= Solution
\b[Experimental evidence]. <Tests of special relativity> check linked predictions about light propagation, clocks and <four-momentum>. Several independent kinds of experiment support those predictions.
* In the <Michelson-Morley experiment>, light follows perpendicular paths in an <interferometer>. Rotating the apparatus would change their round-trip travel-time difference in a classical stationary-aether model. The expected orientation-dependent <interference fringe> shift was absent at the predicted scale, consistent with isotropy of the round-trip <speed of light> in <special relativity>. This experiment alone does not distinguish special relativity from every alternative theory.<On the Relative Motion of the Earth and the Luminiferous Ether by Albert Abraham Michelson and Edward W. Morley>[1887 report, apparatus and predicted versus observed orientation-dependent fringe shifts]{ref}
* The <Fizeau experiment> compares light propagation with and against flowing <water> using an <interference fringe> displacement. Neglecting <optical dispersion>, the <velocity-addition formula> predicts
$$
v_{\rm lab}=\frac{c/n+u}{1+u/(nc)}
\simeq\frac cn+u\left(1-\frac1{n^2}\right).
$$
The observed partial-drag coefficient agrees with this first-order prediction rather than simple Galilean addition of $u$. It tests a different consequence of the <Lorentz transformation> from a particle-lifetime measurement.<The Hypotheses Relating to the Luminous Aether by Hippolyte Fizeau>[1851 report, counter-propagating beams, reversing water flow and resulting fringe displacements]{ref}
* A controlled <muon storage-ring time-dilation experiment> measured circulating <muons> at $\gamma=29.33$. Its measured <mean lifetimes> were about $64.4\,\mathrm{\mu s}$, compared with about $2.2\,\mathrm{\mu s}$ at rest, consistent with $\tau_{\rm lab}=\gamma\tau_0$. This tests <time dilation> with accelerator-selected particles.<Measurements of relativistic time dilatation for positive and negative muons in a circular orbit by J. Bailey and others>[1977 abstract, Lorentz factor and positive- and negative-muon lifetimes]{ref}
\b[Atmospheric muons and the two frame descriptions]. Atmospheric <cosmic rays> create unstable particles whose decay chains produce <muons>. A stopped-muon experiment measures the time between a stopping signal and the subsequent decay electron, obtaining the proper <mean lifetime> $\tau_0\simeq2.2\,\mathrm{\mu s}$. The associated light-travel distance is only $c\tau_0\simeq660\,\mathrm m$, yet substantial numbers of energetic atmospheric <muons> reach detectors after travelling several kilometres.
For an ideal selected cohort produced height $h$ above a detector and moving downward at constant <speed> $v$, the flight time in the terrestrial frame is $h/v$. <Time dilation> predicts a laboratory <mean lifetime> $\gamma\tau_0$, hence survival fraction
$$
\boxed{\frac{N(h\to0)}{N(h)}=\exp\!\left(-\frac{h}{\gamma v\tau_0}\right),\qquad
\gamma=\frac1{\sqrt{1-v^2/c^2}}.}
$$
Without <time dilation>, the predicted exponent would be $-h/(v\tau_0)$ and the surviving population much smaller. Altitude comparisons constitute a <cosmic-ray muon time-dilation experiment>: select comparable <momenta> or <energies> and account for detector acceptance, atmospheric energy loss and backgrounds before interpreting the count ratio.
For illustration, not as a claimed measurement, take $h=10\,\mathrm{km}$ and $v=0.998c$. Then $\gamma=15.82$, the laboratory flight time is $33.4\,\mathrm{\mu s}$, and the predicted surviving fraction is about $0.38$. Omitting <time dilation> would give only about $2.5\times10^{-7}$ on the same idealized constant-speed model.
In the <muon> <rest frame>, its decay still follows $e^{-\Delta\tau/\tau_0}$. The atmosphere and ground move upward. Their separation, measured simultaneously in that frame, is shortened by <length contraction>:
$$
h'=\frac h\gamma,\qquad
\Delta\tau=\frac{h'}v=\frac{h}{\gamma v}.
$$
The same survival fraction follows. The terrestrial description uses a longer-lived moving particle; the muon-frame description uses a shorter atmospheric flight path. They describe the same production and detection events. This consistency illustrates both <time dilation> and <length contraction>, while the survival measurement is not two independent experimental tests merely because it can be described in two frames.
\b[Energy-momentum invariant]. For a particle with rest <mass> $m$,
$$
\boxed{E^2-p^2c^2=m^2c^4.}
$$
Since $E=mc^2+K$, substitution and cancellation of $m^2c^4$ give
$$
(mc^2+K)^2-p^2c^2=m^2c^4,
\qquad
\boxed{K^2+2mKc^2=p^2c^2.}
$$
This <relativistic energy-momentum relation> reduces to $K\simeq p^2/(2m)$ when $K\ll mc^2$.
\b[Laboratory beam energies]. If both final <muons> are at rest in the laboratory, their total <momentum> is zero and their total <energy> is $2m_\mu c^2$. <Four-momentum conservation> therefore requires equal and opposite incoming <momenta>. Because the <Electron> and <Positron> have equal rest <masses>, they then have equal total <energies>:
$$
E_-=E_+=m_\mu c^2.
$$
Subtract each particle's own rest <energy>:
$$
\boxed{K_-=K_+=(m_\mu-m_e)c^2.}
$$
Using the usual handbook rest energies gives approximately $105.1\,\mathrm{MeV}$ for each beam. These are the ideal kinematic threshold values. Just above threshold, the pair has small, equal and opposite <momenta> in this laboratory frame.
\b[Positron momentum]. Apply the <relativistic energy-momentum relation> to $E_+=m_\mu c^2$ and use its stipulated negative-$x$ direction:
$$
p_x^2c^2=m_\mu^2c^4-m_e^2c^4,
\qquad
\boxed{p_x=-c\sqrt{m_\mu^2-m_e^2}.}
$$
The <Electron> has the opposite momentum, $p_{e,x}=+p_0$ with $p_0=c\sqrt{m_\mu^2-m_e^2}$.
\b[Boost to the electron rest frame]. The electron's <velocity> is $v=p_{e,x}c^2/E_-=p_0/m_\mu>0$. Thus $S'$ moves in the positive-$x$ direction, and its <Lorentz factor> satisfies
$$
\gamma=\frac{E_-}{m_ec^2}=\frac{m_\mu}{m_e}.
$$
The positron has the same laboratory total <energy> as the electron, so $vE_+/c^2=vm_\mu=p_0$. The supplied <Lorentz transformation> then gives
$$
p'_x=\gamma\left(p_x-\frac{vE_+}{c^2}\right)
=\gamma(-p_0-p_0),
\qquad
\boxed{p'_x=-2\gamma c\sqrt{m_\mu^2-m_e^2}.}
$$
As a sign check, the same boost gives $p'_{e,x}=\gamma(p_0-vE_-/c^2)=0$, as required for the <Electron> <rest frame>.
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