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 [itex](a \cos(\theta), a \sin(\theta),0)[/itex] while B has 3-velocity [itex](b \cos(\phi), -b \sin(\phi),0)[/itex], where [itex]a, b, \theta, \phi[/itex] are constants.

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

[tex]\cot(\theta)\cot(\phi) = \frac{1}{2}(\gamma(V)+1)[/tex]

Homework Equations

I defined the 4-momentum as follows

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

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

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

[tex]
\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}
[/tex]

The Attempt at a Solution

The equation in the zeroth component gave me

[tex]\gamma(V) + 1 = \gamma(a) + \gamma(b)[/tex]

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

[tex]\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)}[/tex]

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

[tex]\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)}.[/tex]

Any amount of playing around with this fraction, and using the identity [itex]\gamma(u)^2(c^2-u^2)=c^2[/itex] didn't get me any closer to the answer. I was looking to equate the horrible-looking fraction above to [itex]1/2(\gamma(a)+\gamma(b))[/itex] 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:

[tex] a \gamma_a = \frac{v \gamma_v}{\sin(\theta) (cot(\theta)+\cot(\phi))} [/tex]

And a similar equation for b by switching [itex]\theta[/itex] and [itex]\phi[/itex]. Then using:

[tex]\gamma_a=\sqrt{ (a \gamma_a)^2 + 1 } [/tex]

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.