Momentum/Collision Special Relativity Problem

  • #1

Homework Statement


"In this problem, please measure energies in MeV (mega-electron volts), velocities in
units of the speed of light c and rest mass in units of MeV/c^2. Consider a collision
between two particles, each of rest mass m = 0.5MeV/c^2. In this problem I ask
you to compare two reference frames: the center of mass reference frame and the
’lab’ reference frame in which one of the two particles is not moving. Suppose that
in the center of mass reference frame the two particles are initially moving along x
with velocities v = ±0.8c. In the lab frame one particle has velocity v = 0 and the
other particle has a velocity which is directed along negative x.
(a) Suppose that the collision is elastic, and that after the collision the two
particles emerge moving along the ±y direction in the center of mass frame.
(i) Give the final momenta of the two particles, in the center of mass reference
frame
(ii) Give the final momenta of the two particles, in the lab frame.
(b) Suppose now that what collides are a particle and its antiparticle, so that
the two colliding particles annihilate and produce two photons (which have zero
rest mass and velocity c).
(i) Give the total energy of the two photons, in the center of mass frame.
(ii) Give the total energy of the two photons, in the lab frame.

I am concerned right now with part (a) part (ii), but I will give my solution to part i as reference.

Homework Equations


In the following equations, c is the speed of light in vacuum (I do not mean to be condescending, this addendum is just for clarification purposes)

momentum: p = γmv
addition of velocity: v = (v1+v2)/(1+v1v2/c^2)
energy based on momentum:E=√(pc)^2+m^2c^4)
Lorentz Transformation equations given in Textbook: t'=γ(x-vt) and x'=γ(t-vx/c^2)

The Attempt at a Solution


part i) Energy must be conserved, so by the above equaiton p°=±p and v°=±v, just in the y-direction. Thus p=±(.5)(.8c)/(1-.8^2)j=(±2/3 MeV/c)j
part ii) I tried using momentum and energy methods, but get different answers depending upon the method.

Energy methods:
velocity of particle in lab reference frame = v1+v2/(1+v1v2/c^2)=(0.8c+0.8c)/(1+(0.8c*0.8c)/c^2)=0.9756c
E°=Emoving particle + Estationary particle=√(p1c)^2+m^2c^4)+mc^2=√([(0.5 MeV/c^2)(0.97560c)/(1-0.9756^2)]^2*c^2+(0.5 MeV/c^2)^2*c^4)+(0.5 Mev/c^2)c^2=10.63
Since particles have same speed in one reference frame, the Lorentz transformation demands that they have the same speed in the second.
Therefore, E=√(p1c)^2+m^2c^4)+√(p2c)^2+m^2c^4)=2 sqrt((0.5/c^2 v c/(1-v^2))^2 c^2+0.5^2)
Setting the two equal and solving with Wolfram Alpha, we get v~=±1.04836 and v~=±95389. We reject the former answer since a particle cannot travel faster than the speed of light.

Momentum convservation methods:
velocity of particle in motion: v = .8c*2/(1+.8^2)=0.9756c
p°=p
γv°=2γv
0.9756c/(1-0.9756^2) = (2 x)c/(1-x^2)
Using Wolfram Alpha we get x~=0.95181c and x~=-1.05063c. We reject the latter answer and take into account the y-velocity by adding the two using the above equation:
v=(0.95181+.8c)/(1+.8*.95181)=0.9945c

[Remark/Question; However, since the motion is in y-direction wouldn't we need to transform the velocity in the y-direction before adding the velocities above. I am currently leaning towards the energy methods as the correct approach.]

Thus we get different velocities depending upon the method, which should not be the case. I'm relatively certain that the y-velocity is unaffected (see note above) because the center of mass is moving .8c in the -x direction the entire time and thus should not be affected by a reference fram shift.

Furthermore, I would greatly appreciate a mathematical description of how to use the Lorentz Transformation in this context, which I understand conceptually but do not have a concrete mathematical understanding.

Thank you in advance for any help/advice/guidance that you may provide!
 
Last edited:

Answers and Replies

  • #2
BruceW
Homework Helper
3,611
120
Your working is difficult to work through, since you have put in the actual numbers and written 'speed of light squared' in several places, so I haven't checked through it.

You should just use Lorentz transform of the four-momenta of the two particles. It is very simple using this method. And you are right that the y-momentum is not changed by the transform.

Have you been taught how to do Lorentz transform of four-momenta? This is the only method I know of to be able to do this problem. Maybe try google four-momenta. Its very similar to the Lorentz transform of time and space coordinates.
 

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