If I were to travel 1000 light-years, and witness only 1 year elapse in my frame of reference, how much time would have passed in the frame of reference of the Earth?


2 Answers 2


The answer is sort of trivial. If you travel 1000 ly so fast that in your own reference frame it takes one year, then you will have aged by one year in your own reference frame. To do so, you will need a speed of almost the speed of light, so in the reference frame of Earth, you will have spent just a tad more that 1000 yr to travel 1000 ly.

In general, the time dilation is given by the Lorentz factor $\gamma = 1/\sqrt{1-v^2/c^2}$ so to be exact, your speed must be $$ 1000^2 = \frac{1}{1 - v^2/c^2} ⇒ \\ v = (1-10^{-6})^{1/2}\,c = 0.9999995c $$ so your journey will take $$ t = \frac{d}{v} = 1000.0005\,\mathrm{yr}, $$ i.e. 1000 years, 4 hours, and 23 minutes in Earth's reference frame.

"Realistic" acceleration

As David Hammen comments below, this assumes that your spaceship accelerates instantaneously to $v$. There are infinitely many ways to achieve that speed. The proper time $\tau$ (i.e. the time experienced by the traveler) to reach a distance $d$ when traveling at a constant acceleration $a$ is $$ \tau = \frac{c}{a} \cosh^{-1} \left( \frac{ad}{c^2} +1 \right). $$ Solving for $a$ yields the acceleration needed. The most pleasant way would arguably be accelarating at $a\simeq 19.2\,\mathrm{G}$ for half a year, and the decelerate at the same amount for the rest of the journey. A more pleasant way in the beginning, but less pleasant in the end, would be to accelerate at $a\simeq9.6\,\mathrm{G}$ for one year, and then crash into planet WASP-142b, which lies at a distance of roughly 1000 ly.

Your journey, as measured by people on Earth, would then take $$ t(\tau) = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right), $$ which works out to roughly 18 and 36 days more, respectively, than in the instantaneous case. The reason that it doesn't differ that much is that you actually reach relativistic speeds pretty fast at this acceleration.

The largest G-forces can be with endured in the "forward direction", i.e. corresponding to lying on your back and accelerating upwards (accelerating along the direction of your spine tends to break it, and accelerating "backward" makes your eyes pop out). According to Wikipedia, "acceleration pioneer" John Stapp withstood 25 G for 1.1 seconds, so you might want to send him instead of yourself.

  • 1
    $\begingroup$ if you add "in earths reference frame" at the end of your answer, this is a very good answer. $\endgroup$
    – RononDex
    May 15, 2017 at 10:24
  • 4
    $\begingroup$ This assumes an instantaneous acceleration to 99.99995% of the speed of light. The answer is a bit different if the acceleration is constant to the halfway point, then slowing down from the halfway point to the destination. Note well: This will require a constant acceleration of 19 g. $\endgroup$ May 15, 2017 at 11:12
  • $\begingroup$ I totally understand it now. And yes it will be 1000 years in earth reference frame. $\endgroup$
    – lakhassane
    May 15, 2017 at 11:15
  • $\begingroup$ I edited a bit to include the comments by @DavidHammen… $\endgroup$
    – pela
    May 15, 2017 at 13:46
  • $\begingroup$ …and @RononDex. $\endgroup$
    – pela
    May 15, 2017 at 13:46

Alternative method, no Lorentz transformations required.

The spacetime interval measured by $$ (\Delta \tau)^2 = (\Delta t)^2 - (\Delta r)^2\, ,$$ is something that all observers in all inertial frames of reference agree on.

In this case, $\Delta \tau$, the proper time measured on the travellers watch is 1 year.

On the Earth, the change in position is $\Delta r = 1000$ light years; $\Delta \tau =1$ year, and so $$ 1 = (\Delta t)^2 - 1000^2$$ and $$ \Delta t = \sqrt{10^6 + 1} = 1000\ {\rm years,}\ 4\ {\rm hours,}\ 23\ {\rm minutes}$$

This assumes that you are already travelling at uniform velocity when you pass the Earth and when you arrive at your destination. Periods of acceleration and deceleration will increase this time, but not significantly compared with 1000 years.


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