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I have two question,

  1. why group velocity represents energy or information transmission?

  2. what is the relation between phase velocity and Special relativity: why can it exceed C without violation of SR?

I didn't take any classes in special relativity or information theory or quantum mechanics, my background only contains

  1. general physics (basic knowledge about Special relativity)

  2. electromagnetism

  3. signals and systems (basic knowledge about signal transmission),

so please reply in a simple and easy way.

bobie
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user113644
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    It is a common misconception that group velocity represents information transmission. It does not. It is perfectly reasonable to have superluminal group velocities. –  Sep 03 '14 at 12:35
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    Chris, but only if the whole wave is unphysical. If there are waves of a real physical medium or field that have been properly localized or assigned to the points, the group velocity is the same thing as the speed of the packets and be sure that the packets do spread information. – Luboš Motl Sep 03 '14 at 12:47

2 Answers2

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The statement "group velocity represents energy or information transmission" is not entirely accurate. And that is a big can of worms, in fact. But let us start from the beginning.

What is the dispersion relation?
Suppose that you have some quantity $u$ that depends on a coordinate $x$ and on time $t$.

Dispersion relation $\omega(k)$ is basically a statement, that if you have a coordinate dependence in a form of a wave with some wave vector $k$: $$u(x)=Ae^{-ikx},$$ then you can instantly write the time dependence like $$u(x,t)=Ae^{i\omega(k)t-ikx}.$$

What if you have more complex $u(x)$?
In that case you represent it as a sum (or an integral) of such waves. That is what they call "a Fourier transform": $$u(x) = \int\limits_{-\infty}^\infty a(k)e^{-ikx}dk$$ Using the superposition principle, we can then write: $$u(x,t) = \int\limits_{-\infty}^\infty a(k)e^{i\omega(k)t-ikx}dk$$ Now if you assume that $a(k)$ is peaked around some wave vector $k_0$, then you can derive that the resulting wave packet moves with a group velocity. You can look at the derivation here.

But how is that the maximal speed of information transfer?
That is my point. Let us have a $\delta$-signal at $x=0$ -- its Fourier transform is uniform across all the frequencies: $$u(x) = \delta(x) = \frac{1}{2\pi}\int\limits_{-\infty}^\infty e^{-ikx}dk$$ So the time evoultion would be: $$u(x,t) = \frac{1}{2\pi}\int\limits_{-\infty}^\infty e^{i\omega(k)t-ikx}dk$$ And we are supposed to beleive that this expression gives $u(x,t)=0$ for $x>v_gt$ whatever $\omega(k)$ is.

And that is not true?
Yeah. The trick is that if you introduce dispersion and you want to preserve causality -- you must introduce some absorbtion as well. The absorbtion is introduced as a complex component of $\omega(k)$ and the relations between real and complex (disperion and absorbtion) components are called the Kramers Kronig relations.

Check this reference from the wiki for the formal description of the details of how the causality is related to those dispersion-absorbtion laws.

Kostya
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  • The last bit sounds like a circular argument. There hadn't been any reason for w or k to be complex. In the Klein-Gordon equation for instance, we assume all the way through that w^2 - k^2 = m^2, then when we see signals going faster than light, we suddenly say "Oh no, that's the one case that we're not interested in" and side-step it with a contour integral. Is that cricket? – Adrian May Jul 24 '15 at 15:38
  • @AdrianMay I wouldn't say that the argument is circular. The argument is just opposite of what one expects: one doesn't derive causality from $\omega(k)$ -- one constraints $\omega(k)$ with the causality requirement. – Kostya Jul 27 '15 at 12:49
  • Yeah but the KG equation already dictates a w(k). If you want to change it you'll have to change the KG equation. Besides, I couldn't find anything to say that using the KK relations to zero everything at t<0 also has the effect of zeroing the bit where the phase velocity only just arrived. – Adrian May Jul 28 '15 at 14:09
  • I think the real answer is that this really is an objection but by no means the only one. There's also those negative energies to worry about. That's why they invented QFT. In QFT this stuff becomes harmless because the whole field has a zero point excitation anyway and you use commutators to figure out if a poke here can be detected over there. – Adrian May Jul 28 '15 at 14:12
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If we have a wave of a well-defined direction and frequency, the dependence of the field $F$ (something that is waving) on position and time is $$ F = F_0 \cos (\omega t - k_x x - k_y y - k_z z ) $$ Adult physicists would use complex exponentials instead of the cosine but I decided to remove this potentially difficult piece of maths.

The argument of the cosine modulo $2\pi$ is the "phase". The group velocity is extracted from this phase. In the spacetime, one draws the hypersurfaces of constant phase, i.e. $$ \omega t = \vec k \cdot \vec x $$ At a fixed $t$, this is an equation of a plane. If we watch what is happening with this plane as time goes by, it is moving in the transverse direction and the speed of the motion of this plane is $$v_{ph} = \frac{dx}{dt} = \frac{\omega }{ k}$$ That's called the phase velocity because we calculated it from the phase (by looking at which places the phase is constant, and how these places move in time).

In general, the phase velocity can exceed the speed of light because what is actually propagating by the phase velocity is "the plane on which the phase is constant" but this plane isn't a real physical object that carries information. It's just a fictitious place in the space defined by a mathematical property, "constant phase".

The fact that the phase velocity may exceed $c$ is analogous to the observation that if we sit at the center of a large hollow sphere with a lamp and we rotate the lamp, the illuminated trace of the lamp on the distant interior surface of the sphere may move faster than $c$ (it safely does if the radius of the sphere is large enough). But we are not transmitting anything from one place of the surface to another. Instead, the light goes from the center to the various points of the surface.

On the other hand, the group velocity does measure the actual propagation of material objects. We may create a "wave packet" by combining nearby frequencies. It is possible to derive that the center of such a wave packet will propagate by the group velocity $$ v_g = \frac{d\omega}{dk} $$ and because this packet is a localized object that may carry information, relativity prohibits $v_g$ to exceed the speed of light in the vacuum $c$.

Luboš Motl
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    In the example with the lamp, do you mean because distance traced when moving in a cross section is $s=r\theta$ then the speed is: $$ \frac{ds}{dt} = r \frac{d\theta}{dt}$$ which for large $r$ (and some high speed) can exceed the speed of light? – Anton Apr 29 '22 at 11:42
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    Yes, exactly. The speed of the light on the screen scales with the radius, the lamp-screen distance, and the radius may be as high as you can. – Luboš Motl Apr 30 '22 at 17:36
  • If we had a laser instead of a lamp then we would record a single spot in the surface of a sphere. Wouldn't the displacement of the spot (by moving the laser) be considered as transmitting information? – Anton Apr 30 '22 at 18:02
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    Anton, sorry, I don't see any difference between a laser and a lamp. Maybe I should have written "a laser" exactly because it's clear what the position of the dot is. At any rate, the dot is not a real object that may be transmitting any information because the light (and the information if any) is still going from the laser (center) to the dot on the screen, not from the old place of the dot to the new one! The dot is just a "virtual" or gedanken object, not a real one that moves and carries any information. – Luboš Motl May 01 '22 at 19:13