# Lecture summary

Electronics - Devices

Oct 18, 2013 (4 years and 6 months ago)

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A quick introduction to the creation of ElectroMagnetic (EM) radiation.Refer to ﬁgures
distributed.(also check e.g.Hecht’s book “Optics” or similar).
The radial electric ﬁeld E of a stationary electron (or one moving with constant velocity)
gets a “kink” if the electron is accelerated and this disturbance moves with the speed of
light as an EM pulse.The electric ﬁeld in this EM pulse can be derived from the Maxwells
equations.
￿ × B = µj + ￿ µ
∂E
∂t
￿ ∙ B = 0
￿ × E = −
∂B
∂t
￿ ∙ E =
ρ
￿
For a homogeneous,neutral (charge density ρ = 0),non-magnetic (µ = µ
0
) medium where
the current j is zero they can be re-formulated as

2
E
∂ t
2
=
1
￿ µ
0
￿
2
E or

2
B
∂ t
2
=
1
￿ µ
0
￿
2
B
(

2
E
∂t
2
=
1
￿µ
0
￿ ×
∂B
∂t
= −
1
￿µ
0
￿ × (￿ × E) = −
1
￿µ
0
￿
−￿
2
E+￿(￿ ∙ E)
￿
=
=
1
￿µ
0
￿
2
E )
We recognize this as a wave equation with a propagation speed of 1/

(￿ µ
0
).The speed of
an EM wave in vacuum is c = 1/

(￿
0
µ
0
)
Some well-known results (see,e.g.Hecht Chapter 3) are that
• E and B are perpendicular to each other and to the propagation vector k
• E = c B
• Energy density in EM radiation:u = u
E
+ u
M
,i.e.the sum of the energy contained
in the electric and in the magnetic ﬁelds.These two are equally large.u
E
=
1
2
￿
0
E
2
,
u
M
=
1
2 µ
0
B
2
,so u = ￿
0
E
2
=
1
µ
0
B
2
.
• The energy ﬂow per unit area,the Poynting ﬂux,S = c
2
￿ E ×B in the k direction.
This is the instantaneous value.For an EM ﬁeld described by E = E
0
e
iωt
,the time-
aveaged value is c￿
0
< E
2
>=
1
2
c￿
0
E
2
0
.For visual radiation,recall that the electric ﬁeld
will change direction several 10
14
times per second.
• The dipole moment is deﬁned as (the vector) d = qR,i.e the charge times a position
vector from some reference point,R.The acceleration of a charge is then conveniently
described by
¨
d.
1
• The size of the electric ﬁeld vector at a distance r from the accelerating charge is (at
a large distance,close to the charge it is quite complicated...)
|E| =
|
¨
d| sinθ
4π￿
0
c
2
|r|
• Inserting the expression for the electric ﬁeld in the expression for the Poynting ﬂux we
get:
|S| = c￿
0
E
2
=
|
¨
d|
2
sin
2
θ
16π
2
￿
0
c
3
|r|
2
• To get the ﬂux per solid angle we multiply by r
2
dΩ = r
2
dθ sinθdφ.We then integrate
over all directions to get the total (instantaneous) power radiated by an accelerated
charge:
P =
￿￿
all directions
|
¨
d|
2
sin
2
θ
16π
2
￿
0
c
3
dΩ =
￿

0
￿
π
0
|
¨
d|
2
sin
3
θ
16π
2
￿
0
c
3
dθ dφ =
|
¨
d|
2
6π ￿
0
c
3
This relation between the second time derivative of the dipole moment and the total
radiated power is called the Larmor formula.
• Note that the electric ﬁeld decreases linearly with distance from the radiator,the ﬂux
per unit area as the distance squared.The ﬁeld strength varies with sinθ,the power
per solid angle as sin
2
θ.The emitted radiation is polarized with the electric ﬁeld vector
E in the same plane as the acceleration (and
¨
d).
Even nowadays one often ﬁnds these things in cgs (centimeter-gram-second) units instead...
To convert to cgs units “just”:
• replace
q
2
4π￿
0
with q
2
• the elementary charge e = 1.60210
−19
C then becomes 4.803 10
−10
• E = B
• Larmor formula then becomes
P =
2 |
¨
d|
2
3 c
3
If we look at atomic scales,we see that the electric ﬁeld aﬀects molecules with a permanent
dipole moment such that they try to rotate to follow the rapid variations of the E-ﬁeld.This
is successful for radiation frequencies up to,say,10
10
Hz.For atoms and molecules without
a permanent dipole moment,the applied electric ﬁeld will induce a dipole moment.This
induced dipole moment
d = β ￿
0
E
contains the polarizability of the atom,β.(From dimensional arguments we see that it has
the dimension of m
3
;￿
0
= 8.85 10
−12
C
2
/(Nm
2
),E in N/C and d in Cm).
2
The permittivity of the medium,￿,also called the dielectric constant of the medium,is
￿ = ￿
0
(1 +nβ)
where n is the density of the atoms (in m
−3
).
As we just saw,the propagation speed in the medium is 1/

￿µ
0
or
￿
￿
0
/￿ c.
If we write our (plane) wave interacting with the atom as
E(x,t) = E
0
e
−i(k x−ωt)
where E
0
is the amplitude and ω the angular frequency (in radians per second,so that
ω = 2π ν).
We know that the phase of the wave is given by (k x −ωt) it is easy to see that the phase
speed (i.e.the speed of a constant phase,e.g.the maximum) is
v
phase
=
dx
dt
=
ω
k
(kx −ωt = const ⇒ k dx −ωdt = 0)
This is called the dispersion relation.Note that the phase velocity can be both smaller and
larger than c,even if the photons themselves travel with the speed of light.
So,
ω
k
=
￿
￿
0
￿
c
We also saw that the Poynting ﬂux (and thereby the intensity) is proportional to the square
of the electric ﬁeld strength.We can then write
I ∝ |E|
2
= E∙ E

= E
2
0
e
i(k x−ωt)
e
−i(k

x−ωt)
= E
2
0
e
−ix(k

−k)
= E
2
0
e
−2xIm[k]
(the star denotes the complex conjugate)
This looks very much like
I = I
0
e
−τ
= I
0
e
−αx
= I
0
e
−nσ x
So we can identify the absorption coeﬃcient α with 2 Im[k] which we can write as
α = 2 Im[k] = 2 Im[
ω
c
￿
￿
￿
0
] = 2 Im[
ω
c
￿
1 +nβ]
if nβ is a small quantity then

1 +nβ ￿ 1 +
1
2
nβ and
α = 2
ω
c
Im[1 +
1
2
nβ] =
ω
c
nIm[β]
or
σ =
ω
c
Im[β]
We need to ﬁnd the polarizability of the atom in order to know its absorption cross-section.
We do this by looking at a damped harmonic oscillator:
3
Consider an electron in an atom that experience an incoming EM ﬁeld.The electric ﬁeld of
this radiation can be written as E
0
e
iωt
.
Let us call the charge of the electron q.It will react to the force from the radiation ﬁeld
F = q(E+v ×B).We see here that,unless the velocity of the electron is comparable to c,
the inﬂuence from the magnetic part of the EM ﬁeld is small so we neglect it here.
Applying the equation of motion (Newton’s 2nd law) to the motion of an electron bound in
an atom when driven by an external electric ﬁeld we get the well-known equation:
qE
0
e
iωt
−mω
2
0
x = m¨x −mγ ˙x
where the coordinate x is in the direction of E,m is the mass of the particle (electron),γ is a
kind of friction,in radiation theory most often called damping,ω
0
is the resonance frequency
for the electron (if we have a free electron then ω
0
is zero).
On the left hand side we side the forces,the driving force opposed by the restoring force
(cf.weight on a spring);on the right hand side we have mass times acceleration and the
“friction” damping the harmonic oscillations.
Guessing that the electron will follow the driving force we set the solution as x = x
0
e
iωt
,
evaluate ˙x(= iωx),¨x(= −ω
2
x) and get the solution for x:
x =
q
m
E
0
ω
2
0
−ω
2
−iγω
e
iωt
= x
0
e
iωt
and for the dipole moment
d = q ∙ x =
q
2
m
ω
2
0
−ω
2
−iγω
E
0
e
iωt
= β ￿
0
E
0
e
iωt
where the last equality comes fromthe deﬁnition of the induced dipole moment above.Thus,
β becomes (and now replace q with the electron charge,e):
β =
e
2
m￿
0
1
ω
2
0
−ω
2
−iγω
Im[β] =
e
2
m￿
0
γω

2
0
−ω
2
)
2

2
ω
2
And,ﬁnally,
σ =
ω
c
Im[β] =
e
2
mc￿
0
ω
2
γ

2
0
−ω
2
)
2

2
ω
2
For the line absorption coeﬃcient we know that ω is very close to ω
0
and we have

2
0
−ω
2
)
2
= ((ω
0
−ω)(ω
0
+ω))
2
= (ω
0
−ω)
2
(2ω
0
)
2
= 4ω
0

0
−ω)
2
so
σ
ω
=
e
2
2mc￿
0
γ/2

0
−ω)
2
+(γ/2)
2
4
where the σ is marked by an ω to denote that we are using angular frequency units.We know
this proﬁle function from before,it is the Lorentz proﬁle of radiation damping,quadratic
Stark eﬀect and the van der Waals broadening.
Often,when working with spectral lines,we use wavelength units instead.To convert to
these,we know that ω = 2πν = 2π c/λ.
(ω −ω
0
) then is 2π c(λ −λ
0
)/λ
2
0
and we get,after some manipulation,
σ
λ
=
π e
2
4π￿
0
mc
2
λ
2
0
Γ/π

0
−λ)
2

2
where Γ =
λ
2
0
4π c
γ.The proﬁle function (what is given after the λ
2
0
) is normalized to unit
area.
5