= Stern-Gerlach force and flight model
{c}
{title2=$F_z\simeq\mu_z\partial_zB_z,\quad z_s-z_0=\mu_zG\ell(D+\ell/2)/(Mv^2)$}
Let the <atomic beam> travel along $x$, with <velocity> $v>0$, and let $z$ measure displacement along the dominant <magnetic field> <gradient>. For a fixed effective <magnetic moment> component $\mu_z$, the <magnetic dipole potential energy> is $U\simeq-\mu_zB_z$. Taking its negative <partial derivative> gives
$$
F_z=-\frac{\partial U}{\partial z}\simeq\mu_z\frac{\partial B_z}{\partial z}=\mu_zG.
$$
A spatially uniform <magnetic field> can produce <torque> and <Larmor precession>, but cannot produce this translational separation. In a <quantum state> following a local internal <energy eigenvalue> $E_n(B)$, the more general expression is $F_z=-\partial_zE_n$, with effective <magnetic moment> $-\partial E_n/\partial B$. Treating that moment as constant is an approximation with a physical regime, not an extra measured fact. Assume the local <magnetic field> direction stays nearly fixed across the sampled beam, and that internal states follow it without transitions, as in the <adiabatic theorem>. The scalar <gradient> model describes the beam's neighborhood; it omits the full <magnetic field>'s transverse components and fringe regions.
Assume a constant $G$ over magnet length $\ell$, followed by a field-free drift $D$. Neglect transverse motion at entry and changes in longitudinal <velocity>. <Newton's second law> gives transverse <acceleration> $a_z=\mu_zG/M$ for whole-atom <mass> $M$. The magnet transit time is $t_m=\ell/v$, and the drift time is $t_d=D/v$. The magnet supplies both a displacement and an exit transverse <velocity>:
$$
z_m-z_0=\frac{a_zt_m^2}{2},\qquad
v_{z,m}=a_zt_m,\qquad
z_s-z_0=\frac{\mu_zG}{Mv^2}\left(\frac{\ell^2}{2}+\ell D\right).
$$
The $\ell D$ contribution is motion with the acquired transverse <velocity>, not continued <acceleration> outside the magnet. For the plate-at-exit geometry, set $D=0$. For a nonzero entrance transverse <velocity> $w$, add $w(\ell+D)/v$. These terms show why initial <collimation> and flight geometry matter when inferring a <magnetic moment> from a deposit.
Choose hypothetical parameters $M=1.79\times10^{-25}\,\mathrm{kg}$, $G=300\,\mathrm{T\,m^{-1}}$, $\ell=0.035\,\mathrm m$, $D=0.100\,\mathrm m$, and $v=500\,\mathrm{m\,s^{-1}}$. Approximate the two effective <magnetic moments> by $\mu_z=\pm\mu_B$, where the <Bohr magneton> is $\mu_B\simeq9.274\times10^{-24}\,\mathrm{J\,T^{-1}}$. Then $|a_z|\simeq1.55\times10^4\,\mathrm{m\,s^{-2}}$, $t_m=70\,\mu\mathrm s$ and $t_d=200\,\mu\mathrm s$. The positive displacement is about $0.0381\,\mathrm{mm}$ inside the magnet plus $0.2176\,\mathrm{mm}$ during drift: $d=0.2557\,\mathrm{mm}$, with full separation $2d=0.5114\,\mathrm{mm}$. These chosen values illustrate the model and are not reconstructed 1922 measurements.
The exit transverse <velocity> is only about $1.09\,\mathrm{m\,s^{-1}}$, small compared with the chosen forward <velocity>; this checks the small-deflection assumption. Halving $G$ halves $d$, doubling $v$ divides $d$ by four, and setting $D=0$ removes most of this example's separation. A real pole has fringe <magnetic fields> and a position-dependent <gradient>; these require a more complete trajectory calculation.
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