Right, so the correct definition of that would be then $$\frac {\partial (g_{\alpha\beta}A_\mu A^\mu)}{\partial t} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu)}{\partial x} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu}{\partial y} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu}{\partial z}$$No, the derivative is wrt the 4-position, not the ##A## field.
But i thought that's what the 4-position was, and i substituted the betas everywhere relevant?No, first of all, if you expand the sum over ##\beta##, you must substitute ##\beta## everywhere. But even then, what you have is still not 100% correct, because for the derivative to be a contravariant vector ##\partial^\beta## you must differentiate wrt a covariant vector, i.e. ##x_\beta##, and the components of ##x_\beta## are not ##(t,x,y,z)##
Those are the contravariant coordinates of the ##x## 4-vector, you want to take derivatives with respect to the covariant coordinates.But i thought that's what the 4-position was, and i substituted the betas everywhere relevant?
Yeah I'm not planning on expanding ##\beta##. I think i understood now that ##\partial^\beta## can only be operated as $$g^{\beta\alpha} \partial_\alpha = g^{\beta\alpha} \frac{\partial}{\partial x^\alpha}$$Those are the contravariant coordinates of the ##x## 4-vector, you want to take derivatives with respect to the covariant coordinates.
I don't know what you mean by relevant, if you expand ##\beta## you cannot have ##\beta## in the final expression, otherwise you did it wrong.
In any case, you can simplify this expression without any need of expanding any term.
This time the components of ##\partial x^\alpha## are $$\frac {\partial_t}{c} + \frac{\partial}{\partial x} + \frac {\partial}{\partial y} + \frac {\partial}{\partial z}$$Well, you can use the product rule to simplify it a little the derivative, and then collect the terms proportional to ##\beta##, but I don't think this will make a big difference.