I’ve checked baryon no., lepton no., charge, strangeness and isospin, everything seems to be conserved which would suggest it’s an allowed strong interaction, but I feel like it’s wrong. I tried to check charge conjugation conservation since it’s a strong interaction but can’t find the charge conjugation of the proton antiproton pair. Is this process allowed?
$\begingroup$
$\endgroup$
3
-
$\begingroup$ You mean the infamous π(1800)? $\endgroup$– Cosmas ZachosCommented Apr 3, 2020 at 20:15
-
1$\begingroup$ Hahaha, not sure, why is pi(1800) infamous? $\endgroup$– XimenezCommented Apr 3, 2020 at 20:37
-
$\begingroup$ PDG. $\endgroup$– Cosmas ZachosCommented Apr 4, 2020 at 0:37
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
In the center-of-mass frame of the collision, the total momentum is zero. Therefore, the neutral pion must be produced at rest in this frame, or else the interaction violates momentum conservation. But the neutral pion's mass is lower than the mass of the proton and antiproton, so producing it at rest violates energy conservation. Therefore, the interaction is forbidden, as is any interaction of the form
$$A+B\to C$$
where $m_C<m_B+m_A$.
answered Apr 3, 2020 at 19:39
-
1$\begingroup$ That’s perfect, now I get it. Thank you! $\endgroup$– XimenezCommented Apr 3, 2020 at 19:54
-
$\begingroup$ Though this can happen if the total energy of $A$ and $B$ (including kinetic energy) is greater than $m_C$. $\endgroup$ Commented Apr 3, 2020 at 20:00
$\begingroup$
$\endgroup$
1
It's hard to know what you're asking. Protons, of course, are not fundamental particles; a proton is $(u u d)$ and a $\pi^0$ is $\frac{1}{\sqrt{2}}(u \bar{u} - d\bar{d})$, so you can't have $p\, \bar{p} \to \pi^0$ with no other products.
However, $\pi^0$s are produced all the time in $p\, \bar{p}$ collisions such as those in the LHC, so it is certainly possible.
-
$\begingroup$ Okay, I see now what you mean, cheers! $\endgroup$– XimenezCommented Apr 3, 2020 at 20:08