# Special relativity/velocity transformations

1. Feb 16, 2009

1. A cosmic-ray proton streaks through the lab with velocity 0.85c at an angle of 50o with the +x direction (in the xy plane of the lab). Compute the magnitude and direction of the proton’s velocity when viewed from frame S’ moving with β=0.72 in the +x direction.

2. Ux'= Ux-v/1-vUx/c^2
Uy'=Uy/$$\gamma(1-vUx/c^2)$$

3. My attempt at the solution was to simply in an example of 1-D motion. From considering that a proton is moving at 0.85c in the positive x-direction (ignoring the angle), this can be compared when viewed from the S' frame, moving relative to S with 0.72c in the +x direction. The observer in S' measures the proton to be approaching in the +x-direction at 0.85c. Using the first equation U1x' = 0.85c - 0.72c/1-(0.85c)(0.72c)/c^2 = .34c. Introducing a y component makes the problem much more challenging and I'm not quite sure what to do. Do I work out the x and y directions independently? In which case, would the velocity for the y-direction be 0.85c*sin(50 degrees) and the x direction be 0.85c*cos(50 degrees). How would this appear to an observer in S'? That is, how would I calculate the direction as observed in S' and the magnitude?

Thank you.

2. Feb 16, 2009

### xboy

It's not very clear to me what you are asking. The proton has a certain velocity relative to the lab which is given. You can work out the x and y components of that velocity. The other frame's velocity relative to the lab is given. This has only an x component. You know how to calculate relative velocities Ux and Uy in that case. If you know that, can you work out the magnitudes and directions?

3. Feb 17, 2009

Sorry for the ambiguity, the question just asks to compute the magnitude and direction of the proton as observed in S'.

I worked more on the problem and completed the x and y components.

Using the velocity transformations, the x-component was 0.85c*cos(50), doing the calculation out with the above equation for the x-direction I got -0.287c.

The Y-component was 0.85c*sin(50). Doing this calculation out with the above equation for the y-direction, I got 0.569c.

To find the magnitude I used P-theorem. c^2=a^2+b^2. In this example, c^2 =0.569c^2+(-0.287)^2. c=0.637c.

To find the direction I used theta = tan^-1 y'/x'. y' = 0.569c x' = -0.287c. Theta = 63.2 degrees.

Is this correct? Thank you!

4. Jan 6, 2010

### vitruvianman

Seems correct.
But I'd like to calculate it again to make sure :)

5. Jan 6, 2010

### vitruvianman

Seems correct.
But I'd like to calculate it again to make sure :)