# Maximum Momentum of Neutral Pion

I'm considering the reaction $p + p \rightarrow p + p + \pi^0$. To find the maximum momentum that the $\pi^0$ can have after this reaction in the center-of-mass frame, what I am doing is assuming the two protons both get kicked off in the opposite direction of the neutral pion, thus giving the maximum momentum to the neutral pion (and the two-proton system). To get this momentum, I'm treating this as a two-body decay problem. I set the 'rest mass' of the initial system to be the center-of-mass energy, $\sqrt{s}$, the mass of one of the products to be $2m_p$ (twice the mass of a proton), and the other products mass to be $m_{\pi}$ (mass of neutral pion). Using the standard equation for this situation, the maximum momentum would then be

$$p_{max}=\frac{1}{2\sqrt{s}}\left[(s-(2m+m_{\pi})^2)(s-(2m-m_{\pi})^2)\right]^{1/2}$$

However, in a reference book I am using, the equation is supposed to be

$$p_{max}=(\sqrt{s}-2m-m_{\pi})(\sqrt{s}-2m+m_{\pi})$$

I can't see how to get to the answer in my book. First off, I don't see how the units even make sense ($p\propto$ units mass, which isn't the case in the book's answer). Second off, I don't see how my assumptions could be wrong.

Could you please help me out here?

• What reference book? Commented Apr 6, 2015 at 1:19
• Also, if $s\propto p^2$, then wouldn't $s^2\propto p^4$? So unless I'm missing something, how are you adding $s^2$ to $\sim m^2$? Commented Apr 6, 2015 at 1:25
• Oh I'm sorry about the incorrect order. I meant to put $(\sqrt{s})^2$. Commented Apr 6, 2015 at 1:30
• My book is Introduction to Mathematical Physics - Chun Wa Wong (Second edition). Commented Apr 6, 2015 at 1:31
• I don't really know if my assumption for the best case scenario (i.e. most momentum for $\pi^0$ is valid). Commented Apr 6, 2015 at 1:40

## 1 Answer

You're right that the book's answer seems to have the wrong units; most likely the author meant to say $p_{max}^2$ in place of $p_{max}$. Also, both instances of $2m$ in the book's formula should really be $2m_p$.

Assuming these two typos are resolved as just mentioned, it's clear what error the book is probably making. The book seems to be assuming that all the energy besides the mass of the two protons, i.e., $\sqrt{s} - 2m_p$, goes into the pion. This would then give the pion the momentum given by the equation, since we can write $E^2 - m_\pi^2$ as $(E+m_\pi)(E-m_\pi)$, where $E = \sqrt{s} - 2m_p$.

Unfortunately, this does not conserve momentum since the pion momentum is unbalanced.

Conclusion: the book is wrong and you are right (your answer checks out).