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Reversal of velocity in center of mass.

  1. May 22, 2012 #1
    Hi. Just like in classical mechanics, I've read that in relativity, the velocity v of a particle just reverses after an elastic collision in the center of mass. I.e, if v is the velocity before the collison and v' is the velocity after, then v = -v'.

    I wondered how one goes about showing this result in relativity?
     
  2. jcsd
  3. May 22, 2012 #2

    Dale

    Staff: Mentor

    That is only true if the collision is 1D, i.e. the particle is constrained to go forward or backward only. If the collision is 2D or 3D then the speed is the same, but the direction can be any direction.

    They way that you can prove it is simply by the definition of "elastic". An elastic collision is defined as a collision in which KE is the same before and after, so the speed must be the same before and after.
     
  4. May 22, 2012 #3
    It doesn't seem obvious to me that this is a proof. Since it is the total kinetic energy which is the same before and after, speed can distribute itself among the two particles in such a way that the total kinetic energy remains the same, but that the individual speeds does not.
     
  5. May 22, 2012 #4

    Dale

    Staff: Mentor

    Even in classical collisions that isn't true in general for two particles of similar masses. It is only true for one particle colliding with a very large (immovable) object so that all of the KE is due to the one particle.

    The only other time that it is true (in both classical mechanics and in relativity) is in the center of momentum frame, perhaps that is what you mean by the phrase "in the center of mass".
     
  6. May 22, 2012 #5
    What is not true? That total kinetic energy is conserved in elastic collisions?

    I don't know if you understood me correctly. In classical mechanics, it is easy to show trough the equations

    conservation of momentum before and after + center of mass
    [tex] \vec p_1 + \vec p_2 = \vec p_1´+ \vec p_2´ = 0[/tex]

    and conservation of total kinetic energy
    [tex] \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} = \frac{p_1'^2}{2m_1} + \frac{p_2'^2}{2m_2}[/tex]

    that
    [tex] p_1 = - p_1´ \Rightarrow v_1 = - v_1´[/tex]
    is a solution to the equations. I wondered how to show the same result for relativity.
     
  7. May 22, 2012 #6

    Dale

    Staff: Mentor

    Sorry I wasn't clear. It is only true [itex]v_i = -v_f[/itex] in an elastic collision for a collision in the rest frame of a very massive object (immovable) or in the center of momentum frame for two particles.

    Yes, this is the center of momentum frame condition.

    The approach in relativity is pretty much the same. You use the center of momentum condition, the conservation of energy and momentum, and the elastic collision condition maps to a conservation of the invariant mass of the individual particles condition.
     
  8. May 22, 2012 #7
    I must be blind. I can't seem to find a way of showing this. Conservation of invariant mass for the individual particles leads to an equation of the type

    [tex]\frac{E_1 E_2}{c^2} + p_1 p_2 = \frac{E_1' E_2'}{c^2} + p'_1 p'_2[/tex]

    and I know from the center of mass condition that [tex]p_1 = p_2, p_1' = p_2'[/tex], but the equations just seem to get algebraicly horrible if i try to massage them some more.
     
  9. May 22, 2012 #8

    Dale

    Staff: Mentor

    No, you are not blind, the algebra gets messy. To make it somewhat less messy I will use units where c=1 and drop c from the equations.

    We have the relativistic definitions of energy, momentum, and mass
    [itex]\gamma = (1-v^2)^{-1/2}[/itex]
    [itex]E=\gamma m[/itex]
    [itex]p=\gamma m v[/itex]
    [itex]m^2 = E^2 - p^2[/itex]

    And the conditions:
    [itex]p_1+p_2=0[/itex] center of momentum frame
    [itex]p_1'+p_2'=0[/itex] conservation of momentum
    [itex]E_1+E_2=E_1'+E_2'[/itex] conservation of energy
    [itex]m_1=m_1'[/itex] elastic collision
    [itex]m_2=m_2'[/itex] elastic collision

    I will have to show the algebra later.
     
  10. May 22, 2012 #9

    PAllen

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    Science Advisor
    Gold Member

    Picking up from Dalespam's set up, and aiming to bypass the messiest algebra, I come up with the following:

    Noting E^2=p^2+m^2, and that p1=-p2, and p1' = -p2', and the other facts laid out, it is easy to arrive at:


    sqrt(m1^2 + p1^2) + sqrt(m2^2 + p1^2) = sqrt(m1^2 + p1'^2)
    + sqrt(m2^2 + p1'^2)

    We assume collinearity as per the OP (implicitly). Given p1, we have an equation for p1'. The function of p1' on the right is a strictly increasing function of abs(p1'). Therefore the equation can have only one solution for abs(p1'). abs(p1')=abs(p1) is obviously a solution. It is, thus, the only solution. So then we have either p1'=p1 or p1'=-p1. The former represents the objects passing through each other, leaving p1'=-p1 as required for inelastic collision.
     
    Last edited: May 22, 2012
  11. May 23, 2012 #10

    PAllen

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    Science Advisor
    Gold Member

    obviously, last line meant elastic collision, not inelastic.
     
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