# Brownian motion: from wandering to a measurable law This is newly authored explanatory text in modern notation, with editorial review pending. It is not the German source, an English translation, or a complete edition of the paper. The reviewed source faces and pinned facsimile remain in preparation. Section headings identify the argument being discussed, not a completed source inventory. ## Zero average is not no movement What can we measure when left and right cancel? A signed mean answers where the ensemble’s centre has moved. It does not answer how far its members have wandered. For a centred distribution, rightward and leftward contributions balance even while the distribution broadens. $$ \langle x\rangle=0,\qquad\langle x^2\rangle>0 $$ The model’s mean displacement can be zero while its mean-square displacement is positive. To retain the movement, square each displacement before averaging. Taking the square root of that mean square returns a length: the root-mean-square displacement, or RMS. A mean absolute displacement is a different observable, and neither is the length of a wandering trajectory. [Foundation: mean-variance-rms](/foundations/mean-variance-rms/) — Return to the observable that keeps movement when signed displacements cancel. ### Model limits The model expectation need not equal the mean of one small sample. Net displacement is not total path length. ## Why the square grows with time What permits us to add the contributions of many random steps? Write the displacement after n steps as the sum of their increments. Expanding its square exposes both the squared increments and their cross terms. Independence factors each expected cross term into the product of two means; centring makes that product zero. $$ \left\langle\left(\sum_{i=1}^n\Delta_i\right)^2\right\rangle=\sum_{i=1}^n\langle\Delta_i^2\rangle=nl^2 $$ The mean square of the sum of independent centred increments is the sum of their mean squares. $$ t=n\tau,\qquad D=\frac{l^2}{2\tau},\qquad\langle x^2\rangle=2Dt $$ Elapsed time is n tau; defining D as ell squared over two tau gives mean-square displacement two D t. [Foundation: probability-independence](/foundations/probability-independence/) — Return to why the expected cross terms vanish. ### Model limits Correlated steps, bias or an infinite second moment change the argument. This pedagogical walk does not describe fixed physical jumps in a liquid. ## From a step law to a density law How can random individual steps produce a deterministic equation? Let φ(Δ) be the probability density for a displacement Δ during τ. To end at x, a tracer must start at x − Δ and then make that displacement. Adding over all possible increments gives the transition relation, written here in modern notation. $$ p(x,t+\tau)=\int_{-\infty}^{\infty}p(x-\Delta,t)\varphi(\Delta)\,d\Delta $$ The next density is the integral of the previous shifted density times the step density. Expand to first order in time and second order in displacement. Normalization cancels the zeroth-order term. Symmetry removes the first moment. The second moment remains. $$ D=\frac{1}{2\tau}\int_{-\infty}^{\infty}\Delta^2\varphi(\Delta)\,d\Delta $$ D is half the mean-square step divided by the step interval. $$ \frac{\partial p}{\partial t}=D\frac{\partial^2p}{\partial x^2} $$ The retained equation is the diffusion equation. [Foundation: taylor-expansion](/foundations/taylor-expansion/) — Return to the expansion at fixed position or fixed time. ### Model limits A Taylor truncation is an approximation unless justified by a suitable limiting procedure. Physical independence breaks down at sufficiently short times; the limit is not a literal collision movie. ## What the spreading curve predicts How does the density law become a measurable displacement? $$ p(x,t)=\frac{e^{-x^2/(4Dt)}}{\sqrt{4\pi Dt}}\quad(t>0) $$ The point-source solution is the normalized Gaussian density for positive time. Its symmetry gives zero mean. Its second moment is 2Dt. These are ensemble statements: the curve assigns probabilities to intervals, not destinations to individual particles. $$ \lambda_x=\sqrt{\langle x^2\rangle}=\sqrt{2Dt} $$ Coordinate RMS displacement is the square root of two D t. The probability of finding a displacement between a and b is the integral of this density over that interval. At time zero, an interval containing the starting point has probability one; no finite bell represents that state. [Foundation: gaussian-distributions](/foundations/gaussian-distributions/) — Return to normalization and the Gaussian second moment. ### Model limits A finite closed box has a different long-time distribution. At t = 0 the distribution is a point mass, not an ordinary density. Coordinate RMS is not the three-dimensional RMS distance. ## Why viscosity changes the spread What fixes D for a small spherical tracer in a liquid? The displacement law tells us what a given D predicts. A separate model connects D to a tracer’s physical surroundings. Let b be mobility, so a small force F produces mean drift bF. Stokes drag for a sphere gives b = 1/(6πηa). Let c be number density. At isothermal balance the force density cF balances the osmotic-pressure gradient. With ideal osmotic pressure Π = cRT/N, the drift flux cbF becomes b(RT/N) times the density gradient. Equating it with the opposite diffusive flux gives D = bRT/N. $$ D=\frac{RT}{6\pi\eta aN}=\frac{k_BT}{6\pi\eta a} $$ Stokes–Einstein diffusivity is R T over six pi viscosity radius N, or k B T over six pi viscosity radius. The equality kB = R/N relates the two forms. For a prediction using modern constants it is convenient. For an inference of N, using a value of kB derived from that same N would defeat the point. [Foundation: flux-continuity](/foundations/flux-continuity/) — Return to balancing drift and diffusive fluxes. ### Model limits Slip, inertia, interactions, non-Newtonian response and gas corrections are not included. The argument imports constitutive laws; conservation alone does not derive them. ## What would let us count molecules? Which additional measurements turn displacement into an estimate of N? $$ D=\frac{\langle x^2\rangle}{2t},\qquad N=\frac{RTt}{3\pi\eta a\langle x^2\rangle} $$ Diffusivity is mean-square displacement over twice time; molecular number is R T t over three pi viscosity radius mean-square displacement. The first expression refers to the model mean square, or to an estimate obtained from an appropriate sample. The second is an inversion under the stated physical assumptions. An estimate needs uncertainty and checks of those assumptions; rearranging symbols does not remove experimental error. Without an independent radius, the same D can result from many pairs of a and N. The data then select a compatible family rather than a unique molecular number. The existing synthetic laboratories explore this relationship but do not supply a historical measurement. [Foundation: mean-variance-rms](/foundations/mean-variance-rms/) — Return to the observable that must be estimated from displacement data. [Foundation: error-and-inference](/foundations/error-and-inference/) — Return to what a finite sample and independent inputs can identify. ### Model limits A synthetic run generated from an assumed N is not independent evidence for N. Diffusivity alone constrains the product aN when radius is unknown. Measurement noise, drift, exposure and finite sampling require separate treatment. ## Why the apparent speed depends on how you watch Modern teaching equation; review pending. $$ v_{\mathrm{app}} := \frac{\lambda_x}{t} $$ Apparent coordinate speed equals the model coordinate root mean square displacement divided by the observation interval. This quotient depends on the observation interval. It is not instantaneous physical velocity. At zero interval the quotient is undefined, even though the displacement is zero. ## Resistance to motion controls spreading Modern teaching equation; review pending. $$ D = \frac{k_B\,T}{6\,\pi\,\eta\,a} $$ The diffusion coefficient equals the Boltzmann constant times the absolute temperature, divided by six times pi times the dynamic viscosity times the particle radius. Thermal energy competes with viscous drag. Doubling viscosity halves the diffusion coefficient while reducing RMS displacement only by the square root of two. Radius means radius, not diameter. ## From spreading to a measurable distance Modern teaching equation; review pending. $$ \lambda_x = \sqrt{2\,D\,t} $$ The model coordinate root mean square displacement equals the square root of two times the diffusion coefficient times the observation interval. Squaring measures spread without cancellation between directions. Taking the positive square root turns squared distance back into a distance. Four times the observation interval gives twice the model RMS, not four times.