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Baryon decay - maximum energy

  1. Nov 19, 2016 #1
    1. The problem statement, all variables and given/known data
    A Σ0 baryon, travelling with an energy of 2 GeV, decays electromagnetically into a Λ and a photon.
    What condition results in the Λ carrying the maximum possible energy after the decay? Sketch how the decay appears in this case, and calculate this energy. Explain your reasoning.
    [Mass of Σ0 is 1.193 GeV/c2; mass of Λ is 1.116 GeV/c2]

    2. Relevant equations
    E^2 = m^2 + p^2

    3. The attempt at a solution
    in the rest frame of the sigma baryon, E_sigma = m_sigma

    p_lambda = - p_photon
    E^2_lambda = m^2_lambda + p^2 lambda

    E_photon = modulus of p photon.

    im trying to start by solving for the energy of the lambda particle in the rest frame of the sigma baryon but im stuck. so far i have

    E_lam = E_sig - E_photon
     
  2. jcsd
  3. Nov 19, 2016 #2

    mfb

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    In the rest frame, you have energy and momentum of lambda and photon as unknowns, but you have conservation of energy, conservation of momentum and the energy/mass/momentum relation (separately for photon and lambda) - four equations for four unknowns, and you wrote down the four equations already. Plug the equations into each other until you can solve for a single variable.
     
  4. Nov 19, 2016 #3
    thanks. so my four equations are

    E_sig = m_sig

    E^2_photon = modulus p^2_photon

    E^2_lamb = m^2_lamb + p^2_lamb

    p_lamb = - p_photon

    ?
     
  5. Nov 19, 2016 #4

    mfb

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    The first one should be replaced by conservation of energy. The decay products have the same total energy as the original particle.
     
  6. Nov 20, 2016 #5
    so if E_sig^2 = E_lam^2 + p_photon^2 = E_lam^2 + p_lamb^2 then since p_lamb^2 = E^2_lamb - m^2_lamb then i can plug that into the equation to get rid of the momentum which gives me

    E^2_lambda = (E^2_sig + m^2_lamb)/2 = 1.155 Gev?

    is this the condition of maximum energy where they go in opposite directions
     
  7. Nov 20, 2016 #6

    mfb

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    This is the energy of the lambda in the sigma rest frame. In that frame the decay products always go in opposite directions. What leads to the maximal energy of the lambda in the lab frame, where the sigma is moving?
     
  8. Nov 20, 2016 #7
    in the frame where the sigma is moving could I have it so the lambda takes all the momentum and the photon is at rest i.e doesnt exist?
     
  9. Nov 20, 2016 #8

    mfb

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    No, that doesn't work, as you calculated already: the lambda has an energy of 1.155 GeV, which is not identical to 1.193 GeV.

    Not having a photon would violate energy-momentum-conservation.
     
  10. Nov 20, 2016 #9
    im still a little lost with it but in the lab frame the energy of the sigma particle is 2.0Gev so i'll need to use this i guess. could they go the same way? it hasnt yet made sense what makes the sigma energy maximum. The photon is massless so the energy of that will just be the momentum.
     
  11. Nov 20, 2016 #10

    Vanadium 50

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    Why? They didn't ask this. What do they ask?
     
  12. Nov 20, 2016 #11

    mfb

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    Think about the directions the particles can go to.

    This is similar to a classical mechanics problem with a rocket: a rocket (sigma) is flying to the right. It fires its engines (it emits a photon). In which direction should it fire to maximize its speed afterwards?
     
  13. Nov 20, 2016 #12
    to be honest there was another thread on this subject and they said to start from there so thats where i started.

    it should fire the photon to the right
     
  14. Nov 20, 2016 #13

    mfb

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    "Its speed" refers to the rocket (sigma->lambda). If you fire something to the right you slow down the rocket.
     
  15. Nov 20, 2016 #14
    oh i get you now. so the photon would fly out to the left and the lambda would continue right? but in this case they have different momenta?
     
  16. Nov 20, 2016 #15
    so to switch to the moving frame i found the lambda velocity to be 0.034c and doing the velocity equation i calculated that that means the lab frame is moving at 0.034c. im not sure if this was the right step to make
     
  17. Nov 20, 2016 #16

    Vanadium 50

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    Did they ask for this? What did they ask for?
     
  18. Nov 20, 2016 #17

    mfb

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    The velocity didn't change when you switched between frames?

    @Vanadium 50: Where is your point? Calculating the velocity in the sigma rest frame is a perfectly valid intermediate step.
     
  19. Nov 20, 2016 #18

    Vanadium 50

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    He hasn't yet answered the question of the configuration resulting in the maximum energy. Until's he's got this, he shouldn't go calculating things willy-nilly. Yes, they might be useful - but they might not.
     
  20. Nov 20, 2016 #19
    they asked for the configuration which gives the lambda maximum energy and to calculate the energy.

    no, i got the velocity using the relativistic momentum equation then tried to use u' = (u-v)/(1-(uv/c^2)) to switch to the other frame but since its the rest frame i used u=0 so it just cancelled to v
     
  21. Nov 20, 2016 #20

    mfb

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    The other frame is the lab, where the sigma is moving.

    @V 50: I think your questions are more distracting than helpful in this case.
     
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