Hassani S. Mathematical Methods: For Students of Physics and Related Fields

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14.3 Conservative Vector Fields 401

Example 14.3.2. The line integral of the vector ﬁeld of Example 14.1.2 was

independent of the three paths examined there. Could it be that the vector ﬁeld is

conservative? The vector ﬁeld is clearly well behaved everywhere. Therefore, the

vanishing of its curl proves that it is conservative. But

∇ ×A = K



∂

∂x

∂

∂y

∂

∂z

y 0



=(0)

+(0)

+(2xy −2xy)

=0.

So, A is indeed conservative.

Next we ﬁnd the potential of A at a point (x

)inthexy-plane.

Let the

reference point be the origin. Since it does not matter what path we take, we choose

a straight line joining the origin and (x

). A convenient parametric equation is

x = x

t, y = y

t, 0 ≤ t ≤ 1,

which gives dx = x

dt and dy = y

dt.Wenowhave

Φ(x

)=−

)

(0,0)

A · dr

= −K

[(x

t)(y

dt)+(x

t)(y

dt)]

= −2Kx

dt = −

We can now substitute (x, y)for(x

)toobtain

Φ(x, y)=−

The reader may verify that A = −∇Φ.



It should be clear that ∇×A = 0 always implies that A is not conservative.

However, ∇ × A = 0 implies that A is conservative only if the region in

question is contractible to zero.

Example 14.3.3.

Consider the vector ﬁeld

A =

+ y

−

+ y

where k is a constant. Since the components of this vector are independent of z,the

curl of the vector can have only a z-component:

∇ × A =



∂

∂x

∂

∂y

∂

∂z





∂A

∂x

−

∂A

∂y



We completely ignore the z-coordinate because A has no component in that direction.

402 Line Integral and Curl

The reader may easily verify that

∂A

∂x

= −

+ y

+ k

+ y

)

∂A

∂y

+ y

+ k

+ y

)

so that

∂A

∂x

−

∂A

∂y

= −

+ y

+ k

2(x

+ y

)

+ y

)

and ∇ × A =0.

Now take a circle of radius a about the origin and calculate the line integral of

A on this circle. For integration, use the parameterization

x = a cos t, y = a sin t, 0 ≤ t ≤ 2π,

with dx = −a sin tdt and dy = a cos tdt.Then

A · dr = A

dx + A

dy =

k(a sin t)(−a sin tdt)

(a cos t)

+(a sin t)

−

k(a cos t)(a cos tdt)

(a cos t)

+(a sin t)

= −kdt

and, therefore

circ

A · dr = −k

2π

dt = −2πk.

This is an example of a vector ﬁeld whose curl vanishes but yields a nonzero

result for a closed line integral. The reason is, of course, that the region inside the

circle is not contractable to zero: At the origin the vector is inﬁnite.



If the vector ﬁeld is conservative, in principle we can determine its potential

either by direct antidiﬀerentiation or by integration. The following example

illustrates the former procedure.

Example 14.3.4.

Consider the vector ﬁeld

A =(2xy +3z

)

+(x

+4yz)

+(2y

+6xz)

The reader may check that ∇ × A = 0. Thus, since A is well deﬁned everywhere,

it is conservative. To ﬁnd its potential Φ, we note that

∂Φ

∂x

= −A

= −2xy −3z

⇒ Φ=−x

y − 3z

x + g(y, z),

where we have simply antidiﬀerentiated A

with respect to x—assuming that y

and z are merely constants—and added a “constant” of integration: As far as x

diﬀerentiation is concerned, any function of y and z is a constant. Now diﬀerentiate

Φ obtained this way with respect to y and set it equal to −A

−A

= −(x

+4zy)=

∂Φ

∂y

∂

∂y

−x

y − 3z

x + g(y,z)

= −x

∂g

∂y

This gives

∂g

∂y

= −4yz ⇒ g(y, z)=−2y

z + h(z)

Note that our second “constant” of integration has no x-dependence because g(y,z)

does not depend on x. Substituting this back in the expression for Φ, we obtain

Φ=−x

y −3z

x + g(y,z)=−x

y − 3z

x − 2y

z + h(z).

14.3 Conservative Vector Fields 403

Finally, diﬀerentiating this with respect to z and setting it equal to −A

,weobtain

−A

= −(2y

+6xz)=

∂Φ

∂z

∂

∂z

−x

y −3z

x −2y

z + h(z)

= −6xz −2y

This gives

=0 ⇒ h(z)=const.≡ C.

The ﬁnal answer is therefore

Φ(x, y, z)=−x

y − 3z

x − 2y

z + C.

The arbitrary constant depends on the potential reference point, and is zero if we

choose the origin as that point. It is easy to verify that −∇Φ is indeed the vector

ﬁeld we started with.



There are various vector identities which connect gradient, divergence,

and curl. Most of these identities can be obtained by direct substitution. For

example, by substituting the Cartesian components of A×B in the Cartesian

expression for divergence, one can show that

∇ · (A ×B)=B ·∇ × A −A ·∇× B. (14.10)

Similarly, one can show that

∇ · (fA)=A ·∇f + f∇· A,

∇ × (fA)=f∇× A +(∇f) × A (14.11)

A × (∇ × A)=

∇|A|

− (A ·∇)A

We can use Equation (14.10) to derive an important vector integral relation

akin to the divergence theorem. Let B be a constant vector. Then the second

term on the RHS vanishes. Now apply the divergence theorem to the vector

ﬁeld A × B:

A × B · da =

∇ · (A × B) dV.

Using Equation (14.10), the RHS can be written as

RHS =

B · ∇ × A dV = B ·

∇ × A dV.

Moreover, the use of the cyclic property of the mixed triple product (see

Problem 1.15) will enable us to write the LHS as

LHS =

(da × A) · B =

B ·(da × A)=B ·

da × A.

404 Line Integral and Curl

Equating the new versions of the two sides, we obtain

B ·

∇ × A dV = B ·

da × A

B ·

⎛

⎝

∇ × A dV −

da × A

⎞

⎠

=0.

Since the last relation is true of arbitrary B, the vector inside the parentheses

must be zero. This gives the result we are after:

∇ × A dV =

da × A. (14.12)

14.4 Problems

14.1. Evaluate the line integral of

A(x, y, z)=x

+ y

− z

along the path given parametrically by

x = at

,y= bt, z = c sin (πt/2)

from the origin to (a, b, c).

14.2. Evaluate the line integral of

A(x, y, z)=x

−

along the path given parametrically by

x = a cos(πt/2),y= b sin(πt/2),z= ct

from (a, 0, 0) to (0,b,c).

14.3. Evaluate the line integral of

A(x, y)=x

along the closed ellipse given parametrically by

x = a cos t, y = b sin t.

14.4. Show that ∇× (A × r)=2A.

14.4 Problems 405

14.5. Let

A(x, y)=A

(x, y)

+ A

(x, y)

B(x, y)=B

(x, y)

+ B

(x, y)

be vectors in two-dimensions.

(a) Apply the divergence theorem to A using a volume V enclosed by a cylin-

der whose bottom base is an arbitrary closed curve C in the xy-plane and

whose top base is the same curve in a plane parallel to the xy-plane, and

whose lateral side is parallel to the z-axis. Now conclude that

dy − A

dx)=



∂A

∂x

∂A

∂y



dx dy

where R is the region enclosed by C in the xy-plane. This is the divergence

theorem in two dimensions.

(b) Apply Stokes’ theorem to B with C as above and S the region R deﬁned

above. Show that

dx + B

dy)=



∂B

∂x

−

∂B

∂y



dx dy

This is the Stokes’ theorem in two dimensions.

are the same.

14.6. Evaluate the line integral of

A(x, y)=

+3y

+2x

from the origin to the point (1, 2):

(a) along the straight line joining the two points; and

(b) along the parabola passing through the two points as well as the point

(−1, 2).

14.7. Is the vector ﬁeld A(x, y)=xe

cos y

−

sin y

conservative?

If so, ﬁnd its potential.

14.8. A vector ﬁeld is given by

A =





+ x

(x+z)/b

where Φ

and b are constants.

(a) Determine whether or not A is conservative.

(b) Find the potential of A if it is conservative.

14.9. The components of a vector ﬁeld are given by

= V

yze

= V

xze

+ V

k sin ky, A

= V

(a) Determine whether A is conservative or not.

(b) If it is conservative, ﬁnd its potential.

406 Line Integral and Curl

14.10. The Cartesian components of a vector are given by

=2axe

=2aye

= ka(x

+ y

where a and k are constants.

(a) Test whether A is conservative or not.

(b) If A is conservative, ﬁnd its potential.

14.11. Prove Equations (14.10) and (14.11).

14.12. Show that

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

and that

A × (∇× B)=∇(A ·B) −(A ·∇)B

14.13. Verify the vector identity

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

14.14. Verify that for constant A and B

∇[A · (B × r)] = A × B

Chapter 15

Applied Vector Analysis

In the last three chapters, we introduced the operator ∇ and used it to make

vectors out of scalars (gradient), scalars out of vectors (divergence), and new

vector out of old vectors (curl). It is obvious that all these processes can

be combined to form new scalars and vectors. For instance one can create

a vector out of a scalar by the operation of gradient and use the resulting

vector as an input for the operation of divergence. Since almost all equations

of physics involve derivatives of at most second order, we shall conﬁne our

treatment to “double del operations” in this chapter.

15.1 Double Del Operations

We can make diﬀerent combinations of the vector operator ∇ with itself. By

direct diﬀerentiation we can easily verify that

∇ × (∇f)=0. (15.1)

Equation (14.9) states that a conservative vector ﬁeld is the gradient of its

potential. Equation (15.1) says, on the other hand, that if a ﬁeld is the

gradient of a function then it is conservative.

We can combine these two

statements into one by saying that

Box 15.1.1. A vector ﬁeld is conservative (i.e., its curl vanishes) if and

only if it can be written as the gradient of a scalar function, in which case

the scalar function is the ﬁeld’s potential.

Example 15.1.1. The electrostatic and gravitational ﬁelds, which we denote

generically by A,aregivenbyanequationoftheform

A(r) = K

dQ(r



)

|r − r



(r − r



Assuming that the region in which the gradient of the function is deﬁned is contractable

to zero, i.e., the region has no point at which the gradient is inﬁnite.

408 Applied Vector Analysis

Furthermore, the reader may show that (see Problem 12.17)

r − r



|r − r



= −∇



|r − r





. (15.2)

Substitution in the above integral then yields

A(r)=−K

dQ(r



)∇



|r − r





= −∇



dQ(r



)

|r − r





= −∇Φ(r), (15.3)

where Φ, the potential of A,isgivenby

Φ(r) ≡ K

dQ(r



)

|r − r



. (15.4)

Equation (15.3), in conjunction with Equation (15.1), automatically implies that

both the electrostatic and gravitational ﬁelds are conservative.



In a similar fashion, we can directly verify the following identity:

∇ · (∇ × A)=0. (15.5)

Example 15.1.2.

Magnetic ﬁelds can also be written in terms of the so-called

vector potentials. To ﬁnd the expression for the vector potential, we substitute

Equation (15.2) in the magnetic ﬁeld integral:

B =

dq(r



)v(r



) × ( r − r



)

|r − r



= k

dq(r



)v( r



) ×

−∇



|r − r





We want to take the ∇ out of the integral. However, the cross product prevents a

direct “pull out.” So, we need to get around this by manipulating the integrand.

Using the second relation in Equation (14.11), we can write

∇ ×



v(r



)

|r − r





|r − r



  

∇ × v −v × ∇



|r − r





= −v(r



) × ∇



|r − r





We note that ∇ × v =0because∇ diﬀerentiates with respect to (x, y, z)ofwhich

v(r



) is independent. Substituting this last relation in the expression for B,we

obtain

B = k

dq(r



)∇ ×



v( r



)

|r − r





= ∇ ×



dq(r



)v( r



)

|r − r





≡ ∇ × A, (15.6)

wherewehavetaken∇× out of the integral since it diﬀerentiates with respect to

the parameters of integration and Ω is assumed independent of (x, y, z). The vector

potential A is deﬁned by the last line, which we rewrite asvector potential

deﬁned

A = k

dq(r



)v(r



)

|r − r



. (15.7)

15.2 Magnetic Multipoles 409

If the charges are conﬁned to one dimension, so that we have a current loop, then

dq(r



)v(r



)=Idr



and Equation (15.7) reduces to

A = k



|r − r



. (15.8)

An important consequence of Equations (15.6) and (15.5) is Vanishing of

divergence of

magnetic ﬁeld

implies absence of

magnetic charges.

∇ · B =0. (15.9)

Since the divergence of a vector ﬁeld is related to the density of its source, we

conclude that there are no magnetic charges.

This statement is within the context of classical electromagnetic theory. Re-

cently, with the advent of the uniﬁcation of electromagnetic and weak nuclear inter-

actions, there have been theoretical arguments for the existence of magnetic charges

(or monopoles). However, although the theory predicts—very rare—occurrences

of such monopoles, no experimental conﬁrmation of their existence has been

made.



15.2 Magnetic Multipoles

The similarity between the vector potential [Equation (15.8)] and the electro-

static potential motivates the expansion of the former in terms of multipoles

as was done in (10.33). We carry this expansion only up to the dipole term.

Substituting Equation (10.32) in Equation (15.8), we obtain

A = k



·r



+ ···







· r



The reader can easily show that the ﬁrst integral vanishes (Problem 15.5).

To facilitate calculating the second integral, choose Cartesian coordinates

and orient your axes so that

is in the x-direction. Denote the integral by

V.Then

V =

· r



·r



(



We evaluate each component of V separately.



2



end

beginning

because the beginning and end points of a loop coincide.

Now consider the identity



+ y



d(x



)=(x



)



end

beginning

= 0 (15.10)

410 Applied Vector Analysis

with an analogous identity involving x



and z



.Forthey-component of V,

we have



−



  

These add up to nothing!







 

=0 by Equation (15.10)



−







− y



× dr



)



× dr





It follows that



× dr





≡

μ ·

where we have deﬁned the magnetic dipole moment μ as

magnetic dipole

moment

μ ≡



× dr



. (15.11)

A similar calculation will yield

= −



× dr





≡−

μ ·

Therefore,

A = A

+ A

(

μ ·

−

μ ·

)



 

=μ×(

)bybaccabrule

Recalling that

, and that by our choice of orientation of the axes

, we ﬁnally obtain

A =

μ ×

μ ×r

. (15.12)

There is a striking resemblance between the vector potential of a magnetic

dipole [Equation (15.12)] and the scalar potential of an electric dipole [the

second term in the last line of Equation (10.33)]: The scalar potential is

given in terms of the scalar (dot) product of the electric dipole moment and

the position vector, the vector potential is given in terms of the vector product

of the magnetic dipole moment and the position vector.

Example 15.2.1.

Let us calculate the magnetic dipole moment of a circularmagnetic dipole

moment of a

circular current

loop

current of radius a. Placing the circle in the xy-plane with its center at the origin,

we have

μ =



× dr



) × (adϕ



2π

dϕ



= Iπa

So, the magnitude of the magnetic dipole moment of a circular loop of current is

the product of the current and the area of the loop. Its direction is related to the

direction of the current by the right-hand rule.

