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Finding the total energy of a Pi meson

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

    A K meson (an elementary particle with approximately 500 Mev rest mass) traveling through the laboratory breaks up into two Pi mesons (elementary particles with 140 Mev rest energies). One of the Pi mesons is left at rest. What is the total energy of the remaining Pi meson?

    2. Relevant equations

    E=mc^2
    KE=(γ-1)mc^2

    3. The attempt at a solution

    My first step was trying to use energy conservation E(initial)=E(final)
    The K meson is traveling so the total E at the start is E(rest mass of K)+KE(K meson)
    E(initial)=500 Mev+(γ-1)mc^2

    In the end, the K meson breaks into 2 Pi mesons, each with 140 Mev rest mass. One stops and the other keeps on moving.
    So the total E at the end is E(rest mass Pi)+E(rest mass 2nd Pi)+(γ-1)mc^2
    E(final)=140 Mev+140Mev+(γ-1)mc^2

    E(initial)=E(final)
    500 Mev+(γ-1)mc^2=140 Mev+140Mev+(γ-1)mc^2

    But now I don't know what to do next. I don't have a value for gamma.
     
  2. jcsd
  3. Dec 6, 2017 #2

    PeroK

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    What about momentum?
     
  4. Dec 6, 2017 #3
    Then I would use p=γmv but I don't think I have the value for those. I just know that momentum would also be conserved.
     
  5. Dec 6, 2017 #4

    PeroK

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    In general it's messy to use ##\gamma## in this sort of problem. Can you think of a way to tackle the problem without it?
     
  6. Dec 6, 2017 #5
    P=E/C ?
     
  7. Dec 6, 2017 #6

    PeroK

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    That's for a massless particle. Do you know:

    ##E^2 = p^2c^2 + m^2c^4##?
     
  8. Dec 6, 2017 #7
    So am I using this equation to solve for momentum?
     
  9. Dec 6, 2017 #8

    PeroK

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    The general approach to these particle problems is:

    1) Conservation of energy
    2) Conservation of momentum.
    3) For each particle: ##E^2 = p^2c^2 + m^2c^4##

    Hint: stick with ##M## for the initial K meson mass and ##m## for the pi meson mass and leave the numbers out until the end.

    The notation is up to you, but I would use ##E_0, p_0## for the Energy & Momentum of the initial K meson; ##E_1, p_1 (p_1= 0)## for the pi meson that ends up at rest and ##E_2, p_2## for the second pi meson.

    You're trying to solve for ##E_2## of course.
     
  10. Dec 6, 2017 #9
    Using the notation, I have the conservation of energy written as E0=E1+E2
    Then using the energy equation for each particle

    For the K meson
    E02=P02C2+M2C4

    For the Pi stopped
    E12=m2C4 since P1=0

    For Pi moving
    E22=P22C2+m2C4

    Momentum conservation
    P0=P1+P2
    Since P1=0
    P0=P2

    Do I just plug all this into E2=E0-E1?
     
  11. Dec 6, 2017 #10

    PeroK

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    Why not? You may want to square that equation first.
     
  12. Dec 6, 2017 #11
    You mean instead I should use E22=E02-E12?
     
  13. Dec 6, 2017 #12

    PeroK

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    ##(E_0 - E_1)^2 \ne E_0^2 - E_1^2##
     
  14. Dec 6, 2017 #13
    My mistake. Then squaring both sides should be E22=(E0-E1)2
     
  15. Dec 6, 2017 #14

    PeroK

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    ##E_2^2 = (E_0 - E_1)^2 = \dots##?
     
  16. Dec 6, 2017 #15
    (E0-E1)2=E02-2E0E1+E12
     
  17. Dec 6, 2017 #16

    PeroK

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    Okay, but you may need to do more than one line of algebra at a time. I'm going off-line now, but maybe someone else can provide further help if you need it.
     
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