Synchrotron radiation

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 16.

One common instance of an accelerated charge is a charge moving in a circle. In this case the particle’s instantaneous velocity {\mathbf{v}} is always perpendicular to its instantaneous acceleration {\mathbf{a}}. This is known as synchrotron radiation, since it is the radiation given off by particles in a synchrotron particle accelerator, where charged particles move in circular orbits between the poles of a magnet.

We can use the Liénard formula to work out the power radiated by such a charge:

\displaystyle   \frac{dP}{d\Omega} \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}}{16\pi^{2}c^{2}}\frac{\left|\hat{\boldsymbol{\mathfrak{r}}}\times\left(\mathbf{u}\times\mathbf{a}\right)\right|^{2}}{\left(\hat{\boldsymbol{\mathfrak{r}}}\cdot\mathbf{u}\right)^{5}}\ \ \ \ \ (1)
\displaystyle  P \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}\gamma^{6}}{6\pi c}\left[a^{2}-\frac{\left|\mathbf{v}\times\mathbf{a}\right|^{2}}{c^{2}}\right] \ \ \ \ \ (2)

At one instant of time, we can take

\displaystyle   \mathbf{v} \displaystyle  = \displaystyle  v\hat{\mathbf{z}}\ \ \ \ \ (3)
\displaystyle  \mathbf{a} \displaystyle  = \displaystyle  a\hat{\mathbf{x}}\ \ \ \ \ (4)
\displaystyle  \hat{\boldsymbol{\mathfrak{r}}} \displaystyle  = \displaystyle  s_{\theta}c_{\phi}\hat{\mathbf{x}}+s_{\theta}s_{\phi}\hat{\mathbf{y}}+c_{\theta}\hat{\mathbf{z}}\ \ \ \ \ (5)
\displaystyle  \mathbf{u} \displaystyle  = \displaystyle  c\hat{\boldsymbol{\mathfrak{r}}}-\mathbf{v} \ \ \ \ \ (6)

where we’re using our usual shorthand for trig functions: {s_{\theta}\equiv\sin\theta}, {c_{\theta}\equiv\cos\theta} and so on. We can now work out the components of 1:

\displaystyle   \hat{\boldsymbol{\mathfrak{r}}}\cdot\mathbf{u} \displaystyle  = \displaystyle  c\hat{\boldsymbol{\mathfrak{r}}}\cdot\hat{\boldsymbol{\mathfrak{r}}}-vc_{\theta}\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  c-vc_{\theta}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  c\left(1-\beta c_{\theta}\right)\ \ \ \ \ (9)
\displaystyle  \mathbf{u}\times\mathbf{a} \displaystyle  = \displaystyle  c\hat{\boldsymbol{\mathfrak{r}}}\times a\hat{\mathbf{x}}-v\hat{\mathbf{z}}\times a\hat{\mathbf{x}}\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  ca\left(-s_{\theta}s_{\phi}\hat{\mathbf{z}}+c_{\theta}\hat{\mathbf{y}}\right)-av\hat{\mathbf{y}}\ \ \ \ \ (11)
\displaystyle  \hat{\boldsymbol{\mathfrak{r}}}\times\left(\mathbf{u}\times\mathbf{a}\right) \displaystyle  = \displaystyle  \left|\begin{array}{ccc} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}}\\ s_{\theta}c_{\phi} & s_{\theta}s_{\phi} & c_{\theta}\\ 0 & cac_{\theta}-av & -cas_{\theta}s_{\phi} \end{array}\right|\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \hat{\mathbf{x}}\left[-cas_{\theta}^{2}s_{\phi}^{2}-c_{\theta}\left(cac_{\theta}-av\right)\right]+\ \ \ \ \ (13)
\displaystyle  \displaystyle  \displaystyle  \hat{\mathbf{y}}cas_{\theta}^{2}s_{\phi}c_{\phi}+\hat{\mathbf{z}}\left(cas_{\theta}c_{\theta}c_{\phi}-avs_{\theta}c_{\phi}\right)\nonumber

Taking the square of this last vector leads to a lengthy expression which can be simplified by applying {s^{2}+c^{2}=1} repeatedly. We get, using {\beta=v/c}:

\displaystyle   \frac{1}{a^{2}c^{2}}\left|\hat{\boldsymbol{\mathfrak{r}}}\times\left(\mathbf{u}\times\mathbf{a}\right)\right|^{2} \displaystyle  = \displaystyle  s_{\theta}^{4}c_{\phi}^{4}+1+c_{\theta}^{2}\beta^{2}-2s_{\theta}^{2}c_{\phi}^{2}+2s_{\theta}^{2}c_{\phi}^{2}c_{\theta}\beta-2c_{\theta}\beta+\ \ \ \ \ (14)
\displaystyle  \displaystyle  \displaystyle  s_{\theta}^{4}s_{\phi}^{2}c_{\phi}^{2}+s_{\theta}^{2}c_{\phi}^{2}c_{\theta}^{2}-2s_{\theta}^{2}c_{\phi}^{2}c_{\theta}\beta+s_{\theta}^{2}c_{\phi}^{2}\beta^{2}\nonumber

We can simplify this as follows. The first and seventh terms combine to give

\displaystyle  s_{\theta}^{4}c_{\phi}^{4}+s_{\theta}^{4}s_{\phi}^{2}c_{\phi}^{2}=s_{\theta}^{4}c_{\phi}^{2}\left(c_{\phi}^{2}+s_{\phi}^{2}\right)=s_{\theta}^{4}c_{\phi}^{2} \ \ \ \ \ (15)

Combining this with the eighth term:

\displaystyle  s_{\theta}^{4}c_{\phi}^{2}+s_{\theta}^{2}c_{\phi}^{2}c_{\theta}^{2}=s_{\theta}^{2}c_{\phi}^{2}\left(s_{\theta}^{2}+c_{\theta}^{2}\right)=s_{\theta}^{2}c_{\phi}^{2} \ \ \ \ \ (16)

Combining this with the fourth and last terms we get

\displaystyle  s_{\theta}^{2}c_{\phi}^{2}-2s_{\theta}^{2}c_{\phi}^{2}+s_{\theta}^{2}c_{\phi}^{2}\beta^{2}=-\left(1-\beta^{2}\right)s_{\theta}^{2}c_{\phi}^{2} \ \ \ \ \ (17)

The second, third and sixth terms combine to give

\displaystyle  1+c_{\theta}^{2}\beta^{2}-2c_{\theta}\beta=\left(1-\beta c_{\theta}\right)^{2} \ \ \ \ \ (18)

Finally, the fifth and ninth terms cancel, so we’re left with

\displaystyle  \frac{1}{a^{2}c^{2}}\left|\hat{\boldsymbol{\mathfrak{r}}}\times\left(\mathbf{u}\times\mathbf{a}\right)\right|^{2}=\left(1-\beta c_{\theta}\right)^{2}-\left(1-\beta^{2}\right)s_{\theta}^{2}c_{\phi}^{2} \ \ \ \ \ (19)

Putting everything together we get

\displaystyle  \frac{dP}{d\Omega}=\frac{\mu_{0}q^{2}a^{2}}{16\pi^{2}c}\frac{\left(1-\beta c_{\theta}\right)^{2}-\left(1-\beta^{2}\right)s_{\theta}^{2}c_{\phi}^{2}}{\left(1-\beta c_{\theta}\right)^{5}} \ \ \ \ \ (20)

To get the total power, we need to integrate this over all solid angles, so we get

\displaystyle  P=\frac{\mu_{0}q^{2}a^{2}}{16\pi^{2}c}\int_{0}^{\pi}\int_{0}^{2\pi}d\phi d\theta s_{\theta}\frac{\left(1-\beta c_{\theta}\right)^{2}-\left(1-\beta^{2}\right)s_{\theta}^{2}c_{\phi}^{2}}{\left(1-\beta c_{\theta}\right)^{5}} \ \ \ \ \ (21)

The integral over {\phi} is easy, using

\displaystyle  \int_{0}^{2\pi}c_{\phi}^{2}d\phi=\pi \ \ \ \ \ (22)

so we’re left with the integral over {\theta}:

\displaystyle   P \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}a^{2}}{16\pi c}\int_{0}^{\pi}d\theta\frac{2\left(1-\beta c_{\theta}\right)^{2}-\left(1-\beta^{2}\right)s_{\theta}^{2}}{\left(1-\beta c_{\theta}\right)^{5}}s_{\theta}\ \ \ \ \ (23)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}a^{2}}{16\pi c}\int_{0}^{\pi}d\theta\frac{2\left(1-\beta c_{\theta}\right)^{2}-\left(1-\beta^{2}\right)\left(1-c_{\theta}^{2}\right)}{\left(1-\beta c_{\theta}\right)^{5}}s_{\theta} \ \ \ \ \ (24)

This nasty looking integral can be done by using partial fractions, since it is the ratio of two polynomials in {c_{\theta}}. I did the integral using Maple, but if you’re interested in doing it by hand, the partial fraction decomposition is

\displaystyle  \frac{2\left(1-\beta c_{\theta}\right)^{2}-\left(1-\beta^{2}\right)\left(1-c_{\theta}^{2}\right)}{\left(1-\beta c_{\theta}\right)^{5}}=-{\frac{\pi\,\left({\beta}^{4}-2\,{\beta}^{2}+1\right)}{{\beta}^{2}\left(\beta\,\cos\left(\theta\right)-1\right)^{5}}}+2\,{\frac{\pi\,\left({\beta}^{2}-1\right)}{{\beta}^{2}\left(\beta\,\cos\left(\theta\right)-1\right)^{4}}}-{\frac{\pi\,\left({\beta}^{2}+1\right)}{\left(\beta\,\cos\left(\theta\right)-1\right)^{3}{\beta}^{2}}} \ \ \ \ \ (25)

The presence of the extra {\sin\theta} from the solid angle element saves the day, since it multiplies each term in the partial fraction expansion, providing the derivative of {\cos\theta} on the top of each fraction. For example

\displaystyle  \int d\theta\frac{\sin\theta}{\left(\beta\,\cos\left(\theta\right)-1\right)^{5}}=\frac{1}{4\beta\left(\beta\,\cos\left(\theta\right)-1\right)^{4}} \ \ \ \ \ (26)

with the other two terms having similar integrals.

The result of the integral is

\displaystyle   \int_{0}^{\pi}d\theta\frac{2\left(1-\beta c_{\theta}\right)^{2}-\left(1-\beta^{2}\right)\left(1-c_{\theta}^{2}\right)}{\left(1-\beta c_{\theta}\right)^{5}}s_{\theta} \displaystyle  = \displaystyle  \frac{8}{3\left(1-\beta\right)^{2}\left(1+\beta\right)^{2}}\ \ \ \ \ (27)
\displaystyle  \displaystyle  = \displaystyle  \frac{8}{3\left(1-\beta^{2}\right)^{2}}\ \ \ \ \ (28)
\displaystyle  \displaystyle  = \displaystyle  \frac{8\gamma^{4}}{3} \ \ \ \ \ (29)

so we get for the total power

\displaystyle   P \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}a^{2}}{16\pi c}\frac{8\gamma^{4}}{3}\ \ \ \ \ (30)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}a^{2}\gamma^{4}}{6\pi c} \ \ \ \ \ (31)

Liénard’s generalization of the Larmor formula for an accelerating charge

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 15.

For an accelerating charge {q} that is instantaneously at rest, the power radiated is given by the Larmor formula:

\displaystyle  P=\frac{\mu_{0}q^{2}a^{2}}{6\pi c} \ \ \ \ \ (1)

where the acceleration {a} is a function of time. Griffiths shows in his section 11.2.1 that the Larmor formula is the integral of the Poynting vector over a large sphere. Another way of looking at it is that the charge radiates an amount of power {dP} into an element of solid angle {d\Omega=\sin\theta d\theta d\phi} and the formula for this is obtained from the Poynting vector by multiplying by {r^{2}} to make the result independent of distance from the charge.

\displaystyle  \frac{dP}{d\Omega}=\frac{\mu_{0}q^{2}a^{2}}{16\pi^{2}c}\sin^{2}\theta \ \ \ \ \ (2)

We can see from this formula that the angle at which the maximum power is radiated is {\theta_{max}=\frac{\pi}{2}}.

To generalize these formulas to the case where {v\ne0} requires a bit of a slog through the mathematics, but the results are quoted by Griffiths as his equations 11.72 and 11.73:

\displaystyle   \frac{dP}{d\Omega} \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}}{16\pi^{2}c^{2}}\frac{\left|\hat{\boldsymbol{\mathfrak{r}}}\times\left(\mathbf{u}\times\mathbf{a}\right)\right|^{2}}{\left(\hat{\boldsymbol{\mathfrak{r}}}\cdot\mathbf{u}\right)^{5}}\ \ \ \ \ (3)
\displaystyle  P \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}\gamma^{6}}{6\pi c}\left[a^{2}-\frac{\left|\mathbf{v}\times\mathbf{a}\right|^{2}}{c^{2}}\right] \ \ \ \ \ (4)

where {\hat{\boldsymbol{\mathfrak{r}}}} is a unit vector pointing from the charge at the retarded time to the observation point on the enclosing sphere and {\mathbf{u}=c\hat{\boldsymbol{\mathfrak{r}}}-\mathbf{v}}. These formulas are Liénard’s generalization of the Larmor formula.

For a charge with {\mathbf{v}\parallel\mathbf{a}} (at the instant of retarded time), Griffiths shows in his Example 11.3 that

\displaystyle  \frac{dP}{d\Omega}=\frac{\mu_{0}q^{2}a^{2}}{16\pi^{2}c}\frac{\sin^{2}\theta}{\left(1-\beta\cos\theta\right)^{5}} \ \ \ \ \ (5)


where {\beta\equiv v/c}. This formula reduces to 2 when {\beta=0}.

We can find the angle {\theta_{max}} for the case {\beta\ne0} by differentiating 5 and setting the result to 0.

\displaystyle  \frac{d}{d\theta}\frac{dP}{d\Omega}=\frac{\mu_{0}q^{2}a^{2}}{16\pi^{2}c}\left[2\,{\frac{\sin\left(\theta\right)\cos\left(\theta\right)}{\left(1-\beta\,\cos\left(\theta\right)\right)^{5}}}-5\,{\frac{\left(\sin\left(\theta\right)\right)^{3}\beta}{\left(1-\beta\,\cos\left(\theta\right)\right)^{6}}}\right]=0 \ \ \ \ \ (6)

The solution {\theta=0} gives the angle of minimum power, so if we take {\theta\ne0} we can cancel off {\sin\theta} and then multiply through by {\left(1-\beta\,\cos\left(\theta\right)\right)^{6}} to get

\displaystyle  2\,\left(1-\beta\,\cos\left(\theta\right)\right)\cos\left(\theta\right)-5\,\left(1-\left(\cos\left(\theta\right)\right)^{2}\right)\beta=0 \ \ \ \ \ (7)

which has the solution

\displaystyle  \theta_{max}=\arccos\left[\frac{\sqrt{15\beta^{2}+1}-1}{3\beta}\right] \ \ \ \ \ (8)

In the ultrarelativistic case, we can write

\displaystyle  \beta=1-x \ \ \ \ \ (9)

where {x\ll1} and expand in a Taylor series (using Maple to do the heavy lifting):

\displaystyle   \frac{\sqrt{15\beta^{2}+1}-1}{3\beta} \displaystyle  = \displaystyle  \frac{\sqrt{15\left(1-x\right)^{2}+1}-1}{3\left(1-x\right)}\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  1-\frac{x}{4}-\frac{27}{128}x^{2}+\mathcal{O}\left(x^{3}\right) \ \ \ \ \ (11)

Since {x} is small, the series on the right is very close to 1, which means that the argument of the arccos is close to 1, so {\theta_{max}} is close to 0, so we can approximate

\displaystyle   \cos\theta_{max} \displaystyle  \approx \displaystyle  1-\frac{\theta_{max}^{2}}{2}\ \ \ \ \ (12)
\displaystyle  \displaystyle  \approx \displaystyle  1-\frac{x}{4}\ \ \ \ \ (13)
\displaystyle  \theta_{max} \displaystyle  \approx \displaystyle  \sqrt{\frac{x}{2}}\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{1-\beta}{2}} \ \ \ \ \ (15)

To compare the power output at the maximum angles in the two cases, we have from 2

\displaystyle  \frac{dP}{d\Omega}_{rest}=\frac{\mu_{0}q^{2}a^{2}}{16\pi^{2}c} \ \ \ \ \ (16)

and from 5

\displaystyle  \frac{dP}{d\Omega}_{rel}=\frac{\mu_{0}q^{2}a^{2}}{16\pi^{2}c}\frac{\sin^{2}\theta_{max}}{\left(1-\beta\cos\theta_{max}\right)^{5}} \ \ \ \ \ (17)


so the ratio is

\displaystyle  \frac{dP/d\Omega_{rel}}{dP/d\Omega_{rest}}=\frac{\sin^{2}\theta_{max}}{\left(1-\beta\cos\theta_{max}\right)^{5}} \ \ \ \ \ (18)

For {\theta_{max}\approx\sqrt{\frac{1-\beta}{2}}\ll1} we can approximate

\displaystyle   \sin^{2}\theta_{max} \displaystyle  \approx \displaystyle  \frac{1-\beta}{2}\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  \frac{x}{2}\ \ \ \ \ (20)
\displaystyle  1-\beta\cos\theta_{max} \displaystyle  \approx \displaystyle  1-\beta\left(1-\frac{1-\beta}{4}\right)\ \ \ \ \ (21)
\displaystyle  \displaystyle  \approx \displaystyle  1-\left(1-x\right)\left(1-\frac{x}{4}\right)\ \ \ \ \ (22)
\displaystyle  \displaystyle  \approx \displaystyle  \frac{5x}{4} \ \ \ \ \ (23)

Therefore

\displaystyle   \frac{dP/d\Omega_{rel}}{dP/d\Omega_{rest}} \displaystyle  \approx \displaystyle  \frac{4^{5}x}{2\left(5x\right)^{5}}\ \ \ \ \ (24)
\displaystyle  \displaystyle  = \displaystyle  \frac{512}{3125\left(1-\beta\right)^{4}} \ \ \ \ \ (25)

To express this in terms of the relativistic factor {\gamma=1/\sqrt{1-\beta^{2}}} we can approximate {\gamma} for {\beta=1-x}:

\displaystyle   \gamma \displaystyle  = \displaystyle  \frac{1}{\sqrt{1-\left(1-x\right)^{2}}}\ \ \ \ \ (26)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\sqrt{2x-x^{2}}}\ \ \ \ \ (27)
\displaystyle  \displaystyle  \approx \displaystyle  \frac{1}{\sqrt{2x}}\ \ \ \ \ (28)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\left(1-\beta\right)}} \ \ \ \ \ (29)

Therefore

\displaystyle   \frac{1}{\left(1-\beta\right)^{4}} \displaystyle  \approx \displaystyle  16\gamma^{8}\ \ \ \ \ (30)
\displaystyle  \frac{dP/d\Omega_{rel}}{dP/d\Omega_{rest}} \displaystyle  \approx \displaystyle  2.62\gamma^{8} \ \ \ \ \ (31)

Since {\gamma} gets very large for {\beta\approx1}, the power generated by a relativistic charge is enormously greater than that of a charge at rest.

Radiative decay of the Bohr hydrogen atom

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 14.

The instantaneous power radiated by an accelerating point charge is given by the Larmor formula (valid for a charge moving at a speed {v\ll c}):

\displaystyle  P=\frac{\mu_{0}q^{2}a^{2}}{6\pi c} \ \ \ \ \ (1)

One historic application of this formula was to Bohr’s early model of the hydrogen atom as an electron in a classical circular orbit around the proton, with the centripetal force provided by the Coulomb attraction. Since a particle moving in a circle is accelerating, it will radiate away energy, so its orbit should eventually decay until the electron crashes into the proton. This classical instability of atoms was one motivation behind the introduction of quantum theory, but that’s another story. Here, we’ll investigate how long it would take a Bohr hydrogen atom to decay.

First, we need to reassure ourselves that the electron is moving non-relativistically. From equating centripetal and Coulomb forces, we have

\displaystyle   \frac{mv^{2}}{r} \displaystyle  = \displaystyle  \frac{q^{2}}{4\pi\epsilon_{0}r^{2}}\ \ \ \ \ (2)
\displaystyle  v \displaystyle  = \displaystyle  \frac{q}{\sqrt{4\pi\epsilon_{0}mr}}\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  qc\sqrt{\frac{\mu_{0}}{4\pi mr}} \ \ \ \ \ (4)

using {c=1/\sqrt{\mu_{0}\epsilon_{0}}}. Plugging in the numbers we get

\displaystyle  \frac{v}{c}=\frac{5.3\times10^{-8}}{\sqrt{r}} \ \ \ \ \ (5)

The Bohr radius is {a=5\times10^{-11}\mbox{ m}} so at that radius {v/c=0.0075} so we’re safe here. As {r} gets smaller, of course, {v} will increase but since the dependence is on the square root, the rate of increase of {v} is fairly small, so that even when {r=a/100}, {v} has increased only to {0.075c}. So for most of its journey towards the proton, the electron is moving non-relativistically.

To work out how long it takes for the decay to occur, consider the energy radiated during a time {dt}, which is {P\; dt}. From conservation of energy, this must be equal to the amount of energy lost by the electron. The total energy of the electron is its kinetic plus potential energy, so

\displaystyle   E \displaystyle  = \displaystyle  \frac{1}{2}mv^{2}-\frac{q^{2}}{4\pi\epsilon_{0}r}\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \frac{q^{2}c^{2}\mu_{0}}{8\pi r}-\frac{q^{2}c^{2}\mu_{0}}{4\pi r}\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  -\frac{q^{2}c^{2}\mu_{0}}{8\pi r} \ \ \ \ \ (8)

Therefore, the energy lost is

\displaystyle  dE=-\frac{q^{2}c^{2}\mu_{0}}{8\pi r^{2}}dr \ \ \ \ \ (9)

[We’ve taken {dE} as negative, since the electron loses energy.] Putting these results together, we get

\displaystyle   Pdt \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}a^{2}}{6\pi c}dt\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  -\frac{q^{2}c^{2}\mu_{0}}{8\pi r^{2}}dr\ \ \ \ \ (11)
\displaystyle  dt \displaystyle  = \displaystyle  -\frac{3c^{3}}{4r^{2}a^{2}}dr \ \ \ \ \ (12)

We need {a} to solve this, but this is just the centripetal acceleration, so

\displaystyle   a \displaystyle  = \displaystyle  \frac{v^{2}}{r}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \frac{q^{2}c^{2}\mu_{0}}{4\pi mr^{2}} \ \ \ \ \ (14)

Therefore,

\displaystyle   dt \displaystyle  = \displaystyle  -\frac{12\pi^{2}m^{2}}{q^{4}c\mu_{0}^{2}}r^{2}dr\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  -3.159\times10^{20}r^{2}dr\ \ \ \ \ (16)
\displaystyle  \int_{0}^{T}dt \displaystyle  = \displaystyle  -3.159\times10^{20}\int_{a}^{0}r^{2}dr\ \ \ \ \ (17)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{3}\left(3.159\times10^{20}\right)\left(5\times10^{-11}\right)^{3}\ \ \ \ \ (18)
\displaystyle  T \displaystyle  = \displaystyle  1.31\times10^{-11}\mbox{ s} \ \ \ \ \ (19)

Thus in classical electrodynamics, the hydrogen atom is so unstable that it would decay in a tiny fraction of a second. We should be grateful that quantum mechanics saves the universe.

Radiation from a point charge; the Larmor formula

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 13.

We’ve seen that the fields produced by a moving point charge are

\displaystyle   \mathbf{E}\left(\mathbf{r},t\right) \displaystyle  = \displaystyle  \frac{q\mathfrak{r}}{4\pi\epsilon_{0}\left(\boldsymbol{\mathfrak{r}}\cdot\mathbf{u}\right)^{3}}\left[\left(c^{2}-v^{2}\right)\mathbf{u}+\boldsymbol{\mathfrak{r}}\times\left(\mathbf{u\times}\mathbf{a}\right)\right]\ \ \ \ \ (1)
\displaystyle  \mathbf{B}\left(\mathbf{r},t\right) \displaystyle  = \displaystyle  \frac{1}{c}\hat{\boldsymbol{\mathfrak{r}}}\times\mathbf{E}\left(\mathbf{r},t\right) \ \ \ \ \ (2)

where {\boldsymbol{\mathfrak{r}}} is the vector from the charge to the observation point {\mathbf{r}} and {\mathbf{u}=c\hat{\mathfrak{\boldsymbol{r}}}-\mathbf{v}}.

If we’re interested only in the power radiated by a moving point charge, we need keep only those terms in the fields that depend on {\frac{1}{r}} and discard higher order terms such as {\frac{1}{r^{2}}}. This is because the power radiated depends on the product of the fields via the Poynting vector, and any terms in the Poynting vector of order {\frac{1}{r^{3}}} or higher will go to zero when integrated over a large sphere. It’s actually more convenient to use a coordinate system centred on the moving charge, so that {\mathbf{r}} and {\boldsymbol{\mathfrak{r}}} are actually the same, and {\mathfrak{r}} becomes the radius of the enclosing sphere. Looking at 1, we see that there is a factor of order {\frac{1}{\mathfrak{r}^{2}}} out front, and of the two terms inside the square brackets, only the second term contains another factor of {\mathfrak{r}}. Combining these two terms means that the only term that will contribute to radiation from the point charge is the second one, so we have

\displaystyle  \mathbf{E}_{rad}=\frac{q\mathfrak{r}}{4\pi\epsilon_{0}\left(\boldsymbol{\mathfrak{r}}\cdot\mathbf{u}\right)^{3}}\boldsymbol{\mathfrak{r}}\times\left(\mathbf{u\times}\mathbf{a}\right) \ \ \ \ \ (3)

As it depends on the acceleration and results in radiation, this term is known as either the acceleration field or radiation field. The first term in 1 also results in energy flux generated by the moving charge, but as it goes as {1/\mathfrak{r}^{2}} it is a localized flux, and does not contribute to radiation as it drops to zero for large {r}.

From here, the derivation of the formula for the total power radiated is fairly straightforward and is given in detail by Griffiths in his section 11.2.1. The only further assumption that is made is that {\mathbf{v}=0} at the particular instant of time that we’re considering. [Recall that it is possible for the velocity of a particle to be zero while its acceleration is non-zero, as with a mass oscillating on a spring when it reaches the high and low points of its trajectory.] This assumption gives the Larmor formula for the power radiated by a point charge:

\displaystyle  P=\frac{\mu_{0}q^{2}a^{2}}{6\pi c} \ \ \ \ \ (4)

Note that this formula has an implicit time dependence through the acceleration {a=a\left(t_{r}\right)=a\left(t-\frac{\mathfrak{r}}{c}\right)}, although the other parameters are all constants. Since we’re dealing with a point charge, there is only one retarded time that we need to keep track of. The radiation detected at the enclosing sphere, which is centred on the charge (so it moves as the charge moves) is the radiation that left the charge at time {t-\frac{\mathfrak{r}}{c}}.

Although the Larmor formula was derived by assuming that {v=0} it is actually a good approximation for all non-relativistic speeds.

Example As an example, suppose we have an electron moving at a thermal speed of {v_{0}=10^{5}\mbox{ m s}^{-1}} (so {v\ll c}) within a solid, such as a metal, and by colliding with an atom it experiences a constant deceleration so that it comes to rest after travelling {d=3\times10^{-9}\mbox{ m}}. To find the power radiated by the electron using the Larmor formula, we need to know how long the deceleration takes. Since {a} is constant, we have

\displaystyle  t=\frac{v_{0}}{a} \ \ \ \ \ (5)

The total energy radiated over this time is

\displaystyle   E \displaystyle  = \displaystyle  Pt\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}a^{2}}{6\pi c}\frac{v_{0}}{a}\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}av_{0}}{6\pi c} \ \ \ \ \ (8)

The fraction of the electron’s initial kinetic energy radiated away is

\displaystyle   f \displaystyle  = \displaystyle  \frac{2E}{mv_{0}^{2}}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}a}{3\pi mcv_{0}} \ \ \ \ \ (10)

To find {a} we use the formula

\displaystyle   d \displaystyle  = \displaystyle  \frac{1}{2}at^{2}=\frac{v_{0}^{2}}{2a}\ \ \ \ \ (11)
\displaystyle  a \displaystyle  = \displaystyle  \frac{v_{0}^{2}}{2d}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  1.67\times10^{18}\mbox{ m s}^{-2} \ \ \ \ \ (13)

[Yes, that’s an enormous deceleration!]

Plugging in the other constants in 10 we get

\displaystyle  f=2\times10^{-10} \ \ \ \ \ (14)

Thus the amount of energy lost to radiation due to electronic collisions is very small.

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)

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.

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