Relative speed of two particles

[w3390]

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?

 

[SammyS]

In the frame of particle 1, the velocity of the launch location is: (v_L')_x=-u and (v_L')_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"

\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}\,.
..

 

[w3390]

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?

 

[w3390]

Nevermind, I got it. Thanks SammyS.

 

[SammyS]

Nevermind, I got it. Thanks SammyS.

GREAT!

Thanks for the feedback!