<p>Alternative method, no Lorentz transformations required.</p>
<p>The spacetime interval measured by
<span class="math-container">$$ (\Delta \tau)^2 = (\Delta t)^2 - (\Delta r)^2\, ,$$</span>
is something that all observers in all inertial frames of reference agree on.</p>
<p>In this case, <span class="math-container">$\Delta \tau$</span>, the proper time measured on the travellers watch is 1 year.</p>
<p>On the Earth, the change in position is <span class="math-container">$\Delta r = 1000$</span> light years; <span class="math-container">$\Delta \tau =1$</span> year, and so
<span class="math-container">$$ 1 = (\Delta t)^2 - 1000^2$$</span>
and
<span class="math-container">$$ \Delta t = \sqrt{10^6 + 1} = 1000\ {\rm years,}\ 4\ {\rm hours,}\ 23\ {\rm minutes}$$</span></p>
<p>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.</p>

CC BY-SA 4.0
2021-02-19T00:39:24.433Z
2021-02-19T00:39:24.433Z
ProfRob
<p>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.</p>

<p>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.</p>

<strong>"Realistic" acceleration</strong>

<p>As David Hammen comments below, this assumes that your spaceship accelerates instantaneously to $v$. There are infinitely many ways to achieve that speed. 
The <em>proper</em> 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.</p>

<p>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.</p>

<p>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" <a href="https://en.wikipedia.org/wiki/John_Stapp" rel="noreferrer">John Stapp</a> withstood 25 G for 1.1 seconds, so you might want to send him instead of yourself.</p>

CC BY-SA 3.0
2017-05-15T10:24:38.337Z
if you add "in earths reference frame" at the end of your answer, this is a very good answer.
RononDex
CC BY-SA 3.0
2017-05-15T11:12:49.810Z
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*.
David Hammen
CC BY-SA 3.0
2017-05-15T11:15:23.250Z
I totally understand it now. And yes it will be 1000 years in earth reference frame.
lakhassane
CC BY-SA 3.0
2017-05-15T13:46:41.513Z
I edited a bit to include the comments by @DavidHammen…
pela
CC BY-SA 3.0
2017-05-15T13:46:51.053Z
…and @RononDex.
pela
CC BY-SA 3.0
2017-05-15T14:51:28.047Z
Nicely done. I wouldn't deem being subject to over 19 *g* for a year as "pleasant".
David Hammen
CC BY-SA 3.0
2017-05-15T15:30:56.257Z
@DavidHammen: Me neither, but apparently you _can_ survive, at least for a few seconds…
pela
CC BY-SA 3.0
2017-05-19T03:19:20.173Z
classically accelerating at 1g for 0.96yr will get you to 0.99c in 0.47 light years distance (easy math with 1g = 1.03 ly/yr2)
Jack R. Woods
CC BY-SA 4.0
2021-02-18T22:56:10.393Z
Would he gain that one year back if he travelled with the same speed back again to where he came from? Or would that be two years in total?
Constantthin
CC BY-SA 4.0
2021-02-19T08:58:38.423Z
@Constantthin No sorry, no traveling back in time (not in this way, at least). If you travel 1000 lyr at v = 0.9999995 c, and ~instantly return at the same speed, you'll be roughly two years older when you get back. Meanwhile, 2000 yr has passed on Earth. This is the famous twin "paradox" (in quotes, because it's really not a paradox).
pela
CC BY-SA 3.0
2017-05-15T09:09:00.663Z
2017-05-15T15:30:21.283Z
2017-05-15T15:30:21.283Z
pela
<p>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?</p>

<special-relativity>
How old would I be if I travelled 1000 light years in one year
lakhassane
# How old would I be if I travelled 1000 light years in one year

(Asked by: username_0 on 2017-05-14)

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?

**Tags:** special-relativity

---

## 2 answers

### Answer by username_2 on 2017-05-15. Score: 20 | Accepted answer

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](https://en.wikipedia.org/wiki/John_Stapp) withstood 25 G for 1.1 seconds, so you might want to send him instead of yourself.

> **Comments:**
> **username_3**: if you add "in earths reference frame" at the end of your answer, this is a very good answer.  
> **username_4**: 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*.  
> **username_0**: I totally understand it now. And yes it will be 1000 years in earth reference frame.  
> **username_2**: I edited a bit to include the comments by @DavidHammen…  
> **username_2**: …and @RononDex.  
> **username_4**: Nicely done. I wouldn't deem being subject to over 19 *g* for a year as "pleasant".  
> **username_2**: @DavidHammen: Me neither, but apparently you _can_ survive, at least for a few seconds…  
> **username_5**: classically accelerating at 1g for 0.96yr will get you to 0.99c in 0.47 light years distance (easy math with 1g = 1.03 ly/yr2)  
> **username_6**: Would he gain that one year back if he travelled with the same speed back again to where he came from? Or would that be two years in total?  
> **username_2**: @Constantthin No sorry, no traveling back in time (not in this way, at least). If you travel 1000 lyr at v = 0.9999995 c, and ~instantly return at the same speed, you'll be roughly two years older when you get back. Meanwhile, 2000 yr has passed on Earth. This is the famous twin "paradox" (in quotes, because it's really not a paradox).  

---

### Answer by username_1 on 2021-02-19. Score: 0

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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