4
$\begingroup$

I have a doubt about the physical meaning of the complex representation of electric and magnetic fields.

Let's consider an electromagnetic wave, in which both electric and magnetic fields propagate in space and time with a sinusoidal waveform of radian frequency ω.

Now let's consider their phasors E and H: they are simply complex numbers which depend only on the position (since ω is fixed and the time - dependence disappears when passing to phasors). We may represent them in the complex plane, since they have in general both a real and an imaginary part. I was told that their imaginary part represents a shift but, what kind of shift? A time-shift (delay) or a spatial shift? Or a physical rotation of the field?

Moreover, if it is a time-shift, does it coincide with the time-shift we have for instance in a circuit with reactive elements, in which for instance the phase difference between two voltages indicates that one of them is delayed with respect to the other?

$\endgroup$
  • $\begingroup$ It's just representation which makes working and visualizing functions which depend on trigonometric functions easier. In the end, you need to take the real part of the calculation. Look up Euler's formula. Euler's formula makes all those trigonometric identities you had to memorize trivial to compute. It can simplify complex calculations dramatically. $\endgroup$ – Cinaed Simson Jul 3 '19 at 2:16
1
$\begingroup$

Start from the real expression for the physical fields, v.g $$ \vec E=\vec E_0\cos(\omega t-k x-\Delta) $$ and note that an phase shift $\Delta$ can come either from a shift in time, where $\omega t=\omega t'-\Delta$ or a shift in position, $kx=kx'-\Delta$. The key point is that one or the other option makes no difference as far as they actual effect of the phase shift.

In some situations like circuits where the actual position of the elements don't matter, it's convenient to think the shift as a time lag or time lead between the voltage and current (with $\vert \vec E\vert$ related to the conduction current). In other situations like for instance in an interferometer then a phase shift is conveniently thought as a path-length difference.

$\endgroup$
  • $\begingroup$ Thank you, perfect. So when for instance we graph on the complex plane the imaginary and real parts of the electric field of and electromagnetic waves, what do they represent? $\endgroup$ – Kinka-Byo Jul 3 '19 at 15:59
  • $\begingroup$ It depends a bit on the phasor convention but usually the real part of the phasor of $\vec E$ or $\vec B$, after multiplication by $e^{j\omega t}$, is the physical field. (I've never see the convention where the imaginary part is the physical field.) The complex part has no physical significance. Note that, unlike current or voltage phasor (which are scalars), these are vectorial phasors, i.e. the phasors for $\vec E$ or $\vec B$ are vectors since $\vec E$ or $\vec B$ are themselves vectors. $\endgroup$ – ZeroTheHero Jul 3 '19 at 17:23
0
$\begingroup$

The wikipedia page explains this fairly well

https://en.wikipedia.org/wiki/Phasor

Phasor can mean a couple things: the analytic representation for a sinusoidal function $A\cdot e^{i(\omega t+\theta )}$, which is sometimes called a phasor, and the static vector, $Ae^{i\theta }$ which is usually called a phasor. The former is can be plotted in the complex plane to show how the wave is mapped out over time with a changing vector. When multiplying the latter by the former, a phase shift and amplitude change can occur (if A and theta were replaced with different values in one of them). If you plot the static vector in the complex plane it's just a vector, which can just shift the moving vector of the analytic representation when multiplied. So the real and imaginary parts just represent the components of the vector.

But yes the comment above is right in saying complex just makes working with waves and representing them visually easier. Someone saying the imaginary part causes a shift may have been unclear about phasors as analytic representations or complex constants.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.