1. TURBULENCE SPECTRAL INVARIANCE MODEL CLOSURE = SIMCLO© (8/99 summary to John Lumley)
FRED Aston SciPark Birmingham B7’4BJ
(unpublished 1976! - en route from Imp Coll Aero to Cambridge Uni DAMTP)
See Batchelor or Hinze or Tennekes & Lumley for basic considerations leading to spectral stress tensor evolution equation St + T + V = M (homogeneous, at least locally) with S(K, t), K vector wavenumber and t time, T turbulent transfer function, V = 2K2S is viscous transfer function with  kinematic viscosity and M is mean field forcing.

2. From Batchelor (p86), T separates to inertial and pressure components T = I + P satisfying the relation
Pij + KimImj + KjmImi = 0 where Kij = KiKj/K2 are direction cosines (Kkk = 1).
The model must satisfy certain constraints (Hinze, p258) - homogeneity implies Sij = Sji and Tij = Tji
If isotropic then (Hinze, ch 3; Batchelor, ch3) additionally have Pij = 0 and KiIij = 0.
Incompressibility implies Pkk = 0 or KijIij = 0 and
I dv(0;) = 0 where v(K) = 4/3K3.

3. Following Batchelor, also Hinze (p261), define spherical shell averages < > for any variable
<F> = a-1 F da(0;K) and introduce
<F>’ = 1/2 a<F> , <F>’’ = K1/3<F> where a(K) = 4K2.
SIMCLO proposes I be approximated as linear function of <S> taken to be sum of gradient transfer component G and equipartitioning component E with dimensionality met by K and dissipation
D = <V>’kk dK (0;).
Symbolically, I = G + E with E = 0 when
<S>ij = 1/3ij<S>kk with ij Kronecker delta.

4. Simply shown (appendix) that above constraints are satisfied with <I>’ij = CI D1/3K(K5/3 <S>’ij) ,
Gij = ¾ [(im - Kim)<I>mj + (jm - Kjm)<I>mi] and
Eij = CE D1/3K2/3[3(HimRmj + HjmRmi) + 4ijKrsRsr] where
Hij = ij - 1/3ijKkk and Rij = <S>ij - 1/3ij<S>kk with CI and CE are undetermined constants.
The gradient component follows formulations by Pao and Corrsin. Contraction Ers = 6 CE D1/3K2/3KrsRsr indicates equipartitioning role in Tkk (surprising?) although <E>kk  0 as expected.
With Jij = <I>ij - 1/3ij<I>kk ,<P>ij = - 3/10 Jij + 2/5 CE D1/3K2/3Rij so
<P>’ij dK (0;) = 2/5 CE D1/3  K2/3R’ij dK(0;) - see appendix.

5. Length scale LE emerges via
D1/3  K2/3 R’ij dK (0;) = Qkk Nij / 2LE where
Nij = (Qij - 1/3ijQkk) with Qij = 2  <S>’ij dK(0;) such that
 <P>’ij dK(0;) = 1/5CEQkk Nij / LE which compares with Rotta’s 1951) empirical one-point model  Pij dv(0;) = RoQkk Nij / LR , Ro = O(1) constant.
Taken together then Ro = 2/5CE LR / LE . Launder et al (and many others since!) demonstrated Rotta’s proposal with reasonably close confinement of Ro delivers adequate values for Qij over a wide variety of flows, although with some adjustments for best agreement with experiments. Arguably postulate “plausible universality” for CE and regard LR / LE dependence on mean strain structure as source of Ro variability? Small perturbation analysis of axisymmetrically strained turbulence affords access to parameter estimation (Gumilevski’s task – appendix?).

6. For sufficiently small viscosity can break the evolution equation above into St + T = M and T + V = 0 with overlapping (Kolmogorov) mesospectrum T = 0.
Following the SIMCLO outlined above, have as top end approximation Ikk + Vkk = 0 , this I-V interaction delivered via Gkk + Vkk = 0. Within T = 0 (I-subrange) this means Gkk = 0 or <I> = 0 as well as <T> = 0 and <P> = 0 so Rij = 0 , Jij = 0 , Eij = 0 and Iij = Gij = ½ (ij - Kij)<I>kk , hence also Pij = 0.
Throughout I-V subrange ½ (ij - Kij)<I>kk + 2K2Sij = 0 so
Sij = ½ (ij - Kij)<S>kk is isotropic. Also (following Pao)
(<I>’kk + <V>’kk) dK (K;) = 0 with lower limit approximating <I>’kk(K) = 0 recovers Kolmogorov scaling <S>’kk = KoD2/3K-5/3.
Empirical Ko = 3/2 then implies CI = Ko-1 = 2/3.

7. APPENDIX
Note <K>ij = ij and isotropic Sij = ½(ij - Kij)<S>kk with <S>ij = 1/3ij <S>kk so Eij = 0 and <I>ij = 1/3ij <I>kk when
Iij = Gij = ½(ij - Kij)<I>kk .
Also KijGji = 0 and Eij term coefficient 4 selected to satisfy KijEji = 0 so KijIji = 0. Expression for Gij satisfies
<G>ij = <I>ij implying <E>ij = 0 ; see above.
Hence have  I dv (0;) = 2  <G>’dK (0;) = 0 because
 0 ensures <S>’ decreases more rapidly than K-5/3 as
K  .

8. APPENDIX
<KijKmn> = 1/15(ijmn + imjn + ijin) where ijmn = ijmn . Use <P>ij = -<KimImj> - <KjmImi> and I = G + E deducing
KimGmj + KjmGmi = 3/10 (<I>ij - 1/3 ij<I>kk) and
<KimEmj> + <KjmEmi>  - 2/5 Rij respectively to show
<P>ij = - 3/10 Jij + 2/5 CE D1/3K2/3Rij via following manipulations
¾ [<Kim(mn - Kmn)><I>nj + <Kjm(mn - Kmn)><I>ni + <Kim(jn - Kjn)><I>nm + <Kjm(in - Kin)><I>nm] reducing to
¾ [1/3 (imjn + jmin) - 2/15 (imjn + ijmn + injm)]<I>nm = 3/4 (2/5<I>ij - 2/15ij<I>kk) and
3(<KimHmn>Rnj + <KjmHmn>Rni + <KimHjn>Rni + <KjmHnn>Rnm) + 4(mj<KimKrs> + mi<KjnjKrs>)Rrs giving
2/3(inRnj + jnRni) + [1/3 (imjn + jmin) - 2/5 (ijmn+imjn+injm)]Rnm + 8/15 (ijrs+irjs+isjr)Rrs = - 2/5 Rij .