^{4}}{6\pi c} \ \ \ \ \ (27)$

This is exactly twice the power from a single oscillating dipole. Although power doesn’t ordinarily obey the superposition principle since it depends on the product of ${\mathbf{E}}$ and ${\mathbf{B}}$, it does here because the cross terms in 23 average out to zero over a time cycle, since the two perpendicular dipoles are ${\frac{\pi}{2}}$ out of phase. If they were exactly in phase, we would replace ${s_{\omega}}$ by ${c_{\omega}}$ everywhere in the calculation, and then the cross terms wouldn’t average out to zero and the combined power would not be twice the individual power.

## Radiation resistance

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

The model of an oscillating dipole that we’ve been using consists of two charges exchanging charge by passing a current along a wire of length ${l}$ joining the charges. The average total power radiated by the dipole is

$\displaystyle \left\langle P\right\rangle =\frac{\mu_{0}\omega^{4}p_{0}^{2}}{12\pi c} \ \ \ \ \ (1)$

where ${p_{0}}$ is the maximum dipole moment

$\displaystyle p_{0}=q_{0}l \ \ \ \ \ (2)$

Here ${q_{0}}$ is the maximum charge at one end of the dipole, and the charge oscillates according to

$\displaystyle q\left(t\right)=q_{0}\cos\omega t \ \ \ \ \ (3)$

If the energy lost through radiation were to be lost instead by heat generated by the current passing through the wire joining the charges, what would the resistance of the wire need to be? This resistance is known as the radiation resistance. The power generated by a current ${I}$ passing through a resistor ${R}$ is

$\displaystyle P=I^{2}R \ \ \ \ \ (4)$

and in this case, the current is given by

$\displaystyle I=\frac{dq}{dt}=-\omega q_{0}\sin\omega t \ \ \ \ \ (5)$

We’re interested in the average power, so we want the average of ${I^{2}}$ over a single cycle. The average of ${\sin^{2}x}$ over a cycle is ${\frac{1}{2}}$ so we get

 $\displaystyle \left\langle I^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\omega^{2}q_{0}^{2}\ \ \ \ \ (6)$ $\displaystyle \left\langle P\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\omega^{2}q_{0}^{2}R \ \ \ \ \ (7)$

Equating this to 1 we get

 $\displaystyle R$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}\omega^{2}p_{0}^{2}}{6\pi cq_{0}^{2}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}c\omega^{2}q_{0}^{2}l^{2}}{6\pi c^{2}q_{0}^{2}}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}c}{6\pi}\left(\frac{2\pi}{\lambda}\right)^{2}l^{2}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi\mu_{0}c}{3}\frac{l^{2}}{\lambda^{2}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 787\frac{l^{2}}{\lambda^{2}}\;\Omega \ \ \ \ \ (12)$

where in line 3 we used ${\omega/c=2\pi/\lambda}$.

For a typical radio, we can take the length of a wire connecting components to be around ${l=5\mbox{ cm}}$, while radio waves typically have wavelengths around 1 km, so the radiative resistance is around

$\displaystyle R=787\left(\frac{0.05}{10^{3}}\right)^{2}=2\times10^{-6}\;\Omega \ \ \ \ \ (13)$

This is much smaller than typical resistances in a radio’s circuits.

## Fields and radiated power from an oscillating electric dipole

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 2

The potentials for an oscillating dipole at a large distance from the dipole are

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

These formulas apply in the special case where the dipole axis is the ${z}$ axis, so that the dipole moment is

$\displaystyle \mathbf{p}=p_{0}\cos\left(\omega t\right)\hat{\mathbf{z}} \ \ \ \ \ (3)$

We can rewrite these formulas for a dipole pointing in any direction by noting that

$\displaystyle p_{0}\cos\theta=\mathbf{p}_{0}\cdot\hat{\mathbf{r}} \ \ \ \ \ (4)$

so

 $\displaystyle V\left(r,\theta,t\right)$ $\displaystyle =$ $\displaystyle -\frac{\omega}{4\pi\epsilon_{0}rc}\left(\mathbf{p}_{0}\cdot\hat{\mathbf{r}}\right)\sin\left(\omega\left(t-\frac{r}{c}\right)\right)\ \ \ \ \ (5)$ $\displaystyle \mathbf{A}\left(r,\theta,t\right)$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}\omega}{4\pi r}\mathbf{p}_{0}\sin\left(\omega\left(t-\frac{r}{c}\right)\right) \ \ \ \ \ (6)$

The fields can be calculated from the potentials using straightforward differentiation. Griffiths shows the details in his section 11.1.2. After assuming that ${r\gg\frac{c}{\omega}}$ (equivalent to assuming that the observation point is much greater than the wavelength of the radiation) we get from 1 and 2:

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V-\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi}\frac{\sin\theta}{r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\hat{\boldsymbol{\theta}}\ \ \ \ \ (8)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi c}\frac{\sin\theta}{r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\hat{\boldsymbol{\phi}} \ \ \ \ \ (10)$

Note that ${\mathbf{E}}$ and ${\mathbf{B}}$ are perpendicular and in phase, and that ${E/B=c}$ just as with plane waves in vacuum. We can write these equations for general dipole directions by noting that

 $\displaystyle \hat{\mathbf{z}}\times\hat{\mathbf{r}}$ $\displaystyle =$ $\displaystyle \sin\theta\hat{\boldsymbol{\phi}}\ \ \ \ \ (11)$ $\displaystyle \hat{\boldsymbol{\phi}}\times\hat{\mathbf{r}}$ $\displaystyle =$ $\displaystyle \hat{\boldsymbol{\theta}}\ \ \ \ \ (12)$ $\displaystyle \left(\hat{\mathbf{z}}\times\hat{\mathbf{r}}\right)\times\hat{\mathbf{r}}$ $\displaystyle =$ $\displaystyle \sin\theta\hat{\boldsymbol{\theta}} \ \ \ \ \ (13)$

Therefore

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}\omega^{2}}{4\pi r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\left(\hat{\mathbf{p}}_{0}\times\hat{\mathbf{r}}\right)\times\hat{\mathbf{r}}\ \ \ \ \ (14)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}\omega^{2}}{4\pi rc}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\left(\hat{\mathbf{p}}_{0}\times\hat{\mathbf{r}}\right) \ \ \ \ \ (15)$

The energy radiated per unit area per unit time is given by the Poynting vector:

 $\displaystyle \mathbf{S}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{c}\left[\frac{p_{0}\omega^{2}}{4\pi}\frac{\sin\theta}{r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\right]^{2}\hat{\mathbf{r}} \ \ \ \ \ (17)$

For a general dipole direction, this is

$\displaystyle \mathbf{S}=\frac{\mu_{0}}{c}\left[\frac{\omega^{2}}{4\pi}\frac{\left|\mathbf{p}_{0}\times\hat{\mathbf{r}}\right|}{r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\right]^{2}\hat{\mathbf{r}} \ \ \ \ \ (18)$

The intensity is the average of ${\mathbf{S}}$ over a single time cycle (that is, over a time ${2\pi/\omega}$). The average of ${\cos^{2}x}$ over a single cycle is ${\frac{1}{2}}$, so

 $\displaystyle \left\langle \mathbf{S}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{2c}\left[\frac{\omega^{2}}{4\pi}\frac{p_{0}\sin\theta}{r}\right]^{2}\hat{\mathbf{r}} \ \ \ \ \ (19)$

or in direction-independent form

$\displaystyle \left\langle \mathbf{S}\right\rangle =\frac{\mu_{0}}{2c}\left[\frac{\omega^{2}}{4\pi}\frac{\left|\mathbf{p}_{0}\times\hat{\mathbf{r}}\right|}{r}\right]^{2}\hat{\mathbf{r}} \ \ \ \ \ (20)$

There is no radiation along the dipole’s axis, and the maximum radiation occurs perpendicular to the axis.

The average total power radiated is the surface integral of ${\left\langle \mathbf{S}\right\rangle }$ over a sphere of radius ${r}$, so we get from 19

 $\displaystyle \left\langle P\right\rangle$ $\displaystyle =$ $\displaystyle \int\frac{\mu_{0}}{2c}\left[\frac{\omega^{2}}{4\pi}\frac{p_{0}\sin\theta}{r}\right]^{2}r^{2}\sin\theta\hat{\mathbf{r}}\cdot d\mathbf{a}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}\omega^{4}p_{0}^{2}}{32\pi^{2}c}\int_{0}^{\pi}\int_{0}^{2\pi}\sin^{3}\theta d\phi d\theta\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}\omega^{4}p_{0}^{2}}{12\pi c} \ \ \ \ \ (23)$

The result is independent of distance ${r}$ from the dipole, so we see that this power remains constant out to infinity.