Physical degrees of freedom of an Electromagnetic field

[Frank Castle]

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

 

[marcusl]

Yes! See this thread for confirmation:
https://www.physicsforums.com/threads/redundant-degrees-of-freedom-in-em-fields.787007/

 

[Frank Castle]

Yes! See this thread for confirmation:
https://www.physicsforums.com/threads/redundant-degrees-of-freedom-in-em-fields.787007/

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)?

 

[marcusl]

Yes, that's a nice way to put it.

 

[Frank Castle]

Yes, that's a nice way to put it.

Great. Thanks for your help.