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The nerve impulse transmission is specifically a biophysical process. Under a resting stage, the membrane is already polarised (presence of charge on either side leading to a potential difference across it, due to its finite capacitance). Changes in this polarisations mediated by several agents, cause some part of this membrane to be depolarised (inversion of the charges in that localised area). This changes the potential difference across that region and starts ion flow from the depolarised region of the membrane to the adjacent polarised region, inducing the same depolarisation there. Myelinating a neuron is similar to lowering the capacitance of the neuron. Myelinating the neuron causes subsequent action potentials to be spatially separated, and the conduction of voltage between them occurs primarily through ion flow along the neuron. Here are some nerve conduction basics.

Now my actual question. Modelling a neuron as a simple one dimensional membrane/cable, how can a lowered capacitance lead to a faster conduction of the voltage perturbation i.e a change in voltage (depolarisation), which is initially limited to a small localised region? This conduction of the voltage change can occur through the ion flow along the inner side of the membrane and also due to long distance changes in potential due to the altered distribution of charges along the membrane. A very good physical modelling of the neuron is given here and here, but because of the quite complicated mathematical nature of the modified telegrapher's equation which appears as the final answer, I am unable to understand how lowered capacitance increases the speed of voltage conduction?

Tell me if this question is off-topic or too biological to be answered here.

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    $\begingroup$ IMO definitely on topic. In general, lower capacitance raises the propagation speed in the telegrapher's equation: in the lossless case $c = 1/\sqrt{L\,C}$. Intuitively, lower capacitance lowers the time one needs to make a system voltage change: a big capacitance means a great deal of charge has to accumulate to beget a small voltage change. $\endgroup$ Commented Jan 4, 2014 at 13:12
  • $\begingroup$ I also think it is very on-topic. In medical school they showed us the Hodgkin-Huxley equation, which I thought was astonishingly beautiful. Most of my fellow students thought it was just Greek symbols on a blackboard. $\endgroup$
    – DWin
    Commented Jan 4, 2014 at 17:46

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The RC circuit that approximates a myelinated fiber looks like a series of resistors along the length of the fiber with capacitors connected to ground at the gaps between the Schwann cells. The time to reach discharge voltage at the next junction after depolarization at the prior junctions will be determined by the time it takes to raise the capacitor voltage to that threshold for depolarization. Decreasing the capacitance decreases the time to charge a capacitor, so decreasing capacitance shortens the time for the signal to "jump" to the next node.

I did a bit of searching and thought that the cable theory material you found was not entirely on point for myelinated nerves. It looked more germane to brain neuronal transmission. The best synopsis I found was in the 'Nerve Conduction' section of an online biophysics text: http://www.centenary.edu/biophysics/bphy304/book/#/116/

(The course page was taken down, but it's still at the Wayback Machine): https://web.archive.org/web/20160402080329/https://www.centenary.edu/attachments/biophysics/bphy304/11a.pdf

This is the (passive) network model proposed for transmission of signal between nodes in myelinated nerves:

enter image description here

The article has additional analysis and illustrations which contrast cable theory to this model. I think you can also get insights by looking up the Hodgkin-Huxley Model of membrane depolarization which adds an active component at each gap in the myelination along the axon.

enter image description here

Here's the mathematics copied from the Wikipedia article:

Mathematically, the current flowing through the lipid bilayer is written as

${\displaystyle I_{c}=C_{m}{\frac {{\mathrm {d} }V_{m}}{{\mathrm {d} }t}}}$

and the current through a given ion channel is the product of that channel's conductance and the driving potential for the specific ion

${\displaystyle I_{i}={g_{i}}(V_{m}-V_{i})\;}$

where ${\displaystyle V_{i}}V_{i}$ is the reversal potential of the specific ion channel. Thus, for a cell with sodium and potassium channels, the total current through the membrane is given by:

${\displaystyle I=C_{m}{\frac {{\mathrm {d} }V_{m}}{{\mathrm {d} }t}}+g_{K}(V_{m}-V_{K})+g_{Na}(V_{m}-V_{Na})+g_{l}(V_{m}-V_{l})}$

And here's a diagram from that course topic cited above showing the integration fo the active and pasive parts of that model: enter image description here

See also: Discussion of Hopf bifurcation in HH model

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  • $\begingroup$ Does you first sentence mean that one should be thinking of a sequence of RC filters separated by a nonlinear, thresholding circuit? $\endgroup$ Commented Jan 5, 2014 at 13:00
  • $\begingroup$ If I understand you correctly, then definitely. The passive RC networks you sometimes see illustrated do not take into account the active pumping during the "quiescent phase" along the course of the nerve fiber nor the "triggering" of the nerve activation when the depolarization impulse arrives from the upstream portion. I suppose a series of batteries and switches located at each node would a start toward a functional model, but I suspect you are thinking of a more high fidelity model. I was trying to honor the request for understanding the role of capacitance "along the line". $\endgroup$
    – DWin
    Commented Jan 5, 2014 at 16:58
  • $\begingroup$ Is this text describing the saltatory transmission of a myelinated axon amongst exposed nodes? As I understand it, saltatory transmission occurs only when there are exposed nodes on the axon but not when there are no exposed nodes. This theory only described the transmission of action potential along the myelinated line acting as an insulated cable. Where does the pulse come in? $\endgroup$
    – Hans
    Commented May 31, 2014 at 23:54
  • $\begingroup$ The "pulse" hops from depolarized node to next hyperpolarized node. $\endgroup$
    – DWin
    Commented Jun 1, 2014 at 0:16
  • $\begingroup$ @DWin: I am asking for the mathematics describes this hopping of the pulse. I do not see the mathematics there. $\endgroup$
    – Hans
    Commented Jul 12, 2014 at 19:24

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