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Guess solution to fields by minimizing the action.

  1. Jan 3, 2013 #1
    Suppose I want to try and guess the fields of a short laser pulse. We know that fields that satisfy Maxwell's equations minimize the action E^2 - B^2 (say far from charge and current)?

    For a plane wave E^2 - B^2 = 0?

    Will a general solution of maxwell's equations satisfy E^2 - B^2 = 0

    If I have a set of fields (say the approximate solution given by Jackson above) can I be guaranteed that the action for the actual solution will be less then the action for the approximate solution?

    Can I consider E^2 the "kinetic" part of the energy and consider B^2 the "potential" part of the energy?

    Thanks for any help or suggestions!
  2. jcsd
  3. Jan 4, 2013 #2


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    Yes, a single plane wave

    [tex] \mathbf{E} = \hat{\mathbf{i}} E_0 \sin[k (z-ct)] [/tex]

    satisfies [itex] c\mathbf{B} = \hat{\mathbf{k}}\times \mathbf{E}[/itex], so that [itex] |\mathbf{E}|^2 - c^2|\mathbf{B}|^2 = 0[/itex]

    Definitely not in the presence of sources (for which there are additional terms involve the source current and charge in the action). For example, an electric charge distribution at rest has [itex]\mathbf{B}=0[/itex]. However, a perfect superposition of plane waves will also satisfy this relation. By perfect, I mean that they are all in the same plane of polarization, but can have different magnitudes and phases.

    Not if by actual solution you mean real-life case. Take the previous example of a perfect superposition of plane waves. A real laser pulse is not perfectly collimated: different components can have slightly different directions. Suppose we have two plane waves with directions [itex]\hat{\mathbf{e}}_1,\hat{\mathbf{e}}_2[/itex]. Then you can show that

    [tex] |\mathbf{E}_1+\mathbf{E}_2|^2 - c^2|\mathbf{B}_1+\mathbf{B}_2|^2 = (1- \hat{\mathbf{e}}_1\cdot\hat{\mathbf{e}}_2)\mathbf{E}_1\cdot \mathbf{E}_2.[/tex]

    The approximation where the plane waves have the same directions, [itex]\hat{\mathbf{e}}_1\cdot\hat{\mathbf{e}}_2=1[/itex], has vanishing action, whereas the realistic case where they do not has a finite action (which could be positive). In this case, we would conclude that the approximation of the source-free, vacuum field equations is incorrect. Instead, we must more accurately describe the source of the pulse, as well as the nature of the cavity.

    As an idealized mathematical problem, an approximate solution (taken for instance in the sense of perturbation theory) will always have an action that is larger than the exact solution. Some more or less-relevant references can be found at http://www.scholarpedia.org/article/Principle_of_least_action

    From a classical mechanics perspective, this almost makes sense, because E involves time derivatives of the potential, while B involves spatial derivatives. However, if we express the action in terms of relativistic expressions, the entire Maxwell action should be thought of as the kinetic "part" of the action. Because of Lorentz invariance, the time derivatives are not on any different footing than the spatial derivatives. The potential part of the action would involve the source terms that I alluded to above. The correct expression for the EM energy is obtained applying Noether's theorem to the action. I don't believe that Jackson covers this in detail, but perhaps Landau and Lifgarbagez or Goldstein's classical mechanics books do.
  4. Jan 4, 2013 #3
    Thank you fzero for quite a lesson! I just need to learn faster then I forget.
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