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EM Field strength and plane waves

  1. Apr 15, 2009 #1
    Hi !

    I've a question. Where is the connection between the (kinetic) Lagrangian [tex] - \dfrac{1}{4} F_{\mu \nu} F^{\mu \nu} [/tex] and a plane wave of the form [tex] \vec{\varepsilon} exp(i \vec{k} \cdot \vec{x}) / \sqrt{V} [/tex] (the epsilon is a polarization vector) confined in a box with a finite volume V ? I should somehow "motivate" the factor [tex] - \dfrac{1}{4} [/tex] occuring in the Lagrangian by such plane waves. But I absolutely dont't have a clue how to do that. Does anyone have an idea? I hope somebody could help me.
     
  2. jcsd
  3. Apr 15, 2009 #2
    Blegh. It's a helluva long road. Quickest thing to do is to fix a gauge like the Lorenz gauge, and then find out what the Hamiltonian is. The factor of -1/4 will give you a canonical Hamiltonian that is something like [tex]\int\!dx\, \left(\frac{1}{2} \mathbf{E}^2 + \frac{1}{2}|\boldsymbol\nabla \mathbf{A}|^2\right)[/tex] where [tex]E \sim \dot{A}[/tex]. The Laplacian is a vector Laplacian, in case you're wondering. I have also dropped factors of c and [tex]\epsilon_0[/tex], since I can work in units in which they are 1.

    Ok, where were we? You can fourier expand your vector potential as follows:
    [tex]\mathbf{A} = \sum_\lambda \int\!\frac{dk}{(2\pi)^3} a_\lambda(\mathbf{k}) \boldsymbol{\epsilon}(\lambda)e^{i \mathbf{k}\cdot \mathbf{x} }[/tex],
    where the polarization vectors are chosen such that the gauge fixing condition is satisfied (i.e., only two transverse polarizations). If you're doing this inside a box, it's a plain fourier series, instead of an integral. If you write your Hamiltonian in terms of these, you'll obtain something like
    [tex]H = \int\!dk\, \left(\frac{1}{2} p_\lambda^2 + \frac{1}{2} k^2 a_\lambda^2\right)[/tex],
    where [tex]p[/tex] is the momentum conjugate to [tex]a[/tex] i.e., in the Lagrangian terminology, [tex]p = \dot{a}[/tex]. Now you get a bunch of simple harmonic oscillators.

    The factors of two are just convention, really, and keep the equations nice. The original factor of 4 comes about because when you obtain the Euler Lagrange equations, you take a derivative of a square. So to "motivate" all this, you could say that each plane wave is a separate harmonic oscillator. But you'll have to do some math =)
     
  4. Apr 16, 2009 #3
    Thank you! But is it somehow possible to explain the 1/4 factor by considering the dimensions?

    I've seen something like [tex] \left[ F_{\mu \nu} F^{\mu \nu} \right] = 4 [/tex], but I don't really understand why it is equal to 4.
    But if one use this, isn't it possible to somehow justify the 1/4 factor? I need a simple explanation without (much) calculations.
     
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