4 momentum mass non-conservation, special relativity

[Konhbri]

Homework Statement

A particle with mass M and speed v along the positive x-axis hits a stationary mass m. Two particles, each with mass µ, emerge from the collision, at angles with respect to the x-axis.

(a) Write the equation for conservation of the 4-momenta, for arbitrary angles θ_1, θ_2 of exit.

(b) Argue that the angles of exit of the ‘after’ particles must be symmetric about the x-axis;
call the angle θ.

(c) Obtain three independent equations for the velocity u, mass µ, and angle θ of the new particles, in terms of the initial quantities.

(d) Check the non-relativistic limit; for example, expand your expressions in the small parameter v/c. Note the constraint that in this case there is very little energy available for the sum of masses after to be greater than the sum of masses before, while in general it is possible for mass to be created.

Homework Equations

(a)
(Gamma for velocity v is denoted as 'G', gamma for velocity u is denoted as 'g')
GMc(1,v/c,0,0)+mc(1,0,0,0)=
gµc(1,(u/c)*cos(θ_1),(u/c)*sin(θ_1),0)+gµc(1,(u/c)cos(θ_2),(u/c)sin(θ_2),0)

therefore

(GMc+mc, GMc*(v/c),0,0)=2(gµc, gµc*(u/c)*cos(θ),0,0)$ (This is shown in part B)

The Attempt at a Solution

(b) The universe is isotropic and the particles are identical therefore if the particles switch places the situation should be identical, therefore their mass and velocity are the same and the angles are opposite in direction

(c) mass: GMc+mc=2gµc
velocity: GMc*(v/c)= 2gµc*cos(θ)
angle:?? I can't figure this third independent equation out

I haven't attempted part (d) as I have yet to complete (c)

 

[mfb]

(b) The universe is isotropic and the particles are identical therefore if the particles switch places the situation should be identical, therefore their mass and velocity are the same and the angles are opposite in direction

Opposite in direction relative to what?
Your argument doesn't specify a reference frame but the statement is not true in every reference frame.

For (c) you can consider the momentum in two separate dimensions.

mass: GMc+mc=2gµc

That is the energy in the system, not the mass.

 

[PeroK]

(a)
(Gamma for velocity v is denoted as 'G', gamma for velocity u is denoted as 'g')
GMc(1,v/c,0,0)+mc(1,0,0,0)=
gµc(1,(u/c)*cos(θ_1),(u/c)*sin(θ_1),0)+gµc(1,(u/c)cos(θ_2),(u/c)sin(θ_2),0)

therefore

(GMc+mc, GMc*(v/c),0,0)=2(gµc, gµc*(u/c)*cos(θ),0,0)$ (This is shown in part B)

Just to add that you have assumed in a) that the two particles have the same speed. This is equivalent to assuming they have the same angle.(Simply consider the y-momentum.)

I suggest you need some justification that the two particles have the same speed.