Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 79 / 1 / iv / Solution

A passive scalar $\theta$ is transported by the velocity field without feeding back on it:Take its mean to be zero, and define the scalar dissipation rate by the half-scalar variance convention $\chi =\kappa ⟨|\nabla \theta {|}^2⟩$. 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 $\delta u_r\sim (ϵr{)}^{1/3}$ and hence eddy turnover time ${\tau }_r\sim r/\delta u_r\sim {ϵ}^{-1/3}{r}^{2/3}$. Assuming local scalar transfer on this same time scale, $\chi \sim ⟨(\Delta \theta {)}^2⟩/{\tau }_r$. The Obukhov-Corrsin theory therefore gives the scalar analogue: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 $C_{\theta }$.