- #1
ProfDawgstein
- 80
- 1
I was working on an exercise in Ohanian's book.
[Appendix A3, page 484, Exercise 5]
I guess he means charge conservation, but wrote ##j^\nu = 0##.
The Lagrangian was given by ##\mathcal{L}_{em} = -\frac{1}{16\pi} \left( A_{\mu ,\nu} - A_{\nu ,\mu} \right) \left( A^{\mu ,\nu} - A^{\nu ,\mu} \right)##.
Which should be the same as ##\mathcal{L}_{em} = -\frac{1}{16\pi} F_{\mu\nu} F^{\mu\nu}##.
The Euler-Lagrange Equation is ##\frac{\partial}{\partial x^\mu} \frac{\partial \mathcal{L}}{\partial A^{\nu}_{\ ,\mu}} - \frac{\partial \mathcal{L}}{\partial A^\nu} = 0##.
##\frac{\partial \mathcal{L}}{\partial A^\nu} = 0## because there are no ##A^\nu## terms.
Now, because there are different components (covariant, contravariant), I insert some ##\eta_{\mu\nu}## to make them all look the same.
-> making the derivatives easier later
This leads to ##\mathcal{L} = -\frac{1}{16\pi} (\eta_{\mu\alpha} A^{\alpha}_{\ ,\nu} - \eta_{\alpha\nu} A^{\alpha}_{\ ,\mu}) (\eta^{\beta\nu} A^{\mu}_{\ ,\beta} - \eta^{\beta\mu} A^{\nu}_{\ ,\beta})##.
Now just multiply...
Then I get this
##\mathcal{L} = -\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} A^{\alpha}_{\ ,\nu} A^{\mu}_{\ ,\beta} + ... + ... - ...##
Now I am supposed to do
##\frac{\partial \mathcal{L}}{\partial A^{\nu}_{\ ,\mu}}##
which means
##-\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} \frac{\partial}{\partial A^{\nu}_{\ ,\mu}} (A^{\alpha}_{\ ,\nu} A^{\mu}_{\ ,\beta})##
now product rule
...
then I get terms like this
##\frac{\partial A^{\alpha}_{\ ,\nu}}{\partial A^{\nu}_{\ ,\mu}}##
My problem:
1) what about the indices?
-do I need to rename all of ##{\partial A^{\nu}_{\ ,\mu}}## or just the ##,\mu##?
2) what do I get out of that?
-my thoughts were something like ##\delta^a_b##
What I did was
##\frac{\partial A^{\alpha}_{\ ,\nu}}{\partial A^{\nu}_{\ ,\mu}} = \delta^\mu_\nu##
-ignoring the top (contravariant) indices
-just using the ##,\mu## and ##,\nu##
Thank you in advance.
I would be grateful if somebody could show the result of
##-\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} \frac{\partial}{\partial A^{\nu}_{\ ,\mu}} (A^{\alpha}_{\ ,\nu} A^{\mu}_{\ ,\beta})##
[Appendix A3, page 484, Exercise 5]
Show that ##\mathcal{L}_{em}## leads to the field equation ##\partial_\mu \partial^\mu A^\nu - \partial^\nu \partial_\mu A^\mu = 4\pi j^\nu## with ##[\partial_\nu] j^\nu = 0##.
I guess he means charge conservation, but wrote ##j^\nu = 0##.
The Lagrangian was given by ##\mathcal{L}_{em} = -\frac{1}{16\pi} \left( A_{\mu ,\nu} - A_{\nu ,\mu} \right) \left( A^{\mu ,\nu} - A^{\nu ,\mu} \right)##.
Which should be the same as ##\mathcal{L}_{em} = -\frac{1}{16\pi} F_{\mu\nu} F^{\mu\nu}##.
The Euler-Lagrange Equation is ##\frac{\partial}{\partial x^\mu} \frac{\partial \mathcal{L}}{\partial A^{\nu}_{\ ,\mu}} - \frac{\partial \mathcal{L}}{\partial A^\nu} = 0##.
##\frac{\partial \mathcal{L}}{\partial A^\nu} = 0## because there are no ##A^\nu## terms.
Now, because there are different components (covariant, contravariant), I insert some ##\eta_{\mu\nu}## to make them all look the same.
-> making the derivatives easier later
This leads to ##\mathcal{L} = -\frac{1}{16\pi} (\eta_{\mu\alpha} A^{\alpha}_{\ ,\nu} - \eta_{\alpha\nu} A^{\alpha}_{\ ,\mu}) (\eta^{\beta\nu} A^{\mu}_{\ ,\beta} - \eta^{\beta\mu} A^{\nu}_{\ ,\beta})##.
Now just multiply...
Then I get this
##\mathcal{L} = -\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} A^{\alpha}_{\ ,\nu} A^{\mu}_{\ ,\beta} + ... + ... - ...##
Now I am supposed to do
##\frac{\partial \mathcal{L}}{\partial A^{\nu}_{\ ,\mu}}##
which means
##-\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} \frac{\partial}{\partial A^{\nu}_{\ ,\mu}} (A^{\alpha}_{\ ,\nu} A^{\mu}_{\ ,\beta})##
now product rule
...
then I get terms like this
##\frac{\partial A^{\alpha}_{\ ,\nu}}{\partial A^{\nu}_{\ ,\mu}}##
My problem:
1) what about the indices?
-do I need to rename all of ##{\partial A^{\nu}_{\ ,\mu}}## or just the ##,\mu##?
2) what do I get out of that?
-my thoughts were something like ##\delta^a_b##
What I did was
##\frac{\partial A^{\alpha}_{\ ,\nu}}{\partial A^{\nu}_{\ ,\mu}} = \delta^\mu_\nu##
-ignoring the top (contravariant) indices
-just using the ##,\mu## and ##,\nu##
Thank you in advance.
I would be grateful if somebody could show the result of
##-\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} \frac{\partial}{\partial A^{\nu}_{\ ,\mu}} (A^{\alpha}_{\ ,\nu} A^{\mu}_{\ ,\beta})##