If you shine light on a wall, what will be seen is a "patch" with constant intensity. However, if light is viewed as a wave, then it is oscillations of the electromagnetic field changing from 0 to the amplitude and back really fast. So my question is, if I were able to look at the world at extreme slow motion, a quadrillion times slower or so, and I shined a beam of light at a wall, will I see a "patch" with oscillating intensity, with maximum brightness at the peaks of the wave and minimum when the field is 0? If so, is the constant brightness seen normally just our puny mortal eyes capturing only the average of this oscillating brightness?
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4$\begingroup$ Related Have we directly observed the electric component to EM waves? $\endgroup$– FarcherCommented Apr 9, 2019 at 7:03
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3$\begingroup$ What you are talking about is the amplitude. Intensity usually means the average power and that would be constant. And you can't talk about seeing patches in those timescales but observing electric field change. $\endgroup$– DžurisCommented Apr 9, 2019 at 10:23
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2$\begingroup$ If you were able to look at the world at extreme slow motion, a quadrillion times slower or so, then you would have to be using some kind of "vision" that would not, in any way, resemble how your vision actually works. $\endgroup$– Solomon SlowCommented Apr 9, 2019 at 13:14
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$\begingroup$ Possible duplicate of Have we directly observed the electric component to EM waves? $\endgroup$– DK2AXCommented Apr 9, 2019 at 14:07
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4 Answers
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You would need a coherent beam, because in waves it is not only intensity but also phase that makes a difference. In an incoherent beam, as sunlight, you would not get any changes in this thought experiment, because the average intensity would hold even at wavelength distances.
In a coherent laser beam you should see in your thought experiment what is shown towards the end of this youtube video,(at 2' 09") the sinusoidal in time pattern of impinging intensity . The video draws the oscillating electric field E, and the intensity left on the screen will be $=E^2$, which will also be oscillating in time.
After all, mathematics allows us to materialize thought experiments as this one.
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4$\begingroup$ An incoherent beam would mean noise in the electric field magnitude, not constant magnitude. You said yourself "average". If you average a coherent beam over a wavelength, it is also constant intensity. I do agree that the coherent beam makes for the better thought experiment. $\endgroup$– EL_DONCommented Apr 9, 2019 at 7:09
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1$\begingroup$ I would just add that visible light oscillates $10^{14}$ times per second, so the oscillations certainly cannot be seen by eye alone. There have been some ingenious experiments capturing the oscillation though, see science.sciencemag.org/content/305/5688/1267 $\endgroup$– VoidCommented Apr 9, 2019 at 7:26
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3$\begingroup$ Good job not getting sidetracked by the biological impossibility of the question. However, the sum of a set of randomly phase shifted sinusoids of the same frequency still oscillates. Adding waves of different frequencies still makes an oscillating wave, albiet a less regular one. This is because average amplitude will remain zero, while average power will be the sum of the individual waves. Watching such a wave in very slow motion should seem to flash in a very noisy pattern. $\endgroup$– VaelusCommented Apr 9, 2019 at 15:05
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1$\begingroup$ @annav but sometimes the E-field amplitudes of the individual photons would add up to a positive number, and (unless there's a large DC-bias for some reason) sometimes they'd add up to a negative number, and in between they cross zero. So the instantaneous amplitude $I(t)\sim E^2(t)$ also goes to zero. That the beam is incoherent just means the zero-crossings wouldn't be very correlated, and so an incoherent beam would flicker very irregularly, but it would still flicker. $\endgroup$ Commented Apr 10, 2019 at 0:10
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All depends how you "look" at the light.
For example, in a linearly polarized EMW the electric and the magnetic fields oscillate like $\sin(\omega t)$ or $\cos(\omega t)$ at a given point of space. Then the question is: what device do you use for "detecting" the EMF?
For EMF there are two notions expressed via fields: it is the energy density $ \propto E^2+B^2$ and the energy-momentum flux $ \propto \vec{E}\times\vec{B}$ which may oscillate or not at a given point, depending on their phase shifts.
Some devices deal with a light "spot", where there are many points of the space involved, so you must average (sum up, integrate) local things. Some devices have inertial response and effectively average over time the incident wave too.
However, some devices are much more sensitive to the electric field than to the magnetic one (photo effect, for example), so they clearly "feel" oscillations.
There are also some devices (magnetic antennas, for example) that are more sensitive to the magnetic field (some radio-receivers).
In other words, the incident fields get into the equation of motion of the detector charges and currents, and the detector features determine what you get in reality.
answered Apr 9, 2019 at 6:29
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2$\begingroup$ I've heard this before (from an E&M professor, I think), but I don't see how you satisfy Maxwell's equations with it. $\nabla\times E=-\partial B/\partial t$ doesn't work if $E\propto cos(\omega t-k\cdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $E\times B$, so you'd get periodic variation in energy flux, anyway. $\endgroup$– EL_DONCommented Apr 9, 2019 at 6:48
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1$\begingroup$ The fields in a linearly polarized wave are in phase - that is a direct consequence of Maxwell's equations. As such also $|\vec E \times \vec B|$ is oscillating, not a constant. Only when you average the Poynting vector over one period $\langle \vec S \rangle$, the intensity becomes constant. $\endgroup$– DK2AXCommented Apr 9, 2019 at 9:17
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$\begingroup$ @ahemmetter: The Poynting vector is a vector, not an intensity $I$. $\endgroup$ Commented Apr 9, 2019 at 9:30
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1$\begingroup$ The intensity is directly proportional to the magnitude of the Poynting vector. $\endgroup$– DK2AXCommented Apr 9, 2019 at 11:46
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The EM field strength in a linearly polarized, coherent light wave does indeed cross through zero in between + and - peaks, just like the surface of a pond goes through its natural rest height in between going up and down as ripples go by. If you slowed the wave down, you'd change its frequency, which is the same as changing its color. You could change from blue, to red, to infrared, and all the way through radio waves and other invisible colors. So no, you could not see the EM field changing as a flashing light for a slow wave (the slow changes couldn't simulate your vision receptors) , but you could set up an electric field meter and measure the change in field as the (no longer visible) wave went by. What does it really mean for the field to cross through zero? Not much; just like having the surface of a pond cross through its equilibrium height doesn't mean the ripples are gone, neither does a moment of 0 electric field mean the light wave is gone.
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"Light intensity" in my opinion means the number of optical photons absorbed by a light detector. Such detectors resonate with the electric field and absorb the photons with a characteristic time of many periods of oscillation. So you can say that the oscillation is observed by the detector, but is not translated into a rapid oscillation of the detected intensity. The latter varies with the number of photons absorbed instead, which is at a generally much longer timescale. However, extremely short laser pulses can approach the timescale of the light oscillation period.
