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SR: 2 particle collision

  1. Jun 4, 2007 #1
    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 [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]

    2. Relevant 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)

    \gamma(V) c \\
    \gamma(V) V \\
    0 \\
    0 \end{bmatrix}


    c \\
    0 \\
    0 \\
    0 \end{bmatrix}


    \gamma(a) c \\
    \gamma(a) a \cos(\theta) \\
    \gamma(a) a \sin(\theta) \\
    0 \end{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

    [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).
    Last edited: Jun 4, 2007
  2. jcsd
  3. Jun 5, 2007 #2


    User Avatar
    Homework Helper

    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.
    Last edited: Jun 5, 2007
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