This might be an ill-formed question, but maybe you can help clear up my misunderstanding (I think this question was getting at similar concept).

On this website we see a calculator for a rocket to travel half-way to a location under a constant acceleration and then travel the last half under a constant deceleration (opposite direction, same magnitude).

So, let's create a specific example.

  1. Rocket accelerates at $10 \; m/s^2$ for $50 \; lightyears$ and then decelerates at $10m/s^2$ for $50 \; lightyears$.
  2. In doing so, it travels from star $A$ to star $B$, where star $B$ is $100 \; lightyears$ from star $A$.
  3. The rocket starts and ends its journey with $0$ velocity relative to the stars which are at $0$ velocity relative to each other.

From the special theory of relativity, we are able to calculate the following.

  1. The time for the journey as measured on star $A$ is $101.9 \; years$.
  2. The time for the journey as measured from the rocket is $8.9 \; years$.
  3. The maximum velocity is $99.9956 \; %$ of $c$.

So far so good. Obviously, the rocket experiences spatial contraction and time dilation so it does not measure its own velocity as faster than the speed of light relative to, say, a series of milestones along the way.

However, if I were on the rocket and continuously measured the distance to star B ($d_{RB}$) and the time on the rocket ($t$), then I would see the distance shrink from $100 \; lightyears$ to $0 ;\ lightyears$ (including both my changing position and the spatial contraction, but both endpoints would be only from the changed position) and the time increase from $0 \; years$ to $8.9 \; years$. I could, for example, plot the distance and the time on a graph and it would be a continuous graph with an average gradient of $\frac{100}{8.9}c$. So that would mean that the planet was approaching faster than the speed of light in my frame of reference (an observer at star B would see me approach over the course of $101.9 \; years$, so the same restriction wouldn't apply in reverse).

Is this just that looks can be deceiving when spacetime is dilating? For example, there's nothing special about star $B$. If I experienced any amount of spatial contraction, then a distant object would appear to race towards me at a velocity proportional to its distance from me. So spatial contraction is just weird and you could correct for it when doing measurements? Or what?

  • 1
    $\begingroup$ You write: "if I were on the rocket and continuously measured the distance to star B". It's not clear whether such a measurement is possible at all. I propose that you narrow down this thought experiment to a specific assessment procedure. At the point of departure a telescope will detect from the distant star a particular flux of photons per unit of light gathering area. A slow moving starship will, during it's journey, see the photon flux number increase steadily as it approaches the destination. From measured photon flux the instantaneous distance to the destination can be inferred. $\endgroup$
    – Cleonis
    Commented Mar 14, 2021 at 21:17
  • $\begingroup$ Thanks all for the answers and discussions. I regret that I have but one tick to give! $\endgroup$
    – Dr Xorile
    Commented Mar 15, 2021 at 16:18

3 Answers 3


If you make a U-turn in New York, Los Angeles can go from being 3000 miles behind you to being 3000 miles in front of you in a tiny fraction of a second. We could, if we wanted, describe that as Los Angeles moving 6000 miles in essentially no time at all (in a non-inertial frame attached to your car), but that's usually not a terribly useful description.

If you make your U-turn fast enough, you might even want to say that Los Angeles traversed 6000 miles faster than the speed of light. The fact that this is true in the non-inertial frame you stubbornly chose is not a violation of relativity.

Your accelerating rocket is no more inertial than your turning car. If you want to describe the world in a useful way, it is probably best to use your instantaneous inertial frame rather than some non-inertial frame that moves with the rocket. Do that, and you won't see anything moving faster than light.


Is this just that looks can be deceiving when spacetime is dilating?

No, you can certainly choose coordinates where what you described is accurate. There is nothing wrong with that.

You might be uncomfortable with that because of the limitation that nothing can go faster than light. That principle is not violated.

Recall that the second postulate of relativity says that light travels at c in all inertial frames. However, the frame you described is highly non-inertial. So light does not travel at c in that frame. Your spacecraft always moves slower than light, even if it moves faster than c.


The ship can measure its acceleration $g$ all the time. Doppler effect can be used to measure velocity, (redshift of star A and blueshift of star B).

I am not sure how the distance could be measured, but anyway, it is clear by calculation that the star B is approaching faster than $c$, even if the relative velocity with respect to the ship (measured by Doppler effect) is below $c$.

It seems weird because in non relativistic mechanics, $\Delta x(t) \rightarrow v(t)\Delta t$ when $\Delta t \rightarrow 0$. In SR, due to lenght contraction, $\Delta x(t) \rightarrow f(v,t)\neq v(t)\Delta t$. So, the tangent of the graph $x$ and $t$ is not the ship velocity.

  • $\begingroup$ I agree that this is not the ship velocity. My statement is just that the star velocity measured from the ship is higher than $c$. But I think the answer is just that the ship is a non-inertial frame. $\endgroup$
    – Dr Xorile
    Commented Mar 15, 2021 at 16:16

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