Viscous vortex layers subject to more general strain and comparison to isotropic turbulence
aa r X i v : . [ phy s i c s . f l u - dyn ] F e b Viscous vortex layers subject to more general strain andcomparison to isotropic turbulence
Karim Shariff ∗ NASA Ames Research Center
Gerrit E. Elsinga † Laboratory for Aero and Hydrodynamics,Department of Mechanical, Maritime and Materials Engineering,Delft University of Technology, 2628CD Delft, The Netherlands (Dated: March 2, 2021) bstract Viscous vortex layers subject to a more general uniform strain are considered. They includeTownsend’s steady solution for plane strain (corresponding to a parameter a = 1) in which allthe strain in the plane of the layer goes toward vorticity stretching, as well as Migdal’s recentsteady asymmetric solution for axisymmetric strain ( a = 1 /
2) in which half of the strain goesinto vorticity stretching. In addition to considering asymmetric, symmetric and antisymmetricsteady solutions ∀ a ≥
0, it is shown that for a <
1, i.e., anything less than the Townsend case, thevorticity inherently decays in time: only boundary conditions that maintain a supply of vorticityat one or both ends lead to a non-zero steady state. For the super-Townsend case a >
1, steadystates have a sheath of opposite sign vorticity. Comparison is made with homogeneous-isotropicturbulence in which case the average vorticity in the strain eigenframe is layer-like, has wings ofopposite vorticity, and the strain configuration is found to be super-Townsend. Only zero-integralperturbations of the a >
MOTIVATION AND SUMMARY OF RESULTS
Vortex configurations subjected to a spatially uniform strain can be used to model localregions of more complicated flows where the strain represents the local potential velocityinduced by other vortex structures, typically of larger scale than the region being consid-ered. For example, Burgers’ [1] axisymmetrically strained tubular vortex is a good modelfor the high intensity structures of homogeneous isotropic turbulence [2]. Other examplesinclude Townsend’s Gaussian vortex layer subject to plane-strain [3], and the celebratedLundgren spiral which produces Kolmogorov’s k − / energy spectrum [4–7]. While we donot consider the instability of strained layers to wavy perturbations, we mention in passingthat Townsend’s layer is unstable to the formation of concentrated tubular structures [8, 9].This suggests a similar fate for other solutions presented below.Recently, Migdal [10] presented a steady asymmetric vortex layer solution for the case ofaxisymmetric strain ( a = 1 / < a <
1, thoughless and less of a supply of vorticity is needed as a → a = 1 we have Townsend’s plane strain case in which a steady stateis reached with zero boundary conditions on both sides and non-zero integrated vorticity inthe initial condition. For a >
1, steady states exist with inflow of opposite sign vorticity atone or both ends. Only zero integral perturbations of these states relax back to the steadystate; otherwise, they grow.It is known that in homogeneous-isotropic turbulence, vorticity tends to align with thedirection of the intermediate strain rate [11]. However, when the contribution of the straininduced by the local vorticity is removed, it is found that the vorticity is aligned withthe direction of the largest background strain [12]. In our set-up, this corresponds to thesuper-Townsend case a >
1. Elsinga etal. [13] studied the averaged local vorticity structureof homogeneous isotropic turbulence simulations in the strain eigenframe. The structureconsists of a vortex layer and two tube-like vortices adjacent to it. Interestingly, the vorticitycomponent in the direction of the largest principal (background) strain versus the directionnormal to the layer is symmetric and changes sign. This is the type of steady solution weobtain for the super-Townsend cases. The background strain in the turbulence simulationsis also super-Townsend. However, the negative vorticity wings in the turbulent case havea higher amplitude and are more extended (Figure 2d below) than in our steady solution.Finally, the appendix shows that the averaged local flow in the strain eigenframe is governed(apart from one term) by the Reynolds averaged Navier-Stokes equation. This may help tofurther understand its structure.
ANALYSIS
We consider the unidirectional shear flow and associated vorticity u x = U ( z, t ) , ω y ( z, t ) = U ′ ( z, t ) ≡ G ( z, t ) , (1)3ubjected to an irrotational strain written in principal coordinates as ~u strain = α ((1 − a ) x~e x + ay~e y − z~e z ) . (2)Equations (1) and (2) assume that the vorticity is aligned with one of the principal axesof strain. A more general set up would allow the vorticity to be arbitrarily oriented withrespect to the strain axes; in this case the vorticity would undergo a period of alignment.The strain coefficients add up to zero to respect incompressibility. We choose α, a > z and stretching along y . For a = 0all of the straining in the plane ( xy ) of the layer goes into advection and none into vortexstretching. For a = 1 / a = 1 we recoverTownsend’s case of strain in the yz plane (independent of x ). In this case, all of thestraining flow in the plane of the layer goes into vorticity stretching. We mention in passingthat the Townsend case is amenable to conformal mapping in the yz plane for generatingsteady solutions for non-uniform strains [14]. The case a > y stretching than Townsend’s case and has compression along x ; we refer to it as being“super-Townsend.”The only non-trivial component of the vorticity equation is the y -component and it givesthe linear PDE: ∂ t G − αz∂ z G = aαG + ν∂ zz G, (3)where ν is the kinematic viscosity. The second term on the left side of (3) representsadvection while the first term on the right side represents stretching. Setting the timederivative equal to zero gives the ODE: νG ′′ ( z ) + αzG ′ ( z ) + aαG ( z ) = 0 , (4)whose general solution given by Mathematica is G ( z, a ) = c G ( z, a ) + c G ( z, a ) . (5)If we define η = z √ δ with δ ≡ ( ν/α ) / (6)4s a measure of the sheet thickness, then G ( η, a ) = exp (cid:16) − η (cid:17) H a − ( η ) , (7) G ( η, a ) = exp (cid:16) − η (cid:17) F (cid:18) − a , , η (cid:19) . (8)The function F is Kummer’s confluent hypergeometric function and H a − is a Hermitefunction defined as H a − ( η ) = 2 a − √ π (cid:16) − a (cid:17) F (cid:18) − a , , η (cid:19) − η Γ (cid:16) − a (cid:17) F (cid:18) − a , , η (cid:19) . (9)The first term in G (from the first term in eq. (9)) is proportional to G ( z ; a ), i.e., it isnot linearly independent and we may discard it. However, since G ( z, a ) as it stands is thesolution (for a = 1 /
2) presented by Migdal [10], we shall not alter it.Figure 1 plots G ( η, a ) and G ( η, a ) for various strain configurations. The two functionsare asymmetric and symmetric, respectively. The asymmetry of G ( η, a ) comes from the η in the second term of the Hermite function (9). Note that due to symmetry of the ODE(4) under z → − z , a plus sign for the second term instead of the minus sign should be anequally valid solution, i.e, the mirror image of G ( η, a ) should be an equally valid solution.While this can be achieved with a suitable choice of c and c , the z → − z symmetry ismade more obvious if we use the pair of solutions G ± ≡ (cid:16) − a (cid:17) F (cid:18) − a , , η (cid:19) ± η Γ (cid:16) − a (cid:17) F (cid:18) − a , , η (cid:19) , (10)which are mirror images of each other. The functions G + ± G − which are symmetric andantisymmetric, respectively, can also be used as a basis.Consider a finite domain η ∈ [ − L, L ]. The case a = 0 corresponds to plane strainperpendicular to the vorticity so stretching is absent. To achieve a steady state in this casewe need a source of uniform vorticity at one or both ends; for example, a vortex patch (seethe a = 0 case in Figure 1a). As a increases one requires smaller boundary values until for a = 1 (Townsend), the required value is exponentially small in L .For the super-Townsend case ( a > a = 1 .
25 or a = 2 inFigure 1. The symmetric and asymmetric steady solutions show opposite sign vorticityentering one or both boundaries and changing sign before reaching η = 0. Since neitheradvection, stretching, nor dissipation can change the sign of the vorticity, to achieve such a5 η -1-0.500.511.52 G ( η , a ) (a) Asymmetric solution, G ( η , a)a = 0a = 0.25a = 0.5a = 0.75 a = 1 (Townsend)a = 1.25 a = 2 -6 -4 -2 0 2 4 6 η -0.5-0.2500.250.50.7511.25 G ( η , a ) (b) Symmetric solution, G ( η , a)a = 0a = 0.25a = 0.5a = 0.75a = 1 (Townsend) a = 1.25 a = 2 -6 -4 -2 0 2 4 6 η -8-6-4-202468 G ( η , a ) (c) Antisymmetric solution a = 0a = 0.25 a = 0.5a = 0.75a = 0.999a = 1.25 FIG. 1. (a) and (b) Two linearly independent solutions, G ( η ; a ) and G ( η ; a ) to the ODE (4)plotted for various strain configurations a . (c) Antisymmetric solutions constructed from a suitablelinear combination of G ( η ) and G ( η ). steady state requires special initial conditions; this will be discussed in more detail in thenext section.The functions G ( η ) and G ( η ) may be combined to form antisymmetric solutions; seeFigure 1c. One can do this for all a except a = 1 because both G ( η,
1) and G ( η, G ( η, ∝ G ( η,
1) = exp( − η ). This is why in Figure 1c the solutionfor a = 0 .
999 is plotted instead. However, since one can let a → a = 1, an initial condition of zero, and antisymmetric boundary conditions G ( −
7) = − . G (7) = 0 .
5. The run converged to a steady solution as t → ∞ .6
10 20 30 40 50 60 70 x, y or z normalized by η Κ -0.4-0.200.20.40.60.81 ω y (b) Average vorticity in the strain eigenframe of isotropic turbulenceNormal to shear layerAlong vorticityAlong the shear flow( y) ( z) ( x) -60 -40 -20 0 20 40 60 x, y or z normalized by η Κ -1-0.8-0.6-0.4-0.200.20.40.60.81 u x , u y o r u z no r m a li ze d by u K (c) Background straining field in isotropic turbulenceNormal to shear layerAlong vorticityPerpendicular tovorticity u y (y ) u z (z ) u x (x ) z / δ -0.4-0.200.20.40.60.81 P ea k - no r m a li ze d ω y (d) Comparison of symmetric solution with isotropic turbulence.turbulence simulation ( a = 1.31, α = 0.061) Turbulence simulationSymmetric solution for the strain of the
FIG. 2. From the eigenframe analysis of a direct simulation of isotropic turbulence [13] showingthe average local flow in the strain eigenframe. Taylor microscale Reynolds number Re λ = 433.(a) Contours of vorticity ω y in the xz plane. (b) Profiles of ω y along three axes. (c) Profiles ofthe background velocity along the three axes. (d) Comparison of the simulation ω y ( z ) against thesymmetric solution G ( z/ ( √ δ ) , a ) for the strain configuration in the simulation ( α = 0 . , a =1 . CONNECTING TO TURBULENCE
Elsinga etal. [13] studied the average local flow for homogeneous isotropic turbulence sim-ulations in the strain eigenframe. They found that the vorticity consists of a shear layer-likestructure; see Figure 2a. The vorticity component plotted is ω y ; it is the dominant vorticitycomponent in the xz plane, the others being at most 0 . u x = constant = 0 along z = 0 which we call the shear-layer centerline. A7treamline plot, for which we refer the reader to [13], reveals that at each end of the layer,there is an converging spiral indicating a strained tube-like structure. The entire structurehas (full-width-half-maximum) dimensions of (39 . x ) × . y ) × . z )) η K , where η K is theKolmogorov scale. The interesting feature is the presence of a sheath of negative vorticitysimilar to the symmetric solution for the super-Townsend cases. Figure 2b shows profilesof ω y along the three axes; they are symmetric and only one half is plotted. The negativesheath can be observed in the profile (solid line) normal to the layer ( z ). One therefore won-ders whether the background strain is super-Townsend in the simulation. Figure 2c plotsvelocities (in Kolmogorov units) for the background straining flow obtained by removing thecontribution to the strain from the local vorticity using Biot-Savart integration [13]. Thefact that u x ( x ) for the background flow has a negative slope means that 1 − a < a >
1. Evaluating slopes at the origin gives a = 1 .
31 and α = .
061 (in Kolmogorovunits). Note that in Kolmogorov units ǫ = ν = 1 where ǫ is the dissipation rate. Thereforein Kolmogorov units δ ≡ ( ν/α ) / = (1 / . / = 4 .
05 for the simulation. Figure 2d com-pares the simulation profile with the symmetric solution G ( z/ ( √ δ ) , a ) for a = 1 .
31 and δ = 4 .
05. Clearly, in the turbulent simulation the negative sheath has a greater amplitudeand range. In fact, the negative area under the curve is larger than the positive. The resultsof the next section then imply that if the boundary condition and strain were fixed, thenegative vorticity would grow without bound.It would be of interest to investigate the origin of the negative vorticity in the simulationby examining individual fields in the sample. The Appendix shows that the equation gov-erning the average flow in the strain eigenframe is (apart from one term) just the Reynoldsaveraged Navier-Stokes equation. Hence, another approach for understanding the structureof the average flow would be to obtain terms in this equation from the simulation. Thepresent work leads one to expect that divergence of the Reynolds stress and extra term willbe sub-dominant near the origin.The asymmetric solutions G ( z ) are reminiscent of the measured vorticity field at theturbulent-nonturbulent interface of many flows, conditionally averaged with respect to theinterface; see Figs. 5 and 6 in the review article [15] and Fig. 1d in [16]. These interfaces arealso strained [17]. The interfaces of large vorticity voids in isotropic turbulence might besimilar. The asymmetric solutions are also reminiscent of the edge layer at the boundary ofa laminar vortex ring [18]. Vorticity diffusing across the dividing streamline is subject to a8train induced by the fact that the ring is moving as a whole. The edge layer imposes a Robin-type boundary condition on the vorticity, which is similar to Newton’s law of convectivecooling at the boundary of a conducting solid.Note that the structure in Figure 2a has a slight tilt in orientation relative to shear-layer centerline defined earlier: the positive vorticity is tilted clockwise while the negativevorticity is tilted counter-clockwise. This implies an asymmetry in ω y ( z ) profiles whenplotted at x = 0. Hence the tilt could be due to asymmetric vorticity in the backgroundflow.In the discussion section of his paper, Townsend [3] argues that the averaged product ̟ of principal strains in isotropic turbulence must be negative because it is proportionalto the derivative skewness which is known to be negative. For our set-up ̟ = γa ( a − γ, a >
0, we need a < ̟ <
0. That is, the strain must be sub-Townsendaccording to this reasoning. This is contrary to the previous paragraph and the resolutionlikely lies in the fact that the derivative skewness also includes contributions from the localvorticity field. Hence, Townsend’s argument should not be used to infer the configurationof background strain.
TIME-DEPENDENT BEHAVIOR
The unsteady linear PDE (3) was solved numerically. Since there are dimensions of lengthand time, we are free to set α = 2 ν = 1 which makes z = η . The remaining parametersare the strain configuration a and those that involve the initial and/or boundary values ofthe vorticity. Denoting the peak magnitude of the vorticity at any instant as | ω | max ( t ), atime-dependent Reynolds number may be defined asRe ≡ | ω | max ( t ) | δ ν = | ω | max ( t ) , (11)for our choice of units. Since the problem is linear and the vorticity amplitude does notmatter, there is no Reynolds number dependence. The domain is η ∈ [ − L, L ] with L = 7and various Dirichlet boundary conditions are applied.Figure 3 is for axisymmetric strain ( a = 1 /
2) and panel (a) shows relaxation to the exactasymmetric steady state of Migdal [10] with an initial condition of zero and asymmetricboundary conditions G ( −
7) = 0 . G (7) = 0. Keeping in mind that there is advection9 η G ( η ) t = 0.5, 2.5, 4.5, etc.Exact steady Axisymmetric strain (a = 1/2). Initial condition: zero.
Boundary conditions: G(-7) = 0.5, G(7) = 0t = 0.5 t = 2.5etc.(a) (Migdal) -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 η G ( η ) t = 0, 2, 4, 6, 8Exact steady Axisymmetric strain (a = 1/2). Initial condition: Gaussian
Boundary conditions: G(-7) = G(7) = 0(b) time -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 η G ( η ) t = 0.5, 2.5, etc.Exact steady Axisymmetric strain (a = 1/2). Initial condition: zero.
Boundary conditions: G(-7) = G(7) = 0.5t = 0.5 t = 2.5etc.(c) -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 η -1.25-1-0.75-0.5-0.2500.250.50.7511.25 G ( η ) t = 0.5, 2.5Exact steady Axisymmetric strain (a = 1/2). Initial condition: zero.
Boundary conditions: G(-7) = 0.5, G(7) = -0.5t = 0.5 t = 2.5(d)
FIG. 3. Four axisymmetric strain cases ( a = 1 / of vorticity toward the origin, we see that, to maintain Migdal’s asymmetric solution, oneneeds a continual supply of vorticity at one end; otherwise, the solution would eventuallydecay to zero. Figure 3b shows that the solution decays if conditions of zero are applied atboth ends. Panel (c) shows relaxation to the symmetric solution with a symmetric supplyof vorticity at both ends and an initial condition of zero. Figure 3d shows relaxation to theantisymmetric solution in which vorticity of opposite sign enters the domain, is amplifiedby stretching and then annihilates.Finally, we turn attention to some super-Townsend cases (Figure 4, a = 1 . G ( −
7) = G (7) = 0. For this case there is only thetrivial steady solution c = c = 0 and so we expect that an unsteady solution will either10 η -21-18-15-12-9-6-303 G ( η ) Strain: Super-Townsend (a = 1.5). Initial condition: 3 half-cosines
Boundary conditions: G(-7) = G(7) = 0t = 0 t = 1t = 2(a) t = 3t = 4t = 5 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 η -5-4-3-2-1012345 G ( η ) Strain: Super-Townsend (a = 1.5). Initial condition: -sin
Boundary conditions: zerot = 0 t = 2 t = 4(b) t = 6t = 8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 η -5-4-3-2-1012345 G ( η ) Strain: Super-Townsend (a = 1.5). Initial condition: -sin + small line
Boundary conditions: G(-7) = -0.07, G(7) = 0.07t = 0 t = 2 t = 4(c) t = 6t = 8Exact steady -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 η -70-60-50-40-30-20-100102030 G ( η ) Exact steadyt = 0, 0.8, 1.6, etc.
Strain: Super-Townsend (a = 1.5). Initial condition: zero.
Boundary conditions: G(-7) = G(7) = -0.5time Exact steady(d) -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 η -80-60-40-20020406080100 G ( η ) Exact steadyt = 0t = 2t = 4t = 6t = 8
Strain: Super-Townsend (a = 1.5). Initial condition: Exact steady + sin
Boundary conditions: G(-7) = G(7) = -0.5t = 0 Exact steady(e) t = 2 t = 4t = 6t = 8(Exact steady + sin) -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 η -80-60-40-20020 G ( η ) t = 0t = 2t = 4t = 6t = 8t = 10 Strain: Super-Townsend (a = 1.5). Initial condition: Exact steady + cos
Boundary conditions: G(-7) = G(7) = -0.5t = 0(Exact steady +(f) cos perturbation)t = 10
FIG. 4. Super-Townsend cases a = 1 .
5: (a) Exponential growth of negative vorticity with three half-cosine initial condition and zero boundary conditions. (b) Decay with a sine wave initial conditionand zero boundary conditions. (c) Adding a small line to the sine wave leads to relaxation to thesteady solution. (d) Failure to relax to the steady solution. (e) Relaxation back to the steadystate after a conservative perturbation. (f) Failure to relax back to the steady state after a non-conservative perturbation. a = 1) would saturate to a near-Gaussiansteady state. On the other hand, the sine wave initial condition, G ( η,
0) = − sin(2 πη/λ ) , λ = 2 L, (12)which has zero integral, decays to zero (Figure 4b). Next, we make a slight change by addinga small linear function, 0 . η , to (12) and enforce the compatible (antisymmetric) boundaryconditions G ( ±
7) = ± .
07. In this case there is a non-trivial solution for c and c and thesolution relaxes to the corresponding steady state (Figure 4c).The next case is very instructive (Figure 4d). The initial condition is zero and symmetricnegative boundary conditions are applied: G ( −
7) = G (7) = − .
5. The steady solutioncompatible with the boundary conditions is shown as the dashed line. It is obviously notreached as t → ∞ ; instead negative vorticity grows exponentially. Why? In each half of thelayer, η < η = 0 (recallthat we started with no vorticity in the domain). However, neither advection, stretching,nor diffusion can change the sign of the vorticity. To reach the steady solution requires thatthe initial condition already contain the precise amount of vorticity of the opposite sign. Tosee this, consider Figure 4e where we started with the exact steady solution and applied oneperiod of a sine wave as a perturbation: G ( η, t = 0) = G ( η, a = 1 . −
20 sin(2 πη/λ ) , λ = 2 L. (13)The perturbation has zero area and so does not change the total vorticity. As a result,the solution relaxes back to the steady solution. On the other hand, in Figure 4f, theperturbation has non-zero area: it consists of three half-cosines: G ( η, t = 0) = G ( η, a = 1 .
5) + cos(2 πη/λ ) , λ = 2 L/ , (14)and does not disturb the boundary conditions. Note that the perturbation has more negativethan positive area; the solution is unstable and negative vorticity grows exponentially. Itwas verified that when the sign of the perturbation was changed, positive vorticity grew12xponentially. The results of the last two panels, (e) and (f), can be understood as follows.Since the problem is linear, we may consider the evolution of the perturbation separatelyunder homogeneous Dirichlet boundary conditions and that was already done in Figures 4aand b. There we saw that an initial condition with zero total vorticity decays while an initialcondition with a non-zero total vorticity grows.I thank Prof. A. Migdal for sending me his manuscript and for his encouragement. Drs.I. Kiviashvili and A. Wray performed the internal review for which I am grateful. Thedata that support the findings of this study are available from the author upon reasonablerequest. APPENDIX
It is shown that, apart from one term, the equations satisfied by the velocity field inthe local strain eigenframe are the Navier-Stokes equations and the average velocity obeyscorresponding Reynolds-averaged equations.Let us transform the Navier-Stokes equations (with density set to unity) ∂u i ∂t + ∂∂x j ( u i u j ) = − ∂p∂x i , + ν ∂ u i ∂x j , (A.15) ∂u j ∂x j = 0 , (A.16)into strain eigenframe coordinates ~ξ and velocities ~v at a given Eulerian location ~x , whichis taken to be the origin without loss of generality. The transformation is given by x j = A jk ( t ) ξ k , and u j = A jk ( t ) v k , (A.17)where A ( t ) is a rotation matrix which depends only on time for a given ~x . Using the factthat the transpose of a rotation matrix is its inverse, one finds that apart from the timederivative term, the rest of the terms in the Navier-Stokes equations retain their form ineigenframe coordinates: v k ( A − ) ij ˙ A jk + ∂v i ∂t + ∂∂ξ j ( v i v j ) = − ∂p∂ξ i , + ν ∂ v i ∂ξ j , (A.18) ∂v j ∂ξ j = 0 . (A.19)13nsemble averaging over a sample of points ~x and invoking the ergodic hypothesis that thisaverage equals the time average at a fixed ~x in a statistically stationary flow gives, in theusual way, the time-independent Reynolds-averaged equation: v k ( A − ) ij ˙ A jk + ∂∂ξ j ( v i v j ) + ∂R ij ∂x j = − ∂p∂ξ i , + ν ∂ v i ∂ξ j , (A.20) ∂v j ∂ξ j = 0 , (A.21)where R ij ≡ v ′ i v ′ j is the Reynolds stress. It would be of interest to obtain the Reynolds stressterm and the first term in (A.20) from the simulations and compare them with the meanterms. Note that the matrix A ( t ) can be chosen to be the rotation matrix for any desiredframe of interest, not necessarily the strain eigenframe.Lundgren [19] considered the velocity difference ~w ( ~X, ~r, t ) ≡ ~u ( ~X + ~r, t ) − ~u ( ~X, t ), where ~X ( t ) is the position of a Lagrangian particle. Equation (1) in Kolmogorov [20] shows thatthis is also what he had in mind. Lundgren [19] showed that ~w ( ~X, ~r, t ) obeys the Navier-Stokes equation with respect to the separation vector ~r : ∂w i ∂t + ∂∂r j ( w i w j ) = − ∂p∂r i , + ν ∂ w i ∂r j , (A.22) ∂w j ∂r j = 0 . (A.23)It may be useful to consider ~w in the strain eigenframe, its statistics and scaling properties.For that purpose, equation (A.19) is still valid except that ˙ A ( t ) is the time derivative of therotation matrix following a Lagrangian particle.14 Karim.Shariff@nasa.gov † [email protected][1] J. 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