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Electromagnetism of Continuous MediaMathematical Modelling and Applications$

Mauro Fabrizio and Angelo Morro

Print publication date: 2003

Print ISBN-13: 9780198527008

Published to Oxford Scholarship Online: September 2007

DOI: 10.1093/acprof:oso/9780198527008.001.0001

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D Differential operators in curvilinear coordinates

D Differential operators in curvilinear coordinates

Source:
Electromagnetism of Continuous Media
Publisher:
Oxford University Press

D.1 Differentiation

Denote by z = (z 1, z 2, z 3) the rectangular coordinates of a point in the three-dimensional space and by x = (x 1, x 2, x 3) three coordinates so that a coordinate transformation holds, namely,

z k = z k ( x ) , x k = x k ( z ) , k = 1 , 2 , 3

(p.645) The position vector x is expressed in rectangular coordinates as x = zkik. We can then view x as a function of x in the form x(x) = zk(x)ik. The vectors

g k : = x x k = z h x k i h , k = 1 , 2 , 3 ,
are tangential to the coordinate curves. They are taken to be linearly independent and hence they are a possible basis. The inner products
g h k = g h g k = δ j i z j x h z l x k
determine the arc length through dx · dx = gkl dxk dx1. The curvilinear coordinates are said orthogonal if and only if gkl = 0 everywhere when kl. The reciprocal base vectors {gk} are defined uniquely by
g k g l = δ k l .

Since the vectors {gk} are linearly independent, the matrix [ghk] is nonsingular. Let g = det[ghk] ≠ 0. Any vector ν, at any point x, may be expressed through the two bases as

v = v k g k = v k g k .

The components νk are called contravariant, the components νk covariant. It follows that

v k = v g k , v k = v g k .

Letting ghk = gh · gk, we have

δ h l = g h g l = ( g h g k ) g k ( g l g p ) g p = g h k g k l .

Accordingly, the matrix [ghk] is the inverse of [ghk]. Since

d x h d z k i k g p = d x h d z k i k d z l d x p i l = d x h d z k d z k d x p = δ h p ,
it follows that
g h = d x h d z k i k .

In correspondence with curvilinear coordinates, it is convenient to consider physical components, namely, the components relative to an orthonormal basis. Since gkk > 0, let hk = (gkk)1/2, k not summed, and define ek to be the unit vectors

e k = 1 h k g k , k = 1 , 2 , 3.

(p.646) The differentiation in curvilinear coordinates is based on the definition of covariant derivative, νh;k, such that

v x k = v h ; k g h .

By definition,

v h ; k = v x k g h = ( v l g l ) x k g h = v h x h + v l g l x k g h .

Since the bases {gk} and {gk} are reciprocal of each other,

x k ( g l g h ) = 0 , g h g l x k = g l g h x k .

Hence, letting

{ h j k } : = g h g j x k ,
we have
v h ; k = v j x k { j h k } v j
and
g j x k = { j h k } g h .

Likewise,

v ; k h = v h x k + { h j k } v j .

The bracketed terms are known as the Christoffel symbols of the second kind. They are related to the Christoffel symbols of the first kind, [kl, m] = gm · ∂g k/∂x l, by

{ h j k } = g j l [ h k , l ] .

Also,

[ h k , l ] = 1 2 ( g h l , k + g k l , h g h k , l ) .

For second-order tensors T, expressed, for example, as T=T klg kg l, the covariant derivative is defined such that

T x m = T ; m k l g k g l .

(p.647) Hence, because

T x m = T k l x m g k g l T k l g k x m g l + T k l g k g l x m ,
we find that
T ; m k l = T ; m k l + T h l { k h m } + T k h { l h m } .

Setting m = l provides the divergence of a second-order tensor.

D.2 Gradient, divergence, curl and the Laplacian

A scalar function ψ : ℝ3 ⊃ Ω → ℝ is said to be differentiable at x 0 if a vector w exists such that

ψ ( x ) ψ ( x 0 ) = w ( x x 0 ) + ο ( | x x 0 | ) .

If such is the case then w is said to be the gradient of ψ, ▿ψ, at x 0. In rectangular coordinates, we have

ψ ( z ) ψ ( z 0 ) = ψ z k ( z k z 0 k ) + ο ( | z z 0 | ) .

Since

ψ z k ( z k z 0 k ) = ψ z k i k ( z h z 0 h ) i h = ψ z k i k ( x x 0 ) ,
it follows that
ψ = ψ z k i k .

In curvilinear coordinates, letting ψ(z) = ψ (x(z)), we obtain

ψ = ψ z h x h z k i k = ψ z h g h .

Consequently, we have

ψ = ψ z h g h .

This allows ▿ to be viewed as the operator

= g h x h .

Hence, we let the gradient of a vector, the divergence and the curl be defined by

v : = h g h v x h , v : = h g h v x h , × v : = h g h × v x h .

(p.648) Also,

Δ ψ = ψ = h , k g h x h ( g k ψ x k )

Moreover, the definitions give

v = g h v k ; h g k = g h k v k ; h = g h g k v ; h k = v ; h h
and
× v = g h × v k ; h g k = h k m v k ; h g m ,
where
h k m : = g h × g k g m .

D.3 Orthogonal curvilinear coordinates

Let g 1, g 2, g 3 be orthogonal to one another. Hence,

[ g k l ] = diag ( h 1 2 , h 2 2 , h 3 2 ) , g = ( h 1 h 2 h 3 ) 2 , [ g k l ] = diag ( h 1 2 , h 2 2 , h 3 2 ) .

As a consequence,

e k = 1 h k g k = h k g k
and the covariant, contravariant and physical components, respectively, νj, νj and ν^j, of a vector ν are related by
v ^ j = h j v j = ( 1 / h j ) v j .

Accordingly, we can write

ψ = k 1 h k ψ x k e k .

Also, a direct application of the definition shows that

{ l k k } = 1 2 h 1 2 h k 2 x l , { k k l } = ln h k x l , { k k k } = ln h k x k , { l k j } = 0.
(p.649) where equal indices are not summed and j, k, l are unequal. Consequently, we have
v = 1 h 1 h 2 h 3 [ x 1 h 2 h 3 v ^ 1 + x 2 h 1 h 3 v ^ 2 + x 3 h 1 h 2 v ^ 3 ] , × v = 1 h 2 h 3 [ ( h 3 v ^ 3 ) x 2 ( h 2 v ^ 2 ) x 3 ] e 1 + 1 h 3 h 1 [ ( h 1 v ^ 2 ) x 3 ( h 3 v ^ 3 ) x 1 ] e 2 + 1 h 1 h 2 [ ( h 2 v ^ 2 ) x 1 ( h 1 v ^ 1 ) x 2 ] e 3 , Δ ψ = 1 h 1 h 2 h 3 { x 1 [ h 2 h 3 h 1 ψ x 1 ] + x 2 [ h 3 h 1 h 2 ψ x 2 ] + x 3 [ h 1 h 2 h 3 ψ x 3 ] } .

Explicit expressions are now given for cylindrical and spherical coordinates. Cylindrical coordinates. (Fig. D.1). Let x 1 = r, x 2 = θ, x 3 = z; r ∈ ℝ+, θ ∈ [0, 2π), z ∈ ℝ. Hence we have

g k = x k ( r cos θ i 1 + r sin θ i 2 + z i 3 ) ,
whence
h 1 = h 3 = 1 , h 2 = r .

D Differential operators in curvilinear coordinates

Fig. D.1 Cylindrical polar coordinates (left) and spherical polar coordinates (right).

(p.650) Denote the orthonormal basis by er, e θ, ez and the associated (physical) components of ν by νr, νθ, νz. Substitution gives

ψ = ψ r e r + I r ψ θ e θ + ψ z e z , v = 1 r ( r v r ) r + 1 r v θ θ + v z z , × v = ( 1 r v z θ v θ z ) e r + ( v r z v z r ) e θ + [ 1 r ( r v θ ) r 1 r v 0 θ ] e z , Δ ψ = 2 ψ r 2 + 1 r ψ r + 1 r 2 ψ θ 2 + 2 ψ z 2 .

Spherical coordinates. Let x 1 = r, x 2 = θ, x 3 = φ; r ∈ ℝ+, θ ∈ [0, π], φ ∈ [0, 2π). Hence, we have

g k = x k ( r sin θ cos φ i 1 + r sin θ sin φ i 2 + r cos θ i 3 )
whence
h 1 = 1 , h 2 = r , h 3 = r sin θ .

Denote the orthonormal basis by er, e θ, e φ and the associated (physical) components of ν by νr, νθ, νφ. Substitution gives

ψ = ψ r e r + 1 r ψ θ e 0 + 1 r sin θ ψ φ e φ , v = 1 r 2 r ( r 2 v r ) + 1 r sin θ ( v θ sin θ ) θ + 1 r sin θ v φ φ , × v = 1 r sin θ [ ( v φ sin θ ) θ v θ θ ] e r + 1 r [ 1 sin θ v r φ ( v φ r ) r ] e θ + 1 r [ ( r v θ ) r v r θ ] e φ , Δ ψ = 1 r 2 r ( r 2 ψ r ) + 1 r 2 sin θ θ ( sin θ ψ θ ) + 1 r 2 sin 2 θ 2 ψ φ 2 .