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From QED to Maxwell

  1. Apr 14, 2009 #1

    Ben Niehoff

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    I have some confusion as to how to get from the QED Lagrangian to ordinary electrodynamics. Beginning with the QED Lagrangian (density):

    [tex]\mathcal L = \bar{\psi} (i\gamma^{\mu} D_{\mu} - m) \psi - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

    I expand the covariant derivative [itex]D = \partial - iqA[/tex] to obtain

    [tex]\mathcal L = \bar{\psi} i\gamma^{\mu} \partial_{\mu} \psi + q \bar{\psi} \gamma^{\mu} \psi A_{\mu} - m \bar{\psi} \psi - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

    From here, one can either add a total divergence, or take the sum

    [tex]\frac12 (\mathcal L + \mathcal L^{\dagger}) = \frac12 \bar{\psi} i \gamma^{\mu} \overleftrightarrow{\partial_{\mu}} \psi + q \bar{\psi} \gamma^{\mu} \psi A_{\mu} - m \bar{\psi} \psi - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

    which I kinda like because it is more symmetrical. Anyway, using gauge invariance and Noether's theorem, we can get the conserved current

    [tex]j^{\mu} = q \bar{\psi} \gamma^{\mu} \psi[/tex]

    This couples to the EM field in the usual way,

    [tex]\mathcal L = j^{\mu}A_{\mu} - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

    But now the question is, how do I include the kinetic part of the Lagrangian? I would like to end up with something like

    [tex]\mathcal L = m \sqrt{\dot x^{\mu} \dot x_{\mu}} + q \dot x^{\mu} A_{\mu} - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

    but I can't seem to do anything with my current [itex]j^{\mu}[/itex] that resembles the kinetic part of the QED Lagrangian. The 4-divergence identically vanishes (by Noether's theorem), so the only scalar I can form is

    [tex]j^{\mu} j_{\mu} = q^2 (\bar{\psi} \gamma^{\mu} \psi) (\bar{\psi} \gamma_{\mu} \psi)[/tex]

    which I can't see how to simplify. Maybe I am missing something. Maybe what I am trying to do can't actually be done? Basically, I'd like to rewrite the QED Lagrangian in terms of [itex]j^{\mu}[/itex], and (hopefully) recover the classical Lagrangian. Any suggestions?

    Edit:

    Ah, I see one thing I can try. I might be able to make something of it by exploiting

    [tex]i\partial_{\mu} = (i \partial_t, i \vec \nabla) = (E, - \vec p) = p_{\mu}[/tex]

    I'll look into this later.
     
    Last edited: Apr 14, 2009
  2. jcsd
  3. Apr 14, 2009 #2

    Hans de Vries

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    This is possible, but don't expect a square root. This is because we are
    dealing with the Lagrangian density so all terms of the classical Lagrangian
    pick up an extra gamma term due to the Lorentz contraction of the volume.

    You can see this happening in equation 22.31 going to 22.32 here:

    http://physics-quest.org/Book_Chapter_Lagrangian.pdf

    You also might want to read section 22.3

    Now, the kinetical part of the Lagrangian also represents quadratic terms
    in the form of E^2 - p^2, because if you look at.

    [tex] \frac12 \bar{\psi} i \gamma^{\mu} \overleftrightarrow{\partial_{\mu}} \psi [/tex]

    then you see that [itex]\gamma^\mu[/itex] makes it a vector current {p0,px,py,pz} and the
    derivative operating on the imaginary exponential then multiplies these
    into a quadratic kinetic expression {p0^2-px^2-py^2-pz^2}


    Regards, Hans
     
    Last edited: Apr 14, 2009
  4. Apr 17, 2009 #3

    Ben Niehoff

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    OK, that helps, but there is still the mass term

    [tex]m \bar \psi \psi[/tex]

    which I can't seem to do anything with. If psi is a plane wave

    [tex]\psi = u(p,s) e^{-ip \cdot x}[/tex]

    then I get

    [tex]m \bar \psi \psi = m \bar u u = \frac{m^2}{E_p}[/tex]

    leaving me with the Lagrangian

    [tex]L = m \dot x^{\mu} \dot x_{\mu} - \frac{m^2}{E_p} + q \dot x^{\mu} A_{\mu} - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

    However, the Lagrangian I ought to be able to get is

    [tex]L = \frac12 m \dot x^{\mu} \dot x_{\mu} + q \dot x^{\mu} A_{\mu} - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

    But here I'm not sure what to do. If I expand

    [tex]\frac{m^2}{E_p} = \frac{m}{\sqrt{1 + \dot x^{\mu} \dot x_{\mu}}} = m \left( 1 - \frac12 \dot x^{\mu} \dot x_{\mu} + ... \right)[/tex]

    then I get something that is close to what I need. But it doesn't make sense that I would have to make an expansion here...

    Edit: Actually, that last line is wrong. I'll fix it in a bit.
     
  5. Apr 17, 2009 #4
    It seems to me that you have to try

    [tex]
    \psi => e^{-iS(x)}
    [/tex]

    as in the Hamilton-Jacoby approach.

    Bob.
     
  6. Apr 17, 2009 #5
    Hamilton–Jacobi equations...


    The solution I derived based upon the Hamilton–Jacobi equation, specifically the Eikonal approximation and relationship to the Schrödinger equation:

    S represent the phase of a wave with unitless exponential argument:
    [tex]\psi = \psi_{0} e^{\frac{iS}{\hbar}}[/tex]

    Electrodynamic Lagrangian:
    [tex]L = m \dot x^{\mu} \dot x_{\mu} - m \bar \psi \psi + q \dot x^{\mu} A_{\mu} - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

    The Lagrangian mass term Lagrangian Hamilton–Jacobi Eikonal identity:
    [tex]m \bar \psi \psi = m \bar u u = \frac{m^2}{E_p} = \frac{1}{2} m \dot x^{\mu} \dot x_{\mu}[/tex]

    Lagrangian integration by substitution:
    [tex]L = m \dot x^{\mu} \dot x_{\mu} - \frac{1}{2} m \dot x^{\mu} \dot x_{\mu} + q \dot x^{\mu} A_{\mu} - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

    Electrodynamic Lagrangian solution:
    [tex]\boxed{L = \frac{1}{2} m \dot x^{\mu} \dot x_{\mu} + q \dot x^{\mu} A_{\mu} - \frac14 F_{\mu\nu} F^{\mu\nu}}[/tex]


    Reference:
    http://en.wikipedia.org/wiki/Hamilt...elationship_to_the_Schr.C3.B6dinger_equation"
     
    Last edited by a moderator: Apr 24, 2017
  7. Apr 18, 2009 #6

    Ben Niehoff

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    Re: Hamilton–Jacobi equations...

    You lost me here. How do you establish the last equals sign? That's the crucial step.
     
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