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## PHY450H1S. Relativistic Electrodynamics Lecture 21 (Taught by Prof. Erich Poppitz). More on EM fields due to dipole radiation.

Posted by peeterjoot on March 25, 2011

Covering chapter 8 material from the text [1].

Covering lecture notes pp. 147-165: radiated power (154); fields in the “wave zone” and discussions of approximations made (155-159); EM fields due to electric dipole radiation (160-163); Poynting vector, angular distribution, and power of dipole radiation (164-165) [Wednesday, Mar. 16…]

# Where we left off.

For a localized charge distribution, we’d arrived at expressions for the scalar and vector potentials far from the point where the charges and currents were localized. This was then used to consider the specific case of a dipole system where one of the charges had a sinusoidal oscillation. The charge positions for the negative and positive charges respectively were

\begin{aligned}z_{-} &= 0 \\ z_{+} &= \mathbf{e}_3( z_0 + a \sin(\omega t)) ,\end{aligned} \hspace{\stretch{1}}(2.1)

so that our dipole moment $\mathbf{d} = \int \rho(\mathbf{x}') \mathbf{x}'$ is

\begin{aligned}\mathbf{d} = \mathbf{e}_3 q (z_0 + a \sin(\omega t)).\end{aligned} \hspace{\stretch{1}}(2.3)

The scalar potential, to first order in a number of Taylor expansions at our point far from the source, evaluated at the retarded time $t_r = t - {\left\lvert{\mathbf{x}}\right\rvert}/c$, was found to be

\begin{aligned}A^0(\mathbf{x}, t) = \frac{z q}{{\left\lvert{\mathbf{x}}\right\rvert}^3} ( z_0 + a \sin(\omega t_r) ) + \frac{z q}{c {\left\lvert{\mathbf{x}}\right\rvert}^2} a \omega \cos(\omega t_r),\end{aligned} \hspace{\stretch{1}}(2.4)

and our vector potential, also with the same approximations, was

\begin{aligned}\mathbf{A}(\mathbf{x}, t) = \frac{1}{{c {\left\lvert{\mathbf{x}}\right\rvert} }} \mathbf{e}_3 q a \omega \cos(\omega t_r).\end{aligned} \hspace{\stretch{1}}(2.5)

We found that the electric field (neglecting any non-radiation terms that died off as inverse square in the distance) was

\begin{aligned}\mathbf{E} = \frac{a \omega^2 q}{c^2 {\left\lvert{\mathbf{x}}\right\rvert}} \sin( \omega (t - {\left\lvert{\mathbf{x}}\right\rvert}/c) ) \left( \mathbf{e}_3 - \hat{\mathbf{r}} \frac{z}{{\left\lvert{\mathbf{x}}\right\rvert}} \right).\end{aligned} \hspace{\stretch{1}}(2.6)

# Direct computation of the magnetic radiation field

Taking the curl of the vector potential 2.6 for the magnetic field, we’ll neglect the contribution from the $1/{\left\lvert{\mathbf{x}}\right\rvert}$ since that will be inverse square, and die off too quickly far from the source

\begin{aligned}\mathbf{B}&= \boldsymbol{\nabla} \times \mathbf{A} \\ &= \boldsymbol{\nabla} \times \frac{1}{{c {\left\lvert{\mathbf{x}}\right\rvert} }} \mathbf{e}_3 q a \omega \cos(\omega (t - {\left\lvert{\mathbf{x}}\right\rvert}/c)) \\ &\approx - \frac{q a \omega}{c {\left\lvert{\mathbf{x}}\right\rvert} } \mathbf{e}_3 \times \boldsymbol{\nabla} \cos(\omega (t - {\left\lvert{\mathbf{x}}\right\rvert}/c)) \\ &= - \frac{q a \omega}{c {\left\lvert{\mathbf{x}}\right\rvert} } \left( -\frac{\omega}{c} \right)(\mathbf{e}_3 \times \boldsymbol{\nabla} {\left\lvert{\mathbf{x}}\right\rvert}) \sin(\omega (t - {\left\lvert{\mathbf{x}}\right\rvert}/c)),\end{aligned}

which is

\begin{aligned}\mathbf{B} = \frac{q a \omega^2}{c^2 {\left\lvert{\mathbf{x}}\right\rvert} } (\mathbf{e}_3 \times \hat{\mathbf{r}}) \sin(\omega (t - {\left\lvert{\mathbf{x}}\right\rvert}/c)).\end{aligned} \hspace{\stretch{1}}(3.7)

Comparing to 2.6, we see that this equals $\hat{\mathbf{r}} \times \mathbf{E}$ as expected.

# An aside: A tidier form for the electric dipole field

We can rewrite the electric field 2.6 in terms of the retarded time dipole

\begin{aligned}\mathbf{E} = \frac{1}{{c^2 {\left\lvert{\mathbf{x}}\right\rvert}}} \Bigl( -\dot{d}{\mathbf{d}}(t_r) + \hat{\mathbf{r}} ( \dot{d}{\mathbf{d}}(t_r) \cdot \hat{\mathbf{r}} ) \Bigr),\end{aligned} \hspace{\stretch{1}}(4.8)

where

\begin{aligned}\dot{d}{\mathbf{d}}(t) = - q a \omega^2 \sin(\omega t) \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(4.9)

Then using the vector identity

\begin{aligned}(\mathbf{A} \times \hat{\mathbf{r}} ) \times \hat{\mathbf{r}} = -\mathbf{A} + (\hat{\mathbf{r}} \cdot \mathbf{A}) \hat{\mathbf{r}},\end{aligned} \hspace{\stretch{1}}(4.10)

we have for the fields

\begin{aligned}\boxed{\begin{aligned}\mathbf{E} &= \frac{1}{{c^2 {\left\lvert{\mathbf{x}}\right\rvert}}} (\dot{d}{\mathbf{d}}(t_r) \times \hat{\mathbf{r}}) \times \hat{\mathbf{r}} \\ \mathbf{B} &= \hat{\mathbf{r}} \times \mathbf{E}.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(4.11)

# Calculating the energy flux

Our Poynting vector, the energy flux, is

\begin{aligned}\mathbf{S} = \frac{c}{4 \pi} \mathbf{E} \times \mathbf{B} =\frac{c}{4 \pi}\left( \frac{q a \omega^2}{c^2 {\left\lvert{\mathbf{x}}\right\rvert} } \right)^2\sin^2(\omega (t - {\left\lvert{\mathbf{x}}\right\rvert}/c))\left( \mathbf{e}_3 - \hat{\mathbf{r}} \frac{z}{{\left\lvert{\mathbf{x}}\right\rvert}} \right) \times (\hat{\mathbf{r}} \times \mathbf{e}_3).\end{aligned} \hspace{\stretch{1}}(5.12)

Expanding just the cross terms we have

\begin{aligned}\left( \mathbf{e}_3 - \hat{\mathbf{r}} \frac{z}{{\left\lvert{\mathbf{x}}\right\rvert}} \right) \times (\hat{\mathbf{r}} \times \mathbf{e}_3)&=-(\hat{\mathbf{r}} \times \mathbf{e}_3) \times \mathbf{e}_3 - \frac{z}{{\left\lvert{\mathbf{x}}\right\rvert}} (\mathbf{e}_3 \times \hat{\mathbf{r}}) \times \hat{\mathbf{r}} \\ &=-(-\hat{\mathbf{r}} + \mathbf{e}_3(\mathbf{e}_3 \cdot \hat{\mathbf{r}}) ) - \frac{z}{{\left\lvert{\mathbf{x}}\right\rvert}} (-\mathbf{e}_3 + \hat{\mathbf{r}} (\hat{\mathbf{r}} \cdot \mathbf{e}_3)) \\ &=\hat{\mathbf{r}} - \cancel{\mathbf{e}_3(\mathbf{e}_3 \cdot \hat{\mathbf{r}})} + \frac{z}{{\left\lvert{\mathbf{x}}\right\rvert}} (\cancel{\mathbf{e}_3} - \hat{\mathbf{r}} (\hat{\mathbf{r}} \cdot \mathbf{e}_3)) \\ &=\hat{\mathbf{r}}( 1 - (\hat{\mathbf{r}} \cdot \mathbf{e}_3)^2 ).\end{aligned}

Note that we’ve utilized $\hat{\mathbf{r}} \cdot \mathbf{e}_3 = z/{\left\lvert{\mathbf{x}}\right\rvert}$ to do the cancellations above, and for the final grouping. Since $\hat{\mathbf{r}} \cdot \mathbf{e}_3 = \cos\theta$, the direction cosine of the unit radial vector with the z-axis, we have for the direction of the Poynting vector

\begin{aligned}\hat{\mathbf{r}}( 1 - (\hat{\mathbf{r}} \cdot \mathbf{e}_3)^2 )&= \hat{\mathbf{r}} (1 - \cos^2\theta) \\ &= \hat{\mathbf{r}} \sin^2\theta.\end{aligned}

Our Poynting vector is found to be directed radially outwards, and is

\begin{aligned}\mathbf{S} =\frac{c}{4 \pi}\left( \frac{q a \omega^2}{c^2 {\left\lvert{\mathbf{x}}\right\rvert} } \right)^2\sin^2(\omega (t - {\left\lvert{\mathbf{x}}\right\rvert}/c)) \sin^2\theta \hat{\mathbf{r}}.\end{aligned} \hspace{\stretch{1}}(5.13)

The intensity is constant along the curves

\begin{aligned}{\left\lvert{\sin\theta}\right\rvert} \sim r\end{aligned} \hspace{\stretch{1}}(5.14)

PICTURE: dipole lobes diagram with $\mathbf{d}$ up along the z axis, and $\hat{\mathbf{r}}$ pointing in an arbitrary direction.

FIXME: understand how this lobes picture comes from our result above.

PICTURE: field diagram along spherical north-south great circles, and the electric field $\mathbf{E}$ along what looks like it is the $\hat{\boldsymbol{\theta}}$ direction, and $\mathbf{B}$ along what appear to be the $\hat{\boldsymbol{\phi}}$ direction, and $\mathbf{S}$ pointing radially out.

## Utilizing the spherical unit vectors to express the field directions.

In class we see the picture showing these spherical unit vector directions. We can see this algebraically as well. Recall that we have for our unit vectors

\begin{aligned}\hat{\mathbf{r}} &= \mathbf{e}_1 \sin\theta \cos\phi + \mathbf{e}_2 \sin\theta \sin\phi + \mathbf{e}_3 \cos\theta \\ \hat{\boldsymbol{\phi}} &= \sin\theta ( \mathbf{e}_2 \cos\phi - \mathbf{e}_1 \sin\phi ) \\ \hat{\boldsymbol{\theta}} &= \cos\theta ( \mathbf{e}_1 \cos\phi + \mathbf{e}_2 \sin\phi ) - \mathbf{e}_3 \sin\theta,\end{aligned} \hspace{\stretch{1}}(5.15)

with the volume element orientation governed by cyclic permutations of

\begin{aligned}\hat{\mathbf{r}} \times \hat{\boldsymbol{\theta}} = \hat{\boldsymbol{\phi}}.\end{aligned} \hspace{\stretch{1}}(5.18)

We can now express the direction of the magnetic field in terms of the spherical unit vectors

\begin{aligned}\mathbf{e}_3 \times \hat{\mathbf{r}}&=\mathbf{e}_3 \times (\mathbf{e}_1 \sin\theta \cos\phi + \mathbf{e}_2 \sin\theta \sin\phi + \mathbf{e}_3 \cos\theta ) \\ &=\mathbf{e}_3 \times (\mathbf{e}_1 \sin\theta \cos\phi + \mathbf{e}_2 \sin\theta \sin\phi ) \\ &=\mathbf{e}_2 \sin\theta \cos\phi - \mathbf{e}_1 \sin\theta \sin\phi \\ &=\sin\theta ( \mathbf{e}_2 \cos\phi - \mathbf{e}_1 \sin\phi ) \\ &=\sin\theta \hat{\boldsymbol{\phi}}.\end{aligned}

The direction of the electric field was in the direction of $(\dot{d}{\mathbf{d}} \times \hat{\mathbf{r}}) \times \hat{\mathbf{r}}$ where $\mathbf{d}$ was directed along the z-axis. This is then

\begin{aligned}(\mathbf{e}_3 \times \hat{\mathbf{r}}) \times \hat{\mathbf{r}}&=-\sin\theta \hat{\boldsymbol{\phi}} \times \hat{\mathbf{r}} \\ &=-\sin\theta \hat{\boldsymbol{\theta}}\end{aligned}

\begin{aligned}\boxed{\begin{aligned}\mathbf{E} &= \frac{ q a \omega^2 }{c^2 {\left\lvert{\mathbf{x}}\right\rvert}} \sin(\omega t_r) \sin\theta \hat{\boldsymbol{\theta}} \\ \mathbf{B} &= -\frac{ q a \omega^2 }{c^2 {\left\lvert{\mathbf{x}}\right\rvert}} \sin(\omega t_r) \sin\theta \hat{\boldsymbol{\phi}} \\ \mathbf{S} &= \left( \frac{ q a \omega^2 }{c^2 {\left\lvert{\mathbf{x}}\right\rvert}} \right)^2 \sin^2(\omega t_r) \sin^2\theta \hat{\mathbf{r}} \end{aligned}}\end{aligned} \hspace{\stretch{1}}(5.19)

# Calculating the power

Integrating $\mathbf{S}$ over a spherical surface, we can calculate the power

FIXME: remind myself why Power is an appropriate label for this integral.

This is

\begin{aligned}P(r, t)&= \oint d^2 \boldsymbol{\sigma} \cdot \mathbf{S} \\ &= \int \cancel{r^2} \sin\theta d\theta d\phi \frac{c}{4 \pi}\left( \frac{q a \omega^2}{c^2 \cancel{{\left\lvert{\mathbf{x}}\right\rvert}} } \right)^2\sin^2(\omega (t - {\left\lvert{\mathbf{x}}\right\rvert}/c)) \sin^2\theta \\ &=\frac{q^2 a^2 \omega^4}{2 c^3 }\sin^2(\omega (t - r/c))\underbrace{\int \sin^3\theta d\theta}_{=4/3}\end{aligned}

\begin{aligned}P(r, t) = \frac{2}{3} \frac{q^2 a^2 \omega^4}{c^3} \sin^2(\omega (t - r/c)) =\frac{q^2 a^2 \omega^4}{3 c^3} (1 - \cos(2 \omega (t - r/c))\end{aligned} \hspace{\stretch{1}}(6.20)

Averaging over a period kills off the cosine term

\begin{aligned}\left\langle{{P(r, t)}}\right\rangle = \frac{\omega}{2 \pi} \int_0^{2 \pi/\omega} dt P(t) = \frac{q^2 a^2 \omega^4}{3 c^3},\end{aligned} \hspace{\stretch{1}}(6.21)

and we once again see that higher frequencies radiate more power (i.e. why the sky is blue).

We’ve seen now radiation from localized current distributions, and called that electric dipole radiation. There are many other sources of electrodynamic radiation, of which here are a couple.

\begin{itemize}

This will be covered more in more depth in the tutorial. Picture of a positive circulating current $I = I_o \sin \omega t$ given, and a magnetic dipole moment $\boldsymbol{\mu} = \pi b^2 I \mathbf{e}_3$.

This sort of current loop is a source of magnetic dipole radiation.

This is the label for acceleration induced radiation (at high velocities) by particles moving in a uniform magnetic field.

PICTURE: circular orbit with speed $v = \omega r$. The particle trajectories are

\begin{aligned}x &= r \cos \omega t \\ y &= r \sin \omega t\end{aligned} \hspace{\stretch{1}}(7.22)

This problem can be treated as two electric dipoles out of phase by 90 degrees.

PICTURE: 4 lobe dipole picture, with two perpendicular dipole moment arrows. Resulting superposition sort of smeared together.

\end{itemize}

# Energy momentum conservation.

We’ve defined

\begin{aligned}\begin{array}{l l l}\mathcal{E} &= \frac{\mathbf{E}^2 + \mathbf{B}^2}{8\pi} & \mbox{Energy density} \\ \frac{\mathbf{S}}{c^2} &= \frac{1}{{4 \pi c}} \mathbf{E} \times \mathbf{B} & \mbox{Momentum density}\end{array}\end{aligned} \hspace{\stretch{1}}(8.24)

(where $\mathbf{S}$ was defined as the energy flow).

Dimensional analysis arguments and analogy with classical mechanics were used to motivate these definitions, as opposed to starting with the field action to find these as a consequence of a symmetry. We also saw that we had a conservation relationship that had the appearance of a four divergence of a four vector. With

\begin{aligned}P^i = (\mathcal{U}/c, \mathbf{S}/c^2),\end{aligned} \hspace{\stretch{1}}(8.25)

that was

\begin{aligned}\partial_i P^i = - \mathbf{E} \cdot \mathbf{j}/c^2\end{aligned} \hspace{\stretch{1}}(8.26)

The left had side has the appearance of a Lorentz scalar, since it contracts two four vectors, but the right hand side is the continuum equivalent to the energy term of the Lorentz force law and cannot be a Lorentz scalar. The conclusion has to be that $P^i$ is not a four vector, and it’s natural to assume that these are components of a rank 2 four tensor instead (since we’ve got just one component of a rank 1 four tensor on the RHS). We want to know find out how the EM energy and momentum densities transform.

## Classical mechanics reminder.

Recall that in particle mechanics when we had a Lagrangian that had no explicit time dependence

\begin{aligned}\mathcal{L}(q, \dot{q}, \cancel{t}),\end{aligned} \hspace{\stretch{1}}(8.27)

that energy resulted from time translation invariance. We found this by taking the full derivative of the Lagrangian, and employing the EOM for the system to find a conserved quantity

\begin{aligned}\frac{d{{}}}{dt} \mathcal{L}(q, \dot{q}) &=\frac{\partial {\mathcal{L}}}{\partial {q}} \frac{\partial {q}}{\partial {t}}+\frac{\partial {\mathcal{L}}}{\partial {\dot{q}}} \frac{\partial {\dot{q}}}{\partial {t}} \\ &=\frac{d{{}}}{dt} \left( \frac{\partial {\mathcal{L}}}{\partial {\dot{q}}} \right) \dot{q}+\frac{\partial {\mathcal{L}}}{\partial {\dot{q}}} \dot{d}{q} \\ &=\frac{d{{}}}{dt} \left( \frac{\partial {\mathcal{L}}}{\partial {\dot{q}}} \dot{q} \right) \end{aligned}

Taking differences we have

\begin{aligned}\frac{d{{}}}{dt} \left( \frac{\partial {\mathcal{L}}}{\partial {\dot{q}}} \dot{q} -\mathcal{L} \right) = 0,\end{aligned} \hspace{\stretch{1}}(8.28)

and we labeled this conserved quantity the energy

\begin{aligned}\mathcal{E} = \frac{\partial {\mathcal{L}}}{\partial {\dot{q}}} \dot{q} -\mathcal{L} \end{aligned} \hspace{\stretch{1}}(8.29)

## Our approach from the EM field action.

Our EM field action was

\begin{aligned}S = -\frac{1}{{16 \pi c}} \int d^4 x F_{i j} F^{i j}.\end{aligned} \hspace{\stretch{1}}(8.32)

The squared field tensor $F_{i j} F^{i j}$ only depends on the fields $A^i(\mathbf{x}, t)$ or its derivatives $\partial_j A^i(\mathbf{x}, t)$, and not on the coordinates $\mathbf{x}, t$ themselves. This is very similar to the particle action with no explicit time dependence

\begin{aligned}S = \int dt \left( \frac{m \dot{q}^2}{2} + V(q) \right).\end{aligned} \hspace{\stretch{1}}(8.32)

For the particle case we obtained our conservation relationship by taking time derivatives of the Lagrangian. These are very similar with the action having no explicit dependence on space or time, only on the field, so what will we get if we take the coordinate partials of the EM Lagrangian density?

We will chew on this tomorrow and calculate

\begin{aligned}\frac{\partial {}}{\partial {x^k}} \Bigl( F_{i j} F^{i j} \Bigr)\end{aligned} \hspace{\stretch{1}}(8.32)

in full gory details. We will find that instead of finding a single conserved quantity $C^A(\mathbf{x}, t)$, we instead find a quantity that only changes through escape from the boundary of a surface.

# References

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