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 mq = ⋅ E − mω02 q (2.2) With friction the equation of motion reads mq + 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: mq + 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). mq + 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.