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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 the equilibrium generalized Langevin equation with unit mass, the random force has zero mean and the displayed autocorrelation, where $\Theta$ is the equilibrium velocity variance. If the deterministic relaxation $g$ is stable, $g(0)=1$ and $g^{\prime }=-\kappa \ast g$, then the stationary velocity autocorrelation function is $\chi =\Theta g$, equivalentlyOne proof uses the power spectral density. The even force autocorrelation has spectrum $2\Theta \mathrm{Re}\tilde{\kappa }(i\omega )$. With transfer function $H(i\omega )=[i\omega +\tilde{\kappa }(i\omega ){]}^{-1}$, the velocity spectrum is $2\Theta |H{|}^2\mathrm{Re}\tilde{\kappa }=2\Theta \mathrm{Re}H$, the spectrum of the even extension of $\Theta g$. Stability removes initial transients with finite second moment. A valid force autocorrelation must have nonnegative power spectral density; an arbitrary proposed memory kernel does not automatically satisfy this condition.

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.

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.