Massive three particle phase space

  • #1
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If you produce three massive particles with m1=/=m2=/=m3 near threshold (beta -> 0), the cross section of the production is supressed by a factor beta^4, where beta = sqrt(1-(M_tot)^2/s) and s is COM energy. I have been trying to prove this statement, but I can't seem to manage. Could anybody help me?
 

Answers and Replies

  • #2
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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
  • #3
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Did you set up the integrals for the phase space factor and then see where an approximation ##M = s(1-\epsilon)## leads? ##\beta^4 = \epsilon^2## here.
 
  • #4
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Did you set up the integrals for the phase space factor and then see where an approximation ##M = s(1-\epsilon)## leads? ##\beta^4 = \epsilon^2## here.
Yes I have, but my problem is that I have an integral over the square root of a polynomial of degree 3. I dont see it reducing to beta^4 that easily...
 
  • #5
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Can you post the integral you get?
 
  • #6
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Yes:
$$\int {\rm d}s_{23} \sqrt{(s^2 + m_1^4+s_{23}^2-2ss_{23}-2m_1^2s_{23}-2sm_1^2)(s_{23}-m_2^2)}$$
We have ##(m_2+m_3)^2 \leq s_{23} \leq s-m_1^2##, ##s=(p_1+p_2+p_3)^2## and ##s_{23} = (p_2 + p_3)^2##. To get to this form, we have assumed that the combined state ##p_{23}## is at rest.
 
  • #7
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11,510
If p23 is at rest, then particle 3 is also at rest and s23 is fixed, there would be nothing to integrate over.
 

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