My guess is that it’s extremely unlikely that enough energy is concentrated in a standing wave at initialization for it to not dissipate in 20 seconds. By equipartition it should be extremely unlikely for energy to be concentrated in any degree of freedom in any physical system, but I don’t know enough physics to be confident that this argument applies.
Certainly there is no conservation of standing wave amplitude, because with two billiard balls waves can form and dissipate. The question is how long it takes for waves of the tiny amplitudes caused by initialization to dissipate.
Why are you guys talking about waves necessarily dissipating, wouldn’t there be an equal probability of waves forming and dissipating given that we are sampling a random initial configuration, hence in equilibrium w.r.t. formation/dispersion of waves?
If you look at a noise-driven damped harmonic oscillator, the autocorrelation of “oscillator state at time t1” and “oscillator state at time t2” cycles positive and negative at the oscillator frequency [if it’s not overdamped], but with an envelope that gradually decays to zero when t1 and t2 get far enough apart from each other.
This whole thing is time-symmetric—knowing the state at time 0 is unhelpful for guessing the state at very positive timestamps AND unhelpful for guessing the state at very negative timestamps.
But the OP question was about fixing an initial state and talking about later times, so I was talking in those terms, which is more intuitive anyway. I.e., as time moves forward, the influence of the initial state gradually decays to zero (because the wave is damped), while meanwhile the accumulated influence of the noise driver gradually increases.
Yes, it would be more correct to say the question is how long it takes for the probability distribution of the amplitude and phase of a given oscillation mode to be indistinguishable from that of any other random box of gas.
Yes by the equipartition theorem there’s an average of kT of energy in each standing wave mode at any given moment. Might be fun to calculate how many left-right atoms that corresponds to—I think that calculation should be doable. I imagine that for the fundamental mode, it would be comparable to the √(number of atoms in the box) difference that we expect for other reasons.
It’s continuous and exponential. If amplitude of standing wave mode N decays by a factor of 2 in X seconds, then it‚ it’s the same X whether the initial amplitude in that mode is macroscopic versus comparable-to-the-noise-floor. (Well, unless there are nonlinearities / anharmonicities, but that’s probably irrelevant in this context.) But meanwhile, noise is driving the oscillation too. So anyway, I think it really matters how X compares to 20 seconds, which again is something I don’t know.
My guess is that it’s extremely unlikely that enough energy is concentrated in a standing wave at initialization for it to not dissipate in 20 seconds. By equipartition it should be extremely unlikely for energy to be concentrated in any degree of freedom in any physical system, but I don’t know enough physics to be confident that this argument applies.
Certainly there is no conservation of standing wave amplitude, because with two billiard balls waves can form and dissipate. The question is how long it takes for waves of the tiny amplitudes caused by initialization to dissipate.
Why are you guys talking about waves necessarily dissipating, wouldn’t there be an equal probability of waves forming and dissipating given that we are sampling a random initial configuration, hence in equilibrium w.r.t. formation/dispersion of waves?
If you look at a noise-driven damped harmonic oscillator, the autocorrelation of “oscillator state at time t1” and “oscillator state at time t2” cycles positive and negative at the oscillator frequency [if it’s not overdamped], but with an envelope that gradually decays to zero when t1 and t2 get far enough apart from each other.
This whole thing is time-symmetric—knowing the state at time 0 is unhelpful for guessing the state at very positive timestamps AND unhelpful for guessing the state at very negative timestamps.
But the OP question was about fixing an initial state and talking about later times, so I was talking in those terms, which is more intuitive anyway. I.e., as time moves forward, the influence of the initial state gradually decays to zero (because the wave is damped), while meanwhile the accumulated influence of the noise driver gradually increases.
Yes, it would be more correct to say the question is how long it takes for the probability distribution of the amplitude and phase of a given oscillation mode to be indistinguishable from that of any other random box of gas.
Yes by the equipartition theorem there’s an average of kT of energy in each standing wave mode at any given moment. Might be fun to calculate how many left-right atoms that corresponds to—I think that calculation should be doable. I imagine that for the fundamental mode, it would be comparable to the √(number of atoms in the box) difference that we expect for other reasons.
It’s continuous and exponential. If amplitude of standing wave mode N decays by a factor of 2 in X seconds, then it‚ it’s the same X whether the initial amplitude in that mode is macroscopic versus comparable-to-the-noise-floor. (Well, unless there are nonlinearities / anharmonicities, but that’s probably irrelevant in this context.) But meanwhile, noise is driving the oscillation too. So anyway, I think it really matters how X compares to 20 seconds, which again is something I don’t know.