Recent content by Federico

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

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.

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!

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=...