Relativistic addition of orthogonal velocities

[Pi-Bond]

Homework Statement

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: $v{(2-\frac{v^{2}}{c^{2}})}^{1/2}$

Homework Equations

Velocity Transformation
$v_{total}=\sqrt{v_{x}^{2}+v_{y}^{2}}$

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:

$v'_{y}=\large \frac{(v) \sqrt{1-\frac{v^{2}} {c^{2}}} } {1- \frac{(v)(v)}{c^{2}}} = \frac{v}{\sqrt{1-\frac{v^{2}} {c^{2}}}}$

The relative velocity would then be-

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

Which would yield

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

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

 

[Dick]

I think the (v)(v)/c^2 in your formula for v'_y should be the dot product of the two original velocities. That's zero since they are orthogonal vectors.

 

[Pi-Bond]

Is that so? How is the expression with the dot product derived then? I just assumed that the velocity in frame S' is given by $\frac{y'}{t'}$. And from the Lorentz transforms:

$y'=y$
$t'=\gamma (t - \frac {vx}{c^{2}})$

where v is the velocity of frame S' with respect to S. Where does the dot product come in?

 

[vela]

The dot product arises because only the component of velocity in the direction of the boost is contracted. I'm sure you could find a general derivation somewhere.

What I would do, though, is transform the coordinates of the second particle using the Lorentz transformations and express x' and y' as a function of t' (not t). From that expression, you can see what the components of the velocity are to an observer at rest in the moving frame.

 

[Pi-Bond]

Ok, I got the answer following vela's suggestion. In the frame S, the coordinates of the particle going up in the +y direction are given by:

$x=0$
$y=vt$

Transforming these to the frame S' gives the coordinates:

$x'= - \gamma vt$
$y'=vt$

The inverse Lorentz transform from S' to S gives:

$t= \gamma (t'+\frac{vx}{c^{2}})$

Which on substituting x yields:

$t=t' \sqrt{1-\frac{v^{2}} {c^{2}} }$

Substituting this in the original equations for x' and y', and differentiating with respect to t' yields the velocity components of the particle moving in the +y direction in S':

$v'_{x}=-v$ (as expected)
$v'_{y}=v\sqrt{1-\frac{v^{2}} {c^{2}} }$

And these give the required result. I assume you suggested to put x' and y' in terms of t' so that the reading would be at the same time in S', right?

Also, I don't see why the dot product is required in the y-velocity transformation. I assumed that the "v" there was the velocity in the y direction, not any arbitrary direction...none of my books shed light on this.

 

[vela]

I assume you suggested to put x' and y' in terms of t' so that the reading would be at the same time in S', right?

Not really. I suggested it because $v'_x = dx'/dt'$, not $dx'/dt$.

 

[Pi-Bond]

Ok - Thanks for the help!