Special relativity/velocity transformations

  • Thread starter Thread starter QuantumParadx
  • Start date Start date
  • Tags Tags
    Transformations
Click For Summary

Homework Help Overview

The discussion revolves around a problem in special relativity involving the velocity transformation of a cosmic-ray proton moving at a significant fraction of the speed of light. The original poster seeks to compute the magnitude and direction of the proton's velocity as observed from a different inertial frame moving at a specified velocity.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to break down the problem into x and y components of velocity and questions how to apply the velocity transformation equations correctly. Some participants suggest calculating the components independently and question the clarity of the original poster's request.

Discussion Status

Participants have engaged in verifying calculations and discussing the application of the velocity transformation equations. There is a recognition of the need to ensure accuracy in the calculations, but no explicit consensus has been reached regarding the correctness of the final results.

Contextual Notes

The problem involves specific angles and velocities, and participants are navigating the complexities of transforming velocities between frames in special relativity. The original poster has expressed some uncertainty about the calculations and the approach to take, particularly regarding the introduction of the y-component.

QuantumParadx
Messages
3
Reaction score
0
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/[tex]\gamma(1-vUx/c^2)[/tex]


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.
 
Physics news on Phys.org
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?
 
xboy said:
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?

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

Similar threads

Replies
10
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
2K
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K