A Review: Operators involving in Curvilinear Orthogonal Coordinates ∇ and Vector Identities and Product Rules

∇ Markus Selmke Student at the Universität Leipzig, IPSP,[email protected]

22.7.2006, updated 15.02.2008

Contents

Abstract I

1 The Laplacian in Spherical Coordinates - The Tedious Way 1

2 The Angular Momentum Operators in Spherical Coordinates 4

3 Operators in Arbitrary Orthogonal Cuvilinear Coordinate Systems 5 ∇ Summary of Important Formulas 11

4 Examples using the General Expressions 12

5 Proofs for the Product Rules 16 ∇ 6 Gauss Divergence Theorem variants 18

References 19

Abstract

This document provides easily understandable derivations for all important operators applicable to arbitrary orthog- 3 ∇ onal coordinate systems in R (like spherical, cylindrical, elliptic, parabolic, hyperbolic,..). These include the laplacian (of a scalar function and a vector field), the gradient and its absolute value (of a scalar function), the divergence (of a vector field), the curl (of a vector field) and the convective operator (acting on a scalar function and a vector field) (Section 3). For the first two mentioned special coordinate systems the operators are calculated using these expressions along with some remarks on their simplifications in cartesian coordinates (Section 4). Alongside these calculations, some useful derivations are presented, including the volume element and the velocity components. The Laplacian as well as the absolute value of the gradient are also calculated the clumsy way (ab)using the chain-rule (Section 1). Using the same tedious way as in section 1, the angular momentum operators arising in quantum mechanics are derived for spherical coodinates (Section 2) only. Furthermore, all important product rules for the operator, which are utilized throughout the derivations in section 3, are proven using the Levi-Civita-Symbol and the Kronecker-Delta∇ (Section 5).

I 1 THE LAPLACIAN IN SPHERICAL COORDINATES - THE TEDIOUS WAY 1

1 The Laplacian in Spherical Coordinates - The Tedious Way

In this section the Laplacian operator acting on a scalar function is derived for the special case of spherical coordinates only. The way presented here and often described in physic books is very tedious and involved, though only the very basic concepts of partial differential calculus are used here. However, the operator may be calculated way easier using the method decribed in (Section 3) and carried out in (Section 4)! Recall the definition of the Laplacian in cartesian coordonates:

∂ 2 f ∂ 2 f ∂ 2 f ∆f := + + (1) ∂ x 2 ∂ y 2 ∂ z 2

Some things we need and the transforms:

x = r sinθ cosφ y = r sinθ sinφ z = r cosθ (2) ! z y ‹ p 2 2 2 θ = arccos φ = arctan r = x + y + z (3) p 2 2 2 x x + y + z



f = f x r,θ ,φ ,y r,θ ,φ ,z r,θ ,φ (4)

∂ f ∂ f ∂ r ∂ f ∂ θ ∂ f ∂ φ = + + (5) ∂ x ∂ r ∂ x ∂ θ ∂ x ∂ φ ∂ x ∂ f ∂ f ∂ r ∂ f ∂ θ ∂ f ∂ φ = + + (6) ∂ y ∂ r ∂ y ∂ θ ∂ y ∂ φ ∂ y ∂ f ∂ f ∂ r ∂ f ∂ θ ∂ f ∂ φ = + + (7) ∂ z ∂ r ∂ z ∂ θ ∂ z ∂ φ ∂ z    1 € Š 1 1 1 0 1 arccosx x arctanx arctan (8) ( )0 = p p 0 = ( )0 = 2 = 2 − 1 x 2 2px 1 + x x − 1 + x − p  y ‹‹ y  y ‹‹ x ∂ x 2 y 2 z 2 1 2x 0 0 + + arctan = 2 2 arctan = 2 2 = (9) x y y x x x x y ∂ x 2 p 2 2 2 − + + x + y + z Now we begin:

    ∂ f ∂ f  1 2r sinθ cosφ  ∂ f 1 1 2x ∂ f r sinθ sinφ  z    = +  Æ 2 2 − 2
3/2  + 2 € 2 € 2 2 ŠŠ ∂ x ∂ r 2 r ∂ θ − r cos θ 2 r ∂ φ − r sin θ cos φ + sin φ 1 r 2 − ∂ f ∂ f 1 ∂ f sinφ = sinθ cosφ + cosθ cosφ (10) ∂ r ∂ θ r − ∂ φ r sinθ     ∂ f ∂ f  1 2r sinθ sinφ  ∂ f 1 1 2y ∂ f r sinθ cosφ  z    = +  Æ 2 2 − 2
3/2  + 2 € 2 € 2 2 ŠŠ ∂ y ∂ r 2 r ∂ θ − r cos θ 2 r ∂ φ − r sin θ cos φ + sin φ 1 r 2 − ∂ f ∂ f 1 ∂ f cosφ = sinθ sinφ + cosθ sinφ + (11) ∂ r ∂ θ r ∂ φ r sinθ

   1  ∂ f ∂ f 1 2r cosθ ∂ f 1 r r cosθ 2r 2r cosθ ∂ f = +  − 2  + [0] ∂ z ∂ r 2 r ∂ θ − sinθ r ∂ φ ∂ f ∂ f sinθ = cosθ (12) ∂ r − ∂ θ r 1 THE LAPLACIAN IN SPHERICAL COORDINATES - THE TEDIOUS WAY 2

∂ 2 f ∂  ∂ f   ∂  ∂ f  ∂ r  ∂  ∂ f  ∂ θ  ∂  ∂ f  ∂ φ = = + + (13) ∂ x 2 ∂ x ∂ x ∂ r ∂ x ∂ x ∂ θ ∂ x ∂ x ∂ φ ∂ x ∂ x ∂ 2 f ∂  ∂ f   ∂  ∂ f  ∂ r  ∂  ∂ f  ∂ θ  ∂  ∂ f  ∂ φ = = + + (14) ∂ y 2 ∂ y ∂ y ∂ r ∂ y ∂ x ∂ θ ∂ x ∂ y ∂ φ ∂ y ∂ y ∂ 2 f ∂  ∂ f   ∂  ∂ f  ∂ r  ∂  ∂ f  ∂ θ  ∂  ∂ f  ∂ φ = = + + (15) ∂ z 2 ∂ z ∂ z ∂ r ∂ z ∂ z ∂ θ ∂ z ∂ z ∂ φ ∂ z ∂ z Plugging equations (10),(11),(12) into (13),(14),(15) we get:

∂ 2 f ∂  ∂ f ∂ f 1 ∂ f sinφ  2 = sinθ cosφ + cosθ cosφ sinθ cosφ + ... ∂ x ∂ r ∂ r ∂ θ r − ∂ φ r sinθ ∂  ∂ f ∂ f 1 ∂ f sinφ  cosθ cosφ sinθ cosφ + cosθ cosφ + ... ∂ θ ∂ r ∂ θ r − ∂ φ r sinθ r ∂  ∂ f ∂ f 1 ∂ f sinφ  sinφ sinθ cosφ + cosθ cosφ − ∂ φ ∂ r ∂ θ r − ∂ φ r sinθ r sinθ  ∂ 2 f 1 ∂ f sinφ ∂ f  = sinθ cosφ 2 cosθ cosφ 2 + 2 sinθ cosφ + ... ∂ r − r ∂ θ r sinθ ∂ φ  ∂ f 1 ∂ f 1 ∂ 2 f cosθ sinφ ∂ f  cosθ cosφ cosθ cosφ sinθ cosφ + cosθ cosφ 2 + 2 + ... ∂ r − r ∂ θ r ∂ θ r sin θ ∂ φ r  ∂ f 1 ∂ f cosφ ∂ f sinφ ∂ 2 f  sinφ sinθ sinφ cosθ sinφ 2 − − ∂ r − r ∂ θ − r sinθ ∂ φ − r sinθ ∂ φ sinθ 2 2 2 2 ∂ f sinθ cosθ cos φ ∂ f sinφ cosφ ∂ f = sin θ cos φ 2 2 + 2 + ... ∂ r − r ∂ θ r ∂ φ cos2 θ cos2 φ ∂ f sinθ cosθ cos2 φ ∂ f cos2 θ cos2 φ ∂ 2 f cos2 θ cosφ sinφ ∂ f ... 2 + 2 2 + 2 2 + r ∂ r − r ∂ θ r ∂ θ r sin θ ∂ φ sin2 φ ∂ f cosθ sin2 φ ∂ f cosφ sinφ ∂ f sin2 φ ∂ 2 f + + + (16) r ∂ r r 2 sinθ ∂ θ r 2 sin2 θ ∂ φ r 2 sin2 θ ∂ φ2

∂ 2 f ∂  ∂ f ∂ f 1 ∂ f cosφ  = sinθ sinφ + cosθ sinφ + sinθ sinφ + ... ∂ y 2 ∂ r ∂ r ∂ θ r ∂ φ r sinθ ∂  ∂ f ∂ f 1 ∂ f cosφ  cosθ sinφ sinθ sinφ + cosθ sinφ + + ... ∂ θ ∂ r ∂ θ r ∂ φ r sinθ r ∂  ∂ f ∂ f 1 ∂ f cosφ  cosφ sinθ sinφ + cosθ sinφ + ∂ φ ∂ r ∂ θ r ∂ φ r sinθ r sinθ 2 2 2 2 ∂ f cosθ sinθ sin φ ∂ f cosφ sinφ ∂ f = sin θ sin φ 2 2 2 + ... ∂ r − r ∂ θ − r ∂ φ cos2 θ sin2 φ ∂ f sinθ cosθ sin2 φ ∂ f cos2 θ sin2 φ ∂ 2 f cosφ cos2 θ sinφ ∂ f ... 2 + 2 2 2 2 + r ∂ r − r ∂ θ r ∂ θ − r sin θ ∂ φ cos2 φ ∂ f cos2 φ cosθ ∂ f sinφ cosφ ∂ f cos2 φ ∂ 2 f (17) + 2 2 + 2 2 2 r ∂ r r sinθ ∂ θ − r sin θ ∂ φ r sin θ ∂ φ ∂ 2 f ∂  ∂ f ∂ f sinθ  2 = cosθ cosθ + ... ∂ z ∂ r ∂ r − ∂ θ r ∂  ∂ f ∂ f sinθ  sinθ cosθ − ∂ θ ∂ r − ∂ θ r r 2 2 ∂ f sinθ cosθ ∂ f = cos θ + + ... ∂ r 2 r 2 ∂ θ sin2 θ ∂ f cosθ sinθ ∂ f sin2 θ ∂ 2 f + + (18) r ∂ r r 2 ∂ θ r 2 ∂ θ 2 1 THE LAPLACIAN IN SPHERICAL COORDINATES - THE TEDIOUS WAY 3

Using equation (16),(17),(18) we obtain the Laplacian:

∂ 2 f ∂ 2 f ∂ 2 f ∆f = + + ∂ x 2 ∂ y 2 ∂ z 2 ∂ 2 f sinθ cosθ ∂ f cos2 θ ∂ f sinθ cosθ ∂ f cos2 θ ∂ 2 f 1 ∂ f cosθ ∂ f = 2 2 + 2 + 2 2 + + 2 + ... ∂ r − r ∂ θ r ∂ r − r ∂ θ r ∂ θ r ∂ r r sinθ ∂ θ 1 ∂ 2 f sinθ cosθ ∂ f sin2 θ ∂ f cosθ sinθ ∂ f sin2 θ ∂ 2 f + + + + r 2 sin2 θ ∂ φ2 r 2 ∂ θ r ∂ r r 2 ∂ θ r 2 ∂ θ 2 ∂ 2 f 2 ∂ f 1 ∂ 2 f 1 ∂ 2 f cosθ ∂ f = + + + + ∂ r 2 r ∂ r r 2 ∂ θ 2 r 2 sin2 θ ∂ φ2 r 2 sinθ ∂ θ ∂ 2 f 2 ∂ f 1 ∂ 2 f 1  ∂ 2 f ∂ f  = + + + sinθ + cosθ ∂ r 2 r ∂ r r 2 sin2 θ ∂ φ2 r 2 sinθ ∂ θ 2 ∂ θ ∂ 2 f 2 ∂ f 1 ∂ 2 f 1 ∂  ∂ f  = + + + sinθ ∂ r 2 r ∂ r r 2 sin2 θ ∂ φ2 r 2 sinθ ∂ θ ∂ θ

∂ 2 2 ∂ 1 ∂  ∂  1 ∂ 2 ∆ = + + sinθ + (19) ∂ r 2 r ∂ r r 2 sinθ ∂ θ ∂ θ r 2 sin2 θ ∂ φ2 We also note the following relations:    2  2 1 ∂ 2 ∂ f 1 ∂ f 2 ∂ f 2 ∂ f ∂ f r = 2r + r = + (20) r 2 ∂ r ∂ r r 2 ∂ r ∂ r 2 r ∂ r ∂ r 2

2    2  2 1 ∂
1 ∂ ∂ f 1 ∂ f ∂ f ∂ f 2 ∂ f ∂ f r f (r ) = f (r ) + r = + + r = + (21) r ∂ r 2 r ∂ r ∂ r r ∂ r ∂ r ∂ r 2 r ∂ r ∂ r 2 And hence obtain the Laplacian operator in spherical coordinates:

    2 1 ∂ 2 ∂ 1 ∂ ∂ 1 ∂ ∆ = r + sinθ + (22) r 2 ∂ r ∂ r r 2 sinθ ∂ θ ∂ θ r 2 sin2 θ ∂ φ2 Or equivalently: 1 ∂ 2 1 ∂  ∂  1 ∂ 2 ∆ = r + sinθ + (23) r ∂ r 2 r 2 sinθ ∂ θ ∂ θ r 2 sin2 θ ∂ φ2 The absolute value of the gradient in cartesian coordinates is:  2  2  2 2 ∂ f ∂ f ∂ f f = + + (24) |∇ | ∂ x ∂ y ∂ z
2 2 2 2 Using equations (10),(11),(12) and x + y + z = x + y + z + 2x y + 2xz + 2y z we get:

 2  2 2 ∂ f ∂ f 1 ∂ f sinφ ∂ f ∂ f 1 ∂ f cosφ f = sinθ cosφ + cosθ cosφ + sinθ sinφ + cosθ sinφ + + ... |∇ | ∂ r ∂ θ r − ∂ φ r sinθ ∂ r ∂ θ r ∂ φ r sinθ  ∂ f ∂ f sinθ 2 cosθ ∂ r − ∂ θ r  2  2 2  2 2 2 ∂ f 2 2 1 ∂ f sin φ ∂ f = sin θ cos φ + cos θ cos φ + + ... ∂ r r 2 ∂ θ r 2 sin2 θ ∂ φ 2sinθ cosθ cos2 φ ∂ f ∂ f 2cosφ sinφ ∂ f ∂ f 2cosθ cosφ sinφ ∂ f ∂ f 2 + ... r ∂ r ∂ θ − r ∂ r ∂ φ − r sinθ ∂ θ ∂ φ  2  2 2  2 2 2 ∂ f 2 2 1 ∂ f cos φ ∂ f sin θ sin φ + cos θ sin φ + + ... ∂ r r 2 ∂ θ r 2 sin2 θ ∂ φ 2sinθ cosθ sin2 φ ∂ f ∂ f 2sinφ cosφ ∂ f ∂ f 2cosθ sinφ cosφ ∂ f ∂ f + + + ... r ∂ r ∂ θ r ∂ r ∂ φ r 2 sinθ ∂ θ ∂ φ  2 2  2 2 ∂ f cosθ sinθ ∂ f ∂ f sin θ ∂ f cos θ 2 + 2 ∂ r − r ∂ r ∂ θ r ∂ θ  ∂ f 2 1  ∂ f 2 1  ∂ f 2 = + + (25) ∂ r r 2 ∂ θ r 2 sin2 θ ∂ φ 2 THE ANGULAR MOMENTUM OPERATORS IN SPHERICAL COORDINATES 4

2 The Angular Momentum Operators in Spherical Coordinates

Using the same technique we calculate the spherical angular momentum operators:

       ∂  x px x ∂ x  ∂  r p  y   p  i h  y  b b =    y  = ħ    ∂ y   ∂  × z × pz − z × ∂ z

1 ∂ ∂ Lcx = y z i ħh ∂ z − ∂ y −  ∂ ∂ sinθ   ∂ ∂ 1 ∂ cosφ  = r sinθ sinφ cosθ r cosθ sinθ sinφ + cosθ sinφ + ∂ r − ∂ θ r − ∂ r ∂ θ r ∂ φ r sinθ 2 ∂ r sin θ sinφ ∂ ∂ 2 r ∂ r cosθ cosφ ∂ = r sinθ sinφ cosθ r cosθ sinθ sinφ cos θ sinφ ∂ r − r ∂ θ − ∂ r − r ∂ θ − r sinθ ∂ φ ∂ ∂ = sinφ cotθ cosφ − ∂ θ − ∂ φ  ∂ ∂  Lcx = i ħh sinφ cotθ cosφ (26) − − ∂ θ − ∂ φ 1 ∂ ∂ Lcy = z x i ħh ∂ x − ∂ z −  ∂ ∂ 1 ∂ sinφ   ∂ ∂ sinθ  = r cosθ sinθ cosφ + cosθ cosφ r sinθ cosφ cosθ ∂ r ∂ θ r − ∂ φ r sinθ − ∂ r − ∂ θ r ∂ ∂ = cosφ cotθ sinφ ∂ θ − ∂ φ  ∂ ∂  Lcy = i ħh cosφ cotθ sinφ (27) − ∂ θ − ∂ φ 1 ∂ ∂ Lcz = x y i ħh ∂ y − ∂ x −  ∂ ∂ 1 ∂ cosφ  = r sinθ cosφ sinθ sinφ + cosθ sinφ + ... ∂ r ∂ θ r ∂ φ r sinθ −  ∂ ∂ 1 ∂ sinφ  r sinθ sinφ sinθ cosφ + cosθ cosφ ∂ r ∂ θ r − ∂ φ r sinθ 2 ∂ ∂ 2 ∂ = r sin θ cosφ sinφ + sinθ cosφ cosθ sinφ + cos φ ... ∂ r ∂ θ ∂ φ − 2 ∂ ∂ 2 ∂ sin θ sinφ cosφ sinθ sinφ cosθ cosφ + sin φ ∂ r − ∂ θ ∂ φ ∂ = ∂ φ

∂ Lcz = i ħh (28) − ∂ φ

    2 1 ∂ 2 ∂ 1 ∂ ∂ 1 ∂ ∆ = r + sinθ + (29) r 2 ∂ r ∂ r r 2 sinθ ∂ θ ∂ θ r 2 sin2 θ ∂ φ2  ∂ f 2 1  ∂ f 2 1  ∂ f 2 f 2 (30) = + 2 + 2 2 |∇ | ∂ r r ∂ θ r sin θ ∂ φ  ∂ ∂  Lcx = i h sinφ + cotθ cosφ (31) ħ ∂ θ ∂ φ  ∂ ∂  Lcy = i ħh cotθ sinφ cosφ (32) ∂ φ − ∂ θ ∂ Lcz = i ħh (33) − ∂ φ 3 OPERATORS IN ARBITRARY ORTHOGONAL CUVILINEAR COORDINATE SYSTEMS 5 ∇ 3 Operators in Arbitrary Orthogonal Cuvilinear Coordinate Systems ∇ The Laplacian1 (of a scalar function and a vector field), the gradient and its absolute value (of a scalar function), the diver- gence (of a vector field), the curl (of a vector field) and the convective operator are often tabulated in appendices for special chosen ortho-curvilinear coordinate systems like polar and spherical coordinates. In this section we will derive a general expression for all the operators mentioned for any curvilinear orthogonal coordinate system in three dimensions using only basic concepts of algebra and calculus. (No Christoffel Symbols necessary) We will start with some simple observations: The chain rule gives:

∂ x ∂ x ∂ x dx = dq1 + dq2 + dq3 ∂ q1 ∂ q2 ∂ q3 ∂ y ∂ y ∂ y dy = dq1 + dq2 + dq3 ∂ q1 ∂ q2 ∂ q3 ∂ z ∂ z ∂ z dz = dq1 + dq2 + dq3 ∂ q1 ∂ q2 ∂ q3

2 2 2 2 This gives the square of the distance in curvilinear coordinates between two points using x + y + z = x + y + z + 2x y + 2y z + 2xz :

2 2
2 2 (ds ) = (dx) + dy + (dz ) 2
2 2
2 2
2 = Q11 dq1 +Q22 dq2 +Q33 dq3 + 2Q12dq1dq2 + 2Q13dq1dq3 + 2Q32dq3dq2 (34)

With the following notation (related to the scale factors Qi of the coordinate system):  ‚ Œ  ‚ Œ  ‚ Œ 2 ∂ x ∂ x ∂ y ∂ y ∂ z ∂ z Qi j = + + (35) ∂ qi ∂ qj ∂ qi ∂ qj ∂ qi ∂ qj

But we know for any orthogonal system of vectors ek,k = 1...n , ek el = δl k that the Pythagorean Theorem holds: { } · 2 n n X X 2 ek = ek k =1 k =1 k k

And hence only the first three terms of equation (34) contribute and Qi j = 0 for i = j 6 If dqi = 0 and dqj = 0 for all j = i , i.e. when only one of the qi coordinates changes we have: 6 6 È  ∂ x 2  ∂ y 2  ∂ z 2 ds =: dsi =Qi i dqi :=Qi dqi Qi = + + (36) ∂ qi ∂ qi ∂ qi Now we write down the gradient, realizing that any orthogonal coordinate system is cartesian in a infitisimal neighbourhood of a given point and ek,k = 1...3 are the unit vectors associated with this local coordinate system near the given point: { } ∂ f ∂ f ∂ f f = e1 + e2 + e3 ∇ ∂ s1 ∂ s2 ∂ s3 ∂ f ∂ f ∂ f = e1 + e2 + e3 Q1∂ q1 Q2∂ q2 Q3∂ q3 3 X ∂ f = ei (37) Qi ∂ qi i =1

1 proof for Laplacian of a scalar function, for the divergence and gradient are outlined in [1], based on [3]: H. Margenau and G. M. Murphy, ’The Mathe- matics of Physics and Chemistry’, D. Van Nostrand Co., Princeton, 1956, pp. 150-175 3 OPERATORS IN ARBITRARY ORTHOGONAL CUVILINEAR COORDINATE SYSTEMS 6 ∇

To see that this expression is correct we will derive it yet another, though similar way (outlined in [6]). First we formally write
down the unit vectors at some point in space defined by the curvilinear coordinates q = q1,q2,q3 by the following equation:

r q ∂ ( )

∂ qi 1 ∂ r q ∂ r q eqi =: ei = = =Qi ei ∂ r(q) Qi ∂ qi ⇔ ∂ qi ∂ qi

∂ r(q) With the so-called scale factors Qi (They scale the coordinate vectors to unit length) = ∂ qi These scale factors (in the literature often denoted by hi !) coincide with our previous definition:   ∂ x(q) 1/2




 ∂ qi  ‚
2 
2 
2Œ ∂ r q ∂ r x q ,y q ,z q  ∂ y q  ∂ x q ∂ y q ∂ z q Qi  ( )  = = = ∂ qi = + + ∂ qi ∂ qi   ∂ qi ∂ qi ∂ qi  ∂ z (q)  ∂ qi

The unit vectors ei point in the direction of growing qi along the qi -coordinate line.

∂ rq
∂ rq
∂ rq
dr = dq1 + dq2 + dq3 =Q1dq1e1 +Q2dq2e2 +Q3dq3e3 (38) ∂ q1 ∂ q2 ∂ q3
Now we will construct the gradient of a scalar function f = f q with unknown components g i , i = 1,2,3. We first note the following facts and use total differentiation:

f = g 1e1 + g 2e2 + g 3e3 ∇ dr = Q1dq1e1 +Q2dq2e2 +Q3dq3e3 df = f dr = Q1 g 1dq1 +Q2 g 2dq2 +Q3 g 3dq3 (39) ∇∂ f · ∂ f ∂ f df = dq1 + dq2 + dq3 (40) ∂ q1 ∂ q2 ∂ q3

Where the orthogonality was used in equation (39). Now, by comparison of (39), (40) we identify the g i ’s:

! ∂ f ∂ f Qi g i = g i = ∂ qi ⇒ Qi ∂ qi Hence we reconstruct the gradient by plugging in the formely unknown coefficients:

3 e1 ∂ f e2 ∂ f e3 ∂ f X ∂ f f = + + = ei Q1 ∂ q1 Q2 ∂ q2 Q3 ∂ q3 Qi ∂ qi ∇ i =1 ...which is exactly the equation we derived before (37)! From now on we will proceed in a straight-forward manner to get all the other vector operators involving using several tricks, utilizing the right-handedness and orthonormality of the coordinate system for example. The first thing∇ we will find is the divergence expression which is necessary in order to find the Laplacian. Here a simple product rule is used (proof in appendix):

3 3
X
X
g = e1 g 1 + e2 g 2 + e3 g 3 = ei g i = ei g i + g i ei (41) ∇ · ∇ · i =1 ∇ · i =1 · ∇ ∇ · The first term in the sum can be written the following way using eqn. (37) and the orthogonality of the coordinate system:   ∂ g i ∂ g i ∂ g i ∂ g i ei g i = ei e1 + e2 + e3 = (42) · ∇ · Q1∂ q1 Q2∂ q2 Q3∂ q3 Qi ∂ qi 3 OPERATORS IN ARBITRARY ORTHOGONAL CUVILINEAR COORDINATE SYSTEMS 7 ∇

The last term in the sum in eqn (41) is not equal to zero for non-cartesian coordinate systems! We will need some preparing steps in order to compute ei. Using equation (37) with f = qi we get: ∇ · ∂ qi ∂ qi ∂ qi 1 qi = e1 + e2 + e3 = ei (43) ∇ Q1∂ q1 Q2∂ q2 Q3∂ q3 Qi We also need:

e1 = (e2 e3) = e3 e2 e2 e3 (44) ∇ · ∇ · × · ∇ × − · ∇ × Where the right handedness of the system was used and the vector identity (or product rule) (v w) = w v v w as well (see appendix for proof). ∇ · × · ∇ × − · ∇ × We will furthermore encounter the following expression which we derive now:

∂ 1 ∂ 1 ∂ 1 1 Q1 Q1 Q1 ∂ Q1 ∂ Q1 ∂ Q1 = e1 + e2 + e3 = e1 3 e2 2 e3 2 (45) ∇Q1 Q1∂ q1 Q2∂ q2 Q3∂ q3 − Q1∂ q1 − Q1Q2∂ q2 − Q1Q3∂ q3 Now we begin with the following trick and use f = 0 and the product rule: ∇ × ∇  1  1 1 1 1 q1 = 0 = e1 = e1 + e1 = e1 + e1 (46) ∇ × ∇ ∇ × Q1 ∇Q1 × Q1 ∇ × − × ∇Q1 Q1 ∇ × This gives after rearranging:

1 e1 =Q1e1 (47) ∇ × × ∇Q1 Using equation (47) and (45) we get: ‚ Œ ∂ 1 ∂ Q1 ∂ Q1 e1 = Q1e1 e1 3 e2 2 e3 2 ∇ × × − Q1∂ q1 − Q1Q2∂ q2 − Q1Q3∂ q3 e1 e2 ∂ Q1 e1 e3 ∂ Q1 = × × − Q1Q2 ∂ q2 − Q1Q3 ∂ q3 e3 ∂ Q1 e2 ∂ Q1 = + (48) −Q1Q2 ∂ q2 Q1Q3 ∂ q3 Using equation (44) and appropriate cyclic permutations of the previous equation we have:

e1 = e3 ( e2) e2 ( e3) ∇ · · ∇ × − · ∇ ×    e1 ∂ Q2 e3 ∂ Q2 e2 ∂ Q3 e1 ∂ Q3 = e3 + e2 + · −Q2Q3 ∂ q3 Q1Q2 ∂ q1 − · −Q1Q3 ∂ q1 Q2Q3 ∂ q2     1 ∂ Q3 1 ∂ Q2 1 ∂ Q2 ∂ Q3 1 ∂ (Q2Q3) = + = Q3 +Q2 = (49) Q1Q3 ∂ q1 Q1Q2 ∂ q1 Q1Q2Q3 ∂ q1 ∂ q1 Q1Q2Q3 ∂ q1

Or in general:

 Q1Q2Q3   1 ∂ Q e  i  (50) i =   ∇ · Q1Q2Q3 ∂ qi

Now we write down g: ∇ · g = g 1 e1 + g 2 e2 + g 3 e3 + e1 g 1 + e2 g 2 + e3 g 3 ∇ · ∇ ·  ∇ ·  ∇ · · ∇ · ∇ · ∇  g 1 ∂ (Q2Q3) g 2 ∂ (Q1Q3) g 3 ∂ (Q1Q2) ∂ g 1 ∂ g 2 ∂ g 3 = + + + + + Q1Q2Q3 ∂ q1 Q1Q2Q3 ∂ q2 Q1Q2Q3 ∂ q3 Q1∂ q1 Q2∂ q2 Q3∂ q3 


 3   1 ∂ g 1Q2Q3 ∂ g 2Q1Q3 ∂ g 3Q1Q2 1 X ∂ g i = + + = Q1Q2Q3 (51) Q1Q2Q3 ∂ q1 ∂ q2 ∂ q3 Q1Q2Q3 ∂ qi Qi i =1

If we now set g = f we get the Laplacian in curvilinear orthogonal coordinate systems: ∇ ∂ f ∂ f ∂ f ∂ f g = f = e1 + e2 + e3 g i = (52) ∇ Q1∂ q1 Q2∂ q2 Q3∂ q3 ⇔ Qi ∂ qi 3 OPERATORS IN ARBITRARY ORTHOGONAL CUVILINEAR COORDINATE SYSTEMS 8 ∇

 Q2Q3 ∂ f  Q1Q3 ∂ f  Q1Q2 ∂ f   1 ∂ Q ∂ q ∂ Q ∂ q ∂ Q ∂ q g f : f  1 1 2 2 3 3  (53) = = ∆ =  + +  ∇ · ∇ · ∇ Q1Q2Q3 ∂ q1 ∂ q2 ∂ q3

We can rewrite it for convenience in the following way:

       1 ∂ Q2Q3 ∂ f ∂ Q1Q3 ∂ f ∂ Q1Q2 ∂ f ∆f = + + (54) Q1Q2Q3 ∂ q1 Q1 ∂ q1 ∂ q2 Q2 ∂ q2 ∂ q3 Q3 ∂ q3  3 ‚ Œ 1 X ∂ Q1Q2Q3 ∂ f =  2  (55) Q1Q2Q3 ∂ qi Q ∂ qi i =1 i  3 ‚ ‚ Œ 2 Œ 1 X ∂ Q1Q2Q3 ∂ f Q1Q2Q3 ∂ f =  2 + 2 2  (56) Q1Q2Q3 ∂ qi Q ∂ qi Q ∂ q i =1 i i i

The next operator we calculate is the curl of a vector field g:

3 ! 3 X X
g = g i ei = g i ei + g i ei (57) ∇ × ∇ × i =1 i =1 ∇ × ∇ ×
where we used the vector identity g A = g A + g A (see appendix for proof). Considering cyclic permutations of∇ equation × (48)∇ for × ∇e1 (namely× the ones we already used in equation (49)) the only thing left we need is g i ei in order to write down equation∇× (57) explicitely. We use the equation for the gradient developed before (using two different∇ × methods), equation (37), and change the indices (here j is the running index, since i is reserved for the index of g in this calculation) to do the last mentioned . We also use the fact that ei ej = 0 for i = j : ×   3 ∂ g X i  g i ei = ej ei (58)  Q j ∂ qj  ∇ × j =1 × Hence we finally get our searched for expression:

3 3 X X g = g i ei + g i ei ∇ × i =1 ∇ × i =1 ∇ ×     e2 ∂ Q1 e3 ∂ Q1 e3 ∂ Q2 e1 ∂ Q2 = g 1 + g 2 + ... Q1Q3 ∂ q3 − Q1Q2 ∂ q2 Q1Q2 ∂ q1 − Q2Q3 ∂ q3   e1 ∂ Q3 e2 ∂ Q3 g 3 + ... Q2Q3 ∂ q2 − Q1Q3 ∂ q1     ∂ g 1 ∂ g 1 ∂ g 2 ∂ g 2  e e e e   e e e e  ...  2 1 + 3 1 +  1 2 + 3 2 + Q2∂ q2 | {z× } Q3∂ q3 | {z× } Q1∂ q1 | {z× } Q3∂ q3 | {z× } e3 e2 e3 e1  −  − ∂ g 3 ∂ g 3  e e e e   1 3 + 2 3 Q1∂ q1 | {z× } Q2∂ q2 | {z× } e2 e1  −  g 3 ∂ Q3 g 2 ∂ Q2 ∂ g 3 ∂ g 2 = e1 + + ... Q2Q3 ∂ q2 − Q2Q3 ∂ q3 Q2∂ q2 − Q3∂ q3   g 1 ∂ Q1 g 3 ∂ Q3 ∂ g 1 ∂ g 3 e2 + + ... Q1Q3 ∂ q3 − Q1Q3 ∂ q1 Q3∂ q3 − Q1∂ q1   g 2 ∂ Q2 g 1 ∂ Q1 ∂ g 2 ∂ g 1 e3 + Q1Q2 ∂ q1 − Q1Q2 ∂ q2 Q1∂ q1 − Q2∂ q2 

 

 

 e1 ∂ Q3 g 3 ∂ Q2 g 2 e2 ∂ Q1 g 1 ∂ Q3 g 3 e3 ∂ Q2 g 2 ∂ Q1 g 1 = + + Q2Q3 ∂ q2 − ∂ q3 Q1Q3 ∂ q3 − ∂ q1 Q1Q2 ∂ q1 − ∂ q2 3 3 3
X eiQi X ∂
X εi j kQi ∂ Qk g k = εi j k Qk g k = ei (59) Q1Q2Q3 ∂ qj Q1Q2Q3 ∂ qj i =1 j ,k =1 i ,j ,k =1 3 OPERATORS IN ARBITRARY ORTHOGONAL CUVILINEAR COORDINATE SYSTEMS 9 ∇

In the last short formula we realized that the terms belonging to ei represented the i-th coordinate of a cross product of a ∂ formal vector with j-th component and a vector with k-th component being Qk g k . It is Important to realize, that the ∂ qj unsymmetric identities we derive rely heavily on the fact, that e1,e2,e3 form a righthanded system! Thus, one should be careful when identifying the unit vectors in a curvilinear orthogonal{ coordinate} system with the ei’s in the right order and also to identify the corresponding coordinates of the vectors in this space with the g i ’s, or, later on, the Ai ’s and Bi ’s. The last operator will be the Laplacian of a vector field, say g, which is a vector defined as:

2

g =: ∆g := ( )g = g g (60) ∇ ∇ · ∇ ∇ ∇ · − ∇ × ∇ × Where we will use equations (37), (51) and (59) we derived before for the gradient, divergence and the curl of a vector and the

vector identity g = g ( )g . Sometimes the symbol ƒ is used to distinguish the vector Laplacian from the scalar Laplacian.∇ × ∇ Expression × ∇ ∇ (60) · is− somewhat∇ · ∇ nasty to write down and won’t get too much easier writing it this way:

  • ˜‹ h 3  i 3 εk l mQk ∂ Qm g m 3 1 P ∂ g k 3 P ( ) ∂ Q1Q2Q3 ∂ Qk l ,m 1 Q Q Q q X  Q1Q2Q3 k =1 ∂ qk Qk X εi j kQi = 1 2 3 ∂ l  = ei Qi ∂ qi Q1Q2Q3 ∂ qj i =1  − j ,k =1 

 •  ∂ Q2 g 2 ∂ Q1 g 1 ‹˜ •  ∂ Q1 g 1 ∂ Q3 g 3 ‹˜  Q3 ( ) ( ) Q2 ( ) ( ) h 1 P3 ∂  g k i ∂ Q Q ∂ q ∂ q ∂ Q Q ∂ q ∂ q ∂ Q1Q2Q3  1 2 1 2 1 3 3 1 Q1Q2Q3 k =1 ∂ qk Qk  = e1  − + − +  + ... − Q2Q3∂ q2 Q2Q3∂ q3 Q1∂ q1 

 •  ∂ Q3 g 3 ∂ Q2 g 2 ‹˜ •  ∂ Q2 g 2 ∂ Q1 g 1 ‹˜  Q1 ( ) ( ) Q3 ( ) ( ) h 1 P3 ∂  g k i ∂ Q Q ∂ q ∂ q ∂ Q Q ∂ q ∂ q ∂ Q1Q2Q3  2 3 2 3 1 2 1 2 Q1Q2Q3 k =1 ∂ qk Qk  e2  − + − +  + ... − Q1Q3∂ q3 Q1Q3∂ q1 Q2∂ q2 

 •  ∂ Q1 g 1 ∂ Q3 g 3 ˜‹ •  ∂ Q3 g 3 ∂ Q2 g 2 ‹˜  Q2 ( ) ( ) Q1 ( ) ( ) h 1 P3 ∂  g k i ∂ Q Q ∂ q ∂ q ∂ Q Q ∂ q ∂ q ∂ Q1Q2Q3  1 3 3 1 2 3 2 3 Q1Q2Q3 k =1 ∂ qk Qk  e3  − + − +  (61) − Q1Q2∂ q1 Q1Q2∂ q2 Q3∂ q3 

Using equation (37) without f , which is, so to say, the definition for the operator in curvilinear coordinates and ei ej = δi j one gets the so-called convective operator acting on a scalar function f∇: ·   

∂ ∂ ∂ g f = g 1e1 + g 2e2 + g 3e3 e1 + e2 + e3 f · ∇ Q1∂ q1 Q2∂ q2 Q3∂ q3 3 g 1 ∂ f g 2 ∂ f g 3 ∂ f X g i ∂ f = + + = (62) Q1 ∂ q1 Q2 ∂ q2 Q3 ∂ q3 Qi ∂ qi i =1 We can get an explicit formula for the convective operator acting on a vector field A using the following two vector identities (adding eqns. (63)+(64), proofs for both given in the appendix):

(A B) = A( B) B( A) + (B )A (A )B (63) ∇ × × ∇ · − ∇ · · ∇ − · ∇ (A B) = A ( B) + B ( A) + (B )A + (A )B (64) ∇ · × ∇ × × ∇ × · ∇ · ∇ (A B) + (A B) = A( B) B( A) + 2(B )A + A ( B) + B ( A) (65) ∇ × × ∇ · ∇ · − ∇ · · ∇ × ∇ × × ∇ × Hence we get a usable representation:

1 (B )A = [ (A B) + (A B) A( B) + B( A) A ( B) B ( A)] (66) · ∇ 2 ∇ × × ∇ · − ∇ · ∇ · − × ∇ × − × ∇ × In this expression only known and derived expressions like curl of a vector, divergence of a vector and the gradient of a scalar, Ai Bi , are present and could therefore be expressed in a general way. However, it is an insanely lengthy equation for which an easier form will be derived using a different approach, basically by observation. Anyway, finally, just for completeness sake of this script I will write down the formula for the convective operator acting on a vector field A in curvilinear orthogonal coordinates: 3 OPERATORS IN ARBITRARY ORTHOGONAL CUVILINEAR COORDINATE SYSTEMS 10 ∇

3 3 € Š X εi j kQi ∂ (Qk εk l m Al Bm ) X ∂ A j Bj (B )A = ei + ei ... 2Q1Q2Q3 ∂ qj 2Qi ∂ qi · ∇ i ,j ,k =1 i =1 − 3 3   3 3   X 1 X ∂ Bi X 1 X ∂ Ai eiAi Q1Q2Q3 + ei Bi Q1Q2Q3 ... 2Q1Q2Q3 ∂ qi Qi 2Q1Q2Q3 ∂ qi Qi i =1 i =1 i =1 i =1 − 3 3 3 3 X X εi j kQi ∂ (Qk Bk ) X X εi j kQi ∂ (Qk Ak ) egεg hi Ah egεg hi Bh 2Q1Q2Q3 ∂ qj 2Q1Q2Q3 ∂ qj g =1 j ,k =1 − g =1 j ,k =1   3 3 € Š ε Q ∂ Q ε A B ∂ Ap Bp X  X i j k i ( k k l m l m )  = ei + ...  2Q1Q2Q3 ∂ qj 2Qi ∂ qi  i =1 j ,k =1 − 3 3   3  ! X Ai X ∂ Bi Bi X ∂ Ai ei Q1Q2Q3 + Q1Q2Q3 ... 2Q1Q2Q3 ∂ qi Qi 2Q1Q2Q3 ∂ qi Qi i =1 i =1 i =1 −   3 3 3 X X εk l mQk ∂ (Qm Bm ) X εk l mQk ∂ (Qm Am ) ei εi j k A j εi j k Bj  (67)  2Q1Q2Q3 ∂ ql 2Q1Q2Q3 ∂ ql  i =1 l ,m=1 − l ,m=1

For e1 we will write down the term explicitely, using eq. (66), and later on consider cyclic permutations to get the results for the other 3 coordinates and find the general formula.

(2(B )A)1 · ∇ A1 B2 A2 B1 ∂ Q3 A3 B1 A1 B3 ∂ Q2 ∂ (A1 B2 A2 B1) ∂ (A3 B1 A1 B3) = − − + − − ... Q2Q3 ∂ q2 − Q2Q3 ∂ q3 Q2∂ q2 − Q3∂ q3 −     B2 ∂ Q2 B1 ∂ Q1 ∂ B2 ∂ B1 B1 ∂ Q1 B3 ∂ Q3 ∂ B1 ∂ B3 A2 + + A3 + ... Q1Q2 ∂ q1 − Q1Q2 ∂ q2 Q1∂ q1 − Q2∂ q2 Q1Q3 ∂ q3 − Q1Q3 ∂ q1 Q3∂ q3 − Q1∂ q1 −     A2 ∂ Q2 A1 ∂ Q1 ∂ A2 ∂ A1 A1 ∂ Q1 A3 ∂ Q3 ∂ A1 ∂ A3 B2 + + B3 + + ... Q1Q2 ∂ q1 − Q1Q2 ∂ q2 Q1∂ q1 − Q2∂ q2 Q1Q3 ∂ q3 − Q1Q3 ∂ q1 Q3∂ q3 − Q1∂ q1   ∂ (A1 B1 + A2 B2 + A3 B3) A1 ∂ ∂ ∂ (Q2Q3 B1) + (Q1Q3 B2) + (Q1Q2 B3) + ... Q1∂ q1 − Q1Q2Q3 ∂ q1 ∂ q2 ∂ q3   B1 ∂ ∂ ∂ (Q2Q3A1) + (Q1Q3A2) + (Q1Q2A3) Q1Q2Q3 ∂ q1 ∂ q2 ∂ q3

A1 B2 ∂ Q3 A2 B1 ∂ Q3 A3 B1 ∂ Q2 A1 B3 ∂ Q2 A1 ∂ B2 B2 ∂ A1 A2 ∂ B1 B1 ∂ A2 = + + + ... Q2Q3 ∂ q2 − Q2Q3 ∂ q2 − Q2Q3 ∂ q3 Q2Q3 ∂ q3 Q2 ∂ q2 Q2 ∂ q2 − Q2 ∂ q2 − Q2 ∂ q2 − A3 ∂ B1 B1 ∂ A3 A1 ∂ B3 B3 ∂ A1 + + ... Q3 ∂ q3 − Q3 ∂ q3 Q3 ∂ q3 Q3 ∂ q3 − A2 B2 ∂ Q2 A2 B1 ∂ Q1 A2 ∂ B2 A2 ∂ B1 A3 B1 ∂ Q1 A3 B3 ∂ Q3 A3 ∂ B1 A3 ∂ B3 + + + + ... Q1Q2 ∂ q1 Q1Q2 ∂ q2 − Q1 ∂ q1 Q2 ∂ q2 Q1Q3 ∂ q3 − Q1Q3 ∂ q1 Q3 ∂ q3 − Q1 ∂ q1 − B2A2 ∂ Q2 B2A1 ∂ Q1 B2 ∂ A2 B2 ∂ A1 B3A1 ∂ Q1 B3A3 ∂ Q3 B3 ∂ A1 B3 ∂ A3 + + + + + ... Q1Q2 ∂ q1 Q1Q2 ∂ q2 − Q1 ∂ q1 Q2 ∂ q2 Q1Q3 ∂ q3 − Q1Q3 ∂ q1 Q3 ∂ q3 − Q1 ∂ q1 A1 ∂ B1 B1 ∂ A1 A2 ∂ B2 B2 ∂ A2 A3 ∂ B3 B3 ∂ A3 + + + + + ... Q1 ∂ q1 Q1 ∂ q1 Q1 ∂ q1 Q1 ∂ q1 Q1 ∂ q1 Q1 ∂ q1 − A1 ∂ B1 A1 ∂ B2 A1 ∂ B3 A1 B1 ∂ Q3 A1 B1 ∂ Q2 A1 B2 ∂ Q3 A1 B2 ∂ Q1 A1 B3 ∂ Q2 A1 B3 ∂ Q1 + ... Q1 ∂ q1 − Q2 ∂ q2 − Q3 ∂ q3 − Q1Q3 ∂ q1 − Q1Q2 ∂ q1 − Q2Q3 ∂ q2 − Q1Q2 ∂ q2 − Q2Q3 ∂ q3 − Q1Q3 ∂ q3 B1 ∂ A1 B1 ∂ A2 B1 ∂ A3 B1A1 ∂ Q3 B1A1 ∂ Q2 B1A2 ∂ Q3 B1A2 ∂ Q1 B1A3 ∂ Q2 B1A3 ∂ Q1 + + + + + + + + Q1 ∂ q1 Q2 ∂ q2 Q3 ∂ q3 Q1Q3 ∂ q1 Q1Q2 ∂ q1 Q2Q3 ∂ q2 Q1Q2 ∂ q2 Q2Q3 ∂ q3 Q1Q3 ∂ q3

B2 ∂ A1 B3 ∂ A1 A2 B2 ∂ Q2 A2 B1 ∂ Q1 A3 B1 ∂ Q1 A3 B3 ∂ Q3 B1 ∂ A1 = 2 + 2 2 + 2 + 2 2 + 2 (68) Q2 ∂ q2 Q3 ∂ q3 − Q1Q2 ∂ q1 Q1Q2 ∂ q2 Q1Q3 ∂ q3 − Q1Q3 ∂ q1 Q1 ∂ q1 Where in the last line expressions are noted in the order of appereance of non-vanishing twice appearing terms. (You may print out the page and cancel the terms by yourself to check the result) 3 OPERATORS IN ARBITRARY ORTHOGONAL CUVILINEAR COORDINATE SYSTEMS 11 ∇

3     X Bk ∂ A1 A2 ∂ Q1 ∂ Q2 A3 ∂ Q1 ∂ Q3 ((B )A)1 = + B1 B2 + B1 B3 (69) Qk ∂ qk Q1Q2 ∂ q2 ∂ q1 Q1Q3 ∂ q3 ∂ q1 · ∇ k =1 − − 3     X Bk ∂ A2 A3 ∂ Q2 ∂ Q3 A1 ∂ Q2 ∂ Q1 ((B )A)2 = + B2 B3 + B2 B1 (70) Qk ∂ qk Q2Q3 ∂ q3 ∂ q2 Q1Q2 ∂ q1 ∂ q2 · ∇ k =1 − − 3     X Bk ∂ A3 A1 ∂ Q3 ∂ Q1 A2 ∂ Q3 ∂ Q2 ((B )A)3 = + B3 B1 + B3 B2 (71) Qk ∂ qk Q1Q3 ∂ q1 ∂ q3 Q2Q3 ∂ q2 ∂ q3 · ∇ k =1 − − Where cyclic permutations of 123 were used to get the formulas for the second and third coordinate. Summarizing, we can write down the general formula for the convective operator in curvilinear orthogonal coordinates:   3 3 3 ‚ Œ X X Bk ∂ Ai X A j ∂ Qi ∂ Q j  (B )A = ei + Bi Bj (72) Qk ∂ qk QiQ j ∂ qj ∂ qi · ∇ i =1 k =1 j =1 − 

For a curvilinear orthogonal righthanded (only important for the last 3) coordinate system we have

1 ∂ r q1,q2,q3 x = x q1,q2,q3 ; y = y q1,q2,q3 ; z = z q1,q2,q3 ; ei = Qi ∂ qi

È 3  ∂ x 2  ∂ y 2  ∂ z 2 Y Qi = + + ; dV = Qi dqi ∂ qi ∂ qi ∂ qi i =1

s 3 3 X 2 2 X ds = Qi dqi ; dr = Qi dqi ei i =1 i =1 3   3  2 X ∂ f 2 X 1 ∂ f f = ei ; f = 2 Qi ∂ qi Q ∂ qi ∇ i =1 ∇ i =1 i  3 ‚ Œ 3   1 X ∂ Q1Q2Q3 ∂ f 1 X ∂ Ai ∆f =  2 ; A = Q1Q2Q3 Q1Q2Q3 ∂ qi Q ∂ qi Q1Q2Q3 ∂ qi Qi i =1 i ∇ · i =1

3 3 2 X Ai ∂ f 2 X 2 (A ) f = v = Qi q·i Qi ∂ qi · ∇ i =1 i =1 3 ( 3   ) X X Bk ∂ Ai Ak ∂ Qi ∂ Qk (B )A = ei + Bi Bk Qk ∂ qk QiQk ∂ qk ∂ qi · ∇ i =1 k =1 −   3 3 X  Qi X ∂  A = ei εi j k (Qk Ak ) Q1Q2Q3 ∂ qj ∇ × i =1  j ,k =1 

   3 3 ! 3 3 2 ∂ 1 ∂  A  Q ε ∂ εk l mQ ∂ Q A X  X k X i i j k  X k ( m m )  ∆A = ei Q1Q2Q3 Qi ∂ qi Q1Q2Q3 ∂ qk Qk Q1Q2Q3 ∂ qj  Q1Q2Q3 ∂ ql  i =1  k =1 − j ,k =1 l ,m=1  4 EXAMPLES USING THE GENERAL EXPRESSIONS 12

4 Examples using the General Expressions

These examples demonstarte the ease of use of the derived equations, first for spherical coordinates:

¦ © x = r sinθ cosφ; y = r sinθ sinφ; z = r cosθ ; er,eθ ,eφ, (73)

2 2 2 2 2 2 Qr = sin θ cos φ + sin θ sin φ + cos θ = 1 Qr = 1 (74) 2 2 2 2 2 2 2 2 2 ⇒ Qθ = r cos θ cos φ + r cos θ sin φ + sin θ = r Qθ = r (75) 2 2 2 2 2 2 2 2 2 ⇒ Qφ = r sin θ sin φ + r sin θ cos φ = r sin θ Qφ = r sinθ (76) ⇒

¨     ‚ Œ« 1 ∂ QθQφ ∂ f ∂ QrQφ ∂ f ∂ QrQθ ∂ f ∆f = + + (77) QrQθQφ ∂ r Qr ∂ r ∂ θ Qθ ∂ θ ∂ φ Qφ ∂ φ 1  ∂  r 2 sinθ ∂ f  ∂  r sinθ ∂ f  ∂  r ∂ f  = + + r 2 sinθ ∂ r 1 ∂ r ∂ θ r ∂ θ ∂ φ r sinθ ∂ φ     2 1 ∂ 2 ∂ f 1 ∂ ∂ f 1 ∂ f = r + sinθ + (78) r 2 ∂ r ∂ r r 2 sinθ ∂ θ ∂ θ r 2 sin2 θ ∂ φ2

∂ f 1 ∂ f 1 ∂ f f = er + eθ + eφ (79) ∇ ∂ r r ∂ θ r sinθ ∂ φ

€ Š
!
€ Š ! 

 er ∂ Qφ g φ ∂ Qθ g θ eθ ∂ Qr g r ∂ Qφ g φ eφ ∂ Qθ g θ ∂ Qr g r g = + + ∇ × QθQφ ∂ θ − ∂ φ QrQφ ∂ φ − ∂ r QrQθ ∂ r − ∂ θ € Š
!
€ Š ! 

 er ∂ r sinθ g φ ∂ r g θ eθ ∂ g r ∂ r sinθ g φ eφ ∂ r g θ ∂ g r = 2 + + r sinθ ∂ θ − ∂ φ r sinθ ∂ φ − ∂ r r ∂ r − ∂ θ € Š ! € Š ! 
 er ∂ sinθ g φ ∂ g θ eθ ∂ g r ∂ r g φ eφ ∂ r g θ ∂ g r = + + (80) r sinθ ∂ θ − ∂ φ r sinθ ∂ φ − ∂ r r ∂ r − ∂ θ

  1 ∂ € Š ∂ € Š ∂ € Š g = QθQφ g r + QrQφ g θ + QrQθ g φ ∇ · QrQθQφ ∂ r ∂ θ ∂ φ   1 ∂ € 2 Š ∂
∂ € Š = r sinθ g r + r sinθ g θ + r g φ r 2 sinθ ∂ r ∂ θ ∂ φ

1 ∂ € 2 Š 1 ∂
1 ∂ € Š = r g r + sinθ g θ + g φ (81) r 2 ∂ r r sinθ ∂ θ r sinθ ∂ φ

∆g = g + g −∇ × ∇ × ∇ ∇ ·

 •  ∂ r g ‹˜ •  ∂ r g ‹˜  ∂ sinθ ( θ ) ∂ g r ∂ 1 ∂ g r ( φ ) er  r ∂ r ∂ θ r sinθ ∂ φ ∂ r  =  − −  ... − r sinθ  ∂ θ − ∂ φ  −

 •  ∂ sinθ g ‹˜ • ∂ r g ‹˜ ∂ 1 ( φ ) ∂ g θ ∂ ( θ ) ∂ g r eθ  r sinθ ∂ θ ∂ φ ∂ r ∂ θ   − −  ... r  sinθ ∂ φ − ∂ r  −

 • ∂ r g ‹˜‹ •  ∂ sinθ g ‹˜ ∂ ∂ g r ( φ ) ∂ 1 ( φ ) ∂ g θ eφ  sinθ ∂ φ ∂ r r sinθ ∂ θ ∂ φ   − −  + ... r  ∂ r − ∂ θ  4 EXAMPLES USING THE GENERAL EXPRESSIONS 13

h 1 1 1 € Ši ∂ r 2 g
∂ sin g
∂ g ∂ r 2 ∂ r r + r sinθ ∂ θ θ θ + r sinθ ∂ φ φ er + ... ∂ r h 1 1 1 € Ši ∂ r 2 g
∂ sin g
∂ g ∂ r 2 ∂ r r + r sinθ ∂ θ θ θ + r sinθ ∂ φ φ eθ + ... r ∂ θ h 1 1 1 € Ši ∂ r 2 g
∂ sin g
∂ g ∂ r 2 ∂ r r + r sinθ ∂ θ θ θ + r sinθ ∂ φ φ eφ r sinθ ∂ φ  2 2 2  1 ∂ r g r 1 ∂ g r 1 ∂ g r cotθ ∂ g r 2 ∂ g 2 ∂ g φ 2g r 2cotθ e ( ) θ g ... = r 2 + 2 2 + 2 2 2 + 2 2 2 2 2 θ + r ∂ r r ∂ θ r sin θ ∂ φ r ∂ θ − r ∂ θ − r sinθ ∂ φ − r − r  2 2 2  1 ∂ (r g θ ) 1 ∂ g θ 1 ∂ g θ cotθ ∂ g θ 2 cotθ ∂ g φ 2 g θ e ∂ g ∂ θ ... θ 2 + 2 2 + 2 2 2 + 2 2 + 2 r 2 2 + r ∂ r r ∂ θ r sin θ ∂ φ r ∂ θ − r sinθ ∂ φ r − r sin θ ‚ 2 2 2 Œ 1 ∂ (r g φ) 1 ∂ g φ 1 ∂ g φ cotθ ∂ g φ 2 ∂ g r 2cotθ ∂ g θ g φ e φ 2 + 2 2 + 2 2 2 + 2 + 2 + 2 2 2 r ∂ r r ∂ θ r sin θ ∂ φ r ∂ θ r sinθ ∂ φ r sinθ ∂ φ − r sin θ   2 ∂ g θ 2g θ cotθ 2 ∂ g φ = er ∆g r 2 2 2 + ... − r ∂ θ − r − r sinθ ∂ φ   2 ∂ g r g 2cosθ ∂ g φ e g θ ... θ ∆ θ + 2 2 2 2 2 + r ∂ θ − r sin θ − r sin θ ∂ φ   g φ 2 ∂ g r 2cosθ ∂ g φ e g (82) φ ∆ φ 2 2 + 2 + 2 2 − r sin θ r sinθ ∂ φ r sin θ ∂ φ (83)

Using Scharz’ theorem many derivatives dropped out and terms vanished. But still, it would take a lot of time to get the result stated in the last lines. (Forget the before mentioned ease for this one)

  ‚ Œ X  X Bk ∂ Ai X A j ∂ Qi ∂ Q j  (B )A = ei + Bi Bj Qk ∂ qk QiQ j ∂ qj ∂ qi · ∇ i =r,φ,θ k =r,φ,θ j =r,φ,θ −    ∂ Ar Bθ ∂ Ar Bφ ∂ Ar Bθ Aθ + BφAφ = er Br + + + ... ∂ r r ∂ θ r sinθ ∂ φ − r   ∂ Aθ Bθ ∂ Aθ Bφ ∂ Aθ Bθ Ar BφAφ cotθ eθ Br + + + + ... ∂ r r ∂ θ r sinθ ∂ φ r − r   ∂ Aφ Bθ ∂ Aφ Bφ ∂ Aφ BφAr BφAθ cotθ eφ Br + + + + (84) ∂ r r ∂ θ r sinθ ∂ φ r r (85)

For cylindrical coordinates we find the following identities.

¦ © z = z ; x = r cosφ; y = r sinφ; er,eφ,ez, (86) È  ∂ x 2  ∂ y 2  ∂ z 2 Qi = + + (87) ∂ qi ∂ qi ∂ qi

2 2 2 2 2 2 Qr = 0 + r cos φ + r sin φ = 1 Qr = 1 (88) 2 2 2 2 2 2 2 ⇒ Qφ = 0 + r sin φ + r cos φ = r Qφ = r (89) 2 2 2 2 ⇒ Qz = 1 + 0 + 0 = 1 Qz = 1 (90) ⇒

¨   ‚ Œ  « 1 ∂ QφQz ∂ f ∂ QrQz ∂ f ∂ QrQφ ∂ f ∆f = + + QrQzQφ ∂ r Qr ∂ r ∂ φ Qφ ∂ φ ∂ z Qz ∂ z 1  ∂  ∂ f  ∂  1 ∂ f  ∂  ∂ f  1 ∂  ∂ f  1 ∂ 2 f ∂ 2 f = r + + r = r + + (91) r ∂ r ∂ r ∂ φ r ∂ φ ∂ z ∂ z r ∂ r ∂ r r 2 ∂ φ2 ∂ z 2 4 EXAMPLES USING THE GENERAL EXPRESSIONS 14

∂ f 1 ∂ f ∂ f f = er + eφ + ez (92) ∇ ∂ r r ∂ φ ∂ z

€ Š ! 

 € Š
! er ∂ Qz g z ∂ Qφ g φ eφ ∂ Qr g r ∂ Qz g z ez ∂ Qφ g φ ∂ Qr g r g = + + ∇ × QφQz ∂ φ − ∂ z QrQz ∂ z − ∂ r QrQφ ∂ r − ∂ φ
€ Š ! 

 € Š
! er ∂ g z ∂ r g φ ∂ g r ∂ g z ez ∂ r g φ ∂ g r = + eφ + r ∂ φ − ∂ z ∂ z − ∂ r r ∂ r − ∂ φ     € Š ! 1 ∂ g z ∂ g φ ∂ g r ∂ g z 1 ∂ r g φ 1 ∂ g r = er + eφ + ez (93) r ∂ φ − ∂ z ∂ z − ∂ r r ∂ r − r ∂ φ

  1 ∂ € Š ∂ € Š ∂ € Š g = QφQz g r + QrQz g φ + QrQφ g z ∇ · QrQφQφ ∂ r ∂ φ ∂ z   1 ∂
∂ € Š ∂
= r g r + g φ + r g z r ∂ r ∂ φ ∂ z 1 ∂
1 ∂ g φ ∂ g z = r g r + + (94) r ∂ r r ∂ φ ∂ z

∆g = g + g −∇ × ∇ × ∇ ∇ ·   ‹  1 ∂ r g φ 1 ∂ g r    ∂ g  ∂ ( ) ∂ ∂ g r ∂ g z ∂ 1 ∂ r g
1 φ ∂ g z  1 r ∂ r r ∂ φ ∂ z ∂ r  r ∂ r r + r ∂ φ + ∂ z = er  − −  + er ... −  r ∂ φ − ∂ z  ∂ r −   ‹  ∂ g  1 ∂ r g φ 1 ∂ g r  ∂ g  ∂ 1 ∂ g z φ ∂ ( ) ∂ 1 ∂ r g
1 φ ∂ g z  r ∂ φ ∂ z r ∂ r r ∂ φ  1 r ∂ r r + r ∂ φ + ∂ z eφ  − −  + eφ ...  ∂ z − ∂ r  r ∂ φ −

     g    g  r ∂ g r ∂ g z 1 ∂ g z ∂ φ 1 ∂ r g
1 ∂ φ ∂ g z 1 ∂ ∂ z ∂ r 1 ∂ r ∂ φ ∂ z ∂ r ∂ r r + r ∂ φ + ∂ z e  − −  e z   + z r ∂ r − r ∂ φ ∂ z  2 2 2  ∂ g r 1 ∂ g r ∂ g r 1 ∂ g r 2 ∂ g φ g r = er 2 + 2 2 + 2 + 2 2 + ... ∂ r r ∂ φ ∂ z r ∂ r − r ∂ φ − r ‚ 2 2 2 Œ ∂ g φ 1 ∂ g φ ∂ g φ 1 ∂ g φ 2 ∂ g r g φ eφ 2 + 2 2 + 2 + + 2 2 + ... ∂ r r ∂ φ ∂ z r ∂ r r ∂ φ − r  2 2 2  ∂ g z 1 ∂ g z ∂ g z 1 ∂ g z ez + + + ∂ r 2 r 2 ∂ φ2 ∂ z 2 r ∂ r     2 ∂ g φ g r 2 ∂ g r g φ
= er ∆g r 2 2 + eφ ∆g φ + 2 2 + ez ∆g z (95) − r ∂ φ − r r ∂ φ − r

  ‚ Œ X  X Bk ∂ Ai X A j ∂ Qi ∂ Q j  (B )A = ei + Bi Bj Qk ∂ qk QiQ j ∂ qj ∂ qi · ∇ i =r,z ,φ k =r,z ,φ j =r,z ,φ −    ∂ Ar Bφ ∂ Ar ∂ Ar BφAφ = er Br + + Bz + ... ∂ r r ∂ φ ∂ z − r  ∂ Aφ Bφ ∂ Aφ ∂ Aφ BφAr  eφ Br + + Bz + + ... ∂ r r ∂ φ ∂ z r   ∂ Az Bφ ∂ Az ∂ Az ez Br + + Bz (96) ∂ r r ∂ φ ∂ z 4 EXAMPLES USING THE GENERAL EXPRESSIONS 15

An important thing to note is, that only in cartesian coordinates, the Laplacian of a vector field decomposes solemly to the Laplacian of each coordinate times the corresponding unit vector (using eqn. (62)):

∆g = ( )g = g + g = ex∆g x + ey∆g y + ez∆g z (97) ∇ · ∇ −∇ × ∇ × ∇ ∇ · Since we have for the simple cartesian case:

 g x g y g z g x   2 2 2   2 2 2  ∂ (∂ y /∂ ∂ x /∂ ) ∂ (∂ x /∂ ∂ z /∂ ) ∂ g x ∂ g y ∂ g z ∂ g x ∂ g x ∂ g x ∂ −y ∂ −z ∂ x 2 + ∂ x∂ y + ∂ x∂ z ∂ x 2 + ∂ y 2 + ∂ z 2  −   2 ∂ 2 g 2   ∂ 2 g ∂ 2 g ∂ 2 g   ∂ ∂ g x /∂ z ∂ g z /∂ x ∂ ∂ g y /∂ x ∂ g x /∂ y   ∂ g x y ∂ g z   y y y  g ( ) ( ) 2 2 2 2 ∆ =  ∂ −z ∂ −x  +  ∂ x∂ y + ∂ y + ∂ y ∂ z  =  ∂ x + ∂ y + ∂ z  2 2 2 2 2 2 − ∂ ∂ g x /∂ z ∂ g z /∂ x − ∂ ∂ g z /∂ y ∂ g y /∂ z   ∂ g x ∂ g y ∂ g z   ∂ g z ∂ g z ∂ g z  ( ) ( ) 2 2 2 ∂ −x ∂ −y ∂ x∂ z + ∂ y ∂ z + ∂ z 2 ∂ x + ∂ y + ∂ z − Also notice the simplified version of the convective operator in the case of cartesian coordinates.

        B ∂ B ∂ B ∂ A   x ∂ x + y ∂ y + z ∂ z x ∂ ∂ A     X  X   X  X i   ∂ ∂ ∂  (B )A = Bj Ai ei = Bj ei = Bx + By + Bz Ay (98)  ∂ x  ∂ x  ∂ x ∂ y ∂ z  · ∇ i x,y,z  j x,y,z j  i x,y,z j x,y,z j      = = = = B ∂ B ∂ B ∂ A x ∂ x + y ∂ y + z ∂ z z

Looking back at page 6 we can find further interesting facts: For example we get for the differential ds of the line-element, using orthonormality again (ek el = δl k ) and the expression we found for dr: · s 3 2 2 2 2 2 2 2 X 2 2 ds = dr dr = Q1dq1 +Q2dq2 +Q3dq3 ds = Qi dqi (99) · ⇒ i =1 The infitisimal volume element is obviously easily calculable using the parallelepidial product of the scaled parts of the vector dr:

3


 Y dV = Q1dq1e1 Q2dq2e2 Q3dq3e3 =Q1Q2Q3 dq1dq2dq3 = Qi dqi (100) · × i =1

Where we used e1 (e2 e3) = 1. For spherical and cylindrical coordinates we get the well-known expressions which are also derivable via graphicalk · × observationk or, more formally, from the Jacobian due to the general change-of-variables-theorem:

2 dVsph.c. = r sinθ dr dθ dφ (101)

dVcyl.c. = r dr dz dφ (102)

As another application in between we consider the velocity in curvilinear coordinates using again the derived expression for dr and differentiating it with respect to time.

3 3 dr X ∂ r dqi dq1 dq2 dq3 X = r· = v = = Q1e1 +Q2e2 +Q3e3 = Qi q·i ei (103) dt ∂ qi dt dt dt dt i =1 i =1 Hence we get for the absolute value of the velocity (acceleration is more involved!), which is important for expressing the 1 2 kinetic energy 2 mv in dependence on the curvilinear components of the velocity:

s 3 2 2 2 2 2 2 2 2 X 2 v = r· r· = Q1 q·1 +Q2 q·2 +Q3 q·3 v = Qi q·i (104) · ⇒ | | i =1 For spherical and cylindrical coordinates we get, by using common coordinate notation:

2 2 2 2 2 2 2 r· · · vsph.c. = +r θ +r sin θ φ (105) 2 2 2 2 2 r· z· · vcyl.c. = + +r φ (106) 5 PROOFS FOR THE PRODUCT RULES 16 ∇ 5 Proofs for the Product Rules ∇ Throughout this section we will make extensive use of the Einstein summation convention (summation over every repeated index), the Kronecker-Delta δi j (1 for equal indices, otherwise: 0) and the Levi-Civita-Symbol/Pseudo-Tensor (sometimes called the permutation symbol) ε (1 for even permiations, 1 for odd permutations, 0 for at least two equal indices). Defini- tions can be found in [5] for example. −  1 (i j k ) = (123),(231),(312), even permutations ¨  1 i = j εi j k = 1 (i j k ) = (321),(213),(132), odd permutations ; δi j = (107)  − 0 i = j 0 (i j k ) = (112),(221),..., at least 2 equal indices 6

€ Š εi j k := ei ej ek for a right-handed coordinate system · ×  3 3 X X a b e  ε a b  e ε a b = i  i j k j k  = i i j k j k × i =1 j ,k =1 3 X (a b)i = εi j k a j bk = εi j k a j bk × j ,k =1 3 X a b = a i bi = a i bi · i =1 3 X a = eia i = eia i i =1 3 X εi j k εk l m = εi j k εk l m = δi l δj m δi m δj l k =1 − εi j k = εk i j = εj k i = εj i k = εi k j = εk j i − − −

(v w) = w v v w  ∇ · × · ∇ × − · ∇ ×

∂  ∂   ∂  (v w) = i εi j k vj wk = εi j k vj wk = εi j k wk vj + εi j k vj wk ∇ · × ∇ ∂ xi ∂ xi ∂ xi  ∂  ∂ = εi j k wk vj εj i k vj wk ∂ xi − ∂ xi ‚ Œ ‚ Œ ! ∂ ∂ = w ( v) v ( w) = wi εi j k vk vi εi j k wk · ∇ × − · ∇ × ∂ x j − ∂ x j  ∂   ∂   ∂   ∂  = εk i j wk vj εj i k vj wk = εi j k wk vj εj i k vj wk ∂ xi − ∂ xi ∂ xi − ∂ xi

Where renaming of indices and even/odd permutations of the epsilon symbol were used (moving an index left or right by one position changes sign, moving an index by two position left or right does not)

g A = g A + g A  ∇ × ∇ × ∇ ×

‚ Œ

∂
∂ g ∂ Ak g A i = εi j k g Ak = εi j k Ak + g ∇ × ∂ x j ∂ x j ∂ x j !

∂ g ∂ = g A i + g A i = εi j k Ak + g εi j k Ak ∇ × ∇ × ∂ x j ∂ x j 5 PROOFS FOR THE PRODUCT RULES 17 ∇

g = g ( )g  ∇ × ∇ × ∇ ∇ · − ∇ · ∇

   


∂ ∂ € Š ∂ ∂ g i = εi j k j g k = εi j k εk l m g m = δi l δj m δi m δj l g m ∇ × ∇ × ∇ ∇ × ∂ x j ∂ xl − ∂ x j ∂ xl ∂  ∂  ∂ ‚ ∂ Œ ∂ ‚ ∂ Œ ∂ ‚ ∂ Œ = g j g i = g j g i ∂ x j ∂ xi − ∂ x j ∂ x j ∂ xi ∂ x j − ∂ x j ∂ x j


= g i ( )g i ∇ ∇ · − ∇ · ∇ Here, Schwarz’ lemma was used in the second last step.

(A B) = A( B) B( A) + (B )A (A )B  ∇ × × ∇ · − ∇ · · ∇ − · ∇

‚ Œ ∂ € Š ∂ Al ∂ Bm ( (A B))i = εi j k εk l m Al Bm = δi l δj m δi m δj l Bm + Al ∇ × × ∂ x j − ∂ x j ∂ x j ∂ Ai ∂ A j ∂ Bm ∂ Bi = Bm Bi + Ai Al ∂ xm − ∂ x j ∂ xm − ∂ xl ! = (A( B) B( A) + (B )A (A )B)i ∇ · − ∇ · · ∇ − · ∇ ∂ Bj ∂ A j ∂ Ai ∂ Bi = Ai Bi + Bj A j ∂ x j − ∂ x j ∂ x j − ∂ x j (A B) = A ( B) + B ( A) + (B )A + (A )B  ∇ · × ∇ × × ∇ × · ∇ · ∇

∂ € Š ∂ Bj ∂ A j ( (A B))i = A j Bj = A j + Bj ∇ · ∂ xi ∂ xi ∂ xi ! = (A ( B) + B ( A) + (B )A + (A )B)i × ∇ × ∂ × ∇ × · ∇ ∂ · ∇ ∂ ∂ = εi j k A j εk l m Bm + εi j k Bj εk l m Am + Bj Ai + A j Bi ∂ xl ∂ xl ∂ x j ∂ x j € Š ∂ Bm € Š ∂ Am ∂ Ai ∂ Bi = δi l δj m δi m δj l A j + Bj δi l δj m δi m δj l A j + Bj + A j − ∂ xl − ∂ xl ∂ x j ∂ x j ∂ Bm ∂ Bi ∂ Am ∂ Ai ∂ Ai ∂ Bi = Am Al + Bm Bl + Bj + A j ∂ xi − ∂ xl ∂ xi − ∂ xl ∂ x j ∂ x j ∂ Bm ∂ Am = Am + Bm ∂ xi ∂ xi
f A = A f + f A  ∇ · · ∇ ∇ ·

∂
∂ f ∂ Ai f A = f Ai = Ai + f ∇ · ∂ xi ∂ xi ∂ xi
f g = f g + g f  ∇ ∇ ∇

∂
∂ g ∂ f f g i = f g = f + g ∇ · ∂ xi ∂ xi ∂ xi


∆ h f = f (∆h) + 2( h) f + h ∆f  ∇ · ∇

3 2 3  
X ∂
X ∂ ∂ h ∂ f ∆ h f = 2 h f = f + h ∂ x ∂ xi ∂ xi ∂ xi i =1 i i =1 3 3 3 3 X ∂ 2h X ∂ f ∂ h  X ∂ h ∂ f  X ∂ 2 f = f 2 + + + h 2 ∂ x ∂ xi ∂ xi ∂ xi ∂ xi ∂ x i =1 i i =1 i =1 i =1 i


= f (∆h) + (∆h) ∆f + (∆h) ∆f + h ∆f · · 6 GAUSS DIVERGENCE THEOREM VARIANTS 18

6 Gauss Divergence Theorem variants

: To round this article, presented here are some standard textbook variants of Gauss’ Divergence Theorem that may be found in many vector calculus textbooks. R R ψdV ψdS V = ∂ V  ∇ In the following two derivations c is taken to be a constant non-zero arbitrary vector. We use the general form of Gauss’ Divergence theorem: Z Z AdV = A dS (108) V ∇ · ∂ V ·
Now use A = cψ in Gauss’ Divergence Theorem and the previous vector identity ψc = c ψ + ψ c: ∇ · · ∇ ∇ · Z Z Z Z

cψ dV = c ψ + ψ c dV = c ψdV = c ψdV V ∇ · V · ∇ ∇ · V · ∇ · V ∇ Z Z = cψ dS = c ψdS ∂ V · · ∂ V This gives upon comparison of the last terms in both lines:

‚Z Z Œ Z Z c ψdV ψdS = 0 ψdV = ψdS · V ∇ − ∂ V ⇔ V ∇ ∂ V R R BdV dS B V = ∂ V  ∇ × × On the other hand, using A = c B in the Divergence Theorem and the vector identity (c B) = B c c B we obtain the following result: × ∇ · × · ∇ × − · ∇ ×

Z Z Z (c B)dV = (B ( c) c ( B))dV = c BdV V ∇ · × V · ∇ × − · ∇ × − · V ∇ × Z Z Z = c B dS = c B dS = c n BdS ∂ V × · · ∂ V × − · ∂ V × In the last step I used cyclic permutations of the mixed product. Comparison of the last elements of both lines, and noting the arbitrariness of the constant vector c we get the form we wanted to proof. R R φ 2ψ φ ψ
dV φ ψ dS V + = ∂ V  ∇ ∇ · ∇ ∇ · To proove this identity (Green’s first identity) simply put A = φ ψ into Gauss’ Divergence Theorem: ∇ Z Z Z
€ 2 Š φ ψ dV = φ ψ + φ ψ dV = φ ψ dS V ∇ · ∇ V ∇ ∇ ∇ ∂ V ∇ · R R φ 2ψ ψ 2φ
dV φ ψ ψ φ
dS V = ∂ V  ∇ − ∇ ∇ − ∇ · To proove Green’s theorem we interchange the role of φ and ψ and substract them from each other.

Z Z € 2 2 Š
φ ψ + φ ψ ψ φ ψ φ dV = φ ψ ψ φ dS V ∇ ∇ · ∇ − ∇ − ∇ · ∇ ∂ V ∇ − ∇ · REFERENCES 19

Summarizing we have found and prooved the following product rules:

f g = f g + g f (109) ∇ ∇ ∇
f A = A f + f A (110) ∇ · · ∇ ∇ ·

(A B) = A ( B) + B ( A) + (B )A + (A )B (111) ∇ · × ∇ × × ∇ × · ∇ · ∇

(A B) = A( B) B( A) + (B )A (A )B (112) ∇ × × ∇ · − ∇ · · ∇ − · ∇

( A) = ( A) ( )A (113) ∇ × ∇ × ∇ ∇ · − ∇ · ∇
g A = g A + g A (114) ∇ × ∇ × ∇ ×

(A B) = B ( A) A ( B) (115) ∇ · × · ∇ × − · ∇ × 1 (B )A = [ (A B) + (A B) A( B) + B( A) A ( B) B ( A)] (116) · ∇ 2 ∇ × × ∇ · − ∇ · ∇ · − × ∇ × − × ∇ ×


∆ h f = f (∆h) + 2( h) f + h ∆f (117) ∇ · ∇

2 A = ( A) ( A) (118) ∇ ∇ ∇ · − ∇ × ∇ ×

References

[1] Paul Ogren, Earlham College, http://www.earlham.edu/~chem/chem441/overviews/4a_laplacian.pdf [2] Eric W. Weisstein. Vector Laplacian. From MathWorld - A Wolfram Web Resource. http://mathworld.wolfram. com/VectorLaplacian.html ∇ [3] Formula for Laplacian given and discussed in H. Margenau and G.M.Murphey: Mathematics of Physics and Chemistry, D. van Nostrand and Co., Princeton, N.J., 1959, page 175

[4] List of vector identities. (2006, April 12). In Wikipedia, The Free Encyclopedia. Retrieved 11:06, May 25, 2006, from http://en.wikipedia.org/w/index.php?title=List-of-vector-identities [5] Tensors and Tensor Algebra, http://grus.berkeley.edu/~jrg/ay202/node185.html [6] Theoretische Physik Band 1, Walter Greiner; pages 93,95,98