Isotropic tensors - Turbulence in the Atmosphere - page 315

the mean gradient of a scalar, being vectors, vanish in isotropic turbulence. Likewise
any odd-order single-point tensor will change sign under coordinate reflection, so
the isotropic form vanishes.
The situation is different for even orders, as we shall illustrate with the Robertson
technique. Let T ik be an isotropic, second-order, single-point tensor, one that
involves properties at one point in space. Contracting it with two arbitrary vec-
tors A and B yields a scalar that is bilinear in A and B and depends also on the
invariants of T ik :

T ik A i B k = α( invariants )A i B i = αA i B k δ ik . (14.4)

Rewriting this gives

(T ik − αδ ik )A i B k = 0 , ik = αδ ik . (14.5)

Equation (14.5) says that a second-order isotropic single-point tensor has nonzero
components only on the diagonal, and those diagonal components are equal. If the
diagonal components were not equal, they would change under coordinate rotation
(Problem 14.6) , so the tensor would be anisotropic.
The pressure-stress tensor is an application of Eq. (14.5) . A stress tensor σ ik , say,
expresses the force in the i direction per unit area on a surface whose normal is in
the k direction. The force per unit area on the surface is the dot product of the stress
tensor and n i , the unit normal to the surface. Since pressure acts normal to a surface,
whatever the orientation of that surface, the pressure-stress tensor is isotropic. It
must then have the form σ ik =− pδ ik , where the scalar p is the magnitude of

pressure. The pressure force f i is

f i = σ ik n k =− pδ ik n k =− pn i , (14.6)

which is indeed normal to the surface.
To determine the form of the fourth-order isotropic tensor T ij km , we contract
T ij km with four arbitrary vectors and write the resulting scalar as

T ij km A i B j C k D m = αA i B i C j D j + βA i C i B j D j + γA i D i B j C j

= αA i B j δ ij C k D m δ km + βA i C k δ ik B j D m δ jm

(14.7)

+ γA i D m δ im B j C k δ jk

= A i B j C k D m (αδ ij δ km + βδ ik δ jm + γδ im δ jk ).

Since A , B , C , and D are arbitrary we can rewrite this to find

T ij km = αδ ij δ km + βδ ik δ jm + γδ im δ jk . (14.8)

T ij km contains the three unknown scalars α, β ,and γ .