# How to show that the two EM invariants are a complete set?

1. Dec 15, 2015

### bcrowell

Staff Emeritus
In electromagnetism we have these two Lorentz scalars:

$P=B^2-E^2$

$Q=E\cdot B$

WP https://en.wikipedia.org/wiki/Classification_of_electromagnetic_fields claims that these are a complete set of invariants, because "every other invariant can be expressed in terms of these two." How does one prove this? Would the idea be to show that any electromagnetic field tensor can be rendered into one of a set of canonical forms by boosts and rotations? Or maybe you could fiddle with the eigenvalues of the field tensor?

Is the claim only true for invariants that are continuous functions of the field tensor (i.e., continuous functions of its components)?

2. Dec 15, 2015

### bcrowell

Staff Emeritus
I haven't worked this out fully, but the following might do the job. In the general case, where neither P nor Q vanishes, I think you can do a rotation and a boost such that E and B both lie along the x axis. Then clearly there are only two degrees of freedom, corresponding to the x components of the two fields, so there can only be two invariants.

3. Dec 15, 2015

### phyzguy

I don't think this can be the case. If E.B is non-zero, and it is an invariant, then there is no frame where both E and B lie along the x-axis, because then E.B = 0.

4. Dec 16, 2015

### bcrowell

Staff Emeritus
Sorry, I don't follow you. If E and B are parallel and both nonzero, then their dot product is nonzero.

5. Dec 16, 2015

### andresB

6. Dec 16, 2015

### phyzguy

You're right of course. Should teach me to post when I'm tired. Please ignore my comment.

7. Dec 16, 2015

### bcrowell

Staff Emeritus