# Generic Local Electromagnetic field - MTW Ex 4.1

Gold Member
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

Bill_K
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.
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.

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.
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.

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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?

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Gold Member
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

Gold Member
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

#### Attachments

• EXB.pdf
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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.

WannabeNewton