Velocity-averaged Ramsey line ($\overline{P}(\delta )={\int }_0^{\infty }f(v)P(\delta ;\ell /v,L/v)dv$)

Let each drive region have length $\ell$ and the drive-free gap have length $L$, measured edge to edge. A particle of constant speed $v>0$ experiences $\tau (v)=\ell /v$ and $T(v)=L/v$. For a normalized detected-beam speed probability density function $f(v)$, sum the transition probabilities, not the probability amplitudes of different particles:The displayed $f$ is an ideal effusion flux distribution, with speed scale $u$, uniform transmission and speed-independent detection efficiency. It is not a claim about any particular clock's selected beam. The substitution $x=(v/u{)}^2$ gives $f(v)dv=x{e}^{-x}dx$, proving normalization. Actual apertures, state selection and readout can change the effective weights.

For an original example choose $\ell =0.10\mathrm{m}$, $L=1.00\mathrm{m}$, $u=v_{\ast }=200\mathrm{m}{\mathrm{s}}^{-1}$ and $\Omega =2\pi \times 500\mathrm{rad}{\mathrm{s}}^{-1}$. The reference particle has ${\tau }_{\ast }=0.500\mathrm{ms}$, $T_{\ast }=5.00\mathrm{ms}$ and resonant $\pi /2$ pulses. The instantaneous-pulse width would be $100\mathrm{Hz}$. The exact finite-pulse width for that single speed is about $88.7\mathrm{Hz}$; the chosen distribution also reduces the on-resonance average to approximately $0.738$ because not every particle receives a $\pi /2$ pulse.

Different velocities smear side fringes because $\delta L/v$ varies across the beam. Nevertheless every term is even in $\delta$, so this symmetric ideal-model average has no odd line asymmetry that could displace the central lock point. Reduced contrast or altered width alone is not evidence of a shifted line center. A speed-dependent transition shift or unequal drive quantum phases changes that conclusion; atomic clock frequency shifts explains why these distinctions matter experimentally.