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From part (i), the steady mean has $\coth \Lambda -1/\Lambda =1-1/\Lambda +O(e^{-2\Lambda })$. Hence $B{D}_r={\Lambda }^{-1}\to 0$ gives ${\mathbf{V}}_c\to V_s\mathbf{k}$. Part (ii) gives the same mean as $t\to \infty$. ThusBoth limiting distributions concentrate weakly at the upward pole. A common limiting mean does not identify their finite-time distributions or the order in which the limits are taken.

Time scales must be compared at a specified accuracy. The exact deterministic solution approaches the limiting mean with deficit $(4t/B-2)e^{-2t/B}$: its alignment scale is $B$, with exponential late scale $B/2$. For weak but nonzero diffusion, linearize about the upward direction using the two transverse tilt components $\mathbf{q}$. They obey the Ornstein-Uhlenbeck processThe stationary angular variance is $2B{D}_r$, and nearby axisymmetric variance perturbations decay on $B/2$ to leading weak-noise order. This is weak-noise relaxation of gyrotactic alignment: the aligning torque, not the pure rotational-diffusion time $D_r^{-1}$, controls the weak-noise alignment rate.

Nevertheless the full relaxation times are not identical. Finite diffusion leaves mean deficit approximately $B{D}_r$, whereas deterministic alignment has no such floor. Starting from isotropy, reducing the transient deficit to that tiny floor requires $(4t/B-2)e^{-2t/B}\sim B{D}_r$, orwith smaller logarithmic corrections. The two limits have the same mean and the same order-$B$ weak-noise alignment mechanism, but not identical transients or accuracy-dependent settling times. No finite time achieves the exact deterministic point-mass limit; positive diffusion never gives that exact point mass at all.