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Electromagnetic Field Lagrangian - Field Equations

  1. Aug 15, 2013 #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})##
     
  2. jcsd
  3. Aug 15, 2013 #2

    WannabeNewton

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    Perhaps a simpler example will get you up to speed. Consider the Klein-Gordon Lagrangian density in flat space-time ##\mathcal{L} = -\frac{1}{2}(\eta^{ab}\partial_{a}\varphi \partial_{b}\varphi + m^2 \varphi^2)##.

    We have ##\frac{\partial
    \mathcal{L}}{\partial \varphi} = -m^{2}\varphi## and ##\frac{\partial\mathcal{L}}{\partial(\partial_{a} \varphi)} = -\frac{1}{2}\{\eta^{bc}\partial_{b}\varphi \frac{\partial (\partial_{c}\varphi)}{\partial (\partial_{a}\varphi)} + \eta^{bc}\partial_{c}\varphi \frac{\partial (\partial_{b}\varphi)}{\partial (\partial_{a}\varphi)}\}\\ = -\frac{1}{2}\{\eta^{bc}\partial_{b}\varphi\delta^{a}_{c} + \eta^{bc}\partial_{c}\varphi \delta^{a}_{b}\}\\ = -\partial^{a}\varphi##
    where ##\frac{\partial (\partial_{a}\varphi)}{\partial (\partial_{b}\varphi)} = \delta^{b}_{a}## was used. This gives the usual Klein-Gordon field equations ##\partial_{a}\partial^{a}\varphi - m^{2}\varphi = 0 ##. The calculation for the electromagnetic field is extremely similar; you just have a 4-potential ##A_{b}## instead of a scalar potential ##\varphi## so that for example ##\frac{\partial (\partial_{a}A_{b})}{\partial (\partial_{c}A_{d})} = \delta^{c}_{a}\delta^{d}_{b}##.
     
  4. Aug 15, 2013 #3

    dextercioby

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    You're making things more complicated than they are. And you should use indices consistently:

    [tex] L= -\frac{1}{16\pi} F_{\mu\nu}F^{\mu\nu}= -\frac{1}{16\pi}\left(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\right)(...) [/tex]

    So

    [tex]\frac{\partial L}{\partial A_{\alpha}} = 0 [/tex]

    and

    [tex]\frac{\partial L}{\partial \left(\partial_{\beta}A_{\alpha}\right)} = \left[\frac{1}{8\pi}\frac{\partial}{\partial \left(\partial_{\beta}A_{\alpha}\right)} \left(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\right)\right] (...) [/tex]

    Now use

    [tex] \frac{\partial\left(\partial_{\mu}A_{\nu}\right)}{\partial \left( \partial_{\beta}A_{\alpha}\right)} = \delta^{\beta}_{\mu}\delta^{\alpha}_{\nu} [/tex]
     
  5. Aug 15, 2013 #4
    aah, thanks.

    I was wondering if I get 2 ##\delta##'s.

    What should the derivative in the Lagrangian look like (to be compatible with the ##\mathcal{L}## terms)?

    It shouldn't contain any indices, which are used inside the ##\mathcal{L}##, right?

    In my example ##\frac{\partial A^{\alpha}_{\ ,\nu}}{\partial A^{\nu}_{\ ,\mu}}##

    I should get ##\delta^{\alpha}_{\nu} \delta^{\mu}_{\nu}##, right?

    This does not look right, so I think I have to use something like ##\frac{\partial \mathcal{L}}{\partial A^{\sigma}_{\ ,\tau}}## ?

    And in the 2nd part just the opposite?

    [tex] \frac{\partial\left(\partial^{\mu}A^{\nu}\right)}{\partial \left(\partial^{\beta} A^{\alpha}\right)} = \delta^{\mu}_{\beta} \delta^{\nu}_{\alpha} [/tex]
     
  6. Aug 15, 2013 #5

    dextercioby

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    The nu's must be on diagonal, if summed over.
     
  7. Aug 15, 2013 #6
    Using a slightly different Euler-Lagrange Equation:

    ##\frac{\partial}{\partial x^\tau} \frac{\partial \mathcal{L}}{\partial A^{\sigma}_{\ ,\tau}} - \frac{\partial \mathcal{L}}{\partial A^\sigma} = 0##

    Using this Lagrangian:

    ##\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})##

    Then ##\frac{\partial \mathcal{L}}{\partial A^\sigma} = 0##

    Now for example:

    ##-\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} \frac{\partial}{\partial A^{\sigma}_{\ ,\tau}} (A^{\alpha}_{\ ,\nu} A^{\mu}_{\ ,\beta})##

    which leads to

    ##-\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} A^{\alpha}_{\ ,\nu} (\frac{\partial}{\partial A^{\sigma}_{\ ,\tau}} A^{\mu}_{\ ,\beta})##

    and

    ##-\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} (\frac{\partial}{\partial A^{\sigma}_{\ ,\tau}} A^{\alpha}_{\ ,\nu}) A^{\mu}_{\ ,\beta}##

    where

    ##(\frac{\partial}{\partial A^{\sigma}_{\ ,\tau}} A^{\mu}_{\ ,\beta}) = \delta^{\mu}_{\sigma} \delta^{\tau}_{\beta}##

    and

    ##(\frac{\partial}{\partial A^{\sigma}_{\ ,\tau}} A^{\alpha}_{\ ,\nu}) = \delta^{\alpha}_{\sigma} \delta^{\tau}_{\nu}##
     
  8. Aug 15, 2013 #7

    dextercioby

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    Yes, using the partial notation ∂ for derivatives is better, it has more visibility.
     
  9. Aug 15, 2013 #8
    Now I ended up with

    ##-4 k A_{\sigma}^{\ ,\tau} + 3 k A^{\tau}_{\ ,\sigma} + k A^{\mu}_{\ ,\mu}##

    ##k = \frac{1}{16\pi}##

    where the last term looks like a divergence.

    I wonder if this is the correct path...
     
  10. Aug 15, 2013 #9

    WannabeNewton

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    I must echo dexter here and say that you are making this much more complicated than it needs to be. I'm going to drop the ##4\pi## in the denominator and just write ##\mathcal{L} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + A_{\mu}j^{\mu}## (your exercise does it for the case of vanishing 4-current i.e. ##j^{\mu} = 0## but there's no real difference in the computation regardless). Hence ##\frac{\partial \mathcal{L}}{\partial A_{\nu}} = j^{\nu}##. Now ##\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}A_{\nu})} = -\frac{1}{4}\eta^{\alpha \rho}\eta^{\beta \sigma}\{\frac{\partial F_{\alpha \beta}}{\partial (\partial_{\mu}A_{\nu})}F_{\rho\sigma} +\frac{\partial F_{\rho\sigma}}{\partial (\partial_{\mu}A_{\nu})}F_{\alpha\beta} \}## and ##F_{\alpha\beta} = 2\partial_{[\alpha}A_{\beta]}## so just evaluate.
     
    Last edited: Aug 15, 2013
  11. Aug 15, 2013 #10
    I failed on one index...

    Using this Euler-Lagrange:

    ##\frac{\partial}{\partial x^\tau} \frac{\partial \mathcal{L}}{\partial A^{\sigma}_{\ ,\tau}} - \frac{\partial \mathcal{L}}{\partial A^\sigma} = 0##

    Now I have

    ##4 k A^{\tau}_{\ ,\sigma} - 4 k A_{\sigma}^{\ ,\tau}##

    which looks a lot better.

    Now I just raise the ##\sigma## index with ##\eta^{\sigma\alpha}##

    then I get ##4 k ( A^{\tau ,\alpha} - A^{\alpha ,\tau} )##

    which is ##4 k F^{\alpha\tau}##

    doing ##\partial_\tau## from Euler-Lagrange

    leads to ##4 k \partial_\tau F^{\alpha\tau} = 0##.

    -----------------------------------------

    I just wanted to try my way :)

    Correction: He meant "current-free"...
     
    Last edited: Aug 15, 2013
  12. Aug 15, 2013 #11

    WannabeNewton

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    Well as long as it worked out :)
     
  13. Aug 15, 2013 #12
    Two pages and 400 indices later, yes.
    It was a total mess, especially that little index mistake :D

    NOW I will try your way.

    Where

    ##\frac{\partial F_{\rho\sigma}}{\partial (\partial_{\mu}A_{\nu})} = \frac{\partial (\partial_\rho A_\sigma - \partial_\sigma A_\rho )}{\partial (\partial_{\mu}A_{\nu})} = \delta^{\mu}_{\rho} \delta^{\nu}_{\sigma} - \delta^{\mu}_{\sigma} \delta^{\nu}_{\rho}##

    and

    ##\frac{\partial F_{\alpha\beta}}{\partial (\partial_{\mu}A_{\nu})} = \frac{\partial (\partial_\alpha A_\beta - \partial_\beta A_\alpha )}{\partial (\partial_{\mu}A_{\nu})} = \delta^{\mu}_{\alpha} \delta^{\nu}_{\beta} - \delta^{\mu}_{\beta} \delta^{\nu}_{\alpha}##

    Somehow my lazyness prevents me from finishing this :D

    This is indeed much nicer, thanks!

    Thanks to you both!

    [Another apple of knowledge thrown at me by Newton...]
     
    Last edited: Aug 15, 2013
  14. Aug 15, 2013 #13

    WannabeNewton

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    Don't worry about doing it another way; if you got it once then it's fine. If you're interested: ##\eta^{\alpha \rho}\eta^{\beta \sigma}\{\frac{\partial F_{\alpha \beta}}{\partial (\partial_{\mu}A_{\nu})}F_{\rho\sigma} +\frac{\partial F_{\rho\sigma}}{\partial (\partial_{\mu}A_{\nu})}F_{\alpha\beta} \}\\ = 2\eta^{\rho [\mu}\eta^{\nu]\sigma}F_{\rho\sigma} + 2\eta^{\alpha [\mu}\eta^{\nu]\beta}F_{\alpha\beta} = 4F^{\mu\nu}##
    so ##\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}A_{\nu})} = -F^{\mu\nu}## hence ##\partial_{\mu}F^{\nu\mu} = j^{\nu}##.
     
  15. Aug 15, 2013 #14
    Actually I am working on it right now [hahah], but thanks :)

    EDIT:

    done

    ##\partial_\mu F^{\mu\nu} = 0##

    looks good
     
  16. Aug 15, 2013 #15
    I was wondering about the whole thing using ##\delta## notation...

    Using the Lagrangian

    ##\mathcal{L} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}##

    where

    ##\delta F_{\mu\nu} = \delta (\partial_\mu A_\nu - \partial_\nu A_\mu) = \delta (\partial_\mu A_\nu) - \delta (\partial_\nu A_\mu) = \partial_\mu (\delta A_\nu) - \partial_\nu (\delta A_\mu)##

    Now the action

    ##I = \int \mathcal{L} \ d^4x##

    The variation of the action

    ##\delta I = \int \delta (-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}) d^4x = 0##

    where ##F_{\mu\nu}## goes to zero at the surface of the integration volume. [???]

    Using the 'product rule'

    ##\delta I = \int -[(\frac{1}{4} (\delta F^{\mu\nu}) F_{\mu\nu}) + (\frac{1}{4} (\delta F_{\mu\nu}) F^{\mu\nu})] d^4x = 0##

    where

    ##\delta F^{\mu\nu} = \delta (\partial^\mu A^\nu - \partial^\nu A^\mu) = \delta (\partial^\mu A^\nu) - \delta (\partial^\nu A^\mu) = \partial^\mu (\delta A^\nu) - \partial^\nu (\delta A^\mu)##

    now plug in, do some multiplication and then partial integration, that's it?

    The remaining term should be the field equation then?
     
  17. Aug 15, 2013 #16

    WannabeNewton

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    See section E.1 of Wald.
     
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