Proca equation with nonlinear operator A²∂·A

[Gaussian97]

Yes, so how you apply this on #28? Btw, there you forgot the $-\beta$.

 

[Maniac_XOX]

i ended up using the property $g_{\mu\gamma}\delta^\gamma_\alpha=g_{\mu\alpha}$ to come up with this solution:
$$\partial^\beta F_{\beta\alpha} + \partial^\beta(A_\mu A^\mu g_{\rho\beta} \delta^\rho_\alpha) + \mu^2 A_\alpha =(\partial_\rho A^\rho) (g_{\mu\alpha} A^\mu + g_{\gamma\alpha} A^\gamma) + \frac {4\pi}{c} J_\alpha$$which finally becomes$$\partial^\beta F_{\beta\alpha} + \partial^\beta(A_\mu A^\mu g_{\alpha\beta}) + \mu^2 A_\alpha = (A_\alpha + A_\alpha) + \frac {4\pi}{c} J_\alpha$$
so then the right hand side becomes $2A_\alpha + \frac {4\pi}{c} J_\alpha$

 

[Gaussian97]

i ended up using the property $g_{\mu\gamma}\delta^\gamma_\alpha=g_{\mu\alpha}$ to come up with this solution:
$$\partial^\beta F_{\beta\alpha} + \partial^\beta(A_\mu A^\mu g_{\rho\beta} \delta^\rho_\alpha) + \mu^2 A_\alpha =(\partial_\rho A^\rho) (g_{\mu\alpha} A^\mu + g_{\gamma\alpha} A^\gamma) + \frac {4\pi}{c} J_\alpha$$which finally becomes$$\partial^\beta F_{\beta\alpha} + \partial^\beta(A_\mu A^\mu g_{\alpha\beta}) + \mu^2 A_\alpha = (A_\alpha + A_\alpha) + \frac {4\pi}{c} J_\alpha$$
so then the right hand side becomes $2A_\alpha + \frac {4\pi}{c} J_\alpha$

Apart from the $-\beta$ factor, you also lose a $\partial_\rho A^\rho$. But yes essentially is correct. You can still simplify the RHS a little.

 

[Maniac_XOX]

Apart from the $-\beta$ factor, you also lose a $\partial_\rho A^\rho$. But yes essentially is correct. You can still simplify the RHS a little.

Right i forgot those factors while focusing on the other stuff but yeah it's still there. Not sure how i can simplify more but a question i have involving the LHS is:
Does $\partial^\beta(g_{\alpha\beta}A_\mu A^\mu)$
mean the same as $$\frac {\partial (g_{\alpha\beta}A_\mu A^\mu)}{\partial A^\beta}$$ ?

 

[Gaussian97]

No, the derivative is wrt the 4-position, not the $A$ field.

 

[Maniac_XOX]

No, the derivative is wrt the 4-position, not the $A$ field.

Right, so the correct definition of that would be then $$\frac {\partial (g_{\alpha\beta}A_\mu A^\mu)}{\partial t} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu)}{\partial x} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu}{\partial y} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu}{\partial z}$$

 

[Gaussian97]

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?

 

[Gaussian97]

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:
$$\partial^\beta F_{\beta\alpha} -\partial_\alpha (A_\mu A^\mu)\beta + \mu^2 A_\alpha =-2A_\alpha \beta (\partial_\rho A^\rho) + \frac {4\pi}{c} J_\alpha$$

 

[Gaussian97]

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?
$$\partial^\beta F_{\beta\alpha} + 2A_\alpha (\partial_\rho A^\rho) \beta + \mu^2 A_\alpha =+ \partial_\alpha (A_\mu A^\mu)\beta +\frac {4\pi}{c} J_\alpha$$

 

[Gaussian97]

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 :)

 

[Maniac_XOX]

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