Typical drift velocities for electrons in, say, a conducting wire are on the order of $ 10^{-5}$ m/s. This is not the rate that the signal travel, of course, but the net flow of electrons.

As I understand it, the voltage in the wire causes the electrons to accelerate in a certain direction, which shifts the momentum spectrum to one side. Collisions with positive ions cause the electrons to change their momentum as well.

The point is, if you add up all the accelerating and shifting of momentum spectra and colliding with positive ions, you get a net flow of electrons which can be expressed as $$ \vec{v}_{drift} = \frac{ \vec{J} }{nq}$$ with $\vec{J}$ as current density, and n and q as number and current density.

If this is the case, then it seems like eventually (albeit a long eventually), all of the electrons will accumulate at one side, and the signal would be unable propagate.

In this really what would happen?

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    $\begingroup$ The charges move around a loop. $\endgroup$ – Farcher Oct 30 '16 at 17:08

If you're talking about an isolated conductor, then the voltage across the conductor isn't an independent quantity; it depends on the average location of the charge in the conductor. If the net charge is disproportionately towards one end, then this creates a voltage which pushes on the charges such that their movement will neutralize the voltage. So, as the electrons drift, the voltage will decrease, until the voltage has become zero across the conductor.

If you're talking about a wire in a circuit, then the voltage is externally imposed by pulling electrons out of one end of the wire and pushing them into the other end. This causes the electrons to drift, but as they accumulate at one end (and the holes accumulate at the other) then the external circuit will be removing/adding them as necessary to maintain the voltage.


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