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
(ū)i = ui If ū =  v  then u2 = v
w
Matrix (2nd order tensor)
(A)ij = Aij

a11
If A = a21
a31

a13
a23  then A23 = a23
a33
a12
a22
a32
Kronecker delta (2nd order tensor)
1 if i = j
δij = (I)ij =
0 if i 6= j
To indicate operation among tensor we will use Einstein summation convention (summation over repeated
indices)
ui ui =
3
X
ui ui
i is called dummy index (as opposed to free index) and can be renamed
i=1
Example: Kinetic energy per unit volume
1
2
2 ρ|ū|
= 12 ρ(u2 + v 2 + w2 ) = 12 ρui ui
Matrix/Tensor operations
(ā · b̄) = a1 b1 + a2 b2 + a3 b3 = ai bi = δij ai bj = aj bj
(ā b̄)ij = (ā ⊗ b̄)ij = ai bj
(Ab̄)i = Aij bj
(scalar, inner product)
(diadic, tensor product)
(matrix-vector multiplication, inner product)
(AB)ij = (A · B)ij = Aik Bkj
(matrix multiplication, inner product)
(AB)ijkl = (A ⊗ B)ijkl = Aij Bkl
(A : B) = Aij Bij
(diadic, tensor product)
(double contraction)
tr(A) = A11 + A22 + A33 = Aii
(trace)
(A)ij = Aij ⇐⇒ (AT )ij = Aji
(transpose)
1
Permutation symbol

 1 if ijk in cyclic order. ijk = 123, 231 or 312
0 if any two indices are equal
εijk =

−1 if ijk in anticyclic order. ijk = 321, 213 or 132
Vector (Cross) product
ē1 ē2 ē3
ā × b̄ = a1 a2 a3
b1 b2 b3
= ē1 (a2 b3 − a3 b2 ) − ē2 (a1 b3 − a3 b1 ) + ē3 (a1 b2 − a2 b1 ) = εijk ēi aj bk
(ā × b̄)i = εijk aj bk
εijk εilm = δjl δkm − δjm δkl
εijk εijm = {l = j} = 3δkm − δjm δkj = 3δkm − δmk = 2δkm
εijk εijk = {m = k} = 2 · 3 = 6
Example: Rewrite without the cross product:
¯ = (ā × b̄)i (c̄ × d)
¯ i = εijk aj bk εilm cl dm = εijk εilm aj bk cl dm =
(ā × b̄) · (c̄ × d)
¯ − (ā · d)(
¯ b̄ · c̄)
(δjl δkm − δjm δkl ) aj bk cl dm = aj cj bk dk − aj dj bk ck = (ā · c̄)(b̄ · d)
Tensor Invariants
bij , b2ij = bik bkj 6= (bij )2 , b3ij = bik bkl blj 6= (bij )3
Any scalar obtained from a tensor (e.g. bii , b2ii , . . . ) 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
n
o
2
I2 = 21 (bii )2 − b2ii = 12 [tr(B)] − tr(B2 ) = λ1 λ2 + λ2 λ3 + λ1 λ3
I3 = 61 (bii )3 − 12 bii b2jj + 13 b3ii = det(B) = λ1 λ2 λ3
Decomposition of Tensors
Tij = TijS + TijA symmetric and anti-symmetric parts
S
symmetric
TijS = 21 Tij + Tji = Tji
TijA =
1
2
A
Tij − Tji = −Tji
anti-symmetric
The symmetric part of the tensor can be divided further into a trace-less and an isotropic part:
T S = T̄ + T̄¯
ij
ij
T̄ij =
TijS
ij
−
1
3 Tkk δij
T̄¯ij = 13 Tkk δij
trace-less
isotropic
This gives:
2
Tij = TijS + TijA = T̄ij + T̄¯ij + TijA
∂ui
(deformation-rate tensor). The anti-symmetric part
∂xj
describes rotation, the isotropic part describes the volume change and the trace-less part describes the deformation of a fluid element.
In the Navier-Stokes equations we have the tensor
Operators
(∇p)i =
∂
p
∂xi
(gradient, increase of tensor order)
∆p = ∇ · ∇p = ∇2 p =
∇ · ū =
∂
ui
∂xi
(∇ · A)j =
(Laplace operator)
(divergence, decrease of tensor order)
∂
Aij
∂xi
(divergence of a tensor)
(∇ū)ij = (∇ ⊗ ū)ij =
(∇ × ū)i = εijk
∂2
p
∂xi ∂xi
∂
uj
∂xi
∂
uk
∂xj
(gradient of a vector)
(curl)
Gauss theorem (general)
Gauss theorem (divergence theorem):
Z
I
F̄ · n̄ dS =
∇ · F̄ dV
V
S
or with index notation,
I
Z
∂Fi
Fi ni dS =
dV
S
V ∂xi
In general we can write,
I
Z
∂
Tijk dV
Tijk nl dS =
∂x
l
S
V
Example: Put Tijk nl = Tij ul nl
I
Z
∂
(ul Tij ) dV
Tij ul nl dS =
∂x
l
S
V
or,
I
Z
T(ū · n̄) dS =
(∇ · ū)T + (ū · ∇)T dV
S
V
Identities
• Derive the identity
∇ × (F̄ × Ḡ) = (Ḡ · ∇)F̄ + (∇ · Ḡ)F̄ − (∇ · F̄ )Ḡ − (F̄ · ∇)Ḡ
∂
∂
∂
∇ × (F̄ × Ḡ) i = εijk
εklm Fl Gm = εijk εklm
Fl Gm = εkij εklm
Fl Gm =
∂xj
∂xj
∂xj
3
(δil δjm − δim δjl )
Gj
∂
∂
∂
∂Fi
∂Gj
∂Fj
∂Gi
Fl Gm =
Fi Gj −
Fj Gi =
Gj +
Fi −
Gi +
Fj =
∂xj
∂xj
∂xj
∂xj
∂xj
∂xj
∂xj
∂Fi
∂Gj
∂Fj
∂Gi
+
Fi −
Gi − Fj
= [(Ḡ · ∇)F̄ + (∇ · Ḡ)F̄ − (∇ · F̄ )Ḡ − (F̄ · ∇)Ḡ]i
∂xj
∂xj
∂xj
∂xj
• Show that
∇ · (∇ × F̄ ) = 0
∇ · (∇ × F̄ ) =
∂
∂
∂ ∂
εijk
Fk = εijk
Fk
∂xi
∂xj
∂xi ∂xj
Remember that εijk = −εjik and that
εijk
∂ ∂
∂ ∂
Fk =
Fk and thus all terms will cancel.
∂xi ∂xj
∂xj ∂xi
∂ ∂
Fk = 0
∂xi ∂xj
• Similarly, show that
∇ × (∇f ) = 0
∇ × (∇f ) = εijk
∂2
f = 0 according to the same argument as above.
∂xj ∂xk
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