Physics Asked by Philippe Knecht on June 29, 2021
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?
Exclusive & semi-inclusive reactions are commonly as written target(beam, detected)undetected:
$$p(pi^-, pi^+)npi^-$$
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 $npi^+$ and $ppi^-$). That also rules out a lot of other final states, so by detecting the $pi^+$, there is a kinematic region where $X=npi^-$ is the only viable option.
Correct answer by JEB on June 29, 2021
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