M3. Tensor Fields in Curvilinear Coordinates

Mathematical tools M3. Tensor ﬁelds in curvilinear coordinate systems

AleˇsJanka

oﬃce Math 0.107 [email protected] http://perso.unifr.ch/ales.janka/mechanics

November 24, 2010, Universit´ede Fribourg

Mathematical tools M3. Tensor ﬁelds in curvilinear coordinates 1. Curvilinear coordinate system

Position vector of point M (with respect to the origin): −−→OM = x

Let x : R3 R3 be a smooth bijective mapping: → x :(ξ1, ξ2, ξ3)T x(ξ1, ξ2, ξ3) 7→ Curvilinear coordinates of a point: ξ1, ξ2, ξ3. Coordinate curves through a point: parametric curves given by

g 2 x( 1, , 3 ) 2 3 ξ β ξ x1(α): α x(α, ξ , ξ ) 7→ 1 3 M g x2(β): β x(ξ , β, ξ ) x 1 7→ 1 2 x3(γ): γ x(ξ , ξ , γ) 7→ O x( α , ξ 2 , ξ 3 ) 1 2 3 1 2 3 all 3 curves pass through x(ξ , ξ , ξ ) (=point M with ξ , ξ , ξ ﬁxed).

Mathematical tools M3. Tensor ﬁelds in curvilinear coordinates 1. Curvilinear coordinate system: local basis

Local basis: composed of tangents to the coordinate curves: ∂x g = i ∂ξi 3 We suppose here that g1, g2 and g3 are linearly independent in R . Covariant local basis: differential of the position vector x: ∂x dx = dξi = g dξi ∂ξi i “How much x(ξ1, ξ2, ξ3) changes if we perturb ξi by dξi .” Contravariant basis is induced as before so that g gj = δj i · i Metric tensor: g = g g ... (as before) ij i · j Huge difference with what we have seen so far: 1 2 3 gi = gi (x(ξ , ξ , ξ )) ie. the basis is not constant for all points x R3! ∈ Mathematical tools M3. Tensor fields in curvilinear coordinates 1.1 Curvilinear coordinate system: infinitensimal volume

Inﬁnitensimal volume due to coordinate change:

+d1 1 3 dV = (g g ) g dξ1 dξ2 dξ3 x( ξ ξ , β , ξ ) | 1 × 2 · 3| g2 1 3 √g x( ξ , β , ξ )

| {z } g T 1 with g = det[gij ] = det(F F) = Ωx 2 det (F), where F = g g g . +d2 2 3 1| 2| 3 x( α , ξ ξ , ξ ) i O 2 In cartesian coordinates: dx = dx ei x( α , ξ , ξ 3 ) and dV = dx1 dx2 dx3. Volume integral of a scalar ﬁeld f (x) in curvilinear coords:

f (x) dx1 dx2 dx3 = f (x(ξ1, ξ2, ξ3)) √g dξ1 dξ2 dξ3 dV ΩZx ΩZξ dV | {z } | {z } with Ωx = x(Ωξ).

Mathematical tools M3. Tensor fields in curvilinear coordinates 2. Differential (resp. gradient) of a scalar field

Scalar ﬁeld: f : x Ω R3 R ∈ ⊂ 7→ Gradient of f (x) (with respect to the position x): it is a vector f R3 such that for all dx R3: ∇ ∈ ∈ f (x + dx) = f (x) + f (x) dx +o(dx) ∇ · df

here, df is the diﬀerential of f (x) along| {zdx.} Gradient and the directional derivative of f (x) The gradient of f (x) is such a vector f R3 for which ∇ ∈

d 3 [ f (x)] d = f (x + αd) d R . ∇ · dα ∀ ∈ α=0

The definition is independent of the choice of basis g1, g2, g3 Hence, f (x) is a field of tensors of order N = 1 (vector field). ∇ Mathematical tools M3. Tensor fields in curvilinear coordinates 2.1 Coordinates of f in the local basis ∇ f (x+dx) = f (x) + f (x) dx + o(dx) ... definition of gradient ∇ · ∂f = f (x) + dξi + o(dξk ) ... f (x) as a function of ξi ∂ξi Xk ∂f = f (x) + δi dξj + o(dξk ) ∂ξi j Xk ∂f = f (x) + gi g dξj +o(dx) ... cf. the first line ∂ξi · j dx f (x) ∇ | {z } ∂f | {z } Hence, f = gi Covariant components of f (x) are: ∇ dξi ⇒ ∇ ∂f ( f ) = ... They coincide with ∂f ! ∇ i ∂ξi ∂ξi f =( f ) = ∂f is named “the covariant derivative of scalar field f ” ∇i ∇ i ∂ξi The differential df then expressed “in coordinates”: df = f dξi ∇i Mathematical tools M3. Tensor fields in curvilinear coordinates 3. Differential (resp. gradient) of a vector field

Vector ﬁeld u(x) : (e.g. velocity, displacements, el. current, . . . ) ﬁeld of tensors of order N = 1: 3 3 u : x Ω R R ∈ ⊂ → Components of u in the local basis g : h i i i i u(x) = u gi . . . both u and gi depend on x!

Differential du: change in u going from x to x+dx (up to o(dx)): ∂u u(x+dx) u(x) + du = u(x) + dξj = u(x) + u dx ≈ ∂ξj ∇ · i i i From u = u gi , by chain rule (both u and gi depend on x, ie. ξ !): ∂ui ∂g du = dξj g + ui i dξk (1) ∂ξj i ∂ξk Change due to changing local coordinates ui of u Change due to the curvature of the coordinate system Mathematical tools M3. Tensor fields in curvilinear coordinates 3.1 Differential of a vector field: contravariant components

Contravariant components of du: du` = du g`, du = du` g · ` with respect to the local basis g at the point x (not at x+dx!). h i i ∂ui ∂g du = dξj g + ui i dξk g` ∂ξj i ∂ξk ·

Hence, the contravariant components of du:

∂ui ∂g du` = du g` = dξj g g` +ui i g` dξk · ∂ξj i · ∂ξk · ` δi ∂u` ∂g | {z } ∂u` = dξj + ui i g` dξj = + Γ` ui dξj ∂ξj ∂ξj · ∂ξj ij ` Γij where we deﬁne | {z } ∂g Γ` = i g`, ij ∂ξj · the Christoﬀel symbols of the second kind (not a tensor!).

Mathematical tools M3. Tensor fields in curvilinear coordinates 3.2 Differential of a vector field: covariant components

i Analogously to (1), from u = ui g , by chain rule:

∂u ∂gi du = i dξj gi + u dξk ∂ξj i ∂ξk

Change due to changing local coordinates ui of u Change due to the curvature of the coordinate system

Aside diﬀerentiation to get rid of the contravariant basis gi (which is less used):

∂ gi g = δi · ` ` ∂ξj

∂gi ∂g g + gi ` = 0 ∂ξj · ` · ∂ξj ∂gi ∂g Hence, g = gi ` ∂ξj · ` − · ∂ξj Mathematical tools M3. Tensor fields in curvilinear coordinates 3.2 Differential of a vector field: covariant components

Covariant components of du: du = du g , du = du g`: ` · ` ` ∂u ∂gi du = i dξj gi + u dξk g ∂ξj i ∂ξk · `

Hence, by applying the aside diﬀerentiation: ∂u ∂gi du = du g = i dξj gi g +u g dξk ` · ` ∂ξj · ` i ∂ξk · ` i δ` ∂u` ∂g` | {z } ∂u` = dξj u gi dξk = u Γi dξj ∂ξj − i ∂ξk · ∂ξj − i `j i Γ`k | {z }

Mathematical tools M3. Tensor fields in curvilinear coordinates 3.3 Differential of a vector field and covariant derivatives

Perturbations of the position x(ξ1, ξ2, ξ3) by dξ1, dξ2, dξ3 du: −→ ∂u dx = dξi g , du = dξj = u dx = du` g = du g` i ∂ξj ∇ · ` ` Contravariant and covariant coordinates of the differential du: ∂u` du` = + Γ` ui dξj ∂ξj ij u` ∇j | {z } ∂u du = ` Γi u dξj ` ∂ξj − `j i u ∇j ` | {z } where u` is the covariant derivative of contravariant tensor ∇j and u is the covariant derivative of covariant tensor ∇j ` Mathematical tools M3. Tensor fields in curvilinear coordinates 3.4 Covariant derivatives and gradient of a vector field Differential du using covariant derivatives resp. du = u dx: ∇ · ∂u` du = du` g = + Γ` ui g dξj ` ∂ξj ij ` u` ∇j = u` g δj dξk = u` g gj g dξk ∇j ` |k {z ∇j } ` ⊗ · k gj g u dx · k ∇ Hence, the gradient u|{z}is a 2nd order| tensor{z with:} | {z } ∇ ∂u` u = + Γ` ui g gj = u` g gj ∇ ∂ξj ij ` ⊗ ∇j ` ⊗ the covariant derivative u` is in fact the tensor u in ⇒ ∇j ∇ mixed components u ` !! ∇ , j Similarly, u are the 2 covariant components u of u: ⇒ ∇j ` × ∇ `,j ∇ ∂u` i ` j ` j u = Γ ui g g = j u` g g ∇ ∂ξj − `j ⊗ ∇ ⊗ h i h i Mathematical tools M3. Tensor fields in curvilinear coordinates 4. Covariant derivatives of higher order tensor T , N 2 ≥ Remember: covariant deriv. of scalar field = its partial deriv.: ∂f f = ∇k ∂ξk We can exploit it: Multiply T by N arbitrary vector-fields a, b, . . . to form a scalar f . Example for 2nd-order tensor T : ij f = T ai bj Apply the “covariant = partial” trick on f : ∂ T ij a b = T ij a b ∇k i j ∂ξk i j ∂T ij ∂a ∂b = a b + T ij i b + T ij a j ∂ξk i j ∂ξk j i ∂ξk For the arbitrary vector fields a, b, . . . we know how to make a covariant derivative: ∂a ∂a a = ` Γi a i.e. i = a + Γm a ∇j ` ∂ξj − `j i ∂ξk ∇k i ik m

Mathematical tools M3. Tensor ﬁelds in curvilinear coordinates 4. Covariant derivatives of higher order tensor T , N 2 ≥ In (T ij a b ), replace ∂ of a, b,. . . by terms containing : ∇k i j ∂ξk ∇k ∂T ij ∂a ∂b T ij a b = a b + T ij i b + T ij a j ∇k i j ∂ξk i j ∂ξk j i ∂ξk ∂T ij = a b + T ij ( a + Γm a ) b + ∂ξk i j ∇k i ik m j +T ij a b + Γm b i ∇k j jk m ∂T ij = + Γi T `j + Γj T i` a b + ∂ξk `k `k i j +T ij a b + T ij a b ∇k i j i ∇k j Here, we re-indexed conveniently dummy indices in order to regroup terms with “ai bj ”. By analogy with (a b c) = a bc + ab c + abc , the term in brackets is T ij : · · 0 0 0 0 ∇k ∂T ij T ij = + Γi T `j + Γj T i` ∇k ∂ξk `k `k

Mathematical tools M3. Tensor ﬁelds in curvilinear coordinates 5. Divergence of tensor ﬁelds

Divergence of a vector ﬁeld u(x): deﬁnition

u ds 3 3 ∂Ω · div : C(R , R ) R div u = lim → Ω 0 R Ω → | | In cartesian coordinates: ∂ui div u = tr ( u) = ∇ ∂xi In curvilinear coordinates: must replace ∂ by : ∂xi ∇i div u = tr ( u) = ui = δk ui = g ki u ∇ ∇i i ∇k ∇i k Generalization to higher-order tensor-ﬁelds:

div T i = T ki ∇k Mathematical tools M3. Tensor ﬁelds in curvilinear coordinates 6. How to calculate Christoﬀel symbols?

Two ways of calculating Christoffel symbols of 2nd kind: By definition: ∂g Γ` = i g` ij ∂ξj · Through the metric tensor: 1 ∂g ∂g ∂g Γ` = g k` ki + jk ij ij 2 ∂ξj ∂ξi − ∂ξk 2 Γ · k,ij ∂g | {z } Γ = i g are the Christoffel symbols of the 1st kind. k,ij ∂ξj · k Notable property of Christoffel symbols: symmetry in (i, j):

∂g ∂x ∂2x ∂2x Γ` = i g` = g = = g` = g` = Γ` ij ∂ξj · i ∂ξi ∂ξj ∂ξi · ∂ξi ∂ξj · ji

Mathematical tools M3. Tensor ﬁelds in curvilinear coordinates 6.1 Christoﬀel symbols through the metric tensor

∂g ∂x ∂2x ∂2x Γ` = i g` = g = = g` = g g m` ik ∂ξk · i ∂ξi ∂ξk ∂ξi · ∂ξk ∂ξi · m

Γm,ki

∂2x ∂x ∂ ∂x ∂x |∂ {z∂x } ∂x Γ = = = m,ki ∂ξk ∂ξi · ∂ξm ∂ξk ∂ξi · ∂ξm ∂ξi ∂ξk · ∂ξm Hence, by combining half-and-half:

1 ∂ ∂x ∂x ∂ ∂x ∂x Γ =  +  m,ki 2 ∂ξk ∂ξi · ∂ξm ∂ξi ∂ξk · ∂ξm    g gm g gm   i k    |{z} |{z} 1 ∂ | {z } ∂ | {z } ∂g ∂g = (g g ) + (g g ) m g m g 2 ∂ξk m · i ∂ξi m · k − ∂ξk · i − ∂ξi · k  g g  mi km   Mathematical tools M3. Tensor ﬁelds in curvilinear coordinates  | {z } | {z } 6.1 Christoﬀel symbols through the metric tensor

1 ∂ ∂ ∂g ∂g Γ = (g g ) + (g g ) m g + m g m,ki 2 ∂ξk m · i ∂ξi m · k − ∂ξk · i ∂ξi · k  g g  mi km    and | {z } | {z } ∂g ∂g ∂2x ∂2x m g + m g = g + g ∂ξk · i ∂ξi · k ∂ξm ∂ξk · i ∂ξm ∂ξi · k ∂g ∂g ∂ ∂g = k g + i g = (g g ) = ik ∂ξm · i ∂ξm · k ∂ξm i · k ∂ξm Hence, 1 ∂g ∂g ∂g Γ` = g m` Γ = g m` mi + km ik ik m,ik 2 ∂ξk ∂ξi − ∂ξm

Mathematical tools M3. Tensor ﬁelds in curvilinear coordinates