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Relativistic elastic collision

  1. Jan 6, 2017 #1
    1. The problem statement, all variables and given/known data

    VCBC0.png
    2. Relevant equations
    et Em and pm be the energy and momentum of the mass m after the collision. Let p and p' be the momentum of mass M before and after the collision.

    From conservation of 4 momentum:
    [tex] \begin{bmatrix}E+m \\ p\end{bmatrix}=\begin{bmatrix}E_m+E' \\ p'+p_m\end{bmatrix}[/tex]

    We also have our invariants
    E2-p2=M2, etc.

    3. The attempt at a solution
    Squaring the 4-vectors we get: [tex]E^2+2Em+m^2-p^2=E'^2+2E'E_m+E_m^2-p'^2-2p_mp'-p_m^2[/tex]

    Using our invarients, this becomes:
    [tex]M^2+m^2 + 2Em=m^2+m^2+2(E'E_m-p'p_m)
    \implies Em=E'E_m-p'p_m
    [/tex]

    now, using the conservation of energy and momentum:
    [tex]E_m=E+m-E'[/tex] & [tex]p_m=p-p'[/tex]
    Substituting these in and using our invarient again gets us:
    [tex]Em=E'(E+m)-p'p-M^2[/tex]

    I have tried setting [tex]p=\sqrt{E^2-M^2}[/tex]however its gets so algebraically heavy. Is there an easier way??
     
  2. jcsd
  3. Jan 6, 2017 #2

    mfb

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    2016 Award

    Staff: Mentor

    You can make a stronger statement. Energy and momentum are both conserved separately. That way you can avoid p' with its ugly influence on the equations.
     
  4. Jan 6, 2017 #3
    What would I replace it with? p-pm is a bit redundant
     
  5. Jan 6, 2017 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    p-pm is fine. Express pm in terms of Em, then replace it by E' and known parameters. Squaring it at the right place will get rid of any square roots, and then you'll probably have to solve a quadratic equation.
     
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