How to reduce an integral in phase space to a one-dimensional form?

In summary, the conversation discusses the process of reducing a 4-dimensional integral to a one-dimensional integral, specifically for a Higgs decay into an electron pair. The integral is manipulated using delta functions and conservation of momentum to reduce it to an integral over the solid angle. The format of the desired integral is achieved by performing a change of variables from cos ##Theta##.
  • #1
RicardoMP
49
2
I've been trying for a very long time to show that the following integral:
$$ I_D=2{\displaystyle \int} \, {\displaystyle \prod_{i=1}^3} d \Pi_i \, (2\pi
)^4\delta^4(p_H-p_L-p_R) |{\cal M}({e_L}^c e_R \leftrightarrow h^*)|^2
f_{L}^0f_{R}^0(1+f_{H}^0). $$
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!
 
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  • #2
Is this a process of the form (1->2)? Cannot one immediately consider the 2body final state phase space of eL and eR evaluated in h CoM frame? Then apply the same procedure as you did and just deal with the dOmega integral.
 
  • #3
RGevo said:
Is this a process of the form (1->2)? Cannot one immediately consider the 2body final state phase space of eL and eR evaluated in h CoM frame? Then apply the same procedure as you did and just deal with the dOmega integral.
It is indeed a decay process of the form ##1\rightarrow 2##, in particular, a Higgs decaying to an electron pair. You're suggesting using the same procedure of going into the Higgs CoM frame on the last equation? Nonetheless, I have no idea on how to move on from there.
 
  • #4
I am instead suggesting to perform the phase-space sum with i = eL, eR meaning that you will have an integrals over the momenta of those out-going particles only (and not also the Higgs).

This will mean that the manipulations you did to deal with the delta functions before can be applied in the same way. You will then reach the same point with an integral over the solid angle (but not over the Higgs).

Because there is no ##Phi## dependence, d##Omega## = d##Phi## d cos ##Theta##, the ##Phi## integral can be performed trivially (just 2##Pi##) leaving the d cos ##Theta## integral. I suspect the format of the integral you are trying to find is then obtained by performing a change of variables from cos ##Theta##.
 

Related to How to reduce an integral in phase space to a one-dimensional form?

1. What is an integral in phase space?

An integral in phase space is a mathematical concept used in physics to describe the behavior of a system over time. It takes into account both the position and momentum of particles within the system and can be used to calculate various physical quantities such as energy and entropy.

2. Why is it important to reduce an integral in phase space to a one-dimensional form?

Reducing an integral in phase space to a one-dimensional form simplifies the mathematical calculations and makes it easier to analyze the behavior of a system. It also allows for easier comparison with experimental data and helps to identify patterns and relationships within the system.

3. What are the steps involved in reducing an integral in phase space to a one-dimensional form?

The first step is to identify the variables involved in the integral and determine which ones can be integrated out. Then, use appropriate mathematical techniques such as substitution or integration by parts to simplify the integral. Finally, solve for the remaining variables to obtain a one-dimensional form of the integral.

4. Are there any limitations to reducing an integral in phase space to a one-dimensional form?

Yes, there are limitations depending on the complexity of the system and the mathematical techniques used. In some cases, it may not be possible to reduce the integral to a one-dimensional form, or the resulting form may not accurately represent the behavior of the system.

5. How can reducing an integral in phase space to a one-dimensional form help in understanding a physical system?

Reducing an integral in phase space to a one-dimensional form can provide insights into the behavior of a system and help to identify important variables and relationships. It can also aid in making predictions and understanding the underlying mechanisms driving the system.

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