Maxwell’s Equations in Vacuum (1) ∇.E = ρ /εo Poisson’s Equation (2) ∇.B = 0 No magnetic monopoles (3) ∇ x E = -∂B/∂t Faraday’s Law (4) ∇ x B = µoj + µoεo∂E/∂t Maxwell’s Displacement Electric Field E Vm-1 Magnetic Induction B Tesla Charge density ρ Cm-3 Current Density Cm-2s-1 Ohmic Conduction j = σ E Electric Conductivity Siemens (Mho)

Constitutive Relations (1) D = εoE + P Electric Displacement D Cm-2 and Polarization P Cm-2 (2) P = εo χ E Electric Susceptibility χ (3) ε =1+ χ Relative Permittivity (dielectric function) ε (4) D = εo ε E (5) H = B / µo – M Magnetic Field H Am-1 and Magnetization M Am-1

(6) M = χB B / µo Magnetic Susceptibility χB (7) µ =1 / (1- χB) Relative Permeability µ (8) H = B / µ µo µ ~ 1 (non-magnetic materials), ε ~ 1 - 50

Electric Polarisation • Apply Gauss’ Law to right and left ends of polarised dielectric

• EDep = ‘Depolarising field’

• Macroscopic electric field EMac= E + EDep = E - P/εo

σ± surface charge density Cm-2 σ± = P.n n outward normal E+2dA = σ+dA/εo Gauss’ Law E+ = σ+/2εo

E- = σ-/2εo

EDep = E+ + E- = (σ++ σ-)/2εo EDep = -P/εo P = σ+ = σ-

σ-

E

P σ+

E+ E-

Electric Polarisation Define dimensionless dielectric susceptibility χ through

P = εo χ EMac EMac = E – P/εo εo E = εo EMac + P εo E = εo EMac + εo χ EMac = εo (1 + χ)EMac = εoεEMac Define dielectric constant (relative permittivity) ε = 1 + χ

EMac = E /ε E = ε EMac

Typical values for ε: silicon 11.8, diamond 5.6, vacuum 1 Metal: ε →∞ Insulator: ε∞ (electronic part) small, ~5, lattice part up to 20

Electric Polarisation Rewrite EMac = E – P/εo as

εoEMac + P = εoE

LHS contains only fields inside matter, RHS fields outside

Displacement field, D

D = εoEMac + P = εoε EMac = εoE

Displacement field defined in terms of EMac (inside matter, relative permittivity ε) and E (in vacuum, relative permittivity 1). Define

D = εoε E

where ε is the relative permittivity and E is the electric field

Gauss’ Law in Matter • Uniform polarisation → induced surface charges only

• Non-uniform polarisation → induced bulk charges also Displacements of positive charges Accumulated charges

+ + - -

P - + E

Gauss’ Law in Matter Charge entering xz face at y = 0: Px=0∆y∆z Charge leaving xz face at y = ∆y: Px=∆x∆y∆z = (Px=0 + ∂Px/∂x ∆x) ∆y∆z Net charge entering cube: (Px=0 - Px=∆x ) ∆y∆z = -∂Px/∂x ∆x∆y∆z

∆x

∆z

∆y

z

y

x

Charge entering cube via all faces:

-(∂Px/∂x + ∂Py/∂y + ∂Pz/∂z) ∆x∆y∆z = Qpol

ρpol = lim (∆x∆y∆z)→0 Qpol /(∆x∆y∆z)

-∇.P = ρpol

Px=∆x Px=0

Gauss’ Law in Matter Differentiate -∇.P = ρpol wrt time ∇.∂P/∂t + ∂ρpol/∂t = 0 Compare to continuity equation ∇.j + ∂ρ/∂t = 0

∂P/∂t = jpol

Rate of change of polarisation is the polarisation-current density Suppose that charges in matter can be divided into ‘bound’ or polarisation and ‘free’ or conduction charges

ρtot = ρpol + ρfree

Gauss’ Law in Matter Inside matter ∇.E = ∇.Emac = ρtot/εo = (ρpol + ρfree)/εo

Total (averaged) electric field is the macroscopic field -∇.P = ρpol

∇.(εoE + P) = ρfree

∇.D = ρfree

Introduction of the displacement field, D, allows us to eliminate polarisation charges from any calculation. This is a form of Gauss’ Law suitable for application in matter.

Ampère’s Law in Matter

Ampère’s Law

( )

currentssteady -non for t

1

1

0.

0..

≠−=∇⇒

=×∇∇=∇⇒

×∇=⇒=×∇

∂∂ρ

µ

µµ

j

Bj

BjjB

o

oo

Problem!

A steady current implies constant charge density, so Ampère’s law is consistent with the continuity equation for steady currents Ampère’s law is inconsistent with the continuity equation (conservation of charge) when the charge density is time dependent

Continuity equation

Ampère’s Law in Matter Add term to LHS such that taking Div makes LHS also identically equal to zero:

The extra term is in the bracket extended Ampère’s Law

Displacement current (vacuum)

( )

( )0..

0..=∇−∇

=∇+∇

×∇=+

jj?j

B?j

or

1oµ

( ) jEE

EE

...

..

−∇=

∇⇒=∇∴

=∇⇒=∇

ttt oo

oo

∂∂ε

∂∂ρε

∂∂

ρεερ

+=×∇

×∇=

+

t

t

oo

oo

∂∂εµ

µ∂∂ε

EjB

BEj

1

Ampère’s Law in Matter

• Polarisation current density from oscillation of charges in electric dipoles • Magnetisation current density variation in magnitude of magnetic dipoles

PMf jjjj ++=

t∂∂PjP=

tooo ∂∂µεµ EjB +=×∇

M = sin(ay) k

k

i j

jM = curl M = a cos(ay) i

Total current

MjM x ∇=

Ampère’s Law in Matter

∂D/∂t is the displacement current postulated by Maxwell (1862) In vacuum D = εoE ∂D/∂t = εo ∂E/∂t In matter D = εo ε E ∂D/∂t = εo ε ∂E/∂t Displacement current exists throughout space in a changing electric field

( )tt

tt

t1

t

ff

f

PMf

∂∂ε

∂∂

µ

∂∂ε

∂∂

∂∂ε

µ

∂∂µεµ

DjHPEjMB

EPMj

EjjjB

EjB

+=×∇⇒++=

−×∇

++×∇+=

+++=×∇⇒

+=×∇

oo

o

oo

ooo

Maxwell’s Equations

in vacuum in matter ∇.E = ρ /εo ∇.D = ρfree Poisson’s Equation ∇.B = 0 ∇.B = 0 No magnetic monopoles ∇ x E = -∂B/∂t ∇ x E = -∂B/∂t Faraday’s Law ∇ x B = µoj + µoεo∂E/∂t ∇ x H = jfree + ∂D/∂t Maxwell’s Displacement D = εoε E = εo(1+ χ)E Constitutive relation for D H = B/(µoµ) = (1- χB)B/µo Constitutive relation for H

Divergence Theorem 2-D 3-D • From Green’s Theorem

• In words - Integral of A.n dA over surface contour equals integral of div A over surface area

• In 3-D • Integral of A.n dA over bounding surface S equals integral of div V dV

within volume enclosed by surface S • The area element n dA is conveniently written as dS

dV .dA .V

S∫∫∫ ∇= AnA

A.n dA ∇.A dV

Differential form of Gauss’ Law • Integral form

• Divergence theorem applied to field V, volume v bounded by surface S

• Divergence theorem applied to electric field E

∫∫∫ ∇==V

SS

dv .d .dA . VSV nV V.dS ∇.V dv

o

V

d )(d .

ε

ρ∫∫ =

rrS E

S

∫∫

∫∫

=∇

∇=

VV

V

)dv(1dv .

dv ..d

rE

ES E

ρε o

Soε

ρ )( )( . rrE =∇

Differential form of Gauss’ Law (Poisson’s Equation)

Stokes’ Theorem 3-D • In words - Integral of (∇ x A).n dA over surface S equals integral of A.dr

over bounding contour C • It doesn’t matter which surface (blue or hatched). Direction of dr

determined by right hand rule.

(∇ x a) .ndA

n outward normal

dA

local value of ∇ x A

local value of A

dr A. dr ( ) dA . x .d

SC

∫∫∫ ∇= nArA

A

C

Faraday’s Law

( )

( )

form aldifferenti inLaw sFaraday' t

x

.d x t

d . x d .dtd

Theorem Stokes' d . x d .

.ddtd.d

fluxmagnetic of change of rate minus equals emf Induced:law of form Integral

∂∂

−=∇

=

∇+

∂∂

∇=−

∇=

−=

∫

∫∫

∫∫

∫∫

BE

SEB

SES B

SE E

SBE

0S

SS

SC

S

Integral form of law: enclosed current is integral dS of current density j Apply Stokes’ theorem Integration surface is arbitrary Must be true point wise

Differential form of Ampère’s Law

∫∫ ==S

S j B d .d . oenclo µµ I

S

j

B

dℓ

S j d .d =I( )

( ) 0.

.

=×∇

=×∇=

∫

∫∫∫

S

SS

SjB

S jSB B

d -

d .d d .

o

o

µ

µ

jB oµ=×∇

Maxwell’s Equations in Vacuum Take curl of both sides of 3’ (3) ∇ x (∇ x E) = -∂ (∇ x B)/∂t = -∂ (µoεo∂E/∂t)/∂t = -µoεo∂2E/∂t2 ∇ x (∇ x E) = ∇(∇.E) - ∇2E vector identity - ∇2E = -µoεo∂2E/∂t2 (∇.E = 0) ∇2E -µoεo∂2E/∂t2 = 0 Vector wave equation

Maxwell’s Equations in Vacuum Plane wave solution to wave equation E(r, t) = Re Eo ei(ωt - k.r) Eo

constant vector

∇2E =(∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2)E = -k2E ∇.E = ∂Ex/∂x + ∂Ey/∂y + ∂Ez/∂z = -ik.E = -ik.Eo ei(ωt - k.r)

If Eo || k then ∇.E ≠ 0 and ∇ x E = 0 If Eo k then ∇.E = 0 and ∇ x E ≠ 0 For light Eo k and E(r, t) is a transverse wave

λπ

λπ

2k i.e.

r 2r kr k .

0 . . .e wavesD-3e wavesD1- kz

=

=⇒=

=

+=+=→

⊥

||||

||||

||||

r kr k)r(r kr k

k.rii

r r||

r

k Consecutive wave fronts

λ

Plane waves travel parallel to wave vector k Plane waves have wavelength 2π /k

Maxwell’s Equations in Vacuum

Eo

Maxwell’s Equations in Vacuum Plane wave solution to wave equation E(r, t) = Eo ei(ωt - k.r) Eo

constant vector µoεo∂2E/∂t2 = - µoεoω2E µoεoω2 = k2

ω =±k/(µoεo)1/2 = ±ck ω/k = c = (µoεo)-1/2 phase velocity

ω = ±ck Linear dispersion relationship

ω(k)

k

Maxwell’s Equations in Vacuum Magnetic component of the electromagnetic wave in vacuum From Faraday’s law ∇ x (∇ x B) = µoεo ∂(∇ x E)/∂t = µoεo ∂(-∂B/∂t)/∂t = -µoεo∂2B/∂t2 ∇ x (∇ x B) = ∇(∇.B) - ∇2B - ∇2B = -µoεo∂2B/∂t2 (∇.B = 0) ∇2B -µoεo∂2B/∂t2 = 0 Same vector wave equation as for E

Maxwell’s Equations in Vacuum If E(r, t) = Eo ex e

i(ωt - k.r) and k || ez and E || ex (ex, ey, ez unit vectors) ∇ x E = -ik Eo ey e

i(ωt - k.r) = -∂B/∂t From Faraday’s Law

∂B/∂t = ik Eo ey ei(ωt - k.r)

B = (k/ω) Eo ey ei(ωt - k.r) = (1/c) Eo ey e

i(ωt - k.r) For this wave E || ex, B || ey, k || ez, cBo = Eo

More generally

-∂B/∂t = -iω B = ∇ x E

∇ x E = -i k x E

-iω B = -i k x E

B = ek x E / c

Maxwell’s Equations in Matter Solution of Maxwell’s equations in matter for µ = 1, ρfree = 0, jfree = 0

Maxwell’s equations become

∇ x E = -∂B/∂t

∇ x H = ∂D/∂t H = B /µo D = εoε E

∇ x B = µoεoε ∂E/∂t

∇ x ∂B/∂t = µoεoε ∂2E/∂t2

∇ x (-∇ x E) = ∇ x ∂B/∂t = µoεoε ∂2E/∂t2

-∇(∇.E) + ∇2E = µoεoε ∂2E/∂t2 ∇. ε E = ε ∇. E = 0 since ρfree = 0

∇2E - µoεoε ∂2E/∂t2 = 0

Maxwell’s Equations in Matter ∇2E - µoεoε ∂2E/∂t2 = 0 E(r, t) = Eo ex Reei(ωt - k.r)

∇2E = -k2E µoεoε ∂2E/∂t2 = - µoεoε ω2E

(-k2 +µoεoε ω2)E = 0

ω2 = k2/(µoεoε) µoεoε ω2 = k2 k = ± ω√(µoεoε) k = ± √ε ω/c Let ε = ε1 - iε2 be the real and imaginary parts of ε and ε = (n - iκ)2

We need √ε = n - iκ

ε = (n - iκ)2 = n2 - κ2 - i 2nκ ε1 = n2 - κ2 ε2 = 2nκ

E(r, t) = Eo ex Re ei(ωt - k.r) = Eo ex Reei(ωt - kz) k || ez = Eo ex Reei(ωt - (n - iκ)ωz/c) = Eo ex Reei(ωt - nωz/c)e- κωz/c) Attenuated wave with phase velocity vp = c/n

Maxwell’s Equations in Matter Solution of Maxwell’s equations in matter for µ = 1, ρfree = 0, jfree = σ(ω)E

Maxwell’s equations become

∇ x E = -∂B/∂t

∇ x H = jfree + ∂D/∂t H = B /µo D = εoε E

∇ x B = µo jfree + µoεoε ∂E/∂t

∇ x ∂B/∂t = µoσ ∂E/∂t + µoεoε ∂2E/∂t2

∇ x (-∇ x E) = ∇ x ∂B/∂t = µoσ ∂E/∂t + µoεoε ∂2E/∂t2

-∇(∇.E) + ∇2E = µoσ ∂E/∂t + µoεoε ∂2E/∂t2 ∇. ε E = ε ∇. E = 0 since ρfree = 0

∇2E - µoσ ∂E/∂t - µoεoε ∂2E/∂t2 = 0

Maxwell’s Equations in Matter ∇2E - µoσ ∂E/∂t - µoεoε ∂2E/∂t2 = 0 E(r, t) = Eo ex Reei(ωt - k.r) k || ez ∇2E = -k2E µoσ ∂E/∂t = µoσ iω E µoεoε ∂2E/∂t2 = - µoεoε ω2E (-k2 -µoσ ω +µoεoε ω2 )E = 0 σ >> εoε ω for a good conductor

E(r, t) = Eo ex Re ei(ωt - √(ωσµo/2)z)e-√(ωσµo/2)z NB wave travels in +z direction and is attenuated The skin depth δ = √(2/ωσµo) is the thickness over which incident radiation is attenuated. For example, Cu metal DC conductivity is 5.7 x 107 (Ωm)-1 At 50 Hz δ = 9 mm and at 10 kHz δ = 0.7 mm

ωσµωσµωσµ )(121k k2 iio −=−=−=

Bound and Free Charges

Bound charges All valence electrons in insulators (materials with a ‘band gap’) Bound valence electrons in metals or semiconductors (band gap absent/small ) Free charges Conduction electrons in metals or semiconductors

Mion k melectron k Mion Si ion Bound electron pair

Resonance frequency ωo ~ (k/M)1/2 or ~ (k/m)1/2 Ions: heavy, resonance in infra-red ~1013Hz Bound electrons: light, resonance in visible ~1015Hz Free electrons: no restoring force, no resonance

Bound and Free Charges

Bound charges Resonance model for uncoupled electron pairs

Mion k melectron k Mion

tt

t

t

t

t

e Emqe )A(

mk

e Emqx(t)

mk

hereafter) assumed (Re x(t)(t)x x(t)(t)xsolution trial e )ReA(x(t)

Ree Emqx

mkxx

Ree qEkxxmxm

o

o

o

o

ωω

ω

ω

ω

ω

ωωω

ωω

ωω

ω

ii

i

i

i

i

i

i

i

++

+

+

+

+

=

+Γ+−

=

+Γ+−

−=+=

=

=+Γ+

=+Γ+

2

2

2

Bound and Free Charges

Bound charges In and out of phase components of x(t) relative to Eo cos(ωt)

Mion k melectron k Mion ( )

( )

( ) ( )( ) ( ) ( )( )

( )( ) ( )( ) ( ) ( )( )

Γ+−

Γ+

Γ+−

−=

−==

Γ+−

Γ−=

Γ+−

−=

Γ+−=

=+Γ+−

=

+++

22222222

22

22222222

22

22

222

ωωωωω

ωωωωωω

ωωω

ωωωωω

ωωωωωω

ωωωω

ωωωω

ω

ωωω

oo

o

oo

o

o

oo

iii

i

i

t)sin(t)cos(m

qE

)ImeImA()ReeReA()eReA( x(t)

mqE )ImA(

mqE )ReA(

1m

qE )A(

mk1

mqE )A(

o

oo

o

o

ttt

in phase out of phase

Bound and Free Charges Bound charges Connection to χ and ε

( )( ) ( )( ) ( ) ( )( )

( ) ( )( ) ( ) ( )( )

( ) ( )( )function dielectric model

Vmq1)( 1 )(

Vmq )(Im

Vmq )(Re

(t)eERemVq(t)

qx(t)/V volume unit per moment dipole onPolarisati

2

22

o

2t

2222

22

22222222

22

22222222

22

ωωωωωω

εωχωε

ωωωω

εωχ

ωωωωω

εωχ

χεωωω

ωωωω

ωω ω

Γ+−

Γ−−+=+=

Γ+−

Γ=

Γ+−

−=

=

Γ+−

Γ−

Γ+−

−=

=≡

o

o

o

ooo

o

o

o

oo

o

i

i -i EP

1 2 3 4

4

2

2

4

6 ε(ω)

ω/ωο

ω = ωο

Imε(ω)

Reε(ω)

Bound and Free Charges Free charges Let ωο → 0 in χ and ε jpol = ∂P/∂t

( )

( )

tyconductivi Drude

1 V1N qe

mNe

mVq)(

mVq

mVq

mVq)(

LeteVm

qt(t)(t)

eVm

qt(t)(t)

e1Vm

q(t)

tyconductivi (t)(t)density Current

2222

free

222

free

o

2

free

o

2

pol

o

2

t

t

t

Γ≡≡≡≡

Γ=

Γ+Γ+−

=Γ+Γ+−

=Γ+−

=

→Γ+−

+=

∂∂

=

Γ+−+

=∂

∂=

Γ+−=

≡=

+

+

+

ττσ

ωω

ωωωω

ωωωωσ

ωωω

ωε

ε

ωωωω

εε

ωωωεε

σσ

ω

ω

ω

0

0

2224

23

2

2

22

22

iii

ii

ii

ii

oo

o

ooo

ooo

i

i

i

EPj

EPj

EP

Ej

1 2 3 4

4

2

2

4

6

ω

ωο = 0

Reσ(ω)

σ(ω)

Imε(ω)

Drude ‘tail’

Energy in Electromagnetic Waves Rate of doing work on a moving charge W = d/dt(F.dr) F = qE + qv x B = d/dt(q E + q v x B) . dr = d/dt(q E . dr) = qv . E = ∫ j . E d3r j(r) = q v δ(r - r’) W = ∫ j . E d3r fields do work on currents in integration volume Eliminate j using modified Ampère’s law ∇ x H = jfree + ∂D/∂t W = ∫ ∇ x H − ∂D/∂t . E d3r

Energy in Electromagnetic Waves Vector identity ∇.(A x B) = B . (∇ x A) - A . (∇ x B) W = ∫ ∇ x H − ∂D/∂t . E d3r becomes W = ∫ ∇.(H x E) + H . ∇ x E − E . ∂D/∂t d3r = ∫ ∇.(H x E) − H . ∂B/∂t − E . ∂D/∂t d3r ∇ x E = - ∂B/∂t

W = ∫ j . E d3r = -∫∇.(E x H) d3r − ddt∫12 H . B + D . E d3r

U = 1

2 H . B + D . E Local energy density

N = E x H Poynting vector

∂U∂t

+ ∇. N = − j . E Energy conservation

Energy in Electromagnetic Waves Energy density in plane electromagnetic waves in vacuum

( )

( )

HEN

H HE E

ekr keH r keE

E D H B

x Poynting c.f.flux energy mean 21

c abcU

densityenergy mean 21kz) - t(cosc ab

21

cU

c kz) - t(cos c

U

cHcBE kz) - t(cos c

E HcHE21U

kz) - t(cos . .21U

|| . - t(e Re H . - t(e Re E

. .21U

2

2-2

2

2

zyoxo))

==

==

==

==

+=

+=

=

=

+=

H E

H E

H E

τ

ωτ

µεω

µωµ

µµε

ωµε

ωω

oo

oo

ooo

oo

ii