1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Σ neutral decay

  1. Nov 25, 2014 #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
    Relativistic kinematic equations;
    CoM invariant mass

    3. The attempt at a solution
    So im not sure on the condition for maximum energy of the Λ, is it when the photon is travelling with opposite momenta?
    because the photon is a mass less particle it couldn't have 0 velocity right?
     
  2. jcsd
  3. Nov 25, 2014 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    What energy will the ##\Lambda## have in the rest frame of the ##\Sigma##?
     
  4. Nov 25, 2014 #3
    So in the rest frame of the ##\Sigma##
    The energy would be
    $$E_{\Lambda}=E_{\Sigma^o}+p_{photon}c$$
    Right?
    Because if The ##\Sigma## decays at rest then the momenta of the ##\Lambda## and photon will be equal and opposite.
     
    Last edited: Nov 25, 2014
  5. Nov 25, 2014 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    What does energy and momentum conservation tell you?
     
  6. Nov 25, 2014 #5
    If we are looking at the rest frame of the ##\Sigma## energy conservation gave me tee equation above and doesn't momentum conservation just tell me that;
    $$0=p_{photon}-p_{\Lambda}$$
    $$p_{photon}=p_{\Lambda}$$
     
    Last edited: Nov 25, 2014
  7. Nov 25, 2014 #6
    I was wondering should use that $$W^2=(\sum E)^2-(\sum p)^2$$
    for both the lab frame of the ##\Sigma## and then the center of mass frame after the decay?

    which gives
    $$(E_{\Sigma^o})^2-(p_{\Sigma^o})^2=(E_{\Lambda}+E_{photon})^2-(p_{\Lambda}-p_{photon})^2$$
     
    Last edited: Nov 25, 2014
  8. Nov 25, 2014 #7
    Im not quite sure what the condition is to give the ##\Lambda## max energy.
     
  9. Nov 25, 2014 #8

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    What does ##(E_{\Sigma^o})^2-(p_{\Sigma^o})^2## evaluate to? What is the relation between the ##\Lambda## and photon energies and momenta?
     
  10. Nov 25, 2014 #9
    So ##E_{\Sigma}^2-p_{\Sigma}^2=m_{\Sigma}^2##, as for the second question I'm not sure.
     
  11. Nov 25, 2014 #10

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    How about what you just wrote down, but for the ##\Lambda## and for the photon instead?

    Your formula for energy conservation is a bit off and you should correct it.
     
  12. Nov 25, 2014 #11
    So for the photon and ##\Lambda##
    $$(E_{\Lambda}+E_{photon})^2-(p_{\Lambda}+p_{photon})^2$$
    And the Energy conservation should be ##E_{\Sigma}=E_{\Lambda}+p_{photon}c##
    Right? And expanding the relation before you get;
    $$E_{\Lambda}^2+p_{photon}^2+2E_{\Lambda}p_{photon} - p_{\lambda}^2-p_{photon}^2 -2p_{photon}p_{\Lambda}$$?
     
    Last edited: Nov 25, 2014
  13. Nov 25, 2014 #12

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I suggest starting from just the conservation equations without squaring. The only interesting thing at the moment being: What is the energy of the ##\Lambda## in the rest frame of the decaying ##\Sigma##?
     
  14. Nov 25, 2014 #13
    So to consider the problem with the entire problem with everything in the rest frame of ##\Sigma^0##?
    In which case if the energy of the ##\Lambda## is;
    $$E_{\Lambda}=E_{\Sigma^0}-E_{photon}$$
    and The momentum would be;
    $$p_{\Sigma^0}=p_{\Lambda}+p_{photon}=0$$
    Therefore;
    $$p_{\Lambda}=-p_{photon}$$
    Which would give;
    $$E_{\Lambda}=E_{\Sigma^0}-p_{photon}=E_{\Sigma^0}+p_{\Lambda}$$
    because photon has E=pc, so;
    $$E_{\Lambda}-p_{\Lambda}=E_{\Sigma}$$
    is that correct?
    And does this mean that the condition for maximum energy is, to consider the ##\Lambda## in the rest frame with the electron travelling in the opposite direction?
     
    Last edited: Nov 25, 2014
  15. Nov 25, 2014 #14
    I just don't understand where the ##2GeV## energy of the ##\Sigma## comes into it if, we are looking at the rest frame because in that frame it has 0 momentum and ##E_{\Sigma}=mc^2##
     
  16. Nov 25, 2014 #15

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    This equation tells you that the ##\Lambda## and the photon have equal and opposite momenta.

    The signs are probably going to mess you up here. You should keep in mind that momentum is a vector, so it's really ##E_\gamma = \lvert p_\gamma \rvert## for the photon.

    You need to somehow get rid of ##E_\gamma##, ##p_\gamma##, and ##p_\Lambda## because they're all unknowns. You want to solve for ##E_\Lambda## in terms of the masses of the ##\Lambda^0## and ##\Sigma^0##.


    I think you meant photon, not electron. As you found above, conservation of momentum requires that the ##\Lambda^0## and photon go in opposite directions in the rest frame, so no, it's not the condition for maximum energy.

    As Orodruin has suggested, first solve for the energy of the ##\Lambda^0## in the rest frame of the ##\Sigma^0##. This may help give you insight into what the maximum energy condition is.
     
  17. Nov 26, 2014 #16
    Ah sorry yes I did mean photon not electron.
    So when I'm using the fact that the photon has no mass, it should be [itex]E_{\gamma}=|p_{\gamma}|[/itex] because ##E_{\gamma}^2=p_{\gamma}^2+0^2##, and as you said momentum is a vector so the energy isn't, so the energy is just the magnitude of the momentum vector.
    Thank you both so much for your help, I managed to get the solution after a lot of work.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Σ neutral decay
  1. Neutral atoms (Replies: 18)

  2. Neutral electrons. (Replies: 7)

  3. Electrically neutral (Replies: 4)

Loading...