An exercise from Pope, Turbulent Flows
This post discusses exercise 6.11 (p. 205) of Turbulent Flows, the book by S. B. Pope.
There is a database of solutions of this book available online, however the solution to this exercise is not yet available. The problem uses symmetries of statistically homogeneous, isotropic and incompressible flows, to derive some properties of the one-point velocity gradient distribution. The velocity gradient of a fluid with velocity $u_i$ is defined by
\[A_{ij} = \frac{\partial u_i}{\partial x_j}\]Below, the velocities and velocity gradients are always calculated at the same position $\vec{x}$, which is omitted from all formulas.
Let’s start with an example that is simpler than the one in exercise 6.11, the two-gradient correlation function:
\begin{equation} G_{ijkl} = \left\langle \frac{\partial u_i}{\partial x_j} \frac{\partial u_k}{\partial x_l} \right\rangle \label{eq:def_G} \end{equation}
Since the fluid is statistically isotropic, so is this observable. A general isotropic tensor with four indices can be written as
\[G_{ijkl} = a \delta_{ik} \delta_{jl} + b \delta_{il} \delta_{jk} + c \delta_{ij} \delta_{kl}\]where the $\delta_{ij}$ are Kronecker deltas.
If we change the positions of the velocity gradients in Eq.~\ref{eq:def_G}, nothing changes, therefore
\[G_{ijkl} = G_{klij}.\]This is automatically satisfied by the the definition above with the Kronecker deltas.
We also need to satisfy the incompressibility constraint:
\[\frac{\partial u_i}{\partial x_i} = 0\](Notice that Einstein notation is used, and summation over repeated indices is implied). Therefore, we conclude that
\[G_{iikl} = G_{ijkk} = 0,\]which generates the constraint
\[a+b+3c = 0.\]Finally, because of statistical homogeneity, we have
\[\frac{\partial}{\partial x_j} \left\langle u_i \frac{\partial u_j}{\partial x_k} \right\rangle = 0,\]which is equal to
\[G_{ijjk} = 0,\]and also
\[a+3b+c=0.\]This implies a stricter and well-known result:
\[\left\langle \frac{\partial u_i}{\partial x_j} \frac{\partial u_j}{\partial x_i} \right\rangle = 0.\]This is called a Betchov constraint, and is often written in terms of the velocity gradient tensor as
\[\left\langle \mathrm{tr} \, A^2 \right\rangle = 0.\]We have now two equations for three variables. Let us choose $a=2 / \nu$, where $\nu$ is the viscosity, this is a traditional choice in the literature. The result is
\[G_{ijkl} = 2 \delta_{ik} \delta_{jl} - \frac12 \delta_{il} \delta_{jk} - \frac12 \delta_{ij} \delta_{kl}\]In this notation, the mean energy dissipation is
\[\langle \varepsilon \rangle = 2\nu \left\langle S_{ij} S_{ij} \right\rangle = \nu \langle A_{ij} A_{ij} + A_{ij} A_{ji} \rangle = \nu ( G_{ijij} + G_{ijji} ) = 15\]This result has been used in several models of the velocity gradient dynamics, see for instance Refs. 2 and 3
Exercise 6.11
Let’s go to the exercise now: We’re interested in a 6th order tensor,
\[H_{ijklmn} = \left\langle \frac{\partial u_i}{\partial x_j} \frac{\partial u_k}{\partial x_l} \frac{\partial u_m}{\partial x_n} \right\rangle.\]Write it as an isotropic tensor:
\begin{equation} H_{ijklmn} = \sum_{p \in \mathcal{P}\{i,j,k,l,m,n\}} a_p \delta_{p_1,p_2} \delta_{p_3,p_4} \delta_{p_5,p_6}, \label{eq:def_H_iso} \end{equation}
where we call the set of all permutations of the indices $\mathcal{P}\{i,j,k,l,m,n\}$ and we sum over all permutations. Each permutation is named $p$, and each element of it is indicated with an index, for instance:
\begin{equation} p = \{i,k,j,l,m,n\} \ \mathrm{and} \ p_2 = k. \notag \end{equation}
The enumeration of all permutations, as well as the next steps, were done with Mathematica, follow the link for the accompanying code.
To reduce the complexity of Eq.~\ref{eq:def_H_iso}, we appeal to the symmetries of $H$. First we have relabeling symmetry–there are three ways of rearranging the indices that keep the $H$ tensor invariant:
- $H_{ijklmn} = H_{klijmn}$
- $H_{ijklmn} = H_{ijmnkl}$
- $H_{ijklmn} = H_{mnklij}$
This simplifies the tensor to the form given in Eq. 6.95 of the book:
\[\begin{split} H_{ijklmn} &= a_{26} (\delta_{il} \delta_{jm} \delta_{kn} + \delta_{in} \delta_{jk} \delta_{lm}) \\ &+a_{19} (\delta_{in} \delta_{jl} \delta_{km} + \delta_{il} \delta_{jn} \delta_{km} + \delta_{im} \delta_{jl} \delta_{kn} + \delta_{ik} \delta_{jn} \delta_{lm} + \delta_{im} \delta_{jk} \delta_{ln} + \delta_{ik} \delta_{jm} \delta_{ln}) \\ &+ a_{15} \delta_{ij} \delta_{kl} \delta_{mn} + a_{17} (\delta_{in} \delta_{jm} \delta_{kl} + \delta_{ij} \delta_{kn} \delta_{lm} + \delta_{il} \delta_{jk} \delta_{mn}) \\ &+a_{16} (\delta_{im} \delta_{jn} \delta_{kl} + \delta_{ij} \delta_{km} \delta_{ln} + \delta_{ik} \delta_{jl} \delta_{mn}) \end{split}\]The next constraint we have is incompressibility, which means:
\[H_{iiklmn} = H_{ijkkmn} = H_{ijklmm} = 0\]Finally, we have another Betchov constraint (see Ref. 4, eq. 24):
\[\langle A_{ij} A_{jk} A_{ki} \rangle = \frac{\partial}{\partial x_i} \left\langle A_{ij} A_{jk} u_k - \frac12 u_i A_{kj} A_{jk} \right\rangle = 0\]This is a total derivative of a homogeneous quantity, therefore is zero. The relation we obtain from this constraint is
\[a_1 + 3 a_2 + 9 a_3 + 10 a_4 + 12 a_5 = 0,\]instead of Eq. 6.97 of the book. This is mentioned in p. 6 of the errata to the Pope book.
Gathering all the constraints (relabeling, incompressibility and the Betchov constraint), all the free parameters $a_i$ are expressed in terms of a single free-parameter which captures the velocity-gradient skewness.
Equations 6.95 to 6.101 in the Pope book are then verified in the Mathematica code.
References
- Pope, S. B. (2000). Turbulent Flows. Cambridge: Cambridge University Press.
- Chevillard, L., & Meneveau, C. (2006). Lagrangian dynamics and statistical geometric structure of turbulence. Physical review letters, 97(17), 174501.
- Johnson, P. L., & Wilczek, M. (2024). Multiscale velocity gradients in turbulence. Annual Review of Fluid Mechanics, 56(1), 463-490.
- Betchov, R. (1956). An inequality concerning the production of vorticity in isotropic turbulence. Journal of Fluid Mechanics, 1(5), 497-504.