maxwell’s equations in vacuum
DESCRIPTION
Maxwell’s Equations in Vacuum. (1) . E = r / e o Poisson’s Equation (2) . B = 0 No magnetic monopoles (3) x E = - ∂ B / ∂ t Faraday’s Law (4) x B = m o j + m o e o ∂ E / ∂ t Maxwell’s Displacement Electric Field E Vm -1 Magnetic Induction B Tesla - PowerPoint PPT PresentationTRANSCRIPT
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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)
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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
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Electric Polarisation• Apply Gauss’ Law to right and left ends of polarised dielectric
• EDep = ‘Depolarising field’
• Macroscopic electric field EMac= E + EDep = E - P/eo
s± surface charge density Cm-2
s± = P.n n outward normalE+2dA = s+dA/eo Gauss’ LawE+ = s+/2eo
E- = s-/2eo
EDep = E+ + E- = (s++ s-)/2eo EDep = -P/eo P = s+ = s-
s-
E
P s+
E+E-
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Electric PolarisationDefine dimensionless dielectric susceptibility c through
P = eo c EMac
EMac = E – P/eo
eo E = eo EMac + Peo E = eo EMac + eo c EMac = eo (1 + c)EMac = eoeEMac
Define dielectric constant (relative permittivity) e = 1 + c
EMac = E /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
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Electric PolarisationRewrite EMac = E – P/eo as
eoEMac + P = eoE
LHS contains only fields inside matter, RHS fields outside
Displacement field, D
D = eoEMac + P = eoe EMac = eoE
Displacement field defined in terms of EMac (inside matter,relative permittivity e) and E (in vacuum, relative permittivity 1). Define
D = eoe E
where e is the relative permittivity and E is the electric field
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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
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Gauss’ Law in MatterPolarisation charge density
Charge entering xz face at y = 0: Py=0DxDz Cm-2 m2 = CCharge 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
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x
Py=DyPy=0
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Gauss’ Law in MatterDifferentiate -.P = rpol wrt time.∂P/∂t + ∂rpol/∂t = 0Compare 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’ orpolarisation and ‘free’ or conduction charges
rtot = rpol + rfree
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Gauss’ Law in MatterInside matter.E = .Emac = rtot/eo = (rpol + rfree)/eo
Total (averaged) electric field is the macroscopic field-.P = rpol
.(eoE + P) = rfree
.D = rfree
Introduction of the displacement field, D, allows us to eliminatepolarisation charges from any calculation. This is a form of Gauss’ Law suitable for application in matter.
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Ampère’s Law in Matter
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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
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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
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Ampère’s Law in Matter
• Polarisation current density from oscillation of charges in electric dipoles• Magnetisation current density variation in magnitude of magnetic dipoles
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Total current
MjM x
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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
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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 = eoe E = eo(1+ c)E Constitutive relation for D
H = B/(mom) = (1- cB)B/mo Constitutive relation for H
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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
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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
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Differential form of Gauss’ Law(Poisson’s Equation)
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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
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Faraday’s Law
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Theorem Stokes' d . x d .
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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
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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
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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
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Consecutive wave fronts
Plane waves travel parallel to wave vector k
Plane waves have wavelength 2 /k
Maxwell’s Equations in Vacuum
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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
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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
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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 = -iw B = x E
x E = -i k x E
-iw B = -i k x E
B = ek x E / c
ex
ey
ez
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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 = eoe E
x B = moeoe ∂E/∂t
x ∂B/∂t = moeoe ∂2E/∂t2
x (- x E) = x ∂B/∂t = moeoe ∂2E/∂t2
-(.E) + 2E = moeoe ∂2E/∂t2 . e E = e . E = 0 since rfree = 0
2E - moeoe ∂2E/∂t2 = 0
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Maxwell’s Equations in Matter2E - moeoe ∂2E/∂t2 = 0 E(r, t) = Eo ex Re{ei(wt - k.r)}
2E = -k2E moeoe ∂2E/∂t2 = - moeoe w2E
(-k2 +moeoe 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
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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 = eoe E
x B = mo jfree + moeoe ∂E/∂t
x ∂B/∂t = mos ∂E/∂t + moeoe ∂2E/∂t2
x (- x E) = x ∂B/∂t = mos ∂E/∂t + moeoe ∂2E/∂t2
-(.E) + 2E = mos ∂E/∂t + moeoe ∂2E/∂t2 . e E = e . E = 0 since rfree = 0
2E - mos ∂E/∂t - moeoe ∂2E/∂t2 = 0
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Maxwell’s Equations in Matter2E - mos ∂E/∂t - moeoe ∂2E/∂t2 = 0 E(r, t) = Eo ex Re{ei(wt - k.r)} k || ez
2E = -k2E mos ∂E/∂t = mos iw E moeoe ∂2E/∂t2 = - moeoe w2E
(-k2 -mos iw +moeoe w2 )E = 0 s >> eoe 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
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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
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Bound and Free Charges
Bound chargesResonance model for uncoupled electron pairs
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Bound and Free Charges
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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 =
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Energy in Electromagnetic WavesVector identity .(A x B) = B . () - A . () W = becomes
W =
= = -
W = = -
Local energy density
Poynting vector
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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 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
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