Orthogonal, Curvilinear Coordinates Definition We Define a Two Sets Of

Orthogonal, Curvilinear Coordinates Definition We define a two sets of coordinates for each spatial point: (x, y, z), the normal Cartesion form, and (ξ1, ξ2, ξ3), a second reference system. The functions ξi are smooth functions of the xi coordinates, ξi = ξi(x). We define scale factors— the length corresponding to a small displacement dξi by hi(x)=1/|∇ξi| using the ordinary Cartesian form of the gradient. Note that the scale factors depend on position. We next define unit vectors corresponding to the displacements in each of the new coordinate directions by ′ êi = hi∇ξi (1) We will use the following summation convention: indices which appear at least twice on one side of an equation and do not appear on the other are summed over. Thus the i index is not summed in the previous expression. The new coordinate system is orthogonal and right–handed if ′ ′ êi · êj = δij (2) and ′ ′ ′ êi · êj × êk = ǫijk (3) Transformation of a Vector If we have a vector given by its three components in Cartesian coordinates F = Fiêi we can write the components in the new coordinates ′ ′ F = Fi êi using the projection formula ′ ′ ′ Fi = êi · F = êi · êjFj or ′ Fi = γijFj (4) with ′ ∂ξi γij = êi · êj = hi ∂xj so that ′ êi = γijêj We can also transform backwards: ′ Fi = Fjγji 1 This follows from the orthogonality condition ′ ′ δij = êi · êj = γ ê · γ ê ik k jm m (5) = γikγjmδkm = γikγjk We also have a triple product rule γimγjnγklǫmnl = ǫijk (6) The backwards transformation also implies that 1 ∂xj γij = hi ∂ξi and γkiγkj = δij (5a) Gradient We first show that the gradient transforms as a vector: ∂ ∂ξ ∂ φ = j φ ∂xi ∂xi ∂ξj by chain rule. Multiplying and dividing by hj gives ∂ξj 1 ∂ (∇φ)i = hj φ ∂xi hj ∂ξj (7) ′ = γji(∇ φ)j where the ∇′ indicates the gradient in the new coordinate system with components 1 ∂ (grad φ)j = φ (8) hj ∂ξj Multiplying (7) by γki and using (5a), we find ′ (∇ φ)k = γki(∇φ)i as in (4). 2 Divergence Next we consider the transformation of the divergence, ∂ 1 ∂ξj ∂ ′ Fi = hj γmiFm ∂xi hj ∂xi ∂ξj ′ 1 ∂ ′ Fm ∂ = γjiγmi Fm + γji γmi hj ∂ξj hj ∂ξj ′ 1 ∂ ′ Fm ∂ 1 ∂xi = Fj + γji hj ∂ξj hj ∂ξj hm ∂ξm The second term expands to ′ ′ Fm ∂xi ∂ 1 Fm ∂xi 1 ∂ ∂xi γji + 2 hj ∂ξm ∂ξj hm hj ∂ξj hm ∂ξj ∂ξm ′ ′ ∂ 1 Fm ∂ ∂xi ∂xi = Fj + 2 ∂ξj hj 2hmhj ∂ξm ∂ξj ∂ξj We then use the fact that 2 ∂xi = hjγjihjγji ∂ξ Xi j Xi 2 = hj to simplify the last term to ′ Fm ∂ 2 2 hj 2hmhj ∂ξm ′ 1 ∂ 1 2 = Fm ln(hj ) hm ∂ξm 2 Xj ′ 1 ∂ = Fm (h1h2h3) hmh1h2h3 ∂ξm Thus we find 1 ∂ ′ ′ ∂ 1 ′ 1 1 ∂ ∇· F = Fj + Fj + Fj h1h2h3 hj ∂ξj ∂ξj hj h1h2h3 hj ∂ξj 1 ∂ h1h2h3 ′ div F = Fj (9) h1h2h3 ∂ξj hj 3 Curl Next we consider the curl and show that it transforms as a proper vector; i.e., if ∂ Cm ≡ ǫmjk Fk ∂xj then ∂ γimǫmjk Fk ∂xj should be the ith component of the curl, ′ 1 ∂ ′ (curl F)i = Ci ≡ ǫijk (hkFk) (10) hjhk ∂ξj We substitute ′ 1 ∂xm Fk = γkmFm = Fm hk ∂ξk into the last expression to find 2 ′ 1 ∂xm ∂ ∂ xm Ci = ǫijk ( Fm + Fm ) hjhk ∂ξk ∂ξj ∂ξj∂ξk The last term gives zero contribution because it is symmetric in i and j while the ǫijk is antisymmetric. Therefore ′ 1 ∂xm ∂ Ci = ǫijk Fm hj hk ∂ξk ∂ξj 1 ∂xm ∂xn ∂ = ǫijk Fm hj hk ∂ξk ∂ξj ∂xn ∂ = ǫijkγkmγjn Fm ∂xn We thus are asking whether ∂ ∂ ǫijkγkmγjn Fm = γimǫmjk Fk ∂xn ∂xj If we multiply both sides of this by γis and sum, we have ∂ ∂ ǫijkγisγjnγkm Fm = γisγimǫmjk Fk ∂xn ∂xj Using (6) and (5), we find that both sides are equal to ∂ ǫsnm Fm ∂xn 4 So that the curl (10) indeed transforms properly. Advective terms Here we comment that the proper form of the advective terms is found by regarding ∂ uj ui ∂xj as a shorthand for u · u grad − u × curl u 2 with the gradient and curl operations defined by (8) and (10) respectively. We cannot regard the ∇ as an ordinary vector; i.e. ′ 1 ∂ (∇ )i = hi ∂ξi The forms (8), (9), and (11) of the gradient, divergence, and curl are clearly inconsistent with such a definition. Thus the operator u ·∇ does not, in itself, make sense. Rather we must define its forms specifically depending on the operand: u ·∇φ = u · (grad φ) and u · u u ·∇u = grad − u × curl u 2 For advection of a different vector field, we have u · B 1 1 1 1 1 u·∇B = grad + u div B− B div u− u×curl B− B×curl u− curl (u×B) 2 2 2 2 2 2 (see Morse and Feshbach). In terms of the scale factors, we find 1 ∂ u · (grad φ) = ui φ hi ∂ξi ant the ith component of um ∂ umum ∂ um ∂ umui ∂ umum ∂ u ·∇u = hiui − hm = ui + hi − hm hihm ∂ξm hihm ∂ξi hm ∂ξm hihm ∂ξm hihm ∂ξi as part of which we can see a term which looks like the dot product of u with the gradient operator and two terms which depend on the curvature of the coordinate system. 5.

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