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Simplifying KE/Momentum Relativistic Problem

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data

    A D0 meson (with a rest mass of 1.86 GeV/c2), initially at rest, decays into a K0 meson (with a rest mass of .51 GeV/c2) and a [itex]\pi[/itex]0 meson (with a rest mass of .12 GeV/c2). What is the Kinetic Energy of the [itex]\pi[/itex]0 meson?


    2. Relevant equations

    E=[itex]\gamma[/itex]m0c2

    p=[itex]\gamma[/itex]m0v

    3. The attempt at a solution

    Subtracting final rest energy from intial energy its easy to see that the KE(K0+[itex]\pi[/itex]0)=1.23 GeV

    Also: [itex]\gamma[/itex]K0mK0c2+[itex]\gamma[/itex][itex]\pi[/itex]0m[itex]\pi[/itex]0c2 = 1.86 GeV/c^2 (The total energy of the two Mesons must equal the first meson)
    Without going through all the horrible steps me and my friend did, eventually we got a equation that solved: VK0=f(V[itex]\pi[/itex]0) (I don't have the final equation we got on me)


    When then plugged that back into: [itex]\gamma[/itex]K0mK0c2+[itex]\gamma[/itex][itex]\pi[/itex]0m[itex]\pi[/itex]0c2 = 1.86 GeV/c^2


    Now at this point we had a solvable equation, the only variable was V[itex]\pi[/itex]0, but it was a hellish equation. After about 20 minutes working on trying to solve it we ran out of time and had to give up- the algebra was just to hard.
    Had we had enough time, I'm completely confident we could have solved it, as it was simply a matter of foiling again and again (I think we needed to do it three levels down).
    My question is: Is there a way of solving this problem without having to go in and solve for one of the variables in terms of the other, or doing that in a simpler equation?

    Thank you, especially if you made it through this.
     
  2. jcsd
  3. Nov 20, 2011 #2

    jtbell

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    Staff: Mentor

    If you don't need the velocities (i.e. you're not asked for them explicitly), don't take that route to find the energies of the outgoing particles. Instead, use

    [tex]E^2 = (pc)^2 + (m_0 c^2)^2[/tex]

    which is true for each particle, together with conservation of energy and conservation of momentum. You never have to see a single "v".
     
  4. Nov 20, 2011 #3

    vela

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    Staff Emeritus
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    Two suggestions:

    1. Stick with energy (total, not kinetic) and momentum.
    2. Take advantage of [itex]E^2 - (pc)^2 = (mc^2)^2[/itex].

    If you learn how to use four-vectors, you can get the answer in about three or four lines of simple algebra. Read Griffith's chapter on special relativity in his Introduction to Particle Physics.
     
  5. Nov 20, 2011 #4
    Thank you to both of you, while I haven't actually gotten the answer yet, I have gotten numerical values for both the momentum and velocity of one of the particles, I'll be able to get the answer with about 5 more minutes of work.

    In addition, I know the momentum energy four vector, but I've never been given a problem requiring it so using it didn't occur to me, I am going to look at its usefulness more carefully.
     
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