# Kolmogorov scale why $d\sim \left( \frac{\nu^3}{\epsilon} \right)^{1/4}$ and not $d\propto \left( \frac{\nu^3}{\varepsilon} \right)^{1/4}$?

I am looking at the Kolmogorov scale and in numerous sources (e.g. this one). I have seen the following: $$d\sim \left( \frac{\nu^3}{\epsilon} \right)^{1/4}$$ for the Kolmogorov scale. I can see why we must have: $$d\propto \left( \frac{\nu^3}{\varepsilon} \right)^{1/4}$$ on dimension grounds. But to replace $\propto$ with $\sim$ we would have to be sure that the constant of proportionality is of order unity. I have not been able to find justification for this assumption. Does anyone now such a justification?

• Uh...what exactly is the difference between $\sim$ and $\propto$ here? I've known people who use the two interchangably. – ACuriousMind Aug 13 '16 at 16:16
• @ACuriousMind I am assuming that $a \sim b$ means that $a$ and $b$ are of the same order of magnitude and $a \propto b$ means that they are proportional. – Quantum spaghettification Aug 13 '16 at 16:50
• What basis do you have for that assumption? Without further context, I would always read both as "proportional to" first. – ACuriousMind Aug 13 '16 at 17:11
• @ACuriousMind In one of the places I have seen this (which unfortunately isn't publicly available) it uses both $\propto$ and $\sim$ in the two following expressions $E(k)\propto k^{-5/3}\varepsilon^{2/3}$ and $d\sim \left( \frac{\nu^3}{\varepsilon} \right)^{1/4}$ the fact the author uses $\propto$ and $\sim$ close together in the same document, indicates she/he associates a different meaning to them. Which I am assuming is the ones I have given above. – Quantum spaghettification Aug 13 '16 at 17:29
• Either way, you want to know if there's rationale, and what is it, for why the constant of proportionality is of order one. That's a fair question. – Bob Bee Aug 14 '16 at 5:33

I think the most intuitive but maybe not the most mathematically correct interpretation is the implicit assumption that viscous forces start dominating over inertial forces at the Kolmogorov scale. This is easily seen when defining the kolmogorov length and velocity scales: $$l\sim\left(\frac{\nu^{3}}{\epsilon}\right)^{1/4}\quad v\sim\left(\nu\epsilon\right)^{1/4}$$ The Reynolds number at the Kolmogorov scale is then evaluated as: $$\mathrm{Re}=\frac{vl}{\nu}\sim 1$$ For a proportionality sign, the proportionality constant may be much larger than order unity. The Reynolds number would become much larger than order unity as well which would contradict the assumption of viscous forces being of the same order as inertial forces. Hence it suffices to make the constant of order unity and introduce the 'on the order of' sign.

More mathematically, according to the wiki:

The definition of the Kolmogorov time scale can be obtained from the inverse of the mean square strain rate tensor and the definition of the energy dissipation rate per unit mass.

Perhaps from these definitions it follows automatically that the proportionality constant is of order unity, however, i have not investigated this further.

• Hi thanks for your answer. I don't think your answer (correct me if I am wrong) excludes the possibility that constant of proportionality for $l$ be $\sim A$ and that for $v$ be $\sim A^{-1}$ where $A$ can be any (even large) arbitrary constant. Is their a way to show that this cannot be the case e.g. by showing that $l\sim A$ and $v\sim A$? – Quantum spaghettification Aug 14 '16 at 17:28
• @Quantumspaghettification you are right and unfortunately I cannot argue against your case except that in my experience dimensionless proportionality constants are by definition order of unity otherwise the dimensional analysis was not done correctly. Examples are the drag and lift coefficients. I understand that my experience is not a good argument ;), I'll have to think about this some more. – nluigi Aug 14 '16 at 18:00
• Just to let you know I have found a way of doing this (I think :) ). It would have been far to long to put in the comments, so I have put it in the form of an answer. – Quantum spaghettification Aug 15 '16 at 18:17

Throughout this answer $\sim$ means 'of the order of'.

The viscous force takes the form: $$F_V\sim \nu\frac{v}{l^2}$$ So the rate of energy desperation (per unit mass) is: $$\varepsilon \sim \nu\frac{l}{t} \frac{v}{l^2}$$ $$\sim \nu\frac{v^2}{l^2} \tag{1}$$ Now the inertial force is given by: $$F_I \sim \frac{v^2}{l}$$ On the assumption that $F_V \sim F_I$ we have: $$\nu \frac{v}{l^2} \sim \frac{v^2}{l}$$ $$\nu \frac{v^2}{l^2} \sim \frac{v^3}{l}$$ So: $$\varepsilon \sim \frac{v^3}{l} \tag{2}$$ From $(1)^3/(2)^2$ We have: $$\varepsilon \sim \frac{\nu^3}{l^4}$$ Giving us: $$l \sim \left( \frac{\nu^3}{\varepsilon} \right)^{1/4}$$

References

1) Fluid Mechanics: An Introduction to the Theory of Fluid Flows By F.Durst (page 544, link to Google Books).

2) nluigi's answer to this question.

• Are you satisfied with this answer? Does it answer for you why $\tilde$ is used rather than $\propto$? – nluigi Aug 16 '16 at 6:07
• @nluigi to me it does, although I am assuming the order of magnitude of $F_V$ and $F_I$. Whether it is valid or not is another matter... – Quantum spaghettification Aug 16 '16 at 6:32
• It is equivalent to my statement that Reynolds number is order of unity which is generally accepted to be true at the Kolmogorov scale. – nluigi Aug 16 '16 at 6:38