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## GR0877 Problem 40

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (B) Conservation of momentum immediately suggests that after the decay particle $m$ and massless particle have the same momentum $p$. From the relativistic energy-momentum equation (we use units in which $c = 1$) $E^2 - p^2 = m^2$ one gets $E = p$ for massless particle. The conservation of energy: $M = p+ \sqrt{m^2 + p^2}$. Moving $p$ to the left-hand side and squaring results in $(M-p)^2 = m^2+p^2$ or $M^2-2Mp = m^2$, from which one has $p = (M^2-m^2)/(2M)$. Clearly, the decay is possible if $M > m$.

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1. 18.09.2013 at 20:26

I agree with your solution, but I’ve been trying to also do the problem with the typical particle physics method: (p^2)_initial = (p^2)_final, where p = energy-momentum four-vector. This approach should work just as well, but I just can’t get it to work for me. Could you maybe add an alternate solution for doing it this way? (Or could anyone in the comments show me how to do it right?)

Thanks!

2. 18.09.2013 at 20:37

Well, what do you know. Soon as I comment I figure it out. For those who are wondering:

(p_i)^2 = (p_1 + p_2)^2 , p_2 = massless particle
— Evaluate both sides, but DON’T expand the right quadratically, for some reason this kept messing me up. Instead add the four-vectors then square normally.

M^2 = (E_1 + p)^2 – (p – p)^2
— Note that p is the momentum of one particle and (-p) is the momentum of the other, so they cancel out.

M = E_1 + p
— At this point you have the same equation the poster started with, using E_1 = sqrt(m^2 + p^2), here p = momentum 3-vector.

Obviously this approach is totally unnecessary, and the original solution is much faster. However if you recently took particle physics like I did, the itch to always start with p^2 is strong. Hope someone found this helpful.

3. 28.08.2016 at 10:22

Whenever you see any sort of relativistic energy/momentum problem, immediately write down E^2=m^2+p^2 (in natural units). For this problem, you only need that equation and conservation of relativistic momentum (not just the square of the 4-momentum).

4. 05.04.2017 at 16:47

Easy shortcut for the algebra: By previous arguments made, the energy of a massless particle is just p*c. If m=0, then by conservation of momentum, each massless particle has the same momentum p. Then by conservation of energy Mc^2=2*pc for which p=Mc/2. Only solution B satisfies this equation in the limit as m goes to 0.