(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

In a given inertial frame, two particles are shout out simultaneously from a given point, with equal speeds v, in orthogonal directions. What is the speed of each particle relative to the other?

Answer: [itex]v{(2-\frac{v^{2}}{c^{2}})}^{1/2}[/itex]

2. Relevant equations

Velocity Transformation

[itex]v_{total}=\sqrt{v_{x}^{2}+v_{y}^{2}}[/itex]

3. The attempt at a solution

Let the given inertial frame be denoted by S, and let the particles be emitted in the +x and +y directions. Let S' be the rest frame of the particle emitted in the +x direction. These two frames will coincide at t=0 (time of emission), hence the Lorentz transforms are applicable.

In S', the x component of the velocity of the particle going in the +y direction would -v, as that particle has no x component of velocity. The y component would be given by the velocity transformation:

[itex]v'_{y}=\large \frac{(v) \sqrt{1-\frac{v^{2}} {c^{2}}} } {1- \frac{(v)(v)}{c^{2}}} = \frac{v}{\sqrt{1-\frac{v^{2}} {c^{2}}}}[/itex]

The relative velocity would then be-

[itex]v_{rel} = \sqrt{v'_{x}^{2}+v'_{y}^{2}} =\large \sqrt{v^{2}+\frac{v^{2}} {1- \frac{ v^{2} }{ c^{2} } } }[/itex]

Which would yield

[itex]v_{rel}= v{(2-\frac{v^{2}}{c^{2}})}^{1/2} \large \frac{1}{\sqrt{1- \frac{ v^{2} }{ c^{2} }}}[/itex]

So I have an extra factor in my answer. What am I doing wrong?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Relativistic addition of orthogonal velocities

**Physics Forums | Science Articles, Homework Help, Discussion**