I am not following the use of delta function in the 3-body decay process.
In $\gamma^* \to gq\bar{q}$ process, with $\gamma^*$ being a virtual photon, we have a phase space factor $$d^9R_3 = \frac{d^3p_1}{(2\pi)^3(2E_1)}\frac{d^3p_2}{(2\pi)^3(2E_2)}\frac{d^3p_3}{(2\pi)^3(2E_3)},$$ where $p_1$, $p_2$, and $p_3$, are momentums of the three decay particles.
We define dimensionless energy fractions, $x_i = 2E_i/Q$, and neglecting the masses of the decay products we get $p_i\cdot p_j = \frac{1}{2}Q^2(1-x_k)$. ($q$ and $Q$ are the 4-momentum and CM energy of the incoming photon; also note $q^2=Q^2$)
Now integrating over 3-momentum of particle 3 gives
$$\int d^3p_3 \delta^4(q-p_1-p_2-p_3) = \delta (Q-E_1-E_2-E_3).$$
Now my two questions
i) we integrate over the directions of particle 1 & particle 2, the result is
$$\int \int \frac{d^3p_1}{2E_1}\frac{d^3p_2}{2E_2} = \frac{(4\pi)(2\pi)}{4}E_1dE_1 E_2dE_2\int_{-1}^1 dz$$ where $z=\cos \theta_{12}$ is the relative angle between particle 1 and 2.
I don't understand how this integral over $z$ comes out. If the Matrix amplitude does not have a $z$ dependence, the integral just becomes 2 and gives expected result; but I still do not get how this is derived.
ii) Apparently $$\frac{\delta(Q-E_1-E_2-E_3)}{2E_3} = \delta[(q-p_1-p_2-p_3)^2] = \delta[Q^2(1-x_1-x_2+\frac{1}{2}x_1 x_2(1-z))].$$
I don't fully understand how each of equality follows. How is index 3 quantity removed entirely?
Best Answer
part i) answer is just
$\int \int \frac{d^3p_1}{2E_1}\frac{d^3p_2}{2E_2} = \int\int \frac{E_1 dE_1 dcos(\theta_1) d\phi_1}{2}\frac{E_2 dE_2 dcos(\theta_2) d\phi_2}{2}$.
Given that we have in our integrand a piece that only depends on relative angle between $\vec{p_1}$ and $\vec{p_2}$, if we orient our axes in the direction of $\vec{p_1}$ and integrate over the angles of $\vec{p_1}$ and $\phi_2$, then the result follows.