Special relativity/velocity transformations

In summary, we are asked to compute the magnitude and direction of a cosmic-ray proton's velocity when viewed from a frame S' moving with a velocity of 0.72c in the +x direction. To do this, we use the velocity transformations to calculate the x and y components of the proton's velocity in the S' frame. Then, using the Pythagorean theorem, we find the magnitude of the velocity to be 0.637c and the direction to be 63.2 degrees.
  • #1
QuantumParadx
3
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.
 
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  • #2
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
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!
 
  • #4
Seems correct.
But I'd like to calculate it again to make sure :)
 
  • #5
Seems correct.
But I'd like to calculate it again to make sure :)
 

Related to Special relativity/velocity transformations

1. What is special relativity?

Special relativity is a theory developed by Albert Einstein in 1905 that explains the relationship between space and time. It states that the laws of physics are the same for all observers in uniform motion, and the speed of light is constant in all inertial frames of reference.

2. How does special relativity affect our understanding of time?

Special relativity introduces the concept of time dilation, which means that time passes at different rates depending on the relative speeds of the observers. This means that time is not absolute and can be perceived differently by different observers.

3. What is the formula for velocity transformations in special relativity?

The formula for velocity transformations in special relativity is v' = (v + u) / (1 + (vu/c^2)), where v' is the relative velocity between two objects in different inertial frames of reference, v is the velocity of one object, u is the velocity of the other object, and c is the speed of light.

4. How does special relativity explain the twin paradox?

The twin paradox is a thought experiment that demonstrates the effects of time dilation. It involves one twin traveling at high speeds and the other staying on Earth. When the traveling twin returns, they have aged less than the twin who stayed on Earth due to the difference in their relative velocities.

5. What are some real-world applications of special relativity?

Special relativity is essential for understanding and predicting the behavior of particles moving at high speeds, such as in nuclear reactors and particle accelerators. It also plays a crucial role in the development of technologies like GPS, which relies on the precise timing of signals between satellites and receivers on Earth.

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