Velocity in Special Relativity

1. Feb 11, 2009

Kamkazemoose

1. The problem statement, all variables and given/known data
You wish to travel to Pleisdes (at a distance of 130 pc) in 10 years, according to the clock that you carry. How fast do you have to travel to accomplish this (express the velocity as v/c)?
When you reach the Pleisdes, you send a radio signal back to Earth. For someone on Earth, how long has it been between the moment you left and the moment when the radio signal was received?

2. Relevant equations
t=$$\gamma$$t0
d=$$\gamma$$d0
$$\gamma$$=$$\frac{1}{1-\sqrt{1-\frac{v^2}{c^2}}}$$
v=d/t
3. The attempt at a solution
I know that to calculate velocity, you have to use the time and distance in the same inertial frame, so we can't simply insert the given d and the given t, as they are in 2 frames. I tried to come up with an equation for velocityby converting t0 to t, so I had
(using t(naught) as t0, because i can't get it to work within latex, sorry)

v=$$\frac{d}{\gamma*t(naught)}$$

and then, replacing gamma and using algebra etc, I simplified this to

v=$$\frac{d}{\sqrt{t(naught)^2+\frac{d^2}{c^2}}}$$

where t0 = 10yr and d= 130 pc
I got that v=2.9996*108m/s or v=.9997c and $$\gamma$$ = 40.83
then, when I calculate t in the frame of the earth, i get t=408.3 years, but d = 424lightyears, so This should mean that the person is traveling faster than the speed of light, but I found v as less than the speed of light, so something went wrong.

If you could help, did I get the equation for finding velocity wrong, or was there something else, I can't figure it out, thanks. Also, sorry but I don't quite understand latex, so there may be somethings that look ugly.

Last edited: Feb 11, 2009
2. Feb 11, 2009

JoeBonasia

10 light years is the time length according to the moving frame, but what about the distance is that the distance you percieve in the moving frame or is that the distance from the rest frame, according to an observer on Earth?!

If 130pc is the length observed in the moving frame then calculating the velocity is easy because all quantities are in the same frame of reference. If the distance is relative to a 'stationary' observer in the rest frame of the Earth, remember 'moving clocks run slow' so the distance should be a smaller number (just to help as a self check due to length contraction)

The second part would follow as such

Time = time it took you get there(10*gamma) + time it took signal to travel distance of 130pc (if 130 pc was measured in the rest frame, if not then adjust the distance anf find the tiem it took, this is easy since your measuring the distance in light years)

Last edited: Feb 11, 2009