What accelerates the electrons in the thin part of a current-carrying wire?

Consider a thick wire that narrows for some part of its length, connected to the terminals of a battery, hence carrying some current.

Presumably the electrons have to speed up as the wire gets thinner - if the current is constant everywhere, then if the wire cross-section is less, they have to go through faster in order for the same number of electrons to pass any point in a given time.

What accelerates them as the wire narrows?

And what decelerates them again as the wire widens out again after the narrow bit?

• @WaqarAhmad No! Bernoulli's equation has nothing to do with electrical conduction! See my answer below. – codeAndStuff Oct 2 '13 at 20:22
• I think your misunderstanding comes from visualizing an electron starting from one terminal and ending at the other. In reality the current is something like a continuous stadium wave youtube.com/watch?v=H0K2dvB-7WY where people move their hands but the signal travels the stadium. The drift velocity of the electrons is much smaller than the velocity of the electrons and smaller than the signal down the wire : hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html – anna v Nov 2 '13 at 8:42

What leads you to believe they must accelerate? I would argue that they don't. I would also bet that your confusion comes from taking the analogy of incompressible fluid through pipes to the flow of electricity a bit too far.

When electrons (quasiparticles) move through the conduction band of a conductor (semiconductor) you cannot view this as some fluid moving through a pipe and have a complete analogy. For the case of the fluid, it will speed up only if the cross section of the pipe right before the narrow part is 'full' for the fluid flow current to be constant. The cross section of a wire is not 'full' like this in the case of electrical conduction.

Charged particles will accelerate when another force is applied, hence maybe if some segment of wire that has a different potential applied across it one would see an acceleration. I would say that for a single wire connected to two terminals that you should not see this. If the wire got too thin, the resistance would be too great and you would see a change in current.

Side note: acceleration can occur with either $\pm$ sign, do not use the word deceleration. This is something the public (and new physics students) tends to get wrong and can be confusing for people learning physics.

• Thanks for pointing out the ever so common mistaken analogy to see electrons as a flow of fluid in a pipe. – mcodesmart Oct 2 '13 at 20:00
• I'm aware that the fluid/pipes analogy is far from perfect, and I take the point about partially full pipes, but this really isn't an answer to my question, or if it is, I don't understand. I'm trying to understand how the current in the thin section is the same as in the thick section. – ChrisA Oct 2 '13 at 20:18
• What I am saying is that there is plenty of 'space' for the electrons to flow without having to be 'bunched up' and accelerated as is the case when water is constricted. This physical constriction is actually not occurring in this case because there is a very low density of free electrons per unit volume in conductors or semiconductors. Most of the space in the wire is not taken up by electrons so they don't 'feel' the wire getting thinner. This is what I was getting at with the partially full pipes statement. Make sense? – codeAndStuff Oct 2 '13 at 20:26
• In that case, then, why is the resistance of a wire with a thin section higher than the resistance of an otherwise identical wire without the thin section? Or is it the case that the wire has to get much thinner before the resistance increases significantly? Also, is it not the case that all the available charge carriers in the wire experience the same force and hence move along the wire? – ChrisA Oct 2 '13 at 21:52
• The resistance is modeled similar to pipe flow analogy, i.e., the thinner the pipe, the higher the resistance. For the same amount of current, the thinner wire means a higher probability of collisions with the lattice, therefore, the resistance increases much like a pipe wall causes friction and increases the resistance. – mcodesmart Oct 2 '13 at 22:19

"Presumably the electrons have to speed up as the wire gets thinner"

No.

The speed (drift) of the electrons has nothing to do with the area.

$$v = \mu_eE$$

The current density is affected because it is a function of electric field, and mobility and number of electrons. The current is reduced because you have more scattering, and hence higher resistance

$$J = n\mu_e e E$$

• I would say that your statements are correct but do not fully address the original confusion this user has. The question was about how a current could stay constant through a thinner segment of wire, not how it would be reduced. – codeAndStuff Oct 2 '13 at 20:00