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Consider a fixed-target-experiment, where negatively charged pions are shot at protons (the latter being at rest). The kinetic energy of the pions shall be known. One possible reaction is

$$ \pi^- + p \to n + \pi^+ + \pi^- $$

creating a neutron. The neutron apparently remains undetected by measuring devices, and only one of the pions can be measured.

I have two conceptual questions:

(1) Why is this? (I mean, both pions carry charge...)

(2) If the neutron cannot be measured, then how can we be sure that it is there?

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  • $\begingroup$ I presume that the $\pi^{+}$ is the pion that is measured? Note the large flux of $\pi^{-}$ flying around trying to drive the reaction in the first place makes measuring it as a reaction product rather difficult. $\endgroup$
    – Jon Custer
    Commented May 7, 2021 at 14:37
  • $\begingroup$ you should give a link for the experiment. In a bubble chamber both outgoing pions would be seen and the simple answer is the neutron is found by energy and momentum conservation, the invariant mass of the four vector being the mass of the neutron. (few beams enter a bubble chamber) $\endgroup$
    – anna v
    Commented May 7, 2021 at 14:40

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Exclusive & semi-inclusive reactions are commonly as written target(beam, detected)undetected:

$$p(\pi^-, \pi^+)n\pi^-$$

To understand the detection of the final state, we really need more information. While there are detectors that can handle multiple particle final states (https://www.jlab.org/physics/hall-b/clas), this experiment wasn't done there.

More commonly, fixed target experiments put a beam on target and look at reaction products with a spectrometer (e.g., SLAC's 8 GeV spectrometer: https://www.slac.stanford.edu/pubs/slacpubs/5750/slac-pub-5753.pdf). This grabs a small bite of both angular space and momentum space. It uses (a) dipole magnet(s) to select a central momentum, and quadruple magnets to focus particle tracks so that momentum and lab-angle (regardless of out-of-plane angle) can be determined.

The spectrometer can only look a one charged final state, but that can be reversed by reversing the current. Since the beam is $\pi^-$, the $\pi^+$ final state is much cleaner.

The way to separate the reaction from:

$$p(\pi^-, \pi^+)X$$

is the invariant mass of $X$. The initial state's 4-momentum is:

$$p_i^{\mu} = p_p^{\mu} + p_p^{\pi^-} $$ $$p_i^{\mu} = (M_p, 0,0,0) + (E_{\pi^-}, 0,0,p_{\pi^-}) $$

while the final state is:

$$p_f^{\mu} = p^{\mu}_{\pi^+} + p_X^{\mu} $$

Of course, $p_i^{\mu}= p_f^{\mu}$ so that:

$$p_X^{\mu} = p_i^{\mu} - p^{\mu}_{\pi^+} $$

$$p_X^{\mu} = (M_p+E_{\pi^-}, 0,0, p_{\pi^-}) - (E_{\pi^+}, p_{\pi^+}\sin{\theta},0, p_{\pi^+}\cos{\theta}) $$

If you work that all out at a fixed angle, there should be a momentum region that ensures $X$ is a nucleon plus a pion, and since it has charge -1, it has to be a neutron and $\pi^-$. (Another reason to detect the positive pion, as a negative pion in the final state does NOT distinguish between an undetected $n\pi^+$ and $p\pi^-$). That also rules out a lot of other final states, so by detecting the $\pi^+$, there is a kinematic region where $X=n\pi^-$ is the only viable option.

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  • $\begingroup$ Thanks a lot for this thorough explanation. :D $\endgroup$ Commented May 7, 2021 at 21:15
  • $\begingroup$ You're welcome. I didn't actually work out the kinematics, but in my favorite reaction, $d(\gamma, p)n$, which is simpler, the detected proton has more momentum than the electron beam that produced the $\gamma$, which is why I suspect there is similar kinematics. Also: I for got to mention the meson factory at Los Alamos. It was called "LAMPF"...it was a pion beam. $\endgroup$
    – JEB
    Commented May 8, 2021 at 14:38

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