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

user35305
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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.

Emilio Pisanty
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  • 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 $\mathbf{\epsilon}{2}$?! – user35305 May 01 '17 at 12:40
  • Yes, it is correct. – Emilio Pisanty May 01 '17 at 12:42
  • 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)?! – user35305 May 01 '17 at 13:52
  • Yes, that is correct. Note that it is two DOFs per mode, though. – Emilio Pisanty May 01 '17 at 14:38
  • 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 $\mathbf{k}$?! – user35305 May 01 '17 at 16:04
  • There are two DOFs per $\mathbf k$. – Emilio Pisanty May 01 '17 at 17:14
  • Ah ok cool, thanks for the clarification. By mode do you mean this in the sense that the full solution will be an integral over $\mathbf{k}$ of the solutions for each $\mathbf{k}$, $\mathbf{E}{\mathbf{k}}^{0}e^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}$?! Are these solutions $\mathbf{E}{\mathbf{k}}^{0}e^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}$ what are referred to as the modes of $\mathbf{k}$?! – user35305 May 01 '17 at 18:11
  • Please take this kind of back-and-forth to chat. – Emilio Pisanty May 01 '17 at 18:23
  • Let us continue this discussion in chat. – user35305 May 01 '17 at 19:02