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For a freely moving unstable particle of fixed speed $v=\beta c$ and proper mean lifetime ${\tau }_0$, the mean travelled distance before decay is ${\ell }_d=v\gamma {\tau }_0$. Survival beyond distance $L$ has probability $e^{-L/{\ell }_d}$, since elapsed proper time is $L/(\gamma v)$. For rest mass $m$ and three-momentum magnitude $p$, ${\ell }_d=p{\tau }_0/m$.

The proper mean lifetime of a free muon, approximately $2.2\mu \mathrm{s}$. Independent particle decays give survival probability $\exp (-\Delta \tau /{\tau }_{\mu })$. In a frame where the Lorentz factor is constant, $\Delta \tau =\Delta t/\gamma$; interactions of a stopped negative muon in matter can add capture channels.

The laboratory is the parent's rest frame. In this relativistic two-body decay, conservation of momentum gives equal and opposite daughter momenta, of magnitude $p$. Their total relativistic energies satisfySubtracting the relativistic energy-momentum relations eliminates $p$:Solving the two linear equations givesSince three-momentum obeys $p{c}^2=Ev$, the daughter speeds areTheir directions are opposite; the initial state does not specify their common axis. The kinetic energies are $K_1=5m{c}^2/8$ and $K_2=3m{c}^2/8$. Their sum $m{c}^2$ is the loss of rest energy when the total daughter rest mass is $3m$ instead of $4m$.

The proper mean lifetime ${\tau }_0$ is the mean lifetime measured in the unstable particle's own rest frame, or equivalently the mean elapsed proper time before decay. In the exponential decay model, its survival probability to proper time $\tau$ is $e^{-\tau /{\tau }_0}$. For the daughter of rest mass $m$, the Lorentz factor is ${\gamma }_1=E_1/(m{c}^2)=13/8$, so time dilation gives a laboratory mean lifetime of ${\gamma }_1{\tau }_0=3.25\mu \mathrm{s}$.

At constant speed $v_1$, travelling a distance $L$ requires laboratory time $L/v_1$ and proper time $L/({\gamma }_1{v}_1)$. Its mean decay length isfor ${\tau }_0=2\times 10^{-6}\mathrm{s}$. HenceThe mean lifetime sets the scale of an exponential distribution; it is not a fixed deadline, so travelling beyond one mean decay length has a nonzero probability.