Relative speed of two particles

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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 = [tex]\gamma[/tex](vt' + x')

y = [tex]\gamma[/tex](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?
 

Answers and Replies

  • #2
SammyS
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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. [tex]\vec{v}_2=u\hat{j}[/tex]

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

[tex]\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}}\ .[/tex]  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, [tex]\mathbf{v}=-u\hat{i}[/tex] and [tex]\mathbf{u}=u\hat{j}\,.[/tex]
..
 
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  • #3
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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?
 
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Nevermind, I got it. Thanks SammyS.
 
  • #5
SammyS
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Nevermind, I got it. Thanks SammyS.
GREAT!

Thanks for the feedback!
 

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