# Relativistic relative velocity of particles

• scottJH
In summary, using the velocity transformation formula and the given expressions for u'_x and u'_y, the relative velocity of the particles is given by u_R = u(1-3u^2/4c^2)/(1-u^2/c^2).
scottJH

## Homework Statement

In a given inertial frame two particles are shot out simultaneously from a given point with equal speeds u at an angle of 60 degrees with respect to each other. Using the concept of 4-velocity or otherwise, show that the relative speed of the particles is given by

##u_R = u(1-3u^2/4c^2)/(1-u^2/2c^2)##

I have tried this a number of ways but always end up getting

##u_R = u(1-3u^2/4c^2)##

So I guess my question is which answer is correct?

## Homework Equations

##u'_x = (ux-v)/(1-v*ux/c^2)## and ##u'_y = uy/\gamma(1-v*ux/c^2)##

## The Attempt at a Solution

[/B]
I set S' as the stationary state of particle A, moving at a velocity u in the x direction. This meant that particle A was at rest in this frame. Therefore the velocity of B in the S' frame is the relative velocity of the two particles.

For particle B I obtained:

##u'_x = -u/2## and ##u'_y = usin60/\gamma##

Then using those values calculated the relative velocity to be

##u_R = u(1-3u^2/4c^2)##I apologise for the layout of the equations because it is my first time posting.

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You got off on the wrong foot. You need to work out the velocity of particle B in the initial frame, where the angle is valid.

Also, should the answer you're given have ##2c^2# in the denominator?

Hello, and welcome to PF!

scottJH said:

## Homework Statement

show that the relative speed of the particles is given by

##u_R = u(1-3u^2/4c^2)/(1-u^2/c^2)##
I believe there is a misprint in this expression. Should the expression in parentheses in the numerator be raised to the 1/2 power?

## Homework Equations

##u'_x = (ux-v)/(1-v*ux/c^2)## and ##u'_y = uy/\gamma(1-v*ux/c^2)##

## The Attempt at a Solution

[/B]
I set S' as the stationary state of particle A, moving at a velocity u in the x direction. This meant that particle A was at rest in this frame. Therefore the velocity of B in the S' frame is the relative velocity of the two particles.

For particle B I obtained:

##u'_x = -u/2## and ##u'_y = usin60/\gamma##

I do not get these results using your relevant equations. In particular, what happened to the denominators of ##u'_x## and ##u'_y##?

TSny said:
Hello, and welcome to PF!I believe there is a misprint in this expression. Should the expression in parentheses in the numerator be raised to the 1/2 power?

You are correct, the parentheses should be raised to the 1/2 power.

I do not get these results using your relevant equations. In particular, what happened to the denominators of ##u'_x## and ##u'_y##?

For particle B I set the velocity in the x-direction to be ##ucos60## and in the y-direction to be ##usin60##. I then just substituted those values into the equations for u'_x and u'_y. I obtained those values for u'_x and u'_y because both u_x and u_y are zero since the we are considering the rest frame of particle A.

PeroK said:
You got off on the wrong foot. You need to work out the velocity of particle B in the initial frame, where the angle is valid.

Also, should the answer you're given have ##2c^2# in the denominator?

Is the angle still not valid if we consider the frame where particle A is stationary? Surely the relative velocity of particle B in that frame it the overall relative velocity?

No. You need to use the velocity transformation formula, which is more than just ##\gamma## factor.

scottJH said:
For particle B I set the velocity in the x-direction to be ##ucos60## and in the y-direction to be ##usin60##. I then just substituted those values into the equations for u'_x and u'_y. I obtained those values for u'_x and u'_y because both u_x and u_y are zero since the we are considering the rest frame of particle A.

ux and uy are the x and y components of the velocity of B in the original unprimed frame.

Thanks for all your help. I managed to solve it.

## 1. What is the concept of relativistic relative velocity?

The concept of relativistic relative velocity is based on Einstein's theory of relativity, which states that the laws of physics are the same for all observers in uniform motion. It refers to the measurement of the velocity of one particle with respect to another particle in a moving frame of reference.

## 2. How is relativistic relative velocity different from classical relative velocity?

Unlike classical relative velocity, which is calculated using basic vector addition, relativistic relative velocity takes into account the effects of time dilation and length contraction at high speeds. It also follows the principles of special relativity, such as the constancy of the speed of light.

## 3. What is the formula for calculating relativistic relative velocity?

The formula for calculating relativistic relative velocity is v = (u + v)/(1 + (uv/c^2)), where v is the relative velocity, u is the velocity of one particle, and v is the velocity of the other particle, and c is the speed of light.

## 4. Can relativistic relative velocity be greater than the speed of light?

No, according to the principles of special relativity, the speed of light is the fastest speed that can be achieved. Therefore, relativistic relative velocity cannot be greater than the speed of light.

## 5. What are some real-life applications of relativistic relative velocity?

Relativistic relative velocity plays a crucial role in modern physics, especially in fields like particle accelerators and astrophysics. It is also used in GPS systems to accurately calculate the positions of objects in motion. Additionally, it is essential for understanding the behavior of subatomic particles and the movement of matter in the universe.

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