# Determinant of the electromagnetic matrix

1. Apr 15, 2013

### zn5252

hi there,
In this wikipedia article https://en.wikipedia.org/wiki/Electromagnetic_tensor
we have the following invariant :

FαβFμη εαβμη = 8 E*B

However the determinant is the square of this quantity divided by 8, i.e. ( E*B )2 .

Now from the definition of the determinant for a 4x4 matrix , we have :

MiaMjbMkcMid εijkl = εabcd det(M) [D]

Now If I raise the expression 1/8 FαβFμη εαβμη to the power of 2, I would get :

1/8 FαβFμη εαβμη 1/8 FθλFσω εθλσω [E]

Now If I compare this expression with equation D above, I see that some indices do not fall into the right place and also in Equation D, we have the expression for the determinant , however in expression E, we see that we have the matrix F multiplied 4 times much like in expression D (or is it rows or columns that get multiplied).

We should also bear in mind that the magnetic field is the spatial part of F and thhat the electric field is the time part :

Fi0 = E and Fijεijk = Bk

How can we reconcile expression E with D then ? or is this an error perhaps ?

2. Apr 16, 2013

### zn5252

As always, I end up trying to answer my own question :) . So that others benefit .

We have for a skew symmetric tensor :

det(F) = Pf(F)2 , we would need to show that 1/8 εαβγδFαβFγδ = Pf(F)

Now since F is skew symmetric, we would find 2!2!2! = 8 terms which have similar (and counting the levi civita tensor evenness and oddness) indices. There are 3 such terms, namely

8 FαβFγδ - 8 FαγFβδ + 8 FαδFβγ where for Levi civita, the 0123 combination is taken as +1
and so one would obtain the Pf(F). No need to prove that D and E are equivalent.
Hope I got it right.

3. Apr 17, 2013

### zn5252

Typo : that should have been (2!*2!)*2 and not 2!*2!*2! - sorry