# Help with a really nerdy calculation?

1. Jun 2, 2009

### Anticitizen

Apologies in advance for the extremely geeky nature of this post :)

I'm afraid I don't have a background in mathematics so I was hoping someone could help me out.

On the TV show Star Trek: Voyager, the ship is stranded 70,000 light years from Earth. The premise, of course, is to return home.

My question: what if the crew forgot about the prospect of getting back in a timely fashion as far as Earth is concerned, and just wanted to get back home in their OWN lifetimes... and thus switched off the warp drive and accelerated normally, in order to rely on time dilation to make the trip?

Since the starship has magical gizmos called 'inertial dampeners,' let's assume they can accelerate at a solid 10 G's without the crew being pancaked. Let's ignore deceleration time by saying that can use their warp drive to stop suddenly without killing everyone aboard.

So my question is, how long would a journey of 70,000 light years take with a constant acceleration of 10 G-units for the entire journey, relative to the crew's POV? Secondarily, what sort of acceleration would get the crew back home in their own lifetime, if any?

Oh, and what sort of time would pass on Earth in the meantime?

It's all in fun so don't take it to seriously, I'm just scratching an itch :)

2. Jun 2, 2009

### George Jones

Staff Emeritus
3. Jun 3, 2009

### Division

70 000 light years will take a long time, even if they rely on Time Dilation.

Instead of the trip taking thousands of years, from their point of view, it will take just a few, so no one they knew will be ablive when they get back.

I don't know the exact calculation, but that is however what would happen.

4. Jun 3, 2009

### Cyosis

Seeing as I have never done these calculations myself I thought it would be interesting to derive it from the ground up.

Didn't manage to get it to show properly here so I made a pdf of it instead.

http://img34.imageshack.us/img34/7642/rocketship.pdf [Broken]

Inverting equation (19) yields:

$$\tau=\frac{c}{a} \text{arccosh}(a x/c^2)$$

Plugging in x=70 000 ly, a=10*1.032 ly/y^2 and c=1 yields:

$$\tau=1.38\, \text{years}$$

Realize that this means you're accelerating with 10g all the way to earth and then instantly decelerating to zero once you reach earth so I hope those inertial dampeners work properly.

Using equation 17 we can calculate the time that has expired on earth.

$$t=70695 \, \text{years}$$

I hope I haven't made any mistakes.

Last edited by a moderator: May 4, 2017
5. Jun 3, 2009

### George Jones

Staff Emeritus
Very nice.

Don't take any notice of this picking of a nit. It might not be clear to everyone how you went from (4) to (5). What happened to the lower limit of the t integration?

Last edited by a moderator: May 4, 2017
6. Jun 3, 2009

### DrGreg

At the risk of blowing my own trumpet, and if anyone is interested, I'll point out my own derivation of this in the thread Questions about acceleration in SR, posts #13, #14, #15, plus a typographical correction in post #28.

7. Jun 3, 2009

### Cyosis

Thanks, it's hardly nitpicking since it's an error. The lower limit gives -c^2/a this results in the final answer for x being the following expression.

$$x(t)=\frac{c^2}{a}(\cosh \left( \frac{a \tau}{c}\right)-1)$$

I will fix this in the pdf later.

DrGreg I will have a look at your derivation later, I noticed you took an entirely different approach.

8. Jun 3, 2009

### George Jones

Staff Emeritus
This is one way to fix things, but not the usual way. It is possible to get (5) from (3).

9. Jun 3, 2009

### Cyosis

Interesting, I'll have to think about it.

10. Jun 4, 2009

### Anticitizen

Guys, thank you very much! I intend to go over your work and see if I can follow what's going on!