# Kolmogorov scale - dissipation?

The rate at which energy moves down a turbulent cascade is given by: $$\newcommand{\p}{\frac{\partial #1}{\partial #2}} \newcommand{\f}{\frac{ #1}{ #2}} \newcommand{\l}{\left(} \newcommand{\r}{\right)} \newcommand{\mean}{\langle #1 \rangle}\newcommand{\e}{\varepsilon} \e\sim \f{v_l^3}{l}$$ As far as I know $\e$ is a constant, but I am confused by the nature of this expression. Does it hold for any scale $l$ and corresponding velocity scale $v_l$ or only for a particular one? If the former please explain and if the latter what length and velocity scales are chosen and why does it not hold for any?

• Have you looked at introductory turbulence textbooks that derive this expression and where it comes from? It's usually pretty clearly laid out what these variables mean. – tpg2114 Dec 13 '16 at 19:40
• @tpg2114 I've looked in Landau and Lifshitz and several online sources. From what I have found they are all a bit ambiguous about it. e.g. they say $l$ is a 'characteristic vortex length' which could be interpreted as meaning the vortex length I am looking at or a specific one. Hence the confusion. – Quantum spaghettification Dec 13 '16 at 19:43
• You may want to clarify then because "any length scale" is very different from "any of the big scales" or "any of the small scales." A vortex scale is radically different than a Kolmogorov scale and the derivations (and resulting definitions) are different. – tpg2114 Dec 13 '16 at 19:45

According Kolmogorov's theory, that expression holds at any scale in the inertial range, where dissipation rate $\epsilon$ and length scale $l$ are the only parameters that determine what happens at that scale. I think it was G. I. Taylor who later extended above expression to large energy containing scales (integral scale, to be specific), and when done so it indeed becomes a very convenient way of calculating $\epsilon$. But at Kolmogorov scales that expression cannot hold because fluid viscosity becomes important.