Conservation of 4-Momentum in a 2 Particle Collision

[masudr]

Homework Statement

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)

Homework 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)

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

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)},<br /> \,\,\,\,\,\,\,<br /> \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).

 

[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.