I’ve seen a lot of arguing about Bells Spaceship Paradox, so it appears to confuse even some physicists. In Bells Spaceship Paradox, two ships begin simultaneously accelerating at a constant rate relative to frame S. The ships travel in the same direction, one ship behind the other. Acceleration begins at time $t = 0$.

A note on terminology: I use “in S” as a shorthand for “relative to an observer at rest in inertial frame S”. I use “see” to mean something one might see if they could perceive any point in space without light delay (or, equivalently, what they could calculate and later confirm by adjusting for light delay).

In S, for any time t, we can calculate v (velocity), d (distance traveled by the spaceship) and T (the proper time on the ship). The equations can be seen at http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html.

Because acceleration is constant in S, it is not constant in S’, which is the instantaneous inertial frame for the ships at time t. An observer in S would note that the distance between the two ships remains constant and that the ships clocks are synchronized with each other.

Assume that at time t the trailing ship is at $x_1$ and the leading ship is at $x_2$. The formula for calculating L’ (the distance between ships in S’) can be derived as follows (I’ve assumed $c = 1$): $$L=x_2-x_1$$ $$L'=x_2'-x_1'$$ $$x_2'=γ(x_2-tv)$$ $$x_1'=γ(x_1-tv)$$ $$L'=x_2'-x_1'=γ(x_2-tv)-γ(x_1-tv)$$ $$L'=γ(x_2-tv-x_1+tv)$$ $$L'=γ(x_2-x_1 )$$ $$L'=γL$$

What I’m more interested for this question is the derivation of the ship’s time in S’; that is, how the ships would “see” each other’s clocks.

My derivation is: $$t_2'=γ(t-vx_2 )$$ $$t_1'=γ(t-vx_1 )$$ $$t_2'-t_1'=γ(t-vx_2 )-γ(t-vx_1 )$$ $$t_2'-t_1'=γ(t-vx_2-t+vx_1 )$$ $$t_2'-t_1'=γv(-x_2+x_1 )$$ $$t_2'-t_1'=-γv(x_2-x_1 )$$ $$t_2'-t_1'=-γvL$$ $$t_2'=t_1'-γvL$$ $$t_1'=t_2'+γvL$$

It is very odd that the Wikipedia page (https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox) reverses the sign of the delta time from what I have. Another question on this site derives the sign the same as I did. Which is the correct formula?

My second question is about interpreting what this means. Let’s say the spaceships reach destinations separated by L at time t. There are two events: ship 1 arrives at destination 1 at time t (in S) and with the ship's clock reading T. Ship 2 arrives at destination 2 at the same time t (in S) and with their clock also reading T.

The two events are not simultaneous in any instantaneous frame S’. Which occurs first in S’?

When ship 2 arrives at its destination, $t_2'=T=t_1'-γvL$, so $t_1'=T+γvL$. I interpret this as meaning that ship 2 (the leading ship) “sees” a later clock time on ship 1 (the trailing ship), so that for ship 2, ship 1 reached and passed its destination. Similarly, when ship 1 arrives at its destination, $t_1'=T=t_2'+γvL$, so $t_2'=T-γvL$. When ship 1 arrives, it “sees” an earlier time on ship 2’s clock, so ship 2 has not yet reached its destination. Is this correct or do I have it backwards?

EDIT: As Wolfram jonny pointed out, I am mixing apples and oranges.The ships have to stop accelerating (simultaneously in S), to do any of the comparisons that I've done. Given that, though, is there still in error in the Wikipedia page? Is the time difference $-γvL$ or $γvL$?

I would still like to know the distance between ships and the time difference as seen by either ship and without making them stop accelerating, but that should probably be a separate question and only after I think about the problem in light of Wolfram jonny's comments.


I’m going to partially answer my questions. I’m not entering them as the answer, because 1) I am not a physicist so I am never sure my reasoning is sound and 2) because it’s still just a partial answer.

Everything here comes from a page by Michael Weiss written in 1995: http://www.dlugosz.com/files/PhysFAQ-edit/Relativity/SR/spaceship_puzzle.html

First, is the time difference $γvL$ or $–γvL$? I still believe it’s $–γvL$. We are mapping two locations in $S$ to $S’$. But in $S’$, the rear ship’s x axis is tilted, so that the mapped location of the lead ship is in the past. So, yes, $t_2'$ is earlier than $t_1'$, but that is not where the rear ship will see the lead ship. The lead ship will be further ahead and have a later time.

If the ships stop accelerating at time $t$, the formulas I gave should be correct, but if the ship’s are still accelerating, it’s a lot harder (for me, anyway) to determine the correct distance and time. Here’s the procedure I derived from studying Weiss’s page. Any errors are mine.

The spacetime curves for the two ships can be plotted from this formula:


where $k$ is the distance of the ship from the origin of $S$ at time $t=0$ and $a$ is the acceleration. For the rear ship, we can assume $k=0$ and for the lead ship $k=L$.

At any time $t$ in $S$, we can draw the instantaneous spacetime axes for the rear ship in $S’$. For distance $x$, calculate $f(x)$, so we can identify a point $P$ with coordinates $(x, t)$ that lies on the curve for the rear ship.

The time axis is tangent to the curve at point $P$ and the X axis is the reflection of the time axis around the 45° axis that crosses $P$. Calculate the intersection of the X axis with the curve for the lead ship and let’s call that point $Q$. The distance from $P$ to $Q$ is the distance that the rear ship sees to the lead ship.

There are probably multiple ways to calculate the time on the lead ship’s clock, but the easiest might be to use the t coordinate of $Q$:

$$T=\frac{sh^{-1} (at)}{a}$$

My math is super-rusty, but I’ll see if I can find the X axis for point $P$ and its intersection with the spacetime curve of the lead ship. Once I have $Q$, the rest is easy.

  • $\begingroup$ Draw the spacetime diagram and I promise all will be obvious. $\endgroup$
    – WillO
    Commented Oct 17, 2019 at 2:12

2 Answers 2


you defined S' as an inertial reference frame that moves at v relative to the "rest" frame, but in which ships 1 and 2 move at different speeds. Thus $t'_1$ and $t'_2$ are times measured in S', whch is not the same as the reference frame of the two ships. That is why I fail to understand the meaning of substracting two times $t'_1$ and $t'_2$ that should be actually called $t'_1$ and $t''_2$ and do not belong to the same inertial frame.

If you make a minkowski disgram, you will see that ship 2 sees that the event T for ship 1 happens in its future (consistent with the fact that ship 2 is seen as starting to accelerate earlier accordint to both ships, and thus it stops before). So it seems like the oppossite will happen: when ship 1 stops at T in its own frame, ship 2 sees that this even will happen in its future, so $t''_2=t'_1+something)$, and viceverza.

But I would like to derive that from teh equations rather than thh diagram, I might do that tomorrow.

  • $\begingroup$ Actually, the Wikipedia uses a calculation pretty much like mine and comes up with $γvL$. The question Help understanding Bell's spaceship paradox derives the same formula I did (it's buried in the reply by Pulsar, just below the first paragrah. $\endgroup$
    – freixas
    Commented Dec 24, 2018 at 23:57
  • $\begingroup$ Also, the two ships have identical gamma and velocity, so I believe they are in the same frame S'. $\endgroup$
    – freixas
    Commented Dec 25, 2018 at 0:00
  • 1
    $\begingroup$ If they were not accelerating it woulod be true they are in the same s'. That is why in deriving the equation in wikipedia they assume that the ships decelerated and are already at rest relative to each other $\endgroup$
    – user65081
    Commented Dec 25, 2018 at 6:23
  • $\begingroup$ Good point! So is there a way to calculate the distance between the ships while they accelerate? Also, the time difference, what they "see" on each other's clocks. When the trailing rocket "looks at" the leading rocket, does it "see" the leading rocket's clock slow down or speed up? $\endgroup$
    – freixas
    Commented Dec 25, 2018 at 12:29
  • $\begingroup$ To answer my own comment, the link at dlugosz.com/files/PhysFAQ-edit/Relativity/SR/… might answer my questions, but will require some study. $\endgroup$
    – freixas
    Commented Dec 25, 2018 at 15:19

I think your calculation is correct. The Wikipedia article seems to have a sign error: $\gamma(t_B-vx_B/c^2)-\gamma(t_A-vx_A/c^2)$ works out to $-\gamma vL/c^2$, not $\gamma vL/c^2$, given their assumptions.

if the ship’s are still accelerating, it’s a lot harder (for me, anyway) to determine the correct distance and time.

I wouldn't even bother to work this out because it's physically meaningless. There's no sense in which an object "sees"/"observes"/"perceives" things on the spacetime plane perpendicular to its instantaneous velocity in any special way. You're in no way obligated to treat a certain plane of spacetime specially just because you happen to have a certain state of motion. In general it will only make your life more difficult. In certain cases, it may make a problem easier, but when that happens it's not because any particular object has that state of motion, but rather because those coordinates respect some global symmetry of the system you're considering.

You could work out as an exercise what each ship will actually see on the other ship's clock: i.e. the relationship between the proper times of each ship along lightlike paths between their worldlines, rather than spacelike paths. Using subscript 1 for the trailing ship and 2 for the leading ship, you should find for light from 1 to 2 $$τ_1(τ_2) = -(c/a) \log\,(\exp (-aτ_2/c) + aL/c^2)$$ and for light from 2 to 1 $$τ_2(τ_1) = (c/a) \log\,(\exp (aτ_1/c) - aL/c^2).$$ This is an easy calculation if you work with the ships' positions in S as functions of their proper time ($t(\tau)$ and $x(\tau)$), which is the "spacetimier" way of doing it since it treats the coordinates symmetrically.

Note that for light from 1 to 2, $τ_1$ approaches a constant asymptote: there is an event horizon, and ship 2 will never see times on ship 1 later than that. On the other hand, for light from 2 to 1 we have $τ_2(τ_1) \approx τ_1$ when $τ_1 \gg L/c$.

The Doppler shifts are the derivatives of these functions. Light from 1 to 2 will be more and more redshifted as you approach the limit time, while light from 2 to 1 will have essentially no Doppler shift at late times.


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