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A free particle following Langevin dynamics initialized at zero position and velocity has relaxation time $\tau =m/\zeta$ and thermal diffusivity $D=k_{B}T/\zeta$. Its position is the convolution of the random force with $[1-e^{-t/\tau }]/\zeta$. The fluctuation-dissipation relation for a Langevin particle gives force covariance amplitude $2\zeta k_{B}T$, and integrating the squared kernel yields the displayed variance. At short times it grows as $2D{t}^3/(3{\tau }^2)$ because the initial velocity is cold; at long times it grows as $2Dt$. A Maxwellian initial velocity produces a different short-time transient.

For a thermal inertial Langevin equation conditioned on fixed initial position and velocity, the centered displacement is $R_i={\zeta }^{-1}{\int }_0^t[1-e^{-\zeta (t-s)}]A_i(s)ds$. With the fluctuation-dissipation relation for a Langevin particle, $D=k_{B}T/(m\zeta )$ and integrating the squared kernel gives the displayed variance. At short times it is $2D{\zeta }^2{t}^3/3$ per component; at long times it is $2Dt+O(1)$. Averaging initial velocity over thermal equilibrium adds a ballistic contribution and gives $2D[t-(1-e^{-\zeta t})/\zeta ]$ per component.

Langevin dynamics describes a dynamical variable driven by deterministic forces and random noise. Underdamped Langevin dynamics retains inertia, whereas overdamped Langevin dynamics neglects it. At thermal equilibrium, the fluctuation-dissipation relation for a Langevin particle relates Gaussian white noise strength to drag and temperature.

Add the random thermal force to the deterministic optical trap relaxation time model. In one stationary coordinate of a harmonic optical trap, the overdamped Langevin equation isHere $\xi$ is Gaussian white noise understood through its integrals, $\delta$ is the Dirac delta function, and $\zeta$ is the linear friction coefficient. The covariance coefficient has units ${\mathrm{N}}^2\mathrm{s}$; multiplying by $\delta (t-t^{\prime })$, of units ${\mathrm{s}}^{-1}$, gives squared force. This fluctuation-dissipation relation for a Langevin particle makes the stationary variance ${\sigma }^2=k_{B}T/k$, consistent with the equipartition theorem. The solution is an Ornstein-Uhlenbeck process with autocovariance functionIndeed, at positive lag the noise has zero correlation with the initial position, so $\zeta C_x^{\prime }=-k{C}_x$; its initial value is ${\sigma }^2$. This supplies both a decay time and an amplitude, making a stronger calibration test than a position histogram alone.

Specify the Fourier transform convention before using a power spectral density. For frequency $f$ in hertz, define $S_x^{(2)}(f)={\int }_{-\infty }^{\infty }{C}_x(t)e^{-2\pi ift}dt$ and the one-sided density $S_x^{(1)}(f)=2S_x^{(2)}(f)$ for $f>0$. Integration of the exponential givesThe position power spectral density has units ${\mathrm{m}}^2/Hz$; its integral is a variance, not a standard deviation. At $f=f_c$ it is half its zero-frequency plateau. Angular frequency conventions require a corresponding $2\pi$ conversion in both axis and integration measure. If $x=\beta (V-V_0)$ with position sensitivity $\beta$ in $m/V$, then $S_V^{(1)}=S_x^{(1)}/{\beta }^2$ in ${\mathrm{V}}^2/Hz$ before independent detector noise is added.

For an original invented calibration example, choose $a=0.500\mu \mathrm{m}$, dynamic viscosity $\eta =1.00\times 10^{-3}\mathrm{Pa}\mathrm{s}$, $T=300\mathrm{K}$, $k=0.0200pN/nm=2.00\times 10^{-5}N/m$, and $\beta =100nm/V$. The unbounded no-slip Stokes drag law givesThe corresponding voltage standard deviation is $0.14391\mathrm{V}$, and the position power spectral density plateau is $0.390369\mathrm{n}{\mathrm{m}}^2/Hz$, or $3.90369\times 10^{-5}{\mathrm{V}}^2/Hz$ in the detector. Recovering both $k=k_{B}T/{\sigma }^2$ and $k=2\pi \zeta f_c$ is an internal model check; actual agreement also tests the assumed temperature, drag and detector response.

The generator uses the exact stationary update $x_{j+1}=\rho x_j+\sigma \sqrt{1-{\rho }^2}{Z}_j$, with $\rho =e^{-\Delta t/\tau }$ and independent standard normal random variables $Z_j$. For a demeaned block of $N$ samples and its unnormalized discrete Fourier transform $X_j$, the two-sided periodogram is $\Delta t|X_j{|}^2/N$; doubling positive bins except the zero and Nyquist endpoints gives a one-sided density. Summing those bins times $\Delta f=1/(N\Delta t)$ returns the block mean square by discrete Parseval identity. Independent detector noise, finite exposure and aliasing would alter a measured spectrum and must enter a fit before interpreting $f_c$.

Set $v=\dot{x}$, $\zeta =6\pi \mu a$, and $\tau =m/\zeta$. To keep dimensions and the thermal amplitude explicit, write the force covariance as $⟨F(t)F(s)⟩=A\delta (t-s)$. The printed unit coefficient is a choice of noise normalization, not a general dimensional thermal force. Assume the initial velocity is independent of the future white noise. The Underdamped Langevin dynamics has integrating-factor solutionTaking its expectation and using the delta covariance in the double integral givesHere the noise contribution is $A{m}^{-2}{\int }_0^t{e}^{-2(t-s)/\tau }ds$. The equipartition theorem requires the long-time value $⟨m{v}^2/2⟩=k_{B}T/2$. Consequently the fluctuation-dissipation relation for a Langevin particle isWith an initially thermal velocity, $⟨v_0⟩=0$ and $⟨v_0^2⟩=k_{B}T/m$, so the velocity variance is stationary. If one insists on the printed coefficient $A=1$ in dimensional units, the model has temperature fixed by $k_{B}T=1/(2\zeta )$ in those units; it cannot describe arbitrary temperature while retaining that coefficient.

Take $x(0)=0$. Integrating the velocity solution givesThe initial/noise cross term vanishes under the independence assumption. Evaluating the remaining integral yields the general integrated Ornstein-Uhlenbeck displacementUse the thermal initial velocity and the fluctuation-dissipation relation for a Langevin particle. With diffusion coefficient $D=k_{B}T/\zeta$, this simplifies toFor $t\ll \tau$, expansion gives $⟨x^2⟩=(k_{B}T/m)t^2+O(t^3)$: displacement is initially $v_{0}t$, so the particle is ballistic. For $t\gg \tau$, $⟨x^2⟩=2Dt-2D\tau +o(1)$, hence the leading behavior is one-dimensional diffusion with $D=k_{B}T/(6\pi \mu a)$.

The ballistic conclusion depends on the initial state. For a particle prepared exactly at rest, $v_0=0$, the first expression gives $⟨x^2⟩=A{t}^3/(3m^2)+O(t^4)$ at very short times. A thermal initial velocity, or a nonzero initial velocity variance, is required for the stated $t^2$ mean-square law. It should not be replaced silently by a zero-velocity initial condition.

Put $\tau =m/\zeta$. The causal Green function of the displacement equation is $G(t)={\zeta }^{-1}(1-e^{-t/\tau })$ for $t>0$ and zero for $t<0$. Its derivative has the jump needed to give $m{G}^{\prime \prime }+\zeta G^{\prime }=\delta (t)$. For the specified zero position and velocity,For Gaussian white noise these are stochastic convolutions; their first two moments follow from the stated covariance even without assuming Gaussian higher moments. In particular $⟨x⟩=⟨\dot{x}⟩=0$, andThe required equipartition theorem limit therefore fixes the fluctuation-dissipation relation for a Langevin particle:Here $k_B$ denotes the printed Boltzmann constant $k$. Set $D=k_{B}T/\zeta$, the Stokes–Einstein relation. Squaring the displacement kernel givesConsequently ${\lim }_{t\to \infty }⟨x⟩=0$ and ${\lim }_{t\to \infty }⟨x^2⟩=+\infty$, with $⟨x^2⟩=2Dt-3D\tau +o(1)$ for positive temperature. There is no finite equilibrium variance of position in an unconfined fluid. The inertial Brownian displacement with zero initial velocity has short-time variance $2D{t}^3/(3{\tau }^2)+O(t^4)$, rather than the ballistic law associated with an initially thermalized velocity.

For constant $D$, the diffusion equation has fundamental density $P(x,t)=(4\pi Dt{)}^{-1/2}{e}^{-x^2/(4Dt)}$, whose mean is zero and variance is $2Dt$. It describes the large-time, overdamped limit of the Langevin dynamics. Under the usual Gaussian-force assumption, the finite-time position density is Gaussian with the variance above and obeys a time-dependent diffusion equation with coefficient $D(1-e^{-t/\tau }{)}^2$. It does not obey the constant-$D$ equation through the initial inertial transient. The given covariance alone determines the moments, not a Gaussian law; the density statement uses the standard Gaussian thermal-noise model.

Now write $\zeta (x)={\zeta }_0+\epsilon {\zeta }_0^{\prime }x$ and $x=x_0+\epsilon x_1+O({\epsilon }^2)$, with ${\tau }_0=m/{\zeta }_0$ and $D_0=k_{B}T/{\zeta }_0$. For a fixed realization of the additive force, the first-order equation isIf the force amplitude is instead adjusted to local thermal equilibrium, its first-order noise term has zero mean in the inertial Itô formulation and does not change the following first-order mean calculation. The mixed unperturbed moment isLet $M(t)=⟨x_1(t)⟩$. Averaging and solving the scalar forced relaxation equation, with $M(0)=M^{\prime }(0)=0$, givesThus the long-time first-order coefficient and local mean drift areLiterally $\lim M=-\infty$ for ${\zeta }_0^{\prime }>0$, $+\infty$ for ${\zeta }_0^{\prime }<0$, and zero for ${\zeta }_0^{\prime }=0$; the useful finite quantity is $\lim M^{\prime }$. Because the viscosity gradient is only a local expansion, the physical drift result applies in the intermediate window ${\tau }_0\ll t\ll L_{\zeta }^2/D_0$, where $L_{\zeta }={\zeta }_0/(\epsilon |{\zeta }_0^{\prime }|)$. Extrapolating the linear drag law to arbitrarily large displacements could make the drag negative and would invalidate the model.

The isothermal diffusion with position-dependent drag interpretation uses locally matched thermal noise and $D(x)=k_{B}T/\zeta (x)$. Integrating the printed Fickian diffusion equation against $x$, with vanishing boundary fluxes, givesSince $D^{\prime }(x)=-\epsilon k_{B}T{\zeta }_0^{\prime }/{\zeta }_0^2+O({\epsilon }^2)$, this reproduces the same drift after the inertial transient. Equivalently the overdamped Itô diffusion has drift $D^{\prime }$ and noise amplitude $\sqrt{2D}$. Omitting that drift would give $P_t=(DP{)}_{xx}$, which is a different Fokker-Planck equation and not the stated Fickian equation.

Noise conventions cannot be mixed silently. If the force covariance is held at the spatially constant value $F_0=2{\zeta }_0{k}_{B}T$ while drag varies, the local overdamped diffusion coefficient is instead $\tilde{D}=F_0/(2{\zeta }^2)$ and its Itô drift is ${\tilde{D}}^{\prime }/2$. Its density equation is $P_t={\partial }_x[\tilde{D}{P}_x+({\tilde{D}}^{\prime }/2)P]$, not the isothermal Fickian equation. This still gives the same displayed first-order mean drift at $x=0$, but has different spatial diffusion and equilibrium properties. The stated Fickian interpretation therefore requires local thermal-noise matching.

Physically, thermally driven excursions travel farther on the lower-drag side. Their displacement is correlated with velocity, so averaging the position-dependent friction leaves a net drift toward greater diffusivity, without an applied force. This is not a violation of equilibrium: in the isothermal Fickian model the drift and spatially varying noise cancel in the probability flux for a uniform density. It is a local diffusion-induced shift of the centroid of an initially localized particle distribution.

At thermal equilibrium, microscopic reversibility equates the probabilities of a path and its reversed path when both include their Boltzmann distribution initial weights. Denote their endpoint states by $z_1,z_2$, including velocity if needed. Since the Hamiltonian is even under time reversal in classical mechanics,This detailed balance condition and the energy balance for an autonomous Lagrangian yieldThis is the fluctuation-dissipation relation for a Langevin particle: the strength of Gaussian white noise is fixed by the damping and temperature, with $k_B$ the Boltzmann constant.

For an equilibrium coarse-grained variable, the unresolved microscopic states contribute entropy; their statistical weight is encoded in the Helmholtz free energy, rather than in a single microscopic energy. Relative to the same reference measure, ${\rho }_{\mathrm{eq}}(x)\propto e^{-\beta F(x)}$, so microscopic reversibility becomesThis extension assumes an equilibrium coarse-grained description with reversible path statistics; externally driven dynamics need not obey this relation.

For $q\ne 0$ and $\alpha >0$, each Fourier mode is an Ornstein-Uhlenbeck process with decay rate $r_q=\alpha q^2$. Its explicit Ornstein-Uhlenbeck solution isso the Itô isometry givesTo identify the overdamped Langevin dynamics, choose a friction ${\zeta }_q>0$, a harmonic oscillator potential $V_q={\zeta }_q\alpha q^2{h}_q^2/2$, and force noise $f_q={\zeta }_q{\eta }_q$. The fluctuation-dissipation relation for a Langevin particle then defines $k_B{T}_{\mathrm{eff}}={\zeta }_q{\sigma }^2/2$. Its Boltzmann distribution isThis is the stationary spectrum of a linear fluctuating interface, also following from the equipartition theorem. The numerical prefactor uses exactly the mode-noise normalization given in the paper. The nonzero modes can therefore have a stationary Gaussian distribution even though the unpinned zero mode cannot.

Detailed balance says that each equilibrium transition is balanced by its time reverse:With the Boltzmann distribution $P_{\mathrm{eq}}\propto e^{-\beta H}$ and $\Delta H=H_2-H_1$, this requiresParts a and b instead give $\exp [-(2\zeta /{\sigma }^2)\Delta H]$. Equality for every pair of endpoints yields the fluctuation-dissipation relation for a Langevin particle

There is a source error to resolve first. For a drag coefficient defined by $F_{\mathrm{drag}}=-\xi v$, the velocity relaxation time is $\tau =m/\xi$. The mass and drag coefficient in the PDF's exponential are interchanged: $\xi t/m$ is dimensionless, whereas $mt/\xi$ is not. The consistent mean relaxation law is therefore $⟨v(t)⟩=v_0{e}^{-t/\tau }$.

A mean relaxation law alone does not determine a variance. Use the usual thermal Langevin equation, with a bath at temperature $T$ and the fluctuation-dissipation relation for a Langevin particle:Here the calculation is for one component in Cartesian coordinates. With fixed initial position and velocity, integration gives the mean displacementIntegrating the stochastic part of the velocity once more gives the centered displacementThe delta correlation reduces its variance to a single integral:This is the integrated Ornstein-Uhlenbeck displacement for a prescribed initial velocity. Its short-time variance is $2D{t}^3/(3{\tau }^2)+O(t^4)$; at long times it is $2Dt+O(1)$. Thus the diffusion coefficient isIf the requested mean-squared fluctuation instead refers to the conventional equilibrium mean squared displacement, the initial velocity is also averaged over its thermal distribution, with $⟨v_0⟩=0$ and $⟨v_0^2⟩=k_{B}T/m$. Its contribution adds $D\tau (1-e^{-t/\tau }{)}^2$, givingThis has the ballistic limit $(k_{B}T/m)t^2$ and the same diffusive limit $2Dt$. For fixed nonzero $v_0$, the uncentered mean square instead equals the first boxed variance plus $v_0^2{\tau }^2(1-e^{-t/\tau }{)}^2$. In three independent spatial dimensions, the centered variances add, and the long-time result is $⟨|\mathbf{r}-⟨\mathbf{r}⟩{|}^2⟩\sim 6Dt$. Stating the initial-velocity convention is necessary to distinguish these finite-time formulas.

The gap is not a fixed mechanical separation. Thermal fluctuations of the wall compete with drift towards the microtubule under the opposing force. A monomer can be incorporated only after the gap reaches its length. The Smoluchowski equation describes this overdamped drift and diffusion of gap probability density.

For gap potential energy $U(x)=Fx$, the equation and its probability current areThe drift mobility is $D\beta$, the overdamped consequence of the fluctuation-dissipation relation for a Langevin particle. Equivalently, this coefficient is fixed by requiring the zero-current solution to be the Boltzmann distribution $p\propto e^{-\beta U}$. Solve this equation with the appropriate exclusion, incorporation and probability-normalization conditions. The steady flux through the incorporation threshold gives the addition rate $j_0$; the elongation speed is $v=\delta j_0$.

There is a regime distinction in the printed problem. The absorbing boundary imposed in part (c) means instantaneous incorporation once a gap opens, and leads to the diffusion-limited Brownian ratchet with constant load. A genuinely reaction-limited polymerization Brownian ratchet instead assumes fast gap equilibration followed by a finite chemical addition rate, giving an exponential gating factor multiplying that chemical rate. The equations requested in the subsequent parts specify the former model, which is the one solved here.

For a particle of mass $m$ with linear drag $-\zeta v$, the velocity relaxation time is the exponential relaxation time ${\tau }_v=m/\zeta$. Under a constant applied force $F$ and zero-mean random forcing, the averaged Langevin equation is $md⟨v⟩/dt=-\zeta ⟨v⟩+F$. Solving givesThus the initial mean-velocity deviation decays by a factor $e^{-1}$ after time ${\tau }_v$. Since the force-per-velocity coefficient $\zeta$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, $m/\zeta$ has units of time. If the damping parameter is instead defined as a rate $\gamma =\zeta /m$, the same time is $1/\gamma$. For a sphere obeying Stokes's law, ${\tau }_v=m/(6\pi \eta R)$. The fluctuation-dissipation relation for a Langevin particle gives the same decay time for its equilibrium velocity covariance.