Lecture 2 : Curvilinear Coordinates

Lecture 2 : Curvilinear Coordinates Fu-Jiun Jiang October, 200 I. INTRODUCTION A. Definition and Notations In 3-dimension Euclidean space, a vector V can be written as V = e x V x + e y V y + e z V z with e x = (, 0, 0), e y = (0,, 0), e z = (0, 0, ). We call these e x,y,z a basis for the 3-dim Euclidean space. Notice we have e i e i = e i e j = 0 for i j e x ( e y e z ) = > 0. () We call such a basis an orthonormal basis and x, y, z the orthonormal coordinates of the 3-dim Euclidean space. Notice if the first condition is eq. is removed, then such a basis is called a orthogonal basis. Let s assume we have another set of orthogonal coordinates q (x, y, z), q 2 (x, y, z), q 3 (x, y, z) which are functions of the standard coordinates x, y, z. We can also consider the standard coordinates as functions of q, q 2, q 3. Let r(x, y, z) be the position vector with respect to the standard basis e x, e y, e z in 3-dim Euclidean space, namely r(x, y, z) = (x, y, z). Notice if one holds the second and third component fixed and takes the derivative of r with respect to the coordinate x, one reaches Similarly, one has (x, y, z) y=y0,z=z x 0 = (, 0, 0) = e x. (2) (x, y, z) x=x0,z=z y 0 = (0,, 0) = e y, (x, y, z) x=x0,y=y z 0 = (0, 0, ) = e z. (3)

In other word, x, y, z are the standard orthonormal basis of 3-dim Euclidean space. Applying similar arguments to other sets of orthogonal coordinates q i, one arrives at similar conclusions, namely / e qi, i =, 2, 3 (4) q i q i is the orthonormal basis corresponding to the orthogonal coordinates q i. Notice if {q, q 2, q 3 } is not an orthogonal coordinate system, then e qi defined above are still a basis (corresponding to {q, q 2, q 3 }), but they are not an orthogonal basis. Now from eq. 4, one immediately has q i = h i e qi, (5) here h i is called scale factor for each i, 2, 3. To simplify our presentation, we will use e i instead of e qi if there is no any confusion. In terms of the curvilinear coordinates q i, the position vector is denoted by r. Further, the differential distant vector d r takes the form d r = i=,2,3 q i dq i = e h dq + e 2 h 2 dq 2 + e 3 h 3 dq 3. (6) Next, since the idea of inner product between 2 vectors should be independent of the chosen basis, one should have d r d r = d r d r. (7) In other word, we arrive at dx 2 + dy 2 + dz 2 = h 2 dq 2 + h 2 2dq 2 2 + h 2 3dq 2 3. (8) As a result, intutively one can just replace dx, dy and dz by h dq, h 2 dq 2 and h 3 dq 3, respectively in a expression such as a line integral. Remember for 2 different vectors A, B, the oriented area exapnded by A, B is given by A B. Further, for 3 vectors A, B, C not lying on the same 2-dim plane, the vlolume determined by A, B, C is given by C ( A B). Applying these results to {q, q 2, q 3 } (i.e. ( q, q 2, q 3 )), one sees that the unit element of 2

area and volume are given by d σ = d r q,q 3 fixed d r q,q 2 fixed + d r q,q 2 fixed d r q2,q 3 fixed + d r q,q 3 fixed d r q,q 2 fixed = h 2 h 3 dq 2 dq 3 e + h 3 h dq 3 dq e 2 + h h 2 dq dq 2 e 3, dτ = h h 2 h 3 dq dq 2 dq 3, (9) respectively. As a result, the line, surface and volume integrals now become V d r = V i h i dq i i V d σ = V h 2 h 3 dq 2 dq 3 + V 2 h 3 h dq 3 dq + V 3 h h 2 dq dq 2 φdτ = φh h 2 h 3 dq dq 2 dq 3. (0) Example : Let x = 2 (u2 u 2 2), y = u u 2, z = u 3. Then one has u = (u, u 2, 0), u 2 = ( u 2, u, 0), u 3 = (0, 0, ). () Hence {u, u 2, u 3 } is an orthogonal coordinate system. Example : Let x = 3u + u 2 u 3, y = u + 2u 2 + 2u 3, z = 2u u 2 u 3. Then one finds that u = (3,, 2), u 2 = (, 2, ), Hence {u, u 2, u 3 } is not an orthogonal coordinate system. u 3 = (, 2, ). (2) II. DIFFERENTIAL VECTOR OPERATORS AGAIN In general for an orthogonal coordinate system q, q 2, q 3, one has similar definitions for gradient, divengence and curl operators as those of we have learned earlier in our vector analysis section. 3

A. Gradient For a given scalar function Φ(q, q 2, q 3 ) in a orthogonal coordinates q i, let the gradient of Φ be Φ = f e + f 2 e 2 + f 3 e 3, (3) here e, e 2, e 3 is the corresponding orthonormal basis for {q, q 2, q 3 } and f i are some unknown functions of {q i } Now using eq. 6, we have dφ = Φ d r = f h dq + f 2 h 2 dq 2 + f 3 h 3 dq 3. (4) On ther other hand, one can show dφ = Φ q dq + Φ q 2 dq 2 + Φ q 3 dq 3. (5) From eqs. 4 and 5, one immediately finds f = h Φ q, f 2 = h 2 Φ q 2, f 3 = h 3 Φ q 3, (6) namely we have Φ = h Φ q e + h 2 Φ q 2 e 2 + h 3 Φ q 3 e 3. (7) Since eq. 7 holds for any scalar function Φ, one arrives at = h q e + h 2 Notice if we choose Φ = q i for i {, 2, 3}, then q 2 e 2 + h 3 q 3 e 3. (8) From eq. 9, one further finds q = e h, q 2 = e 2 h 2, q 3 = e 3 h 3 (9) q 2 q 3 = e 2 e 3 h 2 h 3 = e h 2 h 3, (20) 4

hence e = h 2 h 3 ( q 2 q 3 ). (2) Similarly, one has e 2 = h 3 h ( q 3 q ), e 3 = h h 2 ( q q 2 ). (22) B. Divergence For Divengence operator in curilinear coordinate, it is most convenient to obtain the relevant expression through its integration definition V (q, q 2, q 3 ) = V S lim d σ. (23) dτ 0 dτ Consider the difference of area integrasl for two faces q = constant (keeping q 2, q 3 fixed) [ V h 2 h 3 + q (V h 2 h 3 )dq ] dq 2 dq 3 V h 2 h 3 dq 2 dq 3 = q (V h 2 h 3 )dq dq 2 dq 3. (24) By adding other 2 similar formulae (surfaces for constant q 2 and q 3 ), we arrive at V (q, q 2, q 3 ) d σ = (V h 2 h 3 ) + (V 2 h 3 h ) + ] (V 3 h h 2 ) dq dq 2 dq 3. (25) q q 2 q 3 Dividing by dτ = h h 2 h 3 dq dq 2 dq 3, we finally obtain V (q, q 2, q 3 ) = (V h 2 h 3 ) + (V 2 h 3 h ) + ] (V 3 h h 2 ). (26) h h 2 h 3 q q 2 q 3 Similarly, the Laplacian in curvilinear coordinate q i is given by φ(q, q 2, q 3 ) = = ( φ e + φ e 2 + φ e 3 ) h q h 2 q 2 h 3 q 3 ( h 2h 3 φ ) + ( h 3h φ ) + ( h h 2 φ ] ) q q 2 q 3 q h q q 2 h 2 q 2 q 3 h 3 q 3 (27) 5

FIG. : rectangular parallelepiped. C. Curl For V, consider a differential surface element in the curvilinear surface q = constant. The from V d σ = e ( V )h 2 h 3 dq 2 dq 3 (28) S and Stokes theorem, we arrive at e ( V )h 2 h 3 dq 2 dq 3 = V d r (29) where the line integral (over c) is along the boundary of the constant surface q = constant. c 6

FIG. 2: Curl in orthogonal coordinates. Notice by the same consideration as we did when deriving the Stoke s theorem (considering the line integrals over the boundaries of S), one can show Hence we finally arrive at c V (q, q 2, q 3 ) d r = V 2 h 2 dq 2 + V 3 h 3 dq 3 = [ V 3 h 3 + ] (V 3 h 3 )dq 2 dq 3 q 2 [ V 2 h 2 + ] (V 2 h 2 )dq 3 dq 2 q 3 ] q 2 (h 3 V 3 ) q 3 (h 2 V 2 ) dq 2 dq 3. (30) V = e (h 3 V 3 ) ] (h 2 V 2 ) h 2 h 3 q 2 q 3 + e 2 (h V ) ] (h 3 V 3 ) h 3 h q 3 q + e 3 (h 2 V 2 ) ] (h V ) h h 2 q q 2 h e h 2 e 2 h 3 e 3 = h h 2 h 3 q q 2 q 3 h V h 2 V 2 h 3 V 3 (3) 7

FIG. 3: Spherical polar coordinate. Source : http://en.citizendium.org III. SPHERICAL POLAR COORDINATE (r, θ, ϕ) The relation between cartesian coordinates (x, y, z) and spherical polar coordinates (r, θ, ϕ) is x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ, (32) or r = (x 2 + y 2 + x 2 ) /2, z θ = arc cos (x 2 + y 2 + z 2 ), /2 ϕ = arc tan y x. (33) The range for spherical polar coordinates are 0 r <, 0 θ π, 0 ϕ 2π. (34) 8

Now using h j = q j, one sees h = h r =, h 2 = h θ = r, h 3 = h ϕ = r sin θ. (35) This gives line and surface (for r = constant) elements d r = e r dr + e θ rdθ + e ϕ r sin θdϕ, d σ θϕ = e r r 2 sin θdθdϕ. (36) as well as the volume element dτ = r 2 dr sin θdθdϕ. One can also express e r, e θ and e ϕ in terms of e x, e y, e z e r = e x sin θ cos ϕ + e y sin θ sin ϕ + e z cos θ, e θ = e x cos θ cos ϕ + e y cos θ sin ϕ e z sin θ, e ϕ = e x sin ϕ + e y cos ϕ. (37) With above expressions of e r, e θ and e ϕ in terms of e x, e y, e z, one can easily verify that e r, e θ and e ϕ indeed form a orthogonal basis. FIG. 4: Orthonormal basis in spherical polar coordinate. Source : http://en.citizendium.org 9

FIG. 5: Spherical polar coordinate. Source : http://www.heliotekinc.com/photometry.htm In spherical polar coordinate,,, and 2 take the form φ = φ r e r + φ r θ e θ + φ r sin θ ϕ e ϕ, V = r 2 sin θ r (V rr 2 sin θ) + θ (V θr sin θ) + ] ϕ (V ϕr), V = = e r r 2 sin θ θ (r sin θv ϕ) ] ϕ (rv θ) + e θ r sin θ ϕ (V r) ] r (r sin θv ϕ) + e ϕ r φ = r 2 sin θ r (r2 sin θ φ r ) + φ (sin θ θ θ ) + ϕ ( sin θ r (rv θ) θ (V r)], φ ] ϕ ). (38) Example : calculated before. Using spherical polar coordinate, we can rederive some formulae we had f(r) = e r df(r) dr, rn = e r nr n, e r f(r) = 2 df f(r) + r dr, e rr n = (n + 2)r n, 2 f(r) = 2 df r dr + d2 f dr, 2 2 r n = n(n + )r n 2, e r f(r) = 0. (39) 0

Example : In term of { e r, e θ, e ϕ }, one has e x = e r sin θ cos ϕ + e θ cos θ cos ϕ e ϕ sin ϕ, e y = e r sin θ sin ϕ + e θ cos θ sin ϕ + e ϕ cos ϕ, e z = e r cos θ e θ sin θ. (40) Example : In term of {,, }, one can show r θ ϕ x = sin θ cos ϕ r + cos θ cos ϕ r θ sin ϕ r sin θ ϕ, y = sin θ sin ϕ r + cos θ sin ϕ r θ + cos ϕ r sin θ ϕ, z = cos θ r sin θ r ϕ. (4) Example : Using above results of expressing {,, } in term of {,, }, one x y z r θ ϕ arrives at ( i x y y ) = i x ϕ. (42) This is the quantum mechanical operator which corresponds to the z-component of orbital angular momentum. IV. CIRCULAR CYLINDRICAL COORDINATES ρ, ϕ, z

FIG. 6: Circular Cylindrical coordinate. Source : http://www.vias.org/ The relation between cartesian coordinate x, y, z and circular cylindrical coordinates is given by ρ = (x 2 + y 2 ) /2, [ y ] ϕ = tan, x z = z (43) with the limits on ρ, ϕ, z being given by 0 ρ <, 0 ϕ 2π, < z <. (44) Or the othter way around x = ρ cos ϕ, y = ρ sin ϕ, z = z. (45) As a result, one sees h = h ρ =, h 2 = h ϕ = ρ, h 3 = h z =. (46) 2

Also the line surface (for ρ = constant) and volume elements are given by d r = e ρ dρ + e ϕ dϕ + e z dz, d σ ϕz = ρdϕdz, dτ = ρdρdϕdz. (47) Similarly, the circular cylindrical basis e ρ, e ϕ, e z can be expressed in terms of e x, e y, e z as follows e ρ = e x cos ϕ + e y sin ϕ, e ϕ = e x sin ϕ + e y cos ϕ, e z = e z. (48) with which one can easily verify that e ρ, e ϕ, e z form a orthorgonal basis. Similarly, in circularl cylindrical coordinate,,, and 2 take the form φ = φ ρ e ρ + φ ρ ϕ e ϕ + φ z e z, V = ρ ρ (V ρρ) + ϕ (V ϕ) + ] z (V zρ), V = = e ρ ρ ϕ (V z) ] z (ρv ϕ) + e ϕ z (V ρ) ] ρ (V z) + e z ρ ρ (ρv ϕ) ] ϕ (V ρ), φ = ρ ρ (ρ φ ρ ) + ϕ ( φ ρ ϕ ) + ] z (ρ φ z ). (49) Example : Let A = z e x 2x e y + y e z. Then from e ρ = e x cos ϕ + e y sin ϕ, e ϕ = e x sin ϕ + e y cos ϕ, e z = e z, (50) 3

we arrive at e x = cos ϕ e ρ sin ϕ e ϕ, e y = sin ϕ e ρ + cos ϕ e ϕ, e z = e z. (5) As a result, one finds A = (z cos ϕ 2ρ cos ϕ sin ϕ) e ρ (z sin ϕ + 2ρ cos 2 ϕ) e ϕ + ρ sin ϕ e z. (52) Example : From e ρ = cos ϕ e x + sin ϕ e y and e ϕ = sin ϕ e x + cos ϕ e y, one reaches d dt e ρ = ϕ e ϕ, d dt e ϕ = ϕ e ρ. (53) Example : A rigid body is rotating about the z-axis with a constant angular velocity ω. Then one sees that v = ω r = ωρ e ϕ. Also one finds v = 2 ω. Example : The magnetic vector potential for a conducting wire along the z-axis which carries a current I is given by Then the corresponding magnetic field B is given by µi A = e z 2π ln( ). (54) ρ B = e ϕ µi 2πρ (55) 4