What does the imaginary part of the exponential of an EM wave represent?

Question

I've read about a bit and some people are saying that either the imaginary part is just forgotten, or that is the orthogonal polarisation.

These don't make sense to me as I thought the imaginary part represented the $B$ field as it is out of phase with the real part with the same fixed amount $\pi /2$

Remember that trigonometric functions can be expressed as sums of complex exponentials (and vice-versa). When you have a plane wave solution $e^{ikx}$, the general solution is then a sum over these plane waves, each with an (at first) arbitrary constant. However, the boundary conditions will fix these constants, and so you'd expect real boundary conditions to fix the coefficients such that the solution is real. TLDR - the solutions can be written in terms of real trigonometric functions. Hopefully someone can give you a more complete answer. — Bedge, Oct 20 '22 at 22:26
Does this answer your question? What is the physical significance of the imaginary part when plane waves are represented as $e^{i(kx-\omega t)}$? — ZeroTheHero, Oct 20 '22 at 22:57

score 1 · Answer 1 · answered Oct 21 '22 at 04:38

Wooga: the magnetic field is NOT out of phase with the electric field in a plane wave.

As far as I can tell, this mistake derives from the common simple explanation of dipole radiation:

When the charges are separated, the E-field is maximum. $\pi/2$ later, the charges are moving through the dipole antenna and are at maximum current: max B-field.

This is true; however, if look at the dipole radiation formula you will see that this term is multiplied by $1/r^n$ with $n>2$: it does not radiate; it is near field.

The term that goes as $1/r^2$ (in power) is the radiation term, and it has E and B in phase.

This is because a changing $E$ does not generate a $B$, and a changing $B$ does not generate a $-E$; rather, distant charges and currents (on the past light cone), create an $E$ and a $B$ such that $dE/dt \propto dB/dz$ (where $z$ is the propagation direction).

In general, in a complex plane wave, $e^{i(kx-\omega t)}$, there is only a real part ($\cos({(kx-\omega t)})$) where we've chosen coordinates such that the phase offset is $0$.

But... in, say, coherent radar, you receive a return signal with a phase offset:

$$\cos({(kx-\omega t)+ \phi})$$, which can be split in a cosine and sine term, which again is a complex exponential.

The detection is done with a stable local oscillator in to a $\phi=0$ signal with a cosine (in phase) and a sine (quadrature) term. This is called I-Q detection.

It's still real though.

Thank you for that, it certainly clears it up more. I realise now my assumption of $\pi /2$ was a stupid one, there are plenty of images where they are in phase — rogerkoulitt, Oct 21 '22 at 09:51
@rogerkoulitt it's not stupid, it's common. Moreover, it's reenforced by the ubiquitous statement "A changing electric field creates a magnetic field", which would put them out of phase. — JEB, Oct 21 '22 at 16:53

What does the imaginary part of the exponential of an EM wave represent?

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