= Germano identity
{c}
{title2=$L_{ij}=T_{ij}-\widehat{\tau}_{ij}$}
The <Germano identity> is an exact relation between stresses at two filtering levels and a stress computable from the resolved field. Let a bar denote a <grid filter> and a hat a <test filter>. With the <subgrid-scale stress> convention
$$
\tau_{ij}=\overline{u_i u_j}-\bar u_i\bar u_j,\qquad
T_{ij}=\widehat{\overline{u_i u_j}}-\widehat{\bar u}_i\widehat{\bar u}_j,
$$
the identity states
$$
L_{ij}:=\widehat{\bar u_i\bar u_j}-\widehat{\bar u}_i\widehat{\bar u}_j
=T_{ij}-\widehat{\tau}_{ij}.
$$
It follows by substituting $\widehat{\tau}_{ij}=\widehat{\overline{u_i u_j}}-\widehat{\bar u_i\bar u_j}$ and cancelling the common term. <Massimo Germano> and collaborators used this relation to construct a <dynamic subgrid-scale model>.<A dynamic subgrid-scale eddy viscosity model by Massimo Germano and others (1990)>[§2, equations 8–10, resolved stress and algebraic two-filter identity]{ref}
A scalar block-average example makes every term visible. Start with four velocity values grouped as $(0,2)$ and $(2,4)$. Averaging within each group gives $\bar u=(1,3)$ and $\overline{u^2}=(2,10)$, hence $\tau=(1,1)$. A test average over the two groups then gives
$$
T=\frac{2+10}{2}-\left(\frac{1+3}{2}\right)^2=2,\qquad
\widehat\tau=1,\qquad
L=\frac{1^2+3^2}{2}-2^2=1.
$$
Thus $L=T-\widehat\tau$ reads $1=2-1$. The resolved group averages contain enough information for $L$, although they alone do not determine the original within-group variances.
The identity has no fitted coefficient. To estimate a missing <subgrid-scale stress>, a closure assumes some model form at both filtering levels and tries to make their difference agree with the known $L_{ij}$. Extracting a varying coefficient from a filter, choosing a filter width and stabilizing a fit are modeling decisions; they do not enter the cancellation proof.
Back to article page
