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PHY450H1S, Relativistic Electrodynamics, Problem Set 4.

Posted by peeterjoot on March 17, 2011

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Problem 1. Energy, momentum, etc., of EM waves.


\item Calculate the energy density, energy flux, and momentum density of a plane monochromatic linearly polarized electromagnetic wave.
\item Calculate the values of these quantities averaged over a period.
\item Imagine that a plane monochromatic linearly polarized wave incident on a surface (let the angle between the wave vector and the normal to the surface be \theta) is completely reflected. Find the pressure that the EM wave exerts on the surface.
\item To plug in some numbers, note that the intensity of sunlight hitting the Earth is about 1300 W/m^2 ( the intensity is the average power per unit area transported by the wave). If sunlight strikes a perfect absorber, what is the pressure exerted? What if it strikes a perfect reflector? What fraction of the atmospheric pressure does this amount to?


Part 1. Energy and momentum density.

Because it doesn’t add too much complexity, I’m going to calculate these using the more general elliptically polarized wave solutions. Our vector potential (in the Coulomb gauge \phi = 0, \boldsymbol{\nabla} \cdot \mathbf{A} = 0) has the form

\begin{aligned}\mathbf{A} = \text{Real} \boldsymbol{\beta} e^{i (\omega t - \mathbf{k} \cdot \mathbf{x}) }.\end{aligned} \hspace{\stretch{1}}(1.1)

The elliptical polarization case only differs from the linear by allowing \boldsymbol{\beta} to be complex, rather than purely real or purely imaginary. Observe that the Coulomb gauge condition \boldsymbol{\nabla} \cdot \mathbf{A} implies

\begin{aligned}\boldsymbol{\beta} \cdot \mathbf{k} = 0,\end{aligned} \hspace{\stretch{1}}(1.2)

a fact that will kill of terms in a number of places in the following manipulations.

Also observe that for this to be a solution to the wave equation operator

\begin{aligned}\frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta,\end{aligned} \hspace{\stretch{1}}(1.3)

the frequency and wave vector must be related by the condition

\begin{aligned}\frac{\omega}{c} = {\left\lvert{\mathbf{k}}\right\rvert} = k.\end{aligned} \hspace{\stretch{1}}(1.4)

For the time and spatial phase let’s write

\begin{aligned}\theta = \omega t - \mathbf{k} \cdot \mathbf{x}.\end{aligned} \hspace{\stretch{1}}(1.5)

In the Coulomb gauge, our electric and magnetic fields are

\begin{aligned}\mathbf{E} &= -\frac{1}{{c}}\frac{\partial {\mathbf{A}}}{\partial {t}} = \text{Real} \frac{-i\omega}{c} \boldsymbol{\beta} e^{i\theta} \\ \mathbf{B} &= \boldsymbol{\nabla} \times \mathbf{A} = \text{Real} i \boldsymbol{\beta} \times \mathbf{k} e^{i\theta}\end{aligned} \hspace{\stretch{1}}(1.6)

Similar to \S 48 of the text [1], let’s split \boldsymbol{\beta} into a phase and perpendicular vector components so that

\begin{aligned}\boldsymbol{\beta} = \mathbf{b} e^{-i\alpha}\end{aligned} \hspace{\stretch{1}}(1.8)

where \mathbf{b} has a real square

\begin{aligned}\mathbf{b}^2 = {\left\lvert{\boldsymbol{\beta}}\right\rvert}^2.\end{aligned} \hspace{\stretch{1}}(1.9)

This allows a split into two perpendicular real vectors

\begin{aligned}\mathbf{b} = \mathbf{b}_1 + i \mathbf{b}_2,\end{aligned} \hspace{\stretch{1}}(1.10)

where \mathbf{b}_1 \cdot \mathbf{b}_2 = 0 since \mathbf{b}^2 = \mathbf{b}_1^2 - \mathbf{b}_2^2 + 2 \mathbf{b}_1 \cdot \mathbf{b}_2 is real.

Our electric and magnetic fields are now reduced to

\begin{aligned}\mathbf{E} &= \text{Real} \left( \frac{-i\omega}{c} \mathbf{b} e^{i(\theta - \alpha)} \right) \\ \mathbf{B} &= \text{Real} \left( i \mathbf{b} \times \mathbf{k} e^{i(\theta - \alpha)} \right) \end{aligned} \hspace{\stretch{1}}(1.11)

or explicitly in terms of \mathbf{b}_1 and \mathbf{b}_2

\begin{aligned}\mathbf{E} &= \frac{\omega}{c} ( \mathbf{b}_1 \sin(\theta-\alpha) + \mathbf{b}_2 \cos(\theta-\alpha)) \\ \mathbf{B} &= ( \mathbf{k} \times \mathbf{b}_1 ) \sin(\theta-\alpha) + (\mathbf{k} \times \mathbf{b}_2) \cos(\theta-\alpha) \end{aligned} \hspace{\stretch{1}}(1.13)

The special case of interest for this problem, since it only strictly asked for linear polarization, is where \alpha = 0 and one of \mathbf{b}_1 or \mathbf{b}_2 is zero (i.e. \boldsymbol{\beta} is strictly real or strictly imaginary). The case with \Beta strictly real, as done in class, is

\begin{aligned}\mathbf{E} &= \frac{\omega}{c} \mathbf{b}_1 \sin(\theta-\alpha) \\ \mathbf{B} &= ( \mathbf{k} \times \mathbf{b}_1 ) \sin(\theta-\alpha) \end{aligned} \hspace{\stretch{1}}(1.15)

Now lets calculate the energy density and Poynting vectors. We’ll need a few intermediate results.

\begin{aligned}(\text{Real} \mathbf{d} e^{i\phi})^2 &= \frac{1}{{4}} ( \mathbf{d} e^{i\phi} + \mathbf{d}^{*} e^{-i\phi})^2 \\ &= \frac{1}{{4}} ( \mathbf{d}^2 e^{2 i \phi} + (\mathbf{d}^{*})^2 e^{-2 i \phi} + 2 {\left\lvert{\mathbf{d}}\right\rvert}^2 ) \\ &= \frac{1}{{2}} \left( {\left\lvert{\mathbf{d}}\right\rvert}^2 + \text{Real} ( \mathbf{d} e^{i \phi} )^2 \right),\end{aligned}


\begin{aligned}(\text{Real} \mathbf{d} e^{i\phi}) \times (\text{Real} \mathbf{e} e^{i\phi}) &= \frac{1}{{4}} ( \mathbf{d} e^{i\phi} + \mathbf{d}^{*} e^{-i\phi}) \times ( \mathbf{e} e^{i\phi} + \mathbf{e}^{*} e^{-i\phi}) \\ &= \frac{1}{{2}} \text{Real} \left( \mathbf{d} \times \mathbf{e}^{*} + (\mathbf{d} \times \mathbf{e}) e^{2 i \phi} \right).\end{aligned}

Let’s use arrowed vectors for the phasor parts

\begin{aligned}\vec{E} &= \frac{-i\omega}{c} \mathbf{b} e^{i(\theta - \alpha)} \\ \vec{B} &= i \mathbf{b} \times \mathbf{k} e^{i(\theta - \alpha)},\end{aligned} \hspace{\stretch{1}}(1.17)

where we can recover our vector quantities by taking real parts \mathbf{E} = \text{Real} \vec{E}, \mathbf{B} = \text{Real} \vec{B}. Our energy density in terms of these phasors is then

\begin{aligned}\mathcal{E} = \frac{1}{{8\pi}} (\mathbf{E}^2 + \mathbf{B}^2)= \frac{1}{{16\pi}} \left( {\left\lvert{\vec{E}}\right\rvert}^2 + {\left\lvert{\vec{B}}\right\rvert}^2 + \text{Real} ({\vec{E}}^2 + {\vec{B}}^2) \right).\end{aligned} \hspace{\stretch{1}}(1.19)

This is

\begin{aligned}\mathcal{E} &=\frac{1}{{16\pi}}\left(\frac{\omega^2}{c^2} {\left\lvert{\mathbf{b}}\right\rvert}^2 + {\left\lvert{\mathbf{b} \times \mathbf{k}}\right\rvert}^2-\text{Real} \left(\frac{\omega^2}{c^2} \mathbf{b}^2 + (\mathbf{b} \times \mathbf{k})^2\right)e^{2 i(\theta - \alpha)} \right).\end{aligned}

Note that \omega^2/c^2 = \mathbf{k}^2, and {\left\lvert{\mathbf{b} \times \mathbf{k}}\right\rvert} = {\left\lvert{\mathbf{b}}\right\rvert}^2 \mathbf{k}^2 (since \mathbf{b} \cdot \mathbf{k} = 0). Also (\mathbf{b} \times \mathbf{k})^2 = \mathbf{b}^2 \mathbf{k}^2, so we have

\begin{aligned}\boxed{\mathcal{E} =\frac{ \mathbf{k}^2 }{8\pi}\left({\left\lvert{\mathbf{b}}\right\rvert}^2 -\text{Real} \mathbf{b}^2 e^{2 i(\theta - \alpha)} \right).}\end{aligned} \hspace{\stretch{1}}(1.20)

Now, for the Poynting vector. We have

\begin{aligned}S = \frac{c}{4 \pi} \mathbf{E} \times \mathbf{B} = \frac{c}{8 \pi} \text{Real} \left( \vec{E} \times \vec{B}^{*} + \vec{E} \times \vec{B} \right).\end{aligned} \hspace{\stretch{1}}(1.21)

This is

\begin{aligned}S &= \frac{c}{8 \pi} \text{Real} \left( -k \mathbf{b} \times (\mathbf{b}^{*} \times \mathbf{k}) + k \mathbf{b} \times (\mathbf{b} \times \mathbf{k} ) e^{2 i(\theta - \alpha)} \right) \\ \end{aligned}

Reducing the terms we get \mathbf{b} \times (\mathbf{b}^{*} \times \mathbf{k}) = -\mathbf{k} {\left\lvert{\mathbf{b}}\right\rvert}^2, and \mathbf{b} \times (\mathbf{b} \times \mathbf{k}) = -\mathbf{k} \mathbf{b}^2, leaving

\begin{aligned}\boxed{S = \frac{c \hat{\mathbf{k}} \mathbf{k}^2 }{8 \pi} \left( {\left\lvert{\mathbf{b}}\right\rvert}^2 - \text{Real} \mathbf{b}^2 e^{2 i(\theta - \alpha)} \right) = c \hat{\mathbf{k}} \mathcal{E}}\end{aligned} \hspace{\stretch{1}}(1.22)

Now, the text in \S 47 defines the energy flux as the Poynting vector, and the momentum density as \mathbf{S}/c^2, so we just divide 1.22 by c^2 for the momentum density and we are done. For the linearly polarized case (all that was actually asked for, but less cool to calculate), where \mathbf{b} is real, we have

\begin{aligned}\mbox{Energy density} &= \mathcal{E} = \frac{ \mathbf{k}^2 \mathbf{b}^2 }{8\pi} ( 1 - \cos( 2 (\omega t - \mathbf{k} \cdot \mathbf{x})) ) \\ \mbox{Energy flux} &= \mathbf{S} = c \hat{\mathbf{k}} \mathcal{E} \\ \mbox{Momentum density} &= \frac{1}{{c^2}} \mathbf{S} = \frac{\hat{\mathbf{k}}}{c} \mathcal{E}.\end{aligned} \hspace{\stretch{1}}(1.23)

Part 2. Averaged.

We want to average over one period, the time T such that \omega T = 2 \pi, so the average is

\begin{aligned}\left\langle{{f}}\right\rangle = \frac{\omega}{2\pi} \int_0^{2\pi/\omega} f dt.\end{aligned} \hspace{\stretch{1}}(1.26)

It is clear that this will just kill off the sinusoidal terms, leaving

\begin{aligned}\mbox{Average Energy density} &= \left\langle{{\mathcal{E}}}\right\rangle = \frac{ \mathbf{k}^2 {\left\lvert{\mathbf{b}}\right\rvert}^2 }{8\pi} \\ \mbox{Average Energy flux} &= \left\langle{\mathbf{S}}\right\rangle = c \hat{\mathbf{k}} \mathcal{E} \\ \mbox{Average Momentum density} &= \frac{1}{{c^2}} \left\langle{\mathbf{S}}\right\rangle = \frac{\hat{\mathbf{k}}}{c} \mathcal{E}.\end{aligned} \hspace{\stretch{1}}(1.27)

Part 3. Pressure.

The magnitude of the momentum of light is related to its energy by

\begin{aligned}\mathbf{p} = \frac{\mathcal{E}}{c}\end{aligned} \hspace{\stretch{1}}(1.30)

and can thus loosely identify the magnitude of the force as

\begin{aligned}\frac{d{\mathbf{p}}}{dt} &= \frac{1}{{c}} \frac{\partial {}}{\partial {t}} \int \frac{\mathbf{E}^2 + \mathbf{B}^2}{8 \pi} d^3 \mathbf{x} \\ &= \int d^2 \boldsymbol{\sigma} \cdot \frac{\mathbf{S}}{c}.\end{aligned}

With pressure as the force per area, we could identify

\begin{aligned}\frac{\mathbf{S}}{c}\end{aligned} \hspace{\stretch{1}}(1.31)

as the instantaneous (directed) pressure on a surface. What is that for linearly polarized light? We have from above for the linear polarized case (where {\left\lvert{\mathbf{b}}\right\rvert}^2 = \mathbf{b}^2)

\begin{aligned}\mathbf{S} = \frac{c \hat{\mathbf{k}} \mathbf{k}^2 \mathbf{b}^2 }{8 \pi} ( 1 - \cos( 2 (\omega t - \mathbf{k} \cdot \mathbf{x}) ) )\end{aligned} \hspace{\stretch{1}}(1.32)

If we look at the magnitude of the average pressure from the radiation, we have

\begin{aligned}{\left\lvert{\frac{\left\langle{\mathbf{S}}\right\rangle}{c}}\right\rvert} = \frac{\mathbf{k}^2 \mathbf{b}^2 }{8 \pi}.\end{aligned} \hspace{\stretch{1}}(1.33)

Part 4. Sunlight.

With atmospheric pressure at 101.3 k Pa, and the pressure from the light at 1300 W/ 3 x 10^8 m/s, we have roughly 4 x 10^-5 Pa of pressure from the sunlight being only \sim 10^-{10} of the total atmospheric pressure. Wow. Very tiny!

Would it make any difference if the surface is a perfect absorber or a reflector? Consider a ball hitting a wall. If it manages to embed itself in the wall, the wall will have to move a bit to conserve momentum. However, if the ball bounces off twice the momentum has been transferred to the wall. The numbers above would be for perfect absorbtion, so double them for a perfect reflector.

Problem 2. Spherical EM waves.


Suppose you are given:

\begin{aligned}\vec{E}(r, \theta, \phi, t) = A \frac{\sin\theta}{r} \left( \cos(k r - \omega t) - \frac{1}{{k r}} \sin(k r - \omega t) \right) \hat{\boldsymbol{\phi}}\end{aligned} \hspace{\stretch{1}}(2.34)

where \omega = k/c and \hat{\boldsymbol{\phi}} is the unit vector in the \phi-direction. This is a simple example of a spherical wave.

\item Show that \vec{E} obeys all four Maxwell equations in vacuum and find the associated magnetic field.
\item Calculate the Poynting vector. Average \vec{S} over a full cycle to get the intensity vector \vec{I} \equiv \left\langle{{\vec{S}}}\right\rangle. Where does it point to? How does it depend on r?
\item Integrate the intensity vector flux through a spherical surface centered at the origin to find the total power radiated.


Part 1. Maxwell equation verification and magnetic field.

Our vacuum Maxwell equations to verify are

\begin{aligned}\nabla \cdot \vec{E} &= 0 \\ \nabla \times \vec{B} -\frac{1}{{c}} \frac{\partial {\vec{E}}}{\partial {t}} &= 0 \\ \nabla \cdot \vec{B} &= 0 \\ \nabla \times \vec{E} +\frac{1}{{c}} \frac{\partial {\vec{B}}}{\partial {t}} &= 0.\end{aligned} \hspace{\stretch{1}}(2.35)

We’ll also need the spherical polar forms of the divergence and curl operators, as found in \S 1.4 of [2]

\begin{aligned}\nabla \cdot \vec{v} &=\frac{1}{{r^2}} \partial_r ( r^2 v_r )+ \frac{1}{{r\sin\theta}} \partial_\theta (\sin\theta v_\theta)+ \frac{1}{{r\sin\theta}} \partial_\phi v_\phi \\ \nabla \times \vec{v} &=\frac{1}{{r \sin\theta}} \left(\partial_\theta (\sin\theta v_\phi) - \partial_\phi v_\theta\right) \hat{\mathbf{r}}+\frac{1}{{r }} \left(\frac{1}{{\sin\theta}} \partial_\phi v_r - \partial_r (r v_\phi)\right) \hat{\boldsymbol{\theta}}+\frac{1}{{r }} \left(\partial_r (r v_\theta) - \partial_\theta v_r\right) \hat{\boldsymbol{\phi}}\end{aligned} \hspace{\stretch{1}}(2.39)

We can start by verifying the divergence equation for the electric field. Observe that our electric field has only an E_\phi component, so our divergence is

\begin{aligned}\nabla \cdot \vec{E}=\frac{1}{{r\sin\theta}} \partial_\phi \left(A \frac{\sin\theta}{r} \left( \cos(k r - \omega t) - \frac{1}{{k r}} \sin(k r - \omega t) \right) \right) = 0.\end{aligned} \hspace{\stretch{1}}(2.41)

We have a zero divergence since the component E_\phi has no \phi dependence (whereas \vec{E} itself does since the unit vector \hat{\boldsymbol{\phi}} = \hat{\boldsymbol{\phi}}(\phi)).

All of the rest of Maxwell’s equations require \vec{B} so we’ll have to first calculate that before progressing further.

A aside on approaches attempted to find \vec{B}

I tried two approaches without success to calculate \vec{B}. First I hoped that I could just integrate -\vec{E} to obtain \vec{A} and then take the curl. Doing so gave me a result that had \nabla \times \vec{B} \ne 0. I hunted for an algebraic error that would account for this, but could not find one.

The second approach that I tried, also without success, was to simply take the cross product \hat{\mathbf{r}} \times \vec{E}. This worked in the monochromatic plane wave case where we had

\begin{aligned}\vec{B} &= (\vec{k} \times \vec{\beta}) \sin(\omega t - \vec{k} \cdot \vec{x}) \\ \vec{E} &= \vec{\beta} {\left\lvert{\vec{k}}\right\rvert} \sin(\omega t - \vec{k} \cdot \vec{x})\end{aligned} \hspace{\stretch{1}}(2.42)

since one can easily show that \vec{B} = \vec{k} \times \vec{E}. Again, I ended up with a result for \vec{B} that did not have a zero divergence.

Finding \vec{B} with a more systematic approach.

Following [3] \S 16.2, let’s try a phasor approach, assuming that all the solutions, whatever they are, have all the time dependence in a e^{-i\omega t} term.

Let’s write our fields as

\begin{aligned}\vec{E} &= \text{Real} (\mathbf{E} e^{-i \omega t}) \\ \vec{B} &= \text{Real} (\mathbf{B} e^{-i \omega t}).\end{aligned} \hspace{\stretch{1}}(2.44)

Substitution back into Maxwell’s equations thus requires equality in the real parts of

\begin{aligned}\nabla \cdot \mathbf{E} &= 0 \\ \nabla \cdot \mathbf{B} &= 0 \\ \nabla \times \mathbf{B} &= - i \frac{\omega}{c} \mathbf{E} \\ \nabla \times \mathbf{E} &= i \frac{\omega}{c} \mathbf{B}\end{aligned} \hspace{\stretch{1}}(2.46)

With k = \omega/c we can now directly compute the magnetic field phasor

\begin{aligned}\mathbf{B} = -\frac{i}{k} \nabla \times \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(2.50)

The electric field of this problem can be put into phasor form by noting

\begin{aligned}\vec{E} = A \frac{\sin\theta}{r} \text{Real} \left( e^{i (k r - \omega t)} - \frac{i}{k r} e^{i(k r - \omega t)} \right) \hat{\boldsymbol{\phi}},\end{aligned} \hspace{\stretch{1}}(2.51)

which allows for reading off the phasor part directly

\begin{aligned}\mathbf{E} = A \frac{\sin\theta}{r} \left( 1 - \frac{i}{k r} \right) e^{i k r} \hat{\boldsymbol{\phi}}.\end{aligned} \hspace{\stretch{1}}(2.52)

Now we can compute the magnetic field phasor \mathbf{B}. Since we have only a \phi component in our field, the curl will have just \hat{\mathbf{r}} and \hat{\boldsymbol{\theta}} components. This is reasonable since we expect it to be perpendicular to \mathbf{E}.

\begin{aligned}\nabla \times (v_\phi \hat{\boldsymbol{\phi}}) = \frac{1}{{r \sin\theta}} \partial_\theta (\sin\theta v_\phi) \hat{\mathbf{r}}- \frac{1}{{r }} \partial_r (r v_\phi) \hat{\boldsymbol{\theta}}.\end{aligned} \hspace{\stretch{1}}(2.53)

Chugging through all the algebra we have

\begin{aligned}i k \mathbf{B} &=\nabla \times \mathbf{E} \\ &=\frac{2 A \cos\theta}{r^2} \left( 1 - \frac{i}{k r} \right) e^{i k r} \hat{\mathbf{r}}- \frac{A\sin\theta}{r } \frac{\partial {}}{\partial {r}} \left( \left( 1 - \frac{i}{k r} \right) e^{i k r} \right)\hat{\boldsymbol{\theta}} \\ &=\frac{2 A \cos\theta}{r^2} \left( 1 - \frac{i}{k r} \right) e^{i k r} \hat{\mathbf{r}}- \frac{A\sin\theta}{r } \left( i k + \frac{1}{{r}} + \frac{i}{k r^2} \right) e^{i k r} \hat{\boldsymbol{\theta}},\end{aligned}

so our magnetic phasor is

\begin{aligned}\mathbf{B} =\frac{2 A \cos\theta}{k r^2} \left( -i - \frac{1}{k r} \right) e^{i k r} \hat{\mathbf{r}}- \frac{A\sin\theta}{r} \left( 1 - \frac{i}{k r} + \frac{1}{k^2 r^2} \right) e^{i k r} \hat{\boldsymbol{\theta}}\end{aligned} \hspace{\stretch{1}}(2.54)

Multiplying by e^{-i\omega t} and taking real parts gives us the messy magnetic field expression

\begin{aligned}\begin{aligned}\vec{B} &=\frac{A}{r} \frac{2 \cos\theta}{k r} \left( \sin(k r - \omega t)- \frac{1}{k r} \cos(k r - \omega t) \right)\hat{\mathbf{r}} \\ &- \frac{A}{r} \frac{\sin\theta}{k r}\left(\sin(k r - \omega t)+ \frac{k^2 r^2 + 1}{k r}\cos(k r - \omega t)\right)\hat{\boldsymbol{\theta}}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.55)

Since this was constructed directly from \nabla \times \vec{E} +\frac{1}{{c}} {\partial {\vec{B}}}/{\partial {t}} = 0, this implicitly verifies one more of Maxwell’s equations, leaving only \nabla \cdot \vec{B}, and \nabla \times \vec{B} -\frac{1}{{c}} {\partial {\vec{E}}}/{\partial {t}} = 0. Neither of these looks particularly fun to verify, however, we can take a small shortcut and use the phasors to verify without the explicit time dependence.

From 2.54 we have for the divergence

\begin{aligned}\nabla \cdot \mathbf{B} &=\frac{2 A \cos\theta}{k r^2 } \frac{\partial {}}{\partial {r}} \left(\left( -i - \frac{1}{k r} \right) e^{i k r} \right)- \frac{A 2 \cos\theta}{r^2} \left( 1 - \frac{i}{k r} + \frac{1}{k^2 r^2} \right) e^{i k r}  \\ &=\frac{2 A \cos\theta}{r^2 } e^{i k r}\left(\frac{1}{{k}}\left( \frac{1}{{k r^2}} + i k \left(-i - \frac{1}{{k r}}\right)\right)-\left( 1 - \frac{i}{k r} + \frac{1}{k^2 r^2} \right) \right) \\ &= 0 \qquad \square\end{aligned}

Let’s also verify the last of Maxwell’s equations in phasor form. The time dependence is knocked out, and we want to see that taking the curl of the magnetic phasor returns us (scaled) the electric phasor. That is

\begin{aligned}\nabla \times \mathbf{B} = - i \frac{\omega}{c} \mathbf{E}\end{aligned} \hspace{\stretch{1}}(2.56)

With only r and \theta components in the magnetic phasor we have

\begin{aligned}\nabla \times (v_r \hat{\mathbf{r}} + v_\theta \hat{\boldsymbol{\theta}}) =-\frac{1}{{r \sin\theta}} \partial_\phi v_\theta\hat{\mathbf{r}}+\frac{1}{{r }} \frac{1}{{\sin\theta}} \partial_\phi v_r \hat{\boldsymbol{\theta}}+\frac{1}{{r }} \left(\partial_r (r v_\theta) - \partial_\theta v_r\right) \hat{\boldsymbol{\phi}}\end{aligned} \hspace{\stretch{1}}(2.57)

Immediately, we see that with no explicit \phi dependence in the coordinates, we have no \hat{\mathbf{r}} nor \hat{\boldsymbol{\theta}} terms in the curl, which is good. Our curl is now just

\begin{aligned}\nabla \times \mathbf{B} &=\frac{1}{{r }} \left( A\sin\theta \partial_r \left( 1 - \frac{i}{k r} + \frac{1}{k^2 r^2} \right) e^{i k r} +\frac{2 A \sin\theta}{k r^2} \left( -i - \frac{1}{k r} \right) e^{i k r} \right) \hat{\boldsymbol{\phi}} \\ &=A \sin\theta \frac{1}{{r }} \left(\partial_r \left( 1 - \frac{i}{k r} + \frac{1}{k^2 r^2} \right) e^{i k r} +\frac{2 }{k r^2} \left( -i - \frac{1}{k r} \right) e^{i k r} \right) \hat{\boldsymbol{\phi}} \\ &=A \sin\theta e^{i k r}\frac{1}{{r }} \left((ik)\left( 1 - \frac{i}{k r} + \frac{1}{k^2 r^2} \right) +\left( \frac{i}{k r^2} - \frac{2}{k^2 r^3} \right) +\frac{2 }{k r^2} \left( -i - \frac{1}{k r} \right) \right) \hat{\boldsymbol{\phi}} \\ &=A \sin\theta e^{i k r}\frac{1}{{r }} \left(i k + \frac{1}{{r}} - \frac{ 4 }{k^2 r^3}\right) \hat{\boldsymbol{\phi}} \\ \end{aligned}

What we expect is \nabla \times \mathbf{B} = - i k \mathbf{E} which is

\begin{aligned}- i k \mathbf{E} =A \sin\theta e^{i k r}\frac{1}{{r }} \left(- i k - \frac{1}{{r}}\right) \hat{\boldsymbol{\phi}} \end{aligned} \hspace{\stretch{1}}(2.58)

FIXME: Somewhere I must have made a sign error, because these aren’t matching! Have an extra 1/r^3 term and the wrong sign on the 1/r term.

Part 2. Poynting and intensity.

Our Poynting vector is

\begin{aligned}\vec{S} = \frac{c}{4 \pi} \vec{E} \times \vec{B},\end{aligned} \hspace{\stretch{1}}(2.59)

which we could calculate from 2.34, and 2.55. However, that looks like it’s going to be a mess to multiply out. Let’s use instead the trick from \S 48 of the course text [1], and work with the complex quantities directly, noting that we have

\begin{aligned}(\text{Real} \mathbf{E} e^{i \alpha}) \times (\text{Real} \mathbf{B} e^{i \alpha}) &= \frac{1}{{4}} ( \mathbf{E} e^{i \alpha} + \mathbf{E}^{*} e^{-i \alpha}) \times ( \mathbf{B} e^{i \alpha} + \mathbf{B}^{*} e^{-i \alpha}) \\ &= \frac{1}{{2}} \text{Real} \left( \mathbf{E} \times \mathbf{B}^{*} + (\mathbf{E} \times \mathbf{B}) e^{2 i \alpha} \right).\end{aligned}

Now we can do the Poynting calculation using the simpler relations 2.52, 2.54.

Let’s also write

\begin{aligned}\mathbf{E} &= A e^{i k r} E_\phi \hat{\boldsymbol{\phi}} \\ \mathbf{B} &= A e^{i k r} ( B_r \hat{\mathbf{r}} + B_\theta \hat{\boldsymbol{\theta}} )\end{aligned} \hspace{\stretch{1}}(2.60)


\begin{aligned}E_\phi &= \frac{\sin\theta}{r} \left( 1 - \frac{i}{k r} \right)  \\ B_r &= -\frac{2 \cos\theta}{k r^2} \left( i + \frac{1}{k r} \right)  \\ B_\theta &= - \frac{\sin\theta}{r} \left( 1 - \frac{i}{k r} + \frac{1}{k^2 r^2} \right) \end{aligned} \hspace{\stretch{1}}(2.62)

So our Poynting vector is

\begin{aligned}\vec{S} &= \frac{A^2 c}{2 \pi} \text{Real}\left(E_\phi \hat{\boldsymbol{\phi}} \times ( B_r^{*} \hat{\mathbf{r}} + B_\theta^{*} \hat{\boldsymbol{\theta}} )+E_\phi \hat{\boldsymbol{\phi}} \times ( B_r \hat{\mathbf{r}} + B_\theta \hat{\boldsymbol{\theta}} ) e^{ 2 i ( k r - \omega t ) }\right) \\ \end{aligned}

Note that our unit vector basis \{ \hat{\mathbf{r}}, \hat{\boldsymbol{\theta}}, \hat{\boldsymbol{\phi}} \} was rotated from \{ \hat{\mathbf{z}}, \hat{\mathbf{x}}, \hat{\mathbf{y}} \}, so we have

\begin{aligned}\hat{\boldsymbol{\phi}} \times \hat{\mathbf{r}} &= \hat{\boldsymbol{\theta}} \\ \hat{\boldsymbol{\theta}} \times \hat{\boldsymbol{\phi}} &= \hat{\mathbf{r}} \\ \hat{\mathbf{r}} \times \hat{\boldsymbol{\theta}} &= \hat{\boldsymbol{\phi}} ,\end{aligned} \hspace{\stretch{1}}(2.65)

and plug this into our Poynting expression

\begin{aligned}\vec{S} &= \frac{A^2 c}{2 \pi} \text{Real}\left(E_\phi B_r^{*} \hat{\boldsymbol{\theta}} -E_\phi B_\theta^{*} \hat{\mathbf{r}} +(E_\phi B_r \hat{\boldsymbol{\theta}} -E_\phi B_\theta \hat{\mathbf{r}} )e^{ 2 i ( k r - \omega t ) }\right) \\ \end{aligned}

Now we have to multiply out our terms. We have

\begin{aligned}E_\phi B_r^{*} &=- \frac{\sin\theta}{r} \frac{2 \cos\theta}{k r^2} \left( 1 - \frac{i}{k r} \right)\left( -i + \frac{1}{k r} \right) \\ &=-\frac{ \sin(2\theta)}{k r^3}\left( -i - \frac{i}{k^2 r^2} \right),\end{aligned}

Since this has no real part, there is no average contribution to \vec{S} in the \hat{\boldsymbol{\theta}} direction. What do we have for the time dependent part

\begin{aligned}E_\phi B_r &=- \frac{\sin\theta}{r} \frac{2 \cos\theta}{k r^2} \left( 1 - \frac{i}{k r} \right)\left( i + \frac{1}{k r} \right) \\ &=-\frac{ \sin(2\theta)}{k r^3}\left( i + \frac{2}{k r} - \frac{i}{k^2 r^2} \right) \end{aligned}

This is non zero, so we have a time dependent \hat{\boldsymbol{\theta}} contribution that averages out. Moving on

\begin{aligned}- E_\phi B_\theta^{*}&= \frac{\sin^2\theta}{r^2} \left( 1 - \frac{i}{k r} \right)\left( 1 + \frac{i}{k r} + \frac{1}{k^2 r^2} \right) \\ &= \frac{\sin^2\theta}{r^2} \left( 1 + \frac{2}{k^2 r^2} - \frac{i}{k^3 r^3}\right).\end{aligned}

This is non-zero, so the steady state Poynting vector is in the outwards radial direction. The last piece is

\begin{aligned}- E_\phi B_\theta&= \frac{\sin^2\theta}{r^2} \left( 1 - \frac{i}{k r} \right)\left( 1 - \frac{i}{k r} + \frac{1}{k^2 r^2} \right) \\ &= \frac{\sin^2\theta}{r^2} \left( 1 - \frac{2i}{k r} - \frac{i}{k^3 r^3}\right).\end{aligned}

Assembling all the results we have

\begin{aligned}\begin{aligned}\vec{S} &= \frac{A^2 c}{2 \pi} \frac{\sin^2\theta}{r^2} \left( 1 + \frac{2}{k^2 r^2} \right) \hat{\mathbf{r}} \\ &\quad +\frac{A^2 c}{2 \pi} \text{Real} \left(\left(-\frac{ \sin(2\theta)}{k r^3} \left( i + \frac{2}{k r} - \frac{i}{k^2 r^2} \right) \hat{\boldsymbol{\theta}}+\frac{\sin^2\theta}{r^2} \left( 1 - \frac{2i}{k r} - \frac{i}{k^3 r^3}\right) \hat{\mathbf{r}} \right) e^{ 2 i ( k r - \omega t ) }\right) \end{aligned}\end{aligned}

We can read off the intensity directly

\begin{aligned}\vec{I} = \left\langle{{\vec{S}}}\right\rangle = \frac{A^2 c \sin^2 \theta}{2 \pi r^2} \left( 1 + \frac{2}{k^2 r^2} \right) \hat{\mathbf{r}} \end{aligned} \hspace{\stretch{1}}(2.68)

Part 3. Find the power.

Through a surface of radius r, integration of the intensity vector 2.68 is

\begin{aligned}\int r^2 \sin\theta d\theta d\phi\vec{I} &= \int r^2 \sin\theta d\theta d\phi \frac{A^2 c \sin^2 \theta}{2 \pi r^2} \left( 1 + \frac{2}{k^2 r^2} \right) \hat{\mathbf{r}} \\ &= A^2 c \left( 1 + \frac{2}{k^2 r^2} \right) \hat{\mathbf{r}} \int_0^\pi \sin^3\theta d\theta \\ &= A^2 c \left( 1 + \frac{2}{k^2 r^2} \right) \hat{\mathbf{r}} {\left.\frac{1}{{12}}( \cos(3\theta) - 9 \cos\theta )\right\vert}_0^\pi.\end{aligned}

Our average power through the surface is therefore

\begin{aligned}\int d^2 \boldsymbol{\sigma} \vec{I} =\frac{4 A^2 c }{3}\left( 1 + \frac{2}{k^2 r^2} \right) \hat{\mathbf{r}}.\end{aligned} \hspace{\stretch{1}}(2.69)

Notes on grading of my solution.

Problem 2 above was the graded portion.

FIXME1: I lost a mark in the spot I expected, where I failed to verify one of the Maxwell equations. I’ll still need to figure out what got messed up there.

What occured to me later, also mentioned in the grading of the solution was that Maxwell’s equations in the space-time domain could have been used to solve for {\partial {\mathbf{B}}}/{\partial {t}} instead of all the momentum space logic (which simplified some things, but probably complicated others).

FIXME2: I lost a mark on 2.68 with a big X beside it. I’ll have to read the graded solution to see why.

FIXME3: Lost a mark for the final average power result 2.69. Again, I’ll have to go back and figure out why.


[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

[2] D.J. Griffith. Introduction to Electrodynamics. Prentice-Hall, 1981.

[3] JD Jackson. Classical Electrodynamics Wiley. John Wiley and Sons, 2nd edition, 1975.

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