A Question about a cross section from PDG

ChrisVer
Science Advisor
Messages
3,372
Reaction score
465
Hi everyone, I was wondering, is the cross section at Eq 50.25 in
http://pdg.lbl.gov/2017/reviews/rpp2017-rev-cross-section-formulae.pdf
correct?

Because I see a term in the denominator with \frac{1}{s\Gamma} whereas in several other references, the propagator term in the matrix element comes with \frac{1}{M \Gamma}.
[eg. eq29 here http://article.sciencepublishinggroup.com/pdf/10.11648.j.ijhep.20150205.11.pdf and I can give a further list]
Thanks.
 
Physics news on Phys.org
I'm a bit surprised to the the center of mass energy there, but note that the two expressions are very similar for ##\Gamma \ll M##: The right term is only relevant if ##\sqrt s \approx M##.
 
Yup. Since I've found instances where the additional part in the propagator (added to s-m^2 ) is written as a function of s: \Pi(s), and through choosing a renormalization scheme they can set it as : \gamma(s = M^2 ) = M , which basically translates to what you've written about the right term.

Additionally I've found that the propagator can be written as: D(s) = \frac{1}{s-m^2 + i \sqrt{s} \Gamma} (after summing up several Feynman diagrams)... as shown on slide7 here: https://www.stfc.ac.uk/files/lecture-7/

Finding multiple different instances for the same thing is confusing indeed...
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top