Distinguishing active motion at the same variance

How long does one push persist?

Describe a real mode by displacement $x$ , stiffness $K$ and friction $\zeta$ . Elastic relaxation has rate $\lambda=K/\zeta$ . Thermal noise comes from a bath of known temperature. Add a zero-mean active force with memory time $\tau_a$ .

An Ornstein–Uhlenbeck process generates this force by relaxing toward zero under random kicks. Its force variance $F_a^2$ and memory time are independent driving parameters. Specify them from a driving model or an independent measurement.

\zeta\dot x=-Kx+f_a+F_{\rm ext}+\xi_T,\qquad \langle\xi_T(t)\xi_T(0)\rangle=2\zeta k_BT\,\delta(t)

Every term on the right has force units. White-noise correlation uses a Dirac delta to represent no finite memory on the modeled time scale.

\langle f_a(t)f_a(0)\rangle=F_a^2e^{-|t|/\tau_a}

Active and thermal forcing are independent here. Spatial correlation, feedback and non-Gaussian forcing would require additional assumptions.

Source: B. Loubet

Match the variances exactly

Set the stationary mean evolution of $x^2$ and $xf_a$ to zero. The resulting variance has a thermal part $k_BT/K$ plus an active part controlled by force strength and memory. At fixed force variance, sufficiently fast kicks cancel before the membrane can respond.

Define a passive comparison with the same $K,\zeta$ and temperature $T_{\rm eff}=K\langle x^2\rangle/k_B$ . It has exactly the same theoretical variance. Temporal correlation provides an additional test of this variance match.

\langle x^2\rangle=\frac{k_BT}{K}+\frac{F_a^2}{K(K+\zeta/\tau_a)}

This stationary result follows from the specified linear dynamics, with no equilibrium assumption for the active noise.

A small push gives an independent observable

Apply a small force $F$ at time zero and hold it. The mean displacement change divided by force is the step response $\chi(t)$ . Zero-mean forcing drops out of the linear mean equation, so changing active variance leaves this deterministic response unchanged.

At thermal equilibrium, correlation and response obey an exact relation. Calculate or measure response independently before comparing them. Generating a supposed active response by inserting active correlations into an equilibrium formula would make the proposed test circular.

\chi(t)=\frac{\Delta\langle x(t)\rangle}{F}=\frac{1-e^{-\lambda t}}{K}

This response comes from the mean dynamics with an applied force small enough to remain in the linear range.

\chi_{\rm eq}(t)=\frac{C(0)-C(t)}{k_BT},\qquad C(t)=\langle x(t)x(0)\rangle

This form assumes stationary thermal equilibrium and a coordinate conjugate to the applied force. Test active correlations against independently obtained response at multiple times.

Worked example with equal variance and different memory

Choose dimensionless units with $K=\zeta=k_BT=1$ , $\tau_a=2$ and $F_a^2=1.5$ . Thermal variance is one and active excess variance is one, for a total of two. A passive comparison at $k_BT_{\rm eff}=2$ also has variance two.

Solving the two-variable linear correlation equations gives $C_{\rm active}(t)=2e^{-t/2}$ for these parameters. The hotter passive process has $C_{\rm hot}(t)=2e^{-t}$ . At $t=1$ the values are $1.21$ and $0.736$ . Equal fluctuation amplitude hides different memory.

Both independent step responses are $1-e^{-1}\simeq0.632$ . Treating the active process as equilibrium at temperature two would instead predict $(2-1.21)/2\simeq0.393$ . The mismatch tests the constant-temperature explanation against a second observable.

The Chapter 8 example has equal variance but different memory. The two lines are analytic model correlations in dimensionless units; response is calculated independently.

Return to the active membrane

Sciortino and colleagues encapsulated microtubules and molecular motors in vesicles to study traveling membrane deformation. Read spatial scales, time correlations and internal-driver images together. Large amplitude by itself does not uniquely specify a driving mechanism.

The animated contour combines linear spherical modes; the correlation–response diagnostic selects one real mode. It does not solve a microtubule network or self-consistent area–tension feedback. The fixed-excess-area theory of Loubet and colleagues additionally solves an area condition for tension. Recalculating that branch requires solving the area condition after changing activity.

Source: A. Sciortino et al. · B. Loubet

Distinguishing active motion at the same variance

How long does one push persist?

Match the variances exactly

A small push gives an independent observable

Worked example with equal variance and different memory

Return to the active membrane

Try the reasoning

Separate equal-variance processes

Exercises and explanations

Sources and calculated scope