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Calculate momentum, kinetic energy, speed etc. for a decay

  1. Oct 20, 2016 #1
    1. The problem statement, all variables and given/known data

    I have the decay

    K+ → μ+ + νμ

    Calculate:
    1. Momentum of μ+
    2. Total energy of νμ
    3. Mass of K+
    4. Speed of μ+
    5. Speed of νμ

    2. Relevant equations

    The only thing we get are the kinetic energy of μ+ so Kμ+ = 152.53 MeV
    and the mass of mμ+ = 105.658 MeV/c2

    3. The attempt at a solution

    1. As I only have the kinetic energy for the kaon and mass of the muon I thought that I could use the relation between momentum and kinetic energy: K = p2 / 2m ⇒ p = (2mK)1/2 ≈ 179.53 MeV/c ??

    Problem is that the teacher did not provide any anwsers even though these questions are just excercises.

    2. The momentum for the neutrino is pνμ2 = Eνμ2 - mνμ2 = Eνμ2 as the neutrino are massless. But the only way I know how to calculate the energy is to know the mass of the kaon. Or to use something similar as in 1. but if so I would need the kinetic energy of the neutrino.

    3. The mass of K+ should equal the sum of Eμ+ and Eνμ. But as stated above the only formula I know of how to calculate the energy in a decay like this require the mass of the kaon so yeah.

    4. I assume I can use p = γmv where γ = 1 if v << c and if the answer is too close to c I use the relativisic formula instead..?
    5. Same, but if this is the case I need to also calculate the momentum of the neutrino.

    I know I haven't come very far in my calculations, but as the teacher do not provide any answers it gets even harder. And I have been going through our course material, but there is no example like this so some tips would be great :)


    Best regards
    Isabelle
     
  2. jcsd
  3. Oct 20, 2016 #2

    mfb

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    1. You'll have to use relativistic formulas here. The nonrelativistic formulas don't work if the kinetic energy is similar to the rest energy.
    2. Neutrinos are not massless, but their mass is negligible here. Assuming the kaon was initially at rest you do not need the mass of the kaon, you can use conservation of momentum.
    4. v<<c is not a valid approximation here.
     
  4. Oct 20, 2016 #3
    Um, could you by any chance elaborate a bit on this?
     
  5. Oct 20, 2016 #4

    mfb

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    Which part is unclear?
     
  6. Oct 20, 2016 #5
    Did a misstake writing it.
     
    Last edited: Oct 20, 2016
  7. Oct 20, 2016 #6

    robphy

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    Can you draw an energy-momentum diagram representing the conservation of 4-momentum in the "collision"?
     
  8. Oct 20, 2016 #7

    mfb

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    Sure. You know the mass and the kinetic energy of the muon, and you have a formula relating energy, mass and momentum. Other particles are irrelevant for the muon momentum.
    The kaon starts at rest, and you know the muon momentum. That allows to find the neutrino momentum.

    For the muon: yes. For the neutrino there is an easier approach if you neglect its mass. What is the speed of a massless particle?
     
  9. Oct 20, 2016 #8
    Well everything if I'm going to be honest. But okey..

    1. Relativistic formulas:
    E2 = p2c2 + m2c4
    E2 - p2 = m2

    The difinition of of kinetic energy is the total energy E minus the rest energy, giving us this:

    Ek = √(p2c2+m2c4) - mc2

    Where the mass and kinetic energy are for the μ+. But can I use this formula for calculating the momenta for μ+? I don't need to consider the neutrino?

    2. So conservation of the momentum is PK+ = pμ+ + pνμ But I'm not sure how to proceed from there.

    4. and 5. So I should use p = mv/√(1-β2) where β = v/c instead?
     
  10. Oct 20, 2016 #9
    Thank you for all your help.

    Well.. a massless particle should have the speed of light of course. - . -

    You don't have any idea how to calculate the mass for K+ ?
     
  11. Oct 20, 2016 #10

    mfb

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    Right.
    Think about the total energy.
     
  12. Oct 20, 2016 #11

    robphy

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    Since your given values have such high precision,
    are you sure that the given kinetic energy Kμ+ = 152.53 MeV is correct?

    Assuming a massless neutrino and the accepted value of the muon mass (mμ+ = 105.658 MeV/c^2, which you gave)
    [and that I calculated correctly, with electronic help],
    I get a K+ mass of 493.767 MeV/c^2, which differs from the accepted value of the kaon mass( mK+ = 493.667 MeV/c^2).
    http://www.wolframalpha.com/input/?i=105.658*exp(arccosh(+152.53/105.658+1))

    If I work backwards from the accepted masses, I get that the kinetic energy should have been 154.482...
    http://www.wolframalpha.com/input/?i=105.658*(cosh(ln(493.667/105.658))-1)

    These formulas come from drawing an energy-momentum diagram of the process, and recognizing that
    with a massless neutrino, the diagram looks like the spacetime diagram of the doppler effect, and thus can be solved almost immediately by analogy. (The ratio of the rest masses is the doppler factor.)
     
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