Take the 2-minute tour ×

Imagine I am on Earth I have a clock which measures time $t$. When my Earth clock reads time $t$, then when I look at a moving spacecraft's clock I see the time $t'$. Let us assume time is measured in years.

Let us suppose the clocks were initially synchronized so that when $t=0$, $t'= 0$.

Now let us suppose that the speed of the rocket continues to increase in the following way (Note that I have increased the speed in discrete time steps for convenience, but I'm sure a continuous speed function could be obtained but just mathematically tedious):

At $t = 0$, $v = c\sqrt{3/4}$. So when $t = 1$, $\delta t'= \frac{\delta t}{\gamma} = 0.5.$
At $t = 1$, $v = c\sqrt{15/16}$. So when $t = 2$, $\delta t'= \frac{\delta t}{\gamma} = 0.25.$
At $t = 2$, $v = c\sqrt{63/64}$. So when $t = 2$, $\delta t'= \frac{\delta t}{\gamma} = 0.125.$

and so on...

If $v$ continues to increase over time in this way, then $t'$ will approach 1, but will never reach 1. It appears from the perspective of the Earthling, that the person in the rocket never ages by over 1 year. From the person in the rocket however, he will experience time as normal.

Now let us assume that after the person in the rocket has aged 10 years, he leaves his rocket ship, and travels back to Earth, leaving his rocket to continue accelerating in the same pattern. Now when the man reaches Earth, will he not see himself in the rocket, where he sees himself not having aged 1 year?

(Ignore optical effects for the purpose of this question.)

share|improve this question
I don't think using special relativity in this regime is valid, since the person on the ship is accelerating. SR is only valid for inertial (non-accelerating) frames. You will probably have to analyze this problem using the general relativistic formulae for time dilation. –  Kitchi Feb 6 '13 at 15:36
I'm assuming a special relativistic universe. Acceleration in special relativity is not forbidden. en.wikipedia.org/wiki/…, where proper time is simply the integral of the lorentz factor, which is what I have done in my analysis, but used discrete steps instead. I'm sure I could create a continous function and integral to demonstrate the same concept. –  Mew Feb 6 '13 at 15:37
"Now when the man reaches Earth, will he not see himself in the rocket," He can't get back to Earth faster than (or even as fast as) light, so no, he won't. By the time he returns the light arriving from the craft will be from after he left. This consideration effectively shows that your discrete approximation to the integral is incorrect. –  dmckee Feb 6 '13 at 15:40
That's where the paradox lies dmckee, because the spacecraft never appears to reach the time he left from the perspective of the Earth. The space craft only is seen to reach 1 year, where as he left at 10 years. So if you calculate that he returns to Earth at t = 1000000 then you will still only be able to calculate t' as <1. –  Mew Feb 6 '13 at 15:41
@dmckee, it shows that the approximation is incorrect OR there is a different resolution to the paradox OR it is a genuine paradox. I will attempt to find a continuous function v(t) so that the above point is still valid. This seems intuitively possible to me, since I can make the time steps arbitrarly small, and choose velocities to give the gamma factors which cause t' to assymptote at 1 –  Mew Feb 6 '13 at 15:50

1 Answer 1

up vote 2 down vote accepted

Your example can't be usefully described because when you change speed you are changing your inertial frame but you are not specifying how you change frames i.e. what acceleration you use.

Special relativity is perfectly capable of describing accelerated motion. See for example John Baez's article on the relativistic rocket, which gives for constant acceleration:

$$ t_{earth} = \frac{c}{a} sinh\left(\frac{a \space t_{rocket}}{c}\right) $$

where $a$ is the acceleration of the rocket must be measured at any given instant in a non-accelerating frame of reference travelling at the same instantaneous speed as the rocket. Since $sinh(x)$ remains finite for all finite $x$, constant acceleration will never give the effect that you want, i.e. for $t_{earth}$ to become infinite at some finite value of $t_{rocket}$. The only way you can achieve this is for the acceleration of the rocket to become infinite, and obviously this isn't physically possible.

However there is an analogy that is close to your question. If you throw someone into a black hole then they will measure a finite time to reach the event horizon while you will never see them reach the horizon i.e. you would have to measure an infinite time before they reach the horizon. The link to your rocket is that the acceleration a shell observer measures for the falling object tends to infinity as the shell observer approaches the event horizon. You can think of this infinite acceleration giving an infinite time dilation just as infinite acceleration would for an accelerating rocket.

But back to the point of your question: although the infalling observer measures a finite time to reach the event horizon, once they've done so an infinite time has passed for the Schwarzschild observers watching them fall. So there is no outside universe to return to. Not that you can return of course, because once you've reached the event horizon all timelike paths lead to the singularity and your inevitable doom.

share|improve this answer
Thanks John for your answer. According to math.SE however, math.stackexchange.com/questions/296358/…, a solution for v(t) does exist so that t' assymptotes at 1, without v(t) ever exceeding c. Does this mean the situation can be described by special relativity? –  Mew Feb 6 '13 at 16:26
@Chris: well yes, but the suggested function has the acceleration going to infinity as $x$ goes to one, and the infinite acceleration causes the infinite time dilation. For the Schwarzschild metric I don't think the acceleration measured by the shell observer has a tanh dependence on $r$ - when time permits I'll have a rummage through my GR textbooks. –  John Rennie Feb 6 '13 at 16:30

Your Answer


