It is my understanding the the phase velocity of a wave can be greater then the speed of light. So imagine we had a wave packet consisting of a single sinusoidal wave; $$y=\sin(\omega t-kx)$$ Then from the definition of phase velocity, this wave will (if I am not mistaken) travel with the phase velocity, which as stated above can be greater then $c$. But in this case information is been passed greater then the speed of light, is it not? If we start the wave packet of at a point $A$ say and let it propagate to another point $B$ it will travel to $B$ faster than the speed of light and pass of the information of its frequency and wavelength. Why is this not breaking special relativity?
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asked Mar 7, 2015 at 8:16
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$\begingroup$ As for matter wave, phase velocity has no physical significance, because the motion of the wave group, not the motion of the individual waves that make up the group, corr. to the motion of the body, & group velocity is less than $c$ as it should be. So , the matter wave doesn't violate special relativity. $\endgroup$– user36790Commented Mar 7, 2015 at 10:31
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$\begingroup$ Same is in the case for your concerned wave. +1 to Rennie's answer. $\endgroup$– user36790Commented Mar 7, 2015 at 10:34
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2 Answers
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The simple answer is that the wave packet travels at the group velocity not the phase velocity, and the group velocity is always less than or equal to $c$.
You might argue that you aren't using a wave packet. For example you might argue that you are just turning the light on and waiting for it to get to the point $B$. However any modulation of the wave intensity, including turning it on and off, will propagate at the group velocity.
answered Mar 7, 2015 at 8:24
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$\begingroup$ Hi, but this wave consists of many 'smaller waves' travelling with the phase velocity, does this not therefore mean that the front of the wave packet will reach $B$ at a time faster then the speed of light. $\endgroup$ Commented Mar 7, 2015 at 8:28
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$\begingroup$ @Joseph: I was just extending my answer when you commented. There's an animation of a wave packet here (and Googling will find many more). The top picture is a phase velocity greater than the group velocity. Note that the carrier wave passes through the packet but doesn't move faster than it. $\endgroup$ Commented Mar 7, 2015 at 8:31
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$\begingroup$ On the top digram in the animation you have shown the wave-packet spreads out, this is due to dispersion. So this means that some waves are travelling faster then the group velocity. My point is the wave-packet is made by the superposition of lots of sinusoidal waves, some of which move faster then then group velocity and others that move slower. The once that move faster will reach then end (due to dispersion) in a shorter time then the slower waves and the wave packet as a whole. $\endgroup$ Commented Mar 7, 2015 at 8:36
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$\begingroup$ @Joseph: yes, the packet does spread and therefore the leading edge moves faster than the average packet velocity. However the leading edge does not move at the phase velocity, it moves at a velocity $\le c$. $\endgroup$ Commented Mar 7, 2015 at 8:57
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1$\begingroup$ I thought only group velocity could exceed c. After reading your reply I was confused so I asked a question on this. Any chance you could help me clarify this? physics.stackexchange.com/questions/541321/… $\endgroup$– Alex P.Commented Apr 3, 2020 at 22:16
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You cant not modulate information on a single frequency wave so there exists no packet here. Turning on and off will add other frequencies to the signal and the group of frequencies will travel with the group velocity which is less than the speed of light.
