Conservative Fields. a Vector Field Is Called Gradient If It Is a Gradient F = Grad Φ of a Scalar Potential. It Is

Lecture 22: Conservative Fields.

A vector field is called gradient if it is a gradient F = grad φ of a scalar potential.

It is called path independent if the line integral dependsZ only onZ the endpoints, i.e. if c1 and c2 are any two paths from P to Q then F · ds = F · ds. c1 c2 This is equivalent to that the line integral along any closed path or loop vanishes.

Th A vector field F in a domain D is gradient if and only if it is path independent. In that case we say that it is conservative Zand the integral is the difference in potential of the endpoint minus initial point: F · ds = φ(Q) − φ(P ). Z c Ex. Evaluate F·ds, where F= yi + xj and c= cos ti + sin tj, 0≤t≤π/4. c Sol. We want to find φ so that ∂φ/∂x = y, ∂φ/∂y = x and ∂φ/∂z = 0. Integration of ∂φ/∂x = y gives φ = xy + g(y, z), where g is any function of y and z. With this φ it follows that ∂φ/∂y = x and ∂φ/∂z = 0 if g(y, z) = C is a constant. Hence φ = xy + C, for any constant C, satisfies grad φ = F. Hence Z √ √ 1 F · ds = φ(1/ 2, 1/ 2, 0) − φ(1, 0, 0) = c 2 Note that F = grad φ is perpendicular to the level surfaces of φ and hence the level surfaces of the potential function are perpendicular to the flow lines of F.

Proof of Th: If F = grad φ then for any curve c from P to Q: Z Z Z b ∂φ dx ∂φ dy ∂φ dz b d F·ds = + + dt = φ(x(t), y(t), z(t)) dt = φ(Q)−φ(P ), c a ∂x dt ∂y dt ∂z dt a dt If the integral is independent of the way then we define

Z (x,y,z) φ(x, y, z) = F · ds, (x0,y0,z0)

where the integral is along any curve from a fixed point (x0, y0, z0) to (x, y, z). This is well-defined since it does not depend on which path we integrate along. In particular we can pick a curve going from (x0, y0, z0) to (a, y, z) and then along a straight line segment (t, y, z), a ≤ t ≤ x, to (x, y, z):

Z (a,y,z) Z (x,y,z) Z (a,y,z) Z x φ(x, y, z) = F · ds + F · ds = F · ds + F1(t, y, z)dt. (x0,y0,z0) (a,y,z) (x0,y0,z0) a The first integral is independent of x and by the fundamental theorem of calculus Z ∂φ ∂ x (x, y, z) = F1(t, y, z)dt = F1(x, y, z). ∂x ∂x a

The proof of that ∂φ/∂y =F2 and ∂φ/∂z =F3 is similar.

Ex. Show that F = xy2i + x3yj is not conservative. Sol. If ∂φ/∂x = xy2 then ∂2φ/∂y∂x = 2xy but if ∂φ/∂y = x3y then ∂2φ/∂x∂y = 3x2y which is a contradiction. 1 2

Conservative fields-Irrotational fields. We have just seen an example of a vector field F = F1i + F2j + F3k that could not be conservative because if there was a potential F = grad φ then ∂2φ/∂x∂y = ∂2φ/∂y∂x etc. so we must have ∂F1/∂y = ∂F2/∂x etc. Hence for a vector field to be conservative we must have ³∂F ∂F ´ ³∂F ∂F ´ ³∂F ∂F ´ curl F = 2 − 3 i + 3 − 1 j + 1 − 2 k = 0 ∂z ∂y ∂x ∂z ∂y ∂x A vector field satisfying this is called irrotational. We have Theorem. A vector field F defined and continuously differentiable throughout a simply connected domain D is conservative if and only if it is irrotational in D. The physical interpretation of this is that the flow lines for a gradient vector field can not curl around in a closed orbit since φ increases in the direction of the gradient. Ex. Show that F = xi + yj is conservative in all space. Sol. By the previous theorem it suffices to show that it is irrotational: curl F=0. Ex. Is F = (−yi + xj)/(x2 + y2) conservative in D ={(x, y, z); (x, y) 6= (0, 0)}. Sol. ∇ × F = ... = 0 but D is not simply connected so it does not follow that it is conservative. In fact, it is not since the line line integral along a circle around the z-axisR is nonvanishingR as we shall see. If x = cos t, y = sinR t and z = 0, 0 ≤ t ≤ 2π 2π 2π 2 2 then C F · ds = 0 (F1dx/dt + F2dy/dt + F3dz/dt)dt = 0 sin t + cos t dt = 2π. Ex. Show that the vector field above in the domain D of space where x > 0 is conservative and find a potential there. Sol. That it is conservative follows from that ∇ × F = 0 and that the domain D is simply connected. φ = tan−1(y/x) is a potential in D, since x 6= 0 in D. Proof of the theorem. That conservative implies irrotational is just the calculation above that ∇ × ∇φ = 0. We shall prove that irrotational implies conservative if the domain is all of space or a rectangular box containing the origin. We define Z x Z y Z z φ(x, y, z) = F1(t, 0, 0) dt + F2(x, t, 0) dt + F3(x, y, t) dt 0 0 0 and we will show that ∇φ=F if ∇×F=0. By the Fundamental Theorem of Calculus ¯ Z z ∂φ d ¯ d (x, y, z) = ¯ φ(x, y, z) = F3(x, y, t) dt = F3(x, y, z) ∂z dz (x,y)−fixed dz 0 Similarly, also using that we can differentiate below an integral sign: Z z ∂φ ∂F3 (x, y, z) = F2(x, y, 0) + (x, y, t) dt ∂y 0 ∂y

If we now also use the fact that ∂F3/∂y = ∂F2/∂z we obtain Z z ¯ ∂φ d ¯ (x, y, z) = F2(x, y, 0) + ¯ F2(x, y, t) dt ∂y 0 dt (x,y)−fixed = F2(x, y, 0) + F2(x, y, z) − F2(x, y, 0) = F2(x, y, z) by the Fundamental Theorem of Calculus. Finally by the same argument Z Z ∂φ y ∂F z ∂F (x, y, z) = F (x, 0, 0) + 2 (x, t, 0) dt + 3 (x, y, t) dt ∂x 1 ∂x ∂x Z0 Z0 y d z d = F1(x, 0, 0) + F1(x, t, 0) dt + F1(x, y, t) dt 0 dt 0 dt = F1(x, 0, 0) + F1(x, y, 0) − F1(x, 0, 0) + F1(x, y, z) − F1(x, y, 0) = F1(x, y, z)