maxwell’s equations in vacuum - trinity college, dublin · maxwell’s equations in vacuum e....
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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
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Bound and Free Charges
Bound charges In and out of phase components of x(t) relative to Eo cos(ωt)
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Bound and Free Charges Bound charges Connection to χ and ε
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Bound and Free Charges Free charges Let ωο → 0 in χ and ε jpol = ∂P/∂t
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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
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