# Relative speed of two particles

• w3390
In summary, the relative speed of one particle to another, where both are shot in different directions at the same speed, is u(2-(u/c)^2)^1/2. This can be found using the velocity addition formula, where u is the velocity in the rest frame, v is the velocity of a frame relative to the rest frame, u(perp) is the component of u perpendicular to v, and u(para) is the component of u parallel to v. In this scenario, v is -u and u is u.

## Homework Statement

One particle is shot in the x direction at speed u and a second is shot in the y direction at speed u as well. Show that the relative speed of one to the other is: u(2-(u/c)^2)^1/2.

## Homework Equations

velocity addition: u = (u' +/- v)/(1 +/- u'*v)

Lorentz Trans.:

x = $$\gamma$$(vt' + x')

y = $$\gamma$$(vt' + y')

## The Attempt at a Solution

So I am getting lost when trying to go from one frame to another. Starting in the rest frame of the particle moving in the x direction (particle 1):

the speed of particle 1 is v'_x = 0 and v'_y = 0

the speed of particle 2 is v'_x = -u and v'_y = u

I do not know what to do from here to try to find the relative velocity of the two particles. Any help?

In the frame of particle 1, the velocity of the launch location is: (vL')x=-u and (vL')y=0

In the frame of the launch location, the velocity of particle 2 has magnitude u, and is in the y direction. $$\vec{v}_2=u\hat{j}$$

So you have a reference frame (the launch position) moving at a velocity of $$-u\hat{i}$$ relative to particle 1 and you have particle 2 moving with a velocity of $$u\hat{j}$$ relative to the reference frame.

The equation, u = (u' ± v)/(1 ± u'*v/c2) gives the velocity u in a rest frame for an object moving at velocity u' relative to a frame moving at v in the rest frame. The velocities here are all parallel to each other. The variables used in this formula may conflict a bit with the variables in the problem.

A more general formula can be found at http://en.wikipedia.org/wiki/Velocity-addition_formula" [Broken]

$$\mathbf{w}=\frac{\mathbf{v}+\mathbf{u}_{\parallel} + \alpha_{\mathbf{v}}\mathbf{u}_{\perp}}{1+(\mathbf{v}\cdot\mathbf{u})/{c^2}},\ \text{ where }\ \alpha_{\mathbf{v}}=\sqrt{1-\frac{\mathbf{v}^2}{c^2}}\ .$$  w is the velocity of the object rel. to the rest frame, v is the velocity of a frame rel. to the rest frame, u is the velocity of the object rel. to the moving frame, u is the component of u perpendicular to v and u is the component of u parallel to v .

So, in your case, $$\mathbf{v}=-u\hat{i}$$ and $$\mathbf{u}=u\hat{j}\,.$$
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Okay, so v dot u in the denominator will be zero, leaving just the 1. However, I'm still confused about u(perp) and u(para). Wouldn't u(perp) just be u and u(para) just be zero?

Nevermind, I got it. Thanks SammyS.

w3390 said:
Nevermind, I got it. Thanks SammyS.
GREAT!

Thanks for the feedback!