Relative speed of two particles

AI Thread Summary
The discussion focuses on calculating the relative speed of two particles moving at speed u in perpendicular directions (x and y). The key equation for velocity addition is provided, which is essential for determining the relative velocity. The user initially struggles with transitioning between reference frames but eventually clarifies the components of the velocities involved. The conversation highlights the importance of understanding parallel and perpendicular components in relativistic contexts. Ultimately, the user expresses gratitude for the assistance received in resolving their confusion.
w3390
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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?
 
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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"

\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!
 
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