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I have a questions about the article Particle decay two-body decay.

$$|\vec{p}_1| = |\vec{p}_2| = \frac{[(M^2 - (m_1 + m_2)^2)(M^2 - (m_1 - m_2)^2)]^{1/2}}{2M}$$

$$(M, \vec{0}) = (E_1, \vec{p}_1) + (E_2, \vec{p}_2)$$

In spherical coordinates

$$d^3 \vec{p} = |\vec{p}\,|^2\, d|\vec{p}\,|\, d\phi\, d\left(\cos \theta \right)$$

I wanted to know how using delta function to perform $d^3\vec{p_{2}}$ and $d |\vec{p_{1}}|$ was calculated $d\Gamma$, which is equal to

$$d\Gamma = \frac{ \left| \mathcal{M} \right|^2}{32 \pi^2} \frac{|\vec{p}_1|}{M^2}\, d\phi_1\, d\left( \cos \theta_1 \right)$$

How then integrate to calculate decay width $\Gamma$? I'll be grateful for step by step explanation.

Klaudia Łapa
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1 Answers1

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It's a straight plug-in! Hint: Your text must have taught you how to evaluate the Lorentz-invariant phase space, of course. The rightmost-hand side of your top equation is what the energy conservation constraint dictates, a number, let's call it a.

Integrate $\vec p_2$ out given its δ-function, and call $|\vec p_1|=p$, and utilize $E_i=\sqrt{m^2_i+ p^2}\leadsto E'_i=p/E_i(p)$ to transform the energy δ-function, mindful of your $E_1(a)+E_2(a)=M$, $$ {d\phi_1 d\cos (\theta_1) \over 16 \pi^2 }\int\!\!dp~~p^2 {\delta (E_1(p)+E_2(p)-M)\over E_1 E_2} \\ = {d\phi_1 d\cos (\theta_1) \over 16 \pi^2 }\int\!\! dp~~a {\delta (p-a)\over E_1(a)+ E_2(a)}= {d\phi_1 d\cos (\theta_1)\over 16 \pi^2 }~~ { a\over M}. $$

Cosmas Zachos
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