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 2u3 (¯u)i = ui If u¯ = 4v 5 then u2 = v w Matrix (2nd order tensor) 2 3 a11 a12 a13 (A)ij = Aij If A = 4a21 a22 a235 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 ρju¯j = 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 8 < 1 if ijk in cyclic order. ijk = 123, 231 or 312 "ijk = 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) = "ijke¯iajbk b1 b2 b3 ¯ (¯a × b)i = "ijkajbk "ijk"ilm = δjlδkm − δjmδkl "ijk"ijm = fl = jg = 3δkm − δjmδkj = 3δkm − δmk = 2δkm "ijk"ijk = fm = kg = 2 · 3 = 6 Example: Rewrite without the cross product: ¯ ¯ ¯ ¯ (¯a × b) · (¯c × d) = (¯a × b)i(¯c × d)i = "ijkaj bk"ilmcldm = "ijk"ilm 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 @ (rp)i = p (gradient, increase of tensor order) @xi @2 ∆p = r · rp = r2p = p (Laplace operator) @xi@xi @ r · u¯ = ui (divergence, decrease of tensor order) @xi @ (r · A)j = Aij (divergence of a tensor) @xi @ (ru¯)ij = (r ⊗ u¯)ij = uj (gradient of a vector) @xi @ (r × u¯)i = "ijk uk (curl) @xj Gauss theorem (general) Gauss theorem (divergence theorem): I Z F¯ · n¯ dS = r · 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 = (r · u¯)T + (¯u · r)T dV S V Identities • Derive the identity r × (F¯ × G¯) = (G¯ · r)F¯ + (r · G¯)F¯ − (r · F¯)G¯ − (F¯ · r)G¯ ¯ ¯ @ @ @ r × (F × G) i = "ijk "klmFl Gm = "ijk"klm Fl Gm = "kij"klm 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¯ · r)F¯ + (r · G¯)F¯ − (r · F¯)G¯ − (F¯ · r)G¯]i @xj @xj @xj @xj • Show that r · (r × F¯) = 0 @ @ @ @ r · (r × F¯) = "ijk Fk = "ijk Fk @xi @xj @xi @xj @ @ @ @ Remember that "ijk = −"jik and that Fk = Fk and thus all terms will cancel. @xi @xj @xj @xi @ @ "ijk Fk = 0 @xi @xj • Similarly, show that r × (rf) = 0 @2 r × (rf) = "ijk f = 0 according to the same argument as above. @xj@xk 4.