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Homework Help: Relativistic Elastic Collision

  1. Sep 29, 2005 #1
    I've done the bulk of this problem (part A) but I still can't figure the last bit (part B):

    "Consider an elastic head-on collision between a projectile with mass m_a, velocity (vector)v_a, energy E_a, and three-momentum (vector)p_a and a stationary target of mass m_b. (That the collision is head-on measn that the two particles emerge from the collision both moving along the line of the incident velocity (vector)v_a.)

    a) What is the final velocity (vector)v_b of the target particle b?"

    To solve this problem I first determined the initial 4-momenta of both a in b the laboratory frame:
    P_ai = (P_t, P_x, P_y, P_z) = (E_a/c, p_a, 0, 0)
    P_bi = (m_b*c, 0, 0, 0)
    Where c is the speed of light.

    Then I did a Lorentz tranformation into a frame where the total 3-momenta is 0 (keeping in mind that I don't know beta or gamma yet because I don't know the speed of this frame relative to the lab frame):
    P'_ai = (y*E_a/c - y*B*p_a, y*p_a - B*y*E_a/c, 0, 0)
    P'_bi = (y*m_b*c, -y*B*m_b*c, 0, 0)
    Here, y is gamma and B is beta.

    Then to find the relative speed of the frames, I add the spatial components of the 4-momenta (i.e. the 3-momenta) and I set them to 0:
    P_total = 0 = y(-B(E_a/c + m_a*c) + p_a)
    B = v_cm/c = p_a*c / (E_a + m_b*c)
    Here, v_cm is the relative velocity of the center-of-momentum frame relative to the lab frame.

    Now I try to find the final 4-momentum of mass m_b. Since we know the collision is elastic, |P'_bi| = |P'_bf|, and
    P'_bf = (y*m_b*c, y*B*m_b*c, 0, 0)
    The 3-momentum simply has the opposite sign.

    Then I transformed P'_bf back into the lab frame by inverse Lorentz transfromation, which gave
    P_bf = (y^2*m_b*c + y^2*B^2*m_b*c, 2*y^2*B*m_b*c, 0, 0) = (E_b/c, p_b, 0, 0)

    Utilizing the relationship, (vector)v_b = p_b*c^2/E_b = 2*y^2*B*m_b*c^3/(y^2*m_b*c^2 + y^2*B^2*m_b*c^2) = 2*y^2*B*m_b*c^3/[y^2*m_b*c^2(1 + B^2)],
    (vector)v_b final = 2*B*c/(1 + B^2), B = v_cm/c = p_a*c / (E_a + m_b*c) from before.

    So that's part a, now comes what seems like the easy part:

    "b) Describe what happens if the 2 masses are equal."

    I suspect that m_a becomes rested and m_b picks up all it's speed, but I'm having a hard time putting it into numbers.

    I tried seeing what happened if I took m_a = m_b into the equation for B, and got y*v/((y + 1)c) but when I try to plug that into (vector)v_b, I can't simplify it and gain no further insight.
  2. jcsd
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