Download Statistische Thermodynamik und Spektroskopie II

Elementary physics of a charged particle on a spring
(Lorentz-Drude-Model)
In this chapter we work in SI.
The Lorentz-Drude model
A particle with mass m is attached to a rigid support by a spring with force constant k.
For example, for optical spectroscopy we may consider a valence electron, and for
infrared spectroscopy a partially charged atom of a molecule. The particle also carries
carries a charge e (which may be different from the elementary charge e0) . When an
electric field E acts upon it, the external force is e E.
The equation of motion
Imagine that at q=0 the dipole moment of the system is µ0. For example the opposite
charge could be sitting on the rigid wall, in which case µ 0 = eR . The dipole moment at
displacement q is µ(q ) = µ 0 + ∆µ = µ 0 + eq , where ∆µ is the induced dipole caused by the
displacement. The energy of the induced dipole with the external field E is
and
the
force
on
the
particle
becomes
Vint eract = −∆µ ⋅ E ,
− ∂Vint eract / ∂q = (∂µ / ∂q ) ⋅ E =  ⋅ E . The internal force at displacement q is given by the
harmonic potential: − ∂V / ∂q . Therefore the total force acting on the particle is
(2.1)
 ⋅ E − kq
(Note that the oscillator strength  has the unit of charge! In quantum-mechanical
calculations it is common to call the dimensionless  / e 0 “oscillator strength.”) The
resonance angular velocity ω0 of the system is given by k = mω02 . If there were no
friction, then the equation of motion would be
mq
=  ⋅ E − mω02 q
But with friction the equation of motion reads
mq + mΓ q =  ⋅ E − mω02 q
(2.2)
(2.3)
Here the “damping constant” Γ has the same dimension as one of the dots on q , hence
[s-1]. Remember that
1
(2.4)
E( t ) = E 0 exp{−i(ωt − kz)} + c.c.
2
where we may set z=0 (the molecule is assumed to rest at the coordinate origin). Here E0
is the field amplitude [V/m]. Let us therefore simplify:
1
E( t ) = E 0 exp{−iωt}
+ c.c.
(2.5)
2
When E0 is real, E(t) behaves as cos(ωt). But if the observed behavior were cos(ωt+Φ)
with a phase Φ, then this phase could be expressed by making E0 complex  please
understand this thoroughly, because it is essential for what follows!
It is customary to rewrite eq. (2.3) so as to separate the coordinate q from the E-field:
mq + mΓ q + mω02 q =  ⋅ E
(2.5)
This is the equation-of-motion for a damped, driven, harmonic oscillator. It can explain
almost all of the observations which chemists make in optical spectroscopy, and this is
why its parts must be thoroughly understood. Note that q(t) because the system is driven
by E(t)=E0 cos{ωt}. For instance, the driving (= externally applied) frequency ω / 2π can
be much lower than the resonance frequency ω0 / 2π . In this case the mass can follow
instantly: q ( t ) = q 0 cos{ωt} , but the (real) amplitude q0 will be small. On the other hand,
when the driving frequency is close to the resonance frequency, the particle will have
very large amplitude motion, however it will lag behind by about T/4 (as we will see).
Quite generally, therefore, we should write the desired solution as
1
q( t ) = Q exp{−iωt}
+ c.c.
(2.6)
2
where the complex oscillation amplitude Q gives both the amplitude Q of the oscillation
as well as its phase Φ relative to the driving field ( Q = Q exp{iΦ} ).
Field-free damped oscillation
If no electric field is present, i.e. for E=0, eq. (2.5) governs the decay of an originally
extended oscillator with q0 = q(0):
1
(2.7)
q ( t ) = q 0 exp{−iω0 t} exp{−αt} + c.c.
2
Here the empirical constant α (because 2.7 is a general ansatz) should be connected to
the damping constant Γ. Imagine that ω0 >> Γ / 2 , ie. many periods fit into the
characteristic damping time T2=2/Γ . In this case we find α ≈ Γ / 2 and can rewrite eq.
2.7:
1
Γ
q( t ) = q 0 exp{−iω0 t} exp{− t} + c.c.
(2.8)
2
2
For example we show q(t) for ν= 100 Hz and Γ=50 s-1, from which T2= 40 ms (indicated
by vertical grid line). In this case only 4 cycles fit into the oscillation decay time T2 .
100 Hz
1
0.75
relative q
0.5
0.25
0
-0.25
-0.5
-0.75
0
0.02
0.04
t
�
s
0.06
0.08
0.1
When the damping is strong such as shown here, two effects can be noted:
• the decay factor exp{-t/T2} pushes the local extrema very slightly to the left, so that
the effective angular velocity is lowered from ω0 to ω0- ∆ω0 (but this is not important)
• the spectral analysis gives a broad lineshape (for example, when looking at the
intensity I(ω) through a spectrometer which is scanned over various ω or,
alternatively, Fourier-transformation of 2.8. For those interested, the Fourier-transform
story is taken up in an accompanying MATHEMATICA notebook.)
An example for decay: The oscillation of a π-electronic charge cloud in a molecule is started by
ππ* excitation, for example in anthracene around 400 nm. The frequency is ν0=750 THz and the
period T=1.33 fs. A typical oscillation decay time is T2≈10 fs, so that 8 cycles fit before.
Solution of equation of motion
Let us return to the situation when the field E(t) acts on the harmonic oscillator. The
equation-of-motion (2.5) is a 2nd order differential equation for q(t). By the general ansatz
(2.6) and for given (2.4), it reduces to an algebraic equation for Q (look only at the
coefficient of exp{−iωt} )
(2.9)
− mω 2 Q − imΓωQ + mω02 Q =  ⋅ E 0
From this we get the complex oscillation amplitude
Q=
 ⋅ E0
,
m D(ω)
ωhere
(2.10)
D(ω) = (ω − ω ) − iΓω
is the complex denominator. This is the most important result of the entire lecture series
because it shows clearly how the charged particle responds the harmonic electric field.
2
0
2
All other phenomena (like absorption or scattering) are then determined by additional
effects which build on (2.8). Let us illustrate what is happening with a few examples.
A macroscopic example: We use an oscillator which has m=10 g, ν0=100 Hz, e=0.015 C,
and Γ=50 s-1. The driving electric field has E0=100 V/m and variable frequency in the
range 0≤ ν ≤ 200 Hz. First we plot the amplitude Q :
m
0.003
Abs Q
�
�
�
0.004
0.002
0.001
0
50
�
100
Hz
nue
150
200
At zero frequency, the constant electric field E produces a constant displacement qstatic =
0.38 mm such that  ⋅ E = kq (remember:  = e in our simple model). The resonance
(peak) occurs at 99.84 Hz, a little below the frequency ν0 = 100 Hz of the undamped
oscillator, with amplitude qpeak= 4.778 mm. Up to about 50 Hz the particle follows the
driving field almost instantaneously (see below) and it reaches about the same maximal
extention. But then it begins to lag behind while the amplitude increases. We see this in
the following few figures where the relative field E(t)/E0 (blue) and amplitude q(t)/qpeak
(red) are shown for ν = 50, 95,100, 105, 150 Hz.
50 Hz
1
50 Hz
The particle follows in phase,
rougly with the static amplitude.
0.5
relative E or q
Before resonance:
0
-0.5

-1
0
0.02
0.04
0
0.02
0.04
t s
95 Hz
0.06
0.08
0.1
0.08
0.1
0.08
0.1
0.08
0.1
0.08
0.1
1
95 Hz
The particle begins to lag behind.
relative E or q
0.5
0
-0.5
-1
On resonance
t

s
0.06
100 Hz
1
0.5
relative E or q
100 Hz
The force is maximum when the
particle goes through q=0 in the
direction of the force. This means
the external work accelerates just
so much that damping is
compensated.
0
-0.5
-1
0
Beyond resonance
0.02
0.04
t

s
0.06
105 Hz
0.5
relative E or q
105 Hz
Now the force begins to work
against the particle momentum
over a section of the period, and
decelarates during this time.
Altogether
the
amplitude
decreases.
1
0
-0.5
-1
0
0.02
0.04
t

s
0.06
150 Hz
1
0.5
relative E or q
150 Hz
By now the decelaration extends
almost over the entire period so
that amplitude can not build up.
The driving field is too fast, the
oscillator follows essentially a
half-period behind.
0
-0.5
-1
0
0.02
0.04
t

s
0.06
The phase Φ (in units of 2π) as function of ν is shown next. The phase build up (particle
lags behind) until resonance at ν0 where Φ = 2π/4. Then the phase grows more slowly
until a half-period has been reached.
0.5
2 Pi
0.3
Fi
�
0.4
0.2
0.1
0
0
50
nue
�
100
Hz
150
200
Dissipated power
Consider the damped oscillator which is being driven by E(t) at ν. The total energy which
at any one time is in the material system is
1
1
(2.11)
H = m q 2 + (mω02 ) q 2
2
2
 = 0 , but for
How does it change over time? In the quasi-stationary stae obviously not, H
the moment let us work on (2.11) formally and build the time derivative
 = m q q + (mω 2 ) qq
(2.12)
H
0
From the equation of motion (2.5) we have for part of the first term
m q =
(2.13)
− mΓ q
− mω02 q +  ⋅ E
Upon substitution into 2.12 the term with (mω02 ) cancels and one has
 = −mΓq 2
(2.14)
H
+ q  ⋅ E = 0
This consists of two terms which balance to zero: the outflowing power Pdiss = −mΓq 2
due to dissipation, and the inflowing power Pabs = +q E due to optical work through the
oscillator strength. We can calculate one or the other; let us consider the disspation part.
We have from (2.6)
q =
1
(−iω) Q exp{−i ω t} + c.c.
2
(2.15)
1 2 2
ω Q . If a time2
average is taken then the oscillatory parts integrate to zero and only the constant part
remains. In this sense we can write
1
2
(2.16)
q 2 = ω 2 Q
2
Substituting Q from equ. (2.10) one finds for Pabs=-Pdiss
Then q 2 has terms which oscillate with ± 2ωt , and a constant part
Pabs
⋅E
1 ω2
= mΓ
2
2 D(ω)
m2
2
(2.17)
where as before
(2.18)
D(ω) = (ω02 − ω2 ) − iΓω
This formula is already complete and usable (see green curve below) but in practice it is
often rearranged. First look at
{
}{
}
D(ω) = (ω02 − ω2 ) − iΓω (ω02 − ω2 ) + iΓω
2
=
(ω02 − ω2 ) 2 + Γ 2 ω2
= (ω0 + ω) 2 (ω0 − ω) 2 + Γ 2 ω2
≈
4ω 2
{(ω
0
− ω) 2 + (Γ / 2) 2
}
(2.19)
The approximation ≈ came about because only frequencies near resonance are effective,
so that ω0 + ω ≈ 2ω . (Actually this means that the asymmetry which we saw in the
amplitude Q(ω) is ignored, which can be done if ω0 is not too low.) With this in (2.17)
one arrives at (and setting Pabs=Pdiss)
⋅E
Γ/2
2π
[W]
Pabs = N
8 π (ω0 − ω) 2 + (Γ / 2) 2
m
It was written such that the central feature
Γ/2
[s]
L(ω) ≡
π{(ω0 − ω) 2 + (Γ / 2) 2 }
is area-normalized, and N is the number of particles which absorb. This is the
Lorentz-Lineshape for absorption. It has its maximum at ω0, and the full width
maximum (fwhm) is
on the ω -axis: Γ, hence
on the ν -axis: Γ/(2 π)
on the ~
ν -axis: Γ/(2 π c)
2
{
This is now shown for our example.
}
(2.19)
(2.20)
famous
at half-
�
2
Pabs
W
1.5
1
0.5
0
0
50
�
100
nue
Hz
150
200
The green line comes from the complete form 2.17&2.18, while the black one is the
approximate equ. 2.19. The peak power is 2.25 W (using N=1, i.e. one particle) which is
an awful lot! But remember that we are driving a very dissipative spring with 100 Hz
resonantly to an amplitude of 4.778 mm. The fwhm on the ν-axis is 7.96 Hz (indicated by
gridlines). By comparison with the figure for Q it is interesting that no power is
absorbed at low frequencies. This comes from the ω2-factor in the power formulas, which
discriminate against Q at low frequencies. At the extreme case ν=0, nothing moves and
no power is dissipated or absorbed, but (as we saw before) q(ν=0) =0.388 mm.