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The phase velocity can be faster than light. Some argue that the phase velocity doesn't convey information, but this doesn't convince me.

  1. We can emit a wave of a single one frequency. Then it will move in the space, but the group velocity will be 0.
  2. We can convey information in the following way: 2Hz means 1, 1 Hz means 0. We emit single-frequency wave for a second. Then we don't emit anything for a second. And so on.
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marked as duplicate by ACuriousMind, Kostya, Gert, MAFIA36790, John Rennie Jan 29 at 6:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The transitions between two frequencies will not move with the phase velocity! Nor will the beginning of transmission!

Only in the steady plane wave parts in the middle, the wave crests will move faster than light. And that's just "appearance" in the sense that bewteen any two times, you can find a maximum of the electric field and pretend the wave moved there. But actually, that's just our interpretation (finding a pattern in the "image"). It's like watching a movie on your screen - you interpret the change of a certain pattern in the image as motion, but actually the pixels don't move at all - they just change intensity so you perceive it as motion. That's all phase velocity tells you: you put in a single sine wave in and observe how fast the pattern goes. But actual propagation of momentum and energy (which is percieved most easily as the speed of the beginning of the beam) is slower than speed of light in the vacuum. What you'll see when you turn it on, is wave crests moving faster than the front, and "disappearing" in the front of the beam.

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In a harmonic oscillator, the maximal of a particle is $v_m = A \omega = A v_p k = 2 \pi A \frac {v_p} {\lambda}$. Since $v_p$ can be bigger than $c$, and $A$ arbitrarily large, we could have the particles of a mechanical wave moving with $v_m > c$ – marmistrz Jan 28 at 9:37
    
The first equality ($v_m = A \omega$) only applies for non-relativistic particles. Consider that the derivation of simple harmonic motion relies on the non-relativistic equation of motion $F = m a$, rather than the relativistically correct $F = \frac{\mathrm{d}p}{\mathrm{d}t}$. – abeboparebop Jan 28 at 18:08

Point a laser to the sky, track where it lands very far away from the Earth, and then move your arm. The point you are tracking in space is moving faster than the speed of light, even though light itself isn't.

This isn't the same as group velocity, but it is very similar. There is no problem in having arbitrary points you imagine going faster than light, and relativity doesn't disallow it, as long as energy doesn't travel faster than it.

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I've also heard this same approach worded as calculating the "speed of darkness," by measuring the speed of the shadow of your own thumb on the moon's surface (presuming you had a bright enough light behind your thumb so that you could see it strike the moon). It's easy to show that no laws are violated when the edge of the shadow actually crosses the surface of the moon faster than the speed of light. – Cort Ammon Jan 28 at 17:16

You have to be careful to interpret your statements 1 and 2 correctly.

In statement 1 you are talking about an infinitely long wave train which is made up of only one single frequency.
As soon as you have a wave train of finite length then the wave train is made up of the sum many frequencies. So in statement 2 what you think of as a pulse of a wave with a frequency 2 Hz is actually a pulse of the superposition of very, very many frequencies which all add up to look like a pulse of a wave with frequency 2 Hz.

This subtle difference does not matter as long as the medium is no dispersive, ie the speed of the wave does not depend on its frequency. If the medium is dispersive then the different frequency components that make up your 2 Hz pulse travel at different speeds and the shape of you pulse changes. As the shape of you pulse changes which part of it do you use to measure its speed?

I think that conceptually this is very difficult.

There are many animations on the Internet which try and show visually what happens.

Here are a couple:
Link 1
Link 2 which gives you more control.

There are many others and I would be glad to hear of any ones which are better.

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Wait, why do we have to have an infinite wave? We could say that that our equation makes sense for $t=0$ and such $(t,z)$ that... Just as we say that out of two solutions of the quadratic equation only one makes sense as far as the physical interpretation is concened. – marmistrz Jan 28 at 9:35
    
Continued: why do we have to superpose the many waves here? (I'm thinking about mechanical waves all the time, they are the easiest to imagine ;) ) – marmistrz Jan 28 at 9:43
    
You are asking about a whole topic called Fourier Analysis. – Farcher Jan 28 at 9:46
    
You are asking about a whole topic called Fourier Analysis. Here is a link which explains how any shaped repetitive wave can be thought of as made up the sum of sine and cosine functions. As a start try this link (there are many others which you might like better) betterexplained.com/articles/… For waves of a finite length it is like Heisenberg uncertainty principle - the more you know about where a wave is the less you know about its frequency. – Farcher Jan 28 at 9:53
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@marmistrz With Fourier analysis, the sines and cosines are the entire functions, not just pieces of them. If you want to represent something that vanishes outside a certain region (even if it looks like a sine wave where it doesn't vanish), you need all sorts of cancellation to happen in the sum, and in fact you can only get the cancellation you want in the limit of infinitely many terms. – Chris White Jan 28 at 17:25

If we take your case of amplitude modulating two frequencies with pulses which are a second wide - then these are not pure frequencies any more - they will have a bandwidth of approx 2Hz (depends on pulse shape) [because there will be two sidebands at +/- 1Hz from carrier]. These pulse therefore have a group velocity.

Only if you emit a wave at a single frequency for all time would it have no group velocity. There would be no information content in such a signal of course.

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Why of course? It's not obvious there is no information here – marmistrz Jan 28 at 15:54
    
For there to be information content there would have to be a change in something to carry the information. – John Jan 28 at 17:46

See Greg Egan's applet. This will clear up all confusion, and ought to be viewed by any reporter who covers a "scientists break light speed" report.

Subluminal shows how a wave composed of a multitude of frequencies moving at different velocities — all less than or equal to c, the speed of light in a vacuum — can appear to have features moving faster than c.

The grid that crosses the screen is moving with a velocity of c, and no individual frequency outpaces it. However, the total wave (the bottom trace, in white) has its strongest peaks where all the individual frequencies are in phase, and the places where that happens shift with time, at a “speed” that is greater than c. Nothing is actually travelling with these peaks, though; they’re just an artifact of the way the different frequencies are slipping in and out of phase.

This illusion of superluminal motion can only occur when the refractive index of the medium falls as the frequency of the light increases, a situation known as anomalous dispersion. If it falls rapidly enough, the group velocity — the speed at which the overall envelope of the wave seems to move — can even become negative.

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protected by Qmechanic Jan 28 at 15:13

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