Gradient, Divergence, Laplacian, and Curl in Non-Euclidean Coordinate Systems Math 225 supplement to Colley’s text, Section 3.4 Many problems are more easily stated and solved using a coordinate system other than rectangular coordinates, for example polar coordinates. It is convenient to have formulas for gradients and Laplacians of functions and divergence and curls of vector fields in terms of other coordinate systems. Coordinate Vector Fields To do this we need analogs of i and j that are adapted to other coordinate systems. These are called coordinate vector fields. What are the important properties of i and j relative to the (x, y) coordinate system? A) They are orthogonal. B) They are unit vectors. C) i points in the direction in which x increases and y is constant, and j points in the direction in which y increases and x is constant. D) i and j represent differentiation of a function by x and y. What this means is that if f is a differentiable function, then ∂f ∂f D f D f. ∂x = i and ∂y = j The function f plays a minor role in these equations. It can be replaced by any differentiable quantity that depends on x and y. To deemphasize its role, we can write the equations as ∂ ∂ D D , ∂x = i and ∂y = j which are equations about differential operators. To indicate that i and j are related to the (x, y) coordinate system, we could write them as vx and vy instead, or as ux and uy to remind ourselves that they are unit vectors. As a hint of things to come, if we think of X =(x, y) as a moving point depending on x and y,we have ∂X ∂X . i = vx = ∂x and j = vy = ∂y (1) 1 Coordinate Vector Fields for Polar Coordinates. Nowsupposewehaveapolarcoor- 2 dinate system (r, θ) in the plane E . What we need are vectors vr and vθ that are related to polar coordinates in the same way that i and j are related to Cartesian coordinates. These will be the coordinate vector fields for polar coordinates. Unfortunately, for an arbitrary coordinate system it is impossible to satisfy all four of the properties above. (It is a fairly deep theorem in differential geometry that the only coordinate systems with coordinate vec- tor fields that satisfy all four properties are Euclidean coordinates.) It turns out that the important ones are C) and D), since these specify how the vectors are related to computing the partial derivatives of a function. When we use polar coordinates, the position X is a function of r and θ,thatis,X = X(r, θ). Taking our cue from (1), we define ∂X ∂X vr vθ . = ∂r and = ∂θ (2) We can verify that these satisfy C) and D) for polar coordinates without even needing their formulas in terms of i and j. To see that they satisfy C), note that the curve θ → X(r, θ) is, by definition, the path taken when θ varies and r is fixed, namely a circle centered at the origin, oriented counter- ∂X θ r clockwise. Thus, its velocity vθ = ∂θ points in the direction in which increases and is ∂X r θ fixed. Similarly, vr = ∂r points in the direction in which increases and is fixed. To verify D) we need the function-curve version of the chain rule, which we saw earlier in the course: if f is differentiable function on En and X(t) is the position of a differentiable curve in En,then d f X t ∇f · X t . dt ( ) = X(t) ( ) X t d f X If it is implicitly understood that is a function of , we might write this as dt ( )= ∇f · dX df ∇f · dX f X dt ,orevenas dt = dt , although the latter form is an abuse of notation since is not really a function of t. Now suppose f : E2 → R is differentiable. Then f is a function of r and θ (a formula for f may or may not be written in terms of r and θ, but we can still talk about f as a function of r and θ since they determine position). If we hold θ fixed, let r vary, and use the chain rule formula, we get ∂f ∂ ∂X f X r, θ ∇fX(r,θ) · r, θ ∇f · vr Dv f. ∂r = ∂r ( ) = ∂r ( )= = r r θ ∂f ∇f · ∂X ∇f · D f Similarly, if we hold fixed and let vary, we get ∂θ = ∂θ = vθ = vθ .The equations ∂f ∂f = Dv f and = Dv f ∂r r ∂θ θ say that vr and vθ represent differentiation of a function by r and θ, which verifies D). To deemphasize the role of f, we can write these as equations of differential operators: ∂ ∂ = Dv and = Dv . ∂r r ∂θ θ 2 To check A) and B) we need formulas for vr and vθ in terms of the related Euclidean coordinate system given by (x, y)=(r cos θ, r sin θ). We have ∂X ∂X θ θ −r θ r θ . vr = ∂r =cos i +sin j and vθ = ∂θ = sin i + cos j From these it is easily verified that vr and vθ are orthogonal, so they satisfy A). However, only one of them is a unit vector, namely vr, so they do not satisfy B). We see that vr is a unit vector pointing directly away from the origin. We have vθ = r. More specifically, we have that vθ = r(vr)⊥. Since unit vectors are useful, we define vθ vθ ur = vr =cosθ i +sinθ j and uθ = = = − sin θ i +cosθ j, vθ r and note that uθ =(ur)⊥. Colley neglected to give these formulas in the section on polar coordinates (pg. 62), but they are the same as two of the three coordinate vector fields for cylindrical coordinates on page 71. You should verify the coordinate vector field formulas for spherical coordinates on page 72. For any differentiable function f we have ∂f 1 1 ∂f Du f = Dv f = and Du f = Dv f = . (3) r r ∂r θ r θ r ∂θ The second formula follows by applying the property Davf = ∇f · (av)=a∇f · v = aDvf to a 1 the case when v = r vθ. This illustrates an important feature of non-Euclidean coordinate systems that is often counterintuitive: The directional derivative of a function in the direction of a unit coordinate vector need not be equal to the partial derivative of the function with respect to the corresponding coordinate. ∂f D f For polar coordinates we have ∂θ = uθ . Leaving f out of the formulas in (3) gives us equations of differential operators: ∂ 1 1 ∂ Du = Dv = and Du = Dv = . (4) r r ∂r θ r θ r ∂θ Coordinate Vector Fields in Non-orthogonal Coordinates (Optional). If (r, s)are coordinates on E2, then position is a function of (r, s), that is, X = X(r, s). The same reasoning as above implies that the coordinate vector fields for this coordinate system are ∂X ∂X vr = ∂r and vs = ∂s . It is very important to note that these are not necessarily unit vectors, and they are not necessarily orthogonal. In the nicest coordinate systems they are orthogonal, and that is enough to make calculations with them relatively simple. Coordinate vector fields in higher dimensions are similar. 3 Example 1. Consider E2 with a Euclidean coordinate system (x, y). On the half of E2 on which x>0 we define coordinates (r, s) as follows. Given point X with Cartesian coordinates (x, y)withx>0, let r = x and s = y/x. Thus the new coordinates of X are its usual x coordinate and the slope of the line joining X and the origin. Solving for x and y we have x = r and y = rs. The formula for X in terms of (r, s), which you can think of as a map from the rs-plane to the xy-plane, is X =(x, y)=X(r, s)=(r, rs). The coordinate vector fields are then ∂X ∂X s r . vr = ∂r = i + j and vs = ∂s = j Notice that even though r = x, the coordinate vector field vr is not i. This is because the coordinate vector fields for a coordinate system depend on all of the coordinates, not just the particular coordinate. The vector vr points in the direction in which s is constant, so it points along the lines through the origin. It points in the direction r increases, and its length reflects how fast you have to go in order for r to increase by one in one unit of time. Note also that vr and vs are not orthogonal. Gradients in non-Euclidean Coordinate Systems If f:E2 → R is differentiable and we express f in Euclidean coordinates (x, y), then the f ∇f ∂f ∂f gradient of is given by = ∂xi + ∂yj. The generalization of this is the following. At any 2 point in E ,letu1 and u2 be orthogonal unit vectors (an orthonormal basis). Then ∇f =(Du1 f)u1 +(Du2 f)u2. (5) If you like to think of ∇ as a operator involving vectors and differential operators, then ∂ ∂ ∇ = u1Du + u2Du , which generalizes ∇ = i + j . 1 2 ∂x ∂y These formulas are actually coordinate-free in the sense that the vectors u1 and u2 are not tied to any particular coordinate system. They can be used to compute the gradient of a function in any coordinate system. Formula (5) is particularly easy to use in orthogonal coordinate systems, that is, coordinate systems in which the coordinate vector fields are orthogonal (which happens for polar, cylindrical, and spherical coordinates).