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

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?

• 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. – Melquíades Apr 13 '14 at 19:56
• The title mentions 1->2 particle decay. Do you have any idea about the answer? Thanks for replying. – Your Majesty Apr 13 '14 at 19:57
$$\Gamma(M\to 1 + 2)=\dfrac{|\vec{p_1}|}{32 \pi^2 M^2} \int \mathrm{d}\Omega |\mathcal{M}|^2=\dfrac{|\vec{p_1}|}{8 \pi M^2}|\mathcal{M}|^2,$$
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.
• 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). – Your Majesty Apr 13 '14 at 20:15
• 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. – Melquíades Apr 13 '14 at 20:22