Do you mean? $$A_\mu \partial_\alpha A^\mu - A_\alpha \partial_\rho A^\rho$$Maniac_XOX said:right, that sorted it, cheers
is there any way that ##A_\mu \partial_\alpha A^\mu - A_\alpha \partial^\rho A^\rho## can be simplified as well or are the indices too different in this case. It seems to me that it can't be simplified
right, that still helped tbh, learned a lot since this whole indices thing has been seriously new to me and trying to learn how to use general relativity has been like trying to start in a new language lol, cheers for the helpPeroK said:Do you mean? $$A_\mu \partial_\alpha A^\mu - A_\alpha \partial_\rho A^\rho$$
You can standardise the dummy indices:$$A_\mu \partial_\alpha A^\mu - A_\alpha \partial_\mu A^\mu$$ But, as one summation is over partial derivatives and one summation is over the components of ##A##, there's nothing meaningful to be done.
Would it make sense to use Lorenz gauge condition ##\partial_\mu A^\mu=0## and turn the equations of motion into $$\Box A_\alpha + \mu^2 A_\alpha = 2\beta A_\mu \partial _\alpha A^\mu + \frac {4\pi}{c} J_\alpha$$ and then use the ##\partial_\alpha## as gauge covariant derivative expanding it into ##\partial_\alpha - i q A_\alpha -i q B_\alpha## ?? Or is that only for ##D_\alpha##? I've seen it being used in both notations on wikipediaPeroK said:Do you mean? $$A_\mu \partial_\alpha A^\mu - A_\alpha \partial_\rho A^\rho$$
You can standardise the dummy indices:$$A_\mu \partial_\alpha A^\mu - A_\alpha \partial_\mu A^\mu$$ But, as one summation is over partial derivatives and one summation is over the components of ##A##, there's nothing meaningful to be done.