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I'm looking for a proof that the electric and magnetic fields in a plane wave are perpendicular that doesn't invoke complex E and B fields. I haven't been able to find one.

If the proof requires complex fields, can you explain why?

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  • $\begingroup$ By complex, do you mean phasor representation? $\endgroup$ – Alfred Centauri Mar 31 '15 at 23:37
  • $\begingroup$ Yes, phasors fall under that category $\endgroup$ – Luke Burns Mar 31 '15 at 23:41
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In one sense the problem is simple, in that any solution with complex numbers (say with phasors) can be literally translated into a real version (just equate the real and imaginary parts of each complex equation as two real equations). However many phasor type setups are designed to only look for solutions of a particular type, so it might be considered question begging.

It is a tricky problem. The reason is that there are many many solutions to Maxwell's equation, and only after specifying the boundary conditions do we get a unique solution. If you specify the correct boundary conditions, then you will get a plane wave, and the electric and magnetic fields will be orthogonal, but it can look like a mathematical chicken-and-egg problem when you try to make a proof.

There are many possible starting points, you might assume that there is a nonzero poynting vector that is uniform. You might assume a sinusoidal solution with surfaces of constant phase. You might specify an initial spatial slice and evolve it into a full 4d spacetime solution. You might specify boundary conditions on just a surface in space, but specify them to be harmonic in time.

And these requirements are necessary. For instance, a constant electric field is unphysical (in the sense that it has an infinite total field energy, but then again so does a perfect plane wave), but a constant and uniform electromagnetic field is a solution to the vacuum Maxwell equation, so you can add it to a vacuum plane wave solution, and it will be another vacuum solution. And if your plane wave is polarized so the electric field goes in the $\pm \hat{x}$ direction and propagates in the $\hat{z}$ direction then you can point your constant electric field in the $\hat{y}$ and have your consnat magnetic field be zero, and then the total electric and total magnetic fields are not orthogonal. So it is not required for a vacuum solution, it is a particular feature of particular solutions.

So if you pick a particular solution you can just note that it has that property. Otherwise you need something that is specific enough to pull out one of those solutions (and not any of the many where the fields are not orthogonal), and then show that it gives you the property you want.

So does the plane wave require those boundary conditions, or do the boundary conditions require the plane wave? There is no free lunch. Any proof of anything has to start with something. Just being a vacuum solution of Maxwell isn't enough to get you that the electric and magnetic fields are orthogonal.

As for proofs with a particular gauge choice. That is not a problem. You are free to pick any gauge to compute with, it doesn't change what fields you get. It doesn't resolve any aspect of question begging or chicken-and-egg issues.

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It is not clear from the question, but let's assume we are in free space/vacuum. Then the Maxwell equations read: $$ \nabla \cdot \mathbf E = 0\\ \nabla \cdot \mathbf B = 0\\ \nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t}\\ \nabla \times \mathbf B = \mu_0 \varepsilon_0 \frac{\partial \mathbf E}{\partial t} $$ A plane wave can be written as $$ \mathbf E = \mathbf {\hat E} \cos(\omega t - \mathbf k \mathbf r + \varphi_e)\\ \mathbf B = \mathbf {\hat B} \cos(\omega t - \mathbf k \mathbf r + \varphi_m) $$ with constant amplitude vectors $\mathbf {\hat E}, \mathbf {\hat B}$.
Now let's put the plane wave in the third Maxwell equation: $$ \nabla \times \mathbf E = \left(\frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z}\right) \mathbf {\hat x} + \left(\frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x}\right) \mathbf {\hat y} + \left(\frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y}\right) \mathbf {\hat z}\\ = \left[ \left( \hat E_z k_y - \hat E_y k_z \right) \mathbf {\hat x} + \left( \hat E_x k_z - \hat E_z k_x \right) \mathbf {\hat y} + \left( \hat E_y k_x - \hat E_x k_y \right) \mathbf {\hat z} \right] \sin(\omega t - \mathbf k \mathbf r + \varphi_e)\\ = -\frac{\partial \mathbf B}{\partial t} = \omega \mathbf {\hat B} \sin(\omega t - \mathbf k \mathbf r + \varphi_m) $$ with the cartesian coordinate vectors $\mathbf {\hat x},\mathbf {\hat y},\mathbf {\hat z}$.
Here we see that the electric and magnetic components have to be in phase and hence can concentrate on the amplitudes $\mathbf {\hat E}, \mathbf {\hat B}$. To check for orthogonality, we evaluate the scalar product of the amplitdues and use the representation of $\mathbf {\hat B}$ in terms of $\mathbf {\hat E}$ we just found: $$ \mathbf {\hat E} \cdot \mathbf {\hat B}/\omega = \\ \left[ E_x \mathbf {\hat x} + E_y \mathbf {\hat y} + E_z \mathbf {\hat z} \right] \cdot \left[ \left( \hat E_z k_y - \hat E_y k_z \right) \mathbf {\hat x} + \left( \hat E_x k_z - \hat E_z k_x \right) \mathbf {\hat y} + \left( \hat E_y k_x - \hat E_x k_y \right) \mathbf {\hat z} \right]\\ = \hat E_x \hat E_z k_y - \hat E_x \hat E_y k_z + \hat E_y \hat E_x k_z - \hat E_y \hat E_z k_x + \hat E_z \hat E_y k_x - \hat E_z \hat E_x k_y = 0 $$ q.e.d.

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  • $\begingroup$ Big... That's why we use complex fields =). +1. $\endgroup$ – Physicist137 Apr 2 '15 at 2:08
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In electrodynamics, one uses complex fields only as a calculation trick, since for instance terms like $$ \exp(i \omega t) $$ are usually more easy to handle as $$ \cos(\omega t) $$ . The same trick can be applied for example to the problem of the driven harmonic oszillator: $$ \ddot x + 2\gamma \dot x + \omega_0^2 x = A \cos(\Omega t) $$. Adding $ i \left( \ddot x_i + 2\gamma \dot x_i + \omega_0^2 x_i \right) = i A \sin(\Omega t) $ this gives $$ \ddot z + 2\gamma \dot z + \omega_0^2 z = A \exp(i\Omega t) $$ with $z = x + i \cdot x_i \in \mathbb C $ leaving the differential equation for $z$ much more easy to solve for complex functions. One can revert to the desired solution $x$ by: $$ x = \textrm{Re}(z) $$ but caution has to be taken for operations like multiplication since in general $$ x^2 \ne \textrm{Re}(z^2) $$. This concepts applies naturally to all linear differential equations like the wave equation in electrodynamics. But just as above, one will be intrested in terms like $$ \textrm{Re}(\vec{E}),\ \textrm{Im}(\vec{E}), \textrm{ etc.} $$

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  • $\begingroup$ This doesn't address my question. Can you modify your answer? $\endgroup$ – Luke Burns Apr 1 '15 at 1:27
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    $\begingroup$ @LukeBurns: "If the proof requires complex fields, can you explain why?" $\rightarrow$ This is what my answer is aiming at. If you want a final conclusion: The proof does not require complex fields in general, but it is a convenient way to do it. $\endgroup$ – image Apr 1 '15 at 9:43
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Let us work in the Coulomb gauge, i.e. $\phi=0$ and $\nabla\cdot\mathbf{A}=0$. Then the electric and magnetic fields are defined as $$\mathbf{E}=-\dot{\mathbf{A}},\quad \mathbf{B}=\nabla\times\mathbf{A}$$ Now one solves the wave equation for $\mathbf{A}$. The constraint $\nabla\cdot \mathbf{A}=0$ tells us that $\mathbf{A}$ is transverse to the wave vector. Furthermore, the electric field is proportional to $\mathbf{A}_0$ times a cosine. Finally, we have the magnetic field proportional to $\mathbf{k}\times\mathbf{A}_0$, which by the rules of vector geometry, is perpendicular to $\mathbf{A}_0$, i.e. perpendicular to the electric field.

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  • $\begingroup$ Doesn't this depend on an arbitrary choice of $\mathbf{A}$? (Refer to physics.stackexchange.com/questions/173499/…) $\endgroup$ – Luke Burns Apr 1 '15 at 1:32
  • $\begingroup$ @LukeBurns Well yes, but everything here is kept arbitrary. It does not change the fact that $\mathbf{E}$ is perpendicular to $\mathbf{B}$. $\endgroup$ – Ryan Unger Apr 1 '15 at 1:58
  • $\begingroup$ Can you justify why this fact is not dependent on your choice of gauge? $\endgroup$ – Luke Burns Apr 1 '15 at 2:01
  • $\begingroup$ @LukeBurns Not in this approach, no. Alternatively, we can follow the proof in here, which does not invoke the gauge at all. $\endgroup$ – Ryan Unger Apr 1 '15 at 2:07

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