Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is the full EM + matter lagrangian?

  1. Mar 14, 2008 #1
    This is a followup to the old thread What are the FULL classical electrodynamic equations? which never really provided a satisfactory answer.

    I have decided to phrase it perhaps in a more straightforward manner. Given that we have the EM field, or the equivalent potential field, and charged matter. What is the lagrangian for this system?

    I will be happy if given in terms of point charges or charge density. But it has to include both the action of the field on the charges and the fact that the sources of the field are those same charges. Writing down the lagrangian for an EM field with fixed sources or point particles influenced by a fixed field is easy and in every textbook.

    Has anyone ever even seen what I am asking for here?

    For an example of what I am talking about, here is the lagrangian density for spin 0 charges + EM field from quantum field theory. (This is just conceptual. Signs and constants might be wrong. H-bar and c are set to 1.)

    [tex] (i\partial_{\mu}\phi^{\dag}-eA_{\mu})(i\partial^{\mu}\phi-eA^{\mu}) + m\phi^{\dag}\phi -\frac{1}{16}F^{\alpha\beta}F_{\alpha\beta}[/tex]

    I am looking for the classical analog to this lagrangian.
    Last edited: Mar 14, 2008
  2. jcsd
  3. Mar 14, 2008 #2


    User Avatar

    [tex] -m \int d\tau \sqrt{- g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu} + \int dx^4 j^\mu A_\mu - \frac{1}{4} \int d^4 x F^{\alpha \beta} F_{\alpha \beta} [/tex]

    with the current being
    [tex] q \int d\tau ~ \delta^4(x-x(\tau)) \frac{dx^\mu (\tau)}{d\tau} [/tex]

    (I'm not sure of all the signs)
    Is that what you were looking for?
  4. Mar 14, 2008 #3
    Yep. That looks like it. But let me ponder it a bit. Might be a couple days. Please check back then so that I can ask you half a dozen questions. :biggrin:

    But, really. Thanks!
    Last edited: Mar 14, 2008
  5. Mar 14, 2008 #4


    User Avatar

    You're very welcome! :smile:
  6. Mar 14, 2008 #5


    User Avatar

    Looking at the other thread I noticed that the Lagrangian had already been given! (by two people)
  7. Mar 16, 2008 #6
    kdv, I haven't much chance for this yet. But I can clarify a couple things first, please. What you have provided is the action not the lagrangian or lagrangian density itself, right?

    So the interaction term written out fully looks like

    [tex]q \int dx'^4 \int d\tau ~ \delta^4(x'-x(\tau)) \frac{dx^\mu (\tau)}{d\tau} A_\mu(x') [/tex]

    correct? I put in a prime on the x as variable of integration because its double use as particle position confused me, esp in the delta function. Am I ok so far?

    I'm trying to get to the point where I could understand how to write the Euler-Lagrange equations from this action.
    Last edited: Mar 16, 2008
  8. Mar 16, 2008 #7


    User Avatar

    This is correct.

    Note that the Dirac delta can be used to do the four-dimensional integral which leaves only a one-dimensional integral over the path of the particle. This si what you see in some books: the interaction part of the action is an integral over the path of the particle, not a four-dimensional integral.
  9. Mar 17, 2008 #8
    I don't know if I would want to do that. It gives the correct value for the action of course, but wouldn't it affect being able to make a variation in the system and looking for the extremum of the action integral?

    The main problem is rather general: how to deal with a system consisting of both discrete particles and continuous fields from a calculus of variations approach? My curiosity is aroused here. I'm sure this situation must arise in other contexts and has been dealt with. I'm going to dig around in my books.

    The other approach is to use continuous charge distributions. Then the interaction is [tex]\int dx^4 j^\mu A_\mu[/tex], period. We would just have to replace the free matter term for point particles with one in terms of j.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook