An oscillator with memory

A moving oscillator disturbs its surroundings. Their delayed response changes how it slows down.

1 · Harmonic bath

Many oscillators form the surroundings

Our oscillator is attached to a continuum of harmonic bath modes. Each mode responds to its motion. Adding their responses gives a force that depends on earlier velocities.

\ddot{x} (t) = - ω_{0}^{2} x (t) - \int_{0}^{t} K (t - s) \dot{x} (s) d s + ξ (t)

ω₀ sets the spring frequency. K describes the memory of past motion. ξ is the fluctuating force from the bath.

We plot the mean motion after release from rest at x = 1. The bath starts in thermal equilibrium around that held position. Its fluctuating force averages to zero. The spring includes the usual bath counterterm, keeping ω₀ fixed as the coupling changes.

2 · Memory in time

How long does a velocity matter?

Choose an exponential memory. A velocity from τ time units ago receives about 37% of the weight of the current velocity.

K (t) = \frac{γ}{τ} e^{- t / τ}, t \geq 0

The area under K is γ. Increasing τ spreads the same area over a longer time. As τ approaches zero, the friction force approaches −γv.

3 · Memory in frequency

How strongly is each frequency damped?

η (ω) = \int_{0}^{\infty} K (t) \cos (ω t) d t = \frac{γ}{1 + {(ω τ)}^{2}}

With memoryInstantaneous: η = γ

The vertical line marks ω₀. This spectrum is even in frequency. The corresponding bath spectral density is J(ω) = ωη(ω) for ω > 0, using K(t) = (2/π)∫₀∞[J(ω)/ω]cos(ωt)dω.

4 · Motion

Release the oscillator

Both curves start at the same displacement, with zero velocity. The comparison uses the same γ and an instantaneous friction force.

With memoryInstantaneous friction

Energy in the mean motion

E_{mean} = \frac{1}{2} {\dot{x}}^{2} + \frac{1}{2} ω_{0}^{2} x^{2}

Memory allows energy to flow back briefly into the oscillator. At equilibrium, thermal fluctuations also contribute energy. The curves here track the energy associated with the mean displacement and velocity.

Analytical result and calculation

The exponential kernel is the Drude harmonic-bath model. Henkel (2021), Appendix A.2, Eqs. A12–A14 gives its transform, spectrum, and memory kernel. The plots evaluate these formulas with unit mass.

How the motion is calculated

Introduce the accumulated friction z. The mean motion obeys three linear equations:

ẋ = v,     v̇ = −ω₀²x − z,     τż = γv − z

Starting from (x, v, z) = (1, 0, 0), the solution is y(t) = exp(At)y(0). The tool evaluates this matrix exponential. Equivalently, the Laplace transform of the displacement is

X (s) = \frac{τ s^{2} + s + γ}{τ s^{3} + s^{2} + (τ ω_{0}^{2} + γ) s + ω_{0}^{2}}

The instantaneous comparison uses the closed-form damped-oscillator solution.

Temperature and fluctuations

In the classical model, ⟨ξ(t)ξ(s)⟩ = kBT K(|t−s|). Temperature controls the fluctuations around the mean. For this linear system, the mean motion shown above follows the same equations at every temperature. At equilibrium the average total oscillator energy is kBT.