Mathematical tools
M3. Tensor fields in curvilinear coordinate systems
Aleš Janka
office Math 0.107
[email protected]
http://perso.unifr.ch/ales.janka/mechanics
November 24, 2010, Université 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 !
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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 ).
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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).
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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
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= ∇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
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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