Section 1.18
1.18 Curvilinear Coordinates: Tensor Calculus
1.18.1 Differentiation of the Base Vectors
Differentiation in curvilinear coordinates is more involved than that in Cartesian coordinates because the base vectors are no longer constant and their derivatives need to be taken into account, for example the partial derivative of a vector with respect to the Cartesian coordinates is
i v vi 1 v v i g i ei but j j g i v j x j x j
The Christoffel Symbols of the Second Kind
First, from Eqn. 1.16.19 – and using the inverse relation,
g x m 2 x m k i e g (1.18.1) j j i m i j m k x this can be written as
g i k g Partial Derivatives of Covariant Base Vectors (1.18.2) j ij k where
2 x m k k , (1.18.3) ij i j x m
k and ij is called the Christoffel symbol of the second kind; it can be seen to be j equivalent to the kth contravariant component of the vector g i / . One then has {▲Problem 1}
g g k k i g k j g k Christoffel Symbols of the 2nd kind (1.18.4) ij ji j i and the symmetry in the indices i and j is evident2. Looking now at the derivatives of the i k k contravariant base vectors g : differentiating the relation g i g i leads to
g k g g i g k mg g k k j i j ij m ij
1 of course, one could express the g i in terms of the e i , and use only the first of these expressions 2 note that, in non-Euclidean space, this symmetry in the indices is not necessarily valid
Solid Mechanics Part III 164 Kelly Section 1.18
and so
g i i g k Partial Derivatives of Contravariant Base Vectors (1.18.5) j jk
Transformation formulae for the Christoffel Symbols
The Christoffel symbols are not the components of a (third order) tensor. This follows from the fact that these components do not transform according to the tensor transformation rules given in §1.17. In fact,
p q k 2 s k k r ij i j r pq i j s
The Christoffel Symbols of the First Kind
The Christoffel symbols of the second kind relate derivatives of covariant (contravariant) base vectors to the covariant (contravariant) base vectors. A second set of symbols can be introduced relating the base vectors to the derivatives of the reciprocal base vectors, called the Christoffel symbols of the first kind:
g g i g j g Christoffel Symbols of the 1st kind (1.18.6) ijk jik j k i k so that the partial derivatives of the covariant base vectors can be written in the alternative form
g i g k , (1.18.7) j ijk and it also follows from Eqn. 1.18.2 that
m k mk ijk ij g mk , ij ijm g (1.18.8) showing that the index k here can be raised or lowered using the metric coefficients as for a third order tensor (but the first two indexes, i and j, cannot and, as stated, the Christoffel symbols are not the components of a third order tensor).
Example: Newton’s Second Law
The position vector can be expressed in terms of curvilinear coordinates, x xi . The velocity is then
dx x di di v g dt i dt dt i
Solid Mechanics Part III 165 Kelly Section 1.18 and the acceleration is
dv d 2i d j g d k d 2i d j d k a g j i g 2 i k 2 jk i dt dt dt dt dt dt dt
Equating the contravariant components of Newton’s second law f ma then gives the general curvilinear expression
i i i j k f m jk ■
Partial Differentiation of the Metric Coefficients
The metric coefficients can be differentiated with the aid of the Christoffel symbols of the first kind {▲Problem 3}:
g ij (1.18.9) k ikj jki
Using the symmetry of the metric coefficients and the Christoffel symbols, this equation can be written in a number of different ways:
gij,k kij jki , g jk,i ijk kij , g ki, j jki ijk
Subtracting the first of these from the sum of the second and third then leads to the useful relations (using also 1.18.8)
1 ijk g jk,i g ki, j gij,k 2 (1.18.10) 1 k g mk g g g ij 2 jm,i mi, j ij,m which show that the Christoffel symbols depend on the metric coefficients only.
Alternatively, one can write the derivatives of the metric coefficients in the form (the first of these is 1.18.9)
gij,k ikj jki (1.18.11) ij im j jm i g ,k g km g km
Also, directly from 1.15.7, one has the relations
g ij g gg , ij gg ij (1.18.12) g ij g
Solid Mechanics Part III 166 Kelly Section 1.18 and from these follow other useful relations, for example {▲Problem 4}
i log g 1 g 1 J ij J (1.18.13) j g j j and
e g ijk g n e n m ijk m ijk mn ijk mn (1.18.14) ijk e ijk 1/ g ijk 1 n ijk n mn e mn m m g
1.18.2 Partial Differentiation of Tensors
The Partial Derivative of a Vector
The derivative of a vector in curvilinear coordinates can be written as
i i v v i g i v vi i g g i v g v j j j j j i j v i v g v i k g or i g i v i g k (1.18.15) j i ij k j i jk i i v g vi g j i j where
i i i k v | j v , j kj v Covariant Derivative of Vector Components (1.18.16) k vi | j vi, j ij vk
The first term here is the ordinary partial derivative of the vector components. The second term enters the expression due to the fact that the curvilinear base vectors are changing. The complete quantity is defined to be the covariant derivative of the vector components. The covariant derivative reduces to the ordinary partial derivative in the case of rectangular Cartesian coordinates.
The vi | j is the ith component of the j – derivative of v. The vi | j are also the components of a second order covariant tensor, transforming under a change of coordinate system according to the tensor transformation rule 1.17.4 (see the gradient of a vector below).
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The covariant derivative of vector components is given by 1.18.16. In the same way, the covariant derivative of a vector is defined to be the complete expression in 1.18.15, v , j , i with v , j v | j g i .
The Partial Derivative of a Tensor
The rules for covariant differentiation of vectors can be extended to higher order tensors. The various partial derivatives of a second-order tensor
ij i j j i i j A A g i g j Aij g g Ai g g j A j g i g are indicated using the following notation:
A Aij |g g A | g i g j A j | g i g A i | g g j (1.18.17) k k i j ij k i k j j k i
Thus, for example,
i j i j i j A ,k Aij,k g g Aij g ,k g Aij g g ,k i j i m j i j m Aij,k g g Aij mk g g Aij g kmg m m i j Aij,k ik Amj jk Aim g g and, in summary,
m m Aij |k Aij,k ik Amj jk Aim ij ij i mj j im A |k A ,k mk A mk A j j j m m j (1.18.18) A i |k Ai ,k mk Ai ik Am i i i m m i A j |k A j,k mk A j jk Am Covariant Derivative of Tensor Components
The covariant derivative formulas can be remembered as follows: the formula contains the usual partial derivative plus for each contravariant index a term containing a Christoffel symbol in which that index has been inserted on the upper level, multiplied by the tensor component with that index replaced by a dummy summation index which also appears in the Christoffel symbol for each covariant index a term prefixed by a minus sign and containing a Christoffel symbol in which that index has been inserted on the lower level, multiplied by the tensor with that index replaced by a dummy which also appears in the Christoffel symbol. the remaining symbol in all of the Christoffel symbols is the index of the variable with respect to which the covariant derivative is taken.
For example,
i i i m m i m i A jk |l A jk,l ml A jk jl Amk kl A jm
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Note that the covariant derivative of a product obeys the same rules as the ordinary differentiation, e.g.
jk jk jk ui A |m ui| |m A ui A |m
Covariantly Constant Coefficients
It can be shown that the metric coefficients are covariantly constant3 {▲Problem 5},
ij g ij |k g |k 0 ,
This implies that the metric (identity) tensor I is constant, I ,k 0 (see Eqn. 1.16.32) – although its components gij are not constant. Similarly, the components of the permutation tensor, are covariantly constant
ijk eijk |m e |m 0 .
In fact, specialising the identity tensor I and the permutation tensor E to Cartesian ij ijk coordinates, one has g ij g ij , eijk e ijk , which are clearly constant.
Specialising the derivatives, gij |k ij,k , eijk |m ijk ,m , and these are clearly zero. From §1.17, since if the components of a tensor vanish in one coordinate system, they vanish in all coordinate systems, the curvilinear coordinate versions vanish also, as stated above.
The above implies that any time any of these factors appears in a covariant derivative, i i they may be extracted, as in g ij u |k g ij u |k .
The Riemann-Christoffel Curvature Tensor
Higher-order covariant derivatives are defined by repeated application of the first-order derivative. This is straight-forward but can lead to algebraically lengthy expressions. For example, to evaluate v i |mn , first write the first covariant derivative in the form of a second order covariant tensor B,
k vi |m vi,m imvk Bim so that
vi |mn Bim |n k k Bim,n in Bkm mn Bik (1.18.19) v k v k v l v k v l v i,m im k ,n in k,m km l mn i,k ik l
3
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The covariant derivative vi |nm is obtained by interchaning m and n in this expression. Now investigate the difference
k k k l k l vi |mn vi |nm vi,m imvk vi,n in vk in vk,m kmvl im vk,n knvl ,n ,m k l k l mn vi,k ik vl nm vi,k ik vl
The last two terms cancel here because of the symmetry of the Christoffel symbol, leaving
v | v | v k v k v v k v k v i mn i nm i,mn im,n k im k ,n i,nm in,m k in k ,m k l k l in vk ,m kmvl im vk ,n knvl
The order on the ordinary partial differentiation is interchangeable and so the second order partial derivative terms cancel,
v | v | v k v k v v k v k v i mn i nm i,mn im,n k im k ,n i,nm in,m k in k ,m k l k l in vk ,m kmvl im vk ,n knvl
After further cancellation one arrives at
j v i |mn v i |nm Rimn v j (1.18.20) where R is the fourth-order Riemann-Christoffel curvature tensor, with (mixed) components
j j j k j k j Rimn in,m im,n in km im kn (1.18.21)
j Since the Christoffel symbols vanish in a Cartesian coordinate system, then so does Rimn . Again, any tensor that vanishes in one coordinate system must be zero in all coordinate j systems, and so Rimn 0 , implying that the order of covariant differentiation is immaterial, v i |mn v i |nm .
From 1.18.10, it follows that
R R R R ijkl jikl ijlk klij Rijkl Riklj Riljk 0
The latter of these known as the Bianchi identities. In fact, only six components of the j Riemann-Christoffel tensor are independent; the expression Rimn 0 then represents 6 equations in the 6 independent components gij .
This analysis is for a Euclidean space – the usual three-dimensional space in which quantities can be expressed in terms of a Cartesian reference system – such a space is
Solid Mechanics Part III 170 Kelly Section 1.18 called a flat space. These ideas can be extended to other, curved spaces, so-called Riemannian spaces (Riemannian manifolds), for which the Riemann-Christoffel tensor is non-zero (see §1.19).
1.18.3 Differential Operators and Tensors
In this section, the concepts of the gradient, divergence and curl from §1.6 and §1.14 are generalized to the case of curvilinear components.
Space Curves and the Gradient
i Consider first a scalar function f x , where x x ei is the position vector, with x i x i j . Let the curvilinear coordinates depend on some parameter s, j j s , so that x(s ) traces out a space curve C.
For example, the cylindrical coordinates j j s, with r a , s / c , z sb / c , 0 s 2c , generate a helix.
From §1.6.2, a tangent to C is
dx x di τ i g ds i ds i so that di / ds are the contravariant components of τ . Thus
df f i f i j g g j . ds i i
For Cartesian coordinates, df /ds f τ (see the discussion on normals to surfaces in §1.6.4). For curvilinear coordinates, therefore, the Nabla operator of 1.6.11 now reads
g i (1.18.22) i so that again the directional derivative is
df f τ ds
The Gradient of a Scalar
In general then, the gradient of a scalar valued function is defined to be
grad g i Gradient of a Scalar (1.18.23) i
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i i and, with dx dx ei d g i , one has
d di dx (1.18.24) i
The Gradient of a Vector
Analogous to Eqn. 1.14.3, the gradient of a vector is defined to be the tensor product of the derivative u / j with the contravariant base vector g j :
u j i j gradu g ui | j g g j Gradient of a Vector (1.18.25) i j u | j g i g
Note that
u u g i u g i u | g i g j u j | g i g i i j i i j so that again one arrives at Eqn. 1.14.7, uT gradu .
Again, one has for a space curve parameterised by s,
T du u u u i τ g i g i τ gradu τ ds i i i
Similarly, from 1.18.18, the gradient of a second-order tensor is
A gradA g k Aij | g g g k k k i j A | g i g j g k ij k Gradient of a Tensor (1.18.26) j i k Ai |k g g j g i j k A j |k g i g g
The Divergence
From 1.14.9, the divergence of a vector is {▲Problem 6}
i u j divu gradu : I u |i g Divergence of a Vector (1.18.27) j
This is equivalent to the divergence operation involving the Nabla operator, divu u. An alternative expression can be obtained from 1.18.13 {▲Problem 7},
Solid Mechanics Part III 172 Kelly Section 1.18
i i i 1 gu 1 Ju divu u |i J g i i
Similarly, using 1.14.12, the divergence of a second-order tensor is
ij A j divA grad A : I A | j g i g j Divergence of a Tensor (1.18.28) j i Ai | j g
Here, one has the alternative definition,
A A g i A g i A ji | g i i j i so that again one arrives at Eqn. 1.14.14, divA A T .
The Curl
The curl of a vector is defined by {▲Problem 8}
u u curlu u g k eijk u | g eijk j g Curl of a Vector (1.18.29) k j i k i k the last equality following from the fact that all the Christoffel symbols cancel out.
Covariant derivatives as Tensor Components
Equation 1.18.25 shows clearly that the covariant derivatives of vector components are themselves the components of second order tensors. It follows that they can be manipulated as other tensors, for example,
im i g um | j u | j and it is also helpful to introduce the following notation:
j mj i j i mj ui | ui |m g , u | u |m g .
The divergence and curl can then be written as {▲Problem 10}
divu u i | u |i i i . ijk j i k curlu e u j |i g k eijk u | g
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Generalising Tensor Calculus from Cartesian to Curvilinear Coordinates
It was seen in §1.16.7 how formulae could be generalised from the Cartesian system to the corresponding formulae in curvilinear coordinates. In addition, formulae for the gradient, divergence and curl of tensor fields may be generalised to curvilinear components simply by replacing the partial derivatives with the covariant derivatives. Thus:
Cartesian Curvilinear Gradient Of a scalar field i grad, / xi ,i |i / of a vector field i gradu ui / x j u | j of a tensor field ij gradT Tij / xk T |k Divergence of a vector field i divu, u ui / xi u |i of a tensor field ij divT Tij / x j T | j Curl of a vector field ijk curlu, u ijk u j / xi e u j |i Table 1.18.1: generalising formulae from Cartesian to General Curvilinear Coordinates
All the tensor identities derived for Cartesian bases (§1.6.9, §1.14.3) hold also for curvilinear coordinates, for example {▲Problem 11}
gradv gradv v grad
divvA v divA A : gradv
1.18.4 Partial Derivatives with respect to a Tensor
The notion of differentiation of one tensor with respect to another can be generalised from the Cartesian differentiation discussed in §1.15. For example:
j g i g j j g i g A Aij Ai
B Bij i j g g g m g n A Amn
A Aij i j m n i j g g g m g n i j g g g m g n A Amn m n g g g m g n
1.18.5 Orthogonal Curvilinear Coordinates
This section is based on the groundwork carried out in §1.16.9. In orthogonal curvilinear systems, it is best to write all equations in terms of the covariant base vectors, or in terms of the corresponding physical components, using the identities (see Eqn. 1.16.45)
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1 1 i ˆ g 2 g i g i (no sum) (1.18.30) hi hi
The Gradient of a Scalar Field
From the definition 1.18.23 for the gradient of a scalar field, and Eqn. 1.18.30, one has for an orthogonal curvilinear coordinate system,
1 1 1 g g g h 2 1 1 h 2 2 2 h 2 3 3 1 2 3 (1.18.31) 1 1 1 ˆ ˆ ˆ 1 g1 2 g 2 3 g 3 h1 h2 h3
The Christoffel Symbols
The Christoffel symbols simplify considerably in orthogonal coordinate systems. First, from the definition 1.18.4,
k 1 g i ij 2 j g k (1.18.32) hk
Note that the introduction of the scale factors h into this and the following equations disrupts the summation and index notation convention used hitherto. To remain consistent, one should use the metric coefficients and leave this equation in the form
g k i g kmg ij j m
Now
g i 2 i g i g i 2 g i 2hi ij j j
2 and g i g i hi so, in terms of the derivatives of the scale factors,
1 h i k i (no sum) (1.18.33) ij ij ki j hi
Similarly, it can be shown that {▲Problem 14}
2 k 2 i 2 j 2 i hk ij hi jk h j ki hi jk when i j k (1.18.34) so that the Christoffel symbols are zero when the indices are distinct, so that there are only 21 non-zero symbols of the 27. Further, {▲Problem 15}
Solid Mechanics Part III 175 Kelly Section 1.18
h h k k i i , i k (no sum) (1.18.35) ii ij i j 2 k hk
From the symmetry condition (see Eqn. 1.18.4), only 15 of the 21 non-zero symbols are distinct:
1 1 1 1 1 1 1 11 , 12 21 , 13 31 , 22 , 33 2 2 2 2 2 2 2 11 , 12 21 , 22 , 23 32 , 33 3 3 3 3 3 3 3 11 , 13 31 , 22 , 23 32 , 33
Note also that these are related to each other through the relation between (1.18.33, 1.18.35), i.e.
2 k hi i ii 2 ik , i k (no sum) hk so that
1 2 3 11 , 22 , 33 2 2 2 1 1 h2 2 1 1 h3 3 2 2 h1 1 12 21 2 11 , 13 31 2 11 , 12 21 2 22 (1.18.36) h1 h1 h2 2 2 2 2 2 h3 3 3 3 h1 1 3 3 h2 2 23 32 2 22 , 13 31 2 33 , 23 32 2 33 h2 h3 h3
The Gradient of a Vector
From the definition 1.18.25, the gradient of a vector is
1 grad v v i g g j v i g g (no sum over h ) (1.18.37) j i 2 j i j j h j
In terms of physical components,
1 v i grad v v k i g g 2 j kj i j h j (1.18.38) 1 v i h i v i i i v k gˆ gˆ j ij kj i j h j hk
The Divergence of a Vector
From the definition 1.18.27, the divergence of a vector is div v v i or {▲Problem 16} i
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1 2 3 vi i 1 v h2 h3 v h1h3 v h1h2 div v i vk ki 1 2 3 (1.18.39) h1h2 h3
The Curl of a Vector
From §1.16.10, the permutation symbol in orthogonal curvilinear coordinates reduceS to
1 eijk ijk (1.18.40) h1h2 h3
ijk where ijk is the Cartesian permutation symbol. From the definition 1.18.29, the curl of a vector is then
1 v j 1 v v curl v g 2 1 g ijk i k 1 2 3 h1h2 h3 h1h2 h3 2 2 1 2 1 v h2 v h1 1 2 g 3 (1.18.41) h1h2 h3 2 1 1 v h2 v h1 h3gˆ 3 h h h 1 2 1 2 3 or
h1gˆ 1 h2gˆ 2 h3gˆ 3 1 curl v 1 2 3 (1.18.42) h1h2 h3 1 2 3 h1v h2v h3v
The Laplacian
From the above results, the Laplacian is given by
1 h h h h h h 2 2 3 1 3 1 2 1 1 2 2 3 3 h1h2 h3 h1 h2 h3
Divergence of a Tensor
From the definition 1.18.28, and using 1.16.59, 1.16.62 {▲Problem 17}
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j j i Ai j m m j i divA Ai | j g j mj Ai ij Am g (1.18.43) h h 1 i ij 1 j im m m mj A mj A ij A gˆ i h j h h h h i j m i j
Examples
1. Cylindrical Coordinates
Gradient of a Scalar Field: 1 ˆ ˆ ˆ 1 g1 2 g 2 3 g 3 2
Christoffel symbols: 1 With h1 1, h2 , h3 1, there are two distinct non-zero symbols: 1 1 22 1 2 2 12 21 1
Derivatives of the base vectors: The non-zero derivatives are g g 1 g 1 2 g , 2 1g 2 1 1 2 2 1 and in terms of physical components, the non-zero derivatives are gˆ gˆ 1 gˆ , 2 gˆ 2 2 2 1 which agree with 1.6.32.
The Divergence (see 1.6.33), Curl (see 1.6.34) and Gradient {▲Problem 18} (see 1.14.18) of a vector:
v 1 v 1 1 v 2 v 3 div v 1 1 1 2 3 1 gˆ 1 gˆ 2 gˆ 3 1 curl v 1 1 2 3 v 1 1v 2 v 3
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v v v grad v 1 gˆ gˆ 2 gˆ gˆ 3 gˆ gˆ 1 1 1 1 2 1 1 3 1
1 v 1 1 v 2 v gˆ gˆ v gˆ gˆ 1 2 2 1 2 1 2 1 2 2
1 v 3 v 1 v 2 v 3 gˆ gˆ gˆ gˆ gˆ gˆ gˆ gˆ 1 2 3 2 3 1 3 3 2 3 3 3 3
The Divergence of a tensor {▲Problem 19} (see 1.14.19): A 11 1 A 12 A 13 A 11 A 22 ˆ divA 1 1 2 3 1 g1 A 21 1 A 22 A 23 A 21 A 12 ˆ 1 1 2 3 1 g 2 A 31 A 31 1 A 32 A 33 ˆ 1 1 1 2 3 g 3
2. Spherical Coordinates
Gradient of a Scalar Field: 1 1 gˆ gˆ gˆ 1 1 1 2 2 1 sin 2 3 3
Christoffel symbols: 1 1 2 With h1 1, h2 , h3 sin , there are six distinct non-zero symbols: 1 1 1 1 2 2 22 ,33 sin 1 2 2 , 2 sin 2 cos 2 12 21 1 33 1 3 3 , 3 3 cot 2 13 31 1 23 32
Derivatives of the base vectors: The non-zero derivatives are g g 1 g g 1 g 1 2 g , 1 3 g , 2 1g 2 1 1 2 3 1 1 3 2 1
g 2 g 3 2 g 3 1 2 2 2 2 cot g 3 , sin g1 sin cos g 2 3 2 3 and in terms of physical components, the non-zero derivatives are gˆ gˆ gˆ 1 gˆ , 1 sin 2gˆ , 2 gˆ 2 2 3 3 2 1
gˆ gˆ 2 cos 2gˆ , 3 sin 2gˆ cos 2gˆ 3 3 3 1 2 which agree with 1.6.37.
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The Divergence (see 1.6.38), Curl and Gradient of a Vector: 1 2 2 1 v 1 1 sin v 2 1 v 3 div v 2 1 1 2 2 1 2 3 1 sin sin 1 1 2 gˆ 1 gˆ 2 sin gˆ 3 1 curl v 2 1 2 3 1 sin 2 v 1 1v 2 1 sin 2v 3
v 1 1 v 1 v 2 grad v gˆ gˆ gˆ gˆ 1 1 1 1 2 1 1 2
1 v 1 v 3 v 2 gˆ gˆ gˆ gˆ 1 2 3 1 1 3 1 2 1 sin
1 v 2 v 1 1 v 2 v 3 gˆ gˆ cot 2 gˆ gˆ 1 2 1 2 2 1 2 3 1 2 3 sin
v 3 1 v 3 gˆ gˆ gˆ gˆ 1 3 1 1 2 3 2
1 v 3 v 1 v 2 cot 2 gˆ gˆ 1 2 3 1 1 3 3 sin
The Divergence of a tensor {▲Problem 20} A 11 1 A 12 1 A 13 2A 11 cot 2 A 12 A 22 A 33 ˆ divA 1 1 2 1 2 3 1 g1 sin A 21 1 A 22 1 A 23 A 12 2A 21 cot 2 A 22 A 33 ˆ 1 1 2 1 2 3 1 g 2 sin A 31 1 A 32 1 A 33 A 13 2A 31 cot 2 A 23 A 32 ˆ 1 1 2 2 1 2 3 1 g 3 sin sin
1.18.6 Problems
k k 1 Show that the Christoffel symbol of the second kind is symmetric, i.e. ij ji , and g that it is explicitly given by k i g k . ij j i j 2 Consider the scalar-valued function Au v Aij u v . By taking the gradient of this function, and using the relation for the covariant derivative of A, i.e. m m Aij |k Aij,k ik Amj jk Aim , show that A u i v j ij A u i v j | , k ij k
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i.e. the partial derivative and covariant derivative are equivalent for a scalar-valued function. 3 Prove 1.18.9: g g ij (i) ij , (ii) g im j g jm i k ikj jki k km km ij i [Hint: for (ii), first differentiate Eqn. 1.16.10, g g kj k .] 4 Derive 1.18.13, relating the Christoffel symbols to the partial derivatives of g and
g g g mn log g . [Hint: begin by using the chain rule j j .] g mn 5 Use the definition of the covariant derivative of second order tensor components, Eqn. ij 1.18.18, to show that (i) g ij |k 0 and (ii) g |k 0 . 6 Use the definition of the gradient of a vector, 1.18.25, to show that i divu gradu : I u |i . 7 Derive the expression divu 1/ g gu i / i k k ijk 8 Use 1.16.54 to show that g u / e u j |i g k . ijk j k k j 9 Use the relation imn m n m n (see Eqn. 1.3.19) to show that
g1 g 2 g 3 curlu v k k k . 1 1k 2 2k 3 3k u 2v 3 u 3v 2 u1v 3 u 3v1 u1v 2 u 2v1 i i ijk j i k 10 Show that (i) u |i ui | , (ii) e u j |i g k eijk u | g 11 Show that (i) gradv gradv v grad , (ii) divvA v divA A : gradv [Hint: you might want to use the relation aTb T : a b for the second of these.] 12 Derive the relation trA / A I in curvilinear coordinates. 13 Consider a (two dimensional) curvilinear coordinate system with covariant base 2 1 vectors g1 e1 2e 2 , g 2 e1 . (a) Evaluate the transformation equations x i x i ( j ) and the Jacobian J. (b) Evaluate the inverse transformation equations i i x j and the contravariant base vectors g i . ij (c) Evaluate the metric coefficients gij , g and the function g: (d) Evaluate the Christoffel symbols (only 2 are non-zero (e) Consider the scalar field 1 2 . Evaluate grad .
2 1 2 (f) Consider the vector fields u g1 g 2 , v g1 2g 2 : (i) Evaluate the covariant components of the vectors u and v (ii) Evaluate divu, divv (iii)Evaluate curlu, curlv (iv) Evaluate gradu, gradv (g) Verify the vector identities
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divu divu grad u curlu curlu grad u divu v v curlu u curl v curlgrad o divcurlu 0 (h) Verify the identities gradv gradv v grad gradu v gradu T v gradv T u
divu v gradu v (div v)u curlu v udivv vdivu gradu v gradv u (i) Consider the tensor field
1 0 i j A 2 g g 2 Evaluate all contravariant and mixed components of the tensor A k 14 Use the fact that g g 0 , k i to show that h 2 k h 2 i . Then permutate the k i k ij i jk ij 2 k 2 i 2 j 2 i indices to show that hk ij hi jk h j ki hi jk when i j k . 15 Use the relation g g 0, i j i i j 2 j hi i to derive ii 2 ij . h j 16 Derive the expression 1.18.39 for the divergence of a vector field v. 17 Derive 1.18.43 for the divergence of a tensor in orthogonal coordinate systems. 18 Use the expression 1.18.38 to derive the expression for the gradient of a vector field in cylindrical coordinates. 19 Use the expression 1.18.43 to derive the expression for the divergence of a tensor field in cylindrical coordinates. 20 Use the expression 1.18.43 to derive the expression for the divergence of a tensor field in spherical coordinates.
Solid Mechanics Part III 182 Kelly