## PHY450HS1. Relativistic electrodynamics. Problem Set 5.

Posted by peeterjoot on April 7, 2011

[Click here for a PDF of this post with nicer formatting]

# Problem 1. Sinusoidal current density on an infinite flat conducting sheet.

## Statement

An infinitely thin flat conducting surface lying in the plane carries a surface current density:

Here is a unit vector in the direction, is the peak value of the current density, and is the theta function: .

\begin{enumerate}

\item Write down the equations determining the electromagnetic potentials. Specify which gauge you choose to work in.

\item Find the electromagnetic potentials outside the plane.

\item Find the electric and magnetic fields outside the plane.

\item Give a physical interpretation of the results of the previous section. Do they agree with your qualitative expectations?

\item Find the direction and magnitude of the energy flux outside the plane.

\item Consider a point at some distance from the plane. Sketch the intensity of the electromagnetic field near this point as a function of time. Explain physically.

\item Consider now a point near the plane. Are the electric and magnetic fields you found continuous across the conducting plane? Explain.

\end{enumerate}

## 1-2. Determining the electromagnetic potentials.

Augmenting the surface current density with a delta function we can form the current density for the system

With only a current distribution specified use of the Coulomb gauge allows for setting the scalar potential on the surface equal to zero, so that we have

Utilizing our Green’s function

we can invert our vector potential equation, solving for

Now a switch to polar coordinates makes sense. Let’s use

This gives us

Since the theta function imposes a

constraint, equivalent to

we can reduce the upper range of the integral and drop the theta function explicitly

Here I got lazy and used Mathematica to help evaluate this integral, for an end result of

## 3. Find the electric and magnetic fields outside the plane.

Our electric field can be calculated by inspection

For the magnetic field we have

which gives us

## 4. Give a physical interpretation of the results of the previous section.

It was expected that the lack of boundary on the conducting sheet would make the potential away from the plane only depend on the components of the spatial distance, and this is precisely what we find performing the grunt work of the integration.

Given that we had a sinusoidal forcing function for our wave equation, it seems logical that we also find our non-homogeneous solution to the wave equation has sinusoidal dependence. We find that the sinusoidal current results in sinusoidal potentials and fields very much like one has in the electric circuits problem that we solve with phasors in engineering applications.

We find that the electric and magnetic fields are oriented parallel to the plane containing the surface current density, with the electric field in the direction of the current, and the magnetic field perpendicular to that, but having energy propagate outwards from the plane.

It’s kind of curious that despite introducing a step function in time for the current, that the end result appears to have no constraints on time , and that we have fields at all points in space and time from this current distribution, even the case (ie: we have a non-causal system). I'd have expected a transient response to switching on the current. Perhaps this is because instantaneously inducing a current on all points in an infinite sheet is not physically realizable?

## 5. Find the direction and magnitude of the energy flux outside the plane.

Our energy flux, the Poynting vector, is

This is

This energy flux is directed outwards along the axis, with magnitude oscillating around an average value of

## 6. Sketch the intensity of the electromagnetic field far from the plane.

I’m assuming here that this question does not refer to the flux intensity , since that is constant, and boring to sketch.

The time varying portion of either the electric or magnetic field is proportional to

We have a sinusoid as a function of time, of period where the phase is adjusted at each position by the factor . Every increase of shifts the waveform back.

A sketch is attached.

## 7. Continuity across the plane?

It is sufficient to consider either the electric or magnetic field for the continuity question since the continuity is dictated by the sinusoidal term for both fields.

The point in time only changes the phase, so let’s consider the electric field at , and an infinitesimal distance . At either point we have

In the limit as the field strength matches on either side of the plane (and happens to equal zero for this case).

We have a discontinuity in the spatial derivative of either field near the plate, but not for the fields themselves. A plot illustrates this nicely

\begin{figure}[htp]

\centering

\includegraphics[totalheight=0.4\textheight]{phy450ps5P1Q7}

\caption{}

\end{figure}

# Problem 2. Fields generated by an arbitrarily moving charge.

## Statement

Show that for a particle moving on a worldline parametrized by , the retarded time with respect to an arbitrary space time point , defined in class as:

obeys

and

\begin{enumerate}

\item Then, use these to derive the expressions for and given in the book (and in the class notes).

\item Finally, re-derive the already familiar expressions for the EM fields of a particle moving with uniform velocity.

\end{enumerate}

## 0. Solution. Gradient and time derivatives of the retarded time function.

Let’s use notation something like our text [1], where the solution to this problem is outlined in section 63, and write

where

From we also have

so if we write

we have

Proceeding in the manner of the text, we have

From 2.20 we also have

so

This and 2.29 gives us

For the gradient we operate on the implicit equation 2.30 again. This gives us

However, we can also use the spatial definition of . Note that this distance is a function of space and time, since is implicitly a function of the spatial and time positions at which the retarded time is to be measured.

We have only this bit to expand, but that’s just going to require a chain rule expansion. This is easier to see in a more generic form

so we have

which gets us close to where we want to be

Putting the pieces together we have only minor algebra left since we can now equate the two expansions of

This is given in the text, but these in between steps are left for us and for our homework assignments! From this point we can rearrange to find the desired result

## 1. Solution. Computing the EM fields from the Lienard-Wiechert potentials.

Now we are ready to derive the values of and that arise from the Lienard-Wiechert potentials. We have for the electric field.

We’ll evaluate

For the electric field we’ll use the chain rule on the vector potential

Similarly for the gradient of the scalar potential we have

Our electric field is thus

For the magnetic field we have

The magnetic field will therefore be found by evaluating

Let’s compare this to

This equals 2.40, verifying that we have

something that we can determine even without fully evaluating .

We are now left to evaluate the retarded time derivatives found in 2.39. Our potentials are

It’s clear that the quantity is going to show up all over the place, so let’s label it . This is justified by comparing to a particle’s boosted rest frame worldline

where we have , so for the remainder of this part of this problem we’ll write

Using primes to denote partial derivatives with respect to the retarded time we have

so the electric field is

Here’s where things get slightly messy.

and messier

then a bit unmessier

Now we are set to plug this back into our electric field expression and start grouping terms

Using

We can verify that

which gets us closer to the desired end result

It is also easy to show that the remaining bit reduces nicely, since all the dot product terms conveniently cancel

This completes the exercise, leaving us with

Looking back to 2.45 where was defined, we see that this compares to (63.8-9) in the text.

## 2. Solution. EM fields from a uniformly moving source.

For a uniform source moving in space at constant velocity

our retarded time measured from the spacetime point is defined implicitly by

Squaring this we have

or

Rearranging to complete the square we have

Taking roots (and keeping the negative so that we have for the case, we have

or with , this is

What’s our retarded distance ? We get

For the vector distance we get (with )

For the unit vector we have

The acceleration term in the electric field is zero, so we are left with just

Leading to , we have

where, following section 38 of the text we write

This gives us

Observe that this equals one when as expected.

We can also compute

Our long and messy expression for the field is therefore

This gives us our final result

As a small test we observe that we get the expected result

for the case.

When this also recovers equation (38.6) from the text as desired, and if we switch to primed coordinates

we recover the field equation derived twice before in previous problem sets

## 2. Solution. EM fields from a uniformly moving source along x axis.

Initially I had errors in the vector treatment above, so tried with the simpler case using uniform velocity along the axis instead. Comparison of the two showed where my errors were in the vector algebra, and that’s now also fixed up.

Performing all the algebra to solve for in

I get

This matches the vector expression from 2.57 with the special case of so we at least started off on the right foot.

For the retarded distance we get

This also matches 2.58, so things still seem okay with the vector approach. What’s our vector retarded distance

So

For , using calculated above, or from 2.73 calculating directly I get

where, as in section 38 of the text, we write

Putting all the pieces together I get

so we have

This matches equation (38.6) in the text.

# Problem 3.

FIXME: TODO.

# Grading notes.

Only the first question was graded (I lost 1.5 marks). I got my units wrong when I integrated 1.11, and my factor in the result is circled. Should have also done a dimensional analysis check. There was also a remark that the integral is zero if , which introduces a function. I think that I incorrectly dropped my function, and should have retained.

FIXME: Revisit both of these and make sure to understand where exactly I went wrong.

# References

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

## Leave a Reply