# Lorentz/velocity transformation

1. Jan 21, 2007

### GreenLRan

1. The problem statement, all variables and given/known data

Two spaceships approach the Earth from opposite directions. According to an observer on the Earth, ship A is moving at a speed of .753c and ship B at a speed of .851c. What is the speed of ship A as observed from ship B? Of ship B as observed from ship A?

2. Relevant equations

v'=(v-u)/(1-u*v/c^2)

3. The attempt at a solution

For the speed of ship A with respect to B I tried
(.851c-.753c)/(1-.851c*.753c/c^2) = .2728c
For the speed of ship B with respect to A i tried
(.851c+.753c)/(1+.851c*.753c/c^2) = .9776c
I am not sure of the correct answers, However I'm pretty sure these are wrong. Thanks.

2. Jan 22, 2007

### Jezuz

Let $$S$$ be a coordinate system fixed on the observer on earth, oriented such that the positive x-axis is in the direction of the motion of ship A. Then in that frame ship A has velocity $$0.753c$$ and ship B has velocity $$-0.851c$$.

Now, we set up a fram $$S'$$ in which ship A is stationary and which has x-axis oriented in the same way as fram $$S$$. Then in your formula $$v'$$ is the velocity of ship B as seen from ship A, $$u$$ is the velocity of ship A w.r.t. $$S$$ (earth) and $$v$$ is the velocity of ship B w.r.t. earth.

Then the velocity of ship B as seen from ship A is

$$v' = \frac{-0.851-0.753}{1+0.753\times 0.851} c$$

If you instead consider an observer in ship B then you will get ($$v'$$ is the velocity of ship A as seen from ship B, $$v$$ is the velocity of ship A as seen from earth and $$u$$ is the velocity of ship B as seen from earth

$$v' = \frac{0.753+0.851}{1+0.753\times 0.851} c$$

and both these are equal to (except for signs) $$0.9776 c$$ .

Last edited: Jan 22, 2007
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