# Time at 1 g acceleration to travel 100 000 light years

How long would it take to go 100,000 light years at a constant 1 g acceleration?

• With or without taking relatavistic effects into account? – Mike Brockington Jun 14 '19 at 15:26
• Note, you can accelerate only on half the way. After that, you will need to decelerate, also with 1G - except if you don't want to explode on arrival like an antimatter-bomb. – peterh - Reinstate Monica Jun 14 '19 at 15:28
• @peterh Antimatter bomb would be at .92 c - this would be so, so much worse... In fact, if you are close enough to c, the mass-equivalent of your kinetic energy may be enough to make you appear as a black hole to an external observer. – Eth Jun 14 '19 at 17:38
• @Eth I think $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=2$ solves to $v=\frac{\sqrt{3}}{2}c$. It is only 0.86c. – peterh - Reinstate Monica Jun 14 '19 at 20:20
• A useful rule of thumb is that 1G acceleration is about the same order of magnitude as 1 light-year in relativistic units. This means that anyone going a lot more than 1 light-year while accelerating at 1G is basically going at the speed of light, while anyone going a lot less than 1 light-year doesn't have to worry too much about relativistic effects. – Micah Jun 15 '19 at 6:54

# Nonrelativistic solution

The variables used will be

• $$x$$ for the distance travelled
• $$v$$ for velocity
• $$a$$ for acceleration ($$1~\mathrm{g}$$)
• $$t$$ for the time
• $$c$$ for the speed of light.

## Non braking

Assuming the velocity you arrive at does not matter we take the equation

$$x = \frac12 a t^2\ .$$

Solve for $$t$$:

$$t = \sqrt{\frac{2x}{a}}\ .$$

(Let’s discard the negative solution here)

Plugging this into Wolfram Alpha gives us

$$1.389 \times 10^{10}~\mathrm{s}\ ,$$ or just over 440 years.

The velocity the object would be arriving at is be calculated by

$$v = a \cdot t \approx 1.362\times 10^{11}~\frac{\mathrm{m}}{\mathrm{s}}\ .$$

About 454.4 times the speed of light.

So no we cannot neglect relativistic effects.

## Braking

If you want to arrive at that location with reasonable speeds you’d have to accelerate half the way and brake the other half. We compute $$t$$ the same way we did above and get

$$9.822 \times 10^9~\mathrm{s}\ ,$$

or just over 311 years. After that time you would only have gone half the way and need to turn your spacecraft around and decelerate which takes the same time again, giving you a total of 622 and a half years. But you would stop next to your target and not shoot past it at extreme speeds. Your maximum speed (at the turning point) would now be

$$9.632\times 10^{10}~\mathrm{s}\ ,$$

just over 321 times the speed of light.

# Relativistic effects

Going anywhere near the speed of light (or close to a great source of gravity like a black hole for that matter) will yield a huge variety of relativistic effects, making time and space not be the same for every person.

The calculation including the relativistic effects is quite complicated.

The important thing to note here is that the traveling object and an external observer will measure the time differently.

## External observer

From the perspective of an external observer (so seen from earth, the target or any other relatively static point in the universe) it would take you 100000 years to travel 100000ly at light speed (that’s kind of the definition), but the object will not be traveling faster than the speed of light:

$$t=\sqrt{\frac{x}{1~\mathrm{ly}}^2 + 2 \frac{ x / 1~\mathrm{ly} }{ a / \frac{\mathrm{ly}}{\mathrm{y}^2} }} \cdot 1~\mathrm{y}\ .$$

I told you it’d be complicated. Plugging it in we get 100001 years. Not surprising: as discussed above: traveling at light speed for about 100000 years plus a bit for accelerating to that speed.

This is the non-braking variant. Braking however would not take a lot longer. About 100002 years in total. So one year of accelerating to the speed of light and one year of braking – simply said.

## Perspective of the traveling object

From the perspective of the object the primary effect to consider here would be length contraction. It makes the distance to the target appear smaller and smaller the faster you are going. From the perspective of that object it would thus take less than duration calculated in the nonrelativistic solution above:

Using the other answer as reference again:

$$t = \frac{c}{a} \mathrm{acosh}\left(\frac{x\cdot a}{c^2} + 1\right)$$

$$3.741\times10^8~\mathrm{s}\ ,$$

Taking braking into account again we get 11.18 year for each half, so about 22.4 years in total.

# Conclusion

Relativistic effects are important here. While it would be totally doable for an astronaut to complete that mission in their lifetime by the time they get there everyone they know will have died 100000 years ago. They would receive that information during the flight. Any communication sent after arrival would have a round trip time of 200000 years.

# Visualisation

The other answer by user Punintended links a wonderful tool you can try out that visualizes these effects. Don't worry about the fuel parts. You can see the rocket become very short (relativistic length contraction) and time progressing at different speeds for the traveller and the observer. When choosing a less extreme example (like $$100~\mathrm{m}$$) you can see the traveller accelerate and decelerate. • @Omnifarious You are completely right. What I was for some reason thinking of was an information sent at the same time as the traveller arrives at the destination. – Hans Jun 15 '19 at 8:25
• @alexdriedger from the perspective of the traveller the entire universe appears to be compressed so the distance to the target becomes smaller. (For the traveller only) – Hans Jun 15 '19 at 8:27
• What I find interesting is that, from the perspective of the rocket, relativistic effects make the trip go by faster than if was working on purely Newtonian physics, even though the latter would be going faster than the speed of light. – nick012000 Jun 15 '19 at 14:27
• Also worth noting: If the onboard navigation system switches to braking mode 5 minutes (board time) too early or too late, you end up 1 ly away from the destination, and it will take you about a year to correct this afterwards - or live with a "slight" deviation from 1g during the non-relativistic deceleration phase – Hagen von Eitzen Jun 16 '19 at 11:33
• @nick012000 The real mind blowing part is that you can get about anywhere in the visible universe in a human lifetime – JollyJoker Jun 17 '19 at 8:31

Welcome to the site!

Using this tool:

• Observer time: 100001 years

• Traveler time: 22.4 years

Edit: Time is fixed, I blame the google calculator

• Your tool actually assumes you want to slow down: take a look at the animation. Also you somehow mistyped the observer time. – Hans Jun 14 '19 at 16:40
• to clarify: choose a shorter distance like 100 meters and see the animation clearly speed up and slow down again. This also is the reason why the rocket expands to its full length again at the end: it is not moving at relativistic speeds anymore, but standing. – Hans Jun 14 '19 at 16:50

Let's start by assuming you don't decelerate halfway. Work in units with $$c=1$$. With a constant acceleration of $$a$$, the rapidity $$\phi=a\tau$$ at a proper time $$\tau$$ after you start from rest, so $$\beta=\tanh a\tau,\,\gamma=\cosh a\tau,\,dx=\beta dt=\beta\gamma d\tau=\sinh a\tau d\tau,$$where $$x$$ is the distance travelled and $$dt=\gamma d\tau$$ is the infinitesimal time that time-dilates to $$d\tau$$. After a proper time $$\tau$$ we have $$t=\int_0^\tau\gamma(\tau^\prime)d\tau^\prime=\frac{1}{a}\sinh a\tau,\,x=\frac{1}{a}\left(\cosh a\tau-1\right)=\sqrt{t^2+\frac{1}{a^2}}-\frac{1}{a}.$$Or if we want to get either $$t$$ or $$\tau$$ from $$x$$,$$t=\sqrt{\left(x+\frac{1}{a}\right)^2-\frac{1}{a^2}},\,\tau=\frac{1}{a}\text{arcosh}(1+ax).$$You can easily put the powers of $$c$$ back in, of course. For example, the time elapsed on the ship after 100 kLy is$$\frac{c}{g}\text{arcosh}\left(1+\frac{gx}{c^2}\right),$$which I'll leave you to calculate. Of course, if you do decelerate halfway you need to double the 50 kLy time, giving $$\frac{2c}{g}\text{arcosh}\left(1+\frac{gx}{2c^2}\right)$$.

When $$gx\gg 2c^2$$, the latter formula approximates $$\frac{2c}{g}\ln\frac{gx}{c^2}$$. But $$c/g$$ approximates $$0.969$$ years, while for $$x$$ equal to one light year $$gx/c^2$$ approximates $$1.03$$. In other words, the time taken in years, if you decelerate halfway, is approximately twice the natural logarithm of the number of light years. This is a convenient rule of thumb due to how the length of Earth's year compares to the Earth's surface gravity.