SR: Finding the speed of particles wrt the laboratory

[rugerts]

Homework Statement

Two particles in a high-energy accelerator experiment approach each other head-on with a relative speed of 0.890c. Both particles travel at the same speed as measured in the laboratory. What is the speed of each particle, as measured in the laboratory?

Homework Equations

for

The Attempt at a Solution

I've interpreted the relative velocity between the two as 0.890c and the velocity of, say particle 2, to be -v wrt laboratory S and -v wrt S'. The answer appears to be 0.611c which is way off from what I've got.

 

[rugerts]

After another try, I've made some progress, but it still doesn't seem to be yielding the correct answer. Can anyone offer any helpful hints for my new work shown below:

 

[rugerts]

So it seems as though using the equation for v prime gives the correct result, 0.611c. Not sure why this is so. My work is shown below... (After the last step shown, I plugged into a Computer Algebra System)

 

[quinoa19]

The key here is what the "relative" speed of the two particles means in this problem. After realizing that it is the speed of one particle in the system of reference where the other one is at rest, everything else is straightforward. This system is one which moves at the same speed as one of the particles (for instance, particle 1). Then using the first equation for the motion of the second particle with respect to this system, $v_x'=-0.89c$, $u=v$, and $v_x=-v$.

 

[rugerts]

The key here is what the "relative" speed of the two particles means in this problem. After realizing that it is the speed of one particle in the system of reference where the other one is at rest, everything else is straightforward. This system is one which moves at the same speed as one of the particles (for instance, particle 1). Then using the first equation for the motion of the second particle with respect to this system, $v_x'=-0.89c$, $u=v$, and $v_x=-v$.

Fair enough. How about using the other equation where the variable solved for is not v' but instead v. Then, rearranging to get v' (this is as opposed to using the equation where v' is the variable solved for). Why doesn't it work both ways?

 

[quinoa19]

Rearranging the second equation and solving for v' will produce the first equation

 

[robphy]

Side comment [possibly too advanced for introductory physics... but really shouldn't be]

There is a geometric analogy that can be exploited in this problem.
The relative velocity $$v_{BA}=\tanh(\theta_B-\theta_A)\qquad v_{BA}=0.89,$$
where (\theta_B-\theta_A) is the relative Minkowski-angle [called rapidity] between the particles' 4-velocities.

The frame which observes both particles with the same speed in opposite directions is the center of mass frame (technically, center of momentum frame) [assuming they have equal masses]. Call this frame O. The 4-velocity of this frame is along the Minkowski-angle bisector. In that frame, A and B have speed
$$v_{BO}=\tanh(\theta_B-\theta_O)=\tanh\left(\frac{\theta_B-\theta_A}{2}\right)=\tanh\left({\rm arctanh}(0.89)\right)$$
https://www.wolframalpha.com/input/?i=tanh(atanh(0.890)/2) gives 0.61128

(Your first equation, the relative-velocity formula, is actually an identity for the hyperbolic-tangent of a difference of angles:
\begin{align*}
\tanh(\theta_B-\theta_A)&=\frac{\tanh\theta_B-\tanh\theta_A}{1-\tanh\theta_B\tanh\theta_A}\\
\tanh((\theta_B-\theta_O)-(\theta_A-\theta_O))&=\frac{\tanh(\theta_B-\theta_O)-\tanh(\theta_A-\theta_O)}{1-\tanh(\theta_B-\theta_O)\tanh(\theta_A-\theta_O)}\\
\end{align*}
.)

For fancier approaches: see my Insight https://www.physicsforums.com/insights/relativity-variables-velocity-doppler-bondi-k-rapidity/