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Optics 4510, Wk. 4 Lecture 7, 8 Emission of EM waves and waves in media
Goldman


Poisson's eqn follows from wave eqn and current-charge continuity eqn.
o
Relates longitudinal current, JL to charge density, : DivJL=-∂t
o
Transverse current, JT, absent because DivJT = 0.

What is JT in the wave eqn for a linear polarizable medium?
o
Look at field pattern produced by given oscillating dipole current, JT.
o
Then see how to produce oscillating dipole currents in polarizable
media by sending a light wave through them.

Transverse electric field lines
radiating from a localized oscillating
electric (dipole) current as it radiates a
classical electromagnetic wave (e.g. radio
waves from an antenna) into empty space.

Electric Field from Dipole Antenna
o
Rotate 2D pattern into 3D
cylindrically symmetric pattern
o
Asymptotically have a plane
wave far from dipolar oscillating current.
o
Why are E-field lines loops?
o
Where is E strongest?
o
What is wave propagation direction?
o
What is local E-field direction?
o
Is E transverse or longitudinal?
o
Plot electric field waveform along
horizontal symmetry axis (for wave traveling in +x)?
o What is asymptotic shape of the wavefront centered at x??
o Why do electric field lines detach from oscillating dipole?
o Where are lines of B according to Faraday's law?
o Only electrons oscillate because they are lighter than ions.
o How might the dipole be made to oscillate as shown?
 Answer- by an oscillating local electric field which is part of an incident electromagnetic
wave. (The wave magnetic force is too small to affect the electrons).
Might expect to need quantum mechanism to understand re-radiation process from electrons in a
bound state in an atomic material such as glass or water or from "free" electrons in a conducting
metal or semiconductor.
o In bound state think (semiclassically) of the electron probability cloud shifted back and forth
relative to the nucleus.
o Current produced by electric field of light incident on an atom can be understood in terms of
virtual atomic electron energy transitions induced by an external electric field.
 Polarization of a two-level atom is very similar to polarization of the vacuum by gammaray virtual pair production in quantum electrodynamics (QED)
2

Feynman diagrams.
Polarization involves virtual transitions
(Non-resonant)
Photon absorption/emission conserves
energy (Resonant)

Classical picture works fine for radio waves from an antenna but can also be applied (with care) to
light waves, by using appropriate classical models either for free electrons (e.g., conduction electrons
in a metal) or for electrons in bound states in an atom.
o Embed the original and re-radiated light waves in a polarizable material E.g., a volume of polarizable atoms or a plasma of polarizable free electrons in which the
electric field of an incident light wave in a vacuum induces an electron (polarization)
current, JT per unit volume. Draw picture. Dielectric such as glass.
o Oscillating dipoles in the material are created by the oscillating electric (light) wave travelling
through it and then each re-radiates to add-to and modify the original light wave.
 One result is that the velocity of the combined incident and polarization fields can be
different from c!
 Another is that surface of the irradiated material can reradiate, leading to a reflected EM
wave in addition to the incident wave .
1. Simplest case is when electrons are free as in a conducting metal.
a. We will learn what gives metals such as silver, copper and gold there apparent different colors
when viewed under white light.
b. Valid to ignore fact that conduction electrons are almost degenerate (temperature and Fermi
energy are of same order of magnitude)
c. Valid to neglect interactions between electrons and ions and between electrons and other
electrons. e·phi/kBT << 1
d. Find "first-order" current (linear in the electric field for linear waves) from fluid eqns for
electrons in a neutral background of ions and electrons subjected to an "external" electric field.
i. Dipole approximation: electron excursion distance small compared to wavelength of
external field.
ii. Magnetic field exerts negligigle force on electrons because v/c is small and
F = -eE + e(v/c) x B
iii. Neglect forces on ions because of their heavy mass and motion proportional to F/Mi
e. Index of refraction and comparison of electric and magnetic field amplitudes.
f. Transverse and longitudinal waves in a plasma
g. Below are fluid eqns for electron and ion fluids in a small oscillating electric field in the dipole
approximation
i. Neglected in the momentum eqn are thermal pressure force and magnetic force.
ii. Fluid variables: ns(r, t), us(r, t). Conservation laws for density and for momentum.
(use Gauss' law to understand)
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Continuity eqn for electrons (s=e) and ions (s=i), ¶t ns + Ñ·( ns u) = 0, Momentum eqn, for electrons and ions,
¶t ( ne meu e ) + Ñ·( ne meu eu e ) = -neeE - g ( ne meu e ) , ∂t ( ni mi ui ) + Ñ·( ni mi ui ·ui ) = ni eE. Poisson: Ñ·E = 4 p e ( ni - ne )
Perturbation theory about neutral fluid with zero imposed field: n s = n0 + n1s ( r,t ) , u s = 0 + u1s ( r,t ) , s = e. i
0-order (unperturbed) "equilibrium" plasma state has no electric field, no current flow (u s 0 = 0) and no net charge
density, q(n0 - n0 ) = 0 so Ñ·E =0 and E is transverse. 0-order continuity eqn and momentum eqns, consistent with
E = 0 in momentum eqn. "First-order" in the density and fluid velocity, n1s and u1s means first-order in E (i.e., µ E).
The k - space first-order eqns are ∂t n1s +in0 k·u1s = 0, ¶t u1s = ( qs / ms ) E - g u1,s , But J e1 = -en0 u1e , so source-term
in EM wave eqn is 4 p (¶t J e1 + g J e1 ) =
4 p n0 e 2
4 p n0 e 2
E = w e2 E , w e2 =
(cgs) is the "plasma frequency," which is
me
me
the frequency of the oscillating free-electron current density in the metal (or plasma). Next, let g = 0 and insert in
transverse wave eqn. to learn what happens to a light wave in a metal:
(c Ñ
2
2
- ¶t2 ) E ( r,t ) = 4 p ¶t J e1 = w e2 E ( r,t )
Wave eqn in k - space: ( c 2 k 2 + ¶t2 ) E ( k,t ) = -4 p ¶t J e1 = -w e2 E ( k,t ) , or, (¶t2 + w e2 + c 2 k 2 ) E ( k,t ) = 0
This is again a harmonic oscillator eqn, but now with frequency w 2 = w e2 + c 2 k 2 rather than w 2 = c 2 k 2 . Sol'n for
monochromatic plane-wave in x-direc.: E ( x,t ) = A k e
ik ( x-vt )
(A k complex), with phase velocity, v =
w
we2
= c + 2.
k
k
Note: v > c is no problem because velocity of a wavepacket of such waves moves with (group) velocity vg < c.
2
Suppose an incident light wave is about to enter this metal with angular frequency, w > w e , speed c and
w
2p c c
wavenumber k inc = , so that the wavelength of the incident light is linc =
= . Since the incident light is
c
w
f
about to enter a medium which is not varying in time macroscopically, the frequency, f, will be the same inside
the medium. But the wave velocity is faster inside and now k =
inside the metal must change to l =
w
v
. This means that the wavelength
2p v v
= = linc which is greater than the incident wavelength!
k
f c
Next suppose the frequency of incident light wave has w < w e and wavenumber k inc =
w
. Inside the metal, k
c
must be imaginary in order to satsify c 2 k 2 = w 2 - w e2 < 0. The light will dim (evanescence) as e -Im kx as it tries to
propagate in the metal (physically it can't grow). It is "reflected" back out of the metal at the surface.
Conventions and notation : J e1 is often called the polarization current, J pol . In our case it is transverse. In general,
J e1 = JT + J L . Related to "polarization field," P, by J pol = ¶t P. Interpret P ( k, w ) as dipole moment per unit volume,
related to E ( k, w ) through susceptibility, c T ( k, w ) : P ( k, w ) =
The index of refraction, n, is the ratio of c to v =
w
k
: Ren =
c T ( k,w )
w2
E ( k, w ) (cgs) For metals, c T ( k, w ) = - e2 .
4p
w
ck
w
. In metals, Ren < 1. In glass and water, Ren > 1


2. Incident light impinging on a metal:
metal


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a. Inside the metal the light
creates oscillating electron dipoles
whose current density contributes to the wave eqn for the E-field of light inside the metal.
b. The E-field of light in the metal is the self-consistent
i. The light in the metal has the same frequency as the incident light but a longer
wavelength and a faster phase velocity but a slower group velocity:
dw k d
ck
c
=
we2 + c 2 k 2 = c 2 2 2 =
<c
dk dk
we + c k
we2
1+ 2 2
ck
ii. Only light with frequency above the plasma frequency can exist inside the metal.
Light below the plasma frequency is extinguished because of destructive
interference between the incident and emitted light waves. (Evanescence)
3. Result with damping,  (gives absorption) and definition of conductivity
4 p n0 e2
we2
i( k·r-wt )
2
4 p (¶t J e1 + g J e1 ) =
E = w e E , For E ( r,t ) = E ( k, w ) e
, so 4πJ e1 ( k, w ) =
E.
me
-iw + g
The wave eqn now becomes, ( -c 2 k 2 + w 2 ) E ( k, w ) =
n=
ck
w
= 1-
we2
w 2 + igw
wwe2
E ( k, w ) , so the index of refraction is,
w + ig
This now has an imaginary part (absorption) even for frequencies above the
plasma frequency. The (complex) conductivity is defined by given by J e1 ( k, w ) = s ( k, w ) E ( k, w ) , so
we2
we2
4π s ( k, w ) =
, (cgs) . In the limit of zero frequency this gives the DC conductivity, limw ®0s =
.
-iw + g
4 pg
4. Additional definitions:
a. Dispersion: Dispersion is said to occur when the wave
phase velocity (i.e., index of refraction) depends on
frequency. This leads to rainbows from prisms and water
droplets. Without dispersion the phase velocity equals the
group velocity!
b. Evanescence: destructive interference between incident
wave field and wave field of oscillating dipole. The
amplitudes of frequencies that don't propagate into the
metal are not damped but are suppressed as a result of
destructive interference.
c. Dielectric function e º1+ c e ,
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5. Complex index of refraction means complex wavenumber. Real part gives wavelength. Imaginary
part (>0) gives spatial attenuation (absorption). Electric field goes as
For complex k = k r + ik , where, the real part of k is k r and the imaginary part of k is k
and both k r and k are > 0 the electric field goes as e (
i kx-wt )
= e ([ r
i k +ik ] x-wt )
= e-k x e ( r
i k x-wt )
6. Group velocity of a wavepacket (inverse Fourier transform or superposiion of plane waves with
different wavelengths)
vgroup º
¶wk
¶
In a metal, this is
we2 + c 2 k 2 =
¶k
¶k
1
1+
we2
c2 k 2
c<c
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7.
There is another, simpler way to treat light propagation in a linearly polarized metal. Look at elec–tron current due to a
single free electron in the light wave in the dipole approximation. Then multiply by the uniform density of free electrons.
(Can add damping)
8. Next consider bound electrons: polarized atoms in a transparent material (a dielectric) such as glass. Model the motion of
bound electrons in an atom as that of a harmonic oscillator with harmonic oscillator frequency, f electron = f0 = (E2 - E1)/h.
Here E2 and E1 are two electron energy levels. Transitions between E2 and E1 account for absorption or emission of a
photon with frequency, f0. or obeys the following equation when in a light wave in the dipole approximation:
7
hf0 photon
energy
electron in E1
energy level
Photon absorption as a
resonant process.
electron in higher energy level,
E2 = E1 + hf0
8
9.
More complete theory of light in a dielectric: For a single electron the oscillator equation is,
9
10.
Graphical derivation of Snell's law for
the angle of refraction in terms of the angle of
incidence and indices of refraction.
a. Angle of reflection equals angle
of incidence can be proven this way.
i
i
r
r
normal to
surface
r
i
surface 
normal to
surface
The hypotenuse of the black triangle equals the hyponenuse of the red triangle, so
li
l
= r .
sin qi sinq r
But on each side of the interface (the medium on the left does not have to be a vaccum),
2 p 2 p vf s
w
c
ls =
=
, where the phase velocity, vf s = . Since the index of refraction is n s º ,
ks
w
ks
vf s
we can write Snell's law in two different ways:
vf r ni sin q r
= =
. When there is a vacuum on
vfi nr sin qi
the left, vfi = c and n i = 1. Since the two angles are also (by geometry) the angles the ray direction
makes to the interface normal we can state the qualitative implication of Snell's law as: the angle q
in the slower medium is always less than the angle in the faster medium. Alternatively, we can say
fast in metal
(n<1)




slow in glass
(n>1)




that a ray going into a slower medium bends towards the normal and a ray going into a faster
medium bends away from the normal.
Reviewing the logic: Light rays entering glass or water at frequencies much smaller than resonance (linetransition) frequencies bend because they are slowed down as a result of the re-radiation from the
oscillating dipoles they drive at frequencies below their resonant frequencies.
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11
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11. Miscellaneous concerning susceptibility and dielectric function
a. How to take square root of complex number without Mathematica or a similar math-processor
-1
w2 æ
w g w2 ö
2
Let n = = 1+ e2 ç1- i
- 2 ÷ = a + b i. Then, ( N + iK ) = a + b i = 1+ c e = e
w
w0 è
w0 w0 w0 ø
ck
where N = Re n and K = Im n.
Equate real and imaginary parts, N 2 - K 2 = a , 2KN = b , so K =
Substitution gives, N 2 -
b2
4N 2
b
2N
.
= a . This is a quadratic for N 2 , N 4 - a N 2 -
b2
4
= 0, with sol'n,
2N 2 = a ± a 2 + b 2 . Only the + sol'n gives real N, so
N=
1
b
a + a 2 + b 2 and K =
. In limit a >> b , N = a , K = 0 and in
2
2 a + a2 + b2
the limit a << b , N =
b
, K=
b
(correct with root i = eip /2 = eip /4 = cos 45° + i sin 45°).
2
2
b. Relation between dielectric function and line width
æ w2 ö æ w g ö
ç1- ÷ + i ç
÷
we2
1
we2 è w02 ø è w 0 w0 ø
e e = 1+ c e = a + b i = 1+ 2
= 1+ 2
, so,
w2
w g
w0
w0 æ w 2 ö2 æ w g ö2
1- 2 - i
ç1- 2 ÷ + ç
÷
w0 w0 w0
è w0 ø è w0 w0 ø
æ w2 ö
æw g ö
ç1- 2 ÷
ç
÷
2
2
we
we
è w0 ø
è w0 w0 ø
a -1 = Rec e = 2
and b = Imc e = 2
w0 æ w 2 ö2 æ w g ö2
w0 æ w 2 ö2 æ w g ö2
ç1- 2 ÷ + ç
÷
ç1- 2 ÷ + ç
÷
è w0 ø è w0 w0 ø
è w0 ø è w0 w0 ø
Ime(/0) is Lorentzian-shaped, representing the line width mainly due to spontaneous emission
at 0 from the bound electron. Plots below are of 1+ Ree(/0) (dashed blue) and Ime(/0)
(red) and of Re(n) (dashed green) and Im(n) (purple) both for 0 = e and /0 = 0.1. (Note
possible incorrect labels in Figs. 2.5 and 2.6 in online textbook.)
2.5
2.0
1.5
1.0
0.5
1.0
1.5
2.0
13
12. Wave energy transport and Poynting flux in real space and Fourier space:
c
1
In cgs units, Ampere's law is cÑ ´ B = 4 p J + ¶t E, or J =
Ñ´B¶t E . J ( r,t ) = J e + J i
4p
4p
Take dot product of both sides with E to get work done by E on J:
c
1
c
1
E·Ñ ´ B E·¶t E =
E·Ñ ´ B ¶t E 2
4p
4p
4p
8p
Faraday's law is 0 = ¶t B + cÑ ´ E . Take dot product of both sides with B and divide by 4π:
J·E =
B·∂t B 1
1
c
=
¶t ( B 2 ) , so, 0 =
¶t ( B 2 ) +
B·Ñ ´ E. Subtract 2nd eqn in red from first to obtain,
4p
8p
8p
4p
E 2 + B2 )
(
c
J·E =
. Vector identity -Ñ·( E ´ B) = E·Ñ ´ B - B·Ñ ´ E. Proof:
( E·Ñ ´ B - B·Ñ ´ E) - ¶t
4p
8p
-Ñ·( E ´ B) = -¶i (e ijk E j Bk ) = -e ijk ¶i ( E j Bk ) = - { Bke ijk ¶i E j + E je ijk ¶i Bk } = -{ Bke kij ¶i E j - E je jik ¶i Bk }
Work/vol by J on E : -J·E = ¶t U + Ñ·S , U ( r,t ) =
E ( r,t ) + B ( r,t )
= wave or field energy density
8p
2
2
c
E ( r,t ) ´ B ( r,t ) = Poynting flux (flux of field or wave energy density). EM transport eqn.
4p
1
1
• Convert from cgs to SI by using rules,
® e 0 , B ® cB and c 2 ®
:
4p
e 0m 0
S ( r,t ) =
U ( r,t ) ®
e 0 E ( r,t ) + m 0-1 B ( r,t )
2
2
2
(SI) ,
S ( r,t ) ®
1
m0
E ( r,t ) ´ B ( r,t ) (SI)
• Work on field shows up as time rate of change of field energy density and divergence of energy density.
flux Use Gauss' Thm to interpret in a volume. Energy flux leaving surface plus change of energy in volume
both result from work done on fields in volume. Field energy becomes particle energy and vice versa.
• For monochromatic wave all terms in EM energy transport eqn. have fast oscillating component usually
removed by time-averaging. All terms are 2nd order in fields and/or currents so be careful about using
e i( k·r-wt ) and taking real part after. Real part afterwards only works for linear eqns such as wave eqn and
1
*
oscillating dipole-approx. eqn. Let, E ( r,t ) = Re éëE ( k, w ) ei( k·r-wt ) ùû = éëE ( k, w ) ei(k·r-wt ) + E ( k, w ) e-i(k·r-wt ) ùû,
2
1é
1
*
*
B ( r,t ) = ëB ( k, w ) ei( k·r-wt ) + B ( k, w ) e-i(k·r-wt ) ùû, and J ( r,t ) = éëJ ( k, w ) ei(k·r-wt ) + J ( k, w ) e-i(k·r-wt ) ùû.
2
2
1
*
Time average over period, 2π/w eliminates oscillations at 2w leaving, S ( r,t ) = J ( k, w ) E ( k, w ) + cc ,
4
2
2
1
U ( r,t ) =
E ( k, w ) + B ( k, w ) .
16 p
{
{
}
}
14
The index of refraction only has an imaginary part when there is damping (g ¹ 0 ) or evanescence.
As long as the index of refraction, n is real, all of the follwoing are also real: n 2 = e ( k, w ) = 1+ c e ( k,w ) .
4 p is ( k, w )
, so s ( k,w ) = i s ( k, w ) must be purely imaginary and therefore
w
i k·r-wt )
J ( k, w ) = s ( k,w ) E ( k, w ) is out of phase with E. Hence, if E(r,t)=Re { E ( k, w ) e (
}
But c e =
(
)
= E ( k,w ) cos ( k·r - wt + q ) for E ( k, w ) = E ( k, w ) eiq ,
{
}
J ( r,t ) = Re i s ( k,w ) E ( k, w ) ei(k·r-wt ) = s ( k, w ) E ( k, w ) sin ( k·r - wt + q ) and the
time average -J ( r,t )·E ( r,t ) t = s ( k, w ) E ( k, w )
2
sin ( k·r - wt + q )·cos ( k·r - wt + q ) t = 0
What is ratio of Poynting flux to energy density for isotropic linear n with no damping or dispersion? In cgs
S ( r,t )
U ( r,t )
c
Re { E ( k, w ) ´ B* ( k, w )}
Re E ( k, w ) ´ n * k̂ ´ E * ( k, w )
= 8p
=c
2
2
2
2
1
E ( k, w ) + n * k̂ ´ E * ( k, w )
E ( k,w ) + B ( k,w )
8p
{
}
{
{
(
)} = ck̂n
}
E ( k, w )
2
(1+ n ) E ( k,w )
2
If magnetic energy density is significantly larger than electric energy density (if n 2 >> 1),
S ( r,t )
U ( r,t )
2
=
»
ck̂n
(1+ n2 )
ck̂ w
= k̂
n k
In a wave packet with dispersion this will become a group velocity which is different from the phase velocity.