Mathematical tools M3. Tensor fields in curvilinear coordinate systems Aleˇs Janka office Math 0.107
[email protected] http://perso.unifr.ch/ales.janka/mechanics
November 24, 2010, Universit´e de 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 : (ξ 1 , ξ 2 , ξ 3 )T 7→ x(ξ 1 , ξ 2 , ξ 3 ) Curvilinear coordinates of a point: ξ 1 , ξ 2 , ξ 3 . Coordinate curves through a point: parametric curves given by
2
g2
3
1
x( ξ , β , ξ 3)
x1 (α) : α 7→ x(α, ξ , ξ ) 1
3
x2 (β) : β 7→ x(ξ , β, ξ ) 1
x
2
x3 (γ) : γ 7→ x(ξ , ξ , γ) 1
2
M
g1
O 3
1
x( α , ξ 2, ξ 3) 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: gi =
∂x ∂ξ i
We suppose here that g1 , g2 and g3 are linearly independent in R3 . Covariant local basis: differential of the position vector x: dx =
∂x i i dξ = g dξ i ∂ξ i
“How much x(ξ 1 , ξ 2 , ξ 3 ) changes if we perturb ξ i by dξ i .” Contravariant basis is induced as before so that gi · gj = δij Metric tensor:
gij = gi · gj
. . . (as before)
Huge difference with what we have seen so far: gi = gi (x(ξ 1 , ξ 2 , ξ 3 )) 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: 1
2
dV = |(g1 × g2 ) · g3 | dξ dξ dξ | {z } √
1
3
g2
g
with g = det[gij ] = det(FT F) = det2 (F), where F = g1 |g2 |g3 . In cartesian coordinates: dx = dx i ei and dV = dx 1 dx 2 dx 3 .
1
g1
1
3
2
3
x( ξ , β , ξ )
Ωx x( α, ξ+dξ,2 ξ )
O
x( α , ξ 2, ξ 3)
Volume integral of a scalar field f (x) in curvilinear coords: Z Z √ 1 2 3 1 2 3 1 2 3 f (x) dx dx dx = f (x(ξ , ξ , ξ )) g dξ dξ dξ | {z } | {z } Ωx
dV
Ωξ
dV
with Ωx = x(Ωξ ). Mathematical tools
3
x( ξ+dξ, β, ξ )
M3. Tensor fields in curvilinear coordinates
2. Differential (resp. gradient) of a scalar field Scalar field: f : x ∈ Ω ⊂ R3 7→ R 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) | {z } df
here, df is the differential of f (x) along dx. Gradient and the directional derivative of f (x) The gradient of f (x) is such a vector ∇f ∈ R3 for which d [∇f (x)] · d = f (x + αd) ∀d ∈ R3 . 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 X ∂f i = f (x) + i dξ + o(dξ k ) . . . f (x) as a function of ξ i ∂ξ k ∂f i j X = f (x) + i δj dξ + o(dξ k ) ∂ξ k
= f (x) +
∂f i g · gj dξ j +o(dx) i ∂ξ | {z } | {z } dx
. . . cf. the first line
∇f (x)
Hence, ∇f =
∂f i g ⇒ Covariant components of ∇f (x) are: dξ i
(∇f )i =
∂f ∂ξ i
. . . They coincide with
∂f ! ∂ξ i
∂f ∇i f = (∇f )i = ∂ξ i is named “the covariant derivative of scalar field f ” The differential df then expressed “in coordinates”: df = ∇i f dξ 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: u : x ∈ Ω ⊂ R3 → R3 Components of u in the local basis hgi i: u(x) = u i gi
. . . both u i and gi depend on x!
Differential du: change in u going from x to x+dx (up to o(dx)): u(x+dx) ≈ u(x) + du = u(x) +
∂u j dξ = u(x) + ∇u · dx ∂ξ j
From u = u i gi , by chain rule (both u i and gi depend on x, ie. ξ i !): ∂u i j ∂gi du = j dξ gi + u i k dξ k ∂ξ ∂ξ
(1)
Change due to changing local coordinates u i 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 hgi i at the point x (not at x+dx!). ∂u i j ∂g i i k · g` dξ du = j dξ gi + u ∂ξ ∂ξ k Hence, the contravariant components of du: du
`
∂u i j ∂gi = du · g = j dξ gi · g` +u i k · g` dξ k | {z } ∂ξ ∂ξ `
δi`
=
∂u ` j i ∂gi ` j dξ + u · g dξ = ∂ξ j ∂ξ j | {z }
∂u ` ` i + Γ ij u ∂ξ j
dξ j
Γ`ij
where we define
∂gi ` · g , ∂ξ j the Christoffel symbols of the second kind (not a tensor!). Γ`ij =
Mathematical tools
M3. Tensor fields in curvilinear coordinates
3.2 Differential of a vector field: covariant components Analogously to (1), from u = ui gi , by chain rule: ∂gi ∂ui j i du = j dξ g + ui k dξ k ∂ξ ∂ξ 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` = δ`i ∂ξ j ∂gi i ∂g` · g + g · j =0 ` ∂ξ j ∂ξ ∂gi i ∂g` · g = −g · j ` ∂ξ j ∂ξ
Hence,
Mathematical tools
M3. Tensor fields in curvilinear coordinates
3.2 Differential of a vector field: covariant components Covariant components of du: du` = du · g` , ∂ui j i ∂gi du = j dξ g + ui k dξ k ∂ξ ∂ξ
du = du` g` : · g`
Hence, by applying the aside differentiation: du`
∂ui j i ∂gi = du · g` = j dξ g · g` +ui k · g` dξ k | {z } ∂ξ ∂ξ δ`i
=
∂u` j ∂g` i dξ − u · g dξ k = i j k ∂ξ ∂ξ | {z }
∂u` − ui Γi`j j ∂ξ
dξ j
Γi`k
Mathematical tools
M3. Tensor fields in curvilinear coordinates
3.3 Differential of a vector field and covariant derivatives Perturbations of the position x(ξ 1 , ξ 2 , ξ 3 ) by dξ 1 , dξ 2 , dξ 3 −→ du: dx = dξ i gi
,
du =
∂u j dξ = ∇u · dx = du ` g` = du` g` j ∂ξ
Contravariant and covariant coordinates of the differential du: ` ∂u du ` = + Γ`ij u i dξ j j ∂ξ {z } | ∇j u ` ∂u` du` = − Γi`j ui dξ j j ∂ξ | {z } ∇j u` where ∇j u ` is the covariant derivative of contravariant tensor and ∇j u` is the covariant derivative of covariant tensor 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`ij u i g` dξ j j ∂ξ {z } | ` ∇j u j ` = ∇j u g` δk dξ k = ∇j u ` g` ⊗ gj · gk dξ k |{z} {z } | {z } | gj ·gk
∇u
dx
Hence, the gradient ∇u is a 2nd order tensor with: ` ∂u ` i j ` j ∇u = + Γij u g` ⊗ g = ∇j u g` ⊗ g ∂ξ j ⇒ the covariant derivative ∇j u ` is in fact the tensor ∇u in ` mixed components ∇u , j !! ⇒ Similarly, ∇ components ∇u `,j of ∇u: j u` are the 2×covariant h i h i ∂u` i ` j ` j ∇u = − Γ`j ui g ⊗ g = ∇j u` g ⊗ g ∂ξ j 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 ∇k f = k ∂ξ We can exploit it: Multiply T by N arbitrary vector-fields a, b, . . . to form a scalar f . Example for 2nd-order tensor T : f = T ij ai bj Apply the “covariant = partial” trick on f : ∂ ij ∇k T ij ai bj = T a b i j ∂ξ k ∂bj ∂T ij ij ∂ai ij a b + T b + T a = i j j i ∂ξ k ∂ξ k ∂ξ k For the arbitrary vector fields a, b, . . . we know how to make a covariant derivative: ∂a` ∂ai ∇j a` = j − Γi`j ai i.e. = ∇k ai + Γm ik am k ∂ξ ∂ξ Mathematical tools
M3. Tensor fields in curvilinear coordinates
4. Covariant derivatives of higher order tensor T , N ≥ 2 In ∇k (T ij ai bj ), replace ij
∇k T ai bj
∂ ∂ξ k
of a, b,. . . by terms containing ∇k :
∂bj ∂T ij ij ∂ai ij = a b + T b + T a i j j i ∂ξ k ∂ξ k ∂ξ k ∂T ij = ai bj + T ij (∇k ai + Γm ik am ) bj + k ∂ξ ij m +T ai ∇k bj + Γjk bm ij ∂T j i `j = + Γ T + Γ T i` ai bj + `k `k k ∂ξ +T ij ∇k ai bj + T ij ai ∇k bj
Here, we re-indexed conveniently dummy indices in order to regroup terms with “ai bj ”. By analogy with (a · b · c)0 = a0 bc + ab 0 c + abc 0 , the term in brackets is ∇k T ij : ∂T ij ∇k T = + Γi`k T `j + Γj`k T i` k ∂ξ ij
Mathematical tools
M3. Tensor fields in curvilinear coordinates
5. Divergence of tensor fields Divergence of a vector field u(x): definition R div : C (R3 , R3 ) → R
div u = lim
∂Ω
Ω→0
u · ds |Ω|
In cartesian coordinates: ∂u i div u = tr (∇u) = ∂x i In curvilinear coordinates: must replace
∂ ∂x i
by ∇i :
div u = tr (∇u) = ∇i u i = δik ∇k u i = g ki ∇i uk Generalization to higher-order tensor-fields: div T
i
Mathematical tools
= ∇k T ki M3. Tensor fields in curvilinear coordinates
6. How to calculate Christoffel symbols? Two ways of calculating Christoffel symbols of 2nd kind: By definition: ∂gi Γ`ij = j · g` ∂ξ Through the metric tensor: ∂g ∂g 1 ∂g jk ij ki Γ`ij = g k` + − 2 ∂ξ j ∂ξ i ∂ξ k | {z } 2· Γk,ij Γk,ij =
∂gi · gk are the Christoffel symbols of the 1st kind. ∂ξ j
Notable property of Christoffel symbols: symmetry in (i, j): ∂g ∂x ∂2x ∂2x i ` ` ` Γij = j · g = gi = i = j i · g = i j · g` = Γ`ji ∂ξ ∂ξ ∂ξ ∂ξ ∂ξ ∂ξ Mathematical tools
M3. Tensor fields in curvilinear coordinates
6.1 Christoffel symbols through the metric tensor ∂g ∂x ∂2x ∂2x i ` ` ` Γik = k · g = gi = i = k i · g = k i · gm g m` ∂ξ ∂ξ ∂ξ ∂ξ ∂ξ ∂ξ {z } | Γm,ki
∂ ∂2x ∂x = k i · m = k ∂ξ ∂ξ ∂ξ ∂ξ
∂x ∂ξ i
Γm,ki
∂x ∂ · m = i ∂ξ ∂ξ
Hence, by combining half-and-half: Γm,ki
=
1 ∂ 2 ∂ξ k
∂x ∂ξ k
·
∂x ∂ξ m
∂x ∂x ∂x ∂x ∂ · · + ∂ξ i ∂ξ m ∂ξ i ∂ξ k ∂ξ m | {z } |{z} | {z } |{z} gm
gi
=
gm
gk
∂ ∂gm ∂gm 1 ∂ · g k (gm · gi ) + i (gm · gk ) − k · gi − k 2 ∂ξ | {z } ∂ξ | {z } ∂ξ ∂ξ i gmi
Mathematical tools
gkm
M3. Tensor fields in curvilinear coordinates
6.1 Christoffel symbols through the metric tensor Γm,ki =
1 ∂ ∂ k (gm · gi ) + i (gm · gk ) − 2 ∂ξ | {z } ∂ξ | {z } gmi
gkm
∂gm ∂gm · g + i ∂ξ i ∂ξ k
· gk
and ∂gm ∂gm ∂2x ∂2x · gi + · gk = · gi + m i · gk ∂ξ i ∂ξ ∂ξ ∂ξ k ∂ξ m ∂ξ k ∂gk ∂gi ∂ ∂gik = · g + · g = (g · g ) = i i k k ∂ξ m ∂ξ m ∂ξ m ∂ξ m Hence, Γ`ik = g m` Γm,ik
1 m` ∂gmi ∂gkm ∂gik = g + − m 2 ∂ξ i ∂ξ ∂ξ k
Mathematical tools
M3. Tensor fields in curvilinear coordinates