= Pion decay at rest in emulsion
{title2=$\pi^+\to\mu^++\nu_\mu$}
<Cecil Frank Powell>'s 1950 lecture illustrates the successive decays of <pions> and <muons> through their tracks and explains why the nearly fixed secondary <muon> range, allowing for <range straggling>, supports a single neutral decay partner. He describes its identification as a <neutrino> tentatively. The neutral particle itself leaves no recorded track. He contrasts this decay with the <nuclear reactions> caused by stopped negative <pions>, connecting the latter with the <nuclear force>.<The Cosmic Radiation by Cecil Frank Powell>[Powell, The cosmic radiation, Nobel lecture, 11 December 1950, printed pp. 151–152, Fig. 5 and paragraphs beginning “When brought to rest” and “When a negative”; inspected in the Columbia-hosted Nobel lecture volume, PDF pp. 151–152]{ref}
Here is a modern <special relativity> derivation for the dominant <leptonic pion decay>, assuming a free <pion> at rest, a massless <muon neutrino>, and no emitted <photon>. Write $m_\pi$ and $m_\mu$ for the <pion> and <muon> masses. By <conservation of momentum> the daughter momenta are opposite with equal magnitude $p$. By <conservation of energy> and the <energy–momentum relation>,
$$
m_\pi c^2=E_\mu+pc,\qquad E_\mu^2=p^2c^2+m_\mu^2c^4.
$$
Substitute $E_\mu=m_\pi c^2-pc$, square, and cancel $p^2c^2$. This gives
$$
p=\frac{(m_\pi^2-m_\mu^2)c}{2m_\pi},\qquad
E_\mu=\frac{(m_\pi^2+m_\mu^2)c^2}{2m_\pi},\qquad
T_\mu=\frac{(m_\pi-m_\mu)^2c^2}{2m_\pi}.
$$
The resulting $p$ is fixed, although the <muon>'s direction is not set by the incoming stopping track. Thus the kink angle relative to that earlier track is not the angle between the two daughters: in the <pion> <rest frame>, the daughters are back to back. A neutral-partner arrow in a diagram represents inferred <momentum>, not an observed string of grains.
For an original calculation, adopt rounded input values $m_\pi c^2=139.57\,\mathrm{MeV}$ and $m_\mu c^2=105.66\,\mathrm{MeV}$. They give $pc=29.79\,\mathrm{MeV}$, $T_\mu=4.119\,\mathrm{MeV}$ and $E_\nu=29.79\,\mathrm{MeV}$. Their sum satisfies $T_\mu+E_\nu=(m_\pi-m_\mu)c^2=33.91\,\mathrm{MeV}$. These are rounded calculation inputs and outputs, not a fit to the 1947 plates.
\Image[/pion-rest-decay-kinematics-and-range-model.png]
{title=<Pion decay at rest in emulsion>: daughter <momentum> balance and a stipulated <particle range> response}
{description=Original <special relativity> calculation and synthetic <histograms>. The dashed <neutrino> arrow is an inference. For illustration only, the range response has arbitrarily chosen mean $1$ and <standard deviation> $0.06$ in dimensionless units; it does not reproduce historical grain measurements, a measured resolution or a calibrated muon range.}
At fixed material and particle identity, a fixed initial <kinetic energy> suggests a narrow <particle range> distribution broadened by <range straggling> and <measurement uncertainty>. The synthetic example convolves a single chosen range with a <normal distribution>; a second example superposes two such chosen ranges. It explains why repeated comparable stopping distances can be informative, without pretending to extract a historical spectrum.
The decay equations do not identify the neutral partner as a <neutrino> on their own: a massless <photon> would satisfy the same energy and momentum equations. Nor does any isolated kink establish the <pion>'s identity. The inference gains force from calibrated <particle range>, <track grain density>, repeated decay patterns and independent particle properties. Historical particle names should not obscure the distinction: a <muon> is a <lepton>, whereas a <pion> is a <meson>.
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