Lab 6.11 · 30 min

Lab: lifetimes by hand

How long does an ion pair last? How long does a solvent molecule stay in a shell? These "lifetimes" are among the most quoted numbers in simulation papers, and among the most definition-dependent. We compute continuous and intermittent correlation functions by hand on a 12-frame on/off sequence, then measure first-shell lifetimes in the LJ liquid and change the cutoff, the saving interval and the tolerance for brief excursions, to see how much each choice moves the answer.

The setup

For each pair (an ion pair, an ion and a solvent molecule, two neighbours) define h(t) = 1 if the pair is "bonded" (closer than a cutoff) at frame t, and 0 otherwise. Two correlation functions are used:

intermittent C_I(t) = ⟨h(0) h(t)⟩ / ⟨h⟩ still (or again) bonded after t; breaks allowed continuous C_C(t) = ⟨h(0) H(t)⟩ / ⟨h⟩ H(t) = 1 only if bonded at every frame from 0 to t lifetime τ = area under C(t) (= decay time if C is exponential)

Data: a 12-frame sequence by hand, and the 3D LJ liquid of Labs 6.9–6.10 (108 atoms, ρ = 0.8, T = 1.0), with "neighbours" defined by the first minimum of g(r) at r = 1.45.

Step 1

Continuous and intermittent by hand

Why: The two functions answer different questions. The continuous one asks "has this pair stayed together without interruption?", the intermittent one asks "is it together again (or still) after time t?". On a short sequence we can count every case.

one pair, frames every 1 ps: 1 1 1 0 1 1 1 1 0 0 0 1 ⟨h⟩ = 8/12 = 0.667; unbroken stretches: 3, 4, 1 lag intermittent continuous 0 8/8 = 1.000 8/8 = 1.000 1 5/7 = 0.714 5/8 = 0.625 2 4/7 = 0.571 3/8 = 0.375 3 3/7 = 0.429 1/8 = 0.125 4 4/7 = 0.571 0/8 = 0.000 intermittent: bonded at t and at t + lag / bonded at t continuous: origins that stay bonded lag frames without a break / all bonded origins (a stretch of length L gives L − lag origins) lifetime = area (lags 0–4, first term halved, × 1 ps): continuous 1.62 ps intermittent 2.79 ps check: continuous area = Σ L² / (2 Σ L) = 26/16 = 1.62 ps

What the numbers say: The continuous function dies as soon as each stretch ends; the intermittent one even rises again at lag 4, because the pair re-forms. The intermittent lifetime is therefore always the longer one (2.79 vs 1.62 ps here). Neither is "right": the continuous lifetime describes how long an uninterrupted contact lasts, the intermittent one how long until the partners have really separated and forgotten each other. The formula Σ L²/(2ΣL) shows that the continuous area depends only on the stretch lengths, which is why anything that cuts a stretch short (next steps) changes it strongly.

Step 2

Forgiving short breaks: the tolerance t*

Why: Thermal motion makes a bonded pair cross the cutoff briefly without really separating. Many analyses therefore forgive breaks shorter than a tolerance t*. It sounds harmless; let us count what it does.

same sequence, breaks of ≤ 1 frame forgiven (t* = 1 ps): 1 1 1 1 1 1 1 1 0 0 0 1 stretches: 8, 1 continuous lifetime = Σ L²/(2 Σ L) = 65/18 = 3.61 ps (was 1.62 ps: 2.2 times longer from one forgiven frame)

What the numbers say: Joining two stretches of 3 and 4 into one of 8 more than doubles the continuous lifetime, because the area grows with L². A tolerance is reasonable physics (a brief excursion is not a real break), but its value must be stated: with different t* two papers can report lifetimes that differ by a factor of 2 for the same simulation. The intermittent function already ignores short breaks, so it hardly changes.

Step 3

First-shell neighbours in the LJ liquid

Why: Now the real case: every pair of the 108-atom liquid, 2000 frames saved every 0.02 time units, neighbours defined by the first minimum of g(r) (r < 1.45). We compute both functions and remove the long-time baseline of the intermittent one: two random atoms are neighbours a fraction ⟨h⟩ of the time anyway.

⟨h⟩ = 0.0991 (10.60 neighbours per atom)
t0.020.10.20.51.02.05.0
continuous0.9840.9230.8560.6930.4960.2570.035
intermittent0.9840.9280.8780.7840.6830.5440.316
intermittent − baseline, normalized0.9830.9200.8640.7600.6480.4930.241
baseline correction: (C_I − ⟨h⟩) / (1 − ⟨h⟩) lifetimes: continuous 1.46, intermittent 3.25 time units (intermittent integrated until it falls below 0.01, at t = 12)

What the numbers say: Within the first 0.2 time units the two functions agree; after that the continuous one falls much faster, because neighbours that step out briefly and come back are lost to it. The intermittent function decays to the baseline ⟨h⟩ = 0.099, not to zero: without subtracting it the lifetime would come out too long (and grow with the integration limit). Limitation of the intermittent lifetime: its tail is long and noisy (here integrated to t = 12), so it needs long trajectories, exactly like the collective quantities of Lab 6.8.

Step 4

How much the definition moves the answer

Why: Three choices enter every lifetime: the cutoff, how often frames are saved, and the tolerance t*. We change each one on the same trajectory.

cutoffneighbourscontinuousintermittent
1.308.120.7232.660
1.45 (g(r) minimum)10.601.4603.251
1.6012.872.1723.519
saving intervalcontinuous lifetimemean unbroken stretch
0.021.4601.241
0.101.5261.356
0.502.0642.210
tolerance t*00.040.100.20
continuous lifetime1.4601.5091.6842.025

What the numbers say: The continuous lifetime triples between cutoffs of 1.30 and 1.60 (0.72 → 2.17); the intermittent one changes by 30 % (2.66 → 3.52). Saving frames less often hides short excursions and makes the continuous lifetime longer (1.46 → 2.06 at 25 times larger intervals); a tolerance does the same on purpose (2.03 with t* = 0.2). So the continuous lifetime is mostly a property of the definition, the intermittent one mostly a property of the liquid. Practical rule: put the cutoff at the minimum of g(r), prefer the intermittent function (with the baseline removed) for "how long do partners stay together", and always state cutoff, saving interval and t*.

Step 5

Compare with diffusion

Why: A lifetime is only "long" or "short" compared with something. The natural yardstick is how long the atoms take to diffuse across the shell or across one diameter, using D from Lab 6.10.

D = 0.0688 (Lab 6.10), t = Δr² / (6D) shell width 0.35: 0.1225 / 0.4128 = 0.30 one diameter 1.0: 1.0000 / 0.4128 = 2.42 intermittent lifetime 3.25 ≈ time to diffuse one diameter continuous lifetime 1.46 ≈ 5 times the shell-crossing time

What the numbers say: In this simple liquid, partners stay together about as long as it takes to diffuse one diameter: nothing special binds them, they just drift apart. In an ionic liquid or electrolyte the interesting question is whether an ion pair lives much longer than this diffusion time (a real, long-lived pair that moves as one neutral unit and lowers the conductivity, Lab 6.7) or not (a transient contact). Comparing τ with σ²/(6D) turns a bare lifetime into that physical statement. Limitation: lifetimes from a single cutoff ignore the shape of the free-energy barrier between bonded and free states; for barrier-controlled processes the PMF of Lab 7.5 and a rate theory give a better-defined number.

Summary

choiceeffect (number)advice
continuous vs intermittent1.46 vs 3.25 (same data)say which; intermittent for "partners"
cutoff 1.30 -> 1.60continuous ×3, intermittent ×1.3cutoff at the g(r) minimum
saving interval ×25continuous 1.46 -> 2.06save often; report the interval
tolerance t* = 0.2continuous 1.46 -> 2.03report t*
intermittent baselinedecays to ⟨h⟩ = 0.099, not 0subtract and normalize
yardstickσ²/(6D) = 2.42 ≈ intermittent 3.25compare τ with diffusion times