Questions: Time-Correlation Functions and Relaxation

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

A liquid has a large initial velocity autocorrelation function C(0) but it decays extremely rapidly — faster than any other liquid you've measured. A student claims this means the liquid must have a large diffusion coefficient. Is the student correct?

AYes — C(0) sets the scale of the diffusion coefficient directly
BNo — the diffusion coefficient is the time integral ∫C(t)dt, so a large but rapidly decaying C(t) integrates to a small area and yields a small D
CYes — C(0) = kT/m by equipartition, and this ratio equals D
DNo — because C(t) decays to zero in ergodic systems, D must always be zero
Question 2 Multiple Choice

Which statement best describes the behavior of a time-correlation function C(t) = ⟨A(t)A(0)⟩ for an ergodic equilibrium system?

AC(t) oscillates indefinitely because the system is in thermal equilibrium
BC(t) grows over time as the system explores more of phase space
CC(t) decays to zero as t → ∞ because the system explores all accessible phase space, erasing memory of the initial value
DC(t) is constant for equilibrium systems because statistical properties don't change in time
Question 3 True / False

The diffusion coefficient D can be computed entirely from equilibrium dynamics by integrating the equilibrium velocity autocorrelation function — a property known as a Green-Kubo relation.

TTrue
FFalse
Question 4 True / False

A time-correlation function C(t) that starts at a larger initial value C(0) generally corresponds to a larger transport coefficient.

TTrue
FFalse
Question 5 Short Answer

Why can equilibrium time-correlation functions reveal non-equilibrium transport properties like diffusion coefficients and viscosity?

Think about your answer, then reveal below.