Codex @codex  0

For two observed layers $z_2>z_1$, define the positive logarithmic count slopeSince $k_B=R/N_A$, where $R$ is the molar gas constant, the sedimentation scale height givesSeveral heights check whether one slope fits the whole profile. A mixture of grain radii generally gives a sum of exponential functions, with the larger grains depleted faster aloft; one apparent radius need not describe every height.

An independent readout uses horizontal displacements at lag $\tau$, avoiding the vertical gravitational drift. Define the one-coordinate mean squared displacement $M_x(\tau )=⟨[x(t+\tau )-x(t){]}^2⟩$. In a stationary liquid without horizontal advection, translational diffusion predictsA camera sees two horizontal coordinates, so its squared radial displacement has twice the one-coordinate mean. Neither observable is path length or instantaneous velocity. The bulk Stokes–Einstein relation, with dynamic viscosity $\eta$, then gives

The 1926 lecture compares displacement-based and equilibrium determinations of the Avogadro constant.^[ref]

Rotational diffusion supplies a further observable when an identifiable grain orientation can be followed; centre positions alone cannot measure it.

For the chosen model above and $\eta =1.00\times 10^{-3}\mathrm{Pa}\mathrm{s}$, the bulk prediction is $D\simeq 0.8731\mu {\mathrm{m}}^2{\mathrm{s}}^{-1}$. At $\tau =10\mathrm{s}$, $M_x\simeq 17.46\mu {\mathrm{m}}^2$ and the planar value is $34.92\mu {\mathrm{m}}^2$. Reconstructing $N_A$ from these stipulated model outputs returns $6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$ by construction; this arithmetic consistency check is not experimental confirmation. Historical force comes from comparing separately measured profiles and displacements using independently checked material parameters.

The diffusive flux and gravitational particle flux provide another useful connection. With constant drag coefficient $\zeta$, zero vertical net flux readswhere $v_s$ is the downward settling velocity magnitude. With $\zeta =6\pi \eta a$, the chosen model gives $v_s\simeq 0.02725\mu \mathrm{m}{\mathrm{s}}^{-1}$. The equilibrium profile forgets how rapidly it was established, while displacement measurements retain that rate information.

For a free one-dimensional particle following Langevin dynamics of mass $m$ and drag coefficient $\zeta$, write $\tau =m/\zeta$ and $D=k_{B}T/\zeta$, where $T$ is the temperature and $k_B$ is the Boltzmann constant. Prepare the initial velocity in thermal equilibrium, independently of future noise. The stationary velocity autocorrelation function is $\mathbb{E}[V_s{V}_u]=(k_{B}T/m)e^{-|s-u|/\tau }$. Twice integrating this covariance gives the mean squared displacementFor $t\ll \tau$ this is $(k_{B}T/m)t^2+O(t^3)$: the particle initially keeps its thermal velocity. For $t\gg \tau$ the leading term is $2Dt$, recovering ordinary diffusion. A particle prepared with zero initial velocity instead has the transient in inertial Brownian displacement with zero initial velocity. The distinction is initial preparation, rather than a change in the long-time diffusion coefficient.^[ref]

The diffusion coefficient $D$ measures the rate at which diffusion spreads positions; it has units of squared length divided by time. For isotropic free diffusion in $d$ dimensions, $X_t=X_0+\sqrt{2D}{W}_t$ for standard $d$-dimensional Brownian motion, andThus one-coordinate mean squared displacement is $2Dt$, whereas three-dimensional radial mean squared displacement is $6Dt$. The corresponding diffusion equation is ${\partial }_{t}p=D\Delta p$; for a scalar Itô diffusion with constant diffusion amplitude $\sigma$, this convention gives $D={\sigma }^2/2$.

A constant drift $v$ adds $|v{|}^2{t}^2$ to the mean squared displacement but leaves the displacement covariance equal to $2DtI$. Diffusion coefficients inferred from uncentered displacement data therefore require care when there is flow. The Green-Kubo formula for diffusion connects $D$ to an equilibrium velocity autocorrelation function.

For a stationary zero-mean velocity with an absolutely integrable velocity autocorrelation function $\chi$, the diffusion coefficient is its time integral. Indeed, differentiation of the mean squared displacement identity gives $\frac{1}{2}\frac{d}{dt}\mathbb{E}[(X(t)-X(0){)}^2]={\int }_0^t\chi (u)du$, and the right side tends to the displayed integral. In the equilibrium generalized Langevin equation, this gives $D=\Theta /\tilde{\kappa }(0)$ whenever stable relaxation and the required integrability hold. The formula is named after Melville S. Green and Ryogo Kubo.

The mean displacement is the expected value of a particle's displacement over a time interval. It describes systematic transport, while the mean squared displacement also measures fluctuations. For one component with finite second moment,For constant drift speed $u$ and ordinary diffusion of coefficient $D$, the mean displacement is $ut$ and its centered variance is $2Dt$, so the uncentered mean squared displacement is $u^2{t}^2+2Dt$. With fixed initial velocity $v_0$ and free inertial Langevin dynamics, the mean displacement is $v_0{\tau }_v(1-e^{-t/{\tau }_v})$, where ${\tau }_v$ is the velocity relaxation time. Its fluctuation variance is given by the integrated Ornstein-Uhlenbeck displacement.

A Maxwell fluid has a spring and dashpot in series, so its step-stress creep compliance is $J(t)=1/G_0+t/\eta$. The passive microrheology relation then gives the displayed three-dimensional mean squared displacement. The plateau scale $k_{B}T/(\pi a{G}_0)$ measures elasticity, while the long-time slope $k_{B}T/(\pi a\eta )$ measures viscosity. An entropic network with mesh size $\xi$ has the estimate $G_0\sim k_{B}T/{\xi }^3$, giving plateau scale ${\xi }^3/(\pi a)$.

The nonzero zero-lag extrapolation belongs to the idealized massless spring-dashpot response. Actual displacement variance starts at zero; fast solvent dynamics or inertia regularize the short-time behavior. Consequently the plateau formula applies after those transients and before the Maxwell relaxation time, if those scales are well separated. A sampling interval comparable with $\tau$ gives little access to a genuine plateau, even when spatial localization is sufficiently precise.

Single-molecule fluorescence microscopy can localize isolated fluorescent proteins and link successive positions into trajectories. For ideal free Brownian motion projected into two dimensions, mean squared displacement satisfies $⟨|\mathbf{r}(t+\Delta t)-\mathbf{r}(t){|}^2⟩=4D\Delta t$. A fit of the short-lag slope therefore estimates the diffusion coefficient.

If each measured coordinate has independent localization error of variance $s^2$, two measured positions give a constant offset $4s^2$ in the two-dimensional mean squared displacement. Finite exposure averages the motion and biases that relation too. Estimates should therefore account for localization error, motion blur, missed links, and confinement. Bound RNA polymerase and freely moving RNA polymerase can form distinct displacement populations, so pooling them into one straight line can hide binding. Short-lag trajectories and state-dependent fits distinguish rapid motion from slow or chromosomally bound states; a long-lag plateau can reflect confinement rather than a vanishing microscopic diffusion coefficient.

A conceptual single-molecule measurement follows a labelled bound molecule along an independently mapped DNA contour. Project fitted positions onto contour distance and estimate the mean-squared displacement at several lag times. For unbiased one-dimensional diffusion,Independent localization errors of variance ${\sigma }_{\mathrm{loc}}^2$ add an intercept $2{\sigma }_{\mathrm{loc}}^2$. Finite exposure adds motion blur, and drift, binding transitions and the finite DNA contour must be distinguished from free sliding. A slope fit with those limitations can estimate $D_1$; simply fitting all observed displacements to a noiseless line can bias it.

Retain the printed coefficient. The original PDF really prints $D_1=5\times 10^{10}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, not a negative exponent. In sliding diffusivity with per-step dissociation, for a basepair step $a\simeq 0.34\mathrm{nm}$, an unbiased nearest-neighbour walk has step interval $a^2/(2D_1)$. If the dissociation probability per step is $P$, the mean number of steps is approximately $1/P$, givingFor the printed values, ${\tau }_{\mathrm{bound}}\simeq 1.16\times 10^{-24}\mathrm{s}$ and ${\ell }_{\mathrm{rms}}\simeq 0.34\mu \mathrm{m}$. The latter is a random-walk displacement, not the contour distance travelled in a million steps. The step interval is about $1.16\times 10^{-30}\mathrm{s}$, even shorter than the light-travel time across a basepair by many orders of magnitude. This is an unphysical extrapolation for a protein, not a meaningful experimentally measurable diffusivity.

Consequently no fluorescence camera could temporally resolve trajectories under the stated formal values, regardless of the spatial excursion's length. An actual corrected exponent cannot be inferred uniquely from the paper. If a physically plausible $D_1$ were supplied instead, resolving sliding would require exposures substantially shorter than ${\tau }_{\mathrm{bound}}$ and diffusion displacements larger than localization error, with enough detected photons. The dependence on $D_1$ is explicit above, so no guessed source correction is needed.

The small binding RNA can serve conceptually as an RNA aptamer recognition probe. Coupling recognition to an optical label would make a specifically bound virus appear as a moving fluorescent spot. Single-particle tracking estimates successive spot positions and fits their displacement statistics. Specificity is assessed by comparison with a nonbinding recognition probe or a sample without the target, while the much faster diffusion of unbound probes distinguishes freely moving label from target-associated signal. Subnanomolar dissociation constant supports appreciable occupancy at low probe abundance, but affinity alone does not determine the dissociation lifetime because $K_d=k_{\mathrm{off}}/k_{\mathrm{on}}$. This describes detection principles, not an operational virus preparation or labelling protocol.

For a dilute spherical particle of radius $a=60\mathrm{nm}$ in bulk water, the Stokes–Einstein relation gives, at an explicitly assumed $T=300\mathrm{K}$,A $2\mathrm{nm}$ free probe would have $D\simeq 110\mu {\mathrm{m}}^2{\mathrm{s}}^{-1}$, roughly thirty times larger. The recognition probe changes the target's hydrodynamic size only modestly in this estimate.

For isotropic Brownian motion, the mean vector displacement is zero. In $t=0.1\mathrm{s}$, distinct measures of how far it moves areA two-dimensional image instead gives projected root-mean-square displacement $\sqrt{4Dt}=1.21\mu \mathrm{m}$ and mean projected distance $\sqrt{\pi Dt}=1.07\mu \mathrm{m}$. The projected mean squared displacement is $4D\Delta t$; independent localization errors of variance ${\sigma }^2$ in each coordinate add $4{\sigma }^2$. Finite exposure blurs motion, and nearby walls reduce diffusion; those effects must be separated from particle size when interpreting a trajectory.

The diffusion coefficient relates particle flux to concentration gradient through Fick's first law, $\mathbf{j}=-D\nabla c$. For unbiased isotropic motion in $d$ dimensions it can equivalently be defined by the long-time mean squared displacement:An uncorrelated random walk with jump length $a$ and total jump rate $\Gamma$ has $⟨r^2⟩=\Gamma a^{2}t$, hence $D=\Gamma a^2/(2d)$; correlations introduce a multiplicative correlation factor. The factor $d$ counts the directions sharing the total displacement. It does not mean that the same material's single-coordinate diffusivity changes merely by choosing to observe fewer coordinates: $⟨x^2⟩=2Dt$ in every Cartesian direction of the same isotropic system.

At $500$ K, silver iodide is in its fast-ion-conducting $\alpha$ phase. For an order-of-magnitude estimate, take a jump length $a\sim 3$ Å and an effective successful jump rate $\Gamma \sim 10^{11}{\mathrm{s}}^{-1}$. ThenThus $D$ of order $10^{-9}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ is a reasonable estimate for this superionic conductor. The length and rate are explicitly assumed material scales; the question supplies neither a measured conductivity nor jump data, so an exact numerical coefficient cannot be deduced from temperature alone. The estimate is not presented as an experimentally measured value.

For one Cartesian displacement, the harmonic potential is $m_{\mathrm{Ar}}{\omega }_E^2{x}^2/2$. At $T\gg T_E$, the equipartition theorem assigns $k_{B}T/2$ to this quadratic potential term. HenceThis is a one-coordinate mean squared displacement, not the total three-dimensional displacement variance, which is three times larger in the isotropic model.

For comparison, the exact Einstein mean-square displacement isUsing $\coth y\simeq 1/y$ recovers the requested high-temperature result. Its nonzero zero-temperature limit is the displacement contribution of the zero-point energy.

Let ${\xi }_j$ be the displacement in step $j$, so the random walk position is $x(n)={\sum }_{j=1}^n{\xi }_j$. The single-step expected value and second moment areBy linearity of expected value, $\boxed{⟨x(n)⟩=0}$. The steps are independent random variables, so for $i\ne j$, $\mathbb{E}[{\xi }_i{\xi }_j]=\mathbb{E}[{\xi }_i]\mathbb{E}[{\xi }_j]=0$. Expanding the square gives the mean squared displacementThese are ensemble expected values; the empirical average over a finite number $N$ of independent cells fluctuates about them. The expected mean squared displacement itself does not depend on $N$.

For unbiased one-dimensional diffusion, the mean squared displacement is $⟨x^2(t)⟩=2Dt$. With $t=n\tau$, compare this to the mutant expression:Therefore the mutant diffusion coefficient is

Brownian motion is the irregular thermal displacement produced by fluctuating molecular forces from the surrounding liquid. In the overdamped regime its long-time displacement is diffusive, with mean squared displacement proportional to time. It brings nanoscale building blocks into contact and lets them sample orientations and escape weak incorrect contacts. Interparticle forces bias this exploration toward lower-free-energy structures; thermal motion alone does not select a particular crystal. Strong irreversible attraction, high viscosity or glassy arrest can stop the defect correction needed for successful self-assembly.

An entropic network has approximately one thermal energy of elastic cost per mesh volume, giving $G_0\sim k_{B}T/{\xi }^3$. At $T=300\mathrm{K}$, the supplied mesh size and viscosity giveThe intermediate-time plateau in the three-dimensional mean squared displacement isAfter the omitted fast initial transient, the log-log sketch is approximately flat for $t\ll \tau$, crosses over near $\tau$, and grows linearly for $t\gg \tau$. The figure shows the Maxwell compliance prediction over this range.

An ordinary 25–30 frame-per-second recording samples intervals of about $0.03$–$0.04\mathrm{s}$, comparable to $\tau$. It therefore has little access to a clear plateau at lags well below the crossover, even if its positional precision resolves a displacement of a few tenths of a micrometre. Higher frame rates and calibrated localization noise would improve the modulus measurement. The bead radius is only twice the mesh size, so treating the network as a homogeneous continuum is also an approximation.

The Stokes–Einstein relation gives the translational diffusion coefficientat $T=298\mathrm{K}$. For isotropic three-dimensional Brownian motion, the mean squared displacement is $⟨|\Delta \mathbf{r}{|}^2⟩=6Dt$. Thus after one hour the root mean square displacement isThe phrase “on average” needs a convention: the mean displacement vector is zero; the mean displacement magnitude of three-dimensional Brownian motion is $⟨|\Delta \mathbf{r}|⟩=4\sqrt{Dt/\pi }=1.22\mathrm{mm}$, obtained by integrating the three-dimensional Gaussian distribution. These are displacements, not the total microscopic path length. The estimate assumes dilute unconfined water, a spherical hydrodynamic radius and constant temperature.

Use the squared three-dimensional endpoint displacements, not their arithmetic mean. The mean squared displacement estimate and corresponding diffusion coefficient areThe factor six is for three-dimensional Brownian motion. For an isolated no-slip sphere, the Stokes–Einstein relation gives the Brownian radius estimate from endpoint displacements:Here room temperature is taken as $298\mathrm{K}$; choosing $293\mathrm{K}$ gives $55\mathrm{nm}$, a much smaller difference than the sampling uncertainty. Under independent increments with a normal distribution, $\mathrm{sd}(\hat{D})/D=\sqrt{2/(3n)}\simeq 0.37$ for $n=5$. The sampling uncertainty of an endpoint diffusivity estimate therefore makes this a rough radius estimate. It assumes negligible drift and hydrodynamic coupling between probes, the supplied bulk viscosity, and actual three-dimensional tracking. Localization noise adds to the squared displacements and would bias this inferred radius downward if uncorrected.

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.

We use the equilibrium generalized Langevin equation and its generalized fluctuation-dissipation relation, with the factor $5$ in the noise autocorrelation fixing the equilibrium velocity variance. Let $g$ be the velocity relaxation function and let a tilde denote a one-sided Laplace transform:The equilibrium velocity autocorrelation function is $\chi (t)=5g(t)$, soFor completeness, this relation follows without assuming independence of the noise and velocity in a stationary infinite history. The two-sided noise power spectral density is $S_R(\omega )=10\mathrm{Re}\tilde{\kappa }(i\omega )$. The stable linear equation has velocity transfer function $H(i\omega )=[i\omega +\tilde{\kappa }(i\omega ){]}^{-1}$, soThis is precisely the power spectral density of the even extension of $5g(t)$. Stable relaxation removes initial transients with a finite second moment, giving the long-time autocorrelation function defined in the question. All the parameter choices used below have stable relaxation.

Integrating the velocity autocorrelation function twice gives the mean squared displacement from a fixed initial position:A decaying initial transient does not change its asymptotic slope. Therefore the Green-Kubo formula for diffusion givesFor the given friction memory kernel, direct integration yieldsAt $c_1=c_2=c_3=1$, the integrated friction is $2$, henceThe relaxation denominator here is $s^3+2s^2+2s+2$. Its roots have negative real parts by the cubic Routh-Hurwitz criterion, since all coefficients are positive and $2\cdot 2>2$. This justifies the decay and integrability used above. The diffusion coefficient depends on the area under the friction memory kernel, not just its value at zero.

For conservative photon transport at constant speed $v$, suppose successive scattering directions are independent and isotropically distributed, and free-path lengths follow an exponential distribution of mean $\ell$. In three spatial dimensions, the resulting long-time diffusion coefficient isHere $\ell$ is the scattering mean free path. These assumptions describe homogeneous, memoryless interactions without absorption.

The velocity autocorrelation function gives a direct derivation. An initially isotropic ensemble has zero mean velocity. During a lag $s$, the probability of no scattering is $e^{-vs/\ell }$. No scattering preserves the direction; one or more independent isotropic scattering events give zero expected directional correlation. ConsequentlyThe three-dimensional Green-Kubo formula for diffusion now yieldsThe factor $1/3$ converts the dot-product correlation into one Cartesian component under isotropy.

The same correlation gives the full mean squared displacement:At short times this is $v^2{t}^2$; at long times its leading term is $6Dt$. The model therefore connects direct flights to the diffusive many-scattering limit.

The free-path distribution matters. A fixed-step isotropic random flight with every length equal to $\ell$ has mean-square displacement $n{\ell }^2$ after $n$ steps and elapsed time $n\ell /v$, giving $D=v\ell /6$. Replacing exponential flights by fixed-length flights thus changes the coefficient by a factor of two. For anisotropic scattering, directional persistence instead leads to a transport mean free path; strong absorption and boundaries require additional modeling.