ENGI 5432 1.5 1.5 Conversions between Coordinate Systems Page 1-32 Conversions between Coordinate Systems In general, the conversion of a vector F Fx ˆi Fy ˆj Fz kˆ from Cartesian coordinates (x, y, z) to another orthonormal coordinate system (u, v, w) in 3 (where “orthonormal” means that the new basis vectors aˆ u , aˆ v , aˆ w are mutually orthogonal and of unit length) is given by F F ˆi F ˆj F kˆ F aˆ F aˆ F aˆ . x y z u u v v w w However, Fu F aˆ u Fx ˆi Fy ˆj Fz kˆ aˆ u ˆi aˆ u Fx ˆj aˆ u Fy kˆ aˆ u Fz . Fv and Fw are defined similarly in terms of the Cartesian components Fx , Fy , Fz . In matrix form ˆi aˆ u ˆj aˆ u kˆ aˆ u F Fu x F ˆi aˆ ˆ ˆ v j aˆ v k aˆ v Fy . v ˆ ˆ ˆ Fw ˆ ˆ F ˆ i a w j a w k a w z The matrices on the right hand side of the equation will contain a mixture of expressions in the new (u, v, w ) and old (x, y, z ) coordinates. This needs to be converted into a set of expressions in (u, v, w ) only. Example 1.5.1 Express the vector F = y i x j + z k in cylindrical polar coordinates. x y plane: coordinates x y plane: basis vectors x ˆi ˆ y ˆj ˆ z kˆ ˆ ENGI 5432 1.5 Conversions between Coordinate Systems Page 1-33 Example 1.5.1 (continued) ˆi ˆ ˆi kˆ ˆj ˆ ˆj kˆ kˆ ˆ kˆ kˆ The coefficient conversion matrix from Cartesian to cylindrical polar is therefore Letting c cos , s sin : Fpolar ENGI 5432 1.5 Conversions between Coordinate Systems Page 1-34 We can also generate the coordinate transformation matrix from Cartesian coordinates (x, y, z ) to spherical polar coordinates (r, , ). [ is the declination (angle down from the north pole, 0 ) and is the azimuth (angle around the equator 0 2 ).] [Vertical] Plane containing z-axis and radial vector r : [Horizontal] Plane z = r cos : r = x, y, z r sin cos , r sin sin , r cos . ENGI 5432 1.5 Conversions between Coordinate Systems Page 1-35 The conversion matrix from Cartesian to spherical polar coordinates is then ˆi rˆ ˆj rˆ kˆ rˆ sin cos ˆi ˆ ˆj ˆ kˆ ˆ cos cos ˆ ˆ ˆ ˆ ˆ ˆ sin i j k sin sin cos sin cos cos sin 0 Example 1.5.2 Convert F y ˆi x ˆj to spherical polar coordinates. Let c1 cos , s1 sin , c2 cos , s2 sin Fr F F F Expressions for the gradient, divergence, curl and Laplacian operators in any orthonormal coordinate system will follow in section 1.7. ENGI 5432 1.5 Conversions between Coordinate Systems Page 1-36 Summary for Coordinate Conversion: To convert a vector expressed in Cartesian components vx ˆi v y ˆj vz kˆ into the equivalent vector expressed in cylindrical polar coordinates v ˆ v ˆ v kˆ , express the Cartesian components vx, vy, vz in terms of , , z using x = cos , y = sin , z z = z ; then evaluate v cos sin 0 vx v sin cos 0 v y vz 0 0 1 vz Use the inverse matrix to transform back to Cartesian coordinates: vx cos sin 0 v v sin cos 0 v y vz 0 0 1 vz To convert a vector expressed in Cartesian components vx ˆi v y ˆj vz kˆ into the equivalent vector expressed in spherical polar coordinates v rˆ v ˆ v ˆ , express the r Cartesian components vx, vy, vz in terms of r, , using x r sin cos , y r sin sin , z r cos ; then evaluate vr sin cos v cos cos v sin sin sin cos sin cos cos sin 0 vx v y vz Use the inverse matrix to transform back to Cartesian coordinates: vx sin cos cos cos sin vr v sin sin cos sin cos v y vz cos sin 0 v Note that, in both cases, the transformation matrix A is orthogonal, so that A1 = AT. This is generally true for transformations between orthonormal coordinate systems. ENGI 5432 1.6 1.6 Basis Vectors in Other Coordinate Systems Page 1-37 Basis Vectors in Other Coordinate Systems In the Cartesian coordinate system, all three basis vectors are absolute constants: d ˆ d ˆ d ˆ i j k 0 dt dt dt The derivative of a vector is then straightforward to calculate: d df df df f1 ˆi f 2 ˆj f 3 kˆ ˆi 1 ˆj 2 kˆ 3 dt dt dt dt But many non-Cartesian basis vectors are not constant. Cylindrical Polar: ˆ ˆ kˆ kˆ Let v dv dt then ˆ ˆ kˆ Therefore if a vector F is described in cylindrical polar coordinates F F ˆ F ˆ Fz kˆ , then F ENGI 5432 1.6 Basis Vectors in Other Coordinate Systems Page 1-38 In particular, the displacement vector is r t t ˆ 0 ˆ z t kˆ , so that the velocity vector is dr d d ˆ dz ˆ v ˆ k dt dt dt dt Example 1.6.1 Find the velocity and acceleration in cylindrical polar coordinates for a particle travelling along the helix x = 3 cos 2t , y = 3 sin 2t , z = t . Cylindrical polar coordinates: x cos , y 2 x 2 y 2 , tan x y sin , zz ENGI 5432 1.6 Basis Vectors in Other Coordinate Systems Spherical Polar Coordinates Vertical plane containing z-axis and radial vector r : rˆ ˆ Equatorial plane ( 0 ): ˆ d rˆ dt d ˆ dt Page 1-39 ENGI 5432 1.6 Basis Vectors in Other Coordinate Systems Page 1-40 Spherical Polar Coordinates (continued) d ˆ dt In particular, the displacement vector is r r rˆ , so that the velocity vector is dr dr d rˆ dr d ˆ d v rˆ r rˆ r ˆ sin dt dt dt dt dt dt dr d ˆ d ˆ v rˆ r r sin dt dt dt It can be shown that the acceleration vector in the spherical polar coordinate system is d 2r d 2 d 2 2 dv a 2 r sin rˆ dt dt dt dt 2 1 d 2 d r d r sin 2 ˆ r dt dt 2 dt 1 d 2 d 2 ˆ sin r r sin dt dt d 2x ˆ d 2 y ˆ d 2z ˆ i j 2k ! Compare this to the Cartesian equivalent a dt 2 dt 2 dt ENGI 5432 1.6 Basis Vectors in Other Coordinate Systems Page 1-41 Example 1.6.2 Find the velocity vector v for a particle whose displacement vector r, in spherical polar coordinates, is given by r = 4, = t , = 2t , (0 < t < ) . r 4 , t , 2t v dr 0, dt d 1, dt d 2 dt dr d ˆ d ˆ rˆ r r sin dt dt dt Summary: Cylindrical Polar: d d ˆ ˆ dt dt d ˆ d ˆ dt dt d ˆ k 0 dt r ˆ z kˆ v ˆ ˆ z eˆ z Spherical Polar: d rˆ d ˆ d ˆ sin dt dt dt d ˆ d d ˆ rˆ cos dt dt dt d ˆ d sin rˆ cos ˆ dt dt r r rˆ v r rˆ r ˆ r sin ˆ ENGI 5432 1.7 1.7 Gradient Operator in Other Coordinate Systems Page 1-42 Gradient Operator in Other Coordinate Systems For any orthogonal curvilinear coordinate system (u1, u2, u3) in 3, 1 r the unit tangent vectors along the curvilinear axes are eˆ i Ti , hi ui r . ui where the scale factors hi The displacement vector r can then be written as r u1eˆ 1 u 2eˆ 2 u3eˆ 3 , where the unit vectors ê i form an orthonormal basis for 3. 0 i j eˆ i eˆ j ij 1 i j The differential displacement vector dr is (by the Chain Rule) r r r dr du1 du2 du3 h1 du1 eˆ1 h2 du2 eˆ 2 h3 du3 eˆ 3 u1 u2 u3 and the differential arc length ds is 2 2 2 ds 2 dr dr h1 du1 h2 du2 h3 du3 The element of volume dV is x, y, z du1du 2 du3 u1 , u 2 , u3 dV h1h2 h3 du1du 2 du3 Gradient operator Gradient Divergence x u1 x u 2 x u 3 y u1 y u 2 y u 3 z u1 z du1du 2 du 3 u 2 z u 3 eˆ eˆ1 eˆ 2 3 h1 u1 h2 u2 h3 u3 eˆ V eˆ V eˆ V V 1 2 3 h1 u1 h2 u2 h3 u3 F 1 h2 h3 f1 h3 h1 f 2 h1 h2 f3 h1 h2 h3 u1 u2 u3 ENGI 5432 1.7 Gradient Operator in Other Coordinate Systems F Curl Laplacian 2V 1 h1 h2 h3 1 h1 h2 h3 h1 eˆ1 u1 h2 eˆ 2 u2 h3 eˆ 3 u3 h1 f1 h2 f 2 h3 f3 Page 1-43 h2 h3 V h3 h1 V h1 h2 V u1 h1 u1 u2 h2 u2 u3 h3 u3 Cartesian: hx = hy = hz = 1 . Cylindrical polar: h hz 1 , h . Spherical polar: hr 1, h r , h r sin . The familiar expressions then follow for the Cartesian coordinate system. In cylindrical polar coordinates, naming the three basis vectors as ˆ , ˆ, kˆ , we have: r ˆ 0ˆ z kˆ , 0, z The relationship to the Cartesian coordinate system is x cos , y sin , z z 2 x 2 y 2 , tan One scale factor is r h In a similar way, we can confirm that h and hz 1 . y x ENGI 5432 1.7 Gradient Operator in Other Coordinate Systems Page 1-44 In cylindrical polar coordinates, dV ds 2 V F F 2V All of the above are undefined on the z-axis ( = 0), where there is a coordinate singularity. However, by taking the limit as 0 , we may obtain well-defined values for some or all of the above expressions. ENGI 5432 1.7 Gradient Operator in Other Coordinate Systems Page 1-45 Example 1.7.1 Given that the gradient operator in a general curvilinear coordinate system is eˆ eˆ eˆ 1 2 3 , why isn’t the divergence of h2 u2 h3 u3 h1 u1 1 F1 1 F2 1 F3 F F1 eˆ 1 F2 eˆ 2 F3 eˆ 3 equal, in general, to ? h2 u2 h3 u3 h1 u1 The quick answer is that the differential operators operate not just on the components F1 , F2 , F3 , but also on the basis vectors eˆ1 , eˆ 2 , eˆ 3 . In most orthonormal coordinate systems, these basis vectors are not constant. The divergence therefore contains additional terms. e1 e e 2 3 F1e1 F2e 2 F3e3 h u h u h u 2 2 3 3 1 1 e1 e1 F1 F1 e e e F F e e e F F e e1 1 2 1 1 1 e2 1 3 1 1 1 e3 1 u1 h2 u2 h2 u2 h3 u3 h3 u3 h1 u1 h1 e1 e2 F2 F2 e e e F F e e e F F e e1 2 2 2 2 2 e2 2 3 2 2 2 e3 2 u1 h2 u2 h2 u2 h3 u3 h3 u3 h1 u1 h1 e1 e3 F3 F3 e e e F F e e e F F e e1 3 2 3 3 3 e2 3 3 3 3 3 e3 3 = u1 h2 u2 h2 u2 h3 u3 h3 u3 h1 u1 h1 1 F1 F1 e F e F e e1 1 1 e2 1 1 e3 1 u1 h2 u2 h3 u3 h1 u1 h1 F2 e 1 F2 F2 e F e e1 2 e2 2 2 e3 2 u1 h2 u2 h2 u2 h3 u3 h1 F3 e F e 1 F3 F3 e e1 3 3 e2 3 e3 3 u1 h2 u2 h3 u3 h3 u3 h1 For Cartesian coordinates, all derivatives of any basis vector are zero, which leaves the familiar Cartesian expression for the divergence. But for most non-Cartesian coordinate systems, at least some of these partial derivatives are not zero. More complicated expressions for the divergence therefore arise. ENGI 5432 1.7 Gradient Operator in Other Coordinate Systems Page 1-46 Example 1.7.1 (continued) For cylindrical polar coordinates, we have 1 F F ˆ F ˆ ˆ F ˆ ˆ ˆ k 1 1 z 1 F ˆ 1 F F ˆ ˆ F ˆ ˆ ˆ k 1 z 1 Fz kˆ Fz ˆ kˆ 1 Fz Fz ˆ kˆ ˆ k 1 z 1 z 1 But none of the basis vectors varies with or z and the basis vector k̂ is absolutely constant. Therefore the divergence becomes 1 F F ˆ 0 ˆ 0 1 1 F F ˆ ˆ 1 Fz 0 0 0 0 0 1 z F F F ˆ ˆ ˆ ˆ But ˆ ˆ and ˆ ˆ F F ˆ ˆ ˆ ˆ 0 So we recover the cylindrical polar form for the divergence, div F F F 1 F Fz z ENGI 5432 1.7 Gradient Operator in Other Coordinate Systems Page 1-47 In spherical polar coordinates, naming the three basis vectors as rˆ , ˆ, ˆ, we have: r r rˆ 0ˆ 0 ˆ r , 0, 0 The relationship to the Cartesian coordinate system is x r sin cos , y r sin sin , z r cos . One of the scale factors is r h In a similar way, we can confirm that hr 1 and h r sin . dV ds 2 V F ENGI 5432 1.7 Gradient Operator in Other Coordinate Systems Page 1-48 F 2V All of the above are undefined on the z-axis (sin = 0), where there is a coordinate singularity. However, by taking the limit as sin 0 , we may obtain well-defined values for some or all of the above expressions. ENGI 5432 1.7 Gradient Operator in Other Coordinate Systems Page 1-49 Example 1.7.2 A vector field has the equation, in cylindrical polar coordinates (, , z), k k F n eˆ n ˆ Find the divergence of F and the value of n for which the divergence vanishes for all 0. ENGI 5432 1.7 Gradient Operator in Other Coordinate Systems Example 1.7.3 In spherical polar coordinates, F r , , f cot rˆ 2 f ˆ g r , ˆ , where f () is any differentiable function of only and g (r, ) is any differentiable function of r and only. Find the divergence of F. Page 1-50 ENGI 5432 1.7 Gradient Operator in Other Coordinate Systems Page 1-51 Example 1.7.4 Find curl sin ˆ ˆ , where , are the two angular coordinates in the standard spherical polar coordinate system. ENGI 5432 1.7 Gradient Operator in Other Coordinate Systems Page 1-52 Central Force Law If a potential function V(x, y, z), (due solely to a point source at the origin) depends only on the distance r from the origin, then the functional form of the potential can be deduced. Using spherical polar coordinates: V (r, , ) = f (r) 1 d df d2 f 2 df 2V 2 r 2 2 r dr dr dr r dr But, in any regions not containing any sources of the vector field, the divergence of the vector field F V (and therefore the Laplacian of the associated potential function V) must be zero. Therefore, for all r 0, ENGI 5432 1.7 Gradient Operator in Other Coordinate Systems Page 1-53 Gravity is an example of a central force law, for which the potential function must be of B the form V r , , A . The zero point for the potential is usually set at infinity: r B lim V lim A A 0 r r r The force per unit mass due to gravity from a point mass M at the origin is F V GM rˆ r2 But, in spherical polar coordinates, V rˆ V ˆ V ˆ V dV B rˆ rˆ 2 r r r sin dr r GM B 2 2 r r B GM Therefore the gravitational potential function is V r GM r The electrostatic potential function is similar, with a different constant of proportionality. END OF CHAPTER 1