Part IB - Electromagnetism · 2020-01-24 · Part IB | Electromagnetism Based on lectures by D. Tong Notes taken by Dexter Chua Lent 2015 These notes are not endorsed by the lecturers,

Part IB — Electromagnetism

Based on lectures by D. TongNotes taken by Dexter Chua

Lent 2015

These notes are not endorsed by the lecturers, and I have modified them (oftensignificantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Electromagnetism and RelativityReview of Special Relativity; tensors and index notation. Lorentz force law. Electro-magnetic tensor. Lorentz transformations of electric and magnetic fields. Currentsand the conservation of charge. Maxwell equations in relativistic and non-relativisticforms. [5]

ElectrostaticsGauss’s law. Application to spherically symmetric and cylindrically symmetric chargedistributions. Point, line and surface charges. Electrostatic potentials; general chargedistributions, dipoles. Electrostatic energy. Conductors. [3]

MagnetostaticsMagnetic fields due to steady currents. Ampre’s law. Simple examples. Vectorpotentials and the Biot-Savart law for general current distributions. Magnetic dipoles.Lorentz force on current distributions and force between current-carrying wires. Ohm’slaw. [3]

Electrodynamics

Faraday’s law of induction for fixed and moving circuits. Electromagnetic energy and

Poynting vector. 4-vector potential, gauge transformations. Plane electromagnetic

waves in vacuum, polarization. [5]

1

Contents IB Electromagnetism

Contents

0 Introduction 3

1 Preliminaries 41.1 Charge and Current . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Forces and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Electrostatics 72.1 Gauss’ Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Electrostatic potential . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2.1 Point charge . . . . . . . . . . . . . . . . . . . . . . . . . 112.2.2 Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2.3 General charge distribution . . . . . . . . . . . . . . . . . 122.2.4 Field lines and equipotentials . . . . . . . . . . . . . . . . 13

2.3 Electrostatic energy . . . . . . . . . . . . . . . . . . . . . . . . . 142.4 Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3 Magnetostatics 183.1 Ampere’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.2 Vector potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.3 Magnetic dipoles . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.4 Magnetic forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4 Electrodynamics 264.1 Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.2 Magnetostatic energy . . . . . . . . . . . . . . . . . . . . . . . . . 294.3 Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.4 Displacement currents . . . . . . . . . . . . . . . . . . . . . . . . 324.5 Electromagnetic waves . . . . . . . . . . . . . . . . . . . . . . . . 334.6 Poynting vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

5 Electromagnetism and relativity 385.1 A review of special relativity . . . . . . . . . . . . . . . . . . . . 38

5.1.1 A geometric interlude on (co)vectors . . . . . . . . . . . . 385.1.2 Transformation rules . . . . . . . . . . . . . . . . . . . . . 405.1.3 Vectors and covectors in SR . . . . . . . . . . . . . . . . . 41

5.2 Conserved currents . . . . . . . . . . . . . . . . . . . . . . . . . . 435.3 Gauge potentials and electromagnetic fields . . . . . . . . . . . . 435.4 Maxwell Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 485.5 The Lorentz force law . . . . . . . . . . . . . . . . . . . . . . . . 49

2

0 Introduction IB Electromagnetism

0 Introduction

Electromagnetism is one of the four fundamental forces of the universe. Apartfrom gravity, most daily phenomena can be explained by electromagnetism. Thevery existence of atoms requires the electric force (plus weird quantum effects)to hold electrons to the nucleus, and molecules are formed again due to electricforces between atoms. More macroscopically, electricity powers all our electricalappliances (by definition), while the magnetic force makes things stick on ourfridge. Finally, it is the force that gives rise to light, and allows us to see things.

While it seems to be responsible for so many effects, the modern classicaldescription of electromagnetism is rather simple. It is captured by merely fourshort and concise equations known as Maxwell’s equations. In fact, in themajority of this course, we would simply be exploring different solutions to theseequations.

Historically, electromagnetism has another significance — it led to Einstein’sdiscovery of special relativity. At the end of the course, we will look at howMaxwell’s equations naturally fit in nicely in the framework of relativity. Writtenrelativistically, Maxwell’s equation look much simpler and more elegant. We willdiscover that magnetism is entirely a relativistic effect, and makes sense only inthe context of relativity.

As we move to the world of special relativity, not only do we get a betterunderstanding of Maxwell’s equation. As a takeaway, we also get to understandrelativity itself better, as we provide a more formal treatment of special relativity,which becomes a powerful tool to develop theories consistent with relativity.

3

1 Preliminaries IB Electromagnetism

1 Preliminaries

1.1 Charge and Current

The strength of the electromagnetic force experienced by a particle is determinedby its (electric) charge. The SI unit of charge is the Coulomb. In this course, weassume that the charge can be any real number. However, at the fundamentallevel, charge is quantised. All particles carry charge q = ne for some integer n,and the basic unit e ≈ 1.6× 10−19 C. For example, the electron has n = −1,proton has n = +1, neutron has n = 0.

Often, it will be more useful to talk about charge density ρ(x, t).

Definition (Charge density). The charge density is the charge per unit volume.The total charge in a region V is

Q =

∫V

ρ(x, t) dV

When we study charged sheets or lines, the charge density would be chargeper unit area or length instead, but this would be clear from context.

The motion of charge is described by the current density J(x, t).

Definition (Current and current density). For any surface S, the integral

I =

∫S

J · dS

counts the charge per unit time passing through S. I is the current, and J isthe current density, “current per unit area”.

Intuitively, if the charge distribution ρ(x, t) has velocity v(x, t), then (ne-glecting relativistic effects), we have

J = ρv.

Example. A wire is a cylinder of cross-sectional area A. Suppose there are nelectrons per unit volume. Then

ρ = nq = −neJ = nqv

I = nqvA.

It is well known that charge is conserved — we cannot create or destroycharge. However, the conservation of charge does not simply say that “the totalcharge in the universe does not change”. We want to rule out scenarios where acharge on Earth disappears, and instantaneously appears on the Moon. So whatwe really want to say is that charge is conserved locally: if it disappears here, itmust have moved to somewhere nearby. Alternatively, charge density can onlychange due to continuous currents. This is captured by the continuity equation:

Law (Continuity equation).

∂ρ

∂t+∇ · J = 0.

4


We can write this into a more intuitive integral form via the divergencetheorem.

The charge Q in some region V is defined to be

Q =

∫V

ρ dV.

SodQ

dt=

∫V

∂ρ

∂tdV = −

∫V

∇ · J dV = −∫S

J · dS.

Hence the continuity equation states that the change in total charge in a volumeis given by the total current passing through its boundary.

In particular, we can take V = R3, the whole of space. If there are nocurrents at infinity, then

dQ

dt= 0

So the continuity equation implies the conservation of charge.

1.2 Forces and Fields

In modern physics, we believe that all forces are mediated by fields (not tobe confused with “fields” in algebra, or agriculture). A field is a dynamicalquantity (i.e. a function) that assigns a value to every point in space and time.In electromagnetism, we have two fields:

– the electric field E(x, t);

– the magnetic field B(x, t).

Each of these fields is a vector, i.e. it assigns a vector to every point in spaceand time, instead of a single number.

The fields interact with particles in two ways. On the one hand, fields causeparticles to move. On the other hand, particles create fields. The first aspect isgoverned by the Lorentz force law:

Law (Lorentz force law).F = q(E + v ×B)

while the second aspect is governed by Maxwell’s equations.

Law (Maxwell’s Equations).

∇ ·E =ρ

ε0

∇ ·B = 0

∇×E +∂B

∂t= 0

∇×B− µ0ε0∂E

∂t= µ0J,

where we have two constants of nature:

– ε0 = 8.85× 10−12 m−3 kg−1 s2 C2 is the electric constant;

5


– µ0 = 4π × 10−6 m kg C−2 is the magnetic constant.

Some prefer to call these constants the “permittivity of free space” and “per-meability of free space” instead. But why bother with these complicated andeasily-confused names when we can just call them “electric constant” and “mag-netic constant”?

We’ve just completed the description of all of (classical, non-relativistic) elec-tromagnetism. The remaining of the course would be to study the consequencesof and solutions to these equations.

6

2 Electrostatics IB Electromagnetism

2 Electrostatics

Electrostatics is the study of stationary charges in the absence of magneticfields. We take ρ = ρ(x), J = 0 and B = 0. We then look for time-independentsolutions. In this case, the only relevant equations are

∇ ·E =ρ

ε0

∇×E = 0,

and the other two equations just give 0 = 0.In this chapter, our goal is to find E for any ρ. Of course, we first start with

simple, symmetric cases, and then tackle the more general cases later.

2.1 Gauss’ Law

Here we transform the first Maxwell’s equation into an integral form, known asGauss’ Law.

Consider a region V ⊆ R3 with boundary S = ∂V . Then integrating the firstequation over the volume V gives∫

V

∇ ·E dV =1

ε0

∫V

ρ dV.

The divergence theorem gives∫V∇ · E dV =

∫S

E · dS, and by definition,Q =

∫Vρ dV . So we end up with

Law (Gauss’ law). ∫S

E · dS =Q

ε0,

where Q is the total charge inside V .

Definition (Flux through surface). The flux of E through the surface S isdefined to be ∫

S

E · dS.

Gauss’ law tells us that the flux depends only on the total charge containedinside the surface. In particular any external charge does not contribute to thetotal flux. While external charges do create fields that pass through the surface,the fields have to enter the volume through one side of the surface and leavethrough the other. Gauss’ law tells us that these two cancel each other outexactly, and the total flux caused by external charges is zero.

From this, we can prove Coulomb’s law:

Example (Coulomb’s law). We consider a spherically symmetric charge densityρ(r) with ρ(r) = 0 for r > R, i.e. all the charge is contained in a ball of radius R.

7


R

E

S

By symmetry, the force is the same in all directions and point outward radially.So

E = E(r)r.

This immediately ensures that ∇×E = 0.Put S to be a sphere of radius r > R. Then the total flux is∫

S

E · dS =

∫S

E(r)r · dS

= E(r)

∫S

r · dS

= E(r) · 4πr2

By Gauss’ law, we know that this is equal to Qε0

. Therefore

E(r) =Q

4πε0r2

and

E(r) =Q

4πε0r2r.

By the Lorentz force law, the force experienced by a second charge is

F(r) =Qq

4πε0r2r,

which is Coulomb’s law.Strictly speaking, this only holds when the charges are not moving. However,

for most practical purposes, we can still use this because the corrections requiredwhen they are moving are tiny.

Example. Consider a uniform sphere with

ρ(r) =

{ρ r < R

0 r > R.

Outside, we know that

E(r) =Q

4πε0r2r

Now suppose we are inside the sphere.

8


R

E

r

Then ∫S

E · dS = E(r)4πr2 =Q

ε0

(r3

R3

)So

E(r) =Qr

4πε0R3r,

and the field increases with radius.

Example (Line charge). Consider an infinite line with uniform charge densityper unit length η.

We use cylindrical polar coordinates:

z

r =√x2 + y2

E

By symmetry, the field is radial, i.e.

E(r) = E(r)r.

Pick S to be a cylinder of length L and radius r. We know that the end caps donot contribute to the flux since the field lines are perpendicular to the normal.Also, the curved surface has area 2πrL. Then∫

S

E · dS = E(r)2πrL =ηL

ε0.

SoE(r) =

η

2πε0rr.

Note that the field varies as 1/r, not 1/r2. Intuitively, this is because we haveone more dimension of “stuff” compared to the point charge, so the field doesnot drop as fast.

9


Example (Surface charge). Consider an infinite plane z = 0, with uniformcharge per unit area σ.

By symmetry, the field points vertically, and the field bottom is the opposite ofthat on top. we must have

E = E(z)z

withE(z) = −E(−z).

Consider a vertical cylinder of height 2z and cross-sectional area A. Now onlythe end caps contribute. So∫

S

E · dS = E(z)A− E(−z)A =σA

ε0.

SoE(z) =

σ

2ε0

and is constant.Note that the electric field is discontinuous across the surface. We have

E(z → 0+)− E(z → 0−) =σ

ε0.

This is a general result that is true for any arbitrary surfaces and σ. We can provethis by considering a cylinder across the surface and then shrink it indefinitely.Then we find that

n ·E+ − n ·E− =σ

ε0.

However, the components of E tangential to the surface are continuous.

2.2 Electrostatic potential

In the most general case, we will have to solve both ∇·E = ρ/ε0 and ∇×E = 0.However, we already know that the general form of the solution to the secondequation is E = −∇φ for some scalar field φ.

Definition (Electrostatic potential). If E = −∇φ, then φ is the electrostaticpotential.

Substituting this into the first equation, we obtain

∇2φ =ρ

ε0.

This is the Poisson equation, which we have studied in other courses. If we arein the middle of nowhere and ρ = 0, then we get the Laplace equation.

There are a few interesting things to note about our result:

10


– φ is only defined up to a constant. We usually fix this by insisting φ(r)→ 0as r →∞. This statement seems trivial, but this property of φ is actuallyvery important and gives rise to a lot of interesting properties. However,we will not have the opportunity to explore this in this course.

– The Poisson equation is linear. So if we have two charges ρ1 and ρ2, thenthe potential is simply φ1 + φ2 and the field is E1 + E2. This is theprinciple of superposition. Among the four fundamental forces of nature,electromagnetism is the only force with this property.

2.2.1 Point charge

Consider a point particle with charge Q at the origin. Then

ρ(r) = Qδ3(r).

Here δ3 is the generalization of the usual delta function for (3D) vectors.The equation we have to solve is

∇2φ = −Qε0δ3(r).

Away from the origin r = 0, δ3(r) = 0, and we have the Laplace equation. Fromthe IA Vector Calculus course, the general solution is

φ =α

rfor some constant α.

The constant α is determined by the delta function. We integrate the equationover a sphere of radius r centered at the origin. The left hand side gives∫

V

∇2φ dV =

∫S

∇φ · dS =

∫S

− αr2

r · dS = −4πα

The right hand side gives

−Qε0

∫V

δ3(r) dV = −Qε0.

So

α =Q

4πε0

and

E = −∇φ =Q

4πε0r2r.

This is just what we get from Coulomb’s law.

2.2.2 Dipole

Definition (Dipole). A dipole consists of two point charges, +Q and −Q atr = 0 and r = −d respectively.

11


To find the potential of a dipole, we simply apply the principle of superpositionand obtain

φ =1

4πε0

(Q

r− Q

|r + d|

).

This is not a very helpful result, but we can consider the case when we are far,far away, i.e. r � d. To do so, we Taylor expand the second term. For a generalf(r), we have

f(r + d) = f(r) + d · ∇f(r) +1

2(d · ∇)2f(r) + · · · .

Applying to the term we are interested in gives

1

|r + d|=

1

r− d · ∇

(1

r

)+

1

2(d · ∇)2

(1

r

)+ · · ·

=1

r− d · r

r3− 1

2

(d · dr3− 3(d · r)2

r5

)+ · · · .

Plugging this into our equation gives

φ =Q

4πε0

(1

r− 1

r+

d · rr3

+ · · ·)∼ Q

4πε0

d · rr3

.

Definition (Electric dipole moment). We define the electric dipole moment tobe

p = Qd.

By convention, it points from -ve to +ve.

Then

φ =p · r

4πε0r2,

and

E = −∇φ =1

4πε0

(3(p · r)r− p

r3

).

2.2.3 General charge distribution

To find φ for a general charge distribution ρ, we use the Green’s function for theLaplacian. The Green’s function is defined to be the solution to

∇2G(r, r′) = δ3(r− r′),

In the section about point charges, We have shown that

G(r, r′) = − 1

4π

1

|r− r′|.

We assume all charge is contained in some compact region V . Then

φ(r) = − 1

ε0

∫V

ρ(r′)G(r, r′) d3r′

=1

4πε0

∫V

ρ(r′)

|r− r′|d3r′

12


Then

E(r) = −∇φ(r)

= − 1

4πε0

∫V

ρ(r′)∇(

1

|r− r′|

)d3r′

=1

4πε0

∫V

ρ(r′)(r− r′)

|r− r′|3d3r′

So if we plug in a very complicated ρ, we get a very complicated E!However, we can ask what φ and E look like very far from V , i.e. |r| � |r′|.We again use the Taylor expansion.

1

|r− r′|=

1

r+ r′ · ∇

(1

r

)+ · · ·

=1

r+

r · r′

r3+ · · · .

Then we get

φ(r) =1

4πε0

∫V

ρ(r′)

(1

r+

r · r′

r3+ · · ·

)d3r′

=1

4πε0

(Q

r+

p · rr2

+ · · ·),

where

Q =

∫V

ρ(r′) dV ′

p =

∫V

r′ρ(r′) dV ′

r =r

‖r‖.

So if we have a huge lump of charge, we can consider it to be a point charge Q,plus some dipole correction terms.

2.2.4 Field lines and equipotentials

Vectors are usually visualized using arrows, where longer arrows represent largervectors. However, this is not a practical approach when it comes to visualizingfields, since a field assigns a vector to every single point in space, and we don’twant to draw infinitely many arrows. Instead, we use field lines.

Definition (Field line). A field line is a continuous line tangent to the electricfield E. The density of lines is proportional to |E|.

They begin and end only at charges (and infinity), and never cross.We can also draw the equipotentials.

Definition (Equipotentials). Equipotentials are surfaces of constant φ. BecauseE = −∇φ, they are always perpendicular to field lines.

Example. The field lines for a positive and a negative charge are, respectively,

13


+ -

We can also draw field lines for dipoles:

+ -

2.3 Electrostatic energy

We want to calculate how much energy is stored in the electric field. Recall fromIA Dynamics and Relativity that a particle of charge q in a field E = −∇φ haspotential energy U(r) = qφ(r).

U(r) can be thought of as the work done in bringing the particle from infinity,as illustrated below:

work done = −∫ r

∞F · dr

= −∫ r

∞E · dr

= q

∫ r

∞∇φ · dr

= q[φ(r)− φ(∞)]

= U(r)

where we set φ(∞) = 0.Now consider N charges qi at positions ri. The total potential energy stored

is the work done to assemble these particles. Let’s put them in one by one.

(i) The first charge is free. The work done is W1 = 0.

(ii) To place the second charge at position r2 takes work. The work is

W2 =q1q2

4πε0

1

|r1 − r2|.

(iii) To place the third charge at position r3, we do

W3 =q3

4πε0

(q1

|r1 − r3|+

q2

|r2 − r3|

)(iv) etc.

14


The total work done is

U =

N∑i=1

Wi =1

4πε0

∑i<j

qiqj|ri − rj |

.

Equivalently,

U =1

4πε0

1

2

∑i 6=j

qiqj|ri − rj |

.

We can write this in an alternative form. The potential at point ri due to allother particles is

φ(ri) =1

4πε0

∑j 6=i

qj|ri − rj |

.

So we can write

U =1

2

N∑i=1

qiφ(ri).

There is an obvious generalization to continuous charge distributions:

U =1

2

∫ρ(r)φ(r) d3r.

Hence we obtain

U =ε0

2

∫(∇ ·E)φ d3r

=ε0

2

∫[∇ · (Eφ)−E · ∇φ] d3r.

The first term is a total derivative and vanishes. In the second term, we use thedefinition E = −∇φ and obtain

Proposition.

U =ε0

2

∫E ·E d3r.

This derivation of potential energy is not satisfactory. The final result showsthat the potential energy depends only on the field itself, and not the charges.However, the result was derived using charges and electric potentials — thereshould be a way to derive this result directly with the field, and indeed there is.However, this derivation belongs to a different course.

Also, we have waved our hands a lot when generalizing to continuous dis-tributions, which was not entirely correct. If we have a single point particle,the original discrete formula implies that there is no potential energy. However,since the associated field is non-zero, our continuous formula gives a non-zeropotential.

This does not mean that the final result is wrong. It is correct, but itdescribes a more sophisticated (and preferred) conception of “potential energy”.Again, we shall not go into the details in this course.

15


2.4 Conductors

Definition (Conductor). A conductor is a region of space which contains lotsof charges that are free to move.

In electrostatic situations, we must have E = 0 inside a conductor. Otherwise,the charges inside the conductor would move till equilibrium. This almostdescribes the whole of the section: if you apply an electric field onto a conductor,the charges inside the conductor move until the external field is canceled out.

From this, we can derive a lot of results. Since E = 0, φ is constant insidethe conductor. Since ∇ · E = ρ/ε0, inside the conductor, we must have ρ = 0.Hence any net charge must live on the surface.

Also, since φ is constant inside the conductor, the surface of the conductor isan equipotential. Hence the electric field is perpendicular to the surface. Thismakes sense since any electric field with components parallel to the surface wouldcause the charges to move.

Recall also from our previous discussion that across a surface, we have

n ·Eoutside − n ·Einside =σ

ε0,

where σ is the surface charge density. Since Einside = 0, we obtain

Eoutside =σ

ε0n

This allows us to compute the surface charge given the field, and vice versa.

Example. Consider a spherical conductor with Q = 0. We put a positive plateon the left and a negative plate on the right. This creates a field from the left tothe right.

With the conductor in place, since the electric field lines must be perpendicularto the surface, they have to bend towards the conductor. Since field lines endand start at charges, there must be negative charges at the left and positivecharges at the right.

− +

We get an induced surface charge.

Example. Suppose we have a conductor that fills all space x < 0. We groundit such that φ = 0 throughout the conductor. Then we place a charge q atx = d > 0.

16


φ = 0 +

We are looking for a potential that corresponds to a source at x = d and satisfiesφ = 0 for x < 0. Since the solution to the Poisson equation is unique, we canuse the method of images to guess a solution and see if it works — if it does, weare done.

To guess a solution, we pretend that we don’t have a conductor. Instead, wehave a charge −q and x = −d. Then by symmetry we will get φ = 0 when x = 0.The potential of this pair is

φ =1

4πε0

[q√

(x− d)2 + y2 + z2− q√

(x+ d)2 + y2 + z2

].

To get the solution we want, we “steal” part of this potential and declare ourpotential to be

φ =

14πε0

[q√

(x−d)2+y2+z2− q√

(x+d)2+y2+z2

]if x > 0

0 if x ≤ 0

Using this solution, we can immediately see that it satisfies Poisson’s equationsboth outside and inside the conductor. To complete our solution, we need to findthe surface charge required such that the equations are satisfied on the surfaceas well.

To do so, we can calculate the electric field near the surface, and use therelation σ = ε0Eoutside · n. To find σ, we only need the component of E in the xdirection:

Ex = −∂φ∂x

=q

4πε0

(x− d|r− d|3/2

− x+ d

|r + d|3/2

)for x > 0. Then induced surface charge density is given by Ex at x = 0:

σ = Exε0 = − q

2π

d

(d2 + y2 + z2)3/2.

The total surface charge is then given by∫σ dy dz = −q.

17

3 Magnetostatics IB Electromagnetism

3 Magnetostatics

Charges give rise to electric fields, currents give rise to magnetic fields. In thissection, we study the magnetic fields induced by steady currents, i.e. in situationswhere J 6= 0 and ρ = 0. Again, we look for time-independent solutions.

Since there is no charge, we obtain E = 0. The remaining Maxwell’s equationsare

∇×B = µJ

∇ ·B = 0

The objective is, given a J, find the resultant B.Before we start, what does the condition ρ = 0 mean? It does not mean that

there are no charges around. We want charges to be moving around to createcurrent. What it means is that the positive and negative charges balance outexactly. More importantly, it stays that way. We don’t have a net charge flowfrom one place to another. At any specific point in space, the amount of chargeentering is the same as the amount of charge leaving. This is the case in manyapplications. For example, in a wire, all electrons move together at the samerate, and we don’t have charge building up at parts of the circuit.

Mathematically, we can obtain the interpretation from the continuity equa-tion:

∂ρ

∂t+∇ · J = 0.

In the case of steady currents, we have ρ = 0. So

∇ · J = 0,

which says that there is no net flow in or out of a point.

3.1 Ampere’s Law

Consider a surface S with boundary C. Current J flows through S. We nowintegrate the first equation over the surface S to obtain∫

S

(∇×B) · dS =

∮C

B · dr = µ0

∫S

J · dS.

So

Law (Ampere’s law). ∮C

B · dr = µ0I,

where I is the current through the surface.

Example (A long straight wire). A wire is a cylinder with current I flowingthrough it.

We use cylindrical polar coordinates (r, ϕ, z), where z is along the directionof the current, and r points in the radial direction.

18


I

S

z

r

By symmetry, the magnetic field can only depend on the radius, and must liein the x, y plane. Since we require that ∇ · B = 0, we cannot have a radialcomponent. So the general form is

B(r) = B(r)ϕ.

To find B(r), we integrate over a disc that cuts through the wire horizontally.We have ∮

C

B · dr = B(r)

∫ 2π

0

r dϕ = 2πrB(r)

By Ampere’s law, we have2πrB(r) = µ0I.

So

B(r) =µ0I

2πrϕ.

Example (Surface current). Consider the plane z = 0 with surface currentdensity k (i.e. current per unit length).

Take the x-direction to be the direction of the current, and the z direction to bethe normal to the plane.

We imagine this situation by considering this to be infinitely many copies ofthe above wire situation. Then the magnetic fields must point in the y-direction.By symmetry, we must have

B = −B(z)y,

with B(z) = −B(−z).Consider a vertical rectangular loop of length L through the surface

L

Then ∮C

B · dr = LB(z)− LB(−z) = µ0kL

19


So

B(z) =µ0k

2for z > 0.

Similar to the electrostatic case, the magnetic field is constant, and the partparallel to the surface is discontinuous across the plane. This is a general result,i.e. across any surface,

n×B+ − n×B− = µ0k.

Example (Solenoid). A solenoid is a cylindrical surface current, usually madeby wrapping a wire around a cylinder.

z

r

C

We use cylindrical polar coordinates with z in the direction of the extension ofthe cylinder. By symmetry, B = B(r)z.

Away from the cylinder, ∇ × B = 0. So ∂B∂r = 0, which means that B(r)

is constant outside. Since we know that B = 0 at infinity, B = 0 everywhereoutside the cylinder.

To compute B inside, use Ampere’s law with a curve C. Note that only thevertical part (say of length L) inside the cylinder contributes to the integral.Then ∮

C

B · dr = BL = µoINL.

where N is the number of wires per unit length and I is the current in each wire(so INL is the total amount of current through the wires).

SoB = µ0IN.

3.2 Vector potential

For general current distributions J, we also need to solve ∇ ·B = 0.Recall that for the electric case, the equation is ∇ · E = ρ/ε0. For the B

field, we have 0 on the right hand side instead of ρ/ε0. This is telling us thatthere are no magnetic monopoles, i.e. magnetic charges.

The general solution to this equation is B = ∇×A for some A.

Definition (Vector potential). If B = ∇×A, then A is the vector potential.

The other Maxwell equation then says

∇×B = −∇2A +∇(∇ ·A) = µ0J. (∗)

This is rather difficult to solve, but it can be made easier by noting that A isnot unique. If A is a vector potential, then for any function χ(x),

A′ = A +∇χ

20


is also a vector potential of B, since ∇× (A +∇χ) = ∇×A.The transformation A 7→ A +∇χ is called a gauge transformation.

Definition ((Coulomb) gauge). Each choice of A is called a gauge. An A suchthat ∇ ·A = 0 is called a Coulomb gauge.

Proposition. We can always pick χ such that ∇ ·A′ = 0.

Proof. Suppose that B = ∇×A with ∇·A = ψ(x). Then for any A′ = A+∇χ,we have

∇ ·A′ = ∇A +∇2χ = ψ +∇2χ.

So we need a χ such that ∇2χ = −ψ. This is the Poisson equation which weknow that there is always a solution by, say, the Green’s function. Hence we canfind a χ that works.

If B = ∇×A and ∇ ·A = 0, then the Maxwell equation (∗) becomes

∇2A = −µ0J.

Or, in Cartesian components,

∇2Ai = −µ0Ji.

This is 3 copies of the Poisson equation, which we know how to solve usingGreen’s functions. The solution is

Ai(r) =µ0

4π

∫Ji(r

′)

|r− r′|dV ′,

or

A(r) =µ0

4π

∫J(r′)

|r− r′|dV ′,

both integrating over r′.We have randomly written down a solution of ∇2A = −µ0J. However, this

is a solution to Maxwell’s equations only if it is a Coulomb gauge. Fortunately,it is:

∇ ·A(r) =µ0

4π

∫J(r′) · ∇

(1

|r− r′|

)dV ′

= −µ0

4π

∫J(r′) · ∇′

(1

|r− r′|

)dV ′

Here we employed a clever trick — differentiating 1/|r−r′| with respect to r is thenegative of differentiating it with respect to r′. Now that we are differentiatingagainst r′, we can integrate by parts to obtain

= −µ0

4π

∫ [∇′ ·

(J(r′)

|r− r′|

)− ∇

′ · J(r′)

|r− r′|

]dV ′.

Here both terms vanish. We assume that the current is localized in some regionof space so that J = 0 on the boundary. Then the first term vanishes since it isa total derivative. The second term vanishes since we assumed that the currentis steady (∇ · J = 0). Hence we have Coulomb gauge.

21


Law (Biot-Savart law). The magnetic field is

B(r) = ∇×A =µ0

4π

∫J(r′)× r− r′

|r− r′|3dV ′.

If the current is localized on a curve, this becomes

B =µ0I

4π

∮C

dr′ × r− r′

|r− r′|3,

since J(r′) is non-zero only on the curve.

3.3 Magnetic dipoles

According to Maxwell’s equations, magnetic monopoles don’t exist. However, itturns out that a localized current looks like a dipole from far far away.

Example (Current loop). Take a current loop of wire C, radius R and currentI.

Based on the fields generated by a straight wire, we can guess that B looks likethis, but we want to calculate it.

By the Biot-Savart law, we know that

A(r) =µ0

4π

∫J(r′)

|r− r′|dV =

µ0I

4π

∮C

dr′

|r− r′|.

Far from the loop, |r− r′| is small and we can use the Taylor expansion.

1

|r− r′|=

1

r+

r · r′

r3+ · · ·

Then

A(r) =µ0I

4π

∮C

(1

r+

r · r′

r3+ · · ·

)dr′.

Note that r is a constant of the integral, and we can take it out. The first 1r

term vanishes because it is a constant, and when we integrate along a closedloop, we get 0. So we only consider the second term.

We claim that for any constant vector g,∮C

g · r′ dr′ = S× g,

where S =∫

dS is the vector area of the surface bounded by C. It follows fromGreen’s theorem for the function f(r′):∮

C

f(r′)dr′ =

∫S

∇f × dS,

22


taking f(r′) = g · r′. Then∮C

g · r′ dr′ =

∫S

g × dS = S× g.

Using this, we have

A(r) ≈ µ0

4π

m× r

r3,

where

Definition (Magnetic dipole moment). The magnetic dipole moment is

m = IS.

Then

B = ∇×A =µ0

4π

(3(m · r)r−m

r3

).

This is the same form as E for an electric dipole, except that there is no 1/r2

leading term here, because we have no magnetic monopoles.

After doing it for a loop, we can do it for a general current distribution:

Example. We have

Ai(r) =µ0

4π

∫Ji(r

′)

|r− r′|dV ′

=µ0

4π

∫ (Ji(r

′)

r− Ji(r

′)(r · r′)r3

+ · · ·)

dV ′.

We will show that the first term vanishes by showing that it is a total derivative.We have

∂′j(Jjr′i) = (∂′jJj)︸︷︷︸

=∇·J=0

r′i + (∂′jr′i)︸︷︷︸

=δij

Jj = Ji.

For the second term, we look at

∂′j(Jjr′ir′k) = (∂′jJj)r

′ir′k + Jir

′k + Jkr

′i = Jkr

′i + Jir

′k.

Apply this trick to ∫Jirjr

′j dV =

∫rj2

(Jir′j − Jjr′i) dV ′,

where we discarded a total derivative ∂′j(Jjr′ir′u). Putting it back in vector

notation, ∫Jir · r′ dV =

∫1

2(Ji(r · r′)− ri(J · r′)) dV

=

∫1

2[r× (J× r′)]i dV.

So the long-distance vector potential is again

A(r) =µ0

4π

m× r

r3,

with

Definition (Magnetic dipole moment).

m =1

2

∫r′ × J(r′) dV ′.

23


3.4 Magnetic forces

We’ve seen that moving charge produces currents which generates a magneticfield. But we also know that a charge moving in a magnetic field experiences aforce F = qr×B. So two currents will exert a force on each other.

Example (Two parallel wires).

d

z

x

y

We know that the field produced by each current is

B1 =µ0I

2πrϕ.

The particles on the second wire will feel a force

F = qv ×B1 = qv ×(µ0I12πd

)y.

But J2 = nqv and I2 = J2A, where n is the density of particles and A is thecross-sectional area of the wire. So the number of particles per unit length isnA, and the force per unit length is

nAF =µ0I1I2

2πdz× y = −µ0

I1I22πd

x.

So if I1I2 > 0, i.e. the currents are in the same direction, the force is attractive.Otherwise the force is repulsive.

Example (General force). A current J1, localized on some closed curve C1, setsup a magnetic field

B(r) =µ0I14π

∮C1

dr1 ×r− r1

|r− r1|3.

A second current J2 on C2 experiences the Lorentz force

F =

∫J2(r)×B(r) dV.

While we are integrating over all of space, the current is localized at a curve C2.So

F = I2

∮C2

dr2 ×B(r2).

24


Hence

F =µ0

4πI1I2

∮C1

∮C2

dr2 ×(

dr1 ×r2 − r1

|r2 − r1|3

).

For well-separated currents, approximated by m1 and m2, we claim that theforce can be written as

F =µ0

4π∇(

3(m1 · r)(m2 · r)− (m1 ·m2)

r3

),

whose proof is too complicated to be included.

Note that we’ve spent a whole chapter discussing “magnets”, but they arenothing like what we stick on our fridges. It turns out that these “real” magnetsare composed of many tiny aligned microscopic dipoles arising from electronspin. However, obviously we do not care about magnets in real life.

25

4 Electrodynamics IB Electromagnetism

4 Electrodynamics

So far, we have only looked at fields that do not change with time. However, inreal life, fields do change with time. We will now look at time-dependent E andB fields.

4.1 Induction

We’ll explore the Maxwell equation

∇×E +∂B

∂t= 0.

In short, if the magnetic field changes in time, i.e. ∂B∂t 6= 0, this creates an E

that accelerates charges, which creates a current in a wire. This process is calledinduction. Consider a wire, which is a closed curve C, with a surface S.

We integrate over the surface S to obtain∫S

(∇×E) · dS = −∫S

∂B

∂t· dS.

By Stokes’ theorem and commutativity of integration and differentiation (assum-ing S and C do not change in time), we have∫

C

E · dr = − d

dt

∫S

B · dS.

Definition (Electromotive force (emf)). The electromotive force (emf) is

E =

∫C

E · dr.

Despite the name, this is not a force! We can think of it as the work done on aunit charge moving around the curve, or the “voltage” of the system.

For convenience we define the quantity

Definition (Magnetic flux). The magnetic flux is

Φ =

∫S

B · dS.

Then we have

Law (Faraday’s law of induction).

E = −dΦ

dt.

This says that when we change the magnetic flux through S, then a currentis induced. In practice, there are many ways we can change the magnetic flux,such as by moving bar magnets or using an electromagnet and turning it on andoff.

The minus sign has a significance. When we change a magnetic field, an emfis created. This induces a current around the wire. However, we also know thatcurrents produce magnetic fields. The minus sign indicates that the inducedmagnetic field opposes the initial change in magnetic field. If it didn’t and theinduced magnetic field reinforces the change, we will get runaway behaviour andthe world will explode. This is known as Lenz’s law.

26


Example. Consider a circular wire with a magnetic field perpendicular to it.

If we decrease B such that Φ < 0, then E > 0. So the current flows anticlockwise(viewed from above). The current generates its own B. This acts to increase Binside, which counteracts the initial decrease.

This means you don’t get runaway behaviour.

There is a related way to induce a current: keep B fixed and move wire.

Example.

d

Slide the bar to the left with speed v. Each charge q will experience a Lorentzforce

F = qvB,

in the counterclockwise direction.The emf, defined as the work done per unit charge, is

E = vBd,

because work is only done for particles on the bar.Meanwhile, the change of flux is

dΦ

dt= −vBd,

since the area decreases at a rate of −vd.We again have

E = −dΦ

dt.

Note that we obtain the same formula but different physics — we used Lorentzforce law, not Maxwell’s equation.

Now we consider the general case: a moving loop C(t) bounding a surfaceS(t). As the curve moves, the curve sweeps out a cylinder Sc.

27


S(t+ δt)

S(t)

Sc

The change in flux

Φ(t+ δt)− Φ(t) =

∫S(t+δt)

B(t+ δt) · dS−∫S(t)

B(t) · dS

=

∫S(t+δt)

(B(t) +

∂B

∂t

)· dS−

∫S(t)

B(t) · dS +O(δt2)

= δt

∫S(t)

∂B

∂t· dS +

[∫S(t+δt)

−∫S(t)

]B(t) · dS +O(δt2)

We know that S(t + δt), S(t) and Sc together form a closed surface. Since∇ ·B = 0, the integral of B over a closed surface is 0. So we obtain[∫

S(t+δt)

−∫S(t)

]B(t) · dS +

∫Sc

B(t) · dS = 0.

Hence we have

Φ(t+ δt)− Φ(t) = δt

∫S(t)

∂B

∂t· dS−

∫Sc

B(t) · dS = 0.

We can simplify the integral over Sc by writing the surface element as

dS = (dr× v) δt.

Then B · dS = δt(v ×B) · dr. So

dΦ

dt= limδ→0

δΦ

δt=

∫S(t)

∂B

∂t· dS−

∫C(t)

(v ×B) · dr.

From Maxwell’s equation, we know that ∂B∂t = −∇×E. So we have

dΦ

dt= −

∫C

(E + v ×B) dr.

Defining the emf as

E =

∫C

(E + v ×B) dr,

we obtain the equation

E = −∂Φ

∂t

for the most general case where the curve itself can change.

28


4.2 Magnetostatic energy

Suppose that a current I flows along a wire C. From magnetostatics, we knowthat this gives rise to a magnetic field B, and hence a flux Φ given by

Φ =

∫S

B · dS,

where S is the surface bounded by C.

Definition (Inductance). The inductance of a curve C, defined as

L =Φ

I,

is the amount of flux it generates per unit current passing through C. This is aproperty only of the curve C.

Inductance is something engineers care a lot about, as they need to createreal electric circuits and make things happen. However, us mathematicians findthese applications completely pointless and don’t actually care about inductance.The only role it will play is in the proof we perform below.

Example (The solenoid). Consider a solenoid of length ` and cross-sectionalarea A (with `�

√A so we can ignore end effects). We know that

B = µ0IN,

where N is the number of turns of wire per unit length and I is the current. Theflux through a single turn (pretending it is closed) is

Φ0 = µ0INA.

So the total flux isΦ = Φ0N` = µ0IN

2V,

where V is the volume, A`. So

L = µ0N2V.

We can use the idea of inductance to compute the energy stored in magneticfields. The idea is to compute the work done in building up a current.

As we build the current, the change in current results in a change in magneticfield. This produces an induced emf that we need work to oppose. The emf isgiven by

E = −dΦ

dt= −LdI

dt.

This opposes the change in current by Lenz’s law. In time δt, a charge Iδt flowsaround C. The work done is

δW = EIδt = −LI dI

dtδt.

SodW

dt= −LI dI

dt= −1

2L

dI2

dt.

29


So the work done to build up a current is

W =1

2LI2 =

1

2IΦ.

Note that we dropped the minus sign because we switched from talking aboutthe work done by the emf to the work done to oppose the emf.

This work done is identified with the energy stored in the system. Recallthat the vector potential A is given by B = ∇×A. So

U =1

2I

∫S

B · dS

=1

2I

∫S

(∇×A) · dS

=1

2I

∮C

A · dr

=1

2

∫R3

J ·A dV

Using Maxwell’s equation ∇×B = µ0J, we obtain

=1

2µ0

∫(∇×B) ·A dV

=1

2µ0

∫[∇ · (B×A) + B · (∇×A)] dV

Assuming that B ×A vanishes sufficiently fast at infinity, the integral of thefirst term vanishes. So we are left with

=1

2µ0

∫B ·B dV.

So

Proposition. The energy stored in a magnetic field is

U =1

2µ0

∫B ·B dV.

In general, the energy stored in E and B is

U =

∫ (ε0

2E ·E +

1

2µ0B ·B

)dV.

Note that while this is true, it does not follow directly from our results for puremagnetic and pure electric fields. It is entirely plausible that when both arepresent, they interact in weird ways that increases the energy stored. However,it turns out that this does not happen, and this formula is right.

4.3 Resistance

The story so far is that we change the flux, an emf is produced, and chargesare accelerated. In principle, we should be able to compute the current. But

30


accelerating charges are complicated (they emit light). Instead, we invoke a neweffect, friction.

In a wire, this is called resistance. In most materials, the effect of resistanceis that E is proportional to the speed of the charged particles, rather than theacceleration.

We can think of the particles as accelerating for a very short period of time,and then reaching a terminal velocity. So

Law (Ohm’s law).E = IR,

Definition (Resistance). The resistance is the R in Ohm’s law.

Note that E =∫

E · dr and E = −∇φ. So E = V , the potential difference.So Ohm’s law can also be written as V = IR.

Definition (Resistivity and conductivity). For the wire of length L and cross-sectional area A, we define the resistivity to be

ρ =AR

L,

and the conductivity is

σ =1

ρ.

These are properties only of the substance and not the actual shape of thewire. Then Ohm’s law reads

Law (Ohm’s law).J = σE.

We can formally derive Ohm’s law by considering the field and interactionsbetween the electron and the atoms, but we’re not going to do that.

Example.

d

z

x

y

Suppose the bar moves to the left with speed v. Suppose that the sliding barhas resistance R, and the remaining parts of the circuit are superconductorswith no resistance.

There are two dynamical variables, the position of the bar x(t), and thecurrent I(t).

If a current I flows, the force on a small segment of the bar is

F = IBy × z

So the total force on a bar isF = IB`x.

31


Somx = IB`.

We can compute the emf as

E = −dΦ

dt= −B`x.

So Ohm’s law givesIR = −B`x.

Hence

mx = −B2`2

Rx.

Integrating once gives

x(t) = −ve−B2`2t/mR.

With resistance, we need to do work to keep a constant current. In time δt,the work needed is

δW = EIδt = I2Rδt

using Ohm’s law. So

Definition (Joule heating). Joule heating is the energy lost in a circuit due tofriction. It is given by

dW

dt= I2R.

4.4 Displacement currents

Recall that the Maxwell equations are

∇ ·E =ρ

ε0

∇ ·B = 0

∇×E = −∂B

∂t

∇×B = µ0

(J + ε0

∂E

∂t

)So far we have studied all the equations apart from the µ0ε0

∂E∂t term. Historically

this term is called the displacement current.The need of this term was discovered by purely mathematically, since people

discovered that Maxwell’s equations would be inconsistent with charge conserva-tion without the term.

Without the term, the last equation is

∇×B = µ0J.

Take the divergence of the equation to obtain

µ0∇ · J = ∇ · (∇×B) = 0.

32


But charge conservation says that

ρ+∇ · J = 0.

These can both hold iff ρ = 0. But we clearly can change the charge density —pick up a charge and move it elsewhere! Contradiction.

With the new term, taking the divergence yields

µ0

(∇ · J + ε0∇ ·

∂E

∂t

)= 0.

Since partial derivatives commute, we have

ε0∇ ·∂E

∂t= ε0

∂

∂t(∇ ·E) = ρ

by the first Maxwell’s equation. So it gives

∇ · J + ρ = 0.

So with the new term, not only is Maxwell’s equation consistent with chargeconservation — it actually implies charge conservation.

4.5 Electromagnetic waves

We now look for solutions to Maxwell’s equation in the case where ρ = 0 andJ = 0, i.e. in nothing/vacuum.

Differentiating the fourth equation with respect to time,

µ0ε0∂2E

∂t2=

∂

∂t(∇×B)

= ∇× ∂B

∂t

= ∇( ∇ ·E︸︷︷︸=ρ/ε0=0

) +∇2E by vector identities

= ∇2E.

So each component of E obeys the wave equation

1

c2∂2E

∂t2−∇2E = 0.

We can do exactly the same thing to show that B obeys the same equation:

1

c2∂2B

∂t2−∇2B = 0,

where the speed of the wave is

c =1

√µ0ε0

Recall from the first lecture that

33


– ε0 = 8.85× 10−12 m−3 kg−1 s2 C2

– µ0 = 4π × 10−6 m kg C−2

Soc = 3× 108 m s−1,

which is the speed of light!We now look for plane wave solutions which propagate in the x direction,

and are independent of y and z. So we can write our electric field as

E(x) = (Ex(x, t), Ey(x, t), Ez(x, t)).

Hence any derivatives wrt y and z are zero. Since we know that ∇ ·E = 0, Exmust be constant. We take Ex = 0. wlog, assume Ez = 0, ie the wave propagatein the x direction and oscillates in the y direction. Then we look for solutions ofthe form

E = (0, E(x, t), 0),

with1

c2∂2E

∂t2− ∂2E

∂x2= 0.

The general solution is

E(x, t) = f(x− ct) + g(x+ ct).

The most important solutions are the monochromatic waves

E = E0 sin(kx− ωt).

Definition (Amplitude, wave number and frequency).

(i) E0 is the amplitude

(ii) k is the wave number.

(iii) ω is the (angular) frequency.

The wave number is related to the wavelength by

λ =2π

k.

Since the wave has to travel at speed c, we must have

ω2 = c2k2

So the value of k determines the value of ω, vice versa.

To solve for B,we use

∇×E = −∂B

∂t.

So B = (0, 0, B) for some B. Hence the equation gives.

∂B

∂t= −∂E

∂x.

34


So

B =E0

csin(kx− ωt).

Note that this is uniquely determined by E, and we do not get to choose ourfavorite amplitude, frequency etc for the magnetic component.

We see that E and B oscillate in phase, orthogonal to each other, andorthogonal to the direction of travel. These waves are what we usually considerto be “light”.

Also note that Maxwell’s equations are linear, so we can add up two solutionsto get a new one. This is particularly important, since it allows light waves topass through each other without interfering.

It is useful to use complex notation. The most general monochromatic takesthe form

E = E0 exp(i(k · x− ωt)),

andB = B0 exp(i(k · x− ωt)),

with ω = c2|k|2.

Definition (Wave vector). k is the wave vector, which is real.

The “actual” solutions are just the real part of these expressions.There are some restrictions to the values of E0 etc due to the Maxwell’s

equations:

∇ ·E = 0⇒ k ·E0 = 0

∇ ·B = 0⇒ k ·B0 = 0

∇×E = −∂B

∂t⇒ k×E0 = ωB0

If E0 and B0 are real, then k,E0/c and B0 form a right-handed orthogonal triadof vectors.

Definition (Linearly polarized wave). A solution with real E0,B0,k is said tobe linearly polarized.

This says that the waves oscillate up and down in a fixed plane.If E0 and B0 are complex, then the polarization is not in a fixed direction.

If we writeE0 = α+ iβ

for α,β ∈ R3, then the “real solution” is

Re(E) = α cos(k · x− ωt)− β sin(k · x− ωt).

Note that ∇ ·E = 0 requires that k ·α = k ·β. It is not difficult to see that thistraces out an ellipse.

Definition (Elliptically polarized wave). If E0 and B0 are complex, then it issaid to be elliptically polarized. In the special case where |α| = |β| and α ·β = 0,this is circular polarization.

We can have a simple application: why metals are shiny.A metal is a conductor. Suppose the region x > 0 is filled with a conductor.

35


Einc

A light wave is incident on the conductor, ie

Einc = E0y exp(i(kx+ ωt)),

with ω = ck.We know that inside a conductor, E = 0, and at the surface, E‖ = 0. So

E0 · y|x=0 = 0.Then clearly our solution above does not satisfy the boundary conditions!To achieve the boundary conditions, we add a reflected wave

Eref = −E0y exp(i(−kx− ωt)).

Then our total electric field is

E = Einc + Eref .

Then this is a solution to Maxwell’s equations since it is a sum of two solutions,and satisfies E · y|x=0 = 0 as required.

Maxwell’s equations says ∇×E = −∂B∂t . So

Binc =E0

cz exp(i(kx− ωt))

Bref =E0

cz exp(i(−kx− ωt))

This obeys B · n = 0, where n is the normal to the surface. But we also have

B · z|x=0− =2E0

ce−iωt,

So there is a magnetic field at the surface. However, we know that inside theconductor, we have B = 0. This means that there is a discontinuity across thesurface! We know that discontinuity happens when there is a surface current.Using the formula we’ve previously obtained, we know that the surface currentis given by

K = ±2E0

µ0cye−iωt.

So shining a light onto a metal will cause an oscillating current. We can imaginethe process as the incident light hits the conductor, causes an oscillating current,which generates a reflected wave (since accelerating charges generate light — cf.IID Electrodynamics)

We can do the same for light incident at an angle, and prove that the incidentangle is equal to the reflected angle.

36


4.6 Poynting vector

Electromagnetic waves carry energy — that’s how the Sun heats up the Earth!We will compute how much.

The energy stored in a field in a volume V is

U =

∫V

(ε0

2E ·E +

1

2µ0B ·B

)dV.

We have

dU

dt=

∫V

(ε0E ·

∂E

∂t+

1

µ0B · ∂B

∂t

)dV

=

∫V

(1

µ0E · (∇×B)−E · J− 1

µ0B · (∇×E)

)dV.

ButE · (∇×B)−B · (∇×E) = ∇ · (E×B),

by vector identities. So

dU

dt= −

∫V

J ·E dV − 1

µ0

∫S

(E×B) · dS.

Recall that the work done on a particle q moving with velocity v is δW = qv·E δt.So the J · E term is the work done on a charged particles in V . We can thuswrite

Theorem (Poynting theorem).

dU

dt+

∫V

J ·E dV︸︷︷︸Total change of energy in V (fields + particles)

= − 1

µ0

∫S

(E×B) · dS︸︷︷︸Energy that escapes through the surface S

.

Definition (Poynting vector). The Poynting vector is

S =1

µ0E×B.

The Poynting vector characterizes the energy transfer.For a linearly polarized wave,

E = E0 sin(k · x− ωt)

B =1

c(k×E0) sin(k · x− ωt).

So

S =E2

0

cµ0k sin2(k · x− ωt).

The average over T = 2πω is thus

〈S〉 =E2

0

2cµ0k.

37

5 Electromagnetism and relativity IB Electromagnetism

5 Electromagnetism and relativity

In this section, we will write Maxwell’s equations in the language of relativity,and show that the two are compatible. We will first start with a review of specialrelativity. We will introduce it in a more formal approach, forgetting about allthat fun about time dilation, twins and spaceships.

5.1 A review of special relativity

5.1.1 A geometric interlude on (co)vectors

Let’s first look at normal, Euclidean geometry we know from IA Vectors andMatrices. We all know what a vector is. A vector is, roughly, a direction. Forexample, the velocity of a particle is a vector — it points in the direction wherethe particle is moving. On the other hand, position is not quite a vector. It is avector only after we pick an “origin” of our space. Afterwards, we can think ofposition as a vector pointing the direction from the origin to where we are.

Perhaps a slightly less familiar notion is that of a covector. A covector issome mathematical object that takes in a vector and spits out a number, and ithas to do this in a linear way. In other words, given a vector space V , a covectoris a linear map V → R.

One prominent example is that the derivative df of a function f : Rn → R at(say) the origin is naturally a covector! If you give me a direction, the derivativetells us how fast f changes in that direction. In other words, given a vector v,df(v) is the directional derivative of f in the direction of v.

But, you say, we were taught that the derivative of f is the gradient, whichis a vector. Indeed, you’ve probably never heard of the word “covector” before.If we wanted to compute the directional derivative of f in the direction of v, wesimply compute the gradient ∇f , and then take the dot product ∇f · v. Wedon’t need to talk about covectors, right?

The key realization is that to make the last statement, we need the notion ofa dot product, or inner product. Once we have an inner product, it is an easymathematical fact that every covector L : V → R is uniquely of the form

L(v) = v ·w

for some fixed w ∈ V , and conversely any vector gives a covector this way.Thus, whenever we have a dot product on our hands, the notion of covector isredundant — we can just talk about vectors.

In special relativity, we still have an inner product. However, the innerproduct is not a “genuine” inner product. For example, the inner product of a(non-zero) vector with itself might be zero, or even negative! And in this case,we need to be careful about the distinction between a vector and a covector, andit is worth understanding the difference more carefully.

Since we eventually want to do computations, it is extremely useful to lookat these in coordinates. Suppose we have picked a basis e1, · · · , en. Then bydefinition of a basis, we can write any vector v as v =

∑viei. These vi are

the coordinates of v in this coordinate system, and by convention, we write theindices with superscript. We also say they have upper indices.

38


We can also introduce coordinates for covectors. If L is a covector, then wecan define its coordinates by

Li = L(ei).

By convention, the indices are now written as subscripts, or lower indices. Usingthe summation convention, we have

L(v) = L(viei) = viL(ei) = viLi.

Previously, when we introduced the summation convention, we had a “rule” thateach index can only appear at most twice, and we sum over it when it is repeated.Here we can refine this rule, and give good meaning to it:

Rule. We can only contract an upper index with a lower index.

The interpretation is that “contraction” really just means applying a covectorto a vector — a very natural thing to do. It doesn’t make sense to “apply” avector to a vector, or a covector to a covector. It also doesn’t make sense torepeat the same index three times, because we can only apply a single covectorto a single vector.

It is common to encounter some things that are neither vectors nor covectors,but we still want to apply the summation convention to them. For example,we want to write v = viei, even though ei is not a covector. It turns out in allcases we encounter, there is one choice of upper or lower indexing that makesit consistent with our summation convention. For example, we should write ei,not ei, so that v = viei works out.

We said previously that the existence of an inner product allows us to converta covector into a vector, and vice versa. So let’s see how inner products work ina choice of basis.

If the basis we picked were orthonormal, then for any vectors v and w, wesimply have v ·w = vTw. Alternatively, we have v ·w =

∑i viwi. If our basis

were not orthonormal (which is necessarily the case in SR), we can define thematrix η by

ηij = ei · ej .We will later say that η is a (0, 2)-tensor, after we defined what that means.The idea is that it takes in two vectors, and returns a number (namely theinner product of them). This justifies our choice to use lower indices for bothcoordinates. For now, we can argue for this choice by noting that the indices onei and ej are already lower.

Using this η, we can compute

v ·w = (viei) · (wjej) = viwj(ei · ej) = wiwjηij .

In other words, we havev ·w = vTηw.

We see that this matrix η encodes all the information about the inner productin this basis. This is known as the metric. If we picked an orthonormal basis,then η would be the identity matrix.

Now it is easy to convert a vector into a covector. The covector (− ·w) isgiven by v ·w = vi(wjηij). We can then read off the coordinates of the covectorto be

wi = wjηij .

39


In general, these coordinates wi are not the same as wi. This is generallytrue only if ηij is the identity matrix, i,e. the basis is orthonormal. Thus,distinguishing between vectors and covectors now has a practical purpose. Each“vector” has two sets of coordinates — one when you think of it as a vector, andone when you turn it into a covector, and they are different. So the positioningof the indices help us keep track of which coordinates we are talking about.

We can also turn a covector wi back to a vector, if we take the inverse of thematrix η, which we will write as ηij . Then

wi = wjηij .

5.1.2 Transformation rules

It is often the case that in relativity, we can write down the coordinates of anobject in any suitable basis. However, it need not be immediately clear to uswhether that object should be a vector or covector, or perhaps neither. Thus,we need a way to identify whether objects are vectors or covectors. To do so, weinvestigate how the coordinates of vectors and covectors when we change basis.

By definition, if we have a change of basis matrix P , then the coordinates ofa vector transform as v 7→ Pv, or equivalently

vi 7→ v′i = P ijvj .

How about covectors? If L is a covector and v is vector, then L(v) is a number.In particular, its value does not depend on basis. Thus, we know that the sumLiv

i must be invariant under any change of basis. Thus, if Li 7→ Li, then weknow

LiPijvj = Liv

j .

Thus, we know LiPij = Li. To obtain Li, we have to invert P ij and multiply Li

with that.However, our formalism in terms of indices does not provide us with a way

of denoting the inverse of a matrix pleasantly. We can avert this problem byfocusing on orthogonal matrices, i.e. the matrices that preserve the metric. Wesay P is orthogonal if

PT ηP = η,

or equivalently,P ijηikP

k` = ηj`.

This implies the inverse of P has coordinates

Pij = (η−1PT η)i

j = ηi`ηjkP `k,

which is the fancy way of describing the “transpose” (which is not the literaltranspose unless η = I). This is just reiterating the fact that the inverse of anorthogonal matrix is its transpose. When we do special relativity, the orthogonalmatrices are exactly the Lorentz transformations.

Thus, we find that if P is orthogonal, then covectors transform as

Li 7→ L′i = PijLj .

We can now write down the “physicists’ ” definition of vectors and covectors.Before we do that, we restrict to the case of interest in special relativity. The

40


reason is that we started off this section with the caveat “in any suitable basis”.We shall not bother to explain what “suitable” means in general, but just do itin the case of interest.

5.1.3 Vectors and covectors in SR

Recall from IA Dynamics and Relativity that in special relativity, we combinespace and time into one single object. For example, the position and time of anevent is now packed into a single 4-vector in spacetime

Xµ =

ctxyz

.

Here the index µ ranges from 0 to 3. In special relativity, we use Greek alphabets(e.g. µ, ν, ρ, σ) to denote the indices. If we want to refer to the spacial components(1, 2, 3) only, we will use Roman alphabets (e.g. i, j, k) to denote them.

As we initially discussed, position is very naturally thought of as a vector, andwe will take this as our starting postulate. We will then use the transformationrules to identify whether any other thing should be a vector or a covector.

In the “standard” basis, the metric we use is the Minkowski metric, definedby

ηµν =

+1 0 0 00 −1 0 00 0 −1 00 0 0 −1

. (∗)

This is not positive definite, hence not a genuine inner product. However, it isstill invertible, which is what our previous discussion required. This means, forexample,

X ·X = (ct)2 − (x2 + y2 + z2),

the spacetime interval.

Definition (Orthonormal basis). An orthonormal basis of spacetime is a basiswhere the metric takes the form (∗). An (orthonormal) coordinate system is achoice of orthonormal basis.

Definition (Lorentz transformations). A Lorentz transformation is a change-of-basis matrix that preserves the inner product, i.e. orthogonal matrices under theMinkowski metric.

Thus, Lorentz transformations send orthonormal bases to orthonormal bases.For example, the familiar Lorentz boost

ct′ = γ(ct− v

cx)

x′ = γ(x− v

cct)

y′ = y

z′ = z

41


is the Lorentz transformation given by the matrix

Λµν =

γ −γv/c 0 0

−γv/c γ 0 00 0 1 00 0 0 1

Other examples include rotations of the space dimensions, which are given bymatrices of the form

Λµν =

1 0 0 000 R0

,

with R a rotation matrix.We can now write down our practical definition of vectors and covectors.

Definition (Vectors and covectors). A vector is an assignment of 4 numbersV µ, µ = 0, 1, 2, 3 to each coordinate system such that under a change of basis byΛ, the coordinates V µ transform as V µ 7→ ΛµνV

ν .A covector is an assignment of 4 numbers Vµ, µ = 0, 1, 2, 3 to each coordinate

system such that under a change of basis by Λ, the coordinates Vµ transform asVµ 7→ Λµ

νVν .

Example. By assumption, the position Xµ is a vector.

Example. Suppose we have a trajectory of a particle Xµ(s) in spacetime. ThenddsX

µ(s) is also a vector, by checking it transforms.

Finally, we would want to be able to talk about tensors. For example, wewant to be able to talk about XµXν . This is an assignment of 16 numbersindexed by µ, ν = 0, 1, 2, 3 that transforms as

XµXν 7→ ΛµρΛνσX

ρXσ.

We would also like to talk about ηµν as a tensor. We can make the followinggeneral definition:

Definition (Tensor). A tensor of type (m,n) is a quantity

Tµ1···µnν1···νn

which transforms as

T ′µ1···µnν1···νn = Λµ1

ρ1 · · ·ΛµmρmΛν1

σ1 · · ·Λνnσn × T ρ1,··· ,ρmσ1,··· ,σn.

As we saw, we can change the type of a tensor by raising and lowering indicesby contracting with ηµν or its inverse. However, the total n + m will not bechanged.

Finally, we introduce the 4-derivative.

Definition (4-derivative). The 4-derivative is

∂µ =∂

∂Xµ=

(1

c

∂

∂t,∇).

42


As we previously discussed, the derivative ought to be a covector. We can alsoverify this by explicit computation using the chain rule. Under a transformationXµ 7→ X ′µ, we have

∂µ =∂

∂Xµ7→ ∂

∂X ′µ=

∂Xν

∂X ′µ∂

∂Xν= (Λ−1)νµ∂ν = Λµ

ν∂ν .

5.2 Conserved currents

Recall the continuity equation

∂ρ

∂t+∇ · J = 0.

An implication of this was that we cannot have a charge disappear on Earthand appear on Moon. One way to explain this impossibility would be that thecharge is travelling faster than the speed of light, which is something related torelativity. This suggests that we might be able to write it relativistically.

We define

Jµ =

(ρcJ

)Before we do anything with it, we must show that this is a 4-vector, i.e. ittransforms via left multiplication by Λµν .

The full verification is left as an exercise. We work out a special case tojustify why this is a sensible thing to believe in. Suppose we have a static chargedensity ρ0(x) with J = 0. In a frame boosted by v, we want to show that thenew current is

J ′µ = ΛµνJν =

(γρ0c−γρ0v

).

The new charge is now γρ0 instead of ρ0. This is correct since we have Lorentzcontraction. As space is contracted, we get more charge per unit volume. Thenthe new current is velocity times charge density, which is J′ = γρ0v.

With this definition of Jµ, local charge conservation is just

∂µJµ = 0.

This is invariant under Lorentz transformation, simply because the indices workout (i.e. we match indices up with indices down all the time).

We see that once we have the right notation, the laws of physics are so shortand simple!

5.3 Gauge potentials and electromagnetic fields

Recall that when solving electrostatic and magentostatic problems, we introducedthe potentials ϕ and A to help solve two of Maxwell’s equations:

∇×E = 0⇔ E = −∇φ∇ ·B = 0⇔ B = ∇×A.

However, in general, we do not have ∇×E = 0. Instead, the equations are

∇× E +∂B

∂t= 0, ∇ ·B = 0.

43


So we cannot use E = −∇φ directly if B changes. Instead, we define E and Bin terms of φ and A as follows:

E = −∇φ− ∂A

∂tB = ∇×A.

It is important to note that φ and A are not unique. We can shift them by

A 7→ A +∇χ

φ 7→ φ− ∂χ

∂t

for any function χ(x, t). These are known as gauge transformations. Plug theminto the equations and you will see that you get the same E and B.

Now we have ended up with four objects, 1 from φ and 3 from A, so we canput them into a 4-vector gauge potential :

Aµ =

(φ/cA

)We will assume that this makes sense, i.e. is a genuine 4-vector.

Now gauge transformations take a really nice form:

Aµ 7→ Aµ − ∂µχ.

Finally, we define the anti-symmetric electromagnetic tensor

Fµν = ∂µAν − ∂νAµ.

Since this is antisymmetric, the diagonals are all 0, and Aµν = −Aνµ. So thisthing has (4× 4− 4)/2 = 6 independent components. So it could encapsulateinformation about the electric and magnetic fields (and nothing else).

We note that F is invariant under gauge transformations:

Fµν 7→ Fµν − ∂µ∂νχ+ ∂ν∂µχ = Fµν .

We can compute components of Fµν .

F01 =1

c

∂

∂t(−Ax)− ∂φ/c

∂x=Exc.

Note that we have −Ax instead of Ax since Fµν was defined in terms of Aµ withindices down. We can similarly obtain F02 = Ey/c and F03 = Ez/c.

We also have

F12 =∂

∂x(−Ay)− ∂

∂y(−Ax) = −Bz.

So

Fµν =

0 Ex/c Ey/c Ez/c

−Ex/c 0 −Bz By−Ey/c Bz 0 −Bx−Ez/c −By Bx 0

44


Raising both indices, we have

Fµν = ηµρηνσFρσ =

0 −Ex/c −Ey/c −Ez/c

Ex/c 0 −Bz ByEy/c Bz 0 −BxEz/c −By Bx 0

So the electric fields are inverted and the magnetic field is intact (which makessense, since moving indices up and down will flip the sign of the spacial com-ponents. The electric field has one, while the magnetic field has two. So themagnetic field is swapped twice and is intact).

Both Fµν and Fµν are tensors. Under a Lorentz transformation, we have

F ′µν = ΛµρΛνσF

ρσ.

For example under rotation

Λ =

1 0 0 000 R0

,

we find that

E′ = RE

B′ = RB.

Under a boost by v in the x-direction , we have

E′x = Ex

E′y = γ(Ey − vBz)E′z = γ(Ez + vBy)

B′x = Bx

B′y = γ(By +

v

c2Ez

)B′z = γ

(Bz −

v

c2Ey

)Example (Boosted line charge). An infinite line along the x direction withuniform charge per unit length, η, has electric field

E =η

2πε0(y2 + z2)

0yz

.

The magnetic field is B = 0. Plugging this into the equation above, an observerin frame S′ boosted with v = (v, 0, 0), i.e. parallel to the wire, sees

E =ηγ

2πε0(y2 + z2)

0yz

=ηγ

2πε0(y′2 + z′2)

0y′

z′

.

45


Also,

B =ηγv

2πε0σ2(y2 + z2)

0z−y

=ηγv

2πε0σ2(y′2 + z′2)

0z′

−y′

.

In frame S′, the charge density is Lorentz contracted to γη. The magnetic fieldcan be written as

B =µ0I′

2π√y′2 + z′2

ϕ′,

where ϕ′ =1√

y′2 + z′2

0−z′y′

is the basis vector of cylindrical coordinates, and

I ′ = −γηv is the current.This is what we calculated from Ampere’s law previously. But we didn’t use

Ampere’s law here. We used Gauss’ law, and then applied a Lorentz boost.

We see that magnetic fields are relativistic effects of electric fields. They arewhat we get when we apply a Lorentz boost to an electric field. So relativity isnot only about very fast objectsTM. It is there when you stick a magnet ontoyour fridge!

Example (Boosted point charge). A boosted point charge generates a current,but is not the steady current we studied in magnetostatics. As the point chargemoves, the current density moves.

A point charge Q, at rest in frame S has

E =Q

4πε0r2r =

Q

4πε20(x2 + y2 + z2)3/2

xyz

,

and B = 0.In frame S′, boosted with v = (v, 0, 0), we have

E′ =Q

4πε0(x2 + y2 + z2)3/2

xγyγz.

.

We need to express this in terms of x′, y′, z′, instead of x, y, z. Then we get

E′ =Qγ

4πε0(γ2(x′ + vt′)2 + y′2 + z′2)3/2

x′ + vt′

y′

z′

.

Suppose that the particle sits at (−vt′, 0, 0) in S′. Let’s look at the electric at

46


t′ = 0. Then the radial vector is r′ =

x′y′z′

. In the denominator, we write

γ2x′2 + y′2 + z′2 = (γ2 − 1)x′2 + r′2

=v2γ2

c2x′2 + r′2

=

(v2γ2

c

2

cos2 θ + 1

)r′2

= γ2

(1− v2

c2sin2 θ

)r′2

where θ is the angle between the x′ axis and r′.So

E′ =1

γ2(1− v2

c2 sin2 θ)3/2 Q

4πε0r2r′.

The factor 1/γ2(

1− v2

c2 sin2 θ)3/2

squashes the electric field in the direction of

motion.This result was first discovered by Lorentz from solving Maxwell’s equations

directly, which lead to him discovering the Lorentz transformations.There is also a magnetic field

B =µ0Qγv

4π(γ2(x′ + vt′)2 + y′2 + z′2)3/2

0z′

−y′

.

Lorentz invariants

We can ask the question “are there any combinations of E and B that allobservers agree on?” With index notation, all we have to do is to contract allindices such that there are no dangling indices.

It turns out that there are two such possible combinations.The first thing we might try would be

1

2FµνF

µν = −E2

c2+ B2,

which works great.To describe the second invariant, we need to introduce a new object in

Minkowski space. This is the fully anti-symmetric tensor

εµνρσ =

+1 µνρσ is even permutation of 0123

−1 µνρσ is odd permutation of 0123

0 otherwise

This is analogous to the εijk in R3, just that this time we are in 4-dimensionalspacetime. Under a Lorentz transformation,

ε′µνρσ = ΛµκΛνλΛραΛσβεκλαβ .

47


Since εµνρσ is fully anti-symmetric, so is ε′µνρσ. Similar to what we did in R3,we can show that the only fully anti-symmetric tensors in Minkowski space aremultiples of εµνρσ. So

ε′µνρσ = aεµνρσ

for some constant a. To figure out what a is, test a single component:

ε′0123 = Λ0κΛ1

λΛ2βΛ3

βεκλαβ = det Λ.

Lorentz transformations have det Λ = +1 (rotations, boosts), or det Λ = −1(reflections, time reversal).

We will restrict to Λ such that det Λ = +1. Then a = 1, and εµνρσ isinvariant. In particular, it is a tensor.

We can finally apply this to electromagnetic fields. The dual electromagnetictensor is defined to be

Fµν =1

2εµνρσFρσ. =

0 −Bx −By −BzBx 0 Ez/c −Ey/cBy −Ez/c 0 Ex/cBz Ey/c −Ex/c 0

.

Why do we have the factor of a half? Consider a single component, say F 12.It gets contributions from both F03 and F30, so we need to average the sum toavoid double-counting. Fµν is yet another antisymmetric matrix.

This is obtained from Fµν through the substitution E 7→ cB and B 7→ −E/c.Note the minus sign!

Then Fµν is a tensor. We can construct the other Lorentz invariant usingFµν . We don’t get anything new if we contract this with itself, since Fµν F

µν =−FµνFµν . Instead, the second Lorentz invariant is

1

4FµνFµν = E ·B/c.

5.4 Maxwell Equations

To write out Maxwell’s Equations relativistically, we have to write them out inthe language of tensors. It turns out that they are

∂µFµν = µ0J

ν

∂µFµν = 0.

As we said before, if we find the right way of writing equations, they look reallysimple! We don’t have to worry ourselves with where the c and µ0, ε0 go!

Note that each law is actually 4 equations, one for each of ν = 0, 1, 2, 3.Under a Lorentz boost, the equations are not invariant individually. Instead,they all transform nicely by left-multiplication of Λνρ.

We now check that these agree with the Maxwell’s equations.First work with the first equation: when ν = 0, we are left with

∂iFi0 = µ0J

0,

where i ranges over 1, 2, 3. This is equivalent to saying

∇ ·(

E

c

)= µ0ρc,

48


or∇ ·E = c2µ0ρ =

ρ

ε0

When ν = i for some i = 1, 2, 3, we get

∂µFµi = µ0J

i.

So after some tedious calculation, we obtain

1

c

∂

∂t

(−E

c

)+∇×B = µ0J.

With the second equation, when ν = 0, we have

∂iFi0 = 0⇒ ∇ ·B = 0.

ν = i gives

∂µFµi = 0⇒ ∂B

∂t+∇×E.

So we recover Maxwell’s equations. Then we now see why the Jν term appearsin the first equation and not the second — it tells us that there is only electriccharge, not magnetic charge.

We can derive the continuity equation from Maxwell’s equation here. Since∂ν∂µF

µν = 0 due to anti-symmetry, we must have ∂νJν = 0. Recall that we

once derived the continuity equation from Maxwell’s equations without usingrelativity, which worked but is not as clean as this.

Finally, we recall the long-forgotten potential Aµ. If we define Fµν in termsof it:

Fµν = ∂µAν − ∂νAµ,

then the equation ∂µFµν = 0 comes for free since we have

∂µFµν =

1

2εµνρσ∂µFρσ =

1

2εµνρσ∂µ(∂ρAσ − ∂σAρ) = 0

by symmetry. This means that we can also write the Maxwell equations as

∂µFµν = µ0J

ν where Fµν = ∂µAν − ∂νAµ.

5.5 The Lorentz force law

The final aspect of electromagnetism is the Lorentz force law for a particle withcharge q moving with velocity u:

dp

dt= q(E + v ×B).

To write this in relativistic form, we use the proper time τ (time experienced byparticle), which obeys

dt

dτ= γ(u) =

1√1− u2/c2

.

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We define the 4-velocity U =dX

dτ= γ

(cu

), and 4-momentum P =

(E/cp

),

where E is the energy. Note that E is the energy while E is the electric field.The Lorentz force law can be written as

dPµ

dτ= qFµνUν .

We show that this does give our original Lorentz force law:When µ = 1, 2, 3, we obtain

dp

dτ= qγ(E + v ×B).

By the chain rule, since dtdτ = γ, we have

dp

dt= q(E + v ×B).

So the good, old Lorentz force law returns. Note that here p = mγv, therelativistic momentum, not the ordinary momentum.

But how about the µ = 0 component? We get

dP 0

dτ=

1

c

dE

dτ=q

cγE · v.

This says thatdE

dt= qE · v,

which is our good old formula for the work done by an electric field.

Example (Motion in a constant field). Suppose that E = (E, 0, 0) and u =(v, 0, 0). Then

md(γu)

dt= qE.

Somγu = qEt.

So

u =dx

dt=

qEt√m2 + q2E2t2/c2

.

Note that u→ c as t→∞. Then we can solve to find

x =mc2

q

(√1 +

q2E2t2

mc2− 1

)

For small t, x ≈ 12qEt

2, as expected.

Example (Motion in constant magnetic field). Suppose B = (0, 0, B). Thenwe start with

dP 0

dτ= 0⇒ E = mγc2 = constant.

So |u| is constant. Then

m∂(γu)

∂t= qu×B.

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So

mγdu

dt= qu×B

since |u|, and hence γ, is constant. This is the same equation we saw in Dynamicsand Relativity for a particle in a magnetic field, except for the extra γ term.The particle goes in circles with frequency

ω =qB

mγ.

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Documents

Part IB - Electromagnetism · 2020-01-24 · Part IB | Electromagnetism Based on lectures by D. Tong Notes taken by Dexter Chua Lent 2015 These notes are not endorsed by the lecturers,