Lorentz-Invariant Phase Space

The Lorentz-invariant phase-space element for the process $2 \rightarrow 2$ is

$$\textrm{dLips}(s;p_3,p_4) = \frac{1}{(2\pi)^2} \delta^4(p_3+p_4-p_1-p_2) \frac{d^3p_3}{E_3} \frac{d^3p_4}{E_4} .$$ (7.119)

Using the three-momentum $\delta$-function, we can eliminate the integral over $d^3p_4$

$$\int \frac{d^3p_4}{E_4} \delta^4(p_3+p_4-p_1-p_2) = \frac{1}{E_4} \delta(E_3+E_4-E_1-E_2) .$$ (7.120)

On the right-hand side $p_4$ and $E_4$ are no longer independent variables but are determined by the conditions

$$p_4 = p_1 + p_2 - p_3 \quad \mathrm{and} \quad E_4 = (\vert\vec{p}_4\vert^2+m_2^2)^{1/2} .$$ (7.121)

Next, convert $d^3p_3$ to angular variables.

$$d^3p_3 = p_3^2 dp_3 d\Omega ,$$ (7.122)

where $p_3$ now stands for the magnitude of the three-momentum. The energy and momentum are related by

$$E_3^2 = p_3^2 + m_1^2$$ (7.123)

so that

$$E_3dE_3 = p_3dp_3 .$$ (7.124)

With all these changes we arrive at the result (valid in any frame)

$$\textrm{dLips}(s;p_3,p_4) = \frac{1}{(2\pi)^2} d\Omega \frac{p_3dE_3}{E_4} \delta(E_3+E_4-E_1-E_2) .$$ (7.125)

We now specialize to the center-of-mass (CM) frame, for which

$$E_3^2 = p^2 + m_1^2 ,$$ (7.126)

$$E_4^2 = p^2 + m_2^2$$ (7.127)

and

$$E_3dE_3 = pdp = E_4dE_4 .$$ (7.128)

Introduce the variable $W^\prime=E_3+E_4$ (since $E_3+E_4$ is only constrained to be equal to $W=E_1+E_2$ after performing the integral over the energy-conserving $\delta$-function). Then

$$dW^\prime = dE_3 + dE_4 = \frac{W^\prime}{E_3E_4} pdp = \frac{W^\prime}{E_4} dE_3 .$$ (7.129)

Thus the factor

$$p_3\frac{dE_3}{E_4}\delta(E_3+E_4-E_1-E_2)$$ (7.130)

becomes

$$\frac{p}{W^\prime} dW^\prime \delta(W^\prime-W) = \frac{p}{W}$$ (7.131)

and we arrive at the important result

$$\textrm{dLips}(s;p_3,p_4) = \frac{1}{(2\pi)^2} \frac{p}{W} d\Omega ,$$ (7.132)

for two-body phase space in the CM frame.