# Generic Local Electromagnetic field - MTW Ex 4.1

1. Sep 28, 2013

### TerryW

Can anyone help me with Ex 4.1 in MTW?

What is a 'generic' field? My expectation is that it would comprise an Electric Field, with arbitrary direction, a Magnetic Field, also with arbitrary direction, and a radiation field (E and B of equal magnitude) , radiating in an arbitrary direction.

Trying to put this together however to get the Poynting density of energy flow and the density of energy is a bit of a mess however, so I suspect my expectation is wrong.

Moving on to the next bit of the problem, if you do define the unit vector n and velocity parameter α as shown, then a solution to this is a radiation field with E along the x axis, B along the y axis, producing a Poynting vector along the z axis. If you then translate to the rocket frame, E X B disappears because B becomes zero, not because B becomes parallel to E. I can't see how B ends up parallel to E.

Can anyone help?

TerryW

2. Sep 28, 2013

### Bill_K

No, just the first two - an arbitrary E field and an arbitrary B field. The "radiation field", more commonly called a null field, is just a special case.

This is one of the three special cases that you were told NOT to consider. Excluding these three cases, you'll get E and B parallel and both nonzero.

3. Sep 28, 2013

### WannabeNewton

A generic EM field is simply one which, relative to a given (in this case inertial) frame, splits into an arbitrary E field and an arbitrary B field.

Why do you think the B field will become zero once we boost to the rocket frame in general i.e. why should the $\alpha$ parameter for the boost velocity necessarily take us to the rest frame of the source of this EM field?

4. Sep 29, 2013

### TerryW

Hi Bill_K and WannabeNewton,

Thanks for clearing up what the generic field is. I've re-read MTW and can see it now but I don't feel it is totally clear (unusually for MTW!)

As for thinking that the B field will become zero, I noticed that a solution to the equation for the ratio of energy flow to energy density is Bxsinhα and Eycoshα with E = B,
giving an E X B in the z direction. So when you translate to the rocket frame travelling along the z axis, B becomes zero. At which point I realised I was well off track.

I've already had a go at resolving the problem based on what you have told me. I haven't cracked it yet and will drop another post if I get terminally stuck.

I appreciate your help so far and hope you will be on hand if I need further help.

Regards

TeryW

5. Oct 9, 2013

### TerryW

Still having problems.

My attempt at proving that the Electric and Magnetic fields become parallel and aligned to the direction of travel of the rocket is attached in a PDF. I've tried to use the invariants E2 - B2 and the equation for tanh2α n to get some more expressions involving the parallel and perpendicular components of E and B which could be used to reduce my expression for EXB to zero but without success.

On the other hand, considering the expression EXB = (E2 + B2)tanh2α n, which is a generic expression for a generic filed...I could say that if we were in the rocket frame, then we are now moving with the flow of energy, so α = 0, leading to the required result, but maybe that is a bit of a cheat?

Regards

TerryW

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6. Jan 8, 2014

### TerryW

Sorted

I've revisited the problem and worked out that my expression for EXB does all cancel out. The first two lines γ[…] and γβ[…] disappear because the parallel components are zero then I use the tanh2α formula to show that γ2[…] + γ2β[…..] - γ2β2[….] reduces to zero.

7. Jan 8, 2014

### WannabeNewton

That's awesome Terry! I'm glad it worked out.

8. Jan 8, 2014

### TerryW

Looking at my original manuscript, if I recognise early on that the parallel components of E and B are zero, I can get to my last expression for ExB more or less straight away!

Much tidier!

TerryW