2.
Elementary physics of a charged particle on a spring
(Lorentz-Drude-Model)
In this chapter we work in SI.
The Lorentz-Drude model
The equation of motion
µ(q ) = µ 0 + ∆µ = µ 0 + eq ,
Energy of the induced dipole with the external field E is
Vint eract = −∆µ ⋅ E ,
Corresponding external force on the particle
− ∂Vint eract / ∂q = (∂µ / ∂q ) ⋅ E =  ⋅ E .
Total force acting on the particle is
 ⋅ E − kq
(2.1)
(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
k = mω02 .
Without friction the equation of motion would be
mq
=  ⋅ E − mω02 q
(2.2)
With friction the equation of motion reads
mq + mΓ q =  ⋅ E − mω02 q
(2.3)
with “damping constant” Γ [s-1].
Remember that
E( t ) =
1
E 0 exp{−i(ωt − kz)} + c.c.
2
(2.4)
For z=0:
1
E( t ) = E 0 exp{−iωt}
2
+ c.c.
(2.5)
When E0 [V/m] is real, E(t) behaves as cos(ωt).
Generally as cos(ωt+Φ) with a phase Φ.
This phase could be expressed by making E0 complex!
Equation-of-motion for a damped, driven, harmonic oscillator:
mq + mΓ q + mω02 q =  ⋅ E
(2.5)
Note that q(t) because the system is driven by E(t)=E0 cos{ωt}.
Quite generally we should write the desired solution as
1
q( t ) = Q exp{−iωt}
2
+ c.c.
(2.6)
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
For E=0, decay of an originally extended oscillator with q0 = q(0):
1
q ( t ) = q 0 exp{−iω0 t} exp{−αt} + c.c.
2
(2.7)
Here the empirical constant α 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
2
(2.8)
For example we show q(t) for ν= 100 Hz and Γ=50 s-1, from which T2= 40 ms.
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:
• the decay factor exp{-t/T2} pushes the local extrema very slightly to the left,
• the spectral analysis gives a broad lineshape.
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).
mq + mΓ q + mω02 q =  ⋅ E
(2.5)
1
q( t ) = Q exp{−iωt}
2
(2.6)
Ansatz
+ c.c.
reduces to an algebraic equation for Q (look only at the coefficient of exp{−iωt} )
− mω 2 Q − imΓωQ + mω02 Q =  ⋅ E 0
(2.9)
From this we get the complex oscillation amplitude
Q=
 ⋅ E0
,
m D(ω)
where
D(ω) = (ω − ω ) − iΓω
2
0
2
where is the complex denominator.
(2.10)
m=10 g, ν0=100 Hz, e=0.015 C, and Γ=50 s-1.
E0=100 V/m, 0≤ ν ≤ 200 Hz.
A macroscopic example:
First we plot the amplitude Q :
m
0.003
Abs Q



0.004
0.002
0.001
0
50

100
nue
Hz
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)
0.5
2 Pi
0.3
Fi

0.4
0.2
0.1
0
0
50
nue

100
Hz
150
200
Dissipated power
damped oscillator driven by E(t) at ν.
The total energy in the material system
H =
1
1
m q 2 + (mω02 ) q 2
2
2
(2.11)
 =0,
How does it change over time? In the quasi-stationary state obviously not, H
but for the moment let us work on (2.11) formally and build the time derivative
 = m q q + (mω 2 ) qq
H
0
(2.12)
From equation of motion (2.5)
m q =
− mΓ q
− mω02 q
(2.13)
+ ⋅E
Upon substitution into 2.12 the term with (mω02 ) cancels and one has
 = −mΓq 2
H
+ q  ⋅ E
=0
(2.14)
This consists of two terms which balance to zero:
Pdiss = −mΓq 2
outflowing power due to dissipation
Pabs = +q E
tinflowing power
due to optical work
through the oscillator strength.
We can calculate one or the other; let us consider the dissipation part. We have from (2.6)
q =
1
(−iω) Q exp{−i ω t} + c.c.
2
(2.15)
1 2 2
ω Q .
2
If a time-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
Then q 2 has terms which oscillate with ± 2ωt , and a constant part
Substituting Q from equ. (2.10) one finds for Pabs=-Pdiss
Pabs
⋅E
1 ω2
= mΓ
2
2 D(ω)
m2
(2.17)
D(ω) = (ω02 − ω2 ) − iΓω
where as before
Rearrange: first look at
{
2
}{
(2.18)
}
D(ω) = (ω02 − ω2 ) − iΓω (ω02 − ω2 ) + iΓω
2
=
(ω02 − ω2 ) 2 + Γ 2 ω2
= (ω0 + ω) 2 (ω0 − ω) 2 + Γ 2 ω2
≈
4ω2
{(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 .
With this in (2.17) one arrives at (and setting Pabs=Pdiss)
Pabs
Γ/2
2π
=N
8 π (ω0 − ω) 2 + (Γ / 2) 2
{
}
⋅E
m
2
[W]
(2.19)
[s]
(2.20)
It was written such that the central feature
L(ω) ≡
Γ/2
π (ω0 − ω) 2 + (Γ / 2) 2
{
}
is area-normalized, and N is the number of particles which absorb. This is the famous
Lorentz-Lineshape for absorption. It has its maximum at ω0, and the full width at halfmaximum (fwhm) is
on the ω -axis: Γ, hence
on the ν -axis: Γ/(2 π)
on the ~
ν -axis: Γ/(2 π c)
This is now shown for our example.

2
Pabs
W
1.5
1
0.5
0
0
50

100
Hz
nue
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.