# Superposition of electromagnetic waves

The superposition of two waves is given by $$\sin(\omega_1 t)+\sin(\omega_2 t)=2\cos\left(\frac{\omega_1-\omega_2}{2}t\right)\sin\left(\frac{\omega_1+\omega_2}{2}t\right).$$ For sound waves, this effect is perceived by the human ear as beats.

Does the human eye perceive a similar effect for electromagnetic waves? Is it possible to produce visible light through the superposition of two microwaves?

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I assume the question has a typo? Perhaps you mean: Is it possible to produce microwaves through the superposition of two optical beams? The answer is yes. –  emarti Dec 23 '12 at 22:03

If you take two coherent light beams with frequencies that are only very slightly different, such as a few cycles per second, there is no reason I can see why the human eye would not be able to perceive electromagnetic beat frequencies as a light source that oscillates in brightness at that same delta.

As for you second question about microwave beat frequencies, no, that simply cannot happen. The reason there is that because waves such as microwaves in free space add together linearly -- meaning just that their local parts add together as simple vector sums -- you can't normally derive higher frequencies from lower ones, and microwaves are vastly lower in frequency than light. Beat frequencies in general cannot be higher than the frequencies that generate them.

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But it is possible to produce audible sound from the beats of two ultrasonic sources - see en.wikipedia.org/wiki/Sound_from_ultrasound - it happens because of nonlinearities in air or in the material the ultrasound hits. So maybe it might be possible to produce visible light from the beat frequencies of UV or X-ray wavelengths by shining them on the right material. Not sure what the applications would be though... –  Nathaniel Jul 24 '12 at 21:29

I agree with Terry Bollinger. I did a simple illustration of superposition of two light waves with the equation S(x,y) = A0+A1cos(2PIr/lamda), where A0/A1=3,A's are wave amplitudes http://www.khanacademy.org/cs/holosim/1245709541 .This is the condition necessary inline Holographic techniques. What happens when two light beams are in the same spatial region, point(x,y), is they interfere if they are coherent. Due to the superposition, regions of bright and dark fringes show the 'beats' created. When you pass a laser beam though a small opening - pinhole - you get a point source. As the light goes through the hole, its diffracted and forms interference fringes within a solid angle. the fringes are as a result of superposition of the reference wave and the diffracted one.

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It seems to me that you actually have two questions. To the first one, namely

"Does the human eye perceive a similar effect for electromagnetic waves?"

the answer is, presumably, "Yes." Two coherent light sources (i.e. lasers) of similar frequency can indeed produce a beat signal when they strike a sensor simultaneously. I don't know if anyone has ever tested this with the human eye as the sensor, but I see no reason it wouldn't work, provided that the two lasers are close enough in frequency for the beat frequency to be perceptible as visible flicker (i.e. less than 20 Hz or so).

"Is it possible to produce visible light through the superposition of two microwaves?"

the answer is "No." That's simply because the beat frequency never exceeds the original frequencies. Specifically, if you look at the equation you quoted,

$$\sin(\omega_1 t)+\sin(\omega_2 t)=2\cos\left(\frac{\omega_1-\omega_2}{2}t\right)\sin\left(\frac{\omega_1+\omega_2}{2}t\right),$$

you can see that superimposing the two original frequencies $\omega_1$ and $\omega_2$ yields a combined waveform that looks like their average frequency $(\omega_1 + \omega_2) / 2$ modulated by half the difference of the frequencies (the "beat tone") $(\omega_1 - \omega_2) / 2$. Neither of these frequencies can ever exceed the higher of the original frequencies.

(I'm assuming here, for simplicity, that $\omega_1 > \omega_2 > 0$. I'll leave it as a simple mathematical exercise to verify that relaxing this assumption won't actually change the conclusion. Keep in mind that negating a frequency just flips the phase of the signal, but leaves the absolute frequency unchanged.)

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Yes, it is very common practice and, if the beat note is a slow enough frequency, it can be seen by the naked eye (I've seen it). This is how we compare two lasers on a photodetector: a photodetector cannot resolve the incredibly rapid oscillation of the electric field (at $10^{14}$ Hz), so instead we interfere two lasers and measure their intensity. For example, I've compared the frequency of two lasers using this beat signal and 'locked' one laser to the other (called frequency offset locking). The general practice is called an optical heterodyne detector.

Measuring the intensity, and not the field, is the required piece of nonlinearity. Fortunately (virtually) all optical detectors measure intensity, including our eye. For microwaves, I have to use a mixer to get the nonlinearity.

However, an essential question is: how do you interfere two waves? You can interfere lasers at a shallow angle, but that just produces intensity fringes (also easily seen by the naked eye) and isn't so useful for a photodetector. For electronic and optical signals, we nearly always use a beamsplitter (optical beamsplitter or splitter for microwaves). That way, when the intensity of one port goes to zero, the intensity of the other is at its peak (see optical heterodyne detector). This is also how an optical interferometer, such as a Michelson interferometer, operates: the two arms are the two oscillators, and a changing pathlengths results in a phase (or frequency) difference and a measurable beat note.

So, in summary, you can indeed produce microwaves from an optical beat note and we often do. (Of course, the other way around the frequencies don't add up!)

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No. Ears hear superposition in this manner because the ear canal effectively takes a Fourier transform on the incoming sound wave -- and decomposes it.

The eyes, on the other hand, detect individual photons with different pigments, thus preventing interference between different frequencies. The addition/subtraction of different colors happens in the processing of the signal.

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zhermes, I hate to disagree, but I'm not aware of any fundamental reason why two laser beams with very slightly different frequencies, e.g. five cycles per second, could not produce a visible beat frequency detectable by the human eye. It would simply change the probabilities at which the photons would be detected. –  Terry Bollinger Apr 23 '12 at 3:29
Hi Terry, disagreement is always welcome! When you 'see' a given color, its based on the amount of signal received from each of the three types of cones in the region the signal is perceived from. 1 source at 6e14 Hz and another at 6e14 + 5 Hz would thus look like (approximately) the average of the two. Lets say the first source is green, the latter slighty bluer (but just call it 'blue'). You would detect a few thousand photons of each, in each your 'green' and 'blue' cones, and interpret a color inbetween. There is no mechanism by which to observe a beat. –  zhermes Apr 23 '12 at 4:02
@zhermes But you can observe different colors, for example in the Doppler effect due to star velocities... –  Pygmalion Apr 23 '12 at 6:46
@Pygmalion I dont see how this relates. We're talking about observing with our eyes in particular, and yes, of course we can see different colors. –  zhermes Apr 23 '12 at 15:29
Hmm... clarification: The beat frequency about which I was speaking is one of intensity. That is, you would not see a change in color, but rather a fading in and out of overall brightness. These are very, very small difference in color by the way, well below the noise levels of color receptors. –  Terry Bollinger Apr 26 '12 at 0:40