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Nuclear physics reaction, minimal energy to make it work

  1. Jan 3, 2015 #1
    1. Consider the reaction that happens when a photon collides with an atomic electron in rest ##\gamma+e^{-} <---> K^{-} + v_{e}##. The masses of all particles can be considered as known. What is the minimal energy that the photon needs to have to make this reaction work?


    2. Conservation of four momentum


    3. Frankly, I am quite lost. At first glance I'd say that ##E_{\gamma}+m_{e}=E_{K}+E_{v}## and then say that the minimal energy required is to create the K and neutrino in rest, and so substituing the masses for the energy in the last expression. This won't work with conservation of momentum though. Hoewever once I consider that it gets harder because of the allowed angles after collisioN.
     
  2. jcsd
  3. Jan 3, 2015 #2

    haruspex

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    There's only one angle to worry about, no? Having set an unknown for the angle the departing neutrino makes to the incoming photon's path, the angle for the K follows by conservation of momentum in the direction orthogonal to the photon's path. Then it's a matter of minimising the energy with respect to the one angle.
     
  4. Jan 3, 2015 #3
    Do you mind elaborating on that a bit? You seem to have seen this easily without calculations so I'm probably overlooking something.
     
  5. Jan 3, 2015 #4

    haruspex

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    Post your momentum equations, then we'll have a reference point for discussion.
     
  6. Jan 3, 2015 #5
    Here you go: http://imgur.com/CpEDxzi

    I just wrote down what I knew, but I wasn't sure where to go from there because there seem to be too much unknown.
     
  7. Jan 3, 2015 #6

    haruspex

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    You've written ##E_\gamma## for both the energy and the momentum of the photon. Is there some convention I don't know, or is that a mistake?
    Your last equation gives a relationship between the two angles. You can take one angle as the independent variable and the other to be a function of it. Differentiate wrt the independent angle to find the minimum ##E_\gamma##.
     
  8. Jan 3, 2015 #7
    In natural units, since the mass of a photon is zero, the momentum and energy are equal. Thanks for the tip I'll try it out and reply with results. In a previous post you mentioned something being perpendicular, will this lead to it or is it a different approacH?
     
  9. Jan 3, 2015 #8

    haruspex

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    OK, I thought it might be something like that. can't help feeling it's a bit confusing though. Dimensional consistency can be a useful check on the algebra.
    Do you mean where I wrote 'orthogonal'? You already did that bit in your eqn 4.
     
  10. Jan 4, 2015 #9
    Tried playing with it for a bit without much luck. It sucks because basically it's an algebra momentum-energy problem which I was supposed to have learned to solve in classical mechanics intro courses. I guess it's because we always worked with given values, not used to having 4 variables in these problems.
     
  11. Jan 4, 2015 #10

    PeroK

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    Can you find a simple argument for both angles being 0?

    Hint: if you can't see it in the original frame, try thinking about the collision in the CM frame.

    Remember you are trying to minimise the energy.
     
  12. Jan 4, 2015 #11
    Oh right! In the CM frame, they can be created in rest without problems, this means an opening angle of 0 in the lab frame! Right?
     
  13. Jan 4, 2015 #12

    PeroK

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    Yes, absolutely.
     
  14. Jan 4, 2015 #13
    Always 0. This means that as long the momentum vectors of the particles are the same and in opposite directions the momentum is conserved. The only worry now is the total energy after the collision which depends only on the magnitude of those vectors. Well if I pick both magnitudes as 0 , I still have total momentum 0 - check , and minimal incoming energy of both particles.
     
  15. Jan 4, 2015 #14
    Thanks a lot for your help, so now I have the expression for the photon energy in function of the known masses in CM frame. Do you now recommend finding an expression for ##\beta## of the CM relative to the lab frame and then transforming the photon energy into that frame? Or just work in in lab frame and assume angle 0?
     
  16. Jan 4, 2015 #15

    PeroK

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    You can do it either way, but if you know how to do energy-momentum transformations, try using the CM frame.
     
  17. Jan 4, 2015 #16
    My attempt here: http://imgur.com/u2izMjJ.

    In the right lower corner it's a derivation for ##\beta_{CM}## in function of the photon energy in the lab frame. The expression will be very ugly, and then that expression will have to be plugged in the other one in the left lower corner. I doubt an elimination of ##E_{\gamma}## will be possible.
     
  18. Jan 4, 2015 #17

    PeroK

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    You're missing a trick. Think about energy-momentum transformations for a system of particles.
     
  19. Jan 4, 2015 #18
    Alright since you mentioned more particles I tried using the fact that the electrons energy transforms as well, this gives me a much better expression:

    http://imgur.com/GvM5u8p

    Oh nvm I made a mistake I'll post an edit in a sec.

    EDIT: correct version - http://imgur.com/kyVNJjk
     
    Last edited: Jan 4, 2015
  20. Jan 4, 2015 #19

    PeroK

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    The trick you're still missing is that energy-momentum transforms for a system, as it does for individual particles:

    ##E_1^2 - P_1^2c^2 = E_2^2 - P_2^2c^2##

    Where the E's and P's are the total energy and momentum of a system of particles in two frames.
     
  21. Jan 4, 2015 #20
    Oh! I never heard of this fact before I'm still in an intro course to relativity. Yeah now it is indeed a quick calculation. Is my final link techincally correct as well?
     
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