+ f f n x n. + (x)

Transcription

1 Math Vector Calculus II Notes 14.5 Divergence, (Grad) and Curl For a vector field in R n, that is F = f 1, f 2,..., f n, where f i is a function of x 1, x 2,..., x n, the divergence is div(f) = f 1 x 1 + f 2 x f n x n. Of course we have already seen this definition in R 2. Just as is the case in R 2, we call a vector field F source-free on a region R if the divergence of F is zero on that region. The del operator is =,,..., x 1 x 2 x n. When we put the del operator next to a scalar function φ, as in φ, it denotes the gradient of φ. We can use the del operator to denote divergence: div(f) = F. Calculate the divergence of each vector field in R 3 : (a) F = y, x, z. Here the divergence is F = ( y) x + (x) y + (z) z = = 1. Note: The rotational part (first two components) of this helical spiral field is source-free (that is, divergence of y, x is 0). (b) F = x 2 y 2, y 2 z 2, z 2 x 2. Here the divergence is F = (x2 y 2 ) x + (y2 z 2 ) + (z2 x 2 ) y z = 2x+2y +2z.

2 (c) The radial vector field F = The divergence is F = ( ) x x (x 2 + y 2 + z 2 ) p/2 r r = 1 x, y, z. p (x 2 + y 2 + z 2 ) p/2 + ( ) y + ( y (x 2 + y 2 + z 2 ) p/2 z z (x 2 + y 2 + z 2 ) p/2 = (x2 + y 2 + z 2 ) p/2 px 2 (x 2 + y 2 + z 2 ) p/2 1 + (x2 + y 2 + z 2 ) p/2 py 2 (x 2 + y 2 + z 2 ) p/2 1 (x 2 + y 2 + z 2 ) p (x 2 + y 2 + z 2 ) p + (x2 + y 2 + z 2 ) p/2 pz 2 (x 2 + y 2 + z 2 ) p/2 1 (x 2 + y 2 + z 2 ) p = 3(x2 + y 2 + z 2 ) p/2 p(x 2 + y 2 + z 2 )(x 2 + y 2 + z 2 ) p/2 1 (x 2 + y 2 + z 2 ) p = (x2 + y 2 + z 2 ) p/2 (3 p) (x 2 + y 2 + z 2 ) p = 3 p r p/2 So if p = 3 the corresponding radial vector field is source-free. ) In R 3 the curl of a vector field F = f, g, h, sometimes denoted curl F is F = h y g z, f z h x, g x f y. Note: The third component g x f y, the so called two-dimensional curl, should make you think of Green s Theorem! If F is a vector field where the curl is zero on a region R in R 3, that is F = 0 on R, then the vector field is irrotational on R. Let s calculate the curl of the vector field F = xy, z, y. F = 2, 0, x.

3 If F = a r for some fixed vector a in R 3 then F is called a general rotational field. This general rotation vector field F rotates around a. Consider F = 1, 0, 1 x, y, z = y, x z, y. Let F be a general rotation field. Then F = a 2 z a 3 y, a 3 x a 1 z, a 1 y a 2 x for some a = a 1, a 2, a 3. Then div F = F = 0. Furthermore, F = 2a 1, 2a 2, 2a 3 = 2a. Consider a vector field F = f, g in R 2. We can easily embed this vector field in R 3 : Then we can take the curl of the field: F = f, g, 0. F = 0, 0, g x f y. The third component is the two-dimensional curl of F (aka scalar curl ).

4 Let s develop some geometric intuition about divergence and curl. Here is the radial vector field F = r with a circle drawn in it. The divergence is positive (2) and the scalar curl is 0 (or, embedded in R 3, we get a curl of 0). The positive divergence can be viewed as the vector field expanding everywhere: If we displace all the points of the circle by the vector field, the resulting curve encloses a bigger region (more area). If the value was larger then that would mean the expansion is greater. If it were negative then it would measure contraction. The curl of 0 can be viewed as the vector field not causing any rotation: Imagine the circle anchored at the center (black dot), but allowed to spin if more of the tangential components of the vector field push the circle counterclockwise than clockwise, or vice-versa (that is, non-zero circulation). Here, by symmetry, that isn t the case. Hence the term irrotational!

5 Here is the rotational vector field F = y, x with a circle drawn in it. The divergence is 0 and the scalar curl is 2 (or, embedded in R 3, we get a curl of 2ˆk). The zero divergence can be viewed as the vector field neither expanding nor contracting anywhere: If we displace all the points of the circle by the vector field, the resulting curve a region with the same area. The curl of 2ˆk can be viewed as the vector field causing (ccw) rotation: Imagine the circle anchored at the center (black dot), but allowed to spin if more of the tangential components of the vector field push the circle counterclockwise than clockwise, or vice-versa (that is, non-zero circulation). Here clearly it spins counter-clockwise. The larger the magnitude of the curl, the faster the spin. The curl vector s direction determines the axis of rotation (in R 3 ) where the spin is determined by right-hand rule: With your right thumb in the direction of the vector the your fingers curl with the orientation of the spin. y, x Things are sometimes counterintuitive: For instance, the rotational field F = x 2 + y has 2 zero curl (so is irrotational). Can you picture why in the macro sense we say it s rotational, but in the micro sense (as measured by curl) it is irrotational? For vector fields in R 3 the same ideas apply to little spheres.

6 The following are simple exercises left for you (ELFY): (F + G) = F + G. (cf) = c F. (φf) = φ F + φ( F). (F + G) = F + G. (cf) = c F. We will prove that the curl of a conservative vector field is 0 and that the divergence of the curl of a vector field is 0. Let F = φ = φ x, φ y, φ z be a conservative vector field with potential function φ (with differentiable partial derivatives). Then by the equality of mixed partials, F = ( φ) = φ zy φ yz, φ xz φ zx, φ yx φ xy = 0. Also by the equality of mixed partials, for a vector field F = f, g, h (with twice-differentiable component functions), ( F) = h y g z, f z h x, g x f y = h yx g zx + f zy h xy + g xz f yz = 0. A vector field that is the curl of another vector field (in R 3 ) is source-free. Consider F = z, 0, y. In which direction n is ( F) n at a maximum? Let n = a, b, c and a 2 + b 2 + c 2 = 1. We have F = 1, 1, 0. So ( F) n = a + b and that is at a maximum when n = a, b, c is in the same direction as F. So we get n = (1/ 2) 1, 1, 0.

7 One of Maxwell s equations of electromagnetism is Ampere s Law, B = C E t, where E, B are the electric and magnetic fields respectively and C is a constant. Let s show that E(z, t) = A sin (kz ωt)î and B(z, t) = A sin (kz ωt)ĵ satisfy this law with ω = k/c: B = x, y, E 0, A sin (kz ωt), 0 = 0 ka cos (kz ωt), 0, 0 = C z t, since E t = ωa cos (kz ωt)î and ω = k/c.