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(In physics.stack I have been suggested to post my question also here.)

In the classical theory of passive neurons (where the action potential is not yet excited), the voltage is successfully described by cable theory. The (unmyelinated) axon is modeled as a series of cylindrical sections in series. Each section is described by an electric circuit (in 1D) with an axial resistance, and the membrane is modeled as a capacitor.

If charges are injected at one end of the cylinder, the "cable equation" gives the transmembrane voltage at steady-state in the shape of an exponentially decreasing function of space. Here the characteristic 1/e decay length of the voltage can be long in large axons, of the order of mm.

Question

My question is about the apparent lack of charge screening in this picture. At physiological conditions (where ions and counter-ions are present), I'd expect that an excess of charge, like the one injected by a micro-pipette, or generated by the opening of channels, is screened very quickly (microseconds), so a few Debye-lengths (nm) apart, the voltage should not be seen.

How can the theory and experiments show a voltage that persists at millimeter distances at physiological conditions where screening should be present?

EDIT from a comment: The electrotonic transmembrane potential is exponentially decreasing along the membrane. If we assume one side isopotential (say the interior), the other side should then have a voltage that decreases along the membrane. So measuring the voltage on two points along the membrane, one should see a drop in voltage, if the points are distant enough. Therefore on that side there is a current, and the charges on that side are not screened.

Is this description correct?

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In biological membranes, the primary physics principle resisting change in voltage is described by capacitance of the membrane, which is a key aspect of the cable equations. If you model some instantaneous change in charge (not particularly biological, but something like it is possible using an amplifier in patch-clamp electrophysiology), the change in membrane potential will be smoothed by the membrane capacitance.

When people talk about voltages in biological systems, they're talking about an electrical potential difference across the membrane, not measuring within nanometers inside the cell. Yes, if you somehow placed two sensors nanometers apart in the cell you'd have some difficulty measuring a voltage between them - for practical intents you can consider the intracellular space to be isopotential over the distances relevant to molecules.

Thinking about other ways to phrase this: maybe it will make sense to you to think about a distinction between charge of a solution and a particle (e.g., a single ion in that solution). When you have more positive than negative ions on one side of a membrane compared to the other, you're going to have a voltage where it's more energetically favorable for positive charge to move on the negative side than vice-versa. Charge-screening refers to how the total charge of the extra positive ions (or lack of negative ions) distributes in solution, such that it's not like you have "islands" of positivity on the positive side around each positive ion (as long as you're viewing from outside Debye lengths), but rather the charge is still somewhere in the solution when you measure compared to the other side - that's the membrane voltage. Note that when you look at the equation for voltage over a capacitor:

V = Q/C

the equation just involves the charge and the capacitance, not the specific locations of the charges.

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  • $\begingroup$ thank you for your answer. The electrotonic (transmembrane) potential is exponentially decreasing along the membrane. If we assume one side isopotential (you suggest the inside one), the other side should have a voltage that decreases along the membrane. So your two point measurement should see a drop in voltage, if the points are distant enough. Therefore on that side there is a current, and the charges on that side are not screened? Does this sound correct? $\endgroup$
    – scrx2
    Dec 15, 2022 at 8:27
  • $\begingroup$ @scrx2 No, it doesn't sound correct, when I'm talking about being isopotential I mean over the short distances where charge screening is relevant. You're not appreciating that charges can't move across the membrane freely, where the actual membrane potential is realized. $\endgroup$
    – Bryan Krause
    Dec 15, 2022 at 15:00
  • $\begingroup$ Maybe here's a different way to think about it: you have net 10 charges on one side of the membrane, and net 10 on the other. Now you move one across so you have 9 charges on 1 side and 11 on the other. If charges now cannot move across the membrane, how would you arrange those charges in space on each side such that there is no voltage across the membrane? $\endgroup$
    – Bryan Krause
    Dec 15, 2022 at 15:17
  • $\begingroup$ yes, there will be a voltage difference across the membrane, following an exponential profile along the membrane (say X). But, because of this exponential profile in X, there should be a potential difference also along X. Along X, as an electrical resistance is present, there will be a current. Do you agree? $\endgroup$
    – scrx2
    Dec 16, 2022 at 19:05
  • $\begingroup$ Are you saying that the extra charge brought, by say one ion pump, from one side to the other cannot be screened? I'd agree $\endgroup$
    – scrx2
    Dec 16, 2022 at 19:08
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So measuring the voltage on two points along the membrane, one should see a drop in voltage, if the points are distant enough. Therefore on that side, there is a current, and the charges on that side are not screened.

You are correct that in theory along an axon membrane for example, the decrease in voltage is exponential in space. However, you need to consider saltatory conduction. There, the action potential propagation occurs from one node of Ranvier to the next. This mechanism increases the conduction velocity of action potentials and also decreases the exponential voltage drop.

How can the theory and experiments show a voltage that persists at millimeter distances at physiological conditions where screening should be present?

The uninsulated nodes of Ranvier are the only places along the axon where ions are exchanged across the axon membrane, regenerating the action potential between regions of the axon that are insulated by myelin, unlike electrical conduction in a simple circuit. That's where cable theory falls short. The saltatory process also allows for maintaining voltages along a vast distance of biomolecular membranes. In addition to increasing the speed of the nerve impulse, the myelin sheath helps in reducing energy expenditure over the axon membrane as a whole, because the amount of sodium and potassium ions that need to be pumped to bring the concentrations back to the resting state following each action potential is decreased. Also, keep in mind that the axon potential propagation is unidirectional because the absolute refractory period (of the channels) prevents the initiation of an action potential in a region of the membrane that has just produced an action potential.

This literature might touch on the physics in more detail: The Membrane Potential and its Representation by a Constant Electric Field in Computer Simulations

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  • $\begingroup$ thank you for your answer. I'm specifically interested in passive conditions, under the firing threshold, to forget the voltage-dependent channels. And as you correctly mention it, I'd add that I'd like to consider unmyelinated axons. Thank you for the ref too, seems interesting! $\endgroup$
    – scrx2
    Dec 16, 2022 at 9:53
  • $\begingroup$ Under the firing threshold, the ion exchange over the membrane is in a steady-state model. In such physiological conditions, a couple of important physical principles need to be considered: The electrochemical potential is NOT in equilibrium. Meaning that Nernst-potentials and cable equations fall short! Meaning a steady-state model is more accurate. In this case, the Goldman-potential or Goldman equation is used to describe the state of no net flow of ions. $\endgroup$ Dec 20, 2022 at 3:29
  • $\begingroup$ But even with this model a couple of assumptions are made: 1. Steady-stateI: on-concentration are in steady-state dc/dt=0. The membrane potential Vm is constant (dV/dt=0) and there is no net current across the membrane (I=0). 2. The membrane has a thickness d and is physically homogeneous 3. The different ion fluxes are uncoupled. 4. We have a distribution equilibrium for an ion i between the solution and membrane interface. 5. Simplification: Only monovalent ions. 6. Flux in the membrane is conservative, nothing gets lost. $\endgroup$ Dec 20, 2022 at 3:33
  • $\begingroup$ I feel like you might want more sophisticated mathematical models so check out Goldman–Hodgkin–Katz flux equation, Hindmarsh–Rose model, Hodgkin–Huxley model, Morris–Lecar model $\endgroup$ Dec 20, 2022 at 3:48

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