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Schwartz QFT problem 3.9

  1. Jul 11, 2014 #1
    1. The problem statement, all variables and given/known data
    I am not sure about the problem set up.

    For (a), Using Equation of motion, need to express Lagrangian in terms of only J?

    I got, [tex]L=-\frac{1}{2 \Box^2 }(\partial_\mu J_\nu)^2 - \frac{{J_\mu}^2}{\Box}[/tex], using lorentz gauge

    (b) [tex] \partial_\mu J_\mu =0[/tex] means [tex]k_\mu J_\mu =0 ? [/tex]

    For (d), I can't understand what it's asking. In momentum space, the term without time derivative is the first term in (a)??

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 12, 2014 #2

    vanhees71

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    How should we help you, if you don't even post the question?
     
  4. Jul 15, 2014 #3

    WannabeNewton

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    Yes.

    Yes.

    ##\omega \rightarrow \partial_t## when going from Fourier space to configuration space. Causally propagating terms have ##\partial_t## to some order e.g. ##\partial^2_t##. Non-causal ones only have spatial derivatives e.g. ##\nabla^2## with no ##\partial^2_t## present. The question is asking why the latter are considered non-causal. What are your thoughts on this? As a simple example, consider interactions in Newtonian gravity ##\nabla^2 \varphi = 4\pi \rho##. In what sense is this non-causal?
     
  5. Jul 21, 2014 #4

    For newton's gravity, the green function wouldn't contain time, which means can't distinguish source and field point for time
     
  6. Jul 24, 2014 #5

    WannabeNewton

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    Yes exactly so there is no retardation of the Green's function. Causality of field propagation is codified by the retarded Green's function of a given field equation. But there is an important distinction to be made between the case of Newtonian gravity and classical EM, the latter of which is considered in the problem. In Newtonian gravity the non-causal mode is physical, albeit incorrect because Newtonian gravity is not an inherently relativistic theory. On the other hand, in EM the non-causal mode, which in this case corresponds to ##\frac{J_0}{k^2} \rightarrow \frac{J_0}{\nabla^2}##, is unphysical because it is just an artifact of the choice of gauge.
     
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