It looks like solutions of the KG eqn travel faster than light, because if $$\omega^2 - k^2 = m^2$$ then $$\mid\ \omega\mid \ > \ \mid k\ \mid$$ and I thought the wave velocity was $\omega / k$.

How do I interpret this?

I heard something about particles following the packet velocity, which isn't too fast, but why can't these fast wave fronts be used for signalling irrespective of what any packets are doing?


In the meantime I've been looking at propagators but it all looks too grandiose to me. To my eyes, the KG equation is nothing but a guitar string where each point on the string is pulled back to the quiescent position by a spring. I could set up a simulation of that and ping it with a tall, narrow gaussian. Sure enough, the waves would go faster with the extra springs because it's all stiffer, and that effect would be stronger at long wavelengths where the springs would dominate the curvature, so we'd have a $d\omega/dk$. But I don't think anything would hit infinity (like this propagator thinks) and I'd be very surprised if the far end of the string stayed completely motionless until the packet velocity arrived (or would it?)

So why all this talk of complex numbers and contour integrals that were never part of the original problem? Even if I promoted the displacement to a complex number, the real and imaginary parts would form two independent systems because all the diffs are second order so there's nothing to mix them up. What am I missing?

The possible dupe doesn't answer this. It tells us what wave packets are but not why the phase velocity can't carry a signal.

  • 3
    $\begingroup$ $\omega/k$ is the phase velocity. What matters is the group velocity, $\frac{d\omega}{dk}=\frac{k}{\sqrt{k^2+m^2}}<1$. $\endgroup$
    – Meng Cheng
    Commented Jul 24, 2015 at 3:55
  • $\begingroup$ That's what I heard, but I still don't see why the phase velocity can't carry a signal. $\endgroup$
    – Adrian May
    Commented Jul 24, 2015 at 5:26
  • $\begingroup$ I guess it could be that the zone in which the signal is waving spreads out at the group velocity, but I'm not seeing from the maths why that would be the case. It still seems to me that I have a lot of choice as to how I wiggle one end of this rope, and there's got to be one way whereby the other end would wiggle, however slightly, as soon as the first (phase) wave front gets there. $\endgroup$
    – Adrian May
    Commented Jul 24, 2015 at 5:32
  • $\begingroup$ possible duplicate of In superluminal phase velocities, what is it that is traveling faster than light? $\endgroup$
    – ACuriousMind
    Commented Jul 24, 2015 at 10:18
  • $\begingroup$ @ACuriousMind That doesn't really answer it though. $\endgroup$
    – Adrian May
    Commented Jul 24, 2015 at 10:37

2 Answers 2


You have a good intuition on the possible answer because it does involve oscillations, but you need to visualize the string differently. The origin of the conundrum is not quantum, but relativistic. Here's why:

Consider an inertial frame O and set up a string spanning the entire $x$ axis, stretching from $x \rightarrow -\infty$ to $x \rightarrow \infty$. Alternatively, you can just set up a uniform scalar field throughout the entire space. Say the string moves only in the y direction and let its displacement from the $x$-axis at point $x$ and time $ct$ be $\phi(x,ct)$ (or for the field, assume it varies only along the $x$-axis). Now somehow make the string oscillate in unison across O$x$ along its entire span according to $$ \phi(x,ct) = \phi_0 \sin(kct),\;\; \text{for}\;\text{some}\;k \in {\mathbb R} $$ The frequency of the oscillation is simply $\omega = ck$.

If we look at this simple setup from another frame O', moving at velocity $v$ along x, the displacement of the string at location x' and time ct' will be $$ \phi'(x',ct') = \phi_0 \sin\left( \gamma k (ct' + \beta x') \right) $$ where $\beta = \frac{v}{c},\;\gamma = {1}{\sqrt{1-\beta^2}}$, $x' = \gamma (x-\beta ct)$, $ct' = \gamma (ct - \beta x)$ as usual, per Lorentz transforms. This is obviously a wave and it must satisfy a wave equation. So let's take a look at the usual ingredients for a wave equation: $$ \frac{\partial \phi'}{\partial x'} = - \beta \gamma k \phi_0 cos\left( \gamma k (ct' + \beta x') \right), \;\;\frac{\partial^2 \phi'}{\partial x'^2} = - \beta^2 \gamma^2 k^2 \phi'(x',ct') \\ \frac{\partial \phi'}{\partial (ct')} = - \gamma k \phi_0 cos\left( \gamma k (ct' + \beta x') \right), \;\;\frac{\partial^2 \phi'}{\partial (ct')^2} = - \gamma^2 k^2 \phi'(x',ct') $$ From this it follows immediately that our wave conveniently satisfies $$ \frac{\partial^2 \phi'}{\partial x'^2} - \frac{\partial^2 \phi'}{\partial (ct')^2} = \gamma^2 k^2 (1 - \beta^2) \phi'(x',ct') $$ or simply the Klein-Gordon equation: $$ \frac{\partial^2 \phi'}{\partial x'^2} - \frac{\partial^2 \phi'}{\partial (ct')^2} = \frac{\omega^2}{c^2} \phi'(x',ct') $$

What we see here is that the oscillations of the string in frame O, which are parallel to the $x$-axis, appear in O' as Klein-Gordon waves traveling along $x'$ in the direction of motion of O and at apparently faster-than-light wavefront velocities.

The reason for the apparently faster-than-light Klein-Gordon wave propagation is very simple: In O a "wavefront" corresponding to time $ct_0$ is simply a line parallel to the $x$-axis at distance $d = \phi_0 sin(kct_0)$ along $y$. All its points belong to a space-like hyperplane and are therefore not causally related. In O' the same front appears not as a wave, but as a single point propagating along $x'$ at faster-than-light "velocity" $-c/\beta$ (the negative sign arises because O' sees the string in O moving in the negative x direction). To see this, look at the coordinates of the front points in O', as given by the Lorentz transform: $$ x' = \gamma (x-\beta ct_0) \\ ct' = \gamma (ct_0 - \beta x) $$ Since $ct_0$ is fixed, each $x$ in O corresponds to a single $x'$ in O', which in turn corresponds to a single time $ct'$. Alternatively, eliminate $x$ or use the reverse Lorentz transform to obtain $$ x' = - \frac{ct'}{\beta} + \frac{ct_0}{\beta\gamma}, \;\; \frac{dx'}{d(ct')} = -\frac{1}{\beta} $$ In other words, O' observes a "propagating point" with faster-than-light velocity, but this "point" is in fact a collection of spatially-separated, causally unrelated points of the O front as observed at consecutive times in O'. Taking into account all wavefronts as generated in O, leaves O' with an observation of a "faster-than-light" Klein-Gordon wave.

And now the answer to your question :

A similar reasoning applies to a scalar field $\phi$. In O, let $\phi$ be a space-wise uniform field that oscillates in time, $\phi({\bf x}, ct) = \phi_0 \sin(\omega t)$. Its wavefront in O at a given moment $ct_0$ becomes the entire 3D-space, which is basically a constant-time, space-like hyperplane. What O' observes of this hyperplane at any moment $ct'$ in his own time is only a 2D plane, $x'=- \frac{ct'}{\beta} + \frac{ct_0}{\beta\gamma}$, perpendicular to the direction of motion of O. When O' follows this plane in time, he observes a wavefront propagating along $x'$ at faster-than-light phase velocity $-c/\beta$. The entire field appears to him as a wave $\phi'(x',ct') = \phi_0 \sin\left( \gamma k (ct' + \beta x') \right)$ satisfying the Klein-gordon equation $\Delta'\phi' - \frac{\partial^2}{\partial (ct')^2}\phi' = \frac{\omega^2}{c^2} \phi'(x',ct')$.

Conversely, we can start with a solution of the Klein-Gordon equation and boost to a frame where $\phi$ appears as a uniform field, then retrace the same reasoning.

Quantum connection:

The difference between the string Klein-Gordon eq. above and the one in relativistic quantum fields is merely the form of the coefficient on the rhs. To obtain the correct term for a field of mass m, one only has to replace the arbitrary frequency $\omega$ with the fundamental frequency (Plank frequency?) for mass m: $$ \frac{\omega^2}{c^2} = \frac{\left(\frac{mc^2}{\hbar}\right)^2}{c^2} = \frac{m^2c^2}{\hbar^2} $$ Everything else you know about "deriving" the Klein-Gordon in relativistic quantum theory is great, but hey, it can be done this way too.

Bottom line:

Leaving aside the string image and considering strictly scalar fields, the correct way to interpret the Klein-Gordon "faster-than-light" waves is that there exists an inertial frame where the Klein-Gordon field is uniform in space and oscillates in time at frequency $\frac{mc^2}{\hbar}$. This frame can be viewed as the "rest frame" of the particle represented by the field. The wavefronts correspond to constant time, space-like hyperplanes of the "rest frame" and necessarily "propagate" faster-than-light because this is how constant time hyperplanes transform under the Lorentz transforms. Everything else follows from the nature of space-time and there is no conflict with causality.

  • $\begingroup$ I don't think this answers the question. Switching frames doesn't add any clarity. Einstein said that we can work in any frame, not that we're obliged to switch. If the propagation is lightlike, spacelike or timelike in any frame, it will be so in all frames. The phase velocity of the KG equation is spacelike, and there's still no reason to deny it the right to carry a signal, so we seem to be in relativistic hell where nothing makes sense anymore - you could order your own abortion before you were born. $\endgroup$
    – Adrian May
    Commented Aug 25, 2015 at 16:28
  • $\begingroup$ @AdrianMay I prefer to be alive as my parents intended a long time ago, thank you very much. As for changing frames, try changing frames for a photon and you'll see the difference. In the m≠0 case however, given that things are all equal between any inertial frames, it can't hurt to get a grip using a frame that provides the simplest picture, namely the one where the field becomes uniform in space. Now try to send a signal from point A to point B using only a uniform field. Otherwise, if connecting my answer to Klein-Gordon plane waves troubles you, you can always ask for details. $\endgroup$
    – udrv
    Commented Aug 25, 2015 at 19:58
  • $\begingroup$ I think we're agreeing on alll but once thing: you say the points on the wave front are spacelike separated and therefore not causally related as per relativity. I agree that they are spacelike separated but unfortunately I can cause them by banging on the string, which conflicts with relativity. $\endgroup$
    – Adrian May
    Commented Aug 27, 2015 at 5:15

have you seen a pulse or train of waves on the surface of water? Look carefully and often you can see the wavecrests rushing from the back to the front then just dissappearing. this is because the phase/crest velocity is double the group/packet velocity. w2=gk gives c2=g/k while v=dw/dk so 2wv=g or v=g/2w or v2=g2/4gk=g/4k=c2/4 so v=c/2 or c=2v. NB...notation c2 equals c to power 2 while 2c equals usual 2 times c. So the motion of crests doesnt equate to motion of information. If you turn a corkscrew the velocity of the imagined crestlike point can advance as fast as you like but it doesnt carry energy or info. Like the shadow or an illuminated spot can travel fast as you like but doesnt carry info along the surface where it is projected. In the QM case its the group packet vel or particle velocity that corresponds to energy and information transfer.


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