11.5: Exercises
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Exercise 11.5.1
Find the time-domain Green’s function of the critically-damped harmonic oscillator (γ=ω0).
Exercise 11.5.2
Consider an overdamped harmonic oscillator (γ>ω0) subjected to a random driving force f(t), which fluctuates between random values, which can be either positive or negative, at each time t. The random force satisfies ⟨f(t)⟩=0and⟨f(t)f(t′)⟩=Aδ(t−t′), where ⟨⋯⟩ denotes an average taken over many realizations of the random force and A is some constant. Using the causal Green’s function, find the correlation function ⟨x(t1)x(t2)⟩ and the mean squared deviation ⟨[x(t+Δt)−x(t)]2⟩.
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For the over-damped oscillator, the Green’s function is G(t,t′)=Θ(t−t′)e−γ(t−t′)Γsinh[Γ(t−t′)],whereΓ=√γ2−ω20. Hence, the response to the force f is x(t)=1mΓ∫t−∞dt′e−γ(t−t′)sinh[Γ(t−t′)]f(t′). From this, we get the following expression for the desired correlation function: ⟨x(t1)x(t2)⟩=1m2Γ2∫t1−∞dt′∫t2−∞dt″e−γ(t1−t′)e−γ(t2−t″)×sinh[Γ(t1−t′)]sinh[Γ(t2−t″)]⟨f(t′)f(t″)⟩. Note that the ⟨⋯⟩ can be shifted inside the integrals, because it represents taking the mean over independent sample trajectories. Now, without loss of generality, let us take t1≥t2. Since ⟨f(t′)f(t″)⟩=Aδ(t′−t″) which vanishes for t′≠t″, the double integral only receives contributions from values of t′ not exceeding t2 (which is the upper limit of the range for t″). Thus, we revise ∫t1dt′ into ∫t2dt′. The delta function then reduces the double integral into a single integral, which can be solved and simplified with a bit of tedious algebra: ⟨x(t1)x(t2)⟩=Am2Γ2e−γ(t1+t2)∫t2−∞dt′e2γt′sinh[Γ(t′−t1)]sinh[Γ(t′−t2)]=A8m2Γ2e−γ(t1+t2)[e−Γt1e(2γ+Γ)t2γ+Γ+eΓt1e(2γ−Γ)t2γ−Γ−e−Γt1e(Γ+2γ)t2+eΓt1e(−Γ+2γ)t2γ]=A8m2Γγ[e−(γ−Γ)(t1−t2)γ−Γ−e−(γ+Γ)(t1−t2)γ+Γ]. Hence, ⟨[x(t+Δt)−x(t)]2⟩=2[⟨x(t)2⟩−⟨x(t+Δt)x(t)⟩]=A4m2Γγ[1−e−(γ−Γ)Δtγ−Γ−1−e−(γ+Γ)Δtγ+Γ].