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 $\mathbf{\epsilon}_{2}$?!


The constraint you're missing is the Faraday law of induction. For a pure plane-wave mode, if you know the amplitude of the electric field, the Faraday law completely determines the magnetic field. Thus, you have two degrees of freedom in the electric field and none in the magnetic field.

  • $\begingroup$ Apologies I think I might have edited my question before realising that you'd posted an answer. With this in mind, is what I put in my OP correct at all? Do all four constraints come from Maxwell's equations such that the general solutions for $\mathbf{E}$ and $\mathbf{B}$ are constructed from a linear combination of two polarisation vectors $\mathbf{\epsilon}_{1}$ and $\mathbf{\epsilon}_{2}$?! $\endgroup$ – user35305 May 1 '17 at 12:40
  • $\begingroup$ Yes, it is correct. $\endgroup$ – Emilio Pisanty May 1 '17 at 12:42
  • $\begingroup$ Ok cool. So it's the Maxwell equations that provide the 4 constraints then meaning that there are two dof remaining that are encoded in the two polarisation vectors that one can construct general solutions from for the E and B fields. Are these what people refer to as the physical dof of the electromagnetic field (encoded in the polarisation vectors)?! $\endgroup$ – user35305 May 1 '17 at 13:52
  • $\begingroup$ Yes, that is correct. Note that it is two DOFs per mode, though. $\endgroup$ – Emilio Pisanty May 1 '17 at 14:38
  • $\begingroup$ Just to clarify, does that mean there are two DOF per $\mathbf{k}$ mode, i.e. two per $\mathbf{E}_{\mathbf{k}}^{0}e^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}$ for each $\mathbf{k}$?! $\endgroup$ – user35305 May 1 '17 at 16:04

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