What is CURL In MATHEMATICS and HOW TO USE IT

CURL, ∇× 

Another possible operation with the vector operator ∇ is to cross it into a vector.We obtain

∇ ×V = ˆx

∂

∂y

Vz −

∂

∂z

Vy

+ ˆy

∂

∂z

Vx −

∂

∂x

Vz

+ ˆz

∂

∂x

Vy −

∂

∂y

Vx

=

ˆx ˆy ˆz

∂

∂x

∂

∂y

∂

∂z

Vx Vy Vz

, (1.69)

which is called the curl of V. In expanding this determinant we must consider the derivative

nature of ∇. Specifically, V × ∇ is defined only as an operator, another vector differential

operator. It is certainly not equal, in general, to −∇ × V.17 In the case of Eq. (1.69) the

determinant must be expanded from the top down so that we get the derivatives as shown

in the middle portion of Eq. (1.69). If ∇ is crossed into the product of a scalar and a vector,

we can show

∇ ×(fV)|x =

∂

∂y

(f Vz)−

∂

∂z

(f Vy )

=

f

∂Vz

∂y +

∂f

∂y

Vz − f

∂Vy

∂z −

∂f

∂z

Vy

= f∇ ×V|x + (∇f )×V|x . (1.70)

If we permute the coordinates x →y,y →z, z→x to pick up the y-component and

then permute them a second time to pick up the z-component, then

∇ ×(fV) = f∇ ×V+ (∇f )×V, (1.71)

which is the vector product analog of Eq. (1.67b). Again, as a differential operator ∇

differentiates both f and V. As a vector it is crossed into V (in each term).

17In this same spirit, if A is a differential operator, it is not necessarily true that A × A = 0. Specifically, for the quantum

mechanical angular momentum operator L=−i(r×∇), we find that L×L = iL. See Sections 4.3 and 4.4 for more details.

44 Chapter 1 Vector Analysis

Example 1.8.1 VECTOR POTENTIAL OF A CONSTANT B FIELD

From electrodynamics we know that ∇·B = 0, which has the general solution B = ∇×A,

where A(r) is called the vector potential (of themagnetic induction), because ∇·(∇×A) = (∇×∇) ·A ≡ 0, as a triple scalar product with two identical vectors. This last identity will

not change if we add the gradient of some scalar function to the vector potential, which,

therefore, is not unique.

In our case, we want to show that a vector potential is A = 1

2 (B×r).

Using the BAC–BAC rule in conjunction with Example 1.7.1, we find that

2∇ ×A = ∇ ×(B×r) = (∇ · r)B−(B · ∇)r = 3B−B = 2B,

where we indicate by the ordering of the scalar product of the second term that the gradient

still acts on the coordinate vector.

Example 1.8.2 CURL OF A CENTRAL FORCE FIELD

Calculate ∇ ×(rf (r)).

By Eq. (1.71),

∇ ×

rf (r)

= f (r)∇ ×r+

∇f (r)

×r. (1.72)

First,

∇ ×r =

ˆx ˆy ˆz

∂

∂x

∂

∂y

∂

∂z

x y z

=

0. (

1.73)

Second, using ∇f (r) = ˆr(df/dr) (Example 1.6.1), we obtain

∇ ×rf (r) =

df

dr ˆr×r = 0. (1.74)

This vector product vanishes, since r = ˆrr and ˆr× ˆr = 0.

To develop a better feeling for the physical significance of the curl, we consider the

circulation of fluid around a differential loop in the xy-plane, Fig. 1.24.

FIGURE 1.24 Circulation around a differential loop.

1.8 Curl, ∇× 45

Although the circulation is technically given by a vector line integral

V · dλ (Section

1.10), we can set up the equivalent scalar integrals here. Let us take the circulation to

be

circulation1234 =

1

Vx(x, y) dλx +

2

Vy(x, y) dλy

+

3

Vx(x, y) dλx +

4

Vy(x, y) dλy . (1.75)

The numbers 1, 2, 3, and 4 refer to the numbered line segments in Fig. 1.24. In the first

integral, dλx =+dx; but in the third integral, dλx =−dx because the third line segment

is traversed in the negative x-direction. Similarly, dλy =+dy for the second integral, −dy

for the fourth. Next, the integrands are referred to the point (x0, y0) with a Taylor expansion18

taking into account the displacement of line segment 3 from 1 and that of 2 from 4.

For our differential line segments this leads to

circulation1234 = Vx(x0, y0)dx +

Vy(x0, y0)+

∂Vy

∂x

dx

dy

+

Vx(x0, y0)+

∂Vx

∂y

dy

(−dx)+Vy(x0, y0)(−dy)

=

∂Vy

∂x −

∂Vx

∂y

dx dy. (1.76)

Dividing by dx dy, we have

circulation per unit area = ∇ ×V|z. (1.77)

The circulation19 about our differential area in the xy-plane is given by the z-component

of ∇ × V. In principle, the curl ∇ × V at (x0, y0) could be determined by inserting a

(differential) paddle wheel into the moving fluid at point (x0, y0). The rotation of the little

paddle wheel would be a measure of the curl, and its axis would be along the direction of

∇ ×V, which is perpendicular to the plane of circulation.

We shall use the result, Eq. (1.76), in Section 1.12 to derive Stokes’ theorem. Whenever

the curl of a vector V vanishes,

∇ ×V = 0, (1.78)

V is labeled irrotational. The most important physical examples of irrotational vectors are

the gravitational and electrostatic forces. In each case

V = C ˆr

r2 = C

r

r3 , (1.79)

where C is a constant and ˆr is the unit vector in the outward radial direction. For the

gravitational case we have C =−Gm1m2, given by Newton’s law of universal gravitation.

If C = q1q2/4πε0, we have Coulomb’s law of electrostatics (mks units). The force V

18Here, Vy (x0 + dx,y0) = Vy (x0, y0) + (

∂Vy

∂x )x0y0 dx + ··· . The higher-order terms will drop out in the limit as dx →0.

A correction term for the variation of Vy with y is canceled by the corresponding term in the fourth integral.

19In fluid dynamics ∇ ×V is called the “vorticity.”

46 Chapter 1 Vector Analysis

given in Eq. (1.79) may be shown to be irrotational by direct expansion into Cartesian

components, as we did in Example 1.8.1. Another approach is developed in Chapter 2, in

which we express ∇×, the curl, in terms of spherical polar coordinates. In Section 1.13 we

shall see that whenever a vector is irrotational, the vector may be written as the (negative)

gradient of a scalar potential. In Section 1.16 we shall prove that a vector field may be

resolved into an irrotational part and a solenoidal part (subject to conditions at infinity).

In terms of the electromagnetic field this corresponds to the resolution into an irrotational

electric field and a solenoidal magnetic field.

For waves in an elastic medium, if the displacement u is irrotational, ∇ × u = 0, plane

waves (or spherical waves at large distances) become longitudinal. If u is solenoidal,

∇ · u = 0, then the waves become transverse. A seismic disturbance will produce a displacement

that may be resolved into a solenoidal part and an irrotational part (compare

Section 1.16). The irrotational part yields the longitudinal P (primary) earthquake waves.

The solenoidal part gives rise to the slower transverse S (secondary) waves.

Using the gradient, divergence, and curl, and of course the BAC–CAB rule, we may

construct or verify a large number of useful vector identities. For verification, complete

expansion into Cartesian components is always a possibility. Sometimes if we use insight

instead of routine shuffling of Cartesian components, the verification process can be shortened

drastically.

Remember that ∇ is a vector operator, a hybrid creature satisfying two sets of rules:

1. vector rules, and

2. partial differentiation rules—including differentiation of a product.

Example 1.8.3 GRADIENT OF A DOT PRODUCT

Verify that

∇(A · B) = (B · ∇)A +(A · ∇)B +B×(∇ ×A)+A×(∇ ×B). (1.80)

This particular example hinges on the recognition that ∇(A · B) is the type of term that

appears in the BAC–CAB expansion of a triple vector product, Eq. (1.55). For instance,

A×(∇ ×B) = ∇(A · B) −(A · ∇)B,

with the ∇ differentiating only B, not A. From the commutativity of factors in a scalar

product we may interchange A and B and write

B×(∇ ×A) = ∇(A · B) −(B · ∇)A,

now with ∇ differentiating only A, not B. Adding these two equations, we obtain ∇ differentiating

the product A · B and the identity, Eq. (1.80). This identity is used frequently

in electromagnetic theory. Exercise 1.8.13 is a simple illustration.

1.8 Curl, ∇× 47

Example 1.8.4 INTEGRATION BY PARTS OF CURL

Let us prove the formula

C(r) · (∇ × A(r)) d3r =

A(r) · (∇ × C(r)) d3r, where A or

C or both vanish at infinity.

To show this, we proceed, as in Examples 1.6.3 and 1.7.3, by integration by parts after

writing the inner product and the curl in Cartesian coordinates. Because the integrated

terms vanish at infinity we obtain

C(r) ·

∇ ×A(r)

d3r

=

 

Cz

∂Ay

∂x −

∂Ax

∂y

+ Cx

∂Az

∂y −

∂Ay

∂z

+Cy

∂Ax

∂z −

∂Az

∂x

d3r

=

 

Ax

∂Cz

∂y −

∂Cy

∂z

+Ay

∂Cx

∂z −

∂Cz

∂x

+Az

∂Cy

∂x −

∂Cx

∂y

d3r

=

A(r) ·

∇ ×C(r)

d3r,

just rearranging appropriately the terms after integration by parts.

Exercises

1.8.1 Show, by rotating the coordinates, that the components of the curl of a vector transform

as a vector.

Hint. The direction cosine identities of Eq. (1.46) are available as needed.

1.8.2 Show that u×v is solenoidal if u and v are each irrotational.

1.8.3 If A is irrotational, show that A×r is solenoidal.

1.8.4 A rigid body is rotating with constant angular velocity ω. Show that the linear velocity

v is solenoidal.

1.8.5 If a vector function f(x, y, z) is not irrotational but the product of f and a scalar function

g(x, y, z) is irrotational, show that then

f · ∇ ×f = 0.

1.8.6 If (a) V = ˆxVx(x, y)+ ˆyVy(x, y) and (b) ∇ ×V 

= 0, prove that ∇ ×V is perpendicular

to V.

1.8.7 Classically, orbital angular momentum is given by L = r × p, where p is the linear

momentum. To go from classical mechanics to quantum mechanics, replace p by the

operator −i∇ (Section 15.6). Show that the quantum mechanical angular momentum

48 Chapter 1 Vector Analysis

operator has Cartesian components (in units of ¯h)

Lx =−i

y

∂

∂z −z

∂

∂y

,

Ly =−i

z

∂

∂x −x

∂

∂z

,

Lz =−i

x

∂

∂y − y

∂

∂x

.

1.8.8 Using the angular momentum operators previously given, show that they satisfy commutation

relations of the form

[Lx,Ly] ≡ LxLy −LyLx = iLz

and hence

L×L = iL.

These commutation relations will be taken later as the defining relations of an angular

momentum operator—Exercise 3.2.15 and the following one and Chapter 4.

1.8.9 With the commutator bracket notation [Lx,Ly] = LxLy − LyLx , the angular momentum

vector L satisfies [Lx,Ly] = iLz, etc., or L×L = iL.

If two other vectors a and b commute with each other and with L, that is, [a,b] =

[a,L] = [b,L] = 0, show that

[a · L,b · L] = i(a ×b) · L.

1.8.10 For A = ˆxAx(x, y, z) and B = ˆxBx(x, y, z) evaluate each term in the vector identity

∇(A · B) = (B · ∇)A +(A · ∇)B+ B×(∇ ×A)+ A×(∇ ×B)

and verify that the identity is satisfied.

1.8.11 Verify the vector identity

∇ ×(A×B) = (B · ∇)A− (A · ∇)B−B(∇ · A)+A(∇ · B).

1.8.12 As an alternative to the vector identity of Example 1.8.3 show that

∇(A · B) = (A×∇)×B+(B×∇)×A+A(∇ · B)+ B(∇ · A).

1.8.13 Verify the identity

A×(∇ ×A) =

1

2

∇

A2

−(A · ∇)A.

1.8.14 If A and B are constant vectors, show that

∇(A · B×r) = A×B.

1.9 Successive Applications of ∇ 49

1.8.15 A distribution of electric currents creates a constant magnetic moment m = const. The

force on m in an external magnetic induction B is given by

F = ∇ ×(B×m).

Show that

F = (m· ∇)B.

Note. Assuming no time dependence of the fields,Maxwell’s equations yield ∇×B = 0.

Also, ∇ · B = 0.

1.8.16 An electric dipole of moment p is located at the origin. The dipole creates an electric

potential at r given by

ψ(r) =

p · r

4πε0r3 .

Find the electric field, E=−∇ψ at r.

1.8.17 The vector potential A of a magnetic dipole, dipole moment m, is given by A(r) =

(μ0/4π)(m×r/r3). Show that the magnetic induction B = ∇ ×A is given by

B =

μ0

4π

3ˆr(ˆr ·m)−m

r3 .

Note. The limiting process leading to point dipoles is discussed in Section 12.1 for

electric dipoles, in Section 12.5 for magnetic dipoles.

1.8.18 The velocity of a two-dimensional flow of liquid is given by

V = ˆxu(x, y) − ˆyv(x, y).

If the liquid is incompressible and the flow is irrotational, show that

∂u

∂x =

∂v

∂y

and

∂u

∂y =−

∂v

∂x

.

These are the Cauchy–Riemann conditions of Section 6.2.

1.8.19 The evaluation in this section of the four integrals for the circulation omitted Taylor

series terms such as ∂Vx/∂x, ∂Vy/∂y and all second derivatives. Show that ∂Vx/∂x,

∂Vy/∂y cancel out when the four integrals are added and that the second derivative

terms drop out in the limit as dx→0, dy→0.

Hint. Calculate the circulation per unit area and then take the limit dx→0, dy→0.