Experimental bounds on Lorentz violation in the neutrino sector and baryonic asymmetry
Lorentz Oscillator model of dilute matter Free...
Transcript of Lorentz Oscillator model of dilute matter Free...
Lorentz Oscillator model of dilute matter
electron bound to nucleus by spring (spring const Ks)
N oscillators per unit volume
x describes the position of an e− in a 1-D potential well V (x)
γ is a frictional damping term, force prop to velocity Fd = mγv
FE is the forcing function from an optical E-field
FE
m=
e
m
E0
2(e−iωt + eiωt)
Force is gradient
of potential
V(x)= ω0x212
Linear restoring force for quadratic potential well V (x) = 12ω20x
2
Fs = ma = mdV
dx= mω2
0x = Ksx ⇒ a =dV
dx= ω2
0x ≡ acceleration
Eqn of motion derived by balancing forces ma + Fd + Fs = FE
x + γx + ω20x =
e
m
E0
2(e−iωt + eiωt)
ma+mγv+mω20x=eE0 cosω0t F = qE
Kelvin Wagner, University of Colorado Physical Optics 2018 1
Free Oscillation
x + γx + ω20x = 0
soln is of the form of damped exponentialx = Aeκt x = κAeκt x = κ2Aeκt
⇒ (κ2 + γκ + ω20)Ae
κt = 0
For nontrivial soln ( ) = 0 ω′ frequency pulling
κ = −b
2± 1
2
√b2 − 4ac = −γ
2± 1
2
√γ2 − 4ω2
0 = −γ
2± i
︷ ︸︸ ︷√ω20 − γ2/4 ≈ −γ
2± iω2
0
x(t) = A1e−γ
2 teiω′t + A2e
−γ2 te−iω′t
ω′ =√ω20 − γ2/4 is pulled frequency and initial conditions used to find A1 and A2.
Q = ω0γ is ratio of frequency to damping where Q/4π cycles reduces amplitude to e−1
Kelvin Wagner, University of Colorado Physical Optics 2018 2
Harmonic Oscillator Solution
Proper quantum mechanical treatment of spontaneous emission is quite involved. Re-quires frictional force for scattering to be proportional to
γ =e2ω2
0
6πǫomc3≡ 1
tspont
Periodic driving function eiωt gives periodic solution x(t) = X(ω)eiωt
−ω2X + iωγX + ω20X = − e
mE0
Solve for frequency domain response
X(ω) =− e
mE0
ω20 − ω2 + iωγ
For a real driving field two harmonic sideband solutions add giving x(t) = ℜX(ω)eiωt
x(t) =emE0
ω20 − ω2 + iωγ
eiωt
2+ cc
Kelvin Wagner, University of Colorado Physical Optics 2018 3
Induced Polarization vector field
microscopic dipole moment
p = −ex =e2
m(ω20 − ω2 + iωγ)
E0eiωt + cc
atomic polarizability relates ~p to ~E
~p = ǫoαrE = α~E = α~E
Linearly DisplacedElectron Cloud
less than 1% radius
Nucleus
Unperturbed Electron cloud
Large displacementcan deform cloud
α =e2
m(ω20 − ω2 + iωγ)
αr =e2
ǫom(ω20 − ω2 + iωγ)
Macroscopic Polarization vector field ~P(~r, t) proportional to ~p and density of dipolesper unit volume
~P = N~p = Nα~E = Nǫoαr~E = ǫoχ~E = ǫoχ~E
⇒ χ =Ne2/m
ǫoD0(ω)
where resonant denominator is abbreviated as D0(ω) = ω20 − ω2 + iωγ
Remember D = ǫE = ǫoεrE = ǫon2E = ǫo(1 + χ)E = ǫoE + P so P = ǫoχE
Kelvin Wagner, University of Colorado Physical Optics 2018 4
Index and absorption resonances
n2 = 1 + χ = 1 +Nα
ǫo= 1 +
Ne2
ǫom(ω20 − ω2 − iωγ)
Dilute media: use√1 + δ = 1 + δ
2
n = 1 +Ne2
2ǫom(ω20 − ω2 − iωγ)
= n + iκ
= 1 +Ne2(ω2
0 − ω2)
2ǫom[(ω20 − ω2)2 + ω2γ2)]
+ iNe2γω
2ǫom[(ω20 − ω2)2 + ω2γ2)]
Wave eqn soln for plane wave
~E(~r, t) = E0pe−i(ωt−Kz) + cc = pE0e
−i(ωt−kz)e−2πλ κz + cc
where complex wavevector K = ωcN = 2π
λ(n + iκ) is frequency dependent
Intensity attenuatuation
I(z) =|E(z)|22η
= e−αIzI(0) where αI = 22π
λκ is attenuation coefficient
Complex propagation const K = ~k + i~α |~k| = 2πλ |~α| = αI
2
Kelvin Wagner, University of Colorado Physical Optics 2018 5
Two Driving Frequencies
F =e
m
∑
k
Ek
2(eiωkt + e−iωkt)
Harmonic Oscillator is linear so individual solns will add
x + γx + ω20x = F
x(t) = em
∑k
Ek2 eiωkt
ω20−ω2k+iωkγ+ cc =
∑k ck cos(ωkt− φk)
x(t) = em
∑k
iωkEk2 eiωkt
ω20−ω2k+iωkγ+ cc =
∑k ckωk sin(ωkt− φk)
x(t) = em
∑k
−ω2kEk2 eiωkt
ω20−ω2k+iωkγ+ cc =
∑k −ω2
kck cos(ωkt− φk)
ak =em
Ek2
ω20 − ω2
k + iωkγck =
em Ek
[(ω2k − ω2
0) + ω2kγ
2]1/2φk = tan−1 γωk
ω2k − ω2
0
plug in ∑
k
(−ω2kak + iωkγak + ω2
0ak)eiωkt + cc =
e
m
∑
k
Ek
2(eiωkt + cc)
LHS = RHS ∀k
Kelvin Wagner, University of Colorado Physical Optics 2018 6
Multiple Resonances
nDC = 1+Ne2
mǫo
∑
q
fqω2q
Visible transparencywindow in betwee IRand UV resonances
When Medium consists of N oscillators per unit volume divided into fractions fq withfrequencies ωq and damping γq
χ(ω) =Ne2
ǫom
∑
q
fqw2
q − ω2 − iωγq
Oscillator sum rules∑
q fq = 1∑
qp fqp = Z #electronsQuantum Theory gives
fqp =2m(ωq − ωp)
3~
∣∣∣∣∫
Ψ∗qDΨpdV
∣∣∣∣2
where electron position operator D =∑Z
m=1 rmR. Loudon, The Quantum Theory of Light, Oxford 1973. Ch 4.
Kelvin Wagner, University of Colorado Physical Optics 2018 7
Susceptibility χ(ω)
Macroscopic polarization ~P proportional to ~p and density of dipoles N
P (t) = Np = NαE = NǫoαrE = ǫoχE(t)
At a particular frequency P (t) = ℜP (ω)eiωt
P (ω) = −NeX(ω) =Ne2E0
m(ω20 − ω2 + iωγ)
= ǫoχ(ω)E(ω)
Solving for the susceptibility
χ(ω) =Ne2/ǫom
ω20 − ω2 + iωγ
= χ′(ω)− iχ”(ω) =Ne2
ǫom
ω20 − ω2 − iωγ
(ω20 − ω2)2 + ω2γ2
Introduce complex index N = n + iκ so that N 2 = n2 − κ2 + i2nκ
εr = N 2 = 1 + χ = 1 + (χ′ − iχ”)
1 + χ′ = n2 − κ2 = 1 +(ω2
0 − ω2)ω20S
(ω20 − ω2)2 + ω2γ2
χ” = 2nκ =ωγω2
0S
(ω20 − ω2)2 + ω2γ2
where S = Ne2
ǫomω20is dimensionless oscillator strength
Kelvin Wagner, University of Colorado Physical Optics 2018 8
Lorentzian approximation
when ω0 ≫ γ can approximate resonance with ω0 = ω in all terms except ω0 − ω(ω2
0 − ω2)2 = [(ω0 − ω)(ω0 + ω)]2 ≈ (ω0 − ω)2(2ω)2
χ” =Sω0γ/4
(ω0 − ω)2 + (γ/2)2=
Ne22πν0ǫom(2πν0)2
∆ν2π/4
(2π)2[(ν − ν0)2 + (∆ν/2)2]
=Ne2
ǫomν016π2
∆ν
(ν − ν0)2 + (∆ν/2)2Lorentzian
χ′ =Sω0(ω0 − ω)/2
(ω0 − ω)2 + (γ/2)2=
Ne2
ǫom8π2ν0
ν0 − ν
(ν0 − ν)2 +(∆ν2
)2 =2(ν0 − ν)
∆νχ”(ν)
Kelvin Wagner, University of Colorado Physical Optics 2018 9
Experimental quantities: Index andabsorption
N 2 = 1 + χ = n2 − κ2 + i2nκ = 1 + χ′ − iχ”
χ′ =2(ν0 − ν)
∆νχ”(ν)
2nκ = χ” ⇒ κ = χ”2n
n2 − κ2 = 1 + χ′ = n2 − χ”2
(2n)2⇒ n4 − n2(1 + χ′)− χ”/4 = 0
n2 =1 + χ′
2± 1
2
√(1 + χ′) + 4χ”/4
Simplify for dilute media
N = 1 +Ne2
2ǫom
1
(ω20 − ω2 − iγω)
=
(1 +
Ne2
2ǫom
ω20 − ω2
(ω20 − ω2)2 + γ2ω2
)+ i
(Ne2
2ǫom
γω
(ω20 − ω2)2 + γ2ω2
)= n + iκ
Kelvin Wagner, University of Colorado Physical Optics 2018 10
K = Nk0 = k0√1 + χ = k0
√1 + χ′ − iχ” ≈
(1 +
χ′
2− i
χ”
2
)k0
=2π
λ(n + iκ) = β − i
α
2Dilute media
n = 1 +χ′
2α/2 = k0χ”/2
Plane wave solnE = eE0e
i(ωt−Kz) = eE0ei(ωt−βz)e−αz/2
Intensity
I(z) =|E(z)|22η
= I0e−αz
Conservation of energy requires
dI
dz= −power absorbed
volume=
ωǫo2χ”|E|2 = ωǫo
2χ”
2
ǫocnI = −αI
Thus α = −ωχ”/cn = −k0χ”/n
Kelvin Wagner, University of Colorado Physical Optics 2018 11
Local Field – Dielectric Constant of DenseMedia
Electrostatics: Charge can not act on itselfField it sees is what would be seen at its position if it were removed without alteringother chargesHowever if it were removed other charges would realign. Difference is local field
Examine by cutting a slot in a dielectric capacitor
C
++++++++++++++++++++++++++++++
--------------------------------------------------
E
Field on two sides of narrow slotcancels since there is no charge
∇× ~E = 0 ⇒∮
C
~E · d~l = 0
~El = ~E
SE ------------------
+++++++++++
++++++++++++++++++++++++++++++
--------------------------------------------------
Gauss’ Lawx
S
~E · d~l = σpol = ~P
ǫo~El = ǫo~E + ~P = ~D~El = ~E + ~P/ǫo
Kelvin Wagner, University of Colorado Physical Optics 2018 12
Atom fits in a spherical hole in the dielectricFeynman, Lectures on Physics, vol 2,ch 11.
E++++
- - - -
~El = ~E +~P
3ǫo
E P++++
- - - -
Uniform E field inside sphereDipole field outside
Fields due to uniformly polarized charged sphere
~Es = −~P
3ǫo~E = ~El + ~Es
θ
σ=P n
Charge density per unit area of cavity surface = −P cos θ
charge on annular ring dQ = −P cos θ 2πr2 sin θdθ =~P·~rrdS
Coulombs law ∇ · ~D = ρ gives field z · dE2 = −dQ cos θ/r24πǫototal field
E2 =
∫ π
0
P cos2 θ sin θ
2ǫodθ =
P
2ǫo
[−1
3cos3 θ
]π
0
=P
3ǫo
Two ways a slot can be parallel to ~E and one perpendicular. Av-erage gives local field correction P
3ǫo
Kelvin Wagner, University of Colorado Physical Optics 2018 13
Local Field and Lorentz-Lorenz correction
microscopic polarization induced by ~Elocal
~p = αrǫo~El = α~El
Macroscopic polarization given by N number density of microscopic dipoles
~P = N~p = Nαrǫo
(~E +
~P
3ǫo
)= ǫoχ~E
~P
(1− Nαr
3
)= Nαrǫo~E ⇒ χ =
Nαr
1−Nαr/3= ǫr − 1 = n2 − 1
In terms of dielectric constant
ǫ = ǫoεr = ǫo(1 + χ) = ǫo1 + 2
3Nαr
1− 13Nαr
= n2ǫo
solving for microscopic polarizability αr
(1− Nαr
3
)εr = 1 +
2
3Nαr ⇒ εr − 1 =
Nαr
3((2 + εr))
αr =3
N
εr − 1
εr + 2=
3
N
n2 − 1
n2 + 2=
α
ǫoLorentz-Lorenz fomula (Clausius-Mosotti)
Kelvin Wagner, University of Colorado Physical Optics 2018 14
Lorentz-Lorenz fomula (Clausius-Mosotti)
n2 − 1
n2 + 2=
αrN
3=
αN
3ǫo=
n2 − 1
n2 − 1 + 3⇒ n2 − 1 =
αrN
3(n2 − 1 + 3)
(n2 − 1)(1− αrN
3) = αrN n2 − 1 =
αrN
1− αrN3
=3αrN
3− αrN
Index of liquids and gases
Gases are dilute
εr − 1 = Nαr
Liquids must be local field corrected
εr − 1 =Nαr
1−Nαr/3
Gas LiquidSubstance εr Nαr Density Density Nαr εr (predict) εr (exp)
CS2 1.0029 0.0029 .00339 1.293 1.11 2.76 2.64O2 1.000523 .000523 .00143 1.19 0.435 1.509 1.507
Kelvin Wagner, University of Colorado Physical Optics 2018 15
Index of Mixtures: Refractive Index ofSugar (molec wt 342) in Water (molec wt
18)Feynman, Lectures on Physics, vol 2,ch 32.Consider two materials of index n1 and n2 and polarizabilities α1 and α2.
mixed with molar fractions N1/N0 and N2/N0 (N0 = 6.23× 1023 Avogadros number)
Is the overall index just the weighted average
n?=
N1
N0n1 +
N2
N0n2 no!
Instead it is given by the sum of the polarizabilities
3
(n2 − 1
n2 + 2
)=∑
j
Njαrj = N1α
r1 +N2α
r2
Assume αrH2O
(molec wt 18) is const and solve for αrsugar (molec wt 342) in sugar water.
Data from Handbook AB342
(1−A)B18
from C F (0) EE(0)
F (1) DD(1)
sum
A B C D E F G H G+H
wt Fraction density index N2N0
moles N1N0
moles 3(n2−1n2+2
)N1α
r1 N2α
r2 compare N0α
r2
sucrose n589 sucrose/l water/l to F H/D
0 .9982 1.333 0 998.2/18 0.617 0.617 0 0.617 –
0.30 1.127 1.3811 .970 43.8 0.698 0.487 .211 0.698 .211
0.50 1.2296 1.4200 1.798 34.15 0.759 0.379 0.380 0.759 .211
0.85 1.4454 1.5033 3.59 12.02 0.886 0.1335 0.752 0.8855 .210
1.00 1.588 1.5577 4.64 0 0.960 0 0.96 0.96 .208
Kelvin Wagner, University of Colorado Physical Optics 2018 16
Local Field Correction Including Dispersion
α =αr
ǫo=
e2
ǫom(ω20 − ω2 − iγω)
=e2
ǫom
1
D0(ω)
Lorentz-Lorenz formula becomes
n2 − 1
n2 + 2=
N
3
e2
ǫom
1
D0(ω)
In the presence of multiple resonances
n2 − 1
n2 + 2=
Ne2
3ǫom
∑
k
fkDk(ω)
=Ne2
3ǫom
∑
k
fk(ω2
k − ω2 − iγkω)
Rearranging as before and decompose by partial fractions using roots of denominator
n2 − 1 =3N e2
ǫom
∑k
fkDk(ω)
3−N e2
ǫom
∑k
fkDk(ω)
=∑
k
ρkν2k − ν2
=∑
k
Akλ20
λ20 − λ2
k︸︷︷︸Bk
ω0 =√ω20 − ω2
p/3 ν0 = ω0/2π λ0 =cν0
ωp =Ne2
ǫom
Or Taylor expand in transparency regime in terms of frequency ν or wavelength λ = c/ν
n2 − 1 ≈ A +Bν2 + Cν4 + · · · − B′
ν2− C ′
ν4≈ A +B
c2
λ2+ C
c2
λ4+ · · · − B′λ2
c2− C ′λ4
c4Kelvin Wagner, University of Colorado Physical Optics 2018 17
Sellmeier Equations to approximate indexvs wavelength across limited band of
transparencySellmeier = Sellmeir = Selmeir = Selmeier
Sellmeir formulas occur in a wide variety of forms, but most inlude only 1-2 terms or atmost 3. For visible sellmeiers, terms account for UV resonances and Far IR resonance.
n2 = 1 +∑
k
fkν2k − ν2
= 1 +∑
k
bkλ2k − λ2
= 1 +∑
k
Akν2k −
∑
j
Bj
ν2j
where the coefficients for a thin film, glass, or crystal may depend on composition orgrowth condition as well as temperature.
Also note, index is temperature dependent so some formulas include T dependence
∆n ≈ 10−6/C glasses
∆n ≈ 10−5/C crystals
∆n ≈ 10−4/C plastics
Kelvin Wagner, University of Colorado Physical Optics 2018 18
Sellmeier Formulas for various glasses takea variety of forms
Cauchy
n = A +B
λ2+
C
λ4
Conrady
n = A +B
λ+
C
λ7/2
Resonance
n2=A+B
λ2 − C2+
D
λ2 − E2+
F
λ2 −G2
Binomial Expansion gives 4 term approximation or more accurate in IR 6 term formula
n2 = aλ2 + b +c
λ2+
d
λ4≡ fλ6 + eλ4 + aλ2 + b +
c
λ2+
d
λ4
Herzberger’s formula used combination with resonance at λ0 = .187
n = A +Bλ2 +C
λ2 − λ20
+D
(λ2 − λ20)
2
Schott Catalog gives expersssions good out to 1µm (very good in UV and blue)
n2 = A0 + A1λ2 +
A2
λ2+
A3
λ4+A4
λ6+
A5
λ8
Bausch and Lhomb gives 7 term formula
n2 = a + bλ2 + cλ4 +d
λ2+
eλ2
(λ2 − f) + gλ2/(λ2 − f)
Kelvin Wagner, University of Colorado Physical Optics 2018 19
Sellmeir for Silica
Fit for index, wavelength in µm
n(λ0) = C0 + C1λ2 + C2λ
40 +
C3
λ20 − l
+C4
(λ20 − l)2
+C5
(λ20 − l)3
C0 = 1.4508554 C1 = −.0031268 C2 = −.0000381 C3 = −.0031268C4 = .0030270 C5 = .0000018 l = .035
√l = .187µm
Form of Sellmeier for index squared
n2(λ0)− 1 =A1λ
2
λ2 − λ21
+A2λ
2
λ2 − λ22
+A3λ
2
λ2 − λ23
=∑
k
Akλ2
λ2 − λ2k
material A1 λ1 A2 λ2 A3 λ3
SiO2 .6961663 .0684043 .4079426 .1162414 .8974794 9.8961613.5% GeO2 .711040 .064270 .451885 .129408 .704048 9.425478
Kelvin Wagner, University of Colorado Physical Optics 2018 20
Dispersion can be found from derivative ofSellmeier
d(n2 − 1)
dλ= 2n
dn
dλ
d
dλ
λ2
λ2 − λ2j
=f ′g − g′f
g2=
2λ(λ2 − λ2j)− 2λλ2
(λ2 − λ2j)
2=
−2λλ2j
(λ2 − λ2j)
2
dn
dλ=
−2λ
2n
(∑
k
Ajλ2j
(λ2 − λ2j)
2
)
Group index
ng = n− λdn
dλGroup Velocity Dispersion found from derivative of group index
dng
dλ=
dn
dλ− 1 · dn
dλ− λ
d2n
dλ2
= −λ
[(1 · n− dn
dλ· λ
n2
)∑ Ajλ2j
(λ2 − λ2j)
2− λ
n
∑−2(λ2 − λ2j)2λ
(λ2 − λ2j)
4Ajλ
2j
]
= −λng
nSS2
(−
ZZn
λ
dn
dλ
)− 4λ3
n
∑ Ajλ2j
(λ2 − λ2j)
3
Kelvin Wagner, University of Colorado Physical Optics 2018 21
Dispersion and GDD for Silica493
Figure D.1. Linear refractive index n and group index ng for fused silica.The zero dispersion wavelength zd = 1:27 m divides the boundary betweenthe normal dispersion regime ( < zd) and the anomalous dispersion regime( > zd). The curves are marked for their values at 1.55 m.is also referred to here as the phase delay coecient, since it is related to theinverse phase velocity. Of more relevance to this thesis, though, are the groupvelocity and the group delay coecient. The group velocity is the velocity atwhich a quasi-monochromatic wavepacket travels, which is in general dierentthan the phase velocity, and is most easily calculated from the group delaycoecient. The group delay coecient is the frequency derivative of the phasedelay: k0 = @k@! = 1c "n+ !@n@!# ngc ; (D.2)
495
Figure D.2. Group-delay dispersion k00 for fused silica. The zero dispersionwavelength zd = 1:27 m divides the boundary between the normal dispersionregime ( < zd) and the anomalous dispersion regime ( > zd). The curveis marked for its value at 1.55 m.Higher-order dispersion becomes important for broad temporal band-widths. The simulations of spatio-temporal solitons in this thesis use thehigher-order dispersion coecients because such broad bandwidths are present.Third-order dispersion is calculated fromk000 = @3k@!3 = 1c "3@2n@!2 + !@3n@!3# (D.7)= 442c3 "3@2n@2 + @3n@3 # :Fourth-order dispersion is calculated fromk0000 = @4k@!4 = 1c "4@3n@!3 + !@4n@!4# (D.8)Kelvin Wagner, University of Colorado Physical Optics 2018 22
Group Velocity and Dispersion
Consider optical pulses (no longer simply harmonic in time)For simplicity, start with a pulsed plane wave in free space
A(~r, t) = e−i(kxx′+kyy
′+kzz′)s
(t−
~k′ ·~rv
)eiω0t
Rotate coordinates to align z with direction of propagation
A(z, t) = e−i(kz−ωt)s(t− z
v
)
Now put into dispersive medium n(ω), thus v = ω(k)k = c
n(ω). Phase velocity is
vp =ω
k=
2πν
2πn(ω)/λ=
c
n(ω)
Utilize a Fourier representation of the frozen spatial representation of pulse s() at t = 0
A(z, 0) = eik0zs(−vz) =
∫ ∞
−∞S(k)eikzdk
Kelvin Wagner, University of Colorado Physical Optics 2018 23
Fourier propagation of each spectralcomponent
Thus, pulse at later time is given by propagating each Fourier component with appro-priate ω(k)
A(z, t) =
∫ ∞
−∞S(k)e−ik(vt−z)dk =
∫ ∞
−∞S(k)e−i(ω(k)t−kz)dk
Now ω(k) = k cn(ω) since k(ω) =
n(ω)c ω
Taylor expand about central wavevector k0
ω(k) = ω|k0+dω
dk
∣∣∣∣k0
(k−k0)+d2ω
dk2
∣∣∣∣k0
(k − k0)2
2+· · · = ω0+vg(k−k0)+D
−112(k−k0)
2+. . .
Plug back in Fourier integral
A(z, t) =
∫ ∞
−∞S(k)e−i([ω0+vg(k−k0)+D−11
2(k−k0)2+... ]t−(k−k0)z−k0z)dk
≈ e−i(ω0t−k0z)
∫ ∞
−∞S(k)e−i[ω′t−z](k−k0)dk
︸ ︷︷ ︸= e−i(ω0t−k0z)︸ ︷︷ ︸E(z − ω′t)︸ ︷︷ ︸
function of z − ω′t phase factor Envelope
Kelvin Wagner, University of Colorado Physical Optics 2018 24
Phase and Group Velocity
Phase Velocity
vp =ω0
k0=
c
n
Group Velocity
vg = ω′ =dω
dk
∣∣∣∣k0
Since we represent k as a function of ω rather than vice versa, it’s easier to calculate
dk
dω=
d
dω
[ωcn(ω)
]=
n(ω)
c+ω
c
dn
dωvg =
c
n(ω) + ωdn(ω)dω
=dω
dk=
(dk
dω
)−1
In 3-D A(~r, t) = A(~r, t)e−i(ωt−gω(~r))
vp =1∣∣∇ gω
∣∣ vg =1∣∣∣∇∂g∂ω
∣∣∣ω
∣∣∣where ω is the mean frequencyfor a plane wave g = ~k ·~r, thus
∇g
ω= ∇n(ω)ω/c
ωk ·~r = n
ck
Kelvin Wagner, University of Colorado Physical Optics 2018 25
Group Velocity, Dispersion, GVD
In terms of free space wavelength λ0 =cν= 2πc
ω
ωdn
dω= ω
dn
dλ0
dλ0
dω= ω
dn
dλ0
−2πc
ω2=
dn
dλ0
−2πc
2πc/λ0= −λ0
dn
dλ0
1
vg=
1
c
(n(λ0)− λ0
dn
dλ0
) PropagationTime τ =
L
vg=
L
c
(n− λ0
dn
dλ0
)
︸ ︷︷ ︸group index = ng
Dispersion is due to variation of propagation time with λ0 (or ω)
∆τ =dτ
dλ∆λ=
L
c
(
dn
dλ0− 1 ·
dn
dλ0− λ0
d2n
dλ20
)∆λ=−λ0
L
c
d2n
dλ20
∆λ=−L
c
(λ20
d2n
dλ20
)∆λ
λ
Material dispersion proportional to L and ∆λ. Dispersion parameter
Dm =∆τ
L∆λ= − 1
λ0c
(λ20
d2n
dλ20
)[ s
m m
]=
1012ps
10−3km 109nm︸ ︷︷ ︸×109 ps
nm km with λ0 (µm), c(kms )
Kelvin Wagner, University of Colorado Physical Optics 2018 26
Group Velocity near Lorentzian
Michelson measured difference between vp and vg.CS2: n = 1.64 using refractometer and ng = 1.76
ω =c
nk
dω = c
(dk
n− k
n2dn
)
dk
dω=
1
c
[n + ω
dn
dω
]
dω
dk=
c
n + ω dndω
For large ω where n < 1 and dndω
> 0 the denominator > 1 and vg < c
Anomalous dispersion where dndω < 0 and where n < 1 ⇒ vg > 1
Near ω0 absorption is highest. Trailing edge of the pulse is preferentially absorbed,causing the peak to be displaced forward, so that energy in the leading edge travels ata velocity less than c. But the peak travels faster than c.
Kelvin Wagner, University of Colorado Physical Optics 2018 27
Propagation of Gaussians and complexGaussians
Gaussian pulse and its Fourier decomposition
E(0, t) = e−t2/σ2
︸ ︷︷ ︸ eiω0t︸︷︷︸ = eiω0t
∫ ∞
−∞
σ√4π
e−Ω2σ2/4eiΩtdΩ
Each frequency component propagates with its own phase factor
E(z, t) = ei(ω0t−β0z)
∫ ∞
−∞e−iβ(Ω)z σ√
4πe−Ω2σ2/4eiΩtdΩ
And expand the remaining terms in β around the center freq
β(ω0 + Ω) = β(ω0) +dβ
dω
∣∣∣∣ω0
Ω + 12
d2β
dω2
∣∣∣∣ω0
Ω2 = β0 + v−1g Ω + 1
2DΩ2
Leads to an integral with a complex Gaussian, which is a chirped Gaussian pulseUse the very useful definite integral
∫ ∞
−∞e−ax2+bxdx =
√π
aeb
2/4a
and do not worry that the constants a or b may be complex
Kelvin Wagner, University of Colorado Physical Optics 2018 28
Complex Gaussian propagation
chirp
Ec(0, t) = e−t2/σ2︷ ︸︸ ︷e−ibct
2eiω0t = e
−t2( 1σ2
+ibc)eiω0t
= eiω0t1√
4π(1/σ2 + ibc)
∫ ∞
−∞e− Ω2
4(1/σ2+ibc)eiΩtdΩ
Instantaneous frequency b = 2bc2π
[GHzns
]
fi = I
1
2π
dφ
dt
=
−2bct
2π+
ω0
2π= f0 + bt
Now can proceed as before to propagate the Fourier decomposition by just replacing thewidth σ2 with the complex “width” 1
1/σ2+ibc
E(z, t) = ei(ω0t−β0z)1√
4π(1/σ2 + ibc)
∫ ∞
−∞e−iβ(Ω)ze
− Ω2
4(1/σ2+ibc)eiΩtdΩ
Solve integral as before, or just generalize the previous time domain result.Must isolate resulting real and imaginary parts of quadratic exponential to identify widthof complex Gaussian.
Kelvin Wagner, University of Colorado Physical Optics 2018 29
Dielectric Response Theory andKramers-Kronig relations
Frohlich, Theory of Dielectrics, Oxford 1958, Appendix A.1
Electric field driven polarization response of material is1) Linear2) Time Invariant3) Causal
u = t− v v = t−u when u = (−∞, t) → v = (∞, 0)
P (t) = ǫ0
∫ t
−∞E(u)χ(t− u)du = −ǫ0
∫ 0
∞E(t− v)χ(v)dv
D(t) =
∫ t
−∞E(u)ǫ(t− u)du = ǫ0
∫ ∞
0
E(t− v)εr(v)dv
4) Finite response time
χ(t), εr(t) → 0 as t → ∞
instantaneous response component
ǫ(ω) → ǫ∞ + i0 ω → ∞
static component
ǫ(ω) → ǫs + i0 ω → 0
E(t)
P(t) χ Es
8χ E
Kelvin Wagner, University of Colorado Physical Optics 2018 30
Harmonically time varying field
E(t) = E0 cosωt = E0e−iωt
2+ cc
x= t−u dt=−dxt→0
−∞→∞ cosω(t− x) = cosωt cosωx− sinωt sinωx
D(t)=E0ǫo
∫ t
−∞cos(ωu)εr(t−u)du+ ǫ∞E(t) = ǫoE0
∫ ∞
0
︷ ︸︸ ︷cosω(t− x) εr(x)dx + ǫ∞E(t)
= ǫoE0 cosωt
[∫ ∞
0
εr(x) cosωx dx + εr∞
]+ ǫoE0 sinωt
∫ ∞
0
εr(x) sinωx dx
= ǫ′E0 cosωt︸ ︷︷ ︸+ ǫ”E0 sinωt︸ ︷︷ ︸in-phase quadrature
Complex dielectric constant at ω
ǫ(ω) = ǫ′(ω) + iǫ”(ω) = ǫo(ε′r(ω) + iε”r(ω))
ǫ′(ω) =
∫ ∞
0
ǫ(t) cosωt dt + ǫ∞ (1)
ǫ”(ω) =
∫ ∞
0
ǫ(t) sinωt dt (2)
Kelvin Wagner, University of Colorado Physical Optics 2018 31
Evaluate Fourier transforms
Take Fourier transform (cos transform of 1, sin transform of 2 )
ǫ(t) =2
π
∫ ∞
0
[ǫ′(ω)− ǫ∞] cosωt dω (3)
ǫ(t) =2
π
∫ ∞
0
ǫ”(ω) sinωt dω (4)
Substitute 4 into 1
ǫ′(ω)− ǫ∞ =
∫ ∞
0
cosωt2
π
∫ ∞
0
ǫ”(ω′) sinω′t dω′ dt
=2
πlimR→∞
∫ ∞
0
ǫ”(ω′)
∫ R
0
cosωt sinω′t dt dω′
=2
π
∫ ∞
0
ǫ”(ω′) limR→∞
1
2
(1− cos(ω′ + ω)R
ω′ + ω+1− cos(ω′ − ω)R
ω′ − ω
)dω′
=2
π
∫ ∞
0
ǫ”(ω′)1
2
[1
ω′ + ω+
1
ω′ − ω
]dω′ =
2
π
∫ ∞
0
ǫ”(ω′)1
2
(ω′ − ω) + (ω′ + ω)
ω′2 − ω2dω′
Although taking the limit and just ignoring the oscillating component seems unjustified
Kelvin Wagner, University of Colorado Physical Optics 2018 32
Kramers-Kronig relations
When frequency data is available only for measured (positive) frequencies
ǫ′(ω)− 1 =2
π
∫ ∞
0
ǫ”(ω′)ω′
ω′2 − ω2dω′
ǫ”(ω) =2
π
∫ ∞
0
(ǫ′(ω′)− 1)ω
ω′2 − ω2dω′
Suprisingly good agreement when only a band of frequencies is available
Alternatively the Hilbert transform pairs integrating across all frequencies
ǫ′(ω)− 1 =1
πPV
∫ ∞
−∞
ǫ”(ω′)
ω′ − ωdω′
ǫ”(ω) =1
πPV
∫ ∞
−∞
ǫ′(ω′)− 1
ω′ − ωdω′
Similar relationships exist for other pairs of real and imaginary parts of frequency re-sponses for linear systems that are causal
n(ω)Kramers−−−−−−−−−−Kronig
κ(ω) χ′(ω)Kramers−−−−−−−−−−Kronig
χ”(ω) ln ρ(ω)Kramers−−−−−−−−−−Kronig
φ(ω)
Kelvin Wagner, University of Colorado Physical Optics 2018 33
Complex Analysis
f(z) is analytic in domain D if it is differentiable at every point in D
w = f(z) = f(x + iy) = u(x, y) + iv(x, y)
f ′(z) = u′x + iv′x = v′y − iu′yCauchy-Rieman eqn
∂u
∂x=
∂v
∂y
∂u
∂y= −∂v
∂x
Complex IntegrationCauchy Thm If f(z) analytic on and inside contour C
∮
C
f(z)dz = 0
Cauchy Integral Formula (f(z) analytic on and inside contour C)
f(a) =1
2πi
∮
C
f(z)
z − adz
Kelvin Wagner, University of Colorado Physical Optics 2018 34
Proof of Cauchy-Rieman condition foranalyticity
f ′(z) = lim∆z→0
f(z +∆z)− f(z)
∆zwhere f(z) = u(x, y) + iv(x, y)Now the path of the infinitesimal difference can be along real or imaginary axis incomplex plane (or along any direction)∆z = ∆x
f ′(z) =u(x +∆x, y) + iv(x + ∆x, y)− [u(x, y) + iv(x, y)]
∆x=
∂u
∂x+ i
∂v
∂x
∆z = i∆y
f ′(z)=u(x, y+∆y) + iv(x, y+∆y)− [u(x, y) + iv(x, y)]
i∆y=1
i
[∂u
∂y+i
∂v
∂y
]=−i
∂u
∂y+∂v
∂y
Setting these two results equal gives Cauchy Rieman condition for analyticty
∂u
∂x=
∂v
∂y
∂v
∂x= −∂u
∂y
Kelvin Wagner, University of Colorado Physical Optics 2018 35
Alternative proof
z = x + iyz∗ = x− iy
x =z + z∗
2⇒ ∂x
∂z= 1
2y =
z − z∗
2i⇒ ∂y
∂z=
1
2iChain rule
∂f
∂z=
∂f
∂x
∂x
∂z+∂f
∂y
∂y
∂z= 1
2
(∂f
∂x+1
i
∂f
∂y
)
∂f
∂z∗=
∂f
∂x
∂x
∂z∗+∂f
∂y
∂y
∂z∗= 1
2
(∂f
∂x− 1
i
∂f
∂y
)= 0 since f(z) not z∗
⇒ ∂f
∂x= −i
∂f
∂y
ux + ivx = −i(uy + ivy)
which gives by setting real and imaginary parts equal ux = vy vx = −uy
And in polar coordinates ∂u
∂r=
1
r
∂v
∂θ
∂v
∂r= −1
r
∂u
∂θ
Kelvin Wagner, University of Colorado Physical Optics 2018 36
Residues
Resz=af(z) = c−1 Laurent expansion f(z) =∑∞
n=−∞ cn(z − a)n
Residue Theorem1
2πi
∮
C
f(z)dz =n∑
k=1
Resz=akf(z)
Calculation of Residues
1. Determine c−1 from Laurent expansion
2. Simple poles Resz=af(z) = limz→a(z − a)f(z)
Use l’Hopital’s rule limx→af(x)g(x)
= limx→af ′(x)g′(x)
eg f(z) = n(z)d(z) n(z) , d(z) analytic. n(a) 6= 0, d(a) = 0, d′(a) 6= 0
Resz=af(z)
g(z)=
f(a)
g′(a)
3. Poles of Order m
Resz=af(z) = limz→a
1
(m− 1)!
(d
dz
)m−1
[(z − a)mf(z)]
Kelvin Wagner, University of Colorado Physical Optics 2018 37
Examples of poles and residues
functions zeroes polesz = x + iy 0 m = 1 ∞ (m = 1)z2 = x2 − y2 + i2xy 0 m = 2 ∞ (m = 2)1/z = x/r2 − iy/r2 ∞ m = 1 0 (m = 1)1/z2 ∞ m = 2 0 (m = 2)log z = ln r = iθ + 2nπ 1 m = 1(principal branch) 0,∞ branch points√z 0, branch point ∞, branch point
ez = ex cos y + iex sin y ∞ essential singularity
Example: Definite integral using CCI f(z) = eiz
z2+a2= eiz
(z+ia)(z−ia)
I =
∫ ∞
−∞
cosx
x2 + a2dx = R
∫ ∞
−∞
eix
x2 + a2dx a > 0
Resz=ia
f(z) = limz→ia
(z−ia)eiz
z2+a2= e−a lim
z→ia
z−iaz2+a2
l′Hop= e−a lim
z→ia
12z
= e−a
2ia⇒ I=R
2πie
−a
2ia
=πe−a
a
f(z) analytic in UHP except finite ak I =∫∞−∞ f(x)dx = 2πi
residues∑k
Resz=ak
f(z)
Kelvin Wagner, University of Colorado Physical Optics 2018 38
Lorentzian Complex Plane
Write Lorentzian using partial frac-tions
χ(ω) =f0Ne2
ǫom
1
ω20 − ω2 − iΓω
=ω2p
ω1 − ω2
(1
ω − ω1− 1
ω − ω2
)
=ω2p[ω − ω2 − (ω − ω1)]
(ω1 − ω2)(ω − ω1)(ω − ω2)
=ω2p
(ω − ω1)(ω − ω2)
Where the poles are at
ω1 = −12i[Γ− (Γ2 − 4ω2
0)12 ]
ω2 = −12i[Γ + (Γ2 − 4ω2
0)12 ]
ε′ ε” n(ω)
Kelvin Wagner, University of Colorado Physical Optics 2018 39
Dielectric Constant and Index in theComplex Plane
Kelvin Wagner, University of Colorado Physical Optics 2018 40
Multiple Lorentzians in Dispersion relation
Kelvin Wagner, University of Colorado Physical Optics 2018 41
Relative Dielectric Constants and ComplexIndex
N2 = n2 − κ2 + i2nκ = 1 + χ′ + iχ′′ N =√1 + χ =
√ǫr
Kelvin Wagner, University of Colorado Physical Optics 2018 42
Dispersion between UV and IR Lorentzianresonances
ng = n + ωdn
dω= n− λ
dn
dλ k =ω
cn(ω) vp =
ω0
k0>
dω
dk
∣∣∣∣k0
= vg
Kelvin Wagner, University of Colorado Physical Optics 2018 43
Kramers-Kronig with Complex ContourIntegration
Kramers-Kronig dispersion relation can be derived from the contour integral(Cauchy Principal value Integral)
P∫ ∞
−∞
χ(ω′)
ω′ − ωdω′ = lim
δ→0
(∫ ω−δ
−∞+
∫ ∞
ω+δ
χ(ω′)
ω′ − ωdω′)
=
∫
A
−∫
B
−∫
C
C
BA
ω2ω1
ωω−δ ω+δ 88−
∫A = 0 since it contains no poles∫B = 0 integrand → 1
ω′2 for large radius length ≈ π|ω′|∫C
χ(ω′)ω′−ω
dω′ = −i2π2residue = −iπχ(ω)
iπχ(ω) = P∫ ∞
−∞
χ(ω′)
ω′ − ωdω′
iπ (χ′(ω) + iχ”(ω)) = P∫ ∞
−∞
χ′(ω) + iχ”(ω)
ω′ − ωdω′
Kelvin Wagner, University of Colorado Physical Optics 2018 44
Kramers-Kronig
Equate real and imaginary parts
χ”(ω) =−1
πP∫ ∞
−∞
χ′(ω′)
ω′ − ωdω′ χ′(ω) =
1
πP∫ ∞
−∞
χ”(ω′)
ω′ − ωdω′
Since ~P(t) and ~E(t) are real, χ(t) is real
⇒ χ(−ω) = χ∗(ω) χ′(−ω) + iχ”(−ω) = χ′(ω)− iχ”(ω)
χ′(−ω) = χ′(ω) real part is evenχ”(−ω) = −χ”(ω) imaginary part is odd
χ(ω) =1
iπP∫ 0
−∞
χ∗(ω′)
ω′ + ωdω′ +
1
iπP∫ ∞
0
χ(ω′)
ω′ − ωdω′
χ′(ω)+iχ”(ω) =−i
πP∫ ∞
0
1
ω′2 − ω2[− (χ′(ω)− iχ”(ω)) [ω′ − ω] + (χ′(ω) + iχ”(ω)) [ω′
χ′(ω)[ω − ω′ +
ω′ + ω] + iχ”(ω)[ω′ − ω + ω′ + ω]
χ′(ω) =2
πP∫ ∞
0
ω′χ”(ω′)
ω′2 − ω2dω′ χ”(ω) =
−2
πP∫ ∞
0
ωχ′(ω′)
ω′2 − ω2dω′
Kelvin Wagner, University of Colorado Physical Optics 2018 45
Propagation of a pulse through a dispersivemedium
Following Jackson, Classical Electrodynamics, sec 7.11
Represent a pulse by its Fourier spectrumui(0, t) incident on surface
A(ω) =1
2π
∫ ∞
−∞ui(0, t)e
iωtdt
note this is using alternative sign defn of FT!Fresnel transmission t01 =
2n0n0+n1
= 21+n(ω)
n2=n0n1(ω)n0=1
z=0
Propagation described by different frequency components propagating at different phasevelocities with different k(ω) = ω
cn(ω) = kR + ikI . For this sign, kI > 0 is absorption
u(z, t) =
∫ ∞
−∞t01(ω)A(ω)e
i(k(ω)z−ωt)dω
output into nondispersive free space t12 =2n1
n1+n2= 2n(ω)
n(ω)+1
u(z, t) =
∫ ∞
−∞
4n(ω)
[1 + n(ω)]2A(ω)ei(k(ω)z−ωt)dω (5)
Kelvin Wagner, University of Colorado Physical Optics 2018 46
Approximate the wavevector by its Taylor expansion about the central frequency ω0
k(ω) = k|ω0+∂k
∂ω
∣∣∣∣ω0
(ω−ω0)+∂2k
∂ω2
∣∣∣∣ω0
12(ω−ω0)
2+· · · = k0+v−1g (ω−ω0)+D
12(ω−ω0)
2+. . .
v−1g = 1
c(n + ω ∂n∂ω) D = 1
c
[2∂n∂ω + ω ∂2n
∂ω2
]
Assume ui(0, t) = 0 t < 0⇒ A(ω) analytic in upper ω-plane
UHP ω = α + iβ, β > 0 eiωt → e−βt
t < 0 (divergent unless u = 0)t > 0 (convergent for u finite)
ui smoothly turns on
ui(0, t) = atm
m!H(t) t → 0
A(ω) → a
2π
(i
ω
)m+1
x as |w| → ∞
t3
21
ttt 0
t
u(t)
t=0
integration in time ⇒ 1iω in Fourier domain using standard FT, 1
−iω = iω using + sign
Kelvin Wagner, University of Colorado Physical Optics 2018 47
Medium with a single Lorentzian resonance
εr(ω) = n2(ω) = 1 +ω2p
ω20 − ω2 − iγω
ω2p =
Ne2
mǫoplasma freq, as ω → ∞ εr(ω) = 1− ω2p
ω2
when ω±z = ±ω1 − iγ/2 with ω2
1 = ω20 + ω2
p − γ2/4 thus (ω±z )
2 = ω21 + γ2/4∓ 2ω1
iγ2
n(ω±z ) = 0 = 1 +
ω2p
ω20 − (
ω20 + ω2
p − γ2
4 +γ2
4 ∓ HHHHH2ω1
iγ2 )∓
HHHHHiω1γ −γ2
2
= 1− 1 = 0
so ω±z are the zeroes of εr(ω) and n(ω)
when ω±p = ±ω2 − iγ/2 with ω2
2 = ω20 − γ2/4
n(ω±p ) → ∞ give poles
0 =ω20 − (
ω20 −
AAAAA
γ2
4∓ ω2iγ −
AAAAA
γ2
4)∓ iω2γ −
AAAAA
γ2
2
n2(ω) =(ω − ω+
z )(ω − ω−z )
(ω − ω+p )(ω − ω−
p )
iγ/2
Analytic inupper half plane
pole zero
branchcut
ω1ω2−ω1 −ω2
Kelvin Wagner, University of Colorado Physical Optics 2018 48
Propagation Velocity Limit
n(ω) =
√(ω − ω+
z )(ω − ω−z )
(ω − ω+p )(ω − ω−
p )
choose branch cuts so that n(ω) → 1 as |ω| → ∞Evaluate 5 by contour integration ω = ωr + iωi n(ω) → 1 as |ω| → ∞
ei(k(ω)z−ωt) = eiω(n(ω)c z−t
)→ ei
ωc (z−ct) = e−ωi(z−ct)eiωr(z−ct)
z > ct use upper half plane contour
z < ct use lower half plane contour
no poles in upper half plane for either n(ω) or A(ω) ⇒∮= 0 z > ct
u(z, t) = 0 z > ct
Kelvin Wagner, University of Colorado Physical Optics 2018 49
Fourier representations
sinω0t ⇐⇒ iδ(ω − ω0)− iδ(ω + ω0)
Π
(t
T
)⇐⇒ iT sinc(T (ω − ω0))− iT sinc(T (ω + ω0))
f(t) =
0 t < 0sinω0t t > 0
=−ω
2π
∫e−iωt
ω2 − ω20
dω
t < 0 ∮→ 0
e−iωt → e−ωIt → 0ωΙ
ωRω0−ω0t > 0∮
= 2πi∑
residues = ω0ie−iω0t − ie+iω0t
2ω0= sinωt
e−iωt → e−ωIt → 0ωΙ
ωR
ω0−ω0
cancels
Dispersive propagation
f(x, t) =−ω
2π
∫ei(k(ω)z−ωt) dω
ω2 − ω20
ωΙ
ωR
ω0−ω0
cancels
−ωp −ωz ωpωz
Kelvin Wagner, University of Colorado Physical Optics 2018 50
Stationary Phase
Integrals of the form
I =
∫F (ω)eiφ(ωdω
F (ω) slowly varying with ω
φ(ω) phase that results in rapid variation of eiφ(ω)
integrand averages to zero when phase varies rapidly except when φ(ω) is stationary
φ′s =
dφdω
∣∣∣ωs
= 0 ωs is stationary point
⇒ φ(ω) = φs + 0 · (ω − ωs) +12φ′′s(ω − ωs)
2 + . . . nearby to ωs
Since F (ω) is slowly varying it is constant near ωs
Is = F (ωs)eiφs
∫eiφ
′′s(ω−ωs)
2/2dω = F (ωs)eiφs
√2πi
φ′′s
more than one stationary point yields a sum of terms
When φ(ω) is real [φ = ωcn− ωt ] stationary phase works, but in regions of absorption
where N = n + iκ it can produce erroneous results. Use steepest descent instead
Kelvin Wagner, University of Colorado Physical Optics 2018 51
Propagation near dispersive resonance
φ(ω) = k(ω)z − ωt where k(ω) = ωcn(ω)
Stationary points (k = ωtz )
∂φ∂ω = k′z − t ⇒ cdkdω = ct
z for ct > z t0 =zc
n2(ω) = 1 +ω2p
ω20 − ω2 − iωγ
|ω|→∞−−−−→ 1− ω2p
ω2
⇒ n(ω) → 1− ω2p
2ω2|ω| → ∞
⇒ k(ω) → ω
c
(1− ω2
p
2ω2
)|ω| → ∞
cdk
dω=
(1− ω2
p
2ω2
)−ω(−2)
ω2p
2ω−3 = 1+
ω2p
2ω2
1
n(0)
ω0+ωp2 2
dkdω
c
n(ω)ωs
ωs
SommerfeldPrecursor
BrilluoinPrecursor
ct/z
ω0
stationary point ωs when cdk(ωs)dω
= ctz= 1 +
ω2p2ω2
ct− z
z=
ω2p
2ω2s
⇒ ωs = ωp
√z/c
2(t− z/c)= ωp
√t0
2(t− t0)
Kelvin Wagner, University of Colorado Physical Optics 2018 52
Sommerfeld and Brillouin Precursor
u(z, t) =
∫ ∞
−∞
2
1 + n(ω)A(ω)ei(k(ω)z−ωt)dω =
2
1 + n(ωs)A(ωs)e
i[k(ωs)z−ωst]
√2πi
φ′′s
Complex contour integration and bessel integral gives
u(z, t) = a
(t− t0ω2pt0/2
)m/2
Jm
[2
√ω2pt0
2(t− t0)
]
Sommerfeld precursor: pushes x-raysthrough media at c
Brilluion precursor: low frequenciespropagate through at DC phase velocity
since this is a point where φ′′ = 0 must ex-pand to φ′′′. Leads to Airy integral
t
zt=z/c
tB=zn(0)/c t=zn(ω0)/c t=zng/c
very low freqs
x-rays
These precursors are due to presence of frequency components at all frequencies due tostep in time, with all frequencies present.More realistic laser pulses are bandlimited ⇒ no presursors observed.
Kelvin Wagner, University of Colorado Physical Optics 2018 53
Dispersing Prism
A
αβ γ δ
A
Snell’s law at inputsinα = n(λ) sinβ
Prism anglesβ + γ + (180− A) = 180 =⇒ γ = A− β
Snell’s law at outputsin δ = n(λ) sin γ
Total deviation angleD = α + δ − A
Kelvin Wagner, University of Colorado Physical Optics 2018 54
Dispersion in a thin prism
B=AL
W
A θ
nAW
W(A+θ)
s
L
A is apex angle. For prism height L base is B = 2L tan A2 ≈ AL
∂D
∂λ=
∂δ
∂λ=
∂δ
∂n
∂n
∂λ= A
∂n
∂λ
nAL = W (A + θ)
Solving for θ and for the thin prism approximating W ≈ L
θ =1
W(nAL−WA) ≈ (n− 1)A
Taking the derivative wrt λ gives
∂θ
∂λ= A
∂n
∂λSo the variation of deviation angle D with wavelength λ is apex angleA× disperison, and ∂δ
∂n= A
While in a thick prism at minimum deviation dδdλ = dδ
dndndλ
For instance at minimum deviation with n = sin δ/ sin γ thus dδdn
= sin γcos δ
For refraction at both faces with γ = A/2 and expressing in terms of lengths
dδ
dn= 2
sin γ
cos δ=
2 sin A2
cos δ=
2s sin A2
s cos δ=
B
WKelvin Wagner, University of Colorado Physical Optics 2018 55
Deviation Angle
sin δ = n(λ) sin γ = n sin(A− β) = n(λ)[sinA cos β − cosA sin β]
n sin β = sinα
n cos β =√n2 − sin2 α
sin δ = sinA√n2 − sin2α− cosA sinα
D = α + δ − A
αβ
∂D∂α = 1 + ∂δ
∂α − 0 = 0 at minimum deviation
∂δ
∂α= −1 =
∂δ
∂γ
∂γ
∂β
∂β
∂α=
n cos γ
cos δ· (−1) · cosα
n cos β=⇒ cosα cos γ = cos β cos δ
=⇒ α = δ β = γ = 12A Symmetric Ray
Kelvin Wagner, University of Colorado Physical Optics 2018 56
Dispersion
Index measured at spectroscopic line
Atom Line wavelength Crown FlintK-50 BaFN-10
H C 656.3 1.51992 1.6658Na D 589.3He d 587.6 1.52257 1.67003H F 486.1 1.52861 1.68000
522602 670472
Abbe
number V
V =nd − 1
nF − nC60.1 47.2
UV
IR
Visible
Absorption
Index
486
H
587
He
656
H
nF n d n c
λ
λ
Glass designation3 digits of index (d-line) -1 + 3 digits of V-number[(nd − 1) ∗ 1000] ∗ 1000 + V ∗ 10
Kelvin Wagner, University of Colorado Physical Optics 2018 57
Prism Dispersion
Change of refractive index with wavelength
∂δ
∂λ=
∂δ
∂n
∂n
∂λsince
sin δ = sinA√n2 − sin2α− cosA sinα
1.5
2.0
2.5
3.0
400 500 600 700
TiO2
FusedSilica
Flint
∂ sin δ
∂n= cos δ
∂δ
∂n= sinA
1
2(n2 − sin2 α)−1/2 · 2n =
sinA
cos β
=⇒ ∂δ
∂n=
sinA
cosβ cos δ
dispersion =∂δ
∂λ=
sinA
cosβ cos δ
∂n
∂λ
For a 60 prism at minimum deviation sinAcos β = sin 2β
cos β = 2 sin β = 1
dispersion = 1cos δ
∂n∂λ
∆n∆λ ≈ 2.7−2.3
.6−.4 = 2.0µm−1
Kelvin Wagner, University of Colorado Physical Optics 2018 58
Simple resolution argument
λλ+δλ
λa a
B
One extra wave OPD gives Rayleigh limit
λ : 2a = nB
λ + δλ : 2a + λ = (n + δn)B
subtracting: λ = δn · B
− λ
δλ= B
dn
dλ
Kelvin Wagner, University of Colorado Physical Optics 2018 59
Prism Spectroscope
Source
Slit
B
Dispersing Prism
Collimating Lens
Focusing lens
Detector Plane
α
γ
δ
• Source must be focussed through a narrow slit. Slit width limits resolution.
Image of slit determines resolution. Diffraction limited or geometrical?
• Radiometric throughput of incoherent source dramatically limited by slit width
•Wide slit ⇒ Photon efficient.
• Narrow slit ⇒ High resolution.
• Angles of refraction determined by Snell’s law and glass dispersion. k-space.
• Blue refracted by larger angle than red, nb > nr
Kelvin Wagner, University of Colorado Physical Optics 2018 60
Resolving Power of a Prism
λλ+δλ
λa a
BSlit
D2
D1
W
F2
S
SW
δ
Each wavelength gives rise to an image of the slit broadened by diffraction
∆θ =λ
W∆x = F2∆θ
Just resolved spectral lines λ, λ + dλ give separation dδ
λW
= ∂δ∂λ
· dλ =⇒ Resolving Power = λdλ
= W ∂δ∂λ
Dispersion - Change of angle with wavelength B
∂δ
∂λ=
sinA
cos β cos δ
∂n
∂λ=⇒ λ
dλ=
W
cos δ
sinA
cos β
∂n
∂λ=
︷ ︸︸ ︷S · 2 sin A
2
∂n
∂λ
Kelvin Wagner, University of Colorado Physical Optics 2018 61
Minimum Deviation Resolving Power
At minimum deviation A = β/2
sinAcos β = sin 2β
cos β = 2 sin β
2∆ = 2 · S sin A2 = D2 −D1
2D /2
D /21
∆
SA/2
Resolving Power
RP =λ
dλ= (D2 −D1)
∂n
∂λ
To resolve Na D lines (588.9 and 589.3) required RP > 980
D2 −D1 > 980/.12µm−1 = 8170µm ≈ 9mm
Kelvin Wagner, University of Colorado Physical Optics 2018 62
Prisms, dispersion, and pulse tilt
Angular dispersion gener-ates tilted pulses
Real and momentum spaceprofiles of tilted pulses
2δ
1δA
x
Tilted Pulse
Dispersion broadened
θ
n
l(x)
x
z3
3
z2
x2
1z
x
1
Prism
Input Pulse
Kishore Y, Department of Electrical Engineering, University of Colorado, Boulder -p.18/40Kelvin Wagner, University of Colorado Physical Optics 2018 63
Titled pulses out of prisms with 1st orderdispersion and including either 2nd order
dispersion or 3rd order dispersion
Kelvin Wagner, University of Colorado Physical Optics 2018 64
Prism curved slit image
Kelvin Wagner, University of Colorado Physical Optics 2018 65
Grating Spectroscopy
θλ
d
θλ
d
sin θ =λ
d
d
x2x1
dθiθ
d
x2x1
dθiθ
sin θi =x1d
sin θi =x2d
Resonace condition: OPD = integer number of wavelengths
x1 + x2 = kλ = d(sin θi + sin θd) order k
Dispersion
d(kλ)
dλ=
d
dλ(d(sin θi + sin θd)) ⇒ k = d cos θd
dθddλ
dθddλ
=k
d cos θdKelvin Wagner, University of Colorado Physical Optics 2018 66
Grating Resolution
Diffracted beam at angle θd has width W = Nd cos θd = L cos θdN=no. grating periods illuminated, L = Nd= total illuminated widthDiffraction of uniform illuminated width leads to a beam spreading angle
∆θ =λ
W=
λ
Nd cos θdAngle between two adjacent wavelengths sparated by by ∆λ
∆θd =k
d cos θd∆λ
When spectral components are separated by ∆θthey are barely resolved
∆θ =λ
Nd cos θd≡ k
d cos θd∆λ = ∆θd
Resolving PowerR =
λ
∆λ= kN
Kelvin Wagner, University of Colorado Physical Optics 2018 67
Grating Resolution and Higher Orders
• One extra wave OPD gives just resolvable angle
∆θ across beam width Nd
• Lens focuses just resolvable beams to just resolv-
able spots
• Higher order lead to higher resolution
• Just resolvable in m = 1
• Well resolved in m ≫ 1Jenkins and White, Fundamentals of Optics, 1950.2001
Kelvin Wagner, University of Colorado Physical Optics 2018 68
Array Theorem
*
comb
window
element
# zeroes =N-1 # subpeaks =N-2To represent diffraction gratings and other finite width periodic structures
Consider a finite width periodic structure
g(x) = w(x) ·[f(x) ∗ 1
Xcomb
( x
X
)] G(u) = W (u) ∗ [F (u) · comb (uX)]
2D
g(x, y) = w(x, y) ·[f(x, y) ∗ ∗ 1
XYcomb
( x
X,y
Y
)]
G(u, v) = W (u, v) ∗ [F (u, v) · comb (uX, vY )]
Kelvin Wagner, University of Colorado Physical Optics 2018 69
Square Wave Amplitude Grating
1/2+tm
1/2-tm
1.0
0
1/2
L
x
Square Wave Amplitude Transmission
Amplitude square wave of period L grating can be represented in various ways
t(x) = comb L(x) ∗[(12 − tm)Π
(xL
)+ 2tmΠ
(xL
)]
Thus the FT is given by
T (u) =1
Lcomb 1/L(u)
[(12− tm)Lsinc(Lu) + 2tmL/2sinc
(Lu
2
)]
With Fourier orders
cn = (12− tm)sinc(n) + tmsinc
(n2
)
Fraction of light power diffracted into each first order is given by
|c1|2 =∣∣∣∣(12 − tm)
sin 1π
1π+ tm
sinπ/2
π/2
∣∣∣∣2
=
∣∣∣∣(12 − tm)0
1π+ tm
1
π/2
∣∣∣∣2
=
(2tmπ
)2
< 10.13%
Kelvin Wagner, University of Colorado Physical Optics 2018 70
Square Wave Amplitude Grating
T (u) =1
Lcomb 1/L(u)
[(12 − tm)Lsinc(Lu) + tmLsinc
(Lu
2
)]
tm = 0.5 tm = 0.2
Kelvin Wagner, University of Colorado Physical Optics 2018 71
Square Wave Phase Grating
1
L
ϕ1
iϕe
real
ima
g
-1
1
When ϕ=π, becomes a bipolar square wave
Phase grating jumps back and forth between 1 (phase 0) and phase φ
Unit amplitude Square Wave Fourier coefficients
cn =1
LFΠ
(x
L/2
)∣∣∣∣fx=n/L
= 12sinc
(n2
)
Square wave phase grating
t(x) = 1− [(1− eiφ)× square wave] = 1−[(1− eiφ)×
∞∑
n=−∞cne
i2πnx/L
]
FT is given by
T (u) = δ(fx)− (1− eiφ)
∞∑
n=−∞
12sinc
(n2
)δ(fx −
n
L
)
To maximize DE into 1st order choose cosφ = −1 ⇒ φ = π
η±1 = −212
sin(1π/2)
1π/2=
2
π⇒ |η±1|2 =
4
π2= 40.5%
Kelvin Wagner, University of Colorado Physical Optics 2018 72
Square Wave Phase Grating
T (u) = δ(fx)− (1− eiφ)
∞∑
n=−∞
12sinc
(n2
)δ(fx −
n
L
)
φ = π φ = π/2
Kelvin Wagner, University of Colorado Physical Optics 2018 73
Sinusoidal Phase grating
t(x) = eim2 sin(2πf0x)Π
(xw
)
Use Bessel identity
eim2 sin(2πf0x) =
∞∑
n=−∞Jq
(m2
)ei2πqf0x
T (u)=
∞∑
n=−∞Jq
(m2
)w2sinc(w(u− qf0))
PhaseGrating
corrugatedwavefron
resolved intoplane wave k-space
Λ=
2π ΛK =---t θ
Fraunhoffer diffraction: U(x, y) = eikz
iλz ei k2z (x
2o+y2o)T
(xλz ,
yλz
)
I(x) =1
λ2z2
∞∑
n=−∞J2q
(m2
)w4sinc2(w(u− qf0))
Kelvin Wagner, University of Colorado Physical Optics 2018 74
Grating Orders and Blazing to IncreaseEfficiency
β δ
Nd
Transmission Grating
• Twin symmetric ± orders
• 25% limit to Efficiency
• red diffracted more than blue
Blazed Transmission Grating
0 d x
OPD
λ0
0th order
+1 order
+2 order
-2 order
-1 order
BlazedTransmission
Grating
• Less common than reflection
• Steers light into desired order
• Efficiency peaks at blaze λbl
•Higher order: echelons
Kelvin Wagner, University of Colorado Physical Optics 2018 75
Blazed Reflection Grating
β
βm=0
m=1
δd
Reflection Grating (Blazed)
Nd
k-spacekx
kz
Kelvin Wagner, University of Colorado Physical Optics 2018 76
Blazed Gratings in k-spaceand the Littrow condition
Element
Function
Kelvin Wagner, University of Colorado Physical Optics 2018 77
Blazed grating Diffraction Efficiency andOrders
In either reflection or transmission geometry, linear phase across element function
t(x) = combL(x) ∗ Π[xL
]eiφ0x/L
φ
x
L/2-L/2
φ0/2
−φ0/2Fourier spectrum gives DE of various orders
T (u) = 1Lcomb1
L(u)Lsinc[L (u + φ0/2πL)] =
∑
n
δ(u− nL)sinc
[Lu + φ0
2π
]
=∑
n
δ(u− nL) sinc
[n + φ0
2π
]
︸ ︷︷ ︸an
DE into nth order ηn = |an|2 goes to 100% when φ0
is 1 wave
|T (u)|2 =∑
n
δ(u− nL)sinc
2[n + φ0
2π
]
φ0=2π⇒∑
n
δ(u− nL)sinc
2[n + 1]
n=−1⇒ 100%
Kelvin Wagner, University of Colorado Physical Optics 2018 78
Monochrometer: Rotatable grating and aslit
MonochromaterRotatable Grating
Entrance Slit
Exit Slit
Kelvin Wagner, University of Colorado Physical Optics 2018 79
High resolution grating: skimming incidence
Kelvin Wagner, University of Colorado Physical Optics 2018 80
Zero dispersion line geometry
Zer
o D
ispe
rsio
nge
omet
ry
Kelvin Wagner, University of Colorado Physical Optics 2018 81
Mickelson’s Echelon
Michelson’s Transmission Echelon, m=20000,N=20 => R=400,000
t=1cm
Entrance Slit, H or V
SpectrometerEntrance slitcan be H or V
ReflectionEchelon
Gasinlet
Win
dowPressure tuned
Kelvin Wagner, University of Colorado Physical Optics 2018 82
Echelle Grating
Tilt can be in or out of plane of dispersion
Echelle Gratingin reflection
100λ
m=100,N=100-1000
Littrow MountedBlazed Grating
λ/2
m=1 or 2N=10 -10
4 5
Convention: Grating tiltangles have integer tan45, 63.43, 71.56, 75.96, 78.69
45
63.4
71.5675.9678.7
• Steps order of 100λ in Echelle, 1λ in blazed grating
Massively overlapping spectral orders
• Litrow mounting in reflection is similar geometry
conventional Echelle at angles with integer tan θ = 2− 8, 63.4,71.5,76,78.7,80.5,81.9
• 31.6,79,158,316Groovesmm ⇒ corresponds to 10,25,50,100 waves at 6328nm per face
• resolution at blaze angle of 63 degree 90% of max achievable
• sorting overlapping orders leads to crossed dispersion prism or grating
Kelvin Wagner, University of Colorado Physical Optics 2018 83
Crossed Echelles for High-ResolutionSpectroscopy: Keck HIRES Solar Spectrum
Kelvin Wagner, University of Colorado Physical Optics 2018 84
Blackbody Distribution
Spectral Energy per unit wavelength
Bλ(T ) =2hc2
λ5
1
ehc/λkBT − 1W/Sr/cm2/nm
Spectral Energy per unit frequency
Bν(T ) =2hν3
c21
ehν/kBT − 1W/Sr/cm2/Hz
Power in differential bandwidths are equal Bλdλ=Bνdν and∫∞0 Bλdλ=
∫∞0 Bνdν=
σT 4
π
integrating over hemisphere solid angle removes 2π half sided integral gives 12
and σ = 5.6704× 10−8Jm−2s−1K−4 is Stefan Boltzman constant
Since ν = cλ, thus dν
dλ= − c
λ2, giving Bλdλ = −Bν
cλ2dλ or Bν = −Bλ
λ2
c
In terms of number of photons of energy hν = hc/λ per unit bandwidth
Bλ(N, T ) =2c
λ4
1
ehc/λkBT − 1hν/Sr/cm2/nm
Kelvin Wagner, University of Colorado Physical Optics 2018 85
Temperature variation of BlackbodyDistributions vs different scales
B.H. Soffer and D.K.Lynch, Some paradoxes, error, and resolutions concerning the spectral optimization of human vision, Am. J. Phys., 67,946, 1999
As the temperature heats up the peak wavelength shifts to higher energy (Wien’s law)But the peak is different when represented as a distribution in different units!In wavelength the solar peak for T=5800K is 500nm (near 560nm eye peak sensitivity)In freqency the solar peak for T = 5800K is at 3.5× 1014 corresponding to 880nmIn photons/sec/nm peak for T = 5800K is at 633nmIn log λ or log ν the peak is at 720nm ( d(log λ) = −d(log ν) )
This discrepancy is due to the Jacobian of the transformationUnitless filter transmission functions dont need Jacobian transformation (ratios cancel)
Kelvin Wagner, University of Colorado Physical Optics 2018 86
Transformation of Spectral Energy Distribution
Density Distribution 6= UnitlessTransmission
Spectra measured as a function of wavelength using grating with linear wavelength dispersion can be
transformed to nonequally spaced frequency (or energy or wavenumber scale), but must also include
Jacobian Bλdλ=Bνdν λ = c/ν ⇒ dλ = −c/ν2dν
Kelvin Wagner, University of Colorado Physical Optics 2018 87
Blackbody Spectral Energy DistributionExample of Solar blackbody
Wavelength Peak at 500nm is only coincidentally related to yellow color
Integral of Energy dλ gives same value as integral across Jacobian weighted dν
Kelvin Wagner, University of Colorado Physical Optics 2018 88
Spectral Energy per Fractional BandwidthDistribution. Solar blackbody at T=5800
Dividing by photon energy hν = hc/λ gives photon number
Fractional bandwidth resolution (eg grating lines) should be plotted on d log λ scale
Kelvin Wagner, University of Colorado Physical Optics 2018 89
Pulse propagation Equations
3+1 dimensional vector wave equation(
∂2
∂x2+
∂2
∂y2+
∂2
∂z2− 1
c2∂2
∂t2
)~E(x, y, z, t) = µ0
∂2
∂t2~P(x, y, z, t)
Describes diffraction, dispersion, refraction, linear loss and gain. Breaking up ~P = ~PL+~PNL allows inclusion of nonlinear effects such as nonlinear absorption and refraction,harmonic generation, Raman and Brilluoin processes and more. We will neglect thesenonlinear terms since they only become important for very high fields.
Linearly polarized field propagating along z leads to reduced 1+1D wave eqn(
∂2
∂z2− 1
c2∂2
∂t2
)E(z, t) = µ0
∂2
∂t2P (z, t)
Now the linear polarization can be represented in the frequency domain as a susceptibility
P (ω, z) = ǫoχ(ω)E(ω, z)
Or by the convolution thm (and Kramers-Kronig causality conditions between χ′(ω)and χ”(ω))
P (t, z) = ǫo
∫ t
−∞χ(t′)E(z, t− t′)dt′
Kelvin Wagner, University of Colorado Physical Optics 2018 90
1D Wave eqn in temporal frequency domain
Finite response time leads to dispersion in the frequency domain
χ(ω) =
∫ ∞
−∞χ(t)e−iωtdt
temporal Fourier transform of 1+1D wave eqn leads to the wave eqn for each spectralcomponent [
∂2
∂z2+
ω2
c2
]E(z, ω) = −µ0ǫoχ(ω)E(z, ω)
ǫ(ω)[∂2
∂z2+ω2
c2
︷ ︸︸ ︷(1 + χ(ω))
]E(z, ω) = 0
Solution of the 2nd order wave eqn leads to ±z propagating solutions
E(z, ω) = E(0, ω)e±ik(ω)z
where propagation constant k(ω) is given by the square root of
k2(ω) =ω2
c2εr(ω) =
ω2
c2n2(ω)
Kelvin Wagner, University of Colorado Physical Optics 2018 91
We will Taylor expand k(ω) about the center frequency ωo
k(ω) = k0 +∂k
∂ω
∣∣∣∣ωo
(ω − ωo) +∂2k
∂ω2
∣∣∣∣ωo
(ω − ωo)2
2!+ · · · = k0 + δk
and write equation for +z propagating monochromatic solution with rapidly varyingpart separated out
E(z, ω) = E(0, ω)e−ik(ω)z = E(0, ω)e−iδk(ω)z
︸ ︷︷ ︸ e−ik0z
E(z, ω) is slowly varying partwith k20 = ω2o n
2(ωo)/c2
For a small bandwidth∣∣∣∆kk0
∣∣∣≪ 1 the envelope varies slowly with z∣∣∣∣d
dzE(z, ω)
∣∣∣∣≪ k0|E(z, ω)|
So that in one wavelength of propagation the envelope varies sufficiently slowly. FT intotime domain
E(t, z) =1
2π
[∫ ∞
−∞E(0, ω)e−iδk(ω)zei(ω−ωo)tdω
]ei(ωot−k0z) = 1
2E(t, z)ei(ωot−k0z)
Kelvin Wagner, University of Colorado Physical Optics 2018 92
Taylor expanding χ(ω) (or ǫ(ω)) about ωo
χ(ω) = χ(ωo) +∞∑
n=1
1
n!
∂nχ
∂ωn
∣∣∣∣ωo
(ω − ωo)n ǫ(ω) = ǫ(ωo) +
∞∑
n=1
1
n!
∂nǫ
∂ωn
∣∣∣∣ωo
(ω − ωo)n
Gives the polarization in the frequency domain
P (ω, z) = ǫo
(ǫ(ωo)− 1 +
∞∑
n=1
1
n!
∂nǫ
∂ωn
∣∣∣∣ωo
(ω − ωo)n
)E(ω, z)
Which can be transformed back to the time domain using ∂n
∂tn (−iω)n
P (t, z) = 12
[ǫo[ǫ(ωo)− 1]E(t, z) + ǫo
∞∑
n=1
(−i)nǫ(n)(ωo)
n!
∂n
∂tnE(t, z)
]ei(ωot−k0z)
where ǫ(n)(ωo) =∂nǫ∂ωn
∣∣ωo. Now substitute this representation into the polarization driven
wave eqn.
Kelvin Wagner, University of Colorado Physical Optics 2018 93
(∂2
∂z2− 1
c2∂2
∂t2
)1
2E(t, z)ei(ωot−k0z) = µo
∂2
∂t2P (t, z)
1
2
(∂2E∂z2
− 2ik0
∂E∂z
−k20E − 1
c2
(SSSSS
∂2E∂t2
+ 2iωo
∂E∂t
−ω2
o E))
ei(ωot−k0z)
=1
2
1
c2
[(ǫ(ω)− SS1)
∂2E(t, z)∂t2
+∂2
∂t2
( ∞∑
n=1
(−i)nǫ(n)(ωo)
n!
∂nE(t, z)∂tn
ei(ωot−k0z)
)]
=1
2
1
c2
[ǫ(ω)
∂2E(t, z)∂t2
+ (−i)ǫ(1)(ωo)
1!
(∂3E∂t3
+ 2iωo
∂2E∂t2
− ω2o
∂E∂t
)
+(−i)2ǫ(2)(ωo)
2!
(∂4E∂t4
+2iωo
∂3E∂t3
−ω2o
∂2E∂t2
)+
∂2
∂t2
(∞∑
n=3
(−i)nǫ(n)(ωo)
n!
∂nE(t, z)∂tn
)]ei(ωot−k0z)
Simplifying to second order
−ik0
(∂E∂z
+(c−1+
ω2o ǫ
(1)(ωo)
k02c2
)
︸ ︷︷ ︸
∂E∂t
)+1
2
∂2E∂z2
=1
2
1
c2
(ǫ(ωo)+2ωoǫ
(1)(ωo)+ǫ(2)(ωo)
2ω2o
)
︸ ︷︷ ︸
∂2E∂t2
k′ k0k”
Kelvin Wagner, University of Colorado Physical Optics 2018 94
Group Velocity Frame
Transform to a coordinate system moving with group velocity vg =(dkdω
)−1
Z = z T = t− z
vg
Partial derivatives become1 0
∂
∂t=
︷︸︸︷∂T
∂t
∂
∂T+
︷︸︸︷∂Z
∂t
∂
∂Z=
∂
∂T
− 1vg
1
∂
∂z=
︷︸︸︷∂T
∂z
∂
∂T+
︷︸︸︷∂Z
∂z
∂
∂Z=
∂
∂Z− 1
vg
∂
∂T
Leading term becomes simple evolution in stationary coordinate system(∂
∂z+
1
vg
∂
∂t
)E =
(∂
∂Z−
1
vg
∂
∂T
)E +
1
vg
∂
∂TE =
∂E∂Z
Kelvin Wagner, University of Colorado Physical Optics 2018 95
Slowly Varying Envelope Approximation
Equation for forward envelope implicitly neglecting reflections becomes
∂
∂ZE − i
2k0”
∂2
∂T 2E +
i
3k0c2
∞∑
n=3
(−i)n
n!
[ω2o ǫ
(n)(ωo) + 2nωoǫ(n−1)(ωo)
+n(n− 1)ωoǫ(n−2)(ωo)
]∂nE∂T n
= − i
2k0
∂
∂Z
(∂
∂Z− 2
vg
∂
∂T
)E
Group Velocity Dispersion (GVD) parameter is
k0” =∂2k
∂ω2
∣∣∣∣ωo
= − 1
v2g
∂vg∂ω
∣∣∣∣ωo
=1
2k0
[2
v2g− 2
c2ǫ(ωo)−
4ωo
c2ǫ(1)(ωo)−
ω2o
c2ǫ(2)(ωo)
]
Slowly varying conditions∣∣∣dE(t)dt
∣∣∣≪ ωo |E(t)| and∣∣∣dE(ω,z)dz
∣∣∣≪ k0
∣∣E(ω, z)∣∣ become
∣∣∣∣1
k0
(∂
∂Z− 2
vg
∂
∂T
)E∣∣∣∣ =
∣∣∣∣1
k0
(∂
∂z− 1
vg
∂
∂t
)E∣∣∣∣≪ |E|
Thus this SVEA allows us to neglect right hand side and gives first order evolution eqn.Neglecting 3rd and higher order dispersions gives GVD dispersion eqn
∂
∂ZE − i
2k0”
∂2
∂T 2E = 0
Kelvin Wagner, University of Colorado Physical Optics 2018 96
Gaussian Pulse Propagation Snapshots
Kelvin Wagner, University of Colorado Physical Optics 2018 97
Rectangular Pulse Propagation Snapshots
Kelvin Wagner, University of Colorado Physical Optics 2018 98
Gaussian and Rectangular PulsePropagation in space z
Kelvin Wagner, University of Colorado Physical Optics 2018 99
Gaussian and Rectangular PulsePropagation in reduced group velocity
coordinate frame (Z, T )
Kelvin Wagner, University of Colorado Physical Optics 2018 100
Gaussian and Rectangular PulsePropagation
in space and time Z, T )
Kelvin Wagner, University of Colorado Physical Optics 2018 101
Dispersive Pulse Propagation Equations
Dispersive vector wave eqn
∇2~E−∇(∇ · ~E)− 1
c2∂2~E
∂t2= µ0
∂2~P
∂t2
First need to expand linear dispersive Polarization response of medium
Pj(~r, t) = ǫo
∫ ∞
0
Rjk(~r, τ )Ek(~r, t− τ )dτ
Linear causal response function Rjk(~r, τ ) anisotropic and spatially varyingAssume quasi-monochromatic SSB Electric Field input centered at ω0 (and −ω0)
~E(~r, t) = 12
[~A(~r, t)e−iω0t + ~A∗(~r, t)eiω0t
]
Fourier transform in time
~E(~r, ω) = 12
[~A(~r, ω − ω0) + ~A∗(~r, ω + ω0)
]
This allows causal temporal response to be written as a product in the frequency domainand inverse FT
Pj(~r, t) = ǫo1
2π
∫χjk(~r, ω)Ek(~r, ω)e+iωtdω
Kelvin Wagner, University of Colorado Physical Optics 2018 102
Polarization in frequency and time domain
Taylor expand χjk(~r, ω) about center freq of pulse ω0
Pj(~r, ω) = ǫoχjk(~r, ω)Ek(~r, ω) =ǫo2χjk(~r, ω)
[Ak(~r, ω − ω0) + A∗
k(~r, ω + ω0)]
=ǫo2
[χjk(~r, ω0) +
∞∑
s=1
[ω − ω0]s
s!
∂sχjk(~r, ω)
∂ωs
∣∣∣∣∣ω=ω0
]Ak(~r, ω − ω0)
+ǫo2
[χjk(~r,−ω0) +
∞∑
s=1
[ω + ω0]s
s!
∂sχjk(~r, ω)
∂ωs
∣∣∣∣∣ω=−ω0
]A∗
k(~r, ω + ω0)
Consider freqs near ω0 as Ω′=ω−ω0 and Ω=ω+ω0. Inverse FT back to time domain ( ∂
n
∂tn ⇐⇒ (−iω)n)
Pj(~r, t) =ǫo2
1
2π
∫ 0
−∞
χjk(~r, ω0) +
∞∑
s=1
[Ω]s
s!
∂sχjk(~r, ω)
∂ωs
∣∣∣∣∣ω=−ω0
Ak(~r,Ω)e
+i(
ω︷ ︸︸ ︷Ω− ωo)tdΩ + cc
=ǫo2
χjk(~r, ω0) +
∞∑
s=1
∂sχjk(~r, ω)
∂ωs
∣∣∣∣∣ω=−ω0
12π
∫ 0
−∞
[ iiΩ]s
s!Ak(~r,Ω)e
+iΩtdΩ
e−iωot + cc
=ǫo2
χjk(~r, ω0)Ak(~r, t) +
∞∑
s=1
is
s!
∂sχjk(~r, ω)
∂ωs
∣∣∣∣∣ω=ω0
∂sAk(~r, t)
∂ts
e−iω0t + cc
+ǫo2
[χjk
(~r, ω0 + i ∂
∂t
)Ak(~r, t)
]e−iω0t + cc
Kelvin Wagner, University of Colorado Physical Optics 2018 103
Vector Helmholtz Eq
Spatial Derivatives
∇×∇× ~E = ∇2~E−∇(∇ · ~E) =[
∂2
∂x2∂2
∂y2∂2
∂z2
]Ex
Ey
Ez
−
∂∂x∂∂y∂∂z
[
∂∂x
∂∂y
∂∂z
]Ex
Ey
Ez
Components
∇2Ej −∂
∂xj
∑
k
∂Ek
∂xk
vector Helmholtz eqn for Single Side Band (SSB) quasi monochromatic envelope
∇2Aj(~r, t)e−iω0t − ∂
∂xj
∑
k
∂Ak(~r, t)
∂xke−iω0t
=1
c2∂2
∂t2[Aj(~r, t)e
−iω0t + χjk(~r, ω0 − Ω)Ak(~r, t)e−iω0t
]
=1
c2∂2
∂t2[Aj(~r, t)e
−iω0t + χjk
(~r, ω0 + i ∂∂t
)Ak(~r, t)e
−iω0t]
Only for fractional BW η = ∆ωFWHMω0
< 1
Kelvin Wagner, University of Colorado Physical Optics 2018 104
SpatioTemporal Vector Helmholtz OperatorEquation
Using the linear dielectric tensor ǫjk(~r, ω) = 1 + χjk(~r, ω)
∇2Aj(~r, t)e−iω0t − ∂
∂xj
∑
k
∂Ak(~r, t)
∂xke−iω0t =
1
c2∂2
∂t2(ǫjk(~r, ω0 + i ∂∂t
)Ak(~r, t)e
−iω0t)
=1
c2
[∂2
∂t2− 2iω0
∂
∂t− ω2
0
] ǫjk(~r, ω0 + i ∂
∂t
)Ak(~r, t)
e−iω0t
= − 1
c2
[ω0 + i
∂
∂t
]2 ǫjk(~r, ω0 + i ∂∂t
)Ak(~r, t)
e−iω0t
In a weakly absorptive isotropic medium with possibly inhomogeneous variation[k(~r, ω) + i
α(~r, ω)
2
]2≡ ω2ǫ(~r, ω)
c2⇒ k2(ω) + ik(ω)α(ω) ≈ ω2
c2[ǫR(ω) + iǫI(ω)]
∇2Aj(~r, t)e−iω0t − ∂
∂xj
∑
k
∂Ak(~r, t)
∂xke−iω0t =
−[k(~r, ωo + i ∂
∂t
)[k(~r, ωo + i ∂
∂t
)+ iα
(~r, ωo + i ∂
∂t
)]Aj(~r, t)
]e−iω0t
Kelvin Wagner, University of Colorado Physical Optics 2018 105
Operator expansion
Instead of plane wave use wave packet centered around ~k0 = (0, 0, k0)
E(x, y, z, t) = B(x, y, z, t)ei(~k0·~r−ω0t)
Evaluation of partial derivatives yields 2 terms that can be regrouped as operator
∂
∂tB(x, y, z, t)ei(k0z−ω0t) = −iωoe
i(k0z−ω0t)B(x, y, z, t) + ei(k0z−ω0t)∂
∂tB(x, y, z, t)
= ei(k0z−ω0t)(−i)[ωo + i ∂∂t
]B(x, y, z, t)
∂
∂xB(x, y, z, t)ei(k0z−ω0t) = ei(k0z−ω0t) (i)
[0 − i ∂
∂x
]B(x, y, z, t)
∂
∂yB(x, y, z, t)ei(k0z−ω0t) = ei(k0z−ω0t) (i)
[0 − i ∂
∂y
]B(x, y, z, t)
∂
∂zB(x, y, z, t)ei(k0z−ω0t) = ei(k0z−ω0t) (i)
[k0 − i ∂
∂z
]B(x, y, z, t)
Kelvin Wagner, University of Colorado Physical Optics 2018 106
Operator equations
Helmholtz eqn∇2E + k2E = 0
Using these operator representation frequency domain dispersion operator
D(kx, ky, kz, ω)B(kx, ky, kz, ω) =
(k2x + k2y + k2z −
n2(ω)ω2
c2
)B(kx, ky, kz, ω) = 0
will become back in the time domain
D(−i ∂
∂x,−i ∂∂y , k0 − i ∂
∂z , ωo + i ∂∂t
)B(x, y, z, t)
=
(−i ∂
∂x
)2+(−i ∂
∂y
)2+(k0 − i ∂
∂z
)2−([
ωo + i ∂∂t]n(ωo + i ∂∂t
)
c
)2B(x, y, z, t) = 0
Algebraic expansion of
1
c2[ωo + i ∂∂t
]2=
1
c2
[ω2o + 2iωo
∂∂t − ∂2
∂t2
]=[k20 +
2ik0c
∂∂t − 1
c2∂2
∂t2
]
Kelvin Wagner, University of Colorado Physical Optics 2018 107
Interpret the Operator
interpret the dispersive wave number operator
k2(ωo + i ∂
∂t
)=
([ωo + i ∂∂t
]n(ωo + i ∂∂t
)
c
)2=
[ωo + i ∂∂t
]2n2(ωo + i ∂∂t
)
c2
with the Fourier domain Taylor expansion about the center frequency Ω = ω − n0
k(ω) = k0 +11!k
′0Ω + 1
2!k′′0Ω
2 + 13!k
′′′0 Ω
3 + 14!k
′′′′0 Ω4 + . . .
Square it
k2(ω)=k20+2k0k′0Ω+(k
′20 +
22!k0k
′′0 )Ω
2+( 22!k′0k
′′0+
23!k0k
′′′0 )Ω
3+( 12!2k′′20 + 2
3!k′0k
′′′0 +
14!k′′′′0 )Ω4+. . .
Fourier relation ∂∂t −iω gives ∆ω +i ∂∂t transforms back into time domain as
k2(ωo + i ∂
∂t
)= k20 + 2ik0k
′0∂∂t
k′0 = 1/vg−(k′20 + k0k
′′0 )
∂2
∂t2GVD
−i2
3(3k′0k
′′0 + k0k
′′′0 )
∂3
∂t3TOD
+1
12(3k′′20 + 4k′0k
′′′0 + k0k
′′′′0 ) ∂
4
∂t4+ . . . FOD
Kelvin Wagner, University of Colorado Physical Optics 2018 108
Expansion of kz(kx, ky, ω)
Fourier expanding the z-component of the wavevector can be extended to both the tem-poral frequency and transverse spatial frequnecies and will include all mixed derivativesas well
kz(kx, ky, ω) = kz(0, 0, ωo) +∂kz∂kx
∣∣∣0kx +
∂kz∂ky
∣∣∣0ky +
∂kz∂ω
∣∣∣ωo(ω − ωo)
+12
∂2kz∂k2x
∣∣∣0k2x +
12
∂2kz∂k2y
∣∣∣0k2y +
∂2kz∂ω2
∣∣∣ωo(ω − ωo)
2
+∂2kz
∂kx∂ky
∣∣∣0,ωo
kxky +∂2kz∂kx∂ω
∣∣∣0,ωo
kx(ω − ωo) +∂2kz∂ky∂ω
∣∣∣0,ωo
ky(ω − ωo)
= kz(0, 0, ωo) +∇kkz · ~kt +12∇k∇k : ~k~k
Terms can be identified as spatial walkoff, group velocity, diffraction, dispersion,anisotropic mixed diffraction (tilted ellipse) and space-time coupling.
Kelvin Wagner, University of Colorado Physical Optics 2018 109
Anisotropic Nonlinear Wave Eqn
∇2~E− 1
c2~E−∇(∇ · ~E) = 1
ǫoc2
(~PL + ~PNL
)
[∂2
∂x2+
∂2
∂y2+
∂2
∂x2− 1
c2∂2
∂t2
]~E−
∂∂x∂∂y∂∂z
[
∂∂x
∂∂y
∂∂z
]Ex
Ey
Ez
= 1
ǫoc2
(χ~E + χ
(2)~E~E + χ
(3)~E~E~E
)
Linear term leads to determinental eqn for plane wave SSB
~E(x, y, z, t) = A0ei(~k·~r−ωt)
∂∂x → ikx
∂∂y → iky
∂∂z → ikz
∂∂t → −iω
−k2x−k2y−k2z+
ω2
c2−
kxkx kxky kxkzkykx kyky kykzkzkx kzky kzkz
− (−ω)2
ǫoc2
χxx
χyy
χzz
Ex
Ey
Ez
= 1
ǫoc2∂2
∂t2
(χ
(2)~E~E)
Determinental soln leads to quadratic for k as fnc of m ~k = km = ωcn(ω)m.
In an arbitrary coordinate system this gives
n4~k · ǫ · ~k + n2(~k · [adjǫ− tr[adjǫ]I ] · ~k
)+ |ǫ| = 0
Kelvin Wagner, University of Colorado Physical Optics 2018 110
Uniaxial medium
ǫ = ǫoI + (ǫe − ǫo)cc
This factors into two terms
(~k2 − k20n2o)(
~k · ǫ · ~k− k20n2on
2e) = 0
Two rootsk+ = k0no
k− =k0none√m · ǫ · m
=k0none√
n2o(m× c)2 + n2
e(m · c)2Booker Quartic: Planar boundary value solutions
~k = ~kT + qn
Ordinary
(~kT + qn) · (~kT + qn)− k20n2o = 0 |~kT |2 + q2 − k20n
2o = 0 q2 = k20n
2o − k2T
q = ±√k20n
2o − k2T This is a circle. Walkoff given by derivative
∂q
∂kx= 1
2
−2kx√k20n
2o − k2T
=−kxkz
Kelvin Wagner, University of Colorado Physical Optics 2018 111
Extraordinary wave
~k · ǫ · ~k = k20n2on
2e
(~kT + qn) · ǫ · (~kT + qn)− k20n2on
2e = 0
⇒ q2 n · ǫn︸ ︷︷ ︸+q(n · ǫ · ~kT + ~kT · ǫ · n︸ ︷︷ ︸) +~kT · ǫ · ~kT − k20n
2on
2e︸ ︷︷ ︸ = 0
A B C
q =−B ±
√B2 − 4AC
2Atilted//ellipse
For a near normal incidence this can be Taylor expanded
kez = kz(0) +∂kz∂kx
∣∣∣0kx +
∂kz∂ky
∣∣∣0ky +
∂2kz∂k2x
∣∣∣0
k2x2+∂2kz∂k2y
∣∣∣0
k2y2+
∂2kz∂kx∂ky
∣∣∣0
kxky2
For an ellipse tilted by θ n = (0, sin θ, cos θ) ~kT = (kx, kt cos θ,−kt sin θ)
(n2o sin
2 θ+n2e cos
2 θ)q2+(n2o−n2
e)kt sin 2θq+k2t (n2o cos
2 θ+n2e sin
2 θ)+n2okt−k20n
2on
2e = 0
Kelvin Wagner, University of Colorado Physical Optics 2018 112