Channel noise blurs the threshold and the spike time

The threshold becomes a probability

The deterministic model has a sharp threshold at 6.5 nC/cm². With a finite number of channels the same shock sometimes works and sometimes fails. We repeated each charge 150 times for each area. On a 1000 square micrometre patch, shocks of 5, 6, 6.5, 7 and 8 nC/cm² fire in 0.05, 0.32, 0.53, 0.69 and 0.94 of trials: a smooth S-shaped curve centred near the deterministic threshold.

Shrink the patch to 100 square micrometres and the curve flattens: the same five charges give 0.31 to 0.60. With no shock at all, $0.17\pm0.03$ of trials still fire, triggered by the noise itself. At 30 square micrometres every charge gives between 0.41 and 0.55, and the shock barely matters.

The result does not depend on the step size. With 300 trials at each step on 100 square micrometres and 6.5 nC/cm², the firing fraction is 0.477 ± 0.029 at 0.01 ms and 0.467 ± 0.029 at 0.005 ms, a difference of 0.25 standard errors.

Source: J. H. Goldwyn and E. Shea-Brown · A. L. Hodgkin and A. F. Huxley

Spike-time jitter follows the square root only on large patches

Raise the shock to 15 nC/cm² and nearly every trial fires, but the spikes come at different times. With 200 trials per area, the standard deviation of the first-spike time is $0.285\pm0.015\ \mathrm{ms}$ at 100 square micrometres, $0.134\pm0.007$ at 300, $0.0602\pm0.0030$ at 1000, $0.0336\pm0.0017$ at 3000 and $0.0190\pm0.0010\ \mathrm{ms}$ at 10000.

When the noise is small it acts as a linear perturbation of the deterministic path: the shift in the threshold-crossing time is proportional to the noise amplitude, which shrinks as $N^{-1/2}$, so the jitter should fall as area to the power $-1/2$. The local slopes between neighbouring areas are $-0.69\pm0.07$, $-0.66\pm0.06$, $-0.53\pm0.06$ and $-0.47\pm0.06$. The first two are steeper than $-1/2$ by nearly 3 standard errors; above 1000 square micrometres they agree with $-1/2$.

On small patches the noise is no longer a small perturbation and the linear argument fails. So a straight line through data at a few hundred square micrometres cannot be extrapolated to large patches.

The larger the patch, the closer the average path comes to the deterministic equations. At 10000 square micrometres the mean peak is 105.41 mV, the deterministic value; the mean latency is $0.009\ \mathrm{ms}$ earlier than the deterministic one, an error of the first-order voltage update in the patch simulation, of the size of its 0.01 ms step.

Source: J. H. Goldwyn and E. Shea-Brown · A. L. Hodgkin and A. F. Huxley

Noise alone can fire a spike

With no stimulus at all, small patches still fire. Over 20 simulated seconds each, the rates are $49.1\pm0.7\ \mathrm{s^{-1}}$ at 3 square micrometres, $38.4\pm0.4$ at 10, $27.1\pm0.6$ at 30 and $11.1\pm0.6\ \mathrm{s^{-1}}$ at 100.

Fig. 1 of Goldwyn and Shea-Brown shows a Markov-chain simulation of a 100 square micrometre patch with no current that fires once on its own near 70 ms. That is a single sample path and fixes only the order of magnitude; our 11 Hz means one spike every 90 ms on average, which agrees.

This firing belongs to the same family of problems as a bead in an optical trap that thermal kicks occasionally push over a barrier: one continuous variable is driven by many small random events and now and then crosses a critical point. The difference is that the noise here comes from a finite number of channels switching at random.

Source: J. H. Goldwyn and E. Shea-Brown · A. L. Hodgkin and A. F. Huxley

Worked example: predict the jitter of a 10000 square micrometre patch

Problem. The latency standard deviation of a 3000 square micrometre patch is $0.0336\pm0.0017\ \mathrm{ms}$. What does the square-root law predict for 10000 square micrometres? What if you extrapolate instead from $0.285\ \mathrm{ms}$ at 100 square micrometres? Each estimate uses 200 trials; roughly how large is the relative error of a standard deviation?

Model. Extrapolate with the left formula. The relative statistical error of a standard deviation is about $1/\sqrt{2(n-1)}$.

Numbers. From 3000 square micrometres: $0.0336\times\sqrt{3000/10000}=0.0184\ \mathrm{ms}$ with an error of $0.0009$. The measured value is $0.0190\pm0.0010$, a difference of 0.0006 ms, about 0.5 combined standard errors. From 100 square micrometres: $0.285\times\sqrt{100/10000}=0.0285\ \mathrm{ms}$, 50% above the measurement. With 200 trials the relative error is $1/\sqrt{398}=5.0\%$, matching the errors listed above.

Explanation. The square-root law is the asymptotic behaviour at large channel numbers. Starting from data already in that range it predicts well; starting from a small, noise-dominated patch it overestimates systematically.

Check. 3000 square micrometres hold 180000 sodium channels and 10000 hold 600000. Both give a mean peak of 105.41 mV and mean latencies that differ by less than 0.003 ms, so both patches are already close to the deterministic path, with only small fluctuations left.

Source: J. H. Goldwyn and E. Shea-Brown · A. L. Hodgkin and A. F. Huxley