Beta+ meson, decay - Relativistic kinematics

  • #1
Matt atkinson
116
1

Homework Statement



The ##B^+## meson decays through the weak interaction. One of its decay channels is
hw8eqn6.gif
. If a ##B^+## is traveling 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.

Homework 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

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 I am 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 doesn't help too much because i still don't know ##p_{\bar{D}}##
 
Last edited:

Answers and Replies

  • #2
Orodruin
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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.
 
  • #3
Matt atkinson
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Is the way I am 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?
 
  • #4
Orodruin
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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.
 
  • #5
Matt atkinson
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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.
 
  • #6
Matt atkinson
116
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so i tried the way you suggested with rest mass, and that seemed to work much better (if I've 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?
 
  • #7
Matt atkinson
116
1
but when i do this i just get that ##E_D'<m_D## which can't work, that was using the gamma and beta from the rest frame, should it be the gamme and beta from lab frame?
 

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