Say I have a laser. If I spin the laser so that the beam sweeps in an arc along a very distant object, could that dot travel faster than the speed of light?

In Diagram form: laser beam thought experiment


marked as duplicate by Qmechanic Apr 11 '13 at 13:37

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Yes, the dot would travel at a speed faster than light. Firstly, note that this wouldn't be immediate. if $r$ is the radius of the arc, it would take $\frac r c$ time before the dot started moving (in the meantime, the beam of light, if "frozen" in time1, would look bent from most reference frames)

This doesn't violate causality. Why? Because the dot cannot relay information along the arc. The motion of the dot is controlled by the person at the center, and he is the only one who can send information by moving it. Since there is a delay of $\frac r c$ seconds before the dot moves, this is perfectly in line with causality.

Remember, the dot isn't a physical object. It's a certain area having some properties (point of intersection of distant object and laser beam)

A simple argument to justify that the dot will go FTL:

At time $t_0$, I shine my laser pointer at point A, on the left of the arc. I sweep it around, and at time $t_1$ it points towards B on the opposite end of the arc. Note that in my reference frame, $t_0$ and $t_1$ can be quite close, while $r$ can be arbitrarily large.

In my reference frame, the photons from my initial point reach A (remember, a photon does not change its direction in special relativity) at time $t_0+\frac r c$, and the photons from my final position reach B at $t_1+\frac r c$

So now, the dot has crossed a distance of $r\theta$ in a time $t_1-t_0$. Note that $r$ can be arbitrarily large, and $t_1-t_0$ is only constrained by the rate of production of photons. Which means that we can cover an arbitrary distance in a fixed time interval, so there will be some $r$ for which $\frac{r\theta}{t_1-t_0}>c$.

Note that since the events of the dot reaching the two points are spacelike separated (assuming we choose $r$ such that $\frac{r\theta}{t_1-t_0}>c$) in my reference frame, they are spacelike separated in all reference frames, and the dot moves faster than light for all.

1. To specify "frozen in time" rigorously: let there be an array of photon detectors. They are synchronized such that they will all record the existence of photons simultaneously in some given reference frame (the frame of the laser pointer or the frame of somebody on some given point on the arc). Now, if that person checks which detectors have recorded a photon, he will get a curved locus.

  • $\begingroup$ OP asked about FTL speed, you answered the speed would be exactly $c$, so the answer should be 'No'. However is there a simple argument to show that the dot speed would be $c$, not a certain fraction of it? $\endgroup$ – Yrogirg Apr 11 '13 at 12:45
  • $\begingroup$ @Yrogirg: Oops, I wasn't paying attention as I typed, thanks for pointing that out :) The simple argument is that photons travel in a straight line and with a constant speed in any frame. (I'll edit that in) $\endgroup$ – Manishearth Apr 11 '13 at 12:47

Yes, it can travel faster than light, but it is not a physical object, merely the locus of intersection of the beam and the screen, so no violation of causality is achieved.

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    $\begingroup$ Could you please address a (popular) concern that the beam would bent, just like a water from a hose? $\endgroup$ – Yrogirg Apr 11 '13 at 12:16
  • $\begingroup$ @Yrogirg: The beam does bend, but that doesn't affect the speed of the dot, it just makes the movement of the dot delayed, right? $\endgroup$ – Manishearth Apr 11 '13 at 12:27
  • $\begingroup$ @Yrogirg That's a separate question, but to answer it you'd first have to define what you mean by "the beam". Any photon travels in a straight line, but I suspect you want to define the beam as the locus of the furthest emitted photon at any time in your selected frame. Or maybe you want to say what a beam looks like in which case you'd have to account for time delays of the photons scattering into your retina. The point I'm making is that it needs careful phrasing of the question! $\endgroup$ – twistor59 Apr 11 '13 at 12:39
  • $\begingroup$ @Yrogirg: Just posted one :P I know that it's a statement that adds nothing to the speed issue :) $\endgroup$ – Manishearth Apr 11 '13 at 12:40

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