# Special Relativity Question

1. Dec 8, 2005

### IntellectIsStrength

The Andromeda glaxaxy is two million light-years from earth, measured in the common rest frame of the earth and Andromeda. Suppose you took a fast spaceship to Andromeda, so it got you there in 50 years measured on the ship. If you sent a radio message home as soon as you reached Andromeda, how long after you left earth would it arrive, according to timekeepers on earth? Note that radio waves travel at the speed of light.

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With my own calculations I got a negative imaginary number. What I did was I found out the speed of the spaceship, and I got it to be 40000c. I'm not sure if this is right. Any help would be greatly appreciated.

2. Dec 8, 2005

### dicerandom

The spaceship cannot possibly travel faster than light, that's one of the consequences of Special Relativity's postulate that all inertial observers see the speed of light as constant, therefore its velocity relative to the Earth-Andromeda system must be some fraction of c. You need to use the proper Lorentzian time and distance formulas.

3. Dec 8, 2005

### emptymaximum

with a speed of 40000c you've either found a space ship with warp drive, or you're wrong.
my assumption is the latter.

how did you find this speed?

4. Dec 8, 2005

### IntellectIsStrength

Thanks for the replies.
For the speed I used a light-years (ly) equation.
ly = cy (y is years)
c= ly / y
c= 2 000 000 / 50 years
c= 40 000
My hunch is also that this is wrong but I don't know what else to do.
dicerandom, do you mean Ls = gamma Lm? How do I use that in this situation? I need the speed for the gamma; do I not?

5. Dec 8, 2005

### emptymaximum

no. c is a constant. it is ≈ 300 000m/s (in vaccuum).
also, you used thetime as measured ON THE SHIP. the ships clock runs slow from the PoV (Point of View) of the earth. how much slower?
that is up to you to find out.
note that the distance to Andromeda is also from the PoV of the earth.

6. Dec 8, 2005

### IntellectIsStrength

It is slower by a factor of gamma, right? The equation is t = gamma t' (t' being the 50 years). But I need a speed to solve for the gamma. How can I find out the speed of the spaceship?

7. Dec 8, 2005

### emptymaximum

use the lorentz transformation for velocity.

8. Dec 8, 2005

### IntellectIsStrength

Is there any other way of doing it? My class hasn't learned the Lorentz Transformation.

9. Dec 8, 2005

### dicerandom

Your factor should be $$1/\gamma$$, not $$\gamma$$. This is assuming that your book is using the standard definition of $$\gamma$$:

$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$$

You then have the following transformation equations:

$$\Delta t^\prime = \frac{\Delta t}{\gamma} = \sqrt{1-v^2/c^2} \Delta t$$
$$\Delta x^\prime = \gamma \Delta x = \frac{\Delta x}{\sqrt{1-v^2/c^2}}$$

Think of the problem this way. You have two events, one of which is that the spaceship departed earth and the second is that the spaceship arrived at the andromeda galaxy. Think about where and when both of these events occured in both refrence frames and I think you'll find that you can use the equations above to reduce the problem to a system of two equations with two unknowns.

Last edited: Dec 8, 2005
10. Dec 8, 2005

### IntellectIsStrength

Thanks dicerandom but I have never come across the transformation equations that you mentioned. So shouldn't here be an alternative method of doing this problem?

11. Dec 8, 2005

### emptymaximum

special relativity problem and you haven't come across these equations?
are you sure?

12. Dec 8, 2005

### dicerandom

There are many ways to solve it, I chose the method by which I first learned the theory since I thought perhaps your book followed a simmilar approach. Your book should have some simmilar equations to transform intervals and events between different refrence frames, read through the relevant section and use the equations that they provide.

What book are you using?

13. Dec 8, 2005

### IntellectIsStrength

Can I use the Galilean transformations for this?

14. Dec 8, 2005

### IntellectIsStrength

Oh nevermind dicerandom you changed your equations... I understand the ones you put up now... Thanks.. So let me give it a try.

15. Dec 8, 2005

### dicerandom

Sorry, I screwed up the TeX the first time around :)

16. Dec 8, 2005

### dicerandom

I just tried working the problem out and it occured to me that you need the full event transformations, not just the interval transformations. That was my mistake, sorry, I didn't pay enough attention when thinking it through the first time.

Here are the equations you'll need:

$$t^\prime=\gamma\left(t-\frac{v}{c^2}x\right)$$
$$x^\prime=\gamma(x-vt)$$

$$t=\gamma\left(t^\prime+\frac{v}{c^2}x^\prime\right)$$
$$x=\gamma(x^\prime+vt^\prime)$$

17. Dec 8, 2005

### IntellectIsStrength

dicerandom I'm not understanding this equation:
$$t^\prime=\gamma\left(t-\frac{v}{c^2}x\right)$$

Isn't the time equation just t' = t / gamma ?

18. Dec 8, 2005

### dicerandom

When you're transforming elapsed intervals of time between refrence frames, yes, that's correct. That equation is for transforming events between refrence frames, it's more general as it allows for both temporal and spatial seperation. You can derive the first equation from that one.

19. Dec 8, 2005

### Tide

Obviously, the spaceship would have to travel very close to the speed of light so 4 MILLION years would be a very good approximation to the elapsed time for an earthbound observer.

20. Dec 8, 2005

### emptymaximum

and in the 2 million years on the way there, there's a good bet they'll invent warp drive and beat you there anyways :)