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In this article, wikipedia describes a constantly accelerated rocket, assuming special relativity : $$ x(\tau) \;=\; \frac{c^2}{a} \left(\cosh \frac{a \ \tau}{c} -1 \right) $$ The proper time $\tau$ is less than $\frac{d_0}{c}$, where $d_0$ is the distance to the foreign star. For example, Alpha centauri is 4.37 light-years away from Earth, and the constantly accelerated rocket arrives there in 3.6 years (including the deceleration on the second half of the trip).

Doesn't this mean that, from the rocket's perspective, Alpha centauri moves faster than light ?

Let $d(\tau)$ be the distance from the rocket to Alpha centauri, as perceived by the rocket. That's the length of the spacelike geodesic orthogonal to the rocket's 4-speed, joining the rocket and the star. $d(0)$ is 4.37 light-years, when the rocket leaves Earth and $d(3.6)$ is zero, when the rocket reaches the star. By the mean value theorem, at some proper time $\tau$, $|\frac{dd}{d\tau}|>c$.

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    $\begingroup$ "Doesn't this mean that, from the rocket's perspective, Alpha centauri moves faster than light ?" - I'm not sure how you conclude this. Consider the simpler case that the rocket's speed is uniform 0.8c relative to Alpha Centauri. In the spaceship's frame, the star has speed 0.8c but the trip from Earth to Alpha Centauri takes just 3.28 yr according to the spaceship's clock. $\endgroup$
    – Alfred Centauri
    Commented Jun 27, 2017 at 11:50
  • $\begingroup$ @AlfredCentauri I conclude it by the following paragraph, where I define a frame $(\tau,d)$ in which Alpha centauri sometimes has $dd/d\tau < -c$. This frame is accelerated though, in contrast to your comment below about the ICRF, where you remind that $c$ only limits speeds in frames at rest. I wonder which frame most accurately represents what the rocket experiences, its ICRF or $(\tau,d)$. $\endgroup$
    – V. Semeria
    Commented Jun 27, 2017 at 12:53
  • $\begingroup$ Imagine that, from here to AC, there were regularly spaced 'mile posts', at rest w.r.t. AC, each marked with the number of posts between that point and AC. Relative to the rods on the spacecraft, the spacing between the posts is less due to length contraction and, further, the spacing changes with spacecraft time. The instantaneous velocity of AC relative to the spacecraft is measured in the instantaneous (or momentary) co-moving reference frame. (continued) $\endgroup$
    – Alfred Centauri
    Commented Jun 27, 2017 at 13:25
  • $\begingroup$ The spacing of the mile posts is constant in this MCRF frame. Now, it appears to me that you're trying add to this the rate of change (in the accelerated frame) of the spacing of the mileposts with respect to spacecraft time. Do I understand correctly? $\endgroup$
    – Alfred Centauri
    Commented Jun 27, 2017 at 13:26
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    $\begingroup$ Differentiation mixed with a function named $d()$ is just asking for confusion. Try using $x()$ or some other letter next time. $\endgroup$
    – StephenG - Help Ukraine
    Commented Jul 17, 2017 at 9:11

2 Answers 2

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While $dx/dt$ is subluminal, $dx/d\tau$ can be superluminal. This is due to time dilation.

To take a simpler example, suppose a ship travelled to a distant star at a constant speed satisfying $\beta \gamma >1$. Then $c\beta \gamma $ is the ratio of the traversed distance from the perspective of people on Earth to the journey's duration from the perspective of people on the ship. Those watching may die of old age even if the crew doesn't.

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  • $\begingroup$ That's why I avoided $\frac{dx}{d\tau}$ and considered $\frac{dd}{d\tau}$ instead. In the latter, everything is from the ship's perspective, there's no reference to Earth's referential. $\endgroup$
    – V. Semeria
    Commented Jun 27, 2017 at 10:54
  • $\begingroup$ @VSemeria At any given moment, the ship's motion relative to the star contracts the perceived total Earth-star distance, and hence also the ship-star distance; and if you want speeds $<c$, you need $d/dt$, not $d/d\tau$. $\endgroup$
    – J.G.
    Commented Jun 27, 2017 at 11:04
  • $\begingroup$ So then it's true, from the ship's perspective, Alpha centauri moves faster than light ? $\endgroup$
    – V. Semeria
    Commented Jun 27, 2017 at 11:21
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    $\begingroup$ @V.Semeria, no it's not true; from he spacecraft's instantaneous co-moving rest frame (ICRF), Alpha Centauri always has speed less than c. $\endgroup$
    – Alfred Centauri
    Commented Jun 27, 2017 at 12:09
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No.

Just as in classical mechanics, in relativistic mechanics objects other than you move with respect to you with the same speed as you are seen moving with respect to them.

The way in which the "you" onboard the rocket reconciles this is the phenomenon of length contraction: both the distance between the objects as well as the objects themselves are shrunken in the direction of your motion by the Lorentz factor $\gamma$. Thus while they are not moving toward you any faster than $c$, you have less distance to cover to get to them from your point of view. You cannot see "external" objects as moving faster than light any more than someone outside can see you as moving faster than light, as that would violate the symmetry mandated by the absolute relativity of motion, creating a distinct state of absolute motion and absolute rest that could be distinguished by determining who sees things moving faster than light and who doesn't.

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