- #1

Aleolomorfo

- 73

- 4

## Homework Statement

In an electromagnetic filed, the elctric field ##\vec{E}## forms an angle ##\theta## with the magnetic field ##\vec{B}##, and ##\theta## is invariant for all inertial observers. Finding the value of ##\theta##.

## Homework Equations

Tranformations of fields perpendicular to the boost:

$$\vec{E'}=\gamma(\vec{E}+(\vec{v}\times\vec{B})_\perp)$$

$$\vec{B'}=\gamma(\vec{B}-(\vec{v}\times\vec{E})_\perp)$$

## The Attempt at a Solution

I know that the only ##\theta## that is invariant in ##\frac{\pi}{2}## or ##\frac{3\pi}{2}##, so when the fields are perpendicular. I have always showed it in this way: I use the invariance of ##\vec{E}\cdot\vec{B}=\vec{E'}\cdot\vec{B'}##, so ##EB\cos{\theta}=E'B'\cos{\theta'}##. If ##\theta'=0##, ##\theta## must also be ##0## to hold the equality. Consequently if they are perpendicular in one frame, they are perpendicular in all the other, otherwise the invariance of ##\vec{E}\cdot\vec{B}## does not hold.

I know this is not the best proof, but I think it is in some way logically. However, the exercise wants a rigorous proof.

I have tried using ##\vec{E}\cdot\vec{B}## better but it is a vicious circle. I have aligned the x-axis along ##\vec{E}## and y-axis in order to have the two vectors in the ##xy## plane.

$$\vec{E}=E\vec{x}$$

$$\vec{B}=B(\cos{\theta}\vec{x}+\sin{\theta}\vec{y})$$

I have written ##E'## and ##B'## using the fields-transfromations for a boost along z:

$$E'_x=\gamma(E-vB\sin{\theta})$$

$$E'_y=\gamma v B\cos{\theta}$$

$$B'_x=\gamma B\cos{\theta}$$

$$B'_y=\gamma(B\sin{\theta}-vE)$$

But then everything I have done is inconclusive. I do not know how to conclude the demonstration.