# Kolmogorov scaling the meaning of the inertial subrange?

I am reading Landau and Lifshitz (Fluid Mechanics) and they say something that contradicts my current understanding of the inertial subrange. I thought (and have read in numerous sources) that the inertial subrange is the range for which: $$v_\lambda \propto (\varepsilon \lambda)^{1/3} \tag{1}$$ holds. In this expression $\varepsilon$ is the rate of energy dissipation per unit mass, $\lambda$ is the size of the vortices under consideration and $v_\lambda$ the size of the velocity variation. In Landau and Lifshitz however it is stated that: $$\varepsilon \sim (\Delta u)^3/l \tag{2}$$ where $l$ is the characteristic size of the external scale and $\Delta u$ is the velocity variation at this scale. Equations (1) and (2) are essentially the same (expect the $\propto$ and $\sim$ which does not concern my question) thus the external scale should, under the definition given above be within the inertial subrange. However it is said that in the inertial subrange $\lambda \lt \lt l$ so how is this contradiction reconciled?

• Are you sure the authors mean that $\Delta u$ is velocity variation at a scale lying in the inertial range, and not at the larger energy containing scale? Eqn (2) is used for calculating dissipation rate by using parameters at large energy containing scales. See A first course in turbulence by Tennekes & Lumley. – Deep Dec 23 '16 at 5:01
• I would second what @Zero suggested at the end -- Tennekes and Lumley is by far the best introduction to turbulence I am aware of. It's easy to read, insightful, and doesn't get bogged down in the really obscure math and figures like newer books such as Pope's do. – tpg2114 Dec 23 '16 at 21:18

So, there is no inconsistency in the scaling laws. They are both correct. But you would never say that $\lambda = l$ in Eq. 1 because Eq. 1 is only valid in the inertial range, while $l$ is the integral scale.
Be careful, because there is another scale $\lambda$ that is typically called the Taylor microscale and represents the eddy sizes at the end of the inertial range where viscosity becomes important, but these scales are larger then the scale at which energy is dissipated. Thus $l > \lambda > \eta$.
• you say 'Eq. 1 is only valid within the inertial subrange' and then 'They are both correct' aren't these two statements contradictory since equation is basically equation 1 using a $\lambda$ outside the inertial subrange? – Quantum spaghettification Dec 22 '16 at 22:00
• @Quantumspaghettification I don't follow your logic there. Eq gives a estimate of the velocity fluctuation associated with an eddy of size $\lambda$ in the inertial range. This is based on the dissipation rate. Eq. 2 gives the value of the dissipation rate based on the integral scales in the flow and the large-scale velocity fluctuation. $\Delta u \neq u_\lambda$ because $l \neq \lambda$. And you can't try to assign an equivalency between them because they aren't valid expressions at the same time – tpg2114 Dec 22 '16 at 22:03
• But if you wanted to, you could do something like $u_\lambda \propto (\Delta u^3 \lambda/l)^{1/3}$ if you wanted to. And then you would find that the ratio of $u_\lambda/\Delta u \propto (\lambda/l)^{1/3}$. And that's a fine expression, and even asymptotes to the correct answer if you say that $\lambda = l$. – tpg2114 Dec 22 '16 at 22:05
• Please could you expand on your comment that 'you can't try to assign an ...'? why not? $\lambda$ is a variable after all? In between the scale $l$ and the inertial subrange does equation 2 (or 1) not hold? – Quantum spaghettification Dec 22 '16 at 22:09
• @Quantumspaghettification Well, I think $\propto$ is stronger because it implies that there exists a $k$ such that $y \propto x \rightarrow y = kx$ while $y \sim x$ just says that $y$ is of the order of $x$ and behaves like it, but it could be non-linear or any other imaginable relationship. So I read those expressions as Eq 1 is stronger than Eq 2 and the additional restrictions are only valid in the inertial range. But I'm trying to find in my turbulence texts if they make the argument any clearer. – tpg2114 Dec 22 '16 at 22:25