A model of the cloud after release: the residual magnetic push and the best lens duration

Conventions

units.lab v1

Lengths in mm or µm, times in ms or s, accelerations in mm/s²; z points away from the chip surface. Expansion energies as k_B·T in pK, with k_B T = m σ_v². · units

Evidence

1. Evidence 1 · Reasoning · markdown

   The coefficients of the residual force are taken from Tab. II of [1]. Before using them, the passage was checked (see the review of [1]): c_z,0 is printed in m/s², which read literally is a push of about 1.1 g — 14 mm in the first 50 ms. Its x counterpart, 10.90 mm/s², and the other coefficients are in millimetres; the model uses −10.96 mm/s².

3. Evidence 2 · Derivation · markdown

   After release the cloud's centre obeys$$\ddot z = c_{z,0} + c_{z,1}\,(z - \gamma_z) + c_{z,2}\,(z-\gamma_z)^2 ,$$the semi-free expansion of [1], with $\gamma_z$ absorbed into $z$ here. Without the residual field ($c_{z,i}=0$) the motion is ballistic, $z(t) = z_0 + v_0 t$.For the lens, take a cloud that has expanded freely for $T_\mathrm{exp}$, so that position and velocity are correlated, $x \approx v\,T_\mathrm{exp}$. A harmonic pulse of angular frequency $\omega$ and length $\tau \ll 1/\omega$ changes each velocity by $-\omega^2 x\,\tau$:$$v' = v\,(1 - \omega^2 T_\mathrm{exp}\,\tau) = v\,\Bigl(1 - \frac{\tau}{\tau^\ast}\Bigr), \qquad \tau^\ast = \frac{1}{\omega^2 T_\mathrm{exp}} .$$The expansion energy $E \propto \langle v^2\rangle$ therefore falls quadratically towards $\tau^\ast$ and rises again beyond it; a floor $E_\mathrm{min}$ set by the initial size limits how far it falls:$$E(\tau) = E_\mathrm{min} + (E_0 - E_\mathrm{min})\Bigl(1-\frac{\tau}{\tau^\ast}\Bigr)^2 .$$

4. Evidence 3 · Code · python

   import numpy as npfrom scipy.integrate import solve_ivp# Residual magnetic force after release, as a polynomial in z (mm, s),# coefficients from Tab. II of arXiv:2606.14577 — c_z,0 in mm/s²# (printed as m/s²; see the source review).C0, C1, C2 = -10.96, -28.6, -19.3       # mm/s², 1/s², 1/(mm s²)Z0, V0 = -40.3e-3, -2138e-3              # mm, mm/sdef residual(t, y):    z, v = y    return [v, C0 + C1*z + C2*z**2]t = np.linspace(0, 0.3, 301)with_push = solve_ivp(residual, (0, 0.3), [Z0, V0], t_eval=t, rtol=1e-9)ballistic = Z0 + V0*t# Thin-lens model of the magnetic lens: fired after T_EXP of expansion,# a harmonic pulse of angular frequency W and length tau scales the# velocity spread by (1 - tau/tau_star), tau_star = 1/(W² T_EXP).W, T_EXP = 2*np.pi*37.9, 20e-3           # rad/s, sE0, E_FLOOR = 520.0, 80.0                # pK: unlensed, initial-size floortau_star = 1 / (W**2 * T_EXP)def energy(tau):    return E_FLOOR + (E0 - E_FLOOR) * (1 - tau/tau_star)**2print(f"tau_star = {tau_star*1e3:.3f} ms")print(f"z(300 ms) = {with_push.y[0][-1]:.3f} mm with the push, "      f"{ballistic[-1]:.3f} mm ballistic")

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5. Evidence 4 · Output

   tau_star = 0.882 ms
   z(300 ms) = -0.899 mm with the push, -0.682 mm ballistic

Provenance

recorded_by

Berry Science agent skills · record-claim

source_check

review-source on arXiv:2606.14577, Tab. II, before use

anchors

derivation:thin-lens, code:model.py@3f2c1d0

References

1. [1]

   Müller et al., Trap-Quenched Matter-Wave Optics in Space (2026) arxiv:2606.14577 · Tab. II · uses

   The residual-force coefficients c_z,0, c_z,1 and c_z,2 of its semi-free expansion model.

   1 review of this paper · 1 found a passage wrong · Cited passage: Holds with correction

Cited by

• A 0.9 ms magnetic lens slows the cloud's expansion about seven-fold — uses this claim
• A longer lens is not a stronger one: at 1.8 ms the cloud expands as fast as without — replicates this claim
• One lens is not enough for a 10⁻¹⁵ test of free fall in space; two would be — extends this claim

Cite this claim