# Drift speed of an electron

Bipolarity
So in Resnick & Halliday, it explains that the drift speed, or the average speed of charge moving in a wire under the influence of an electric field E is defined by the equation ## J = (ne)v_{d} ##.
Now if J,n,e are all constants then the drift speed ##v_{d}## will also be a constant. But this appears to not make sense to me given the fact that electrons are expected to constantly accelerate since the electric field continues to act on the electron as it is conducted along the wire. For instance, if a force constantly acts on you in some direction, you won't just drift in that direction, you will accelerate! So how can the drift speed not be dependent on the time you have spent in motion?

Under this argument it would make sense that ## v_{d} = ta ## where a is the acceleration of the electron, which is constant if the field E is constant, and t where t is the total time the electron has spent in motion.

But it turns out t is actually the mean free time (average time between two electron collisions). So what's the deal here?

BiP

Ok, you can say that the electronic speed shows some saw-tooth like variation with time where it increases linearly during collisions and is reset to 0 in collisisons. However if you average over the times when the collisions take place you will get a constant drift speed. Usually, the drift speed is of interest as an average property of all electrons, so the timing of the individual collisions aren't of interest anyways.

Gold Member
The simple ballistic model of an accelerating electron is ok for the situation in a CRT (vacuum) but the conduction electrons are not strictly operating in this way and are better described in terms of waves, I believe, when in this semi-bound state in the lattice.
It's all a bit approximate to apply Newton's laws. For instance, your expression for drift time would really be twice that value as it's mean speed that counts(?).
The drift speed, as an idea, doesn't need to involve such matters because it's a macroscopic idea which doesn't say anything about the nature of the electron or what it's actually doing. It assumes a continuum of charge and uniform speed - just like water in a pipe.

Staff Emeritus
The simple ballistic model of an accelerating electron is ok for the situation in a CRT (vacuum) but the conduction electrons are not strictly operating in this way and are better described in terms of waves, I believe, when in this semi-bound state in the lattice.
It's all a bit approximate to apply Newton's laws. For instance, your expression for drift time would really be twice that value as it's mean speed that counts(?).
The drift speed, as an idea, doesn't need to involve such matters because it's a macroscopic idea which doesn't say anything about the nature of the electron or what it's actually doing. It assumes a continuum of charge and uniform speed - just like water in a pipe.

For the level that the OP is asking (i.e. Halliday and Resnick), the simplistic model is sufficient. After all, open the first couple of chapters of Ashcroft and Mermin, and you are dealing with the Drude model that has a good enough success that we can derive Ohm's Law.

Zz.

Electrons feel a drag due to collisions. The drift velocity is a balance between the accelerating field and the drag force. The electrical energy is converted into heat. The resistance of a wire is caused by these collisions.

Collisions between electrons won't affect the average electron speed. But collisions between electrons traveling in one direction and the heavy ions will slow down the electrons.

For DC circuits, you don't really care about the time it takes for the electrons to reach the drift velocity because the time it takes to reach the equilibrium is very short. But for AC circuits, the time dependence does matter, and it can cause the resistance to depend on frequency, and it is more appropriate to talk about the impedance.

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Staff Emeritus
Electrons feel a drag due to collisions. The drift velocity is a balance between the accelerating field and the drag force. The electrical energy is converted into heat. The resistance of a wire is caused by these collisions.

Collisions between electrons won't affect the average electron speed. But collisions between electrons traveling in one direction and the heavy ions will slow down the electrons.

For DC circuits, you don't really care about the time it takes for the electrons to reach the drift velocity because the time it takes to reach the equilibrium is very short. But for AC circuits, the time dependence does matter, and it can cause the resistance to depend on frequency, and it is more appropriate to talk about the impedance.

There's a lot of things incorrect here.

The resistance in the wire is caused by a number of factors: electron-electron scattering, electron-ion scattering, electron-impurity scattering, etc. Electron-ion dominates over large range of temperatures.

Electron-ion collision preserves the electron's kinetic energy, and thus, its speed. This is a mostly elastic collision because the ions are so big, so the electron just simply bounce back. Electron-electron collision is energy-absorbing for the electron that has the energy speed or energy. That is why the escape depth of electrons in a semiconductor tends to be LONGER than the escape depth of electrons in metals.

Zz.

Zz, I don't see how that can be correct. Granted, I haven't studied wires and I'm kind of guessing, but the situation seems analogous to a plasma. What is wrong with the following argument?

The current is the average electron velocity times e (charge). By conservation of momentum, electron-electron collisions can't actually change the average electron velocity. Therefore, electron-electron collisions don't change the current. Electron-ion collisions can change the average electron velocity. Even if the collision is elastic the electron changes direction so the average electron velocity isn't the same anymore.

Staff Emeritus
Zz, I don't see how that can be correct. Granted, I haven't studied wires and I'm kind of guessing, but the situation seems analogous to a plasma. What is wrong with the following argument?

The current is the average electron velocity times e (charge). By conservation of momentum, electron-electron collisions can't actually change the average electron velocity. Therefore, electron-electron collisions don't change the current. Electron-ion collisions can change the average electron velocity. Even if the collision is elastic the electron changes direction so the average electron velocity isn't the same anymore.

Electron-electron collision has a tendency to remove energy from the energetic ones. Try it yourself. Throw a ball at a wall and look at the speed after collision. Then throw a ball at another identical ball, and then look at the speed of the original ball afterwards. We do this in intro physics classes!

I've given you EVIDENCE, i.e. via the observed escaped depth, of what I've said. I can also cite the fact that photocathodes made of metals have a lower quantum efficiency than photocathodes made of semiconductors. Why? Because one of the main reason for this is that the excited electrons will have a higher probability of colliding with the free electrons in a conductor, and will lose its energy (lower speed) in such a collision such that it will no longer have enough energy to escape.

Here's a Masters Thesis for you to read. (http://psec.uchicago.edu/library/photocathodes/loch_thesis_CsTe.pdf) In particular, read the paragraph starting at the bottom of Pg. 7:

In metallic photocathodes the electrons are excited into the conduction band. However, in a metal there is an abundance of free electrons in the conduction band due to gradual transitions of thermally excited electrons from the valence band that is overlapping the conduction band. Thus when photoexcited electrons move to the vacuum interface, they encounter a high number of scattering events with free electrons from the Fermi sea and
thus thermal equilibrium is approached within a short distance. Only the small part of the electrons that are excited very close to the surface and which have not thermalized when reaching the surface, still posses sufficient energy to escape across the surface potential. This short distance (typically a few nm for metals) is
called the escape depth. In contrast to metals, the conduction band in semiconductors is almost empty before photoexcitation, such that electron-phonon scattering is the dominant loss process. The corresponding losses are much smaller per collision compared to electron-electron scattering. Consequently, the photoelectron lifetime (time when thermal equilibrium is reached) is longer and electrons from deeper regions can reach the surface. Thus the escape depth, which is the average distance from where excitedelectrons have a 1/e probability to escape, in semiconductors may be several times larger than the mean free path, which is the average distance between two sequential collisions.

You are responding to this via your "intuition", which in this case isn't correct! I strongly suggest you look up what I've given you (I've given you TWO separate experimental evidence already).

Zz.

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Ok, I looked into it a little more and I am more convinced that you are incorrect, Zz.

The evidence you gave (about photocathode emission) is not appropriate to the problem of resistivity in a metal. The problem of photocathode emission is different because there is a large work function that the electron has to overcome. Electron-electron collisions are fast at thermalizing the electron energy distribution function so that very few electrons have the energy to escape the metal. For resistivity in a metal, there is no work function because the electrons aren't escaping the metal. It is the average velocity that matters, not the tail energy distribution.

In the Drude model of resistivity (which is too simplistic, but sometimes gives ok results), the conductivity scales with the mean time between electron-ion collisions. https://en.wikipedia.org/wiki/Drude_model
Unfortunately, I can't read Drude's papers, which are in German. But the Drude model doesn't care about electron-electron collisions.

There are more advanced models of resistivity, such as the Drude-Sommerfeld model, but it becomes harder to define what an electron-ion collision is, since it is a quantum mechanical treatment of an electron in a periodic potential (which is ultimately generated by the ions. So it is still an electron-ion collision, but treated more properly).

From what I know, at least at room temperature, electron-impurity scattering and electron-phonon scattering are the most important sources of resistance. Electron-electron scattering is hardly effective in ordinary metals. The reason is given within the Landau theory of Fermi liquids, which is rather subtle. Basically, very near the Fermi energy, there is no phase space available for electron-electron scattering, even if the electron-electron interaction is strong.

Staff Emeritus
Ok, I looked into it a little more and I am more convinced that you are incorrect, Zz.

The evidence you gave (about photocathode emission) is not appropriate to the problem of resistivity in a metal. The problem of photocathode emission is different because there is a large work function that the electron has to overcome. Electron-electron collisions are fast at thermalizing the electron energy distribution function so that very few electrons have the energy to escape the metal. For resistivity in a metal, there is no work function because the electrons aren't escaping the metal. It is the average velocity that matters, not the tail energy distribution.

In the Drude model of resistivity (which is too simplistic, but sometimes gives ok results), the conductivity scales with the mean time between electron-ion collisions. https://en.wikipedia.org/wiki/Drude_model
Unfortunately, I can't read Drude's papers, which are in German. But the Drude model doesn't care about electron-electron collisions.

There are more advanced models of resistivity, such as the Drude-Sommerfeld model, but it becomes harder to define what an electron-ion collision is, since it is a quantum mechanical treatment of an electron in a periodic potential (which is ultimately generated by the ions. So it is still an electron-ion collision, but treated more properly).

You don't appear to understand the relevant material I was citing here. We are talking about the energy loss upon scattering!

When an electron has been excited to the vacuum state, it then must migrate to the surface to escape the metal. This process is similar to ANY charge transport process! It undergoes all the same scattering mechanism that a conduction electron has to go through.

Remember, the argument here is which type of collision destroys the high energy/speed that an electron has. You argued that electron-electron collision preserves most of the original electron's energy, while I disagree and argued that the electron-ion/phonon collision is the one that does that. The paragraph I cited (and the fact that electrons have longer escaped depth in a semiconductor than in metals) are evidence to show this!

You seem to have lost the point of this argument.

I'm just surprised that you've never done this simple experiment in school. Try colliding two objects together. Have two different targets: one with a mass similar to the incoming object, and the other with a mass many times bigger than the incoming object. Which of the collision will "absorb" more of the velocity of the incoming object? Think about it! If that second object is a wall, what do you think will happen? It is basic, first year physics!

Zz.

ZapperZ, your argument seems to be all right for electrons excited high above the Fermi level, but I thought the OP was asking about the average drift speed of electrons in a wire under a (probably very small) external field? In the latter case, electron-electron collisions are ineffective.
But even for electrons well above the Fermi level, electron-electron scattering is not very effective in reducing the average current (as opposed to reducing the speed of a single excited electron), as the scattered electrons will move preferentially in the same direction as the highly exited electron before a collision. Only collisions of the Umklapp-type are effective in reducing the current, but I think that these are very rare in generic situations. This maybe the argument Kashishi is referring to.

Bipolarity
I am satisfied as far as the responses are concerned, but this thread has derailed off-topic. Thank you all for your replies. I request that a moderator close the thread.

BiP

Staff Emeritus
ZapperZ, your argument seems to be all right for electrons excited high above the Fermi level, but I thought the OP was asking about the average drift speed of electrons in a wire under a (probably very small) external field? In the latter case, electron-electron collisions are ineffective.
But even for electrons well above the Fermi level, electron-electron scattering is not very effective in reducing the average current (as opposed to reducing the speed of a single excited electron), as the scattered electrons will move preferentially in the same direction as the highly exited electron before a collision. Only collisions of the Umklapp-type are effective in reducing the current, but I think that these are very rare in generic situations. This maybe the argument Kashishi is referring to.

Note what I was originally replying to in this line of discussion, which was post #5.

Electrons feel a drag due to collisions. The drift velocity is a balance between the accelerating field and the drag force. The electrical energy is converted into heat. The resistance of a wire is caused by these collisions.

Collisions between electrons won't affect the average electron speed. But collisions between electrons traveling in one direction and the heavy ions will slow down the electrons.

For DC circuits, you don't really care about the time it takes for the electrons to reach the drift velocity because the time it takes to reach the equilibrium is very short. But for AC circuits, the time dependence does matter, and it can cause the resistance to depend on frequency, and it is more appropriate to talk about the impedance.

Also note that in Post #6, I specifically wrote the scattering mechanism that would affect the resistivity of a typical metal, which is exactly what you wrote here.

The problem is the notion that " Collisions between electrons won't affect the average electron speed. But collisions between electrons traveling in one direction and the heavy ions will slow down the electrons... " The electron-electron collision "thermalizes" the speed, i.e. it tends to slow electrons having higher speeds and it tends to "pick up" those moving slower than the average. The collisions between ions don't do that, i.e. it tends to preserve the spread in speeds of an electron gas.

Zz.

Gold Member
Sometimes it is a good idea to step away from the specifics and look at things from a macroscopic point of view.

If you can bring yourself to accept that the current must be related to the rate of flow of charge through a section of the wire then, whatever ideas you want to introduce about the fine detail of the statistics of the speeds of electrons, the mean velocity is what it is. (That is assuming that there are no other charge carriers about - and I can't think of any likely candidates for that role.)

dipole
The drift velocity is an average - the electrons in a wire don't flow freely, but at scattered in various ways (and the specific details of how they are scattered and by what aren't that important here). So yes, at anyone moment an electron is accelerating, but because they bounce around like pin balls, their forward movement is greatly impeded, and you can work out what the average velocity at which they move in the direction of the applied field will be.

Imagine you have a bunch of bb's falling downward through a very thick lattice of sticks and plates with small holes in them - the bb's are constantly accelerated due to gravity, but there is an average speed at which they'll move through the lattice, because the faster they move due to gravity, the more strongly they'll be scattered.