# Intensity of EMR and time dependence

I imagine an electromagnetic wave as the propagation of electric and magnetic fields.

Consequently the amplitude of the field vectors are time-dependent. I also know that the intensity is proportional to the square of the electric field vector. Does this mean that light when it hits a screen is actually flickering? That the frequency is just way too high, so that we can't see it? Unfortunately I am not yet familiar with the Poynting-Vector, so I couldn't completely follow the derivation.

I also read that one usually calculates the time-averaged intensity. This really seems to imply that my idea is correct. Is it?

The Poynting vector is just an intensity vector, it is just $\vec S = \vec E \times \vec B$, one might need some factors of $c$ in there, I have been in $c = 1$ land too long ….

Due to the time dependence, the amplitudes of the electric fields are time dependent. If one would call that flickering, it would be the frequency of the light. I don't think that this is a good idea. To me, flickering is something that happens after averaging over at least one wavelength. The oscillation is mandatory for propagation. Constant fields to not propagate.

So what you want is the average of $E^2 + B^2$ over at least one wavelength. That is your intensity.

• Thanks for the quick answer. With flickering I mean that if we were too see light of very low frequencies (say 5 Hz, I know it's absurd), so that we could see light on the screen going on and off, due to the oscillation of the E-vector. I guess that E^2+B^2 is the magnitude of the S-vector, which is equal to the intensity? Jul 11, 2016 at 18:10

Your idea is basically correct. It all depends on what type of measurement is being done. If you have an instrument with a pointer that moves left or right in proportion to the electric field amplitude, then, yes, it will oscillate left and right when a wave impacts on it, as long as the wave frequency is low enough to give the instrument time to respond.

Our eyes do not work like that however. They respond by accumulating energy over a time which is short for us, but long in comparison with the period of the light waves.

Note that whenever considering measurement of something which is changing with time, you always have to take into consideration the temporal response of the detector (whether a human sense-organ or an instrument of some kind).