# Proca Lagrangian with extra operator - find Laws of motion

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Maniac_XOX
No, the derivative is wrt the 4-position, not the ##A## field.
Right, so the correct definition of that would be then $$\frac {\partial (g_{\alpha\beta}A_\mu A^\mu)}{\partial t} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu)}{\partial x} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu}{\partial y} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu}{\partial z}$$

Homework Helper
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)##

Maniac_XOX
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)##
But i thought that's what the 4-position was, and i substituted the betas everywhere relevant?

Homework Helper
But i thought that's what the 4-position was, and i substituted the betas everywhere relevant?
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.

Maniac_XOX
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.
Yeah I'm not planning on expanding ##\beta##. I think i understood now that ##\partial^\beta## can only be operated as $$g^{\beta\alpha} \partial_\alpha = g^{\beta\alpha} \frac{\partial}{\partial x^\alpha}$$

The most i have managed to simplify so far is to this point:
$$\partial^\beta F_{\beta\alpha} -\partial_\alpha (A_\mu A^\mu)\beta + \mu^2 A_\alpha =-2A_\alpha \beta (\partial_\rho A^\rho) + \frac {4\pi}{c} J_\alpha$$

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Homework Helper
Ok, now that's correct (except for some index that has raised magically)

Maniac_XOX
true i corrected it, is that as far as i can simplify it?
$$\partial^\beta F_{\beta\alpha} + 2A_\alpha (\partial_\rho A^\rho) \beta + \mu^2 A_\alpha =+ \partial_\alpha (A_\mu A^\mu)\beta +\frac {4\pi}{c} J_\alpha$$

Homework Helper
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.

Maniac_XOX
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.
This time the components of ##\partial x^\alpha## are $$\frac {\partial_t}{c} + \frac{\partial}{\partial x} + \frac {\partial}{\partial y} + \frac {\partial}{\partial z}$$

Found that the equation could also become
but it could also become: $$\partial^\beta F_{\beta\alpha}+\mu^2 A_\alpha =+ \beta(A_\nu \partial_\alpha A^\nu + A_\mu \partial_\alpha A^\mu -2A_\alpha \partial_\rho A^\rho) +\frac {4\pi}{c} J_\alpha$$

Any further tips to simplify? or is this alright :)

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Maniac_XOX
[EDIT] all solved now finally, thank you for the help :)