Complete two-body phase space and coefficient flux map

Let $a=(a_0,\mathbf{a})$ be any future unit vector, so $a_0>0$ and $a_0^2-\mathrm{\mid }\mathbf{a}{\mathrm{\mid }}^2=1$. If $\mathbf{a}=0$, put $B=I$. Otherwise put $r=\mathrm{\mid }\mathbf{a}\mathrm{\mid }$, $\mathbf{n}=\mathbf{a}\mathrm{/}r$, and define $B$ to act by $\left(\begin{array}{cc}{a}_0& r\\ r& a_0\end{array}\right)$ on $\mathrm{span}(e_0,\mathbf{n})$ and by the identity on its spatial orthogonal complement. In both cases $B$ preserves $h$, has determinant one, preserves the future cone, and sends $e_0$ to $a$. For any proper orthochronous $h$-Lorentz map $\mathrm{\Lambda }$, apply this construction to $a=\mathrm{\Lambda }{e}_0$; then $B^{-1}\mathrm{\Lambda }$ fixes $e_0$ and restricts to a proper spatial rotation on $e_0^{\perp }$. Thus every such $\mathrm{\Lambda }$ is a boost followed by a spatial rotation. The construction also gives, for every future unit vector $a$, an inverse boost taking $a$ to $e_0$.

A spatial rotation has Jacobian one and leaves $E$ fixed. By step 1.1, it remains to check a boost directed along the first spatial axis. For that boost, $p_1^{\mathrm{\prime }}=a_0{p}_1+rE$ and $E^{\mathrm{\prime }}=a_{0}E+r{p}_1$, so $\mathrm{\partial }{p}_1^{\mathrm{\prime }}\mathrm{/}\mathrm{\partial }{p}_1=a_0+r{p}_1\mathrm{/}E=E^{\mathrm{\prime }}\mathrm{/}E>0$; the other spatial coordinates are unchanged. The inverse boost is obtained by replacing $r$ with $-r$, so this spatial map is a $C^1$ diffeomorphism. By [F3], $d^3{\mathbf{p}}^{\mathrm{\prime }}=(E^{\mathrm{\prime }}\mathrm{/}E)d^3\mathbf{p}$ in every nonnegative Borel integral, and therefore $d^3{\mathbf{p}}^{\mathrm{\prime }}\mathrm{/}(2E^{\mathrm{\prime }})=d^3\mathbf{p}\mathrm{/}(2E)$. This proves shell-measure invariance under the boost and, using the rotation factorization, under every proper orthochronous $h$-Lorentz map.

Apply the boost construction in step 1.1 to the future unit vector $P\mathrm{/}\sqrt{s}$; its inverse takes $P\mathrm{/}\sqrt{s}$ to $e_0$. Any two maps taking $P\mathrm{/}\sqrt{s}$ to $e_0$ differ by a Lorentz map fixing $e_0$, hence by a spatial rotation on $e_0^{\perp }$; orthogonal invariance in [F4] therefore makes ${\mathrm{\Phi }}_{2,P}$ independent of the chosen map. Under the chosen map, write $\tilde{f}(k_1,k_2)=f({\mathrm{\Lambda }}_P^{-1}{k}_1,{\mathrm{\Lambda }}_P^{-1}{k}_2)$. Step 2.1 preserves each shell measure, and $\mathrm{\mid }\det {\mathrm{\Lambda }}_P\mathrm{\mid }=1$ preserves the four-dimensional delta, so the integral reduces to the rest-frame calculation with $\tilde{f}$. Integrating the spatial delta first sets ${\mathbf{k}}_2=-{\mathbf{k}}_1$ and gives the remaining radial-angular measure $\frac{\tilde{f}((E_1(r),r\mathbf{n}),(E_2(r),-r\mathbf{n}))}{4(2\pi {)}^2{E}_1(r)E_2(r)}{r}^{2}drd\mathrm{\Omega }\delta (\sqrt{s}-E_1(r)-E_2(r))\mathrm{.}$ Here $d^3{\mathbf{k}}_1=r^{2}drd\mathrm{\Omega }$ by [F4].

For $r>0$, $E_1(r)+E_2(r)$ is strictly increasing, with derivative $r(1\mathrm{/}{E}_1+1\mathrm{/}{E}_2)>0$. The threshold assumption gives a unique root $k_{\ast }>0$. At this root, $E_1+E_2=\sqrt{s}$ and $E_1-E_2=({\mathsf{M}}_1^2-{\mathsf{M}}_2^2)\mathrm{/}\sqrt{s}$, so $E_1=(s+{\mathsf{M}}_1^2-{\mathsf{M}}_2^2)\mathrm{/}(2\sqrt{s})$, $E_2=(s-{\mathsf{M}}_1^2+{\mathsf{M}}_2^2)\mathrm{/}(2\sqrt{s})$, and $k_{\ast }^2=E_1^2-{\mathsf{M}}_1^2=\frac{[s-({\mathsf{M}}_1+{\mathsf{M}}_2{)}^2][s-({\mathsf{M}}_1-{\mathsf{M}}_2{)}^2]}{4s}\mathrm{.}$ The derivative at the root is $k_{\ast }\sqrt{s}\mathrm{/}(E_1{E}_2)$. Applying [F3] to the one-dimensional increasing change of variable in step 3.1, for each fixed $\mathbf{n}$, gives the angular density $\tilde{f}((E_1^{\ast },k_{\ast }\mathbf{n}),(E_2^{\ast },-k_{\ast }\mathbf{n}))k_{\ast }d\mathrm{\Omega }\mathrm{/}(16{\pi }^2\sqrt{s})$. Integrating this nonnegative Borel density in the stipulated order gives exactly $\int fd{\mathrm{\Phi }}_{2,P}$ by its pushforward definition. In the center-of-momentum frame, this is the pushforward $({\mathrm{\Psi }}_2{)}_{\mathrm{\#}}{\mathrm{\Phi }}_2$ from the Definition, where ${\mathrm{\Phi }}_2$ is a measure on $S^2$. To compute the angular total, use $X(\vartheta ,\phi )=(\sin \vartheta \cos \phi ,\sin \vartheta \sin \phi ,\cos \vartheta )$ for $0<\vartheta <\pi$ and $0<\phi <2\pi$; its chart-area density is $\mathrm{\mid }{\mathrm{\partial }}_{\vartheta }X\times {\mathrm{\partial }}_{\phi }X\mathrm{\mid }=\sin \vartheta$. The omitted meridian and poles are null in the two-dimensional chart measure, so $\mathrm{\Omega }(S^2)={\int }_0^{\pi }{\int }_0^{2\pi }\sin \vartheta d\phi d\vartheta =4\pi$. It follows that ${\mathrm{\Phi }}_2(S^2)=k_{\ast }\mathrm{/}(4\pi \sqrt{s})$. For equal daughter energies $\mathsf{M}$, this gives $k_{\ast }=\sqrt{s}\beta \mathrm{/}2$ and ${\mathrm{\Phi }}_2(S^2)=\beta \mathrm{/}(8\pi )$, where $\beta =\sqrt{1-4{\mathsf{M}}^2\mathrm{/}s}$.

In the incoming center-of-momentum frame from step 3.1, write $p_1=(E_1,{\mathbf{p}}_{\ast })$ and $p_2=(E_2,-{\mathbf{p}}_{\ast })$, so $E_1+E_2=\sqrt{s}$. The mass-shell equations give $p_1\cdot p_2=E_1{E}_2+p_{\ast }^2$ and ${\mathsf{M}}_{i,\mathrm{i}\mathrm{n}}^2=E_i^2-p_{\ast }^2$. Direct expansion yields $(p_1\cdot p_2{)}^2-{\mathsf{M}}_{1,\mathrm{i}\mathrm{n}}^2{\mathsf{M}}_{2,\mathrm{i}\mathrm{n}}^2=(E_1{E}_2+p_{\ast }^2{)}^2-(E_1^2-p_{\ast }^2)(E_2^2-p_{\ast }^2)=p_{\ast }^2(E_1+E_2{)}^2=s{p}_{\ast }^2,$ so [F2] gives $F=4p_{\ast }\sqrt{s}$. If the incoming rest energies are equal to ${\mathsf{M}}_{\mathrm{i}\mathrm{n}}$, then $E_1=E_2=\sqrt{s}\mathrm{/}2$ and $p_{\ast }^2=s\mathrm{/}4-{\mathsf{M}}_{\mathrm{i}\mathrm{n}}^2$; since $p_{\ast }>0$, $p_{\ast }=\sqrt{s}{\beta }_{\mathrm{i}\mathrm{n}}\mathrm{/}2$ and $F=2s{\beta }_{\mathrm{i}\mathrm{n}}$. Combining the phase-space density from step 4.1 with [F2]'s coefficient map yields $\frac{d{\sigma }_{\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{f}\mathrm{f}}}{d\mathrm{\Omega }}=\frac{(\mathrm{\hslash }c{)}^2}{64{\pi }^{2}s}\frac{k_{\ast }}{p_{\ast }}\frac{\overline{\mathrm{\mid }\mathcal{A}{\mathrm{\mid }}^2}}{S_f}\mathrm{.}$

For a parent at rest with energy ${\mathsf{M}}_0>2\mathsf{M}$ and two equal-mass daughters, step 4.1 gives $\int d{\mathrm{\Phi }}_2=\sqrt{1-4{\mathsf{M}}^2\mathrm{/}{\mathsf{M}}_0^2}\mathrm{/}(8\pi )$. The decay coefficient map in [F2] divides $\mathrm{\mid }{G}_E{\mathrm{\mid }}^2\int d{\mathrm{\Phi }}_2$ by $2\mathrm{\hslash }{\mathsf{M}}_0{S}_f$; for identical daughters $S_f=2$, this is $\mathrm{\mid }{G}_E{\mathrm{\mid }}^2\sqrt{1-4{\mathsf{M}}^2\mathrm{/}{\mathsf{M}}_0^2}\mathrm{/}(32\pi \mathrm{\hslash }{\mathsf{M}}_0)$. Its units are ${\mathrm{J}}^2\mathrm{/}(\mathrm{J}\mathrm{s}\mathrm{J})={\mathrm{s}}^{-1}$. No survival law enters this coefficient calculation.