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I am able to prove in a few lines that the electrodynamic field vectors $\vec{E}$, $\vec{H}$ and $\vec{S}$ are all orthogonal to each other considering that $\vec{E}$ and $\vec{H}$ are coherent plane waves with constant complex amplitudes propagating in the $\vec{k}$ direction. See for example JACKSON J.D. Classical Electrodynamics (§7.1, Plane Waves in a Nonconducting Medium, 1st ed. (1962), pp. 204-205, eqs. 7.9-7.15).

My question is: is there any way to prove the same but considering waves of amplitudes varying with space $(x,y,z)$?

I haven't seen this even mentioned in any of the books I've consulted, they all consider plane waves with constant amplitudes. I tried to prove it on my own using the different methods I've seen for constant amplitudes but they all fail.

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$\vec{S}$ is certainly orthogonal to both $\vec{E}$, $\vec{H}$ because it is defined as their vector product. But the orthogonality of $\vec{E}$ and $\vec{H}$ is far less trivial. I can't see any simple proof now but I believe that one can show their orthogonality. – Ondřej Černotík Nov 22 '13 at 16:32
Yes, the problem is really with showing orthogonality between $\vec{H}$ and $\vec{E}$. If you happen to come across a proof for this, I'd really appreciate it if you report back! – Mike D. Nov 22 '13 at 16:48
I must mention, in this discussion, that the answer is likely to be negative. The inner product $\mathbf E\cdot \mathbf B$ is a fundamental invariant of the field, $\frac14F_{\mu\nu}{}^\ast F^{\mu\nu}$. As the Wikipedia article makes clear, the radiation fields you're interested in will have that equal to zero, but you need to introduce a suitably strong hypothesis that distances you from the non-null case. – Emilio Pisanty Nov 25 '13 at 11:42
up vote 1 down vote accepted

As Ondřej points out, $\mathbf{S}$ is orthogonal to both $\mathbf{E}$ and $\mathbf{B}$ as it's defined as their cross product. However, $\mathbf{E}$ and $\mathbf{B}$ are in general not orthogonal to each other.

This is obvious in the general case: you can make static fields using Helmholtz coils and electrodes, completely independent of each other, for which $\mathbf{E}\cdot\mathbf{B}$ is completely arbitrary. In a sense this is trivially true and not interesting to the (radiation) regime you're asking about. However, it does make it clear that you need to be very precise with what sort of fields you are allowing, and what general premises you're starting with, because if they are too broad then the result will not be true.

Also, I'd like to point out that there's no such thing as "plane waves of amplitudes varying with space". Plane waves are plane waves, period. The most general plane wave possible has electric field $$\mathbf{E}(\mathbf{r},t)=\rm{Re}\left(\mathbf{E}_0e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}\right)$$ and it is specified uniquely by a real wavevector $\mathbf{k}$ and a (possibly complex) polarization vector $\mathbf{E}_0$ such that $\mathbf{E}_0\cdot\mathbf{k}=0$. The amplitude stays constant and the wave spans all of space.

What I think you're after, though, is monochromatic fields, which are much more general, and which can be written in the form $$\mathbf{E}(\mathbf{r},t)=\rm{Re}\left(\mathbf{E}_0(\mathbf{r})e^{-i\omega t}\right),$$ with a similar equation for $\mathbf{B}$, and for which the Maxwell equations are the standard ones if you replace $\mathbf{E}$ and $\mathbf{B}$ with the possibly complex $\mathbf{E}_0$ and $\mathbf{B}_0$, and all temporal partial derivatives with $-i\omega$ (and where you need to impose, because of the above counterexample, the nontriviality condition $\omega\neq0$). If this is indeed the régime you're after, then I'll try to think of a proof or a counter-example.

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My regime is indeed monochromatic fields, I thought plane waves could have non-constant amplitudes and still be called plane waves. To be precise I'm considering optical TE or TM modes in a waveguide. $\vec E\left( {\vec r,t} \right) = \vec E\left( {\vec r} \right)\exp \left( { - i\omega t} \right) = E\left( x \right)\hat y\exp \left\{ {i\left( {\beta z - \omega t} \right)} \right\}$ – Mike D. Nov 22 '13 at 17:55
I think that the interesting question is : "are the electric and magnetic fields produced by a unique source always orthogonal ?". That way, we rule out the rather trivial counter-example of two different sources. – Adam Nov 22 '13 at 18:20
@Adam I would think it will be quite a challenge to formulate in any precise way the fact that a radiation field is due to "two sources" instead of one. The concept of a field is precisely a way to abstract out the source of the EM forces and simply have a local property. – Emilio Pisanty Nov 24 '13 at 2:04
@EmilioPisanty: I see your point. Though I think that one can reformulate the problem in this way : given the movements of a point like charge, can on prove that the local EM field at a given time are orthogonal or not ? But I agree that it might not be a very well defined question and a challenge to answer anyway. – Adam Nov 24 '13 at 6:25

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