Gondolo-Gelmini Change of Variables

Question

In the article Cosmic abundances of stable particles: Improved analysis, P. Gondolo and G. Gelmini, Nucl. Phys. B 360 (1991), p. 145-179, they convert $\rm{d}^3p_1\rm{d}^3p_2=2\pi^2p_1p_2\rm{d}E_1\rm{d}E_2\rm{d}\cos{\theta}$ (eq.3.2) into $\rm{d}^3p_1\rm{d}^3p_2=2\pi E_1E_2\rm{d}E_+\rm{d}E_-\rm{d}s$ (eq. 3.4) with the following change of variables:

$$E_+=E_1+E_2, \quad E_-=E_1-E_2, \quad s=2m^2+2E_1E_2-2p_1p_2\cos{\theta}, \quad \rm{(eq. 3.3)}$$

When I try to derive the second expresion of $\rm{d}^3p_1\rm{d}^3p_2$ using these new variables $E_+,\ E_-, s$, I always get second order diferentials that in theory should vanish. I don't know how one could get $\rm{d}^3p_1\rm{d}^3p_2=2\pi E_1E_2\rm{d}E_+\rm{d}E_-\rm{d}s$.

They also define new limits of integration due to this change of variables: $\{ E_1>m,\ E_2>m, \ |\cos{\theta}|\leq 1 \}$ changes to $\{ s\geq 4m^2,\ E_+\geq \sqrt{s}, \ |E_-|\leq\sqrt{1-4m^2/s}\sqrt{E_+^2-s} \}$ (eq. 3.5). I get the first two, but I don't know how to compute $E_-$. The limit cases give some idea on how the expresion should be, but this is not enough to know the exact expresion, something like: $$\rm{If} \quad E_+^2=s \implies E_-=0$$ $$\rm{If} \quad s=4m^2 \implies E_-=0$$ $$\rm{Then (?):} \quad |E_-| \leq \sqrt{1-\frac{4m^2}{s}}\sqrt{E_+^2-s}$$

Answer 1

Assume some general $A \bar{A}$ states where $\theta = \theta_{A \bar{A}}$.

The bound on $E_-$ is given by $|\cos \theta_{A\bar{A}}| \leq 1$ :

\begin{aligned} \cos \theta_{A \bar{A}} = & \frac{2m_A^2 + 2E_{A}E_{\bar{A}} - s}{2 |\vec{p}_A| |\vec{p}_{\bar{A}}|} \\ = & \frac{4m_A^2 + E_+^2 - E_-^2 - 2s}{4 |\vec{p}_A| |\vec{p}_{\bar{A}}|} \\ = & \frac{4m_A^2 + E_+^2 - E_-^2 - 2s}{4\sqrt{(E_A^2 - m_A^2)(E_{\bar{A}}^2 - m_A^2)}} \\ = & \frac{4m_A^2 + E_+^2 - E_-^2 - 2s}{4\sqrt{E_A^2 E_{\bar{A}}^2 - m_A^2(E_A^2 + E_{\bar{A}}^2) + m_A^4}} \\ = & \frac{4m_A^2 + E_+^2 - E_-^2 - 2s}{4\sqrt{\left(\frac{E_+^2 - E_-^2}{4} \right)^2 - m_A^2 \frac{E_+^2 + E_-^2}{2} + m_A^4}} \\ = & \frac{4m_A^2 + E_+^2 - E_-^2 - 2s}{\sqrt{(E_+^2 - E_-^2)^2 - 8m_A^2 (E_+^2 + E_-^2) + 16m_A^4}} \end{aligned}

Using Mathematica's Reduce function we get \begin{equation} E_-^2 \leq \frac{1}{s} (E_+^2 - s) (s - 4m_{A}^2) \end{equation}

Answer 2

Since nobody has answered and I figured it out some time ago, I'll respond to my own post. You have to use the Jacobian to change variables:

$$ J_{ij}=\frac{\partial y_i}{\partial x_j} = \begin{pmatrix} \frac{\partial E_1}{\partial E_+} & \frac{\partial E_1}{\partial E_+} & \frac{\partial E_1}{\partial s}\\ \frac{\partial E_2}{\partial E_+} & \frac{\partial E_2}{\partial E_-} & \frac{\partial E_2}{\partial s}\\ \frac{\partial \cos{\theta}}{\partial E_+} & \frac{\partial \cos{\theta}}{\partial E_-} & \frac{\partial \cos{\theta}}{\partial s} \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 1 & 2E_2\\ 1 & -1 & 2E_1\\ 0 & 0 & -2 p_1 p_2 \end{pmatrix}^{-1}.$$ Knowing that the determinant of the inverse matrix is the inverse of the determinant of the original matrix, we get $\det{J}=(4p_1 p_2)^{-1}$. $$ \mathrm{d}^3p_1 \mathrm{d}^3p_2 = \left(4\pi p_1 \mathrm{d}E_1\right) \left(4\pi p_2 \mathrm{d}E_2\right) \left(\frac{1}{2}\mathrm{d}\cos{\theta}\right) = 8\pi^2 p_1 p_2 \frac{1}{4p_1 p_2}\mathrm{d}E_+ \mathrm{d}E_- \mathrm{d}s = 2\pi^2\mathrm{d}E_+ \mathrm{d}E_- \mathrm{d}s. $$ I guess the integrating limits logic is good enough. If anybody has something else to add there, please point it out.