 Solution

ID: past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2013/3/d/solution

Γ > 0

, the time dependence of each spatial normal coordinate follows its damping regime:

If $ω_{0} > γ$ , it is underdamped: $u_{k} (t) = e^{- γ t} [A cos (Ω t) + B sin (Ω t)]$ , with $Ω = ω_{0}^{2} - γ^{2}$ . The amplitude envelope has decay time $1/ γ = 2 m /Γ$ ; cycle-averaged energy decays as $e^{- 2 γ t}$ .
If $ω_{0} = γ$ , it is critically damped: $u_{k} (t) = (A + Bt) e^{- γ t}$ .
If $0 < ω_{0} < γ$ , it is overdamped: $u_{k} (t) = A e^{- [γ - γ^{2} - ω_{0}^{2}] t} + B e^{- [γ + γ^{2} - ω_{0}^{2}] t}$ . Both terms decay without oscillation, with two different rates.

For any fixed positive damping, sufficiently small nonzero

k

is overdamped because

ω_{0} ≃ c_{s} ∣ k ∣

. The slow branch then has

ω ≃ - i (m c_{s}^{2} /Γ) k^{2}

, a diffusive long-wavelength relaxation. At exactly

k = 0

u_{0} (t) = A + B e^{- Γ t / m}

and the constant displacement need not decay. Thus saying every amplitude simply decays as

e^{- Γ t / (2 m)}

is valid only for the underdamped envelope, not for all wave numbers.

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