The voltage and the open doors

Miraia S. Chiou

Reading · Neuroscience

The voltage and the open doors

The same voltage can conceal different numbers of open channels, and different responses.

2026-09-08 · 12 min read · 1 interactive demos

Picture a room with doors opening onto three terraces. Each terrace sits at a different height. Opening a door lets the room exchange water with that terrace. The room’s eventual water level depends on both the terrace heights and how widely the doors open.

A neuron’s membrane offers an electrical version of this scene. Its voltage describes an electrical potential difference. Its ion channels provide conducting paths. Each path pulls toward a particular reversal potential, the voltage at which that path carries no net current.

Here is the puzzle: two rooms can have the same water level even though one has far more open doors. If you pour in the same extra cup, do they rise by the same amount? How quickly do they settle?

This is where keeping both direction and amount, the recurring habit of Epistemic Physics, becomes useful.

First, follow the current

Take one passive membrane compartment: a region whose voltage we treat as spatially uniform. Let $C$ be its capacitance, the amount of electrical charge it stores per unit voltage. Let $g_{E}$ , $g_{I}$ and $g_{L}$ be nonnegative conductances for excitation, inhibition and leak. Their reversal potentials are $E_{E}$ , $E_{I}$ and $E_{L}$ .

Using inward current as positive, the model is

C \frac{d V}{d t} = g_{E} (E_{E} - V) + g_{I} (E_{I} - V) + g_{L} (E_{L} - V) + I (t) .

Here $I (t)$ is an externally supplied current. The dependence on the difference $E - V$ matters: the same opened channel can pass different currents at different membrane voltages. This conductance formulation and its effective time constant are developed in Neuronal Dynamics, §13.6.3.

For the next calculation, hold all conductances and reversal potentials fixed. Collect the terms:

G = g_{E} + g_{I} + g_{L}, V_{\infty} = \frac{g_{E} E_{E} + g_{I} E_{I} + g_{L} E_{L}}{G} .

Assume $G > 0$ . With no external current, $V_{\infty}$ is the equilibrium voltage. The entire equation becomes

C \frac{d V}{d t} = - G (V - V_{\infty}) .

The weighted average locates the destination. The total conductance helps determine how the membrane gets there.

How far and how fast

Start at equilibrium and switch on a constant current $I_{0}$ . Write $Δ V (t) = V (t) - V_{\infty}$ , the change above equilibrium. Solving the equation gives

Δ V (t) = \frac{I_{0}}{G} (1 - e^{- t G / C}) .

There are three experimentally distinct quantities hiding in this expression:

What you measure	Formula	What increasing $G$ does, with $C$ and $I_{0}$ fixed
Eventual displacement	$I_{0} / G$	Makes the displacement smaller
Time constant	$C / G$	Makes settling faster
Initial slope	$I_{0} / C$	Leaves the first instant unchanged

A time constant is the time needed to complete a fraction $1 - e^{- 1}$ , about 63%, of the eventual change. A membrane can therefore be difficult to displace and quick to settle. Those properties fit together beautifully once we stop using “responsiveness” for both of them.

Predict before touching: choose the two same-voltage presets below. Which curve ends higher? Which reaches most of its own eventual height sooner? Look closely at their first instant.

Open the membrane's channels

Same voltage. A different response.

Drag the two channel controls and watch both cells respond to the same brief current pulse. The two “same voltage” presets have identical resting output, but one is four times as conductive.

Chosen cellLow-conductance referenceCurrent pulse is on

Equilibrium voltage: -40.0 mV
Total conductance G: 24.0 nS
Response time C / G: 8.3 ms
Centered voltage v: 0.000

The chosen cell moves 2.48 mV during the pulse; the reference moves 6.99 mV. More conductance can mean a smaller response that settles faster.

A passive cell with fixed conductances: C = 200 pF, leak = 2 nS, reversal potentials 0, −80 and −60 mV. The current is 60 pA for 40 ms. Curves show changes above each cell's equilibrium. Active channels and spikes are outside this model.

Open the membrane experiment in the playground.

The low-conductance preset has $G = 6$ nanosiemens; the high-conductance preset has $G = 24$ nanosiemens. Both rest at $- 40$ millivolts. With $C = 200$ picofarads, their time constants are about $33.3$ and $8.3$ milliseconds. A sustained $60$ picoampere current would eventually displace them by $10$ and $2.5$ millivolts. The displayed pulse ends after $40$ milliseconds, so neither curve is being measured at infinite time.

At pulse onset, both slopes are $60 / 200 = 0.3$ millivolts per millisecond. After the current stops, the accumulated displacement decays exponentially with the same cell-specific time constant.

Open the evidence-coordinate door

Suppose the reversal potentials obey $E_{I} < E_{L} < E_{E}$ . Mark the leak’s relative location between the inhibitory and excitatory endpoints:

p_{L} = \frac{E_{L} - E_{I}}{E_{E} - E_{I}} .

Choose a positive conductance unit $g_{*}$ . Define two dimensionless weights:

α = \frac{g_{E} + p_{L} g_{L}}{g_{*}}, β = \frac{g_{I} + (1 - p_{L}) g_{L}}{g_{*}} .

Now align the electrical and evidence descriptions:

Passive membrane	Two-pool coordinates
Excitatory-side weight $(g_{E} + p_{L} g_{L}) / g_{*}$	$α$
Inhibitory-side weight $(g_{I} + (1 - p_{L}) g_{L}) / g_{*}$	$β$
Total conductance $G / g_{*}$	Total shape amount $A = α + β$
Normalized equilibrium $(V_{\infty} - E_{I}) / (E_{E} - E_{I})$	Mean coordinate $μ = α / A$
Centered equilibrium $2 (V_{\infty} - E_{I}) / (E_{E} - E_{I}) - 1$	Direction $v = (α - β) / A$

These equalities follow by substitution. In particular, setting $S = α - β$ gives the cone identity

A^{2} - S^{2} = 4 α β \geq 0.

For strictly positive weights, we could also attach a Beta distribution with shapes $α, β$ . Its mean would equal the normalized equilibrium. The voltage coordinate here is the mean, not the Beta mode. When both shapes exceed one, the mode would instead be $(α - 1) / (A - 2)$ .

The algebra leaves $g_{*}$ free. Changing that scale changes a proposed Beta’s variance while leaving the physical equilibrium untouched. To interpret that distribution as biological variability, we would need a specified noise mechanism and a way to determine the scale. The exact coordinate map already lets us study equilibrium and conductance together.

The leak makes an asymmetric baseline

In a Beta evidence model, write $α = α_{0} + Y$ and $β = β_{0} + N$ , where $Y, N$ are added supporting and opposing trial amounts and $α_{0}, β_{0}$ are baseline shapes. Our membrane dictionary suggests

Y = \frac{g_{E}}{g_{*}}, N = \frac{g_{I}}{g_{*}}, α_{0} = \frac{p_{L} g_{L}}{g_{*}}, β_{0} = \frac{(1 - p_{L}) g_{L}}{g_{*}} .

This is a useful prior-like role for leak: it supplies a baseline location and amount. It need not be symmetric. In the demo, $E_{I} = - 80$ , $E_{E} = 0$ and $E_{L} = - 60$ millivolts, so $p_{L} = 1 / 4$ . The baseline weights have ratio $1 : 3$ .

For a Jeffreys Beta baseline, both shapes would have to equal $1 / 2$ . Our mapping produces that particular choice only when the leak lies at the midpoint and its total conductance equals $g_{*}$ . A neuron does not acquire that prior merely by having a leak.

Also notice the different clocks. Conductance describes channels open now. A trial ledger usually records retained observations. Translating a history into present conductance requires a storage or decay rule. A continuing supply of synaptic events could maintain open channels; a long-term synaptic change could alter how strongly future events open them. These are distinct mechanisms to explore.

Shunting: change the doors without moving the level

Imagine adding a conductance $h$ whose reversal potential equals the current equilibrium $V_{\infty}$ . At that voltage its net current is zero, so the resting level stays fixed. Nevertheless,

G ⟶ G + h, τ ⟶ \frac{C}{G + h}, Δ V_{steady} ⟶ \frac{I_{0}}{G + h} .

This illustrates shunting: an extra conducting path reduces the voltage effect of other injected current. A shunt can change the membrane’s behavior even when it barely changes its resting voltage.

In two-pool coordinates, adding an amount at the existing mean preserves the balance and increases the total. That is the same weighted-average operation. If the added path pulls toward a different reversal potential, the balance changes too.

Scaling every conductance by the same positive factor is another exact way to preserve equilibrium while changing time constant and current sensitivity. Scaling only excitation and inhibition while holding an asymmetric leak fixed generally changes the equilibrium; preserving the voltage then requires choosing the two increments appropriately.

What might a branching neuron compute?

Let’s give our room a corridor into a second room. In a two-compartment passive model, compartment $i$ has voltage $V_{i}$ and capacitance $C_{i}$ . A coupling conductance $k$ joins it to compartment $j$ :

C_{i} \frac{d V_{i}}{d t} = G_{i} (E_{i}^{*} - V_{i}) + k (V_{j} - V_{i}) .

Here $G_{i}$ collects the local membrane conductances, and $E_{i}^{*}$ is their weighted local reversal potential. For a temporarily fixed neighboring voltage, the instantaneous equilibrium is another weighted average:

V_{i}^{*} = \frac{G_{i} E_{i}^{*} + k V_{j}}{G_{i} + k} .

Now the interesting part: the neighboring voltage moves too. The whole system has coupled relaxation modes, patterns of joint settling, rather than one universally sufficient $C / G$ clock.

This suggests an adventurous model-building route. Let separate dendritic branches, the neuron’s receiving extensions, retain different two-pool states. Let their conductances govern how strongly they influence a shared cell body. Then ask whether matching the cell body’s mean voltage can hide different branch-level disagreements and different pulse responses.

A source lineage matters here. If one presynaptic event affects two branches, its two appearances need not represent two independent trials. A tracelet could retain that common origin while a chosen electrical rule distributes its effects. The compartment equation supplies the coupling; the lineage model supplies the identity accounting.

The next question is wonderfully concrete: can two cells look identical at rest and reveal different hidden branch structure when nudged in different places? Our little room with open doors has become a house worth exploring.

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