InterviewSolution
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InterviewSolution
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A relativistic particle with rest mass `m` cillides with a stationary particle of mass `M` and activities at reaction leading to formation of new particles: `m+Mrarrm_(1)+m_(2)+…..`, where the rest masses of newly formed particles are written on the right-hand side. Making use of invariance of the quantity `E^(2)-p^(2)c^(2)`, demonstrate that the threshold kinetic energy of the particle `m` required for this reaction is defined by Eq.(6.7c). |
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Answer» With particle masses standing for the names of particles, the reaction is `m+Mrarrm_(1)+m_(2)+…..` On R.H.S. let the energy momenta be `(E_(1),cvec(p)_(1)),(E_(2),cvec(p)_(2))` etc. On the left the energy momentum of the particle `m is (E,cvec(p))` and that of the particle is `(Mc^(2),vec(O))`, where ofcourse, the usual relations `E^(2)-c^(2)vec(p)^(2)=m^(2)c^(4)eta` hold. From the conservation of energy momentum we see that `(E+Mc^(2))^(2)-c^(2)vec(p)^(2)=(SigmaE_(i))^(2)-(SigmaE_(i))^(2)-(Sigmacvec(p)i)^(2)` Left hand side of `m^(2)c^(4)+M^(2)c^(4)+2Mc^(2)E` We evaluate the R.H.S. in the frame where `Sigmavec(pi)=0` (`CM` frame of the decay product). Then R.H.S `=(SigmaE_(i))^(2)=(SigmaE_(i))^(2) ge(Sigmam_(i)c^(2))^(2)` beacuse all energies are `+ve`. Therefore we have the result `E ge((Sigmam_(i))^(2)-m^(2)-M^(2))/(2M)c^(2)` M or Since `E=mc^(2)+T`, we see that `T geT_(th)` where `T_(th)=((Sigmam_(i))^(2)-(m+M)^(2))/(2M)c^(2)` |
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