Fluid Mechanics, SG2214, HT2013 September 4, 2013

Exercise 1: Tensors and Invariants

Tensor/Index Notation

Scalar (0th order tensor), usually we consider scalar fields function of space and time p = p(x, y, z, t)

Vector (1st order tensor), defined by direction and magnitude u (¯u)i = ui If u¯ = v  then u2 = v w

Matrix (2nd order tensor)   a11 a12 a13 (A)ij = Aij If A = a21 a22 a23 then A23 = a23 a31 a32 a33

Kronecker delta (2nd order tensor)  1 if i = j δ = (I) = ij ij 0 if i 6= j

To indicate operation among tensor we will use Einstein summation convention (summation over repeated indices)

3 X uiui = uiui i is called dummy index (as opposed to free index) and can be renamed i=1

Example: Kinetic energy per unit volume 1 2 1 2 2 2 1 2 ρ|u¯| = 2 ρ(u + v + w ) = 2 ρuiui

Matrix/Tensor operations ¯ (¯a · b) = a1b1 + a2b2 + a3b3 = aibi = δijaibj = ajbj (scalar, inner product)

¯ ¯ (¯a b)ij = (¯a ⊗ b)ij = aibj (diadic, tensor product)

¯ (Ab)i = Aijbj (matrix-vector multiplication, inner product)

(AB)ij = (A · B)ij = AikBkj (matrix multiplication, inner product)

(AB)ijkl = (A ⊗ B)ijkl = AijBkl (diadic, tensor product)

(A : B) = AijBij (double contraction)

tr(A) = A11 + A22 + A33 = Aii (trace)

T (A)ij = Aij ⇐⇒ (A )ij = Aji (transpose)

1 Permutation symbol   1 if ijk in cyclic order. ijk = 123, 231 or 312 εijk = 0 if any two indices are equal −1 if ijk in anticyclic order. ijk = 321, 213 or 132

Vector (Cross) product

e¯1 e¯2 e¯3 ¯ a¯ × b = a1 a2 a3 =e ¯1(a2b3 − a3b2) − e¯2(a1b3 − a3b1) +e ¯3(a1b2 − a2b1) = εijke¯iajbk

b1 b2 b3

¯ (¯a × b)i = εijkajbk

εijkεilm = δjlδkm − δjmδkl

εijkεijm = {l = j} = 3δkm − δjmδkj = 3δkm − δmk = 2δkm

εijkεijk = {m = k} = 2 · 3 = 6

Example: Rewrite without the cross product: ¯ ¯ ¯ ¯ (¯a × b) · (¯c × d) = (¯a × b)i(¯c × d)i = εijkaj bkεilmcldm = εijkεilm aj bk cl dm = ¯ ¯ ¯ ¯ (δjlδkm − δjmδkl) aj bk cl dm = ajcjbkdk − ajdjbkck = (¯a · c¯)(b · d) − (¯a · d)(b · c¯)

Tensor Invariants

2 2 3 3 bij, bij = bikbkj 6= (bij) , bij = bikbklblj 6= (bij) 2 Any scalar obtained from a tensor (e.g. bii, bii, . . . ) is invariant, i.e. independent of the coordinate system. The principal invariants are defined by (λi are the eigenvalues of B)

I1 = bii = tr(B) = λ1 + λ2 + λ3

1  2 2  1 n 2 2 o I2 = 2 (bii) − bii = 2 [tr(B)] − tr(B ) = λ1λ2 + λ2λ3 + λ1λ3

1 3 1 2 1 3 I3 = 6 (bii) − 2 biibjj + 3 bii = det(B) = λ1λ2λ3

Decomposition of Tensors

S A Tij = Tij + Tij symmetric and anti-symmetric parts   S 1 S Tij = 2 Tij + Tji = Tji symmetric   A 1 A Tij = 2 Tij − Tji = −Tji anti-symmetric

The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: S ¯ ¯ Tij = Tij + Tij ¯ S 1 Tij = Tij − 3 Tkkδij trace-less ¯ 1 Tij = 3 Tkkδij isotropic

This gives:

2 S A ¯ ¯ A Tij = Tij + Tij = Tij + Tij + Tij ∂u In the Navier-Stokes equations we have the tensor i (deformation-rate tensor). The anti-symmetric part ∂xj describes rotation, the isotropic part describes the volume change and the trace-less part describes the defor- mation of a fluid element.

Operators

∂ (∇p)i = p (gradient, increase of tensor order) ∂xi ∂2 ∆p = ∇ · ∇p = ∇2p = p (Laplace operator) ∂xi∂xi ∂ ∇ · u¯ = ui (divergence, decrease of tensor order) ∂xi ∂ (∇ · A)j = Aij (divergence of a tensor) ∂xi ∂ (∇u¯)ij = (∇ ⊗ u¯)ij = uj (gradient of a vector) ∂xi ∂ (∇ × u¯)i = εijk uk (curl) ∂xj

Gauss theorem (general)

Gauss theorem (divergence theorem): I Z F¯ · n¯ dS = ∇ · F¯ dV S V or with index notation, I Z ∂Fi Fi ni dS = dV S V ∂xi

In general we can write, I Z ∂ Tijk nl dS = Tijk dV S V ∂xl

Example: Put Tijk nl = Tij ul nl I Z ∂ Tij ul nl dS = (ul Tij) dV S V ∂xl or, I Z T(¯u · n¯) dS = (∇ · u¯)T + (¯u · ∇)T
dV S V Identities

• Derive the identity

∇ × (F¯ × G¯) = (G¯ · ∇)F¯ + (∇ · G¯)F¯ − (∇ · F¯)G¯ − (F¯ · ∇)G¯

¯ ¯
∂ ∂ ∂ ∇ × (F × G) i = εijk εklmFl Gm = εijkεklm Fl Gm = εkijεklm Fl Gm = ∂xj ∂xj ∂xj

3 ∂ ∂ ∂ ∂Fi ∂Gj ∂Fj ∂Gi (δilδjm − δimδjl) Fl Gm = Fi Gj − Fj Gi = Gj + Fi − Gi + Fj = ∂xj ∂xj ∂xj ∂xj ∂xj ∂xj ∂xj

∂Fi ∂Gj ∂Fj ∂Gi Gj + Fi − Gi − Fj = [(G¯ · ∇)F¯ + (∇ · G¯)F¯ − (∇ · F¯)G¯ − (F¯ · ∇)G¯]i ∂xj ∂xj ∂xj ∂xj

• Show that ∇ · (∇ × F¯) = 0 ∂ ∂ ∂ ∂ ∇ · (∇ × F¯) = εijk Fk = εijk Fk ∂xi ∂xj ∂xi ∂xj ∂ ∂ ∂ ∂ Remember that εijk = −εjik and that Fk = Fk and thus all terms will cancel. ∂xi ∂xj ∂xj ∂xi ∂ ∂ εijk Fk = 0 ∂xi ∂xj

• Similarly, show that ∇ × (∇f) = 0

∂2 ∇ × (∇f) = εijk f = 0 according to the same argument as above. ∂xj∂xk

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