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  $\mathrm{d} \Omega_{CM}$ for a $1\rightarrow 2$ particle decay?

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The differential solid angle is described in e.g. Srednicki's QFT text but only for the case of scattering. Because in the case of scattering it's defined with respect to the incoming three-momentum ${\bf k_1}$ and it's scattered three-momentum ${\bf k_1'}$. This definition would not make much sense for a particle decay (I can't at least see how it would easily carry over).

What about the case of particle decay? What is $$\mathrm{d} \Omega_{CM}$$ in the case of particle decay? Any good references where it is properly defined?

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user Love Learning
asked Apr 13, 2014 in Theoretical Physics by Love Learning (165 points) [ no revision ]
It is a solid angle, but its definition depends on the particular case. How many particles you have in the final state? Check "Elementary Particles and Their Interactions", Quang Ho-Kim.

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user JSchwinger
The title mentions 1->2 particle decay. Do you have any idea about the answer? Thanks for replying.

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user Love Learning
Ok, sorry. See the answer and comment below. There is nothing special about this angle.

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user JSchwinger

1 Answer

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Since the 4-momentum must be conserved, there is no angular dependence in a 2-body decay in the CM frame. But If you are asking yourself what is the phase-space of such process, you should take a look at equation 4.72 of the book I mentioned in the comments:

$$\Gamma(M\to 1 + 2)=\dfrac{|\vec{p_1}|}{32 \pi^2 M^2} \int \mathrm{d}\Omega |\mathcal{M}|^2=\dfrac{|\vec{p_1}|}{8 \pi M^2}|\mathcal{M}|^2,$$

where $\mathcal{M}$ is the amplitude of the process, $M$ is the mass of the initial particle and $|\vec{p_1}|=|\vec{p_2}|$ is the final 3-momentum of one of the particles.

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user JSchwinger
answered Apr 13, 2014 by JSchwinger (0 points) [ no revision ]
Yes I know it's $4\pi$ in this case, but I mean, how is it defined. Because in the case of scattering it's defined with respect to the incoming three-momentum ${\bf k_1}$ and it's scattered three-momentum ${\bf k_1'}$. This definition would not make much sense for a particle decay (I can't at least see how it would carry over in any sense).

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user Love Learning
It is the angle of one of the particles with respect to a given axis. Since the amplitude does not depend on this angle, you have only $4\pi$. Think in terms of the general formula for the phase-space: $$\mathrm{d}\phi_f \propto \frac{\mathrm{d}^3 p_1}{2 E_1}\frac{\mathrm{d}^3 p_2}{2 E_2} \delta^4(p_1+p_2-p).$$ We integrate the momentum of one of the particles to get rid of the distribution $\delta^3$ and at the end we'll have only a 3D integral. To do this integral, you have to define an angle.

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user JSchwinger

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