# 4 Body Relativistic Collision (2D)

1. Sep 18, 2008

### neutrino2063

1. The problem statement, all variables and given/known data

See Pic.

Update: Pic works, take a look

Since the pic is not loading correctly yet, I shall try to described the situation. Ball 1 is coming in on the x-axis and impacts two balls (2 & 3) which are arranged vertically, centered at x=0 and the point at which they touch is the origin. So the incoming ball impacts the two stationary balls so that the triangle drawn between their centers is equilateral which means the impact angle is 30 degrees (or -30 for the bottom ball). The fourth ball is situated at 30 degrees above the top stationary ball so that the collision should be head on once the stationary ball has been scattered.

Incoming ball is along x-axis with speed v=.8c. Assume collision is still at 30 degrees (let's not assume Lorentz Contraction) and since the speed is "low enough" scattered balls continue you at 30 degrees. All balls have radius=1 and mass=m
(except ball 4 which has 3m). Assume ball 1 & 3 collide once.
Calculate all final velocities, momenta, and kinetic energies.

2. Relevant equations

$$\gamma=\frac{1}{\sqrt{1-\beta^2}}$$
$$\beta=\frac{\overrightarrow{v}}{c}$$
$$\overrightarrow{p}=\gamma m\overrightarrow{v}$$
$$E=\gamma mc^2$$

3. The attempt at a solution

For the first collision:
We can show that energy conservation can be simplified to (dividing by mc^2):
$$\gamma_{0}+2=\gamma_{1}+\gamma_{2}+\gamma_{3}$$

And that momentum conservation can be simplified to:

$$\overrightarrow{\beta_{0}}\gamma_{0}=\overrightarrow{\beta_{1}}\gamma_{1}+\overrightarrow{\beta_{2}}\gamma_{2}+\overrightarrow{\beta_{3}}\gamma_{3}$$

Symmetry would say that: $$\gamma_{2}=\gamma_{3}$$

Assuming ball 1 continues only along the x-axis then ball 2 & 3 will conserve momentum in the y-direction by having equal and opposite components.

If we assume that $$v_{1}$$ is only in the x direction then we know:
(Components and magnitudes written without vector arrow)

$$\gamma_{0}v_{0}=\gamma_{1}v_{1}+2\gamma_{2}v_{x_{2}}$$

Where $$v_{x_{2}}=v_{2}\cos{\frac{\pi}{6}}$$

Then the x momentum becomes can be transformed by using that relation and dividing by c to:

$$\beta_{0}\gamma_{0}=\beta_{1}\gamma_{1}+2\gamma_{2}\beta_{2}\cos{\frac{\pi}{6}}$$

Then by substituting for $$\beta=\sqrt{1-\frac{1}{\gamma^2}}$$
to get momentum in terms of $$\gamma$$ only, we can combine this with the energy conservation equation to solve for either $$\gamma_{2}$$ or $$\gamma_{1}$$ in terms of the initial $$\gamma_{0}=\frac{5}{3}$$

Doing this I found that $$\gamma_{2}=1$$, which makes absolutely no sense. The problem seems as though it should be straight-forward and my assumptions I think are pretty decent. Any ideas or critiques would be appreciated. I feel like once I find one good value for a gamma everything else in the problem should be easy.

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