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I'm actually studying the famous paper of Kolmogorov (1941) about incompressible turbulence (https://royalsocietypublishing.org/doi/abs/10.1098/rspa.1991.0075), but something struck me and I can't get my head around it.

Kolmogorov considers the velocity $u(x)$ at each point $x$ to be random variables, and he introduces the velocity increment : \begin{equation} w(y)=u(x_0+y)-u(x_0). \end{equation} He assumes that for a fixed $y$, the probability law of $w(y)$ is independent from $x_0$ and $u(x_0)$.

Then, that means that : \begin{equation} \langle u(x_0)\cdot w(y) \rangle=\langle u(x_0)\rangle\cdot\langle w(y)\rangle=0 \end{equation} since $\langle w(y)\rangle=0$.

Computing the mean kinetic energy at, say, point $x \equiv x_0+y$ we get : \begin{align} \langle u(x)^2\rangle&=\langle (u(x_0)+w(y))^2\rangle \\ &=\langle u(x_0)^2\rangle + \langle w(y)^2\rangle. \end{align} That simple result seems very odd, since it implies that under Kolmogov's hypothesis, we cannot have true homogeneity in the flow, and there is a particular $x_0$ where kinetic energy is minimal.

The last point does not hold, since we assumed that the probability law does not depend on $x_0$. Indeed, if we set $x_0'\equiv x$ and compute this time the kinetic energy at $x_0 = x_0' - y$ : \begin{align} \langle u(x_0)^2\rangle &= \langle (u(x_0')+u(x_0'-y)-u(x_0'))^2 \rangle \\ &=\langle (u(x_0')+w(-y))^2 \rangle \\ &=\langle u(x_0')^2\rangle + \langle w(-y)^2 \rangle \\ &=\langle u(x)^2\rangle + \langle w(-y)^2 \rangle \end{align} since $u(x_0'-y)-u(x_0')=w(-y)$ is by hypothesis independent of the choice of $x_0'$ and $u(x_0')$. This results clearly contradicts the first one, and this where I'm stuck.

What am I actually missing? Am I misinterpreting the independence hypothesis from Kolmogorov?

Thanks for your help.

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  • $\begingroup$ I read the first pages of Kolmogorov's paper and I think the problem may be in the fact that he assumes fixed value of $\mathbf u^{(0)}$. Then $\mathbf w$ actually is correlated with $\mathbf u^{(0)}$ (see the definition of $\mathbf w$). He only assumes that probability function for $\mathbf w$ is independent of $\mathbf u^{(0)}$, but not that random variable $\mathbf w$ is independent of $\mathbf u^{(0)}$. If this is right, then it is not possible to transform the mixed term $\langle \mathbf u^0 \cdot \mathbf w\rangle$ into $\langle \mathbf u^0\rangle \cdot \langle \mathbf w\rangle$. $\endgroup$ – Ján Lalinský May 23 '19 at 21:40
  • $\begingroup$ But isn't it the same ? He basically says that the conditional probability $P(w(y)=a|u(x_0)=b)$ is independent of $b$. Then we have : $P(w(y)=a)=\sum_{b}P(w(y)=a|u(x_0)=b)P(u(x_0)=b)=P(w(y)|u(x_0)=b_0)$ where $b_0$ can take any value we want. And that means that random variables $u(x_0)$ and $w(y)$ are independent. $\endgroup$ – Ixeday May 24 '19 at 22:11
  • $\begingroup$ You are right, that made no sense. But there is a different thing, see my answer below. $\endgroup$ – Ján Lalinský May 25 '19 at 12:51
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I think what Kolmogorov's definition means is that he wants to consider a model where statistical properties of fluid velocity field as observed from inertial frame of any mass point of the fluid are the same, whatever the value of velocity of this point is in the laboratory reference frame.

When considering all velocity fields in the lab frame that obey the condition $u(x_0) = u_0$ for fixed constant $u_0$, and averaging over these fields, we obtain

$$ \langle u(x)^2\rangle =u_0^2 + \langle w(x-x_0)^2\rangle $$

which means average value of $u^2$ at other points $x\neq x_0$ is always either the same or higher than $u_0^2$. So the point $x_0$ is special in that it has lowest expectation value of velocity squared. This holds no matter the magnitude of $u_0$.

This result is not entirely absurd, as occurence of velocity field with extremely large magnitude of $u(x_0)$ makes same and higher velocities at other nearby points more probable.

We could swap the role of the points, fix the value $u(x)=u_1$ and evaluate average for $u(x_0)$. Then we obtain

$$ \langle u(x_0)^2\rangle =u_1^2 + \langle w(x_0-x)^2\rangle $$ But this does not contradict the first result, as now we are averaging over different subset of velocity fields.

So far there is no absurdity, but what happens when we do averaging of both sides of the equation

$$ u(x)^2 = (u(x_0) + w(x-x_0))^2 $$

over all velocity fields, with no constraint on the value of velocity squared at any point? The only way how this could work seems to be that both averages are infinite...

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