M3_curvilin-handout2.pdf

  • Uploaded by: Jorge Yarasca
  • 0
  • 0
  • October 2019
  • PDF

This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA


Overview

Download & View M3_curvilin-handout2.pdf as PDF for free.

More details

  • Words: 2,587
  • Pages: 9
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

More Documents from "Jorge Yarasca"