# Maxwell Tensor Symmetries Problem

1. Jul 4, 2012

### Federico

Hi community:

I'm Federico and I'm new user here!

I'm trying to show that the Electromegnetic Field Tensor

$F_{ab}$ = 2A(r) $(e_{0})_{[a}(e_{1})_{b]}$ + 2B(r) $(e_{2})_{[a}(e_{3})_{b]}$

where $(e_{0},e_{1},e_{2},e_{3})$ is the tetrad basis associated with the metric

$ds^2= -f(r)dt^2+h(r)dr^2+r^2dθ^2+r^2sin^2(θ)d\varphi^2$

has the same symmetries that this metric (static and spherical symmetry).

I`ve tried using the Lie Derivative in the direction of the Killing fields of this metric, but the algebra becomes a little complicated.

Any ideas on this issue?

Thanks a lot!

2. Jul 4, 2012

### Muphrid

Hello Federico, I'm familiar with tetrads, but could you elaborate in the meaning of the square brackets you used in your expression for the EM tensor?

3. Jul 4, 2012

### Federico

Yes, no problem: the square brackets means antisymmetric

$(e_{0})_{[a}(e_{1})_{b]}$=$\frac{1}{2}[(e_{0})_{a}(e_{1})_{b} - (e_{0})_{b}(e_{1})_{a}]$

thanks!

4. Jul 4, 2012

### Muphrid

All right, that makes sense. U can't give an in-depth analysis right this moment, but do you have an idea of the tensor that converts between the tetrad and coordinate bases? I suspect if you choose this to be have a specific form, the properties you want will hold.

5. Jul 4, 2012

### Federico

yes, looking the metric, it let me know that a good choice is:

$(e_{0})_{a}=\sqrt{f}(dt)_{a}$
$(e_{1})_{a}=\sqrt{h}(dr)_{a}$
$(e_{2})_{a}=r(d\theta)_{a}$
$(e_{3})_{a}=rsin(\theta)(d\varphi)_{a}$

I mean, the tensor is diagonal.

6. Jul 4, 2012

### Muphrid

Yeah, I mean, I know that's not the only gauge choice that gives the metric, but it's easy and it works. Nothing is a function of time, so I think you're okay there. How about applying a rotation matrix in theta or phi and verifying by hand that spherical symmetry is still manifest?

7. Jul 4, 2012

### Federico

Ok, I'll try with that and let you know later. Thanks a lot for the idea!