# How to find the magnitude of electric and magnetic fields in an arbitary inertial frame?

In the context of special relativity, and using proper Lorentz transformations:

If the electromagnetic fields have an uniform value [= constant in space and time] in an inertial frame, how will the magnitude of these fields change in a different (arbitrary) inertial frame?

For example, an inertial frame which makes an angle $\alpha$, or moves with a different velocity $v'$ with respect to the original one.

• There was some useful information in the comments here, but the discussion was getting rather long, so I've deleted them. I'll just leave here the link to the help center, and also mention that Physics Meta or Physics Chat are the places to ask for clarification on how the site works. – David Z Dec 1 '15 at 7:45

The relevant transformations, where ${\bf v}$ is a velocity (vector) of a "primed" frame with respect to the stationary "lab frame" and $\gamma$ is the Lorentz factor at speed $v$, are
$$\mathbf {{E}}' = \mathbf{{E}_{\parallel}} + \gamma \left( \mathbf {E}_{\bot} + \mathbf{ v} \times \mathbf {B} \right)$$ $$\mathbf {{B}}' = \mathbf{{B}_{\parallel}} + \gamma \left( \mathbf {B}_{\bot} -\frac{1}{c^2} \mathbf{ v} \times \mathbf {E} \right),$$ where the parallel and perpendicular subscripts refer to components of the E- and B-field that are either parallel or perpendicular to ${\bf v}$. i.e. The components parallel to ${\bf v}$ end up unchanged but the perpendicular components are transformed.