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Maxwell Lagrangian

  1. Nov 14, 2012 #1
    Hello,

    Where can I find a good explanation (book) of the derivation via Noether's theorem of the three momentum and angular momentum operators of the usual maxwell lagrangian ?

    Thank you!
     
  2. jcsd
  3. Nov 14, 2012 #2

    dextercioby

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    This is standard QFT (actually QED) material, any thorough book should have it. Check out a nice treatment in Chapter 2 of F. Gross' "Relativistic Quantum Mechanics and Field Theory", Wiley, 1999.

    In purely classical context (no operators), advanced electrodynamics books should also have this.
     
  4. Nov 14, 2012 #3
    I've been watching the book and yes, the book treats it but don't deduce them. He just announces and perform some calculations with them
     
  5. Nov 14, 2012 #4

    dextercioby

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    Can you calculate [itex] T^{\mu\nu} [/itex] and [itex] M^{\lambda}_{~~\mu\nu} [/itex] from the Lagrangian and the general Noether formula which for the energy-momentum 4 tensor reads:

    [itex] T^{\mu}_{~~\nu} [/itex] = ([itex] \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}A_{\rho})}[/itex] [itex] -\mathcal{L}\delta^{\mu}_{\lambda} [/itex]) X [itex] \frac{\partial x'^{\lambda}}{\partial\epsilon^{\nu}} [/itex],

    where

    [tex] x'^{\mu} = x^{\mu} + \epsilon^{\mu} [/tex]
     
    Last edited: Nov 14, 2012
  6. Nov 14, 2012 #5
    I'll try it.
     
  7. Nov 14, 2012 #6
    [tex] T^{\mu\nu}=-F^{\mu\nu}\partial^{\nu}A_{\rho}+\frac{1}{4}F^{2}g^{\mu\nu}[/tex]

    And now? How I relate this to the momentum and total angular momentum operators ?
     
  8. Nov 14, 2012 #7

    dextercioby

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    The momentum should be [itex] T^{0i} [/itex], just like energy is [itex] T^{00} [/itex]. For angular momentum, you should derive the general formula using the linearized version of a general Lorentz transformation (i.e. a linearized space-time rotation):

    x'μ=xμμ ν xν, where

    ϵμν = - ϵνμ

    A minor change

    Tμν=−FμρνAρ+1/4 F2gμν
     
  9. Nov 20, 2012 #8

    Meir Achuz

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    I too would be interested in seeing this for EM angular momentum.
    Every place, I have looked seems to use the result in some form without actually deriving it.
     
  10. Nov 25, 2012 #9
    Maggiore "Modern introduction in QFT"
     
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