= Solution
A <passive scalar> $\theta$ is transported by the <velocity field> without feeding back on it:
$$
\partial_t\theta+\mathbf u\cdot\nabla\theta=\kappa\Delta\theta.
$$
Take its mean to be zero, and define the <scalar dissipation rate> by the half-<scalar variance> convention $\chi=\kappa\langle|\nabla\theta|^2\rangle$. In a statistically equilibrated <energy cascade> of scalar fluctuations, this is also the flux of half-<scalar variance> toward small scales.
The <Kolmogorov two-thirds law> gives a typical <velocity increment> $\delta u_r\sim(\epsilon r)^{1/3}$ and hence <eddy turnover time> $\tau_r\sim r/\delta u_r\sim\epsilon^{-1/3}r^{2/3}$. Assuming local scalar transfer on this same time scale, $\chi\sim\langle(\Delta\theta)^2\rangle/\tau_r$. The <Obukhov-Corrsin theory> therefore gives the scalar analogue:
$$
\boxed{\langle(\Delta\theta)^2\rangle=C_\theta\chi\epsilon^{-1/3}r^{2/3}}.
$$
This applies to separations at which both direct forcing and molecular <diffusion> are negligible and the transporting <velocity increments> lie in the <inertial range>. For very large or very small ratios $\nu/\kappa$, the scalar and velocity cutoff scales differ, so this common range must be checked. Defining $\chi$ as dissipation of the full <scalar variance> instead simply changes the convention for $C_\theta$.
Back to article page
