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In the formula for the decay width of $\Upsilon(4S)$ to B-mesons from $\text{e}^+\text{e}^-$ collisions:

$$\Gamma_{\Upsilon(4S)\to B\bar{B}}=\frac{\left|\underline{P}_B \right|}{8\pi M_{\Upsilon(4S)}^2}\left| \mathcal{M}_{\Upsilon(4S)\to B\bar{B}} \right|^2$$

Isn't the units of the LHS energy and the RHS inverse energy?

To make the $\Upsilon(4S)$ which has a mass of $10.58 \text{ GeV}$, in the centre of mass frame, $\sqrt{s}=10.58 \text{ GeV}$ at a minimum?

Say the B-mesons all have the same mass of approximately 5.28 GeV, and $\Gamma_{\Upsilon(4S)\to B\bar{B}}\approx20 \text{ MeV}$, doesn't this make the matrix element larger than one, which, as a probability squared is not allowed?

Maybe I have calculated $\underline{P}_B$ wrong?

In the comments section:

$$\Gamma=\frac{S \left|\underline{p} \right|}{8\pi \hbar cm_1^2}\left|\mathcal{M}\right|^2$$ I think there is an energy term that goes missing by setting $c=\hbar=1$

$\left|\underline{p}\right| = \frac{c}{2m_1}\sqrt{m_1^4+m^4_2+m_3^4-2m_1^2 m_2^2-2m_1^2m_3^2-2m_2^2m_3^2}$

And $m_2=m_3=m_B$, $m_1=M_{\Upsilon(4S)}$ and $S=1$. Any ideas?

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    $\begingroup$ Please mention where you got the formula from. I have some doubts about that formula. $\endgroup$
    – Siva
    May 9, 2013 at 17:19
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    $\begingroup$ Griffiths Introduction to Electrodynamics p. 429: $\Gamma=\frac{S\left|\underline{p}\right|}{8\pi\hbar m_1^2 c}\left|\mathcal{M}\right|^2$ $\endgroup$
    – shilov
    May 9, 2013 at 17:42

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