- #1

- 80

- 1

## Main Question or Discussion Point

I was working on an exercise in Ohanian's book.

[Appendix A3, page 484, Exercise 5]

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]

I guess he means charge conservation, but wrote ##j^\nu = 0##.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##.

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})##