How can a nerve impulse travel along an axon without fading away?

A nerve impulse reaches the far end of a long axon because successive membrane regions regenerate its voltage change. The membrane separates fluids with different ion concentrations, and selective permeability helps create a voltage across it. What travels is a relayed electrical event, rather than one packet of ions moving the entire distance.

When a region is depolarized enough, voltage-sensitive sodium channels open. Sodium enters along its electrochemical gradient, making the local interior less negative. That voltage change drives local electrical currents through the neighboring interior and exterior fluids. These currents change the voltage across nearby membrane.

If the next region reaches threshold, its own sodium channels open and produce a new action potential. The signal is thus relayed by successive local events. Regeneration replaces what passive spread alone would lose, much as a line of triggered devices can transmit an event even though no single device moves down the line.

Behind the active region, sodium channels become temporarily unavailable and potassium-channel activity helps return the voltage toward rest. This refractory period limits immediate re-excitation. For a normally initiated impulse, membrane ahead is ready while recently active membrane behind is recovering, favoring onward propagation. Artificial stimulation in the middle can produce travel in both directions.

Energy-consuming pumps maintain ion gradients over time, while local currents and channel opening carry each impulse forward. Myelin allows voltage to spread between exposed nodes, where action potentials regenerate. The recovery–active–resting illustration neither measures a neuron nor calculates speed; it shows how renewing a local event prevents the traveling signal from simply fading away.