Pion Decay into Gamma rays angle

cooev769
Messages
114
Reaction score
0
So i have a question regarding a homework question I'm working on which suggest a neutral pion traveling with velocity v decays into two gamma rays of equal theta to the normal and they of course have velocity c. It then asks to prove that

v= cos theta

Which kind of confuses me. I mean if we assume that the gamma rays travel off at an angle and the distance they travel in time t is at a right angle to the distance the pion would have traveled at velocity v, we can form a right handed triangle with adjacent length vt and hypotenuse ct, but how can we just assume that you can just go north of vt and it will find ct right at the end of the hypotenuse and hence use this trigonometric rule.

Thanks.
 
Physics news on Phys.org
Oh wait is it due to conservation of momentum so the vertical components of the c's must cancel and therefore the horizontal components must be the same length? Cheers.
 
Oh wait is it due to conservation of momentum so the vertical components of the c's must cancel and therefore the horizontal components must be the same length? Cheers.
Exactly.
 
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}...

Similar threads

Back
Top