# Time dilation of a rocket

1. Nov 2, 2006

### Tonyt88

A rocket flies between two planets that are one light-year apart. What should the rocket's speed be so that the time elapsed on the captain's watch is one year?

I figure I need to use the Lorentz factor:

1/(sqrt(1 - v^2/c^2))

But exactly how would i go about solving it.

Last edited: Nov 2, 2006
2. Nov 2, 2006

### Galileo

Assuming constant speed: time = distance/speed.

3. Nov 4, 2006

### AstroGuy

Possible help here...

I have 3 possible helps...

1) No rocket can go at the speed of light, so the question is pointless.

2) Off the top of my head, I would figure that you'd set the rocket speed for c and in one year, the captains watch would read 1 year, when you get to the next planet- his watch would be time dilated too...

3) Check out http://en.wikipedia.org/wiki/Time_dilation
Look under 'Overview"
it could be that you are looking for
The formula for determining time dilation in special relativity is:

Δ t = (gamma)Δ t0
where
Δ t is a time interval measured by an observer in a stationary frame of reference,
Δ t0 is that same time interval as measured by an observer in the moving frame,
and (gamma) is the Lorentz factor, ( your formula of 1/(sqrt(1 - v^2/c^2))

v is the relative speed between the clock and the stationary system, and
c is the speed of light.

If you are looking for the time interval as measured by an observer in the moving frame (the space ship captain) set Δ t0
to 1 year, but ...hmmm....
gamma times 1=gamma. Damn. You would also need to know the time elapsed for the stationary observer in that case to solve for v.

i say go with option #2.
Hope this helps more than hinders!

Last edited: Nov 4, 2006
4. Nov 4, 2006

### OlderDan

Galileo's hint may have been a bit too obsscure for you, but it is what you need. The rest of what you need is that the distance between the planets as seen by the captain depends on how fast he is going. Use the Lonrentz factor to find the distance in terms of the velocity and use time = distance/velocity and you can solve for velocity.

5. Nov 4, 2006

### bob1182006

I agree with option #2 but i don't believe the captain's watch would be affected by time dilation since the captain is presumed to be within the ship.

6. Nov 4, 2006

### dimachka

you have to talk about reference frames when dealing with special relativity, however in this case, the requirement that the ship travel a light-year in one year means it would have to go at the speed of light, which is impossible since I am assuming the ship and crew would have non-zero mass...

7. Nov 5, 2006

### Galileo

The ship travels one light-year relative to the planet's reference frames. The duration should be one year as seen from the ship's reference frame.

The formula (time = distance/speed), is really all you need. The only 'catch' is that you should consider in which reference frame you do the calculation. Since you need the time in the ship's frame to be one year you could take that frame. In that case the distance between the planets is lorentz-contracted (shorter by a factor gamma).
Or you could take the planet's reference frame, in which case the distance is 1 lightyear, but the time elapsed on the ship's frame is time-dilated (by a factor of gamma).

Whichever way you choose, the formulas will be the same.

Last edited: Nov 5, 2006
8. Nov 5, 2006

### dimachka

if in any referance frame you are moving at the speed of light, then you are moving at the speed of light in all reference frames....

9. Nov 5, 2006

### Galileo

Yes, but you don't have to go at the speed of light. The distance between the planets are 1 lightyear apart as seen from the planets frame. It has to take one hour as seen from the ship's frame.

10. Nov 5, 2006

### dimachka

Sure, but if in the planet's reference frame you have moved 1 light year in 1 year, you were going at the speed of light. This necessarily means you would have to go at the speed of light in any frame, if you dont believe me, plug it into your trusty lorentz transformations.

11. Nov 5, 2006

### nrqed

You are mixing quantities measured in different frames! The question never says that it took one year in the planet's frame. The one light year distance is in the planet's frame and the duration of one year for the trip is in the ship's frame.

The ship does not have to move at the speed of light. Actually, one could even ask how fast must be the speed of the ship to get there in 2 months (as measured in the ship's frame).

Patrick

12. Nov 5, 2006

### nrqed

Galileo and OlderDan already suggested the solution. I will simply be a bit more specific. Let's call "t" the time measured in the ship's frame and "d" the distance measured between the two planets in the ship's frame. Then t= one year and "d" is unknown.

You get that $$v = {d \over t} = {d \over {\rm one ~year}}$$
which contains two unknowns. Now use length contraction
$$d = {1~ ly \over \gamma}$$
where $\gamma$ is the usual gamma factor which obviously contains the speed "v", and "ly" stands for light-year.

Now you have two equations for two unknowns ("d" and "v") and you can solve.

There are other ways to go about it (for example by relating the time in the two frames instead of the distances) but that's probably the most straighforward.

Hope this makes sense.

Patrick

Last edited: Nov 5, 2006
13. Nov 5, 2006

### dimachka

I'm sorry, but that is just plain wrong, If you use that logic it is possible to go an arbitrarily long distance in an arbitrarily short amount of time by just going closer to the speed of light. What i said was correct.

14. Nov 5, 2006

### nrqed

And I am sorry too, but I stand 100% behind all I said. And indeed, it is true that, assuming a constant speed, special relativity allows a massive object to travel between two planets (say) separated by arbitraily long distances measured in the frame of the planets in arbitrarily small (but non zero) time measured in the frame of the ship .

This follows from special relativity. Now, in the more realistic case where
one starts from rest and one has to aceclerate and then decelerate and one has a finite amount of energy available, then a time constraint will arise. But here I am considering the case of a spaceship already moving at constant speed as it passes the first planet and which keeps this speed all the way to the other planet.

I am sorry to say that if you disagree with this you need to go back and study special relativity more carefully. I am not going to enter a long debate with you on this point so I will leave it at this. It's clear at this point that you have made your position clear and I have made mine clear too so there is no point arguing. Hopefully others (Galileo, OlderDan and anybody else with a good knowledge of SR) will jump in and confirm what I said. Meanwhile, I want to say to the original poster that my post showing a couple of equations gives the way to solve the problem.

Best regards

Patrick

15. Nov 5, 2006

### OlderDan

It is possible to go an arbitrarily long distance in an arbitrarily short amount of time, if you are talking about distance measured in one reference frame and time in another reference frame. This is the basis of the famous twin problem in Special Relativity. The travelling twin can get to a planet that is 20 light years away as measured by the stationary twin, then turn around and come back, all in one year of the travelling twin's time. He can do this because from his frame of reference the distant planet is only half a light year away, and after he turns around he is only half a light year from earth.

Last edited: Nov 5, 2006
16. Nov 5, 2006

### dimachka

brain fart: nrqed is indeed correct on this one.