Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What effects are bounding a missing mass plot? (with figure)

  1. Feb 17, 2016 #1

    so for example in the figure below the missing mass squared has been calculated under the assumption of the detection of a pi+ particle decaying from a K+. What I am wondering is what is it that's bounding all the decays at 0.122 GeV2?

    And for the decays at all, what is e.g. bounding the pi+pi+pi- (black line) decay? I figured the lowest value, around 0.078, is around the squared mass of two pions? Not sure though.

    Appreciate any help.

    Last edited: Feb 17, 2016
  2. jcsd
  3. Feb 17, 2016 #2
    O nevermind I figured it out :) I was completely disregarding how I calculated the missing mass. They are of course bounded by 0.122 GeV2=(m(K+)-m(pion))^2, or simply the mass that's missing compared to the assumed mass.
    Last edited: Feb 17, 2016
  4. Feb 18, 2016 #3
    So I have been thinking a little more about the problem I had and I'd like to share it for future reference.

    The missing mass squared we define using the 4-momenta P as (Pπ-PK)2 where we investigate it under a so called π+ hypothesis, meaning we assume the mass of the particle we detected is that of m(π+)~0.139 GeV/c2.

    1. Data is collected from a measurement system which can output the momentum of the particle.
    2. We construct the 4-momenta of the initial K+, with an assumed constant absolute momentum of 75 GeV/c and mass of ~0.494 GeV/c2 (this can be refined if you can properly match the decay registered in the detector we use to measure the pion with its mother particle upstream).
    3. Likewise we construct the 4-momenta of the particle decays registered in our detector.
    4. We calculate (Pπ-PK)2.
    5. We construct a histogram with the data.

    Few points on the analysis:
    -All of the K+ decays are bounded by (m(π+)-m(K+)2=0.122 (GeV/c2)2, which is the response when all decayed particles are measured in our detector.
    -The π0 particle rapidly decays, usually into two photons, which commonly escape the detector system, thus the peak at m(π0)2.
    -The negative valued peak arise from the μ+ having less mass (~106 GeV/c2) than π+, which under the π+ hypothesis results in the negative value. I assume the maximum value should be located at -(m(μ+))2 ~ -0.0112 (GeV/c2)2, but I am not sure. It could also be at -(m(π0)-m(μ+))2 ~ -0.0011 (GeV/c2)2
    -The lower boundaries on K+→π+π0π0 and K+→π+π+π- equal the actual missing mass from only detecting one particle, so (m(2⋅π0))2 and (m(2⋅π+/-))2 respectively. m(π0)~0.135 GeV/c2.
    -For the decay K+→π+[itex]\nu \bar{\nu}[/itex] I am also not completely sure, my best guess is that it is bounded in the lower limit by no signal detection or low momentum, and in the upper limit by the full kaon momentum being transferred to the particle and that the neutrinos went off 180 degrees perpendicular to each other. Reasoning like this makes me think that the lower limit would be the latter suggestion, but that wouldn't conform with the earlier hypothesis I had, which is why I will need to investigate this further.

    Any clarification on the subject is appreciated.

  5. Feb 18, 2016 #4


    User Avatar
    2017 Award

    Staff: Mentor

    You always measure the charged pion only.
    In other words: "if the kaon decayed to the measured pion plus a single other particle, what would be the mass of this single other particle?".

    The maximal mass is clearly the difference between kaon and pion mass, but then the pion and the other (imaginary) particle have to have the same velocity as the initial kaon. That is possible in three-body decays only.

    If you have a two-body decay, your assumption about a two-body decay is right, so apart from measurement issues your reconstruction will always lead to the mass of the second particle (e. g. the pi0 in K-> pi+ pi0).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook