Maxwell Tensor Symmetries Problem - Federico

[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!

 

[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?

 

[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!

 

[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.

 

[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.

 

[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?

 

[Federico]

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