Gradient operator. In our calculation of dφ along the vector ds, we see that it can be described as the scalar product

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1 Gradient operator In our calculation of dφ along the vector ds, we see that it can be described as the scalar product ( φ dφ = x î + φ y ĵ + φ ) z ˆk ( ) u x dsî + u y dsĵ + u z dsˆk We take dφ = φ ds = ds φ u and hence define the gradient operator (in a Cartesian system) φ = gradφ = φ x î + φ y ĵ + φ z ˆk We then define the directional derivative as dφ ds = φ u

2 Example of gradient operators and directional derivatives As an example, consider Section 6, problems 11 and 12 When φ(x, y, z) is the electrostatic potential, the electric field E(x, y, z) = φ For φ(x, y) = x 2 y 2, compute the field E(x, y) (There is no dependence on z) E(x, y) = φ(x, y) = φ x î φ ĵ = 2xî + 2yĵ y Sketch the equipotential lines, and the direction and magnitude of the vector field at various points

3 Example, continued Find the rate of change if φ with distance ds at (1, 2) along direction 3î ĵ The direction as a unit vector is u = 3 10 î 1 10 ĵ At x = 1, y = 2, φ = 2î 4ĵ, so dφ ( ) ds = 2î 4ĵ ( 3 10 î 1 10 ĵ ) = 2 10

4 Gradient operator in polar coordinates We can determine the gradient operator in a polar coordinate system, so φ(r, θ) In polar coordinates, ds = drê r + rdθê θ We have that dφ = φ φ r dr + θ dθ = φ ds We take for φ, Then we can see φ = φ r êr + 1 φ r θ êθ dφ = φ ds = ( φ r êr + 1 ) φ r θ êθ (drê r + rdθê θ ) = φ φ dr + r θ dθ

5 Example: Section 6, Problem 17 Find r with r = x 2 + y 2, and write the answer in polar and Cartesian coordinates and verify that they are the same We can take φ(r, θ) = r and use φ = φ r êr + 1 φ r θ êθ, r = ê r Now in Cartesian coordinates, take φ(x, y) = x 2 + y 2 and use φ(x, y) = φ x î + φ y ĵ x 2 + y 2 = xî + yĵ x 2 + y 2

6 Example continued Now we can prove that ê r = xî+yĵ x 2 +y 2 First use that ê r = cos θî + sin θĵ x Next we notice that cos θ = and sin θ = y x 2 +y 2 ê r = cos θî + sin θĵ = xî + yĵ x 2 + y 2 x 2 +y 2

7 Another example, Section 6, problem 18 Evaluate x in using the defined in a Cartesian and polar coordinate systems and prove they are the same In Cartesian coordinates, it is very easy to show x = î To do this in polar coordinates, use that x = r cos θ, and then φ(r, θ) = r cos θ φ(r, θ) = φ r êr + 1 φ r θ êθ = cos θê r sin θê θ Then we use that ê r = cos θî + sin θĵ and ê θ = sin θî + cos θĵ cosθê r sin θê θ = î

8 The vector operator We take for the vector operator the definition in Cartesian coordinate system = x î + y ĵ + z ˆk Notice this operator is linear, and also has vector properties (contrast with d dx or x ) We have so far considered only the gradient of a scalar field φ(x, y, z) which gives a vector field, but there are other possibilities

9 Divergence of a vector field Consider a vector field defined in a Cartesian system at each point x, y, and z V (x, y, z) = V x (x, y, z)î + V y (x, y, z)ĵ + V z (x, y, z)ˆk We take the divergence of V (x, y, z) and get a scalar, V = divv = V x x + V y y + V z z

10 Laplacian in Cartesian system Since φ(x, y, z) (the gradient of a scalar field) results in a vector field, we can imagine taking φ (the divergence of the vector field φ) The linear operator acts on a scalar field and is called the Laplacian. It can be conveniently expressed in Cartesian coordinates as 2 φ = φ = 2 φ x φ y φ z 2

11 Curl operator Another linear operator is curl, or, which operates on a vector field and results in a vector field We take for the curl in Cartesian coordinates. V = curlv î ĵ ˆk = x y z V x V y V = z Completing the determinant we see, V = ( Vz y V ) ( y Vx î + z z V ) ( z Vy ĵ + x x V ) x ˆk y

12 Example: Section 7, Problem 6 Compute the divergence and curl of the vector field V = x 2 yî + y 2 xĵ + xyzˆk First we identify the components V x = x 2 y, V y = y 2 x, and V z = xyz, and then V = V = V x x + V y y + V z z = 2xy + 2xy + xy = 5xy ( Vz y V ) ( y Vx î + z z V ) ( z Vy ĵ + x x V ) x ˆk y Again using the components determined above, we get the vector field V = xyî yzĵ

13 Computing a line integral, work and potential energy dw = F d r To compute the total work along some path, integrate along the path W = F d r Change in potential energy due to some (conservative) force is minus the work du = dw = F d r Notice that the scalar potential U for a conservative force is unique only up to a constant U = U 0 F d r

14 Electric field and electrostatic potential difference We saw before that E = φ where E is the electric field and φ is the electrostatic potential We can also show that using ds = dxî + dyĵ + dzˆk we can express this equivalently dφ = E ds = ( φ x î + φ y ŷ + φ ) z ˆk ( ) dxî + dyĵ + dzˆk Then as we expect we get dφ = φ φ φ x dx + y dy + z dz To get the potential difference between to points A and B we integrate B A B dφ = φ B φ A = E ds A

15 Conservative fields In the example of electrostatics, B A dφ = φ B φ A = B A E ds does not depend on the path. Instead it only depends on the end points A and B Then E said to be a conservative field, which in this case arises because we defined E = φ The function dφ is an exact differential of the function φ(x, y, z) (dφ = φ φ φ x dx + y dy + z dz) When E ds = dφ, then E ds is also an exact differential When E ds = dφ, it turns out to be easy to show that E = 0

16 Demonstrating the E = 0 E = φ = φ x î φ y ĵ φ z ˆk E î ĵ ˆk = φ = x y z = 0 Note that E 0 when there is a time-varying magnetic flux! In this case, we can not just assuming E = φ and a line integral of E ds around a closed path does not give zero! This is Faraday s law and is a subject left for electromagnetism courses. φ x φ y φ z

17 More on conservative fields Consider a field F and the path integral from A to B W = B A F d r If the integral does not depend on the particular path from A to B, then F is a conservative field If F d r is an exact differential: F d r = U d r = U x dx U y U dy dz = du z then F is a conservative field, and B B W = F d r = du = [U(B) U(A)] A A Key point: Work is independent of the path between A and B

18 Closed path integrals for conservative fields For two paths (path I and path II) between A and B with F conservative, we have [ B ] [ B ] F d r = F d r A pathi A pathii This means that [ B A ] [ A ] F d r + F d r pathi B pathii = 0 Since this forms a closed path (A to B and then back to A), F d r = 0 Key point: Work done by a conservative field around a closed path is zero

19 Characteristics of a conservative field Any one of the following implies the others, and also that F is conservative: F = 0

20 Characteristics of a conservative field Any one of the following implies the others, and also that F is conservative: F = 0 F d r = 0

21 Characteristics of a conservative field Any one of the following implies the others, and also that F is conservative: F = 0 F d r = 0 The line integral B A F d r is independent of path between A and B

22 Characteristics of a conservative field Any one of the following implies the others, and also that F is conservative: F = 0 F d r = 0 The line integral B A F d r is independent of path between A and B F d r is an exact differential

23 Characteristics of a conservative field Any one of the following implies the others, and also that F is conservative: F = 0 F d r = 0 The line integral B A F d r is independent of path between A and B F d r is an exact differential F = W = U

24 Green s theorem in the plane We have seen how to do double integrals, now let s do them in the xy-plane over an area A b a dx yu P(x, y) y l y dy = b a [P(x, y u ) P(x, y l )] dx = Pdx C The integral C Pdx means a counterclockwise integral around the curve that bounds area A We could also do the same thing but integrate first with respect to y

25 Continued, Green s theorem in the plane d c dy xr Q(x, y) x l x dx = d c [Q(x r, y) Q(x l, y)] dy = Qdy C Putting these together we get Green s theorem in the plane, ( Q A x P ) dxdy = (Pdx + Qdy) y A