pep 332: mathematical methods for physicists this is one

PEP 111-S MechanicsDepartment of Physics

Fall 2015

Lecture Times: Tuesday 3:00pm - 4:40pm & Thursday 3:00pm - 4:40pm Classroom Location: ??Instructor: Jeff Grube Contact Info: [email protected] Office Hours: Burchard room 709, Thursday 12:00pm - 1:00pm Prerequisite(s): none Corequisite(s): MA 115 Cross-listed with: none

COURSE DESCRIPTIONThis is the foundation physics course for Pinnacle Scholars with science and engineering majors, and/or for Physics majors. Upon successful completion of the course the student will have an improved conceptual understanding of the mechanical description of nature as codified in the three Newtonian laws of motion. The student will also be able to apply these laws to analyze simple mechanical systems.

LEARNING OBJECTIVESAfter successful completion of this course, students will be able to…

- Develop improved conceptual understanding of the nature of force and acceleration.- Distinguish between acceleration and velocity.- Become familiar with the concept of conservation laws and be able to apply the conservation of momentum and the conservation of energy to the analysis of simple mechanical systems.- Become familiar with the quantitative description of rotational motion and related concepts including torque, moment of inertia and angular momentum.- Use free body diagrams to analyze simple mechanical systems.- Apply Newton’s Laws of Motion to analyze simple mechanical systems.- Manipulate vector quantities, including vector addition, resolution of vectors into components, dot

product and cross product.

FORMAT AND STRUCTUREThis course is comprised of two lectures each week. In addition, in-class quizzes are being scheduled, and there is one final exam.

1

Math Methods (Hassani 2009)

Ch 15 Applied Vector Analysis double del operations magnetic multipoles Laplacian Maxwell’s equations

PEP 332: Mathematical Methods for Physicists

15.4 Maxwell’s Equations 415

Example 15.3.3. Stellar equilibrium A star is a large mass of fluid heldtogether by gravitational attraction. If the star is in equilibrium, its fluid has nomotion and (15.26) becomes

∇p = ρg or ∇p = −ρ∇Φ

where Φ is the gravitational potential. Dividing this equation by ρ, and taking thedivergence of both sides, we obtain

∇ ·(

∇pρ

)= −∇2Φ or ∇ ·

(∇pρ

)= 4πGρ

where we used the Poisson equation (15.14). For a spherically symmetric star, only equation for stellarequilibriumthe radial coordinate enters in the equation above, and borrowing from the next

chapter the expressions (16.7) for gradient and (16.12) for divergence in sphericalcoordinates, the equation above takes the form

1r2

ddr

(r2

ρdpdr

)= 4πGρ

This is one of the fundamental equations of astrophysics. !

15.4 Maxwell’s Equations

No treatment of vector analysis is complete without a discussion of Maxwell’sequations. Electromagnetism was both the producer and the consumer ofvector analysis. It started with the accidental discovery by Orsted in 1820that an electric current produced a magnetic field. Subsequently, an intensesearch was undertaken by many physicists such as Ampere and Faraday tofind a connection between electric and magnetic phenomena. By the mid-1800s, a fairly good theory of electromagnetism was attained which, in thecontemporary language of vectors is translated in the following four equations: the four equations

that Maxwellinherited inintegral form(1)

∫ ∫

S

E · da =Q

ϵ0; (2)

∫ ∫

S

B · da = 0;

(3)∮

CE · dr = −dφm

dt; (4)

∮

CB · dr = µ0I. (15.27)

The first integral, Gauss’s law (or Coulomb’s law in disguise), states that theelectric flux through the closed surface S is essentially the total charge Q inthe volume surrounded by S. The second integral says that the correspond-ing flux for a magnetic field is zero. The fact that this holds for an arbitrarysurface implies that there are no magnetic charges. The third equation, Fara-day’s law, connects the electric field to the rate of change of magnetic fluxφm. Finally, the last equation, Ampere’s law, states that the source of themagnetic field is the electric current I. The constant ϵ0 and µ0 arise from aparticular set of units used for charges and currents.


(1) ∇ · E =ρ

ϵ0; (2) ∇ ·B = 0;

(3) ∇ × E = −∂B∂t

; (4) ∇ × B = µ0J + µ0ϵ0∂E∂t

. (15.29)

It was a great moment in the history of physics and mathematics whenMaxwell, prompted solely by the forces of logic and pure deduction, intro-duced the second term in the last equation. Such moments were rare prior mathematics and

the force of logicand humanreasoning unravelone of the greatestsecrets of Nature!

to Maxwell, and with the exception of Copernicus’s introduction of the he-liocentric theory of the solar system and Descartes’s introduction of analyticgeometry, deductive reasoning was the exception rather than the rule. The-ories and laws were empirical (or inductive); they were introduced to fit thedata and summarize, more or less directly, the numerous observations made.Maxwell broke this tradition and set the stage for deductive reasoning which,after a great deal of struggle to abandon the inductive tradition, became thenorm for modern physics.

Today, we aptly call all four equations in (15.29) Maxwell’s equations,although his contribution to those equations was a “mere” introduction ofthe second term on the RHS of the last equation. However, no other “small”contribution has ever affected humankind so enormously. This very “small”contribution was responsible for Maxwell’s prediction of the electromagneticwaves which were subsequently produced in the laboratory in 1887—only eightyears after Maxwell’s premature death—and put to technological use in 1901in the form of the first radio. Today, Maxwell’s equations are at the heart ofevery electronic device. Without them, our entire civilization, as we know it,would be nonexistent.

15.4.2 Electromagnetic Waves in Empty Space

Let us look at some of the implications of Maxwell equations. Taking the curl from Maxwell’sequations to waveequation

of the third Maxwell’s equation and using (15.19) and the first and fourthequations of (15.29), we obtain for the LHS

LHS = ∇ × (∇ × E) = ∇(∇ ·E) −∇2E =1ϵ0

∇ρ −∇2E,

and for the RHS

RHS = −∇ ×(

∂B∂t

)= − ∂

∂t(∇ × B) = − ∂

∂t

(µ0J + µ0ϵ0

∂E∂t

).

In particular, in free space, where ρ = 0 = J, these equations give

∇2E− µ0ϵ0∂2E∂t2

= 0. (15.30)

Ch 15.1 Double Del Operations

Chapter 15

Applied Vector Analysis

In the last three chapters, we introduced the operator ∇ and used it to makevectors out of scalars (gradient), scalars out of vectors (divergence), and newvector out of old vectors (curl). It is obvious that all these processes canbe combined to form new scalars and vectors. For instance one can createa vector out of a scalar by the operation of gradient and use the resultingvector as an input for the operation of divergence. Since almost all equationsof physics involve derivatives of at most second order, we shall confine ourtreatment to “double del operations” in this chapter.

15.1 Double Del Operations

We can make different combinations of the vector operator ∇ with itself. Bydirect differentiation we can easily verify that

∇ × (∇f) = 0. (15.1)

Equation (14.9) states that a conservative vector field is the gradient of itspotential. Equation (15.1) says, on the other hand, that if a field is thegradient of a function then it is conservative.1 We can combine these twostatements into one by saying that

Box 15.1.1. A vector field is conservative (i.e., its curl vanishes) if andonly if it can be written as the gradient of a scalar function, in which casethe scalar function is the field’s potential.

Example 15.1.1. The electrostatic and gravitational fields, which we denotegenerically by A, are given by an equation of the form

A(r) = K

∫∫

Ω

dQ(r′)|r − r′|3 (r− r′).

1Assuming that the region in which the gradient of the function is defined is contractableto 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′|3 = −∇(

1|r − r′|

). (15.2)

Substitution in the above integral then yields

A(r) = −K

∫∫

Ω

dQ(r′)∇(

1|r − r′|

)= −∇

(K

∫∫

Ω

dQ(r′)|r − r′|

)

= −∇Φ(r), (15.3)

where Φ, the potential of A, is given by

Φ(r) ≡ K

∫∫

Ω

dQ(r′)|r − r′| . (15.4)

Equation (15.3), in conjunction with Equation (15.1), automatically implies thatboth 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-calledvector potentials. To find the expression for the vector potential, we substituteEquation (15.2) in the magnetic field integral:

B =

∫∫

Ω

kmdq(r′)v(r′) × ( r− r′)|r − r′|3 = km

∫∫

Ω

dq(r′)v( r′) ×−∇

(1

|r − r′|

).

We want to take the ∇ out of the integral. However, the cross product prevents adirect “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′|

)=

1|r − r′|

=0︷︸︸︷∇ × v−v× ∇

(1

|r − r′|

)

= −v(r′) × ∇(

1|r − r′|

).

We note that ∇ × v = 0 because ∇ differentiates with respect to (x, y, z) of whichv(r′) is independent. Substituting this last relation in the expression for B, weobtain

B = km

∫∫

Ω

dq(r′)∇ ×(

v( r′)|r − r′|

)= ∇ ×

(km

∫∫

Ω

dq(r′)v( r′)|r − r′|

)

≡ ∇ × A, (15.6)

where we have taken ∇× out of the integral since it differentiates with respect tothe parameters of integration and Ω is assumed independent of (x, y, z). The vectorpotential A is defined by the last line, which we rewrite asvector potential

defined

A = km

∫∫

Ω

dq(r′)v(r′)|r − r′| . (15.7)

PEP 332 Mathematical Methods for Physicists

Homework 5: Problems from Ch 14, 15 (Hassani 2009) due Wednesday 10/21

14.4 Problems 405

14.5. Let

A(x, y) = Ax(x, y)ex + Ay(x, y)ey

B(x, y) = Bx(x, y)ex + By(x, y)ey

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 andwhose top base is the same curve in a plane parallel to the xy-plane, andwhose lateral side is parallel to the z-axis. Now conclude that

∮

C(Axdy − Aydx) =

∫∫

R

(∂Ax

∂x+

∂Ay

∂y

)dx dy

where R is the region enclosed by C in the xy-plane. This is the divergencetheorem in two dimensions.(b) Apply Stokes’ theorem to B with C as above and S the region R definedabove. Show that

∮

C(Bxdx + Bydy) =

∫∫

R

(∂By

∂x− ∂Bx

∂y

)dx dy

This is the Stokes’ theorem in two dimensions.(c) Show that in two dimensions the Stokes’ theorem and divergence theoremare the same.

14.6. Evaluate the line integral of

A(x, y) =(x2 + 3y

)ex +

(y2 + 2x

)ey

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).(c) Is A conservative?

14.7. Is the vector field A(x, y) = xex2cos y ex − 1

2ex2sin y ey conservative?

If so, find its potential.

14.8. A vector field is given by

A =Φ0

b2

[y

(1 +

x

b

)ex + xey +

xy

bez

]e(x+z)/b,

where Φ0 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

Ax = V0k3yzek2xy, Ay = V0k

3xzek2xy + V0k sinky, Az = V0kek2xy.

(a) Determine whether A is conservative or not.(b) If it is conservative, find its potential.


physics. Solving these equations Maxwell predicted the existence of electromagneticwaves and the fact that these waves propagate at the speed of light (1862). Heproposed that the phenomenon of light is therefore an electromagnetic phenomenon.

Maxwell left King’s College, London, in the spring of 1865 and returned tohis Scottish estate. He made periodic trips to Cambridge and, rather reluctantly,accepted an offer from Cambridge to be the first Cavendish Professor of Physics in1871. He designed the Cavendish laboratory and helped set it up.

15.5 Problems

15.1. Show that the curl of the gradient of a function is always zero.

15.2. Show that the divergence of the curl of a vector is always zero.

15.3. Verify Equation (15.19) component by component.

15.4. Provide the details of Example 15.3.1:(a) Compute the three components of L and verify Equation (15.20).(b) Calculate L2

xf , L2yf , L2

zf and show that you obtain the expressions givenin the example.(c) Verify that L2f is as given in Equation (15.21).(d) Show that A = (r ·∇)2f − r · (∇f) and obtain (15.22). Here A is definedby the sum of the expressions in the two pairs of parentheses in Equation(15.21)

15.5. By taking each component of dr′ separately in a convenient coordinatesystem show that its integral round any closed loop vanishes.

15.6. Recall that the total magnetic force on a current loop is given bytotal magneticforce on a currentloop in a constantmagnetic field iszero.

F = I

∮dr × B.

Show that the total force on a current loop located in a homogeneous magneticfield is zero.

15.7. Derive the differential form of Maxwell’s last equation from the corre-sponding integral form.

15.8. Starting with Maxwell’s equations, show that the magnetic field satis-fies the same wave equation as the electric field. In particular, that it, too,propagates with the same speed.

15.9. Consider E = E0ei(ωt−k·r) and B = B0ei(ωt−k·r), where i =√−1, E0,

B0, k, and ω are constants. The E and the B so defined represent plane wavesmoving in the direction of the vector k.(a) Show that they satisfy Maxwell’s equations in free space if:

(1) k ·E0 = 0; (2) k ·B0 = 0;

(3) k × E0 = ωB0; (4) k × B0 = − ω

c2E0.




15.5 Problems





xf , L2yf , L2




F = I

∮dr × B.






(1) k ·E0 = 0; (2) k ·B0 = 0;

(3) k × E0 = ωB0; (4) k × B0 = − ω

c2E0.



14.4 Problems 405

14.5. Let




∮

C(Axdy − Aydx) =

∫∫

R

(∂Ax

∂x+

∂Ay

∂y

)dx dy


∮

C(Bxdx + Bydy) =

∫∫

R

(∂By

∂x− ∂Bx

∂y

)dx dy



A(x, y) =(x2 + 3y

)ex +

(y2 + 2x

)ey






A =Φ0

b2

[y

(1 +

x

b

)ex + xey +

xy

bez

]e(x+z)/b,









15.5 Problems





xf , L2yf , L2




F = I

∮dr × B.






(1) k ·E0 = 0; (2) k ·B0 = 0;

(3) k × E0 = ωB0; (4) k × B0 = − ω

c2E0.




15.5 Problems





xf , L2yf , L2




F = I

∮dr × B.






(1) k ·E0 = 0; (2) k ·B0 = 0;

(3) k × E0 = ωB0; (4) k × B0 = − ω

c2E0.




r − r′

|r − r′|3 = −∇(

1|r − r′|

). (15.2)


A(r) = −K

∫∫

Ω

dQ(r′)∇(

1|r − r′|

)= −∇

(K

∫∫

Ω

dQ(r′)|r − r′|

)

= −∇Φ(r), (15.3)


Φ(r) ≡ K

∫∫

Ω

dQ(r′)|r − r′| . (15.4)



∇ · (∇ × A) = 0. (15.5)


B =

∫∫

Ω

kmdq(r′)v(r′) × ( r− r′)|r − r′|3 = km

∫∫

Ω

dq(r′)v( r′) ×−∇

(1

|r − r′|

).


∇ ×(

v(r′)|r − r′|

)=

1|r − r′|

=0︷︸︸︷∇ × v−v× ∇

(1

|r − r′|

)

= −v(r′) × ∇(

1|r − r′|

).


B = km

∫∫

Ω

dq(r′)∇ ×(

v( r′)|r − r′|

)= ∇ ×

(km

∫∫

Ω

dq(r′)v( r′)|r − r′|

)

≡ ∇ × A, (15.6)


defined

A = km

∫∫

Ω

dq(r′)v(r′)|r − r′| . (15.7)



r − r′

|r − r′|3 = −∇(

1|r − r′|

). (15.2)


A(r) = −K

∫∫

Ω

dQ(r′)∇(

1|r − r′|

)= −∇

(K

∫∫

Ω

dQ(r′)|r − r′|

)

= −∇Φ(r), (15.3)


Φ(r) ≡ K

∫∫

Ω

dQ(r′)|r − r′| . (15.4)



∇ · (∇ × A) = 0. (15.5)


B =

∫∫

Ω

kmdq(r′)v(r′) × ( r− r′)|r − r′|3 = km

∫∫

Ω

dq(r′)v( r′) ×−∇

(1

|r − r′|

).


∇ ×(

v(r′)|r − r′|

)=

1|r − r′|

=0︷︸︸︷∇ × v−v× ∇

(1

|r − r′|

)

= −v(r′) × ∇(

1|r − r′|

).


B = km

∫∫

Ω

dq(r′)∇ ×(

v( r′)|r − r′|

)= ∇ ×

(km

∫∫

Ω

dq(r′)v( r′)|r − r′|

)

≡ ∇ × A, (15.6)


defined

A = km

∫∫

Ω

dq(r′)v(r′)|r − r′| . (15.7)

14.3 Conservative Vector Fields 403

Finally, differentiating this with respect to z and setting it equal to −Az, we obtain

−Az = −(2y2 + 6xz) =∂Φ∂z

=∂∂z

(−x2y − 3z2x− 2y2z + h(z)

)= −6xz − 2y2 +

dhdz

.

This givesdhdz

= 0 ⇒ h(z) = const. ≡ C.

The final answer is therefore

Φ(x, y, z) = −x2y − 3z2x − 2y2z + C.

The arbitrary constant depends on the potential reference point, and is zero if wechoose the origin as that point. It is easy to verify that −∇Φ is indeed the vectorfield we started with. !

There are various vector identities which connect gradient, divergence,and curl. Most of these identities can be obtained by direct substitution. Forexample, by substituting the Cartesian components of A×B in the Cartesianexpression 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) = 12∇|A|2 − (A · ∇)A

We can use Equation (14.10) to derive an important vector integral relationakin to the divergence theorem. Let B be a constant vector. Then the secondterm on the RHS vanishes. Now apply the divergence theorem to the vectorfield A× B: ∫ ∫

S

A× B · da =∫ ∫

V

∫∇ · (A × B) dV.

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

RHS =∫ ∫

V

∫B · ∇ × A dV = B ·

∫ ∫

V

∫∇ × A dV.

Moreover, the use of the cyclic property of the mixed triple product (seeProblem 1.15) will enable us to write the LHS as

LHS =∫ ∫

S

(da × A) · B =∫ ∫

S

B · (da × A) = B ·∫ ∫

S

da × A.




r − r′

|r − r′|3 = −∇(

1|r − r′|

). (15.2)


A(r) = −K

∫∫

Ω

dQ(r′)∇(

1|r − r′|

)= −∇

(K

∫∫

Ω

dQ(r′)|r − r′|

)

= −∇Φ(r), (15.3)


Φ(r) ≡ K

∫∫

Ω

dQ(r′)|r − r′| . (15.4)



∇ · (∇ × A) = 0. (15.5)


B =

∫∫

Ω

kmdq(r′)v(r′) × ( r− r′)|r − r′|3 = km

∫∫

Ω

dq(r′)v( r′) ×−∇

(1

|r − r′|

).


∇ ×(

v(r′)|r − r′|

)=

1|r − r′|

=0︷︸︸︷∇ × v−v× ∇

(1

|r − r′|

)

= −v(r′) × ∇(

1|r − r′|

).


B = km

∫∫

Ω

dq(r′)∇ ×(

v( r′)|r − r′|

)= ∇ ×

(km

∫∫

Ω

dq(r′)v( r′)|r − r′|

)

≡ ∇ × A, (15.6)


defined

A = km

∫∫

Ω

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, thendq(r′)v(r′) = I dr′ and Equation (15.7) reduces to

A = kmI

∮dr′

|r − r′| . (15.8)

An important consequence of Equations (15.6) and (15.5) is Vanishing ofdivergence ofmagnetic fieldimplies absence ofmagnetic charges.

∇ · B = 0. (15.9)

Since the divergence of a vector field is related to the density of its source, weconclude 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—occurrencesof such monopoles, no experimental confirmation of their existence has beenmade. !

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 multipolesas 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 = kmI

∮ (1r

+er · r′

r2+ · · ·

)dr′ =

kmI

r

∮dr′

︸︷︷︸=0

+kmI

r2

∮er · r′dr′.

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 er is in the x-direction. Denote the integral byV. Then

V =∮

er · r′dr′ =∮

ex · r′dr′ =∮

x′dr′ =∮

x′(ex dx′ + ey dy′ + ez dz′).

We evaluate each component of V separately.

Vx =∮

x′dx′ = 12

∮d

(x′2) = 1

2x′2∣∣∣end

beginning= 0

because the beginning and end points of a loop coincide.Now consider the identity

∮(x′dy′ + y′dx′) =

∮d(x′y′) = (x′y′)

∣∣∣end

beginning= 0 (15.10)

Ch 15.2 Magnetic Multipoles

10.5 Multipole Expansion 297

is indeterminate. Using Equation (10.30) yields

limx→0

ln[h(x)] = limx→0

1/x−1/x2

= limx→0

(−x) = 0.

Therefore, limx→0 h(x) = e0 = 1. So, we have the interesting result limx→0 xx = 1.The limit of x2/(1 − cos x) as x goes to zero is obtained as follows:

limx→0

x2

1 − cos x= lim

x→0

2xsin x

= limx→0

2cos x

= 2.

Here we had to differentiate twice because the ratio of the first derivatives was alsoindeterminate. !

It is instructive for the reader to verify all limits in Example 10.4.1 usingL’Hopital’s rule to appreciate the ease of the Taylor expansion method.

10.5 Multipole Expansion

One extremely useful application of the power series representation of func-tions is in potential theory. The electrostatic or gravitational potential canbe written as

Φ(r) = K

∫∫

Ω

dQ(r′)|r − r′| , (10.31)

where K is ke for electrostatics and −G for gravity. Similarly, Q representseither electric charge or mass. In some applications, especially for electrostaticpotential, the distance of the field point P from the origin is much larger thanthe distance of the source point P ′ from the origin. This means that r >> r′

and we can expand in the powers of the ratio r′/r which we denote by ϵ. Thekey to this expansion is a power series expansion of 1/|r− r′|. First write

1|r − r′| =

1√r2 + r′2 − 2r · r′

=1

r√

1 + ϵ2 − 2ϵer · er′

=1r

(1 + ϵ2 − 2ϵer · er′

)−1/2.

Next use the binomial expansion (10.15) with x = ϵ2 − 2ϵer · er′ and α = − 12 .

Up to second order in ϵ, this yields

1|r− r′| =

1r

1 − 1

2

(ϵ2 − 2ϵer · er′

)+ 3

8

(ϵ2 − 2ϵer · er′

)2 + · · ·

=1r

1 + ϵer · er′ + ϵ2

[− 1

2 + 32 (er · er′)2

]+ · · ·

=1r

+er · r′

r2+

r′2

r3

[− 1

2 + 32 (er · er′)2

]+ · · · . (10.32)



A = kmI

∮dr′

|r − r′| . (15.8)


∇ · B = 0. (15.9)





A = kmI

∮ (1r

+er · r′

r2+ · · ·

)dr′ =

kmI

r

∮dr′

︸︷︷︸=0

+kmI

r2

∮er · r′dr′.



V =∮

er · r′dr′ =∮

ex · r′dr′ =∮

x′dr′ =∮



Vx =∮

x′dx′ = 12

∮d

(x′2) = 1

2x′2∣∣∣end

beginning= 0


∮(x′dy′ + y′dx′) =

∮d(x′y′) = (x′y′)

∣∣∣end




A = kmI

∮dr′

|r − r′| . (15.8)


∇ · B = 0. (15.9)





A = kmI

∮ (1r

+er · r′

r2+ · · ·

)dr′ =

kmI

r

∮dr′

︸︷︷︸=0

+kmI

r2

∮er · r′dr′.



V =∮

er · r′dr′ =∮

ex · r′dr′ =∮

x′dr′ =∮



Vx =∮

x′dx′ = 12

∮d

(x′2) = 1

2x′2∣∣∣end

beginning= 0


∮(x′dy′ + y′dx′) =

∮d(x′y′) = (x′y′)

∣∣∣end


magnetic dipole moment a multipole expansion of the magnetic vector potential A gives:

by substituting equation 10.32 into equation 15.8



A = kmI

∮dr′

|r − r′| . (15.8)


∇ · B = 0. (15.9)





A = kmI

∮ (1r

+er · r′

r2+ · · ·

)dr′ =

kmI

r

∮dr′

︸︷︷︸=0

+kmI

r2

∮er · r′dr′.



V =∮

er · r′dr′ =∮

ex · r′dr′ =∮

x′dr′ =∮



Vx =∮

x′dx′ = 12

∮d

(x′2) = 1

2x′2∣∣∣end

beginning= 0


∮(x′dy′ + y′dx′) =

∮d(x′y′) = (x′y′)

∣∣∣end


Ch 15.2 Magnetic Multipolesmagnetic dipole moment

Ch 15.2 Magnetic Multipolesmagnetic dipole moment410 Applied Vector Analysis

with an analogous identity involving x′ and z′. For the y-component of V,we have

Vy =∮

x′dy′ = 12

∮x′dy′ + 1

2

∮x′dy′ + 1

2

∮y′dx′ − 1

2

∮y′dx′

︸︷︷︸These add up to nothing!

= 12

(∮x′dy′ +

∮y′dx′

)

︸︷︷︸=0 by Equation (15.10)

+ 12

(∮x′dy′ −

∮y′dx′

)

= 12

∮(x′dy′ − y′dx′) = 1

2

∮(r′ × dr′)z = 1

2

(∮r′ × dr′

)· ez.

It follows that

Ay =kmI

r2Vy =

kmI

2r2

(∮r′ × dr′

)· ez ≡ km

r2µ · ez,

where we have defined the magnetic dipole moment µ asmagnetic dipolemoment

µ ≡ I

2

∮r′ × dr′. (15.11)

A similar calculation will yield

Az =kmI

r2Vz = −kmI

2r2

(∮r′ × dr′

)· ey ≡ −km

r2µ · ey.

Therefore,

A = Axex + Ayey + Azez =km

r2( ey µ · ez − ez µ · ey )︸︷︷︸=µ×(ey×ez) by bac cab rule

.

Recalling that ey × ez = ex, and that by our choice of orientation of the axeser = ex, we finally obtain

A =kmµ × er

r2=

kmµ × rr3

. (15.12)

There is a striking resemblance between the vector potential of a magneticdipole [Equation (15.12)] and the scalar potential of an electric dipole [thesecond term in the last line of Equation (10.33)]: The scalar potential isgiven in terms of the scalar (dot) product of the electric dipole moment andthe position vector, the vector potential is given in terms of the vector productof 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 acircular currentloop

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

µ =I2

∮r′ × dr′ =

I2

∮(aeρ′) × (a dϕ′eϕ′) =

Ia2

2

∫ 2π

0

dϕ′ez = Iπa2ez.

So, the magnitude of the magnetic dipole moment of a circular loop of current isthe product of the current and the area of the loop. Its direction is related to thedirection of the current by the right-hand rule. !



Vy =∮

x′dy′ = 12

∮x′dy′ + 1

2

∮x′dy′ + 1

2

∮y′dx′ − 1

2

∮y′dx′


= 12

(∮x′dy′ +

∮y′dx′

)

︸︷︷︸=0 by Equation (15.10)

+ 12

(∮x′dy′ −

∮y′dx′

)

= 12

∮(x′dy′ − y′dx′) = 1

2

∮(r′ × dr′)z = 1

2

(∮r′ × dr′

)· ez.

It follows that

Ay =kmI

r2Vy =

kmI

2r2

(∮r′ × dr′

)· ez ≡ km

r2µ · ez,


µ ≡ I

2

∮r′ × dr′. (15.11)


Az =kmI

r2Vz = −kmI

2r2

(∮r′ × dr′

)· ey ≡ −km

r2µ · ey.

Therefore,



.


A =kmµ × er

r2=

kmµ × rr3

. (15.12)




µ =I2

∮r′ × dr′ =

I2

∮(aeρ′) × (a dϕ′eϕ′) =

Ia2

2

∫ 2π

0

dϕ′ez = Iπa2ez.


Ch 15.2 Magnetic Multipolesmagnetic dipole moment



Vy =∮

x′dy′ = 12

∮x′dy′ + 1

2

∮x′dy′ + 1

2

∮y′dx′ − 1

2

∮y′dx′


= 12

(∮x′dy′ +

∮y′dx′

)

︸︷︷︸=0 by Equation (15.10)

+ 12

(∮x′dy′ −

∮y′dx′

)

= 12

∮(x′dy′ − y′dx′) = 1

2

∮(r′ × dr′)z = 1

2

(∮r′ × dr′

)· ez.

It follows that

Ay =kmI

r2Vy =

kmI

2r2

(∮r′ × dr′

)· ez ≡ km

r2µ · ez,


µ ≡ I

2

∮r′ × dr′. (15.11)


Az =kmI

r2Vz = −kmI

2r2

(∮r′ × dr′

)· ey ≡ −km

r2µ · ey.

Therefore,



.


A =kmµ × er

r2=

kmµ × rr3

. (15.12)




µ =I2

∮r′ × dr′ =

I2

∮(aeρ′) × (a dϕ′eϕ′) =

Ia2

2

∫ 2π

0

dϕ′ez = Iπa2ez.


Ch 15.2 Magnetic Multipoles



14.4 Problems 405

14.5. Let




∮

C(Axdy − Aydx) =

∫∫

R

(∂Ax

∂x+

∂Ay

∂y

)dx dy


∮

C(Bxdx + Bydy) =

∫∫

R

(∂By

∂x− ∂Bx

∂y

)dx dy



A(x, y) =(x2 + 3y

)ex +

(y2 + 2x

)ey






A =Φ0

b2

[y

(1 +

x

b

)ex + xey +

xy

bez

]e(x+z)/b,









15.5 Problems





xf , L2yf , L2




F = I

∮dr × B.






(1) k ·E0 = 0; (2) k ·B0 = 0;

(3) k × E0 = ωB0; (4) k × B0 = − ω

c2E0.




15.5 Problems





xf , L2yf , L2




F = I

∮dr × B.






(1) k ·E0 = 0; (2) k ·B0 = 0;

(3) k × E0 = ωB0; (4) k × B0 = − ω

c2E0.




15.5 Problems





xf , L2yf , L2




F = I

∮dr × B.






(1) k ·E0 = 0; (2) k ·B0 = 0;

(3) k × E0 = ωB0; (4) k × B0 = − ω

c2E0.



14.4 Problems 405

14.5. Let




∮

C(Axdy − Aydx) =

∫∫

R

(∂Ax

∂x+

∂Ay

∂y

)dx dy


∮

C(Bxdx + Bydy) =

∫∫

R

(∂By

∂x− ∂Bx

∂y

)dx dy



A(x, y) =(x2 + 3y

)ex +

(y2 + 2x

)ey






A =Φ0

b2

[y

(1 +

x

b

)ex + xey +

xy

bez

]e(x+z)/b,









15.5 Problems





xf , L2yf , L2




F = I

∮dr × B.






(1) k ·E0 = 0; (2) k ·B0 = 0;

(3) k × E0 = ωB0; (4) k × B0 = − ω

c2E0.




15.5 Problems





xf , L2yf , L2




F = I

∮dr × B.






(1) k ·E0 = 0; (2) k ·B0 = 0;

(3) k × E0 = ωB0; (4) k × B0 = − ω

c2E0.

Ch 15.3 Laplaciandefinition of the Laplacian

15.3 Laplacian 411

15.3 Laplacian

The divergence of the gradient is an important and frequently occurring op-erator called the Laplacian: Laplacian of a

function

∇ · (∇f) ≡ ∇2f =∂2f

∂x2+

∂2f

∂y2+

∂2f

∂z2. (15.13)

Laplacian occurs throughout physics, in situations ranging from the waves on Laplacian is foundeverywhere!a drum to the diffusion of matter in space, the propagation of electromagnetic

waves, and even the most basic behavior of matter on a subatomic scale, asgoverned by the Schrodinger equation of quantum mechanics.

We discuss one situation in which the Laplacian occurs naturally. Theresult of the example above and Theorem 13.2.4 can be combined to obtainan important equation in electrostatics and gravity called the Poisson equa-tion: ∇ · (−∇Φ) = 4πKρQ, or Poisson equation

∇2Φ(r) = −4πKρQ(r). (15.14)

This is a partial differential equation whose solution determines the potentialat various points in space.2 In many situations the density in the region ofinterest is zero. Then the RHS vanishes and we obtain an important specialcase of the above equation called Laplace’s equation: Laplace’s equation

∇2Φ(r) = 0. (15.15)

Consider a fixed point P in space with Cartesian coordinates (x0, y0, z0)and position vector r0. Take another (variable) point with Cartesian coordi-nates (x, y, z) and position vector r. By direct differentiation, one can verifythat

∇ ·(

r− r0

|r − r0|3

)= 0

at all points of space except at r = r0 for which the vector is not defined.Moreover, if S is any closed surface bounding a volume V , we have

∫ ∫

S

(r − r0

|r − r0|3

)· da ≡ ΩS

P =

4π if P is in V,

0 if P is not in V,

by Theorem 12.1.2. On the other hand, the divergence theorem relates theLHS of this equation with the volume integral of divergence. Thus,

∫ ∫

V

∫∇ ·

(r − r0

|r− r0|3

)dV =

4π if P is in V,

0 if P is not in V.(15.16)

2The reader should consider this, and any other differential equation, as a local equation,meaning that the derivatives on the LHS and the quantities on the RHS are to be evaluatedat the same point.

15.3 Laplacian 411

15.3 Laplacian

The divergence of the gradient is an important and frequently occurring op-erator called the Laplacian: Laplacian of a

function

∇ · (∇f) ≡ ∇2f =∂2f

∂x2+

∂2f

∂y2+

∂2f

∂z2. (15.13)

Laplacian occurs throughout physics, in situations ranging from the waves on Laplacian is foundeverywhere!a drum to the diffusion of matter in space, the propagation of electromagnetic

waves, and even the most basic behavior of matter on a subatomic scale, asgoverned by the Schrodinger equation of quantum mechanics.

We discuss one situation in which the Laplacian occurs naturally. Theresult of the example above and Theorem 13.2.4 can be combined to obtainan important equation in electrostatics and gravity called the Poisson equa-tion: ∇ · (−∇Φ) = 4πKρQ, or Poisson equation

∇2Φ(r) = −4πKρQ(r). (15.14)

This is a partial differential equation whose solution determines the potentialat various points in space.2 In many situations the density in the region ofinterest is zero. Then the RHS vanishes and we obtain an important specialcase of the above equation called Laplace’s equation: Laplace’s equation

∇2Φ(r) = 0. (15.15)

Consider a fixed point P in space with Cartesian coordinates (x0, y0, z0)and position vector r0. Take another (variable) point with Cartesian coordi-nates (x, y, z) and position vector r. By direct differentiation, one can verifythat

∇ ·(

r− r0

|r − r0|3

)= 0

at all points of space except at r = r0 for which the vector is not defined.Moreover, if S is any closed surface bounding a volume V , we have

∫ ∫

S

(r − r0

|r − r0|3

)· da ≡ ΩS

P =

4π if P is in V,

0 if P is not in V,

by Theorem 12.1.2. On the other hand, the divergence theorem relates theLHS of this equation with the volume integral of divergence. Thus,

∫ ∫

V

∫∇ ·

(r − r0

|r− r0|3

)dV =

4π if P is in V,

0 if P is not in V.(15.16)

2The reader should consider this, and any other differential equation, as a local equation,meaning that the derivatives on the LHS and the quantities on the RHS are to be evaluatedat the same point.

Ch 15.3 Laplacianfluid dynamics for a volume V of a fluid bounded by a surface S, the outside pressure p is normal to S and pointed into the volume

15.3 Laplacian 413

Adding the three components and using a little algebra, we get

−L2f = r2∇2f −(

x2 ∂2f

∂x2 + y2 ∂2f

∂y2 + z2 ∂2f

∂z2

)

− 2r · (∇f) − 2

(yz

∂2f∂y∂z

+ xz∂2f∂x∂z

+ xy∂2f

∂x∂y

). (15.21)

Let A denote the sum of the two expressions in the large parentheses. We can writeA in a compact form by expanding (r · ∇)(r · ∇f):

(r · ∇)2f ≡ (r · ∇)(r · ∇f) =

(x

∂∂x

+ y∂∂y

+ z∂∂z

) (x

∂f∂x

+ y∂f∂y

+ z∂f∂z

)

= x∂f∂x

+ x2 ∂2f

∂x2 + xy∂2f

∂x∂y+ xz

∂2f∂x∂z︸︷︷︸

comes from x differentiation

+terms from y and z differentiation.

Adding the terms from x, y, and z differentiations we obtain

(r · ∇)2f = r · (∇f) + A or A = (r · ∇)2f − r · (∇f).

Substituting this in (15.21) yields

L2f = −r2∇2f + r · (∇f) + (r · ∇)2f. (15.22)

As a differential operator, L2 is written as

L2 = −r2∇2 + r · ∇ + (r · ∇)2. (15.23)

We shall come back to this discussion in Chapter 17 to show how index manipulationeases the calculation (see Example 17.3.3). !

15.3.1 A Primer of Fluid Dynamics

We have already talked about the flow of a fluid in Section 13.2.3, wherewe derived the continuity equation, which states the conservation of mass inmathematical terms. We now want to take up the dynamics of a fluid, i.e.,the motion of various parts of the fluid due to the forces acting on them.

Consider a volume V of the fluid bounded by a surface S. The pressure pexerted from outside at any point of S in the element of area da is normal toS at that point and pointing into the volume V . Thus, the element of forcedue to pressure is −pda. If pressure is the only source of force on the volumeV of the fluid, then the total force on V is

F = −∫ ∫

S

p da.

Using Equation (13.12), we rewrite this as

F = −∫ ∫

S

p da = −∫ ∫

V

∫∇p dV.


This shows that ∇p is a force density, whose volume integral gives the force.If the density of the fluid is ρ and the mass element dm in V has velocity v,then the “mass time acceleration” is dm dv/dt = ρ dV (dv/dt), and the total“mass time acceleration” is the volume integral of this quantity. If there areother forces acting on the fluid described by a force density f, we can add itto the right-hand side. Thus, Newton’s second law of motion gives

∫ ∫

V

∫ρ(dv/dt) dV = −

∫ ∫

V

∫∇p dV +

∫ ∫

V

∫f dV,

and this holds for any volume V , in particular for an infinitesimal volume forwhich the integrals become the integrand. Hence, the second law of motionfor the fluid is

ρ(dv/dt) = −∇p + f. (15.24)

The total time derivative of velocity is

dvdt

=∂v∂t

+∂v∂x

dx

dt+

∂v∂y

dy

dt+

∂v∂z

dz

dt=

∂v∂t

+ (v · ∇)v.

Substituting this in (15.24) and dividing by ρ yieldsEuler’s equation offluid dynamics

∂v∂t

+ (v · ∇)v =−∇p + f

ρ. (15.25)

This is Euler’s equation and is one of the fundamental equations of fluiddynamics.

The force density f in Euler’s equation is usually that of the gravitationalforce. Since the gravitational force on an element ρ dV is gρ dV , where g isthe gravitational acceleration (or field), the gravitational force density is ρgand (15.25) becomes

∂v∂t

+ (v · ∇)v =−∇p

ρ+ g. (15.26)

Example 15.3.2. In hydrostatic situations with a uniform gravitational field thefluid is not moving and Equation (15.26) becomes

∇p = ρg,

and if g is in the negative z-direction, then

∂p∂x

=∂p∂y

= 0,∂p∂z

= −ρg.

Thus the pressure is independent of x and y, and depends only on height z. Weassume that the fluid (really the liquid) is incompressible, meaning that its densitydoes not depend on the pressure. Then, integrating the z equation gives

p = −ρgz + C.

If the liquid has a free surface at z = h where the pressure is p0, then C = p0 + ρgh,and

p = p0 + ρg(h − z). !

if there are other forces acting on the fluid (given by the force density f) then the total force is given as:



∫ ∫

V


∫ ∫

V

∫∇p dV +

∫ ∫

V

∫f dV,


ρ(dv/dt) = −∇p + f. (15.24)


dvdt

=∂v∂t

+∂v∂x

dx

dt+

∂v∂y

dy

dt+

∂v∂z

dz

dt=

∂v∂t

+ (v · ∇)v.


∂v∂t

+ (v · ∇)v =−∇p + f

ρ. (15.25)



∂v∂t

+ (v · ∇)v =−∇p

ρ+ g. (15.26)


∇p = ρg,


∂p∂x

=∂p∂y

= 0,∂p∂z

= −ρg.


p = −ρgz + C.


p = p0 + ρg(h − z). !

Ch 15.3 Laplacianfluid dynamics



∫ ∫

V


∫ ∫

V

∫∇p dV +

∫ ∫

V

∫f dV,


ρ(dv/dt) = −∇p + f. (15.24)


dvdt

=∂v∂t

+∂v∂x

dx

dt+

∂v∂y

dy

dt+

∂v∂z

dz

dt=

∂v∂t

+ (v · ∇)v.


∂v∂t

+ (v · ∇)v =−∇p + f

ρ. (15.25)



∂v∂t

+ (v · ∇)v =−∇p

ρ+ g. (15.26)


∇p = ρg,


∂p∂x

=∂p∂y

= 0,∂p∂z

= −ρg.


p = −ρgz + C.


p = p0 + ρg(h − z). !



∫ ∫

V


∫ ∫

V

∫∇p dV +

∫ ∫

V

∫f dV,


ρ(dv/dt) = −∇p + f. (15.24)


dvdt

=∂v∂t

+∂v∂x

dx

dt+

∂v∂y

dy

dt+

∂v∂z

dz

dt=

∂v∂t

+ (v · ∇)v.


∂v∂t

+ (v · ∇)v =−∇p + f

ρ. (15.25)



∂v∂t

+ (v · ∇)v =−∇p

ρ+ g. (15.26)


∇p = ρg,


∂p∂x

=∂p∂y

= 0,∂p∂z

= −ρg.


p = −ρgz + C.


p = p0 + ρg(h − z). !



∫ ∫

V


∫ ∫

V

∫∇p dV +

∫ ∫

V

∫f dV,


ρ(dv/dt) = −∇p + f. (15.24)


dvdt

=∂v∂t

+∂v∂x

dx

dt+

∂v∂y

dy

dt+

∂v∂z

dz

dt=

∂v∂t

+ (v · ∇)v.


∂v∂t

+ (v · ∇)v =−∇p + f

ρ. (15.25)



∂v∂t

+ (v · ∇)v =−∇p

ρ+ g. (15.26)


∇p = ρg,


∂p∂x

=∂p∂y

= 0,∂p∂z

= −ρg.


p = −ρgz + C.


p = p0 + ρg(h − z). !

Ch 15.3 Laplacianfluid dynamics



∫ ∫

V


∫ ∫

V

∫∇p dV +

∫ ∫

V

∫f dV,


ρ(dv/dt) = −∇p + f. (15.24)


dvdt

=∂v∂t

+∂v∂x

dx

dt+

∂v∂y

dy

dt+

∂v∂z

dz

dt=

∂v∂t

+ (v · ∇)v.


∂v∂t

+ (v · ∇)v =−∇p + f

ρ. (15.25)



∂v∂t

+ (v · ∇)v =−∇p

ρ+ g. (15.26)


∇p = ρg,


∂p∂x

=∂p∂y

= 0,∂p∂z

= −ρg.


p = −ρgz + C.


p = p0 + ρg(h − z). !

Ch 15.4 Maxwell’s Equationsfour equations in integral form


Example 15.3.3. Stellar equilibrium A star is a large mass of fluid heldtogether by gravitational attraction. If the star is in equilibrium, its fluid has nomotion and (15.26) becomes

∇p = ρg or ∇p = −ρ∇Φ

where Φ is the gravitational potential. Dividing this equation by ρ, and taking thedivergence of both sides, we obtain

∇ ·(

∇pρ

)= −∇2Φ or ∇ ·

(∇pρ

)= 4πGρ

where we used the Poisson equation (15.14). For a spherically symmetric star, only equation for stellarequilibriumthe radial coordinate enters in the equation above, and borrowing from the next

chapter the expressions (16.7) for gradient and (16.12) for divergence in sphericalcoordinates, the equation above takes the form

1r2

ddr

(r2

ρdpdr

)= 4πGρ

This is one of the fundamental equations of astrophysics. !

15.4 Maxwell’s Equations

No treatment of vector analysis is complete without a discussion of Maxwell’sequations. Electromagnetism was both the producer and the consumer ofvector analysis. It started with the accidental discovery by Orsted in 1820that an electric current produced a magnetic field. Subsequently, an intensesearch was undertaken by many physicists such as Ampere and Faraday tofind a connection between electric and magnetic phenomena. By the mid-1800s, a fairly good theory of electromagnetism was attained which, in thecontemporary language of vectors is translated in the following four equations: the four equations

that Maxwellinherited inintegral form(1)

∫ ∫

S

E · da =Q

ϵ0; (2)

∫ ∫

S

B · da = 0;

(3)∮

CE · dr = −dφm

dt; (4)

∮

CB · dr = µ0I. (15.27)

The first integral, Gauss’s law (or Coulomb’s law in disguise), states that theelectric flux through the closed surface S is essentially the total charge Q inthe volume surrounded by S. The second integral says that the correspond-ing flux for a magnetic field is zero. The fact that this holds for an arbitrarysurface implies that there are no magnetic charges. The third equation, Fara-day’s law, connects the electric field to the rate of change of magnetic fluxφm. Finally, the last equation, Ampere’s law, states that the source of themagnetic field is the electric current I. The constant ϵ0 and µ0 arise from aparticular set of units used for charges and currents.

(1) Gauss’ law: electric flux through a closed surface S is the total charge in the volume (2) Gauss’ law for magnetism: the corresponding flux for the magnetic field B is zero (3) Faraday’s law: rate of change of the magnetic flux dΦm/dt is connected to E (4) Ampere’s law: the source of the magnetic field is the current I

four equations in differential form


15.4.1 Maxwell’s Contribution

Equations (15.27) can be cast in differential form as well. The differentialform of the equations is important because it places particular emphasis onthe fields which are the primary objects. The differential form of the equationsabove are:the four equations

that Maxwellinherited indifferential form

(1) ∇ · E =ρ

ϵ0; (2) ∇ ·B = 0;

(3) ∇ × E = −∂B∂t

; (4) ∇ × B = µ0J. (15.28)

We have already derived the first two equations in Theorem 13.2.4 and Equa-tion (15.9). Here we derive the third equation and leave the derivation ofthe last equation—which is very similar to that of the third—to the reader.Stokes’ theorem turns the LHS of the third equation of (15.27) into

LHS =∫ ∫

S

∇ × E · da.

The RHS is

−dφm

dt= − d

dt

∫ ∫

S

B · da =∫ ∫

S

(−∂B

∂t

)· da,

where we have assumed that the change in the flux comes about solely dueto a change in the magnetic field. This makes it possible to push the timedifferentiation inside the integral, upon which it becomes a partial derivativebecause B is a function of position as well. Since the last two equations holdfor arbitrary S, the integrands must be equal. This proves the third equationin (15.28).

Maxwell inherited the four equations in (15.28), and started ponderingabout them in the 1860s. He noticed that while the second and third areMaxwell discovers

the inconsistencyof Equation(15.28) with theconservation ofelectric charge,and modifies thelast equation toresolve theinconsistency.

consistent with other aspects of electromagnetism, the other two equationslead to a contradiction. Let us retrace his argument. By Equation (15.5),the divergence of the LHS of the last equation of (15.28) vanishes. Therefore,taking the divergence of both sides, we get ∇ · J = 0. This contradicts thedifferential form of the continuity equation (13.22) for charges which expressesthe conservation of electric charge. Because of the firm establishment of thecharge conservation, Maxwell decided to try altering the four equations tomake them compatible with charge conservation. The clue is in the firstequation. If we differentiate that equation with respect to time, we obtain

∂

∂t∇ · E =

1ϵ0

∂ρ

∂t⇒ ∇ ·

(∂E∂t

)=

1ϵ0

∂ρ

∂t⇒ ∇ ·

(ϵ0

∂E∂t

)=

∂ρ

∂t

This suggested to Maxwell that, if the four equations are to be consistentwith charge conservation, the fourth equation had to be modified to includeϵ0∂E/∂t. With this modification, the four equations in (15.28) becomethe four Maxwell

equations





(1) ∇ · E =ρ

ϵ0; (2) ∇ ·B = 0;

(3) ∇ × E = −∂B∂t

; (4) ∇ × B = µ0J. (15.28)


LHS =∫ ∫

S

∇ × E · da.

The RHS is

−dφm

dt= − d

dt

∫ ∫

S

B · da =∫ ∫

S

(−∂B

∂t

)· da,





∂

∂t∇ · E =

1ϵ0

∂ρ

∂t⇒ ∇ ·

(∂E∂t

)=

1ϵ0

∂ρ

∂t⇒ ∇ ·

(ϵ0

∂E∂t

)=

∂ρ

∂t


equations

Ch 15.4 Maxwell’s EquationsMaxwell’s contribution for the conservation of electric charge Maxwell noticed in the 1860s a contradiction with the equations





(1) ∇ · E =ρ

ϵ0; (2) ∇ ·B = 0;

(3) ∇ × E = −∂B∂t

; (4) ∇ × B = µ0J. (15.28)


LHS =∫ ∫

S

∇ × E · da.

The RHS is

−dφm

dt= − d

dt

∫ ∫

S

B · da =∫ ∫

S

(−∂B

∂t

)· da,





∂

∂t∇ · E =

1ϵ0

∂ρ

∂t⇒ ∇ ·

(∂E∂t

)=

1ϵ0

∂ρ

∂t⇒ ∇ ·

(ϵ0

∂E∂t

)=

∂ρ

∂t


equations

380 Flux and Divergence

where in the last integral we have emphasized the dependence of various quan-tities on location and time. Now, if Q is a conserved quantity such as energy,momentum, charge, or mass,10 the amount of Q that crosses S outward (i.e.,the flux through S) must precisely equal the rate of depletion of Q in thevolume V .global or integral

form of continuityequation

Theorem 13.2.5. In mathematical symbols, the conservation of a conservedphysical quantity Q is written as

dQ

dt= −

∫ ∫

S

JQ · da, (13.21)

which is the global or integral form of the continuity equation.

The minus sign ensures that positive flux gives rise to a depletion, andvice versa. The local or differential form of the continuity equation can beobtained as follows: The LHS of Equation (13.21) can be written as

dQ

dt=

d

dt

∫ ∫

V

∫ρQ(r, t) dV (r) =

∫ ∫

V

∫∂ρQ

∂t(r, t) dV (r),

while the RHS, with the help of the divergence theorem, becomes

−∫ ∫

S

JQ · da = −∫ ∫

V

∫∇ · JQ dV.

Together they give∫ ∫

V

∫∂ρQ

∂tdV = −

∫ ∫

V

∫∇ · JQ dV

or∫ ∫

V

∫ ∂ρQ

∂t+ ∇ · JQ

dV = 0.

This relation is true for all volumes V . In particular, we can make the volumeas small as we please. Then, the integral will be approximately the integrandtimes the volume. Since the volume is nonzero (but small), the only way thatthe product can be zero is for the integrand to vanish.

Box 13.2.4. The differential form of the continuity equation is

∂ρQ

∂t+ ∇ · JQ = 0. (13.22)

local (differential)form of continuityequation

10In the theory of relativity mass by itself is not a conserved quantity, but mass incombination with energy is.



r − r′

|r − r′|3 = −∇(

1|r − r′|

). (15.2)


A(r) = −K

∫∫

Ω

dQ(r′)∇(

1|r − r′|

)= −∇

(K

∫∫

Ω

dQ(r′)|r − r′|

)

= −∇Φ(r), (15.3)


Φ(r) ≡ K

∫∫

Ω

dQ(r′)|r − r′| . (15.4)



∇ · (∇ × A) = 0. (15.5)


B =

∫∫

Ω

kmdq(r′)v(r′) × ( r− r′)|r − r′|3 = km

∫∫

Ω

dq(r′)v( r′) ×−∇

(1

|r − r′|

).


∇ ×(

v(r′)|r − r′|

)=

1|r − r′|

=0︷︸︸︷∇ × v−v× ∇

(1

|r − r′|

)

= −v(r′) × ∇(

1|r − r′|

).


B = km

∫∫

Ω

dq(r′)∇ ×(

v( r′)|r − r′|

)= ∇ ×

(km

∫∫

Ω

dq(r′)v( r′)|r − r′|

)

≡ ∇ × A, (15.6)


defined

A = km

∫∫

Ω

dq(r′)v(r′)|r − r′| . (15.7)





(1) ∇ · E =ρ

ϵ0; (2) ∇ ·B = 0;

(3) ∇ × E = −∂B∂t

; (4) ∇ × B = µ0J. (15.28)


LHS =∫ ∫

S

∇ × E · da.

The RHS is

−dφm

dt= − d

dt

∫ ∫

S

B · da =∫ ∫

S

(−∂B

∂t

)· da,





∂

∂t∇ · E =

1ϵ0

∂ρ

∂t⇒ ∇ ·

(∂E∂t

)=

1ϵ0

∂ρ

∂t⇒ ∇ ·

(ϵ0

∂E∂t

)=

∂ρ

∂t


equations





(1) ∇ · E =ρ

ϵ0; (2) ∇ ·B = 0;

(3) ∇ × E = −∂B∂t

; (4) ∇ × B = µ0J. (15.28)


LHS =∫ ∫

S

∇ × E · da.

The RHS is

−dφm

dt= − d

dt

∫ ∫

S

B · da =∫ ∫

S

(−∂B

∂t

)· da,





∂

∂t∇ · E =

1ϵ0

∂ρ

∂t⇒ ∇ ·

(∂E∂t

)=

1ϵ0

∂ρ

∂t⇒ ∇ ·

(ϵ0

∂E∂t

)=

∂ρ

∂t


equations

taking the divergence of equation (4) gives since

however the continuity equation (eq 13.22) giving the conservation of electric charge is:

Maxwell differentiated equation (1) with respect to time:


(1) ∇ · E =ρ

ϵ0; (2) ∇ ·B = 0;

(3) ∇ × E = −∂B∂t

; (4) ∇ × B = µ0J + µ0ϵ0∂E∂t

. (15.29)

It was a great moment in the history of physics and mathematics whenMaxwell, prompted solely by the forces of logic and pure deduction, intro-duced the second term in the last equation. Such moments were rare prior mathematics and

the force of logicand humanreasoning unravelone of the greatestsecrets of Nature!

to Maxwell, and with the exception of Copernicus’s introduction of the he-liocentric theory of the solar system and Descartes’s introduction of analyticgeometry, deductive reasoning was the exception rather than the rule. The-ories and laws were empirical (or inductive); they were introduced to fit thedata and summarize, more or less directly, the numerous observations made.Maxwell broke this tradition and set the stage for deductive reasoning which,after a great deal of struggle to abandon the inductive tradition, became thenorm for modern physics.

Today, we aptly call all four equations in (15.29) Maxwell’s equations,although his contribution to those equations was a “mere” introduction ofthe second term on the RHS of the last equation. However, no other “small”contribution has ever affected humankind so enormously. This very “small”contribution was responsible for Maxwell’s prediction of the electromagneticwaves which were subsequently produced in the laboratory in 1887—only eightyears after Maxwell’s premature death—and put to technological use in 1901in the form of the first radio. Today, Maxwell’s equations are at the heart ofevery electronic device. Without them, our entire civilization, as we know it,would be nonexistent.

15.4.2 Electromagnetic Waves in Empty Space

Let us look at some of the implications of Maxwell equations. Taking the curl from Maxwell’sequations to waveequation

of the third Maxwell’s equation and using (15.19) and the first and fourthequations of (15.29), we obtain for the LHS

LHS = ∇ × (∇ × E) = ∇(∇ ·E) −∇2E =1ϵ0

∇ρ −∇2E,

and for the RHS

RHS = −∇ ×(

∂B∂t

)= − ∂

∂t(∇ × B) = − ∂

∂t

(µ0J + µ0ϵ0

∂E∂t

).

In particular, in free space, where ρ = 0 = J, these equations give

∇2E− µ0ϵ0∂2E∂t2

= 0. (15.30)

Maxwell then modified equation (4) to be consistent with charge conservation

pep 332: mathematical methods for physicists this is one

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