ENGI 940 5.0 - Gradient, Divergence, Curl Page 5.0 5. e Gradient Operator A brief review is provided ere for te gradient operator in bot Cartesian and ortogonal non-cartesian coordinate systems. Sections in tis Capter: 5.0 Gradient, Divergence, Curl and Laplacian (Cartesian) 5.0 Differentiation in Ortogonal Curvilinear Coordinate Systems 5.0 Summary able for te Gradient Operator 5.04 Derivatives of Basis Vectors 5.0 Gradient, Divergence, Curl and Laplacian (Cartesian) Let z be a function of two independent variables (x, y), so tat z = f (x, y). e function z = f (x, y) defines a surface in. At any point (x, y) in te x-y plane, te direction in wic one must travel in order to experience te greatest possible rate of increase in z at tat point is te direction of te gradient vector, f ˆ f f i ˆj x y e magnitude of te gradient vector is tat greatest possible rate of increase in z at tat point. e gradient vector is not constant everywere, unless te surface is a plane. (e symbol is usually pronounced del ). e concept of te gradient vector can be extended to functions of any number of variables. If u = f (x, y, z, t), ten f f f f f x y z t. If v is a function of position r and time t, wile position is in turn a function of time, ten by te cain rule of differentiation, d v v dx v dy v dz v dr v v dt x dt y dt z dt t dt t d v v v v dt t wic is of use in te study of fluid dynamics.
ENGI 940 5.0 - Gradient, Divergence, Curl Page 5.0 e gradient operator can also be applied to vectors via te scalar (dot) and vector (cross) products: e divergence of a vector field F(x, y, z) is div F F F F F F F F x y z x y z A region free of sources and sinks will ave zero divergence: te total flux into any region is balanced by te total flux out from tat region. e curl of a vector field F(x, y, z) is ˆi curl F F ˆj F kˆ F F F x y z F F y z x F F F z x y In an irrotational field, curl Wenever or Proof: F 0. F for some twice differentiable potential function, curl F 0 curl grad F F F F x y z 0 ˆ i 0 x x y z z y curl ˆ j 0 y y z x x z ˆ k 0 z z x y y x
ENGI 940 5.0 - Gradient, Divergence, Curl Page 5.0 Among many identities involving te gradient operator is div curl F F 0 for all twice-differentiable vector functions F Proof: div curl F F F F F F F x y z y z x z x y F F F F F F 0 x y x z y z y x z x z y e divergence of te gradient of a scalar function is te Laplacian: div grad f f f f f f x y z for all twice-differentiable scalar functions f. In ortogonal non-cartesian coordinate systems, te expressions for te gradient operator are not as simple.
ENGI 940 5.0 - Curvilinear Gradient Page 5.04 5.0 Differentiation in Ortogonal Curvilinear Coordinate Systems For any ortogonal curvilinear coordinate system (u, u, u ) in te unit tangent vectors along te curvilinear axes are ˆ r e, u were te scale factors i r u i., ˆi i i i e displacement vector r can ten be written as r ueˆ ˆ ˆ ue ue, were te unit vectors ê i form an ortonormal basis for. 0 i j eˆi e ˆ j i j i j e differential displacement vector dr is (by te Cain Rule) r r r dr du du du du eˆ du eˆ du e ˆ u u u and te differential arc lengt ds is given by ds dr dr du du du e element of volume dv is x, y, z dv du du du du du du u, u, u Jacobian x y u u x y u u x y u u z u z u z u Example 5.0.: Find te scale factor θ for te sperical polar coordinate system x, y, z r sin cos, r sin sin, r cos : r x y z r cos cos r cos sin r sin r r r r cos cos cos sin sin r cos cos sin sin cos sin du du du r r r
ENGI 940 5.0 - Gradient Summary Page 5.05 5.0 Summary able for te Gradient Operator eˆ eˆ ˆ e Gradient operator u u u eˆ V eˆ ˆ V e V Gradient V u u u Divergence F F F F u u u Curl F eˆ F u eˆ F u eˆ F u Laplacian V = u V u u V u u V u Scale factors: Cartesian: x = y = z =. Cylindrical polar: = z =, =. Sperical polar: r =, = r, = r sin. Example 5.0.: e Laplacian of V in sperical polars is sin V V r sin V V r sin r r sin or V V V cot V V V r r r r r sin
ENGI 940 5.0 - Gradient Summary Page 5.06 Example 5.0. A potential function V r is sperically symmetric, (tat is, its value depends only on te distance r from te origin), due solely to a point source at te origin. ere are no oter sources or sinks anywere in V r.. Deduce te functional form of V r is sperically symmetric V r,, f r In any regions not containing any sources of te vector field, te divergence of te vector field F V (and terefore te Laplacian of te associated potential function V) must be zero. erefore, for all r 0, div F V V 0 But sin V r sin V r sin r r sin d V dv sin r 0 0 0 r sin dr dr d dv dv dv r 0 r B B r dr dr dr dr V Br A, were A, B are arbitrary constants of integration. erefore te potential function must be of te form B V r,, A r is is te standard form of te potential function associated wit a force tat obeys te inverse square law F. r
ENGI 940 5.04 - Basis Vectors Page 5.07 5.04 Derivatives of Basis Vectors d Cartesian: ˆ d ˆ d i j kˆ 0 dt dt dt r x ˆi y ˆj z kˆ v x ˆi y ˆj z kˆ Cylindrical Polar Coordinates: x cos, y sin, z z d d ˆ ˆ r ˆ z kˆ dt dt d ˆ d ˆ v ˆ ˆ z kˆ dt dt d kˆ 0 [radial and transverse components of v ] dt Sperical Polar Coordinates. e declination angle θ is te angle between te positive z axis and te radius vector r. 0 < θ < π. e azimut angle is te angle on te x-y plane, measured anticlockwise from te positive x axis, of te sadow of te radius vector. 0 < < π. z = r cos θ. e sadow of te radius vector on te x-y plane as lengt r sin θ. It ten follows tat x = r sin θ cos and y = r sin θ sin. d d ˆ ˆ d r sin ˆ dt dt dt r r r ˆ d ˆ d d rˆ cos ˆ dt dt dt v r rˆ r ˆ r sin ˆ d ˆ d sin rˆ cos ˆ dt dt
ENGI 940 5.04 - Basis Vectors Page 5.08 Example 5.04. Find te velocity and acceleration in cylindrical polar coordinates for a particle travelling along te elix x = cos t, y = sin t, z = t. Cylindrical polar coordinates: x cos, y sin, z z y x y, tan x 9cos t 9sin t 9 0 sin t tan tan t t cost r ˆ z kˆ z t z dr v ˆ ˆ zkˆ 0 ˆ ˆ kˆ 6 ˆ kˆ dt [e velocity as no radial component te elix remains te same distance from te z axis at all times.] d v a 6 ˆ kˆ 6 ˆ 0 ˆ dt [e acceleration vector points directly at te z axis at all times.] Oter examples are in te problem sets. END OF CHAPER 5