# Basic special relativity problem

In the laboratory frame, particle 1 moves along the x-axis with a uniform velocity
ux, and particle 2 moves along the y-axis with a uniform velocity vy. Find the
velocity of particle 2 in the rest frame of particle 1

Since we are looking at particle 2 in the laboratory frame from the rest frame of particle 1, I cant figure out whether or not I need to use the lorentz transform on the velocity of particle 2's speed in the y direction. It would make no sense to, as the frame in which particle 2 is moving has no motion in the y direction, and since particle 2 has no component of velocity in the x or -x direction, its component of velocity in the -x direction would be simply -ux

Vx(p2) = -ux
Vy(p2) = vy

So would not the velocity of particle 2 with respect to particle 1 simply be the square root of the sum of the squares of -ux and vy? This approach seems far too simplistic though.

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mfb
Mentor
You need a Lorentz transformation to go from the lab frame to the frame of particle 1.

So would not the velocity of particle 2 with respect to particle 1 simply be the square root of the sum of the squares of -ux and vy?
Simple counterexample: ux=0.9c, vy=0.9c, but the speed of particle 2 in the frame of particle 1 cannot exceed c.

I'm still trying to riddle out question for myself. I'm having trouble finding the correct way to apply the velocity transformation equations. It seems for example, that the equation involving the transformation of the x-component velocity involves the variables $$u_x u'_x v$$ and I find myself in a similar situation while looking at the equation for the y component.
The only three variables I can think of to plug in for these are the velocity of particle one from it's rest frame (0) and then the velocity of particle 2 from particle 1's rest frame ($$-u'_x$$). What do I Use for the v in this equation? they component of the particles velocity in it's own frame? That's the only variable I have left to plug in. I'm confused.

You need a Lorentz transformation to go from the lab frame to the frame of particle 1.

Simple counterexample: ux=0.9c, vy=0.9c, but the speed of particle 2 in the frame of particle 1 cannot exceed c.
So allow me to apply the transformation mathematically, and perhaps tell me where I'm going wrong?

vx' = (vx - V)/(1-(vxV/c^2))

However, since we have no value for vx because particle 2 has no velocity in the x direction with respect to the lab frame, the equation is simplifying to this.

vx' = (0-V)/(1-(0V/c^2)) -> vx' = -V, where V is simply the velocity with which the two frames move apart from each other, which would be -ux if we look at the lab frame from the rest frame of particle 1.

Where am I making my misstep?

mfb
Mentor
vx' looks right, what about vy'?

vx' looks right, what about vy'?
Ahhhh. Thank you very much! I must have misappropriated my variables the first time I ran this calculation. It all works out now.

wouldn't V be bigger than -ux because the particle is moving in the y direction as well as the x direction? isn't V the total velocity with which v2 is moving away from vx? that number is greater than ux because it's moving in the y direction as well?

wouldn't V be bigger than -ux because the particle is moving in the y direction as well as the x direction? isn't V the total velocity with which v2 is moving away from vx? that number is greater than ux because it's moving in the y direction as well?
V should equal -ux, as it is the speed at which the lab frame as a whole is moving away from particle 1, which we are defining to be the point of reference of the prime frame. If we defined particle 2 as being the reference point of the lab frame (which would not make sense, considering that the speed vy is the speed of particle 2 as observed from the lab frame), then V would be > -ux. Think of particle 2 as a moving object within a moving frame, not as the frame itself. This is probably not the best explanation possible, but its the best I can come up with right now in between electrodynamics homework problems.

vela
Staff Emeritus