# SR: 2 particle collision

1. Jun 4, 2007

### masudr

1. The problem statement, all variables and given/known data
Two particles, A and B, have the same rest mass, m. Suppose that, in O, A has 3-velocity (V,0,0) and B is at rest. The particles collide elastically at the origin and after the collision A has 3-velocity $(a \cos(\theta), a \sin(\theta),0)$ while B has 3-velocity $(b \cos(\phi), -b \sin(\phi),0)$, where $a, b, \theta, \phi$ are constants.

Define the 4-momentum of a massive particle. By using conservation of 4-momentum in the above collision, show that

$$\cot(\theta)\cot(\phi) = \frac{1}{2}(\gamma(V)+1)$$

2. Relevant equations
I defined the 4-momentum as follows

$$P^a = m_0 \frac{dx^a}{d\tau} = \gamma(u) m_0 (c, \vec{u}) = (E/c, \vec{p})$$

which defines the energy and momentum as $E= \gamma m_0 c^2 \mbox{ and }\vec{p} =\gamma m_0 \vec{u}$ respectively.

I wrote the conservation of four-momentum equation as (the masses don't appear as they cancel out)

$$\begin{bmatrix} \gamma(V) c \\ \gamma(V) V \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} c \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} \gamma(a) c \\ \gamma(a) a \cos(\theta) \\ \gamma(a) a \sin(\theta) \\ 0 \end{bmatrix} + \begin{bmatrix} \gamma(b) c \\ \gamma(b) b \cos(\phi) \\ -\gamma(b) b \sin(\phi) \\ 0 \end{bmatrix}$$

3. The attempt at a solution
The equation in the zeroth component gave me

$$\gamma(V) + 1 = \gamma(a) + \gamma(b)$$

Aha! I thought. All I now need to show is that the products of the cotangents will be twice the above sum, and I'm done.

To get the cotangent of each angle, I divided the cosine of it by the sine of it. This is fairly straightforward from the equation above. So I got

$$\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{\gamma(V) V - \gamma(b) b \cos(\phi)}{\gamma(b) b \sin(\phi)}, \,\,\,\,\,\,\, \cot(\phi) = \frac{\cos(\phi)}{\sin(\phi)} = \frac{\gamma(V) V - \gamma(a) a \cos(\theta)}{\gamma(a) a \sin(\theta)}$$

So I did what any sane man would do, and multiplied them together to get

$$\cot(\theta)\cot(\phi) = \frac{\gamma(V)^2 V^2 - \gamma(b)\gamma(V) b V \cos(\phi) - \gamma(a)\gamma(V) a V \cos(\theta) + \gamma(a)\gamma(b) a b \cos(\theta)\cos(\phi)}{ab\gamma(a)\gamma(b)\sin(\theta)\sin(\phi)}.$$

Any amount of playing around with this fraction, and using the identity $\gamma(u)^2(c^2-u^2)=c^2$ didn't get me any closer to the answer. I was looking to equate the horrible-looking fraction above to $1/2(\gamma(a)+\gamma(b))$ but have had no luck so far.

Thank you for taking the time to read the somewhat long question. Just for context, this is part of a question from a past paper (3rd year undergraduate Maths at Oxford).

Last edited: Jun 4, 2007
2. Jun 5, 2007

### StatusX

Don't try to just do everything at one. Realize that each equation allows you to eliminate one variable of your choice. Since there are five variables, and 3 of them appear in the final equation, this tells you which ones you need to get rid of: a and b.

From the last two equations you can derive:

$$a \gamma_a = \frac{v \gamma_v}{\sin(\theta) (cot(\theta)+\cot(\phi))}$$

And a similar equation for b by switching $\theta$ and $\phi$. Then using:

$$\gamma_a=\sqrt{ (a \gamma_a)^2 + 1 }$$

You can eliminate a and b in the first equation. From there it's just a lot of algebra that I'll let you work on.

Since the final equation is so simple, I'm guessing there's some shortcut, but I don't see it. Often you won't realize the shortcut until you do it the hard way. Then at least you can use it in the write up, or when a similar problem comes up again.

Last edited: Jun 5, 2007