1 Differential Operators in Curvilinear Coordinates

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1 1 Differential Operators in Curvilinear Coordinates worked out and written by Timo Fleig February/March 2012 Revision 1, Feb. 15, 201 Revision 2, Sep. 1, 2015 Université Paul Sabatier using LaTeX and git v Generalities We will here in particular be interested in transformations of certain classes of (differential) operators from linear to curvilinear coordinates Coordinates Be {α j }, j {1...} an arbitrary set of linear or curvilinear coordinates 1. We express a position vector in a -dimensional vector space as x = and the total differential of the position vector accordingly as d x = α j e j (1) α j dα j (2) Scalar Fields Be f a scalar and differentiable field, allowing us to write the total differential in terms of a set of coordinates 1.2 Cylindrical Coordinates d f = α j dα j () We define a local orthonormal basis of cylindrical coordinates as in Fig. 1.2 which are related to cartesian coordinates as follows: Accordingly we obtain from elementary geometrical considerations: 1 coordinate patches on differentiable manifolds x = ρcosϕ (4) y = ρsinϕ (5) z = z (6) 1

2 z φ z ρ y x Figure 1: Cylindrical coordinates Elementary transformations and position vectors A position vector can therefore be written in terms of cylindrical coordinates as and we ensuingly obtain x = (ρcosϕ) e x + (ρsinϕ) e y + (z) e z (7) = (cosϕ) e x + (sinϕ) e y = (ρsinϕ) e x + (ρcosϕ) e y = e z (8) and therefore the corresponding norms (scaling factors) = 1 := h 1 = ρ := h 2 = 1 := h (9) This allows us to write the orthonormal local tripod in terms of the new cylindrical coordinates as e ρ = e ϕ = e z = = (cosϕ) e x + (sinϕ) e y = (sinϕ) e x + (cosϕ) e y = e z (10) 2

3 Consequently, the original tripod can be expressed as e x = (cosϕ) e ρ (sinϕ) e ϕ e y = (sinϕ) e ρ + (cosϕ) e ϕ Finally we rewrite the position vector from Eq. (7) as and the total differential from Eqs. (2), (8) and (11) as e z = e z (11) x = ρ e ρ + z e z (12) d x = dρ + dϕ + dz = dρ e ρ + ρdϕ e ϕ + dz e z (1) Line Elements Elements for multidimensional integration can be easily obtained from the determined scaling factors, Eq. (9). We obtain for line elements dl ρ = h 1 dρ = dρ dl ϕ = h 2 dϕ = ρdϕ dl z = h dz = dz, (14) surface elements for integration over a slice (orthogonal or longitudinal) or the curved surface of the cylinder, respectively, ds o-slice = dl ρ dl ϕ = h 1 h 2 dρdϕ = ρdρdϕ ds l-slice = dl ρ dl z = h 1 h dρdz = dρdz (15) ds surface = dl ϕ dl z = h 2 h dϕdz = ρdϕdz (16) and finally for the volume element dv = dl ρ dl ϕ dl z = h 1 h 2 h dρdϕdz = ρdρdϕdz. (17) 1.2. The Gradient Operator In terms of cylindrical coordinates the gradient operator grad = can be written as ( ( ) ( ) f = f )ρ e ρ + f e ϕ + f e z (18) ϕ z Since can be related to the total differential of a scalar field according to d f = f d x (which in a slightly improper but intuitive form can be written as d f d x = f ). Using the result from Eq. (1) we obtain ( ) ( ) ( ) f d x = f dρ + ρ f dϕ + f dz (19) ρ ϕ z

4 We now employ the general form of the total differential of the scalar field Eq. () d f = dρ + dϕ + dz (20) and compare coefficients between Eqs. (19) and (20) to obtain ( = f )ρ ( = ρ f )ϕ ( ) = f z This yields the gradient operator in cylindrical coordinates acting on a scalar field f = eρ + e ϕ ρ 1 + e z (21) (22) We realize that the gradient operator in curvilinear coordinates can in general be written as 1 f = j (2) e h j α j where h j = α j are scaling factors in the respective coordinate system (for example in cylindrical coordinates they are given in Eq. (9)). This is also readily verified in cartesian coordinates The Divergence Operator Once the gradient operator in curvilinear coordinates is known, the divergence operator acting on a vector field can be deduced in the following manner: We introduce an auxiliary scalar field g = g(x 1,x 2,x ) = g(ρ,ϕ,x ) and calculate the integral over the product of the test vector field F and gradg F(x 1,x 2,x ) gradg(x 1,x 2,x ) dv = F(ρ,ϕ,x ) gradg(ρ,ϕ,x ) dv (24) V both in cartesian and cylindrical coordinates, remembering that the result must of course be identical. For the left-hand side of Eq. (24) we obtain straightforwardly F gradg dv = V = + F 1 V x 1 dx 1 dx 2 dx + F1 F 1 g dx 2 dx F 2 g dx 1 dx 2 dx x 1 F2 F 2 g dx 1 dx g dx 1 dx 2 dx x 2 x 2 dx 1 dx 2 dx + F x dx 1 dx 2 dx F + F g dx 1 dx 2 g dx 1 dx 2 dx (25) x = (div F)g dx 1 dx 2 dx (26) 4

5 choosing g such that the terms integrated by parts vanish 2. The right-hand-side of Eq. (24) becomes V F gradg dv = = = ( F(ρ,ϕ,z) e ρ + 1 ) ρ e ϕ + e z ρdρdϕdz 1 ρf ρ dρdϕdz + ρ F ϕ ρdρdϕdz + ( ρf ρ g dϕdz F ρ g + ρ F ) ρ dρdϕdz + + Fϕ F ϕ g dρdz ρf z g dρdϕ 1 ( ) = ρfρ g ρdρdϕdz ρ 1 F ϕ ρ g ρdρdϕdz Fz ( 1 = ρ g ρdρdϕdz ( ρfρ ) + 1 ρ g g dρdϕdz ρ F z g dρdϕdz F z ρdρdϕdz F ϕ + F ) z g ρdρdϕdz (27) with vanishing terms integrated by parts. From Eqs. (26) and (27) the identity of the integrands yields the divergence of the vector field F in cylindrical coordinates F = ρ 1 ( ) ρfρ + 1 F ϕ ρ + F z (28) The Curl Operator Here we want to deduce a general expression for the curl operator in any locally orthonormal curvilinear coordinate system, and then extract special cases such as cylindrical coordinates. We express the curl of a vector field F as rot F = F = ( i=1 F j e j ) = (F 1 e 1 ) + (F 2 e 2 ) + (F e ) (29) due to the linearity of the vector product. Replacing f by α k in Eq. (2) we infer αk = 1 α k 1 1 e j = h j α j e j δ jk = e k (0) h j h k 2 TODO: Some more detailed mathematical comments are required on this point. 5

6 and so the first term of Eq. (29) can be written as (F1 e 1 ) = ) (F 1 h 1 α1. (1) We now employ the following identity ( ) ( ) ( ) rot ϕ F = gradϕ F + ϕ rot F (2) which allows us to reformulate Eq. (1) [ ] ( ) ( ) (F1 e 1 ) = (F1 h 1 ) α1 + F 1 h 1 α1. () Since the curl of a gradient vanishes, rot grad f = 0 for any scalar field f, Eq. () reduces to [ ] ( ) (F1 e 1 ) = (F1 h 1 ) α1. (4) We now substitute Eq. (0) into Eq. (4) and obtain [ ] (F1 e 1 ) = (F1 h 1 ) e 1. (5) h 1 We now use the general expression of the gradient given in Eq. (2) rewritten for the scalar field F 1 h 1 to reformulate Eq. (5) to [ ] 1 (F 1 h 1 ) (F1 e 1 ) = e j e 1 h j α j h 1 1 (F 1 h 1 ) 1 1 (F 1 h 1 ) 1 = e 2 e (6) h α h 1 h 2 α 2 h 1 In an analogous fashion the two remaining terms of Eq. (29) are obtained as 1 (F 2 h 2 ) 1 1 (F 2 h 2 ) 1 (F2 e 2 ) = e 1 + e (7) h α h 2 h 1 α 1 h 2 1 (F h ) 1 1 (F h ) 1 (F e ) = e 2 + e 1 (8) h 1 α 1 h h 2 α 2 h Adding the three terms yields the general expression of the curl operator rot F = F = e ( 1 (F h ) (F ) 2h 2 ) h 2 h α 2 α + e ( 2 (F1 h 1 ) (F ) h ) h 1 h α α 1 + e ( (F2 h 2 ) (F ) 1h 1 ) h 1 h 2 α 1 α 2 1 h 1 e 1 h 2 e 2 h e = h 1 h 2 h α 1 α 2 α F 1 h 1 F 2 h 2 F h (9) This equation can easily be proven to be correct in cartesian coordinates. Since we do not want general expressions to depend on coordinate frames, it has to be valid in general. 6

7 where we have written the curl conveniently using a determinant. Note that the term h 1 h 2 h in the prefactor is just the determinant of the Jacobian matrix for the coordinate transformation. Eq. (9) is a powerful and general expression from which the explicit form of the curl operator can be deduced with ease for different coordinate systems. In the present case of cylindrical coordinates, the scaling factors are h 1 = 1, h 2 = ρ, and h = 1 and so the curl of a vector field F becomes ( ) ( ) ( ) 1 (F F = e z ) ρ ρ (F ϕ) (F ρ) + e ϕ (F z) (ρf ϕ) + e z 1ρ (F ρ) in cylindrical coordinates. (40) The Laplacian Operator By definition, the Laplacian operator of a scalar field f is given as f := div( grad f ) (41) Using the expression in Eq. (28) for the divergence of a vector field f := 1 ρ ( ) ρ( grad f ) ρ From Eq. (22) the components of the gradient of f are grad f we obtain + 1 ρ ( grad f ) ϕ + ( grad f ) z. (42) ( grad f ) ρ = ( grad f ) ϕ = 1 ρ ( grad f ) z = and therefore the Laplacian operator becomes f = 1 ( ρ ) ρ ρ ρ + or 1. Spherical Polar Coordinates (4) (44) ( ) f = ρ 1 ρ f + 2 f (45) ρ Central-field problems are typically formulated and solved in spherical polar coordinates. We define from elementary geometrical considerations 7

8 x = r sinϑcosϕ y = r sinϑsinϕ z = r cos ϑ (46) which allows us to write a position vector as: 1..1 Elementary transformations and position vectors It follows x = (r sinϑcosϕ) e x + (r sinϑsinϕ) e y + (r cosϑ) e z (47) r = (sinϑcosϕ) e x + (sinϑsinϕ) e y + (cosϑ) e z ϑ = (r cosϑcosϕ) e x + (r cosϑsinϕ) e y (r sinϑ) e z = (r sinϑsinϕ) e x + (r sinϑcosϕ) e y (48) and therefore the scaling factors become r = 1 := h 1 ϑ = r := h 2 = r sinϑ := h. (49) The orthonormal local tripods can thus be deduced as: e r = e ϑ = e ϕ = r r ϑ ϑ The original tripod is then expressed as: = sinϑcosϕ e x + sinϑsinϕ e y + cosϑ e z = cosϑcosϕ e x + cosϑsinϕ e y sinϑ e z = sinϕ e x + cosϕ e y (50) e x = sinϑcosϕ e r + cosϕcosϑ e ϑ sinϕ e ϕ e y = (sinϑ) 2 e r + sinϕcosϑ e ϑ + cosϕ e ϕ e z = cosϑ e r sinϑ e ϑ (51) 8

9 We obtain for a position vector in spherical polar coordinates x = r e r (52) by using Eqs. (47) and (51). In the following we deduce, in analogy with subsection 1.2. the differential operators in spherical polar coordinates Line Elements Elements for multidimensional integration can be easily obtained from the determined scaling factors, Eq. (49). We obtain for line elements dl r = h 1 dr = dr dl ϑ = h 2 dϑ = r dϑ dl ϕ = h dϕ = r sinϑdϕ. (5) The interesting surface element is the one over the curved sphere, for which we obtain ds surface = dl ϑ dl ϕ = h 2 h dϑdϕ = r 2 sinϑdϑdϕ. (54) The other possibilities are simply slices of the sphere which can be treated by using the calculus from cylindrical coordinates. The volume element finally reads dv = dl r dl ϑ dl ϕ = h 1 h 2 h drdϑdϕ = r 2 sinϑdrdϑdϕ. (55) 1.. The Gradient Operator With the help of the general expression for the gradient operator in curvilinear coordinates, Eq. (2), and the obtained scaling factors in Eqs. (49) the gradient operator can be formulated without additional calculation: 1..4 The Divergence Operator f = er r + e ϑ 1 r ϑ + e ϕ 1 r sinϑ (56) As expounded in subsection we start from an expression div Fg dx 1 dx 2 dx = F(r,ϑ,ϕ) gradg dv (57) ( = F(r,ϑ,ϕ) e r r + 1 r e ϑ ϑ + 1 ) r sinϑ e ϕ r 2 sinϑdrdϑdϕ where the expression in Eq. (56) has been used for writing the gradient. Integrating by parts and comparing with the left-hand side of Eq. (57) we deduce the final expression for the divergence in spherical polar coordinates: F = 1 r 2 r (r2 F r ) + 1 r sinϑ ϑ (sinϑ F ϑ) + 1 r sinϑ F ϕ (58) 9

10 1..5 The Curl Operator We use the second form in Eq. (9) as a starting point for deriving the curl operator. In spherical polar coordinates the scaling factors are determined as h 1 = 1, h 2 = r, and h = r sinϑ, and we immediately find ( ) ( ) ( ) 1 F = e r r sinϑ ϑ (sinϑf ϕ) F ϑ 1 + e 1 F r ϑ r sinϑ r (r F 1 ϕ) + e ϕ r r (r F ϑ) F r ϑ (59) 1..6 The Laplacian Operator We again determine the Laplacian operator as f := div( grad f ). Using the expression for the divergence in Eq. (58) and replacing the vector field by the gradient of a scalar field f, given in Eq. (56) the Laplacian operator is derived as ( ) f = 1 r 2 r r 2 r + 1 r 2 sinϑ ϑ ( ) sinϑ ϑ f (60) r 2 (sinϑ) 2 2 Note that the radial part is often written differently, but the form given here is the one most commonly used in the literature. 10