Lorentz transformation where electric field vanishes

[castlemaster]

Homework Statement

We have an homogeneus electromagnetic field with [tex]E \bullet B=0[/tex] and [tex]E \neq cB[/tex]
Find the velocity of the reference frames in which ony E exists.

Homework Equations

[tex]\mathbf{E}' = \gamma \left( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right ) - \left (\frac{\gamma-1}{v^2} \right ) ( \mathbf{E} \cdot \mathbf{v} ) \mathbf{v}[/tex]

[tex]\mathbf{B}' = \gamma \left( \mathbf{B} - \frac {\mathbf{v} \times \mathbf{E}}{c^2} \right ) - \left (\frac{\gamma-1}{v^2} \right ) ( \mathbf{B} \cdot \mathbf{v} ) \mathbf{v}[/tex]

The Attempt at a Solution

I guess I can't use the transformations for a boost in the x direction, so I guess I have to use the fact that

[tex]E \bullet B[/tex]
[tex]E^2-B^2[/tex]

are invariants under lorentz transformations.
But I don't know how to start. Do I need the EM field tensor for something?

Thanks

 

[grey_earl]

First, if you don't set c = 1, your second invariant has a factor of c² missing (I let you find out where). But you won't need them for the solution to your problem anyway.

For this, you have already your transformation formulae for E and B, so just set B' = 0 ...

 

[castlemaster]

Hi,

Should I use the fact that E.B=E'.B'=0 and merge the equations from E' and B'?
Cause I don't see a simple way of taking the velocity out of there.

Thanks

 

[grey_earl]

Since you know that E.B = 0, surely, as an ansatz, v.B = 0 would greatly simplify the second equation. Then can v be parallel to E? Or must it be perpendicular also?

(If you don't believe in an ansatz, take E = (E, 0, 0) and B = (0, B, 0) and v = (vx,vy,vz). You can always choose your coordinate system that way, but in the end you have to figure out how to describe it without coordinate systems, which can be a bit ugly).

 

[castlemaster]

E can't be parallel to v or B' won't vanish, right?

 

[grey_earl]

Yes. So since now we know that v must be perpendicular to E and B, we can write v = s E x B, with s a scalar constant. Now you just plug in this v into your equation for B', use some vector analysis and find out s. Then, |v| = |s| |E| |B|.

(BTW: v can have a component parallel to E, but that won't contribute in the equation for B'. So what you calculate above is really the minimum velocity you need to make B' vanish. The question is ambiguous.)

 

[castlemaster]

Maybe it easier to see with the equations for B' and E' parallel and perpendicular to the velocity (I've read those are equivalent to the above for E' and B' http://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity)

Thanks