Solving Proca Lagrangian w/ Extra Operator: Find Laws of Motion

  • #31
Yes, so how you apply this on #28? Btw, there you forgot the ##-\beta##.
 
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  • #32
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##
 
  • #33
Maniac_XOX said:
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.
 
  • #34
Gaussian97 said:
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 $$\frac {\partial (g_{\alpha\beta}A_\mu A^\mu)}{\partial A^\beta}$$ ?
 
  • #35
No, the derivative is wrt the 4-position, not the ##A## field.
 
  • #36
Gaussian97 said:
No, the derivative is wrt the 4-position, not the ##A## field.
Right, so the correct definition of that would be then $$\frac {\partial (g_{\alpha\beta}A_\mu A^\mu)}{\partial t} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu)}{\partial x} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu}{\partial y} + \frac {\partial(g_{\alpha\beta}A_\mu A^\mu}{\partial z}$$
 
  • #37
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)##
 
  • #38
Gaussian97 said:
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?
 
  • #39
Maniac_XOX said:
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.
 
  • #40
Gaussian97 said:
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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  • #41
Ok, now that's correct (except for some index that has raised magically)
 
  • #42
true i corrected it, is that as far as i can simplify it?
$$\partial^\beta F_{\beta\alpha} + 2A_\alpha (\partial_\rho A^\rho) \beta + \mu^2 A_\alpha =+ \partial_\alpha (A_\mu A^\mu)\beta +\frac {4\pi}{c} J_\alpha$$
 
  • #43
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.
 
  • #44
Gaussian97 said:
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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  • #45
[EDIT] all solved now finally, thank you for the help :)