Maxwell’s Equations in Vacuum

(1) .E = r /eo Poisson’s Equation

(2) .B = 0 No magnetic monopoles

(3) x E = -∂B/∂t Faraday’s Law

(4) x B = moj + moeo∂E/∂t Maxwell’s Displacement

Electric Field E Vm-1

Magnetic Induction B Tesla

Charge density r Cm-3

Current Density Cm-2s-1

Ohmic Conduction j = s E Electric Conductivity Siemens (Mho)

Constitutive Relations

(1) D = eoE + P Electric Displacement D Cm-2 and Polarization P Cm-2

(2) P = eo c E Electric Susceptibility c

(3) e =1+ c Relative Permittivity (dielectric function) e

(4) D = eo e E

(5) H = B / mo – M Magnetic Field H Am-1 and Magnetization M Am-1

(6) M = cB B / mo Magnetic Susceptibility cB

(7) m =1 / (1- cB) Relative Permeability m

(8) H = B / m mo m ~ 1 (non-magnetic materials), e ~ 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/o

s± surface charge density Cm-2

s± = P.n n outward normal

E+2dA = s+dA/o Gauss’ Law

E+ = s+/2o

E- = s-/2o

EDep = E+ + E- = (s++ s-)/2o

EDep = -P/o P = s+ = s-

s-

E

P s+

E+E-

Electric PolarisationDefine dimensionless dielectric susceptibility c through

P = o c EMac

EMac = E – P/o

o E = o EMac + Po E = o EMac + o c EMac = o (1 + c)EMac = oEMac

Define dielectric constant (relative permittivity) = 1 + c

EMac = E / E = e EMac

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

Electric PolarisationRewrite EMac = E – P/o as

oEMac + P = oE

LHS contains only fields inside matter, RHS fields outside

Displacement field, D

D = oEMac + P = o EMac = oE

Displacement field defined in terms of EMac (inside matter,

relative permittivity e) and E (in vacuum, relative permittivity 1).

Define

D = o E

where 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

Polarisation charge density

Charge entering xz face at y = 0: Py=0DxDz Cm-2 m2 = C

Charge leaving xz face at y = Dy: Py=DyDxDz = (Py=0 + ∂Py/∂y Dy) DxDz

Net charge entering cube via xz faces: (Py=0 - Py=Dy ) DxDz = -∂Py/∂y DxDyDz

Charge entering cube via all faces:

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

rpol = lim (DxDyDz)→0 Qpol /(DxDyDz)

-. P = rpol

Dx

Dz

Dy

z

y

x

Py=DyPy=0

Gauss’ Law in Matter

Differentiate -.P = rpol wrt time

.∂P/∂t + ∂rpol/∂t = 0

Compare to continuity equation .j + ∂r/∂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

rtot = rpol + rfree

Gauss’ Law in Matter

Inside matter

.E = .Emac = rtot/o = (rpol + rfree)/o

Total (averaged) electric field is the macroscopic field

-.P = rpol

.(oE + P) = rfree

.D = rfree

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..

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 MatterAdd 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

1

o

jE

E

EE

...

..

ttt oo

oo

t

t

oo

oo

EjB

BE

j

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

tP

jP

tooo E

jB

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 = eoE ∂D/∂t = eo ∂E/∂t

In matter D = eo e E ∂D/∂t = eo e ∂E/∂t

Displacement current exists throughout space in a changing electric field

tt

tt

t

1t

ff

f

PMf

DjHPEjM

B

EPMj

EjjjB

EjB

oo

o

oo

ooo

Maxwell’s Equations

in vacuum in matter

.E = r /eo .D = rfree Poisson’s Equation

.B = 0 .B = 0 No magnetic monopoles

x E = -∂B/∂t x E = -∂B/∂t Faraday’s Law

x B = moj + moeo∂E/∂t x H = jfree + ∂D/∂t Maxwell’s Displacement

D = eo e E = eo(1+ c)E Constitutive relation for D

H = B/(mom) = (1- cB)B/mo 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 .

rr

S ES

VV

V

)dv(1

dv .

dv ..d

rE

ES E

o

So

)( )( .

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

drA. dr

dA . x .dS

C nArA

A

C

Faraday’s Law

form aldifferenti inLaw sFaraday' t

x

.d x t

d . x d .dt

d

Theorem Stokes' d . x d .

.ddt

dd .

fluxmagnetic of change of rate minus equals emf Induced

:law of form Integral

BE

SEB

SES B

SE E

SB E

0S

SS

SC

SC

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 I

S

j

B

dℓ

S j d .d I

0.

.

S

SS

SjB

S jSB B

d -

d .d d .

o

o

jB o

Maxwell’s Equations in VacuumTake curl of Faraday’s Law

(3) x ( x E) = -∂ ( x B)/∂t

= -∂ (moeo∂E/∂t)/∂t

= -moeo∂2E/∂t2

x ( x E) = (.E) - 2E vector identity

- 2E = -moeo∂2E/∂t2 (.E = 0)

2E -moeo∂2E/∂t2 = 0 Vector wave equation

Maxwell’s Equations in VacuumPlane wave solution to wave equation

E(r, t) = Re {Eo ei(wt - 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(wt - 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

2k i.e.

r 2r k

r k .

0 . . .

e wavesD-3e wavesD1- kz

||||

||||

||||

r k

r k)r(r kr k

k.rii

rr||

r

k

Consecutive wave fronts

l

Plane waves travel parallel to wave vector k

Plane waves have wavelength 2 p /k

Maxwell’s Equations in Vacuum

Eo

Maxwell’s Equations in VacuumPlane wave solution to wave equation

E(r, t) = Eo ei(wt - k.r) Eo

constant vector

moeo∂2E/∂t2 = - moeow2E moeow2 = k2

w =±k/(moeo)1/2 = ±ck w/k = c = (moeo)-1/2 phase velocity

w = ±ck Linear dispersion relationship

w(k)

k

Maxwell’s Equations in Vacuum Magnetic component of the electromagnetic wave in vacuum

From Maxwell-Ampère and Faraday laws

x ( x B) = moeo ∂( x E)/∂t

= moeo ∂(-∂B/∂t)/∂t

= -moeo∂2B/∂t2

x ( x B) = (.B) - 2B

- 2B = -moeo∂2B/∂t2 (.B = 0)

2B -moeo∂2B/∂t2 = 0 Same vector wave equation as for E

Maxwell’s Equations in Vacuum

If E(r, t) = Eo ex ei(wt - k.r) and k || ez and E || ex (ex, ey, ez unit vectors)

x E = -ik Eo ey ei(wt - k.r) = -∂B/∂t From Faraday’s Law

∂B/∂t = ik Eo ey ei(wt - k.r)

B = (k/w) Eo ey ei(wt - k.r) = (1/c) Eo ey e

i(wt - k.r)

For this wave E || ex, B || ey, k || ez, cBo = Eo

More generally

-∂B/∂t = -i w B = x E

x E = -i k x E

-i w B = -i k x E

B = ek x E / c

ex

ey

ez

Maxwell’s Equations in Matter

Solution of Maxwell’s equations in matter for m = 1, rfree = 0, jfree = 0

EM wave in a dielectric with frequency at which dielectric is transparentMaxwell’s equations become

x E = -∂B/∂t

x H = ∂D/∂t H = B /mo D = eo e E

x B = moeo e ∂E/∂t

x ∂B/∂t = moeo e ∂2E/∂t2

x (- x E) = x ∂B/∂t = moeo e ∂2E/∂t2

-(.E) + 2E = moeo e ∂2E/∂t2 . e E = e . E = 0 since rfree = 0

2E - moeo e ∂2E/∂t2 = 0

Maxwell’s Equations in Matter

2E - moeo e ∂2E/∂t2 = 0 E(r, t) = Eo ex Re{ei(wt - k.r)}

2E = -k2E moeo e ∂2E/∂t2 = - moeo e w2E

(-k2 +moeo e w2)E = 0

w2 = k2/(moeoe) moeoe w2 = k2 k = ± w√(moeoe) k = ± √e w/c

Let e = e1 - ie2 be the real and imaginary parts of e and e = (n - ik)2

We need √ e = n - ik

e = (n - ik)2 = n2 - k2 - i 2nk e1 = n2 - k2 e2 = 2nk

E(r, t) = Eo ex Re{ ei(wt - k.r) } = Eo ex Re{ei(wt - kz)} k || ez

= Eo ex Re{ei(wt - (n - ik)wz/c)} = Eo ex Re{ei(wt - nwz/c)e- kwz/c)}

Attenuated wave with phase velocity vp = c/n

Maxwell’s Equations in MatterSolution of Maxwell’s equations in matter for m = 1, rfree = 0, jfree = s(w)E

EM wave in a metal with frequency at which metal is absorbing/reflecting

Maxwell’s equations become

x E = -∂B/∂t

x H = jfree + ∂D/∂t H = B /mo D = eo e E

x B = mo jfree + moeo e ∂E/∂t

x ∂B/∂t = mo s ∂E/∂t + moeo e ∂2E/∂t2

x (- x E) = x ∂B/∂t = mo s ∂E/∂t + moeo e ∂2E/∂t2

-(.E) + 2E = mo s ∂E/∂t + moeo e ∂2E/∂t2 . e E = e . E = 0 since rfree = 0

2E - mo s ∂E/∂t - moeo e ∂2E/∂t2 = 0

Maxwell’s Equations in Matter

2E - mo s ∂E/∂t - moeo e ∂2E/∂t2 = 0 E(r, t) = Eo ex Re{ei(wt - k.r)} k || ez

2E = -k2E mo s ∂E/∂t = mo s iw E moeo e ∂2E/∂t2 = - moeo e w2E

(-k2 -mo s iw +moeo e w2 )E = 0 s >> eo e w for a good conductor

E(r, t) = Eo ex Re{ ei(wt - √(wsmo/2)z)e-√(wsmo/2)z}

NB wave travels in +z direction and is attenuated

The skin depth d = √(2/wsmo) is the thickness over which incident radiation is attenuated. For example, Cu metal DC conductivity is 5.7 x 107 (Wm)-1

At 50 Hz d = 9 mm and at 10 kHz d = 0.7 mm

ooo iii )(12

1k k2

Bound and Free Charges

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

Free chargesConduction electrons in metals or semiconductors

Mion k melectron k MionSi ionBound electron pair

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

Bound and Free Charges

Bound chargesResonance model for uncoupled electron pairs

Mion k melectron k Mion

tt

t

t

t

t

e Em

qe )A(

m

k

e Em

qx(t)

m

k

hereafter) assumed (Re{} x(t)(t)x x(t)(t)x

solution trial }e )Re{A(x(t)

}Re{e Em

qx

m

kxx

}Re{e qEkxxmxm

o

o

o

o

ii

i

i

i

i

i

i

i

2

2

2

Bound and Free Charges

Bound chargesIn and out of phase components of x(t) relative to Eo cos(wt)

Mion k melectron k Mion

22222222

22

22222222

22

22

222

oo

o

oo

o

o

oo

iii

i

i

t)sin(t)cos(

m

qE

})}Im{eIm{A(})}Re{eRe{A(})eRe{A( x(t)

m

qE )}Im{A(

m

qE )}Re{A(

1

m

qE )A(

m

k1

m

qE )A(

o

oo

o

o

ttt

in phase out of phase

Bound and Free Charges

Bound chargesConnection to c and e

function dielectric model

Vm

q1)( 1 )(

Vm

q )}(Im{

Vm

q )}(Re{

(t)eERemV

q(t)

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

2

22

o

2t

2222

22

22222222

22

22222222

22

o

o

o

ooo

o

o

o

oo

o

i

i -i EP

1 2 3 4

4

2

2

4

6 ( )e w

/w wo

= w wo

Im{ ( )e w }

Re{ ( )e w }

Bound and Free Charges

Free chargesLet wo → 0 in c and e jpol = ∂P/∂t

tyconductivi Drude

1

V

1N qe

m

Ne

mV

q)(

mV

q

mV

q

mV

q)(

LeteVm

q

t

(t)(t)

eVm

q

t

(t)(t)

e1

Vm

q(t)

tyconductivi (t)(t)density Current

2222

free

222

free

o

2

free

o

2

pol

o

2

t

t

t

0

0

2224

23

2

2

22

22

ii

i

i

i

i

i

i

i

oo

o

ooo

ooo

i

i

i

EP

j

EP

j

EP

Ej

1 2 3 4

4

2

2

4

6

w

wo = 0

Im{ ( )s w }

( )s w

Re{ ( )e w }

Drude ‘tail’

Energy in Electromagnetic WavesRate 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(r) = q v d(r - r’)

W = fields do work on currents in integration volume

Eliminate j using modified Ampère’s law

x H = jfree + ∂D/∂t

W =

Energy in Electromagnetic WavesVector identity .(A x B) = B . () - A . () W = becomes

W =

= = -

W = = -

Local energy density

Poynting vector

Energy in Electromagnetic WavesEnergy 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 2

1

c ab

cU

densityenergy mean 2

1kz) - t(cosc ab

2

1

cU

c kz) - t(cos c

U

cHcBE kz) - t(cos c

E HcHE

2

1U

kz) - t(cos . .2

1U

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

. .2

1U

2

2-2

2

2

zyoxo))

H E

H E

H E

oo

oo

ooo

oo

ii