Lecture 8. Relativity and electromagnetism

1. Force per unit charge −→ f = q(E+ v ∧B)

Change frame: f → f ′, hence:

Transformation of electromagnetic field

E′∥ = E∥

E′⊥ = γ (E⊥ + v ∧B) ,

B′∥ = B∥

B′⊥ = γ

(B⊥ − v ∧ E/c2

),

where S′ has velocity v in S.

e.g. capacitor, particle beam

2. Correct understanding of J = (ρc, j) and charge conservation.

3. Fields due to a moving point charge:

E′ =γQr′

4πϵ0(γ2(x′)2 + (y′)2 + (z′)2)3/2,

B′ = −v ∧ E′

c2.

S

S

Electrons (dots) and positively charged metal ions in acurrent-carrying wire.

v

Electric field

Magnetic field

Figure 1: “B of the Bang”, a sculpture in Manchester, England, designed by Thomas Heatherwick.The sculpture draws its inspiration from the explosive start of a sprint race at the “Bang” of thestarting pistol, but to a physicist it is also reminiscent of the electric field due to a fast movingcharged particle: perhaps a muon arriving in Manchester from a cosmic ray event. [Photo by NickSmale.]

Lecture 9. 4-vector potential and Maxwell’s equations

1. Lorentz covariance of Maxwell’s equations2. Scalar and vector potential

B = ∇ ∧A,

E = −∇ϕ− ∂A

∂t

−→ automatically satisfy M2, M3.

3. Gauge transformation:

A → A+∇χ,

ϕ → ϕ− ∂χ∂t

4. 4-vector potential

A ≡(ϕ/c

A

), Gauge transformation: A → A+χ

5. Lorenz gauge, Maxwell’s equations in a manifestly covariant form:

Maxwell’s equations (!)

2A =−1

c2ϵ0J, with · A = 0.

6. Potentials of a uniformly moving charged particle (located at r0 = (vt, 0, 0)):

ϕ =q

4πϵ0

γ

(γ2(x− vt)2 + y2 + z2)1/2,

A = vϕ/c2.

Maxwell equations and Lorentz force equation:

∇ · E =ρ

ϵ0(M1)

∇ ·B = 0 (M2)

∇ ∧ E = −dB

dt(M3)

c2∇ ∧B =j

ϵ0+

dE

dt, (M4)

f = q(E + v ∧B)

Maxwell equations and Lorentz force equation:

∇ · E =ρ

ϵ0(M1)

∇ ·B = 0 (M2)

∇ ∧ E = −dB

dt(M3)

c2∇ ∧B =j

ϵ0+

dE

dt, (M4)

f = q(E + v ∧B)

Change frame:ρc

jxjyjz

= Λ−1

ρ′c

j′xj′yj′z

,

E∥ = E′∥

E⊥ = γ (E′⊥ − v ∧B′) ,

B∥ = B′∥

B⊥ = γ(B′

⊥ + v ∧ E′/c2).

∂(· · · )∂x

=∂(· · · )∂t′

∂t′

∂x+

∂(· · · )∂x′

∂x′

∂x+

∂(· · · )∂y′

∂y′

∂x+

∂(· · · )∂z′

∂z′

∂x∂(· · · )∂y

= etc.

Lecture 10. Wave equation and general solution of Maxwell’s equations

1. Poisson equation and its solution (reminder) (‘Green’s method’):

∇2ϕ =−ρ

ϵ0, ϕ(r) =

∫ρ(rs)

4πϵ0|r− rs|dVs.

2. How to calculate ∇2(1/r)

3. Wave equation and its (‘retarded’) solution

−1

c2∂2ϕ

∂t2+∇2ϕ =

−ρ(r, t)

ϵ0, ϕ(r, t) =

∫ρ(rs, t− |r− rs|/c)

4πϵ0|r− rs|dVs.

N.B. ‘source event’, ‘field event’.

Hence

A =1

4πϵ0c

∫J(rs, t− rsf/c)

rsfdVs.

4. Potentials of an arbitrarily moving charged particle

A =q

4πϵ0

U/c

(−R · U). (‘Lienard-Wiechart’ potentials)

ρr

rs

O

ρr

rs

O

.

s

r

rsf

O

(t)

rs

∆ = /crsf

( -t ∆)

advancedretarded

Lecture 11. Electromagnetic radiation

Fields of an accelerated charge:

E =q

4πϵ0κ3

(n− v/c

γ2r2+

n ∧ [(n− v/c) ∧ a]

c2r

)B = n ∧ E/c

where n = r/r, κ = 1− vr/c = 1− n · v/c

In terms of the displacement r0 = r− vr/c from the projected position,

E =q

4πϵ0r30(γ2 cos2 θ + sin2 θ)3/2

(γr0 +

γ3

c2r ∧ [r0 ∧ a]

)1. General features: bound field, radiative field

2. Case of linear motion coming to rest

3. Radiation from slowly moving dipole oscillator

A =q

4πϵ0c2v

(rsf − rsf · v/c),

⇒ far field: B =ω2qx04πϵ0c3

sin θ

rsin(kr − ωt) ϕ,

E = cB

4. The half-wave dipole antenna. Emitted power = I2rms × (73 ohm).

r

a

v

field point

EE0

rad

where the particle is now

s

α

r

vtθ

r0

sP

f

Projected position: vt from source point, where t = r/c

r0 = vector from projected position to field point: r0 = r− vr/c.

Then

R · U = γ(−rc + r · v)= . . .

= r0c(γ2 cos2 θ + sin2 θ

)1/2

r

a

r0

v

s

EE

0

rad

θ

y

x

x

t

Electric field due to a particle that was at rest, then was nudged to the right, andnow is at rest again.

Lecture 12. Radiated power

1. Radiated power (Larmor)

dP = Nr2dΩ =q2

4πϵ0

a2 sin2 θ

4πc3dΩ ⇒ PL =

2

3

q2

4πϵ0

a20c3.

2. Linear particle accelerator: a0 = f0/m = f/m, −→ loss ≃ 0

3. Dipole oscillator

PL =2

3

q2

4πϵ0c3(γ3ω2x0 cosωt)

2, ⇒ PL ≃ 1

3

q2

4πϵ0c3ω4x20

4. Headlight effect:received energy per unit time at the detector, per unit solid angle

dPdΩ

=q2a2

4πϵ0c3sin2 θ

(1− (v/c) cos θ)6for linear motion

5. Circular motion: a0 = f0/m = γf/m = γ2a, ∆E =q2

3ϵ0rγ4(v/c)3.

6. Synchrotron radiation: ‘lighthouse’ pulses with frequency spread ∆ω ≃ γ3ω0.

(7. Self-force and radiation reaction)

a

x=−ct

Electric field of a charge undergoing hyperbolic motion

. . .For a sphere of radius b: Fself =2

3

q2

4πϵ0c2

(−U

b+ Uc2

)− PL

c2U

Lecture 13. Tensors

1. General idea of 3-dimensional tensors such as conductivity, susceptibility, . . . .

2. Outer product ⇒ F′ = ΛFΛT and A = FgB ≡ F · B

3. Vector product, e.g. L = XPT − PXT

4. Transformation of an antisymmetric tensor:

F =

0 ax ay az

−ax 0 bz −by−ay −bz 0 bx−az by −bx 0

, F′ = ΛFΛT ⇒

a′∥ = a∥,

a′⊥ = γ(a⊥ + v ∧ b/c),b′∥ = b∥,

b′⊥ = γ(b⊥ − v ∧ a/c).

5. Differentiation: (i) AT − (AT )T ‘4-curl’this is ∂aAb − ∂bAa in index notation(ii) · T = A divergence of a tensor makes a 4-vectorthis is ∂µT b

µ = Ab in index notation

6. Covariant/contravariant.

X → ΛX “contravariant”(gX) → (Λ−1)T (gX) “covariant”

We have been using contravariant 4-vectors all along, and g is covariant.

AT =

− 1

c∂At

∂t −1c∂Ax

∂t −1c∂Ay

∂t −1c∂Az

∂t

∂At

∂x∂Ax

∂x∂Ay

∂x∂Az

∂x

∂At

∂y∂Ax

∂y∂Ay

∂y∂Az

∂y

∂At

∂z∂Ax

∂z∂Ay

∂z∂Az

∂z

.

a2

b2

v

1

b

a1

Each bi is orthogonal to all the aj except one.

Lecture 14. Index notation

1. Basic idea: ϕ, Aa, Fab, gab, Λab

2. Summation convention: AabXb means3∑

b=0

AabXb . . . ‘dummy’ index

[3. Contravariant/covariant. ATgB = A′Tg′B′ ⇒ g′ = (Λ−1)TgΛ−1

Contravariant: X → ΛXCovariant: (gX) → (Λ−1)T (gX)

]4. Index lowering: Fa ≡ gaµF

µ so U · F = UλFλ

5. Legal tensor operations: sum, outer product, contract.

6. Caution when comparing with matrix notation

AaλBλ ↔ A · Bbut AλaBλ = BλAλa ↔ B · A

7. Invariants: contract down to a scalar. e.g. AλBλ, T λλ, T µνTµν

8. Differentiation. ∂a ≡ ∂∂xa = (1c

∂∂t ,

∂∂x ,

∂∂y ,

∂∂z ). Thus ∂a = a = gaλλ and ∂a = a.

↔ ∂a. e.g. continuity equation ∂λJλ = 0

Product rule: ∂a(U bV c) = (∂aU b)V c + U b(∂aV c)

• Matrix notation can only conveniently handle scalars, vectors and 2nd

rank tensors

• Even at 2nd rank, index notation allows some manipulations that are

hard to express using matrices

• Index notation makes it easy to find invariants

• We will only need some simple examples (explained in lectures).

Tips

1. Name your indices sensibly; make repeated indices easy to spot.

2. Look for scalars. e.g. FλµAabFλµ is sAa

b where s = FλµFλµ.

3. You can always change the names of dummy (summed over) indices; if

there are two or more, you can swap names.

4. The ‘see-saw rule’

AλBλ = AλBλ (works for any rank)

5. In the absence of differential operators, everything commutes.

Lecture 15. Spin; parity inversion symmetry

1. “Dual” Fab =12ϵabµνF

µν; case of symmetric and antisymmetric tensors

2. Total angular momentum Jab = Sab + Lab

Pauli-Lubanski vector Wa ≡ JaλPλ =

1

2ϵaλµνJ

µνPλ ⇒ Wa = (s · p, (E/c)s)

In the rest frame: (0, mcs0) so W · U = 0 and

s∥ = s0∥, s⊥ =s0⊥γ

Hence for a photon, W is null and points along P.

4. Helicity s∥ = s · p/p (= ±~ for a photon)

5.dSa

dτ=

SλUλ

c2Ua when s0 is constant.

6. Mirror reflection; polar and axial vectors

7. Parity inversion: x → −x, y → −y, z → −z

x → −x, p → −p ⇒ x ∧ p → x ∧ p

8. Classical physics ‘covariant’ under parity inversion

x

L

L x

xL L

x

Lecture 16. Lagrangian mechanics (symmetry again!)

1. Reminder of Least Action and Euler-Lagrange equations

Lagrangian L = L(qi, qi, t) ≡ T − V

Action S[q(t)] =

∫ q2,t2

q1,t1

L(q, q, t)dt

stationary for path satisfying Euler-Lagrange equations:d

dt

(∂L∂qi

)=

∂L∂qi

.

∂L∂qi

= “generalized force”,∂L∂qi

= “canonical momentum”

Hamiltonian: H(q, p, t) ≡n∑i

piqi − L(q, q, t)

2. Special Relativity (version 1).Freely moving particle: L = −mc2/γ, p = γmvParticle in an e-m field: L = −mc2/γ + q(−ϕ+ v ·A), p = γmv + qA

Taking a derivative along the worldline:dA

dt=

∂A

∂t+ (v ·∇)A

Hamiltonian H = γmc2 + qϕ =((p− qA)2 c2 +m2c4

)1/2+ qϕ.

3. Special Relativity (version 2), using τ instead of t in the action:

L(X,U) = −mc(−U · U)1/2 + qU · A, S[X(τ)] =

∫ (X2)

(X1)

L(X,U, τ)dτ,

d

dτ

∂L∂Ua

=∂L∂Xa

, Pa = mUa + qAa, mdU

dτ= q( ∧ A) · U

Use of a parameter to minimize the action.

Integrating with respect to proper time means the value of τ at the end

event is different for each path.

Problem!: calculus of variations needs fixed start and end values.

Introduce a parameter λ:∫L(X, X, τ )dτ =

∫ λ2

λ1

Ldτ

dλdλ.

Now the Lagrangian is

L = Ldτ

dλ= L1

c

(−gµν

dXµ

dλ

dXν

dλ

)1/2

.

(using dτ 2 = dt2 − (dx2 + dy2 + dz2)/c2)

giving Euler-Lagrange equations

d

dλ

∂L∂Xa

=∂L∂Xa

in which X = dX/dλ.

Now pick λ = τ along the solution worldline.

For that worldline, and for that worldline only (but it is the only one we areinterested in from now on), we must then find dτ/dλ = 1 and L = L andXa = Ua.

NowdAa

dτ= Uλ∂λAa

so

mdUa

dτ= q ((∂aAλ)− (∂λAa))U

λ

ordP

dτ= q( ∧ A) · U

Lecture 17. Electromagnetic field theory via field tensor F

1.Lorentz covariance,simplicity

⇒

4-force = charge × field × 4-velocity

F = qF · U Field tensor F2. Pure force (F · U = 0) ⇔ F = −FT

⇒ F =

0 Ex/c Ey/c Ez/c

−Ex/c 0 Bz −By

−Ey/c −Bz 0 Bx

−Ez/c By −Bx 0

.

3. propose field equation · F = −µ0ρ0U, i.e. ∂λFλb = −µ0ρ0Ub

4. need a further equation; try F = ∧ A, i.e. Fab = ∂aAb − ∂bAa

⇒ ∂cFab + ∂aFbc + ∂bFca = 0.⇒ The physical world ?

Implications5. Antisymmetric F ⇒ charge conservation: ∂µ∂νFµν = 0 ⇒ ∂λJ

λ = 0

6. Invariants D ≡ 1

2FµνFµν = B2 − E2/c, α ≡ 1

4FµνFµν = B · E/c.

e.g. orthogonal in one frame ⇒ orthogonal in all (α = 0)purely magnetic in one frame ⇒ not purely electric in another (D > 0).

7. Finding the frame (if there is one) in which B or E vanishes.

v

v

f12f2121

Figure 2: Two charged particles move on orthogonal trajectories. At the mo-

ment when one passes directly in front of the other, the forces are not in

opposite directions (the electric contributions are opposed, but the magnetic

contributions are not). Does this mean momentum is not conserved?

v

v

f12f2121

Figure 3: Same as previous figure, but showing the force on (=rate of injection

of momentum into) the field as well as the particles.

Lecture 18. Energy-momentum flow; stress-energy tensor

1. Conservation of energy: −∂u

∂t= ∇ ·N+ E · j.

energy density u = 12ϵ0E

2 + 12ϵ0c

2B2

Poynting vector N = ϵ0c2E ∧B.

follow from Maxwell’s equations

2. Transfer of 4-momentum per unit volume from fields to matter:Let W = (E · j/c, ρE+ j ∧B) then

Wb = − ∂λTλb [ W = − · T ,

and one finds (by using Maxwell’s equations):

Tab =

(u N/c

N/c σij

)where σij = uδij − ϵ0(EiEj + c2BiBj)

Conservation of 4-momentum of matter and field together(E · j/c, ρE+ j ∧B

)= −

(1

c

∂

∂t, ∇·

)(u N/c

N/c σ

)

Poynting’s argument (John Henry Poynting (1852-1914)):

We want to find expressions for u and N, such that −∂u

∂t= ∇ ·N+ E · j.

Using M4 to express j in terms of the fields:

E · j = ϵ0c2E · (∇ ∧B)− ϵ0E · ∂E

∂t.

but for any pair of vectors E, B,

∇ · (E ∧B) = B · (∇ ∧ E)− E · (∇ ∧B).

so

E · j = −ϵ0c2∇ · (E ∧B) + ϵ0c

2B · (∇ ∧ E)− ∂

∂t(12ϵ0E · E).

Now use M3:

E · j = −ϵ0c2∇ · (E ∧B)− ∂

∂t

(12ϵ0c

2B ·B+ 12ϵ0E · E

)Which shows that a possible assignment is

u = 12ϵ0c

2B2 + 12ϵ0E

2 ,

N = ϵ0c2E ∧B.

Figure 4: Freely moving charged sphere. Shading = u, arrows = N.

Figure 5: u and N for a charged sphere moving at constant velocity in a uniformapplied electric field (e.g. a viscous drag opposes electric force qE).

The argument regarding flow of momentum:

Basic idea:(Rate of decrease of

momentum in a volume

)=

(rate of momentum

flow out

)+

(rate of transferto particles

)−∂g

∂tdV =

(divergence

of something

)× dV + (ρE+ j ∧B) dV

Momentum per unit volume in the electromagnetic field is g =N

c2= ϵ0E ∧B

Use M3 and M4:

−∂g

∂t= −ϵ0

(E ∧ ∂B

∂t+

∂E

∂t∧B

)= ϵ0

(E ∧ (∇ ∧ E)− c2(∇ ∧B) ∧B

)+j ∧B

Add ρE− ϵ0(∇ · E)E (= 0 says M1) to the right hand side:

−∂g

∂t= −ϵ0

[(∇ · E)E+ (∇ ∧ E) ∧ E+ c2(∇ ∧B) ∧B

]+ ρE+ j ∧B

Defineσij = 1

2ϵ0(E2 + c2B2)δij − ϵ0(EiEj + c2BiBj),

then by differentiating it one can show that

= −ϵ0[(∇ · E)E+ (∇ ∧ E) ∧ E+ c2(∇ ∧B) ∧B

]i=

(∂σix∂x

+∂σiy∂y

+∂σiz∂z

)

Lecture 18. Energy-momentum flow; stress-energy tensor

1. Conservation of energy: −∂u

∂t= ∇ ·N+ E · j.

energy density u = 12ϵ0E

2 + 12ϵ0c

2B2

Poynting vector N = ϵ0c2E ∧B.

follow from Maxwell’s equations

2. Transfer of 4-momentum per unit volume from fields to matter:Let W = (E · j/c, ρE+ j ∧B) then

Wb = − ∂λTλb [ W = − · T ,

and one finds (by using Maxwell’s equations):

Tab =

(u N/c

N/c σij

)where σij = uδij − ϵ0(EiEj + c2BiBj)

This can also be written

Tab = ϵ0c2

(−FaµF b

µ − 1

2gabD

), where D = 1

2FµνFµν.

[ i.e. T = ϵ0c2

(−F · F− 1

2gD

). ]

Conservation of 4-momentum of matter and field together(E · j/c, ρE+ j ∧B

)= −

(1

c

∂

∂t, ∇·

)(u N/c

N/c σ

)

T11

T TT12 1310

T21

T TT22 2320

T31

T TT32 3330

T01

T TT02 0300

energydensity energy flux

momentumdensity

pressure

momentumflux

sheerstress

v

v

f12f2121

Dust

ρ0c

2

00

0

, T ab = ρ0UaUb

Ideal fluid

ρ0c

2

pp

p

, T ab = (ρ0 + p/c2)UaUb + pgab

Capacitor 12ϵ0E

2

1

−11

1

Solenoid 12ϵ0c

2B2

1

−11

1

Plane wave ϵ0E20 cos

2(kx− ωt)

1 11 1

00

= ϵ0c2(E2/ω2)KaKb

Point chargeq2

16π2ϵ0r6

r2

20 0 0

0 r2

2− x2 −xy −xz

0 −yx r2

2− y2 −yz

0 −zx −zy r2

2− z2

Table 1: Example stress-energy tensors.