- #1

- 49

- 2

$$ I_D=2{\displaystyle \int} \, {\displaystyle \prod_{i=1}^3} d \Pi_i \, (2\pi

)^4\delta^4(p_H-p_L-p_R) |{\cal M}({e_L}^c e_R \leftrightarrow h^*)|^2

f_{L}^0f_{R}^0(1+f_{H}^0). $$

can be reduced to one dimension:

$$

I_D = \frac{m_H T^3h_e^2\gamma^2}{\pi^3}\int_1^\infty d u \frac{e^{um_H/T}}

{(e^{um_H/T}-1)^2}

\! \ln \left( \frac {\textstyle \cosh (\alpha_{e_L} u + \gamma \sqrt{u^2-1})}

{\textstyle \cosh (\alpha_{e_L} u - \gamma \sqrt{u^2-1})}

% \right. \nonumber \\ & \t \left.

\frac {\textstyle \cosh (\alpha_{e_R} u + \gamma \sqrt{u^2-1})}

{\textstyle \cosh (\alpha_{e_R} u - \gamma \sqrt{u^2-1})} \right).$$

where:

$$\alpha_{e_L}\equiv (m_H^2+m_{e_L}^2-m_{e_R}^2)/4m_HT$$

$$\alpha_{e_R}\equiv (m_H^2+m_{e_R}^2-m_{e_L}^2)/4m_HT$$

$$\gamma\equiv \lambda^{\frac{1}{2}}(m_H^2, m_{e_L}^2, m_{e_R}^2)/4m_HT$$

$$\lambda(x,y,z)\equiv (x-y-z)^2-4yz$$

and ##f^0_i=(e^{\beta E_i}\pm 1)^{-1}##, whether the particle species is a fermion(+) or a boson(-).

My attempt started by determining ##|{\cal M}({e_L}^c e_R \leftrightarrow h^*)|^2## which, for the indicated process, gave me the following:

$$I_D=2{\displaystyle \int} \, \frac{d^3p_H}{(2\pi)^32E_H}{\displaystyle \int} \, \frac{d^3p_L}{(2\pi)^32E_L}{\displaystyle \int} \, \frac{d^3p_R}{(2\pi)^32E_R}\, (2\pi

)^4\delta(E_H-E_L-E_R)\delta^3(\vec{p_H}-\vec{p_L}-\vec{p_R}) [2h_e^2(|\vec{p_L}||\vec{p_R}|-\vec{p_L}.\vec{p_R})]\frac{e^{\beta E_H}}{(e^{\beta E_H}-1)(e^{\beta E_L}+1)(e^{\beta E_R}+1)}.$$

I've proceeded by separating the integral and focus on the manifestly Lorentz invariant term ##\vec{p_L}.\vec{p_R}## and on the center of mass frame where the 4-momenta ##p_H =(E_H,\vec{0})##:

$$I_D=\frac{h_e^2}{2(2\pi)^5}{\displaystyle \int} \frac{d^3p_H}{E_H}{\displaystyle \int} \frac{d^3p_L}{E_L}{\displaystyle \int} \, \frac{d^3p_R}{E_R}\delta(m_H-E_L-E_R)\delta^3(\vec{p_L}+\vec{p_R})\vec{p_L}.\vec{p_R}\frac{e^{\beta E_H}}{(e^{\beta E_H}-1)(e^{\beta E_L}+1)(e^{\beta E_R}+1)}.$$

By using some of the delta function properties and the conservation of momentum I easily arrive to:

$$\delta(m_H-E_L-E_R) = \delta(m_H-\sqrt{m_L^2 + |\vec{p_L}|}-\sqrt{m_R^2 + |\vec{p_R}|}) \equiv \delta(f(|\vec{p_R}|)) = \left|\frac{df}{d|\vec{p_R}|}\right|^{-1}_{p*}\delta(|\vec{p_r}|-p*) .$$

$$p* = \frac{1}{2m_H}\lambda^{\frac{1}{2}}(m_H^2, m_{e_L}^2, m_{e_R}^2)$$

$$\left|\frac{df}{d|\vec{p_R}|}\right|^{-1}_{p*}=\frac{1}{p*}\frac{E_RE_L}{E_R+E_L}$$

The delta function on the momenta enabled me to remove one of the integrals (in this case, the integral over ##\int d^3p_L##. The new delta function ##\delta(|\vec{p_r}|-p*)## will allow me to take out the integral over ##\int dp_R## that comes from ##d^3p_R = |\vec{p_R}|^2dp_Rd\Omega## and so:

$$I_D=\frac{h_e^2}{2(2\pi)^5}{\displaystyle \int} \frac{d^3p_H}{E_H}{\displaystyle \int}\frac{p*^3}{E_R+E_L}\frac{e^{\beta E_H}}{(e^{\beta E_H}-1)(e^{\beta E_L}+1)(e^{\beta E_R}+1)}d\Omega.$$

From here on out, I have no idea on what to do in order to arrive to the one dimensional form, as I've no clue on how to fit the natural logarithm of those hyperbolic cosines in the integrand and neither which substitution I should do.

Any help would be welcome!