# Special Relativity Question

1. Apr 19, 2010

### tomeatworld

1. The problem statement, all variables and given/known data
According to observations on Earth, the distance to nearest star is 4.5 light-years. A ship which leaves earth takes 4.25 years (according to onboard clock) 4.25 years to reach this star. Calculate the speed at which this ship travels.

2. Relevant equations
(I think)
x' = $$\gamma$$(x-ut)
t' = $$\gamma$$(t-ux/c2)
L = L0/$$\gamma$$

3. The attempt at a solution
so we know:
x = 4.25 light years, and, t' = 4.25 years.
using length contraction, L (or x') = 4.25 / $$\gamma$$]

So i tried using this in x = $$\gamma$$(x' + ut') but only managed to get 0 = $$\gamma$$ut'.

I can't find any other way of reducing variables to obtaining new variables and would love a push in the right direction.

2. Apr 19, 2010

### Matterwave

If you want to do this question the Lorentz transformation way, you need to have 2 events, one being leaving Earth, and one being arriving at the star.

So event 1: (x,t)=(0,0) (x',t')=(0,0) so that your coordinates are synchronized at the point of departure. This step is sometimes implicit, but you should account for this for completeness.

Event 2: (x,t)=(4.5ly, ?) (x',t')=(0,4.25years)

Your job is then to use the transforms to get the question mark. In the end v=x/t.

A simpler way to answer this IMO, is to just use length contraction. You know that L'=L/gamma. Speed v, then is just v=L'/t'=L/gamma*t'. Gamma has a v in it as well, and you can solve for v.

This method requires you to keep track of what exactly L and L' mean, and what exactly you are trying to solve for. It can be misused...but it's simpler than using the Lorentz transforms.

3. Apr 19, 2010

### tomeatworld

Using the length contraction method, I managed to get:

v = $$\frac{1}{\sqrt{2}}$$c

Any chance of confirmation?

4. Apr 19, 2010

### Matterwave

Let's see:

L=4.5ly t'=4.25 years

$$v=\frac{L}{\gamma t'}=\frac{L\sqrt{1-\frac{v^2}{c^2}}}{t'}$$

$$v^2=\frac{L^2(1-\frac{v^2}{c^2})}{t'^2}$$

Solving for v, I get: v=.529c which doesn't seem to match yours. Perhaps you can show your work?