* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Statistische Thermodynamik und Spektroskopie II
Introduction to gauge theory wikipedia , lookup
Renormalization wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Standard Model wikipedia , lookup
Speed of gravity wikipedia , lookup
Field (physics) wikipedia , lookup
Fundamental interaction wikipedia , lookup
Quantum electrodynamics wikipedia , lookup
Electrostatics wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Classical mechanics wikipedia , lookup
Path integral formulation wikipedia , lookup
Elementary particle wikipedia , lookup
Chien-Shiung Wu wikipedia , lookup
Time in physics wikipedia , lookup
Centripetal force wikipedia , lookup
Lorentz force wikipedia , lookup
History of subatomic physics wikipedia , lookup
Equations of motion wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Work (physics) wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
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 mq = ⋅ E − mω02 q But with friction the equation of motion reads mq + 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: mq + 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.