1. Limited time only! Sign up for a free 30min personal 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!

Beta+ meson, decay - Relativistic kinematics

  1. Dec 8, 2014 #1
    1. The problem statement, all variables and given/known data

    The ##B^+## meson decays through the weak interaction. One of its decay channels is hw8eqn6.gif . If a ##B^+## is travelling with a total energy of 9.90 GeV, calculate the range of possible energies the produced d0bar.gif may have. (i.e. determine the minimum and maximum values of its energy.) Justify why the values you calculate are the minimum and maximum possible.
    2. Relevant equations
    Mass of ##B^+## is 5.279 GeV/c2, mass of d0bar.gif is 1.865 GeV/c2,mass of ##\rho^+##is 0.770 GeV/c2

    3. The attempt at a solution
    max energy => when ##B^+ \parallel \bar{D^o}## and ##\rho^+## anti parallel to ##B^+##
    min energy => when ##B^+ \parallel \bar{D^o}## and ##\rho^+\parallel B^+##

    Max
    conservation of momentum ; ##E_{\beta}=E_{\rho}+E_{\bar{D}}##
    conservation of energy ; ##p_{\beta}=-p_{\rho}+p_{\bar{D}}##
    $$E_{\bar{D}}^2=m_{\bar{D}}^2+p_{\bar{D}}^2$$
    $$E_{\bar{D}}^2=m_{\bar{D}}^2+(p_{\beta}+p_{\rho})^2$$
    $$E_{\bar{D}}^2=m_{\bar{D}}^2+p_{\beta}^2+p_{\rho}^2+2p_{\rho}p_{\beta}$$
    $$E_{\bar{D}}^2=m_{\bar{D}}^2+p_{\beta}^2+E_{\rho}^2-m_{\rho}^2+2p_{\rho}p_{\beta}$$
    $$E_{\bar{D}}^2=m_{\bar{D}}^2+p_{\beta}^2+(E_{\beta}-E_{\bar{D}})^2-m_{\rho}^2+2p_{\rho}p_{\beta}$$
    $$E_{\bar{D}}=\frac{m_{\bar{D}}^2+p_{\beta}^2+E_{\beta}^2-m_{\rho}^2+2p_{\rho}p_{\beta}}{2E_{\beta}}$$
    but after this im not sure how to get rid of the ##p_{\rho}## from the ##2p_{\rho}p_{\beta}}{2E_{\beta}## term.
    also to do the case for minimum energy, i would just have ##p_{\beta}=p_{\rho}+p_{\bar{D}}## for my momentum conservation right?

    edit;using momentum conservation on the ##p_{\beta}=p_{\rho}+p_{\bar{D}}## term;
    $$E_{\bar{D}}=\frac{m_{\bar{D}}^2+m_{\beta}^2-m_{\rho}^2+2p_{\bar{D}}p_{\beta}}{2E_{\beta}}$$
    but this doesnt help too much because i still don't know ##p_{\bar{D}}##
     
    Last edited: Dec 8, 2014
  2. jcsd
  3. Dec 8, 2014 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You can get rid of the rho momentum in favor of the rho energy by the energy-momentum-mass relation for the rho. You can get rid of the rho energy in favor of the D energy by conservation of energy.

    But you are taking a long way around. I would suggest working in the rest frame of the B and then Lorentz boost the result.
     
  4. Dec 8, 2014 #3
    Is the way im doing it correct, but just very long winded?
    so consider ##\beta## at rest and then ##\rho## and ##D## would be moving oppositely? and use lorentz transformations to get my result from that?
     
  5. Dec 8, 2014 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    That is how I would do it (well more or less). You can of course do it just algebraically, but thinking first helps a lot.
     
  6. Dec 8, 2014 #5
    I tried to think about the way I did it, i'm just really struggling with relativistic decays or collisions.
    Not sure why they don't seem too bad when i study them, just really struggle solving the problem.

    My way of doing it, just as you said by changing rho momentum for rho energy and then D energy, resulted in a expression with a ##E_{D}## and a ##\sqrt{E_D}## term which I just figured id done something wrong, didnt think solving an equation that was quadractic in ##\sqrt{E_D}## a good way to solve it.
     
  7. Dec 8, 2014 #6
    so i tried the way you suggested with rest mass, and that seemed to work much better (if ive done it right that is).
    I end up getting ##E_D=\frac{m_D^2+m_{\beta}^2-m_{\rho}^2}{2m_{\beta}}##
    then doing lorentz $$ E_D'=\gamma E_D-\beta \gamma \sqrt{E_D^2+m_{\beta}^2-m_{\rho}^2-2E_Dm_{\beta}}$$
    Where ##p_D=\sqrt{E_D^2+m_{\beta}^2-m_{\rho}^2-2E_Dm_{\beta}}## in the rest frame and ##E_D## is the one i calculated for the rest frame?
     
  8. Dec 9, 2014 #7
    but when i do this i just get that ##E_D'<m_D## which cant work, that was using the gamma and beta from the rest frame, should it be the gamme and beta from lab frame?
     
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: Beta+ meson, decay - Relativistic kinematics
  1. Beta+ Decay (Replies: 13)

  2. Beta decay (Replies: 3)

  3. Meson decay (Replies: 0)

  4. Meson decay (Replies: 0)

Loading...