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Coarse-grained energy dissipation is a spatial average of the local rate $2\nu S_{ij}{S}_{ij}$ over a region of size $r$. Its random value ${ϵ}_r$ has mean $ϵ$ in homogeneous turbulence. Refined similarity uses this local averaged viscous dissipation to scale velocity increments.

Homogeneous turbulence has velocity statistics unchanged by translating all observation points. Mean spatial divergences vanish when the moments exist and averaging commutes with derivatives. Homogeneity is distinct from isotropy: it does not require equal statistics in different directions.

For unforced homogeneous turbulence with isotropic turbulence statistics, put $F=u^{2}f=u^2-S_2/2$ and $S_3=6u^{3}K$. The longitudinal Kármán-Howarth equation is $F_t=r^{-4}{\partial }_r[r^4(S_3/6-\nu S_2^{\prime })]$. Since ${\partial }_t{u}^2=-2ϵ/3$, regularity at zero givesNeglecting the last term in a locally equilibrated small-scale range gives the displayed equation. The Kolmogorov four-fifths law follows where the viscous term is negligible as well.

In mean-field electrodynamics, use Reynolds averaging to write ${\mathbf{U}}_{\mathrm{tot}}=\overline{\mathbf{U}}+\mathbf{u}$ and ${\mathbf{B}}_{\mathrm{tot}}=\overline{\mathbf{B}}+\mathbf{b}$, with zero-mean fluctuations. Assume the averaging operation commutes with the derivatives and that resolved quantities vary little across an averaging region. Averaging the induction equation then givesThis is the mean-field electromotive force, a vector correlation with units of velocity times magnetic field. The closure problem is to express this correlation in terms of the resolved field.

In the kinematic regime the flow statistics are prescribed, so the response is linear in the mean magnetic field, after independently excited fluctuation transients have decayed. If the mean-field scale $L$ is much greater than the turbulent scale $\ell$, and its evolution is slow compared with the response memory, expand that response in the local mean magnetic field and its gradients:Without scale separation one generally has a spatially nonlocal and temporally retarded response rather than constant local coefficients.

For homogeneous turbulence the coefficients are independent of position. Invariance under proper rotations makes the rank-two response proportional to ${\delta }_{ij}$, and the isotropic rank-three gradient response proportional to the alternating tensor. Choosing the diffusivity sign convention givesThe alpha effect can regenerate a mean magnetic field, while $\beta$ supplies turbulent magnetic diffusivity. For constant positive $\beta$, $\nabla \times (-\beta \nabla \times \overline{\mathbf{B}})=\beta {\nabla }^2\overline{\mathbf{B}}$ because the magnetic field is solenoidal. Thus the total diffusivity is $\eta +\beta$.

The electromotive vector is a polar vector, whereas the magnetic field is an axial vector. Therefore $\alpha$ is a pseudoscalar and changes sign under reflection; $\beta$ is an ordinary scalar. A nonzero isotropic alpha effect requires broken reflection symmetry, such as nonzero kinetic helicity. If “isotropic” includes reflection symmetry as well as rotations, $\alpha$ must vanish. The dimensions are $[\alpha ]=L/T$ and $[\beta ]=L^2/T$.

Write $A_{ij}={\partial }_j{u}_i$ and use incompressibility, $A_{ii}=0$. The displayed vector can be expressed asOn taking its divergence, ${\partial }_i{A}_{ij}=0$. Commuting the remaining partial derivatives giveswhich cancels the derivative of the second term of $D_i$. The derivatives falling on $u_k$ leaveIn homogeneous turbulence, averaging commutes with differentiation and $⟨D_i⟩$ is position independent, provided these moments exist. Consequently $⟨\mathrm{tr}(A^3)⟩=0$.

Using the strain-rate tensor and vorticity decomposition supplied in the question now gives the Betchov relation:This uses homogeneity and incompressibility; isotropy is unnecessary.

Use $A=S+\Omega$, where $S$ is the rate-of-strain tensor and $\Omega$ is antisymmetric. For incompressible flow $\mathrm{tr}S=0$. Since ${\Omega }^2=(\mathbf{\omega }{\mathbf{\omega }}^T-{\omega }^{2}I)/4$, taking a trace givesThis uses the positive convention $R=\mathrm{tr}{A}^3/3$ specified here; a convention with the opposite sign for $R$ must not be mixed into the calculation. For a compressible flow there is an additional term $-3{\omega }^2\mathrm{tr}S/4$, so the displayed incompressible formula is not literally valid for every flow.

One can exhibit the required divergence directly. DefineDifferentiating, terms containing ${\partial }_i{u}_i$ vanish, and the two terms with a second derivative cancel after relabeling indices. Thereforein homogeneous turbulence with finite differentiable moments, or under periodic averaging. This is the Betchov relation. The preceding trace identity then impliesThe principal strain rates are the real eigenvalues of the symmetric matrix $S$. Incompressibility gives $a+b+c=0$, and the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ givesNo isotropy is required for this averaged tensor identity: statistical homogeneity and incompressibility are the key assumptions.

Taylor turbulent dispersion describes the absolute displacement of a marked fluid element in stationary homogeneous turbulence. Here $R_L$ is its Lagrangian velocity autocorrelation. Continuous correlation at zero gives ballistic spreading at short times, and an integrable correlation gives diffusive spreading at long times. This is distinct from shear-induced Taylor dispersion and from Richardson pair dispersion.