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I Physical degrees of freedom of an Electromagnetic field

  1. May 1, 2017 #1
    As I understand it, the classical source-free electric, ##\mathbf{E}## and magnetic, ##\mathbf{B}## wave equations are solved by solutions for the electric and magnetic fields of the following form: $$\mathbf{E}=\mathbf{E}_{0}e^{i (\mathbf{k}\cdot\mathbf{x}-\omega t)}$$ $$\mathbf{B}=\mathbf{B}_{0}e^{i (\mathbf{k}\cdot\mathbf{x}-\omega t)}$$

    Naively counting the degrees of freedom (dof) at this point it would appear that the electromagnetic field has 6 dof.

    However, is it correct that Maxwell's equations provide 4 constraints: $$\mathbf{k}\cdot\mathbf{E}_{0}=0 \\ \mathbf{k}\cdot\mathbf{B}_{0}=0$$ $$\mathbf{E}_{0}=-\frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}\mathbf{k}\times\mathbf{B}_{0}$$ and $$\mathbf{B}_{0}=\sqrt{\mu_{0}\varepsilon_{0}}\mathbf{k}\times\mathbf{E}_{0}$$
    Thus reducing the number of physical dof to 2?!

    If the above is correct what do these remaining dof correspond to? Are they simply the two possible polarisation (unit) vectors ##\mathbf{\epsilon}_{1}##, ##\mathbf{\epsilon}_{2}## that one can construct such that $$\mathbf{k}\cdot\mathbf{\epsilon}_{1}=\mathbf{k}\cdot\mathbf{\epsilon}_{2}=0$$ and $$\mathbf{k}\times\mathbf{\epsilon}_{1}=\mathbf{\epsilon}_{2}\\ \mathbf{k}\times\mathbf{\epsilon}_{2}=-\mathbf{\epsilon}_{1}$$ and hence ##\lbrace\mathbf{k},\;\mathbf{\epsilon}_{1},\;\mathbf{\epsilon}_{2}\rbrace## form an orthornormal basis, such that the general solutions for ##\mathbf{E}## and ##\mathbf{B}## are linear combinations of ##\mathbf{\epsilon}_{1}## and ##\mathbf{\epsilon}_{2}##?!
     
  2. jcsd
  3. May 2, 2017 #2

    marcusl

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  4. May 3, 2017 #3
    Ok cool.
    So are the two physical degrees of freedom simply the choice of a component for the polarisation ##\mathbf{\epsilon}_{1}##, and the choice of a component for the polarisation vector ##\mathbf{\epsilon}_{2}## (in principle, they both have 3 dof, but the requirement that ##\mathbf{k}\cdot\mathbf{\epsilon}_{1}=\mathbf{k}\cdot\mathbf{\epsilon}_{2}=0## and ##\mathbf{\epsilon}_{1}\cdot\mathbf{\epsilon}_{2}=0##, reduces their dof to one each)?
     
  5. May 3, 2017 #4

    marcusl

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    Yes, that's a nice way to put it.
     
  6. May 3, 2017 #5
    Great. Thanks for your help.
     
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