Recent content by carpinus

thanks to everyone

I'm pondering over the meaning of the quantity ## ϵ_{abcd} ## occurring in (4.2.22). In his text Wald says ## ϵ_{abcd} ## is (a totally antisymmetric) tensor. Then according to (4.2.22) ## B_a ## should be a tensor (4-vector) as well. However it seems to me ## ϵ_{abcd} ## is considered to be the...

would you like to complete your answer?

I am curious: is this true only for cartesian coordinates or for general coordinates? thank you for the link.

I think you are right I wanted to refer only to the space like part: ##v^b = (1,\vec v), v= | \vec v | =0 ##

now it seems to work. at the beginning some strange things happened. e.g. it was as though I had to log out and log in again to get the code rendered. at some point I used 'save draft' and 'delete draft'. thank you very much for your advice

hello, 1. according to Robert Wald, General Relativity, equation (4.2.22) the magnetic field as measured by an observer with 4-velocity ## v^b ## is given by ## B_a = - \frac {1}{2} {ϵ_{ab}}^{cd} F_{cd} v^b ## where ## {ϵ_{ab}}^{cd}##, the author says, is the totally antisymmetric tensor (for...

sorry, I was unclear I tried using the dollar or the hash sign from the beginning and they don't work

hello, I am trying to use latex in a thread, but for some reason it does not work, eg. v^a is displayed as v^a I am using firefox browser version 99. many thanks for your help

thank you for your answers. As far as I understand, pa and va both relate to the same laboratory frame, whereas E is the energy of the particle in the rest frame of the observer. However, the 3-momentum only enters equation (4.2.8) if the laboratory frame is not the rest frame of the observer...

Hello, this is my first thread. Robert Wald, in General Relativity, equation (4.2.8) says : E = – pa va where E is the energy of a particle, pa the energy-momentum 4-vector and va the 4-velocity of the particle. How can I see this is compatible with the common energy-momentum-relation E2 – p2 =...