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PHY450H1S. Relativistic Electrodynamics Lecture 19 (Taught by Prof. Erich Poppitz). Lienard-Wiechert potentials.

Posted by peeterjoot on April 26, 2011

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Reading.

Covering chapter 8 material from the text [1].

Covering lecture notes pp. 136-146: the Lienard-Wiechert potentials (143-146) [Wednesday, Mar. 9…]

Fields from the Lienard-Wiechert potentials

(We finished off with the scalar and vector potentials in class, but I’ve put those notes with the previous lecture).

To find \mathbf{E} and \mathbf{B} need

\frac{\partial {t_r}}{\partial {t}}, and \boldsymbol{\nabla} t_r(\mathbf{x}, t)

where

\begin{aligned}t - t_r = {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r) }\right\rvert}\end{aligned} \hspace{\stretch{1}}(2.1)

implicit definition of t_r(\mathbf{x}, t)

In HW5 you’ll show

\begin{aligned}\frac{\partial {t_r}}{\partial {t}} = \frac{{\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r) }\right\rvert}}{{\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r) }\right\rvert} - \frac{\mathbf{v}_c }{c} \cdot (\mathbf{x} - \mathbf{x}_c(t_r))}\end{aligned} \hspace{\stretch{1}}(2.2)

\begin{aligned}\boldsymbol{\nabla} t_r = \frac{1}{{c}} \frac{\mathbf{x} - \mathbf{x}_c(t_r) }{{\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r) }\right\rvert} - \frac{\mathbf{v}_c }{c} \cdot (\mathbf{x} - \mathbf{x}_c(t_r))}\end{aligned} \hspace{\stretch{1}}(2.3)

and then use this to show that the electric and magnetic fields due to a moving charge are

\begin{aligned}\mathbf{E}(\mathbf{x}, t) &= \frac{e R}{ (\mathbf{R} \cdot \mathbf{u})^3 } \left( (c^2 - \mathbf{v}_c^2) \mathbf{u} + \mathbf{R} \times (\mathbf{u} \times \mathbf{a}_c) \right) \\ &= \frac{\mathbf{R}}{R} \times \mathbf{E} \\ \mathbf{u} &= c \frac{\mathbf{R}}{R} - \mathbf{v}_c,\end{aligned} \hspace{\stretch{1}}(2.4)

where everything is evaluated at the retarded time t_r = t - {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert}/c.

This looks quite a bit different than what we find in section 63 (63.8) in the text, but a little bit of expansion shows they are the same.

Check. Particle at rest.

With

\begin{aligned}\mathbf{x}_c &= \mathbf{x}_0 \\ X_c^k &= (ct, \mathbf{x}_0) \\ {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert} &= c(t - t_r)\end{aligned}

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{particleAtRestTrCalc}
\caption{Retarded time for particle at rest.}

\end{figure}

As illustrated in figure (\ref{fig:particleAtRestTrCalc}) the retarded position is

\begin{aligned}\mathbf{x}_c(t_r) = \mathbf{x}_0,\end{aligned} \hspace{\stretch{1}}(3.7)

for

\begin{aligned}\mathbf{u} = \frac{\mathbf{x} - \mathbf{x}_0}{{\left\lvert{\mathbf{x} - \mathbf{x}_0}\right\rvert}} c,\end{aligned} \hspace{\stretch{1}}(3.8)

and

\begin{aligned}\mathbf{E} = e \frac{{ {\left\lvert{\mathbf{x} - \mathbf{x}_0}\right\rvert}} }{ (c {\left\lvert{\mathbf{x} - \mathbf{x}_0}\right\rvert})^3 } c^3 \frac{\mathbf{x} - \mathbf{x}_0}{{{\left\lvert{\mathbf{x} - \mathbf{x}_0}\right\rvert}}},\end{aligned} \hspace{\stretch{1}}(3.9)

which is Coulomb’s law

\begin{aligned}\mathbf{E} = e \frac{\mathbf{x} - \mathbf{x}_0}{{\left\lvert{\mathbf{x} - \mathbf{x}_0}\right\rvert}^3}\end{aligned} \hspace{\stretch{1}}(3.10)

Check. Particle moving with constant velocity.

This was also computed in full in homework 5. The end result was

\begin{aligned}\mathbf{E} =e \frac{\mathbf{x} - \mathbf{v} t}{{\left\lvert{\mathbf{x} - \mathbf{v} t}\right\rvert}^3}\frac{1 -\boldsymbol{\beta}^2}{ \left(1 - \frac{(\mathbf{x} \times \boldsymbol{\beta})^2}{{\left\lvert{\mathbf{x} - \mathbf{v} t}\right\rvert}^2} \right)^{3/2} }\end{aligned} \hspace{\stretch{1}}(4.11)

Writing

\begin{aligned}\frac{\mathbf{x} \times \boldsymbol{\beta}}{{\left\lvert{\mathbf{x} - \mathbf{v} t}\right\rvert}}&=\frac{1}{{c}} \frac{(\mathbf{x} - \mathbf{v} t) \times \mathbf{v}}{{\left\lvert{\mathbf{x} - \mathbf{v} t}\right\rvert}}  \\ &=\frac{{\left\lvert{\mathbf{v}}\right\rvert}}{c} \frac{(\mathbf{x} - \mathbf{v} t) \times \mathbf{v}}{{\left\lvert{\mathbf{x} - \mathbf{v} t}\right\rvert} {\left\lvert{\mathbf{v}}\right\rvert}}  \end{aligned}

We can introduce an angular dependence between the charge’s translated position and its velocity

\begin{aligned}\sin^2 \theta = {\left\lvert{ \frac{\mathbf{v} \times (\mathbf{x} - \mathbf{v} t)}{\Abs{\mathbf{v}} \Abs{\mathbf{x} - \mathbf{v} t}} }\right\rvert}^2,\end{aligned} \hspace{\stretch{1}}(4.12)

and write the field as

\begin{aligned}\mathbf{E} =\underbrace{e \frac{\mathbf{x} - \mathbf{v} t}{{\left\lvert{\mathbf{x} - \mathbf{v} t}\right\rvert}^3}}_{{*}}\frac{1 -\boldsymbol{\beta}^2}{ \left(1 - \frac{\mathbf{v}^2}{c^2} \sin^2 \theta \right)^{3/2} }\end{aligned} \hspace{\stretch{1}}(4.13)

Observe that {*} = \text{Coulomb's law measured from the instantaneous position of the charge}.

The electric field \mathbf{E} has a time dependence, strongest when perpendicular to the instantaneous position when \theta = \pi/2, since the denominator is smallest (\mathbf{E} largest) when \mathbf{v}/c is not small. This is strongly \theta dependent.

Compare

\begin{aligned}\frac{{\left\lvert{\mathbf{E}(\theta = \pi/2)}\right\rvert} - {\left\lvert{\mathbf{E}(\theta = \pi/2 + \Delta \theta)}\right\rvert} }{{\left\lvert{\mathbf{E}(\theta = \pi/2)}\right\rvert}}&\approx\frac{\frac{1}{{(1 - \mathbf{v}^2/c^2)^{3/2}}} - \frac{1}{{(1 - \mathbf{v}^2/c^2(1 - (\Delta \theta)^2))^{3/2}}}}{\frac{1}{{(1 - \mathbf{v}^2/c^2)^{3/2}}}} \\ &=1 - \left(\frac{1 - \mathbf{v}^2/c^2}{1 - \mathbf{v}^2/c^2 + \mathbf{v}^2/c^2(\Delta \theta)^2}\right)^{3/2} \\ &=1 - \left(\frac{1}{1 + \mathbf{v}^2/c^2 \frac{(\Delta \theta)^2}{1 - \mathbf{v}^2/c^2}}\right)^{3/2} \\ \end{aligned}

Here we used

\begin{aligned}\sin(\theta + \pi/2) = \frac{e^{i (\theta + \pi/2)} - e^{-i(\theta + \pi/2)}}{2i} = \cos\theta \end{aligned} \hspace{\stretch{1}}(4.14)

and

\begin{aligned}\cos^2 \Delta \theta \approx \left( 1 - \frac{(\Delta \theta)^2}{2} \right)^2 \approx 1 - (\Delta \theta)^2\end{aligned} \hspace{\stretch{1}}(4.15)

FIXME: he writes:

\begin{aligned}\Delta \theta \le \sqrt{1 - \frac{\mathbf{v}^2}{c^2}}\end{aligned} \hspace{\stretch{1}}(4.16)

I don’t see where that comes from.

FIXME: PICTURE: Various \mathbf{E}‘s up, and \mathbf{v} perpendicular to that, strongest when charge is moving fast.

Back to extracting physics from the Lienard-Wiechert field equations

Imagine that we have a localized particle motion with

\begin{aligned}{\left\lvert{\mathbf{x}_c(t_r)}\right\rvert} < l\end{aligned} \hspace{\stretch{1}}(5.17)

The velocity vector

\begin{aligned}\mathbf{u} = c \frac{\mathbf{x} - \mathbf{x}_c(t_r)}{{\left\lvert{\mathbf{x} - \mathbf{x}_c}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(5.18)

doesn’t grow as distance from the source, so from 2.4, we have for {\left\lvert{\mathbf{x}}\right\rvert} \gg l

\begin{aligned}\mathbf{B}, \mathbf{E} \sim \frac{1}{{{\left\lvert{\mathbf{x}}\right\rvert}^2}}(\cdots) + \frac{1}{\mathbf{x}}(\text{acceleration term})\end{aligned} \hspace{\stretch{1}}(5.19)

The acceleration term will dominate at large distances from the source. Our Poynting magnitude is

\begin{aligned}{\left\lvert{\mathbf{S}}\right\rvert} \sim {\left\lvert{\mathbf{E} \times \mathbf{B}}\right\rvert} \sim \frac{1}{{\mathbf{x}^2}} (\text{acceleration})^2.\end{aligned} \hspace{\stretch{1}}(5.20)

We can ask about

\begin{aligned}\oint d^2 \boldsymbol{\sigma} \cdot \mathbf{S} \sim R^2 \frac{1}{{R^2}} (\text{acceleration})^2 \sim (\text{acceleration})^2 \end{aligned} \hspace{\stretch{1}}(5.21)

In the limit, for the radiation of EM waves

\begin{aligned}\lim_{R\rightarrow \infty} \oint d^2 \boldsymbol{\sigma} \cdot \mathbf{S} \ne 0\end{aligned} \hspace{\stretch{1}}(5.22)

The energy flux through a sphere of radius R is called the radiated power.

References

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

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