# Speed of current

1. Jun 2, 2005

### primal schemer

Hi,

How fast does electrical current travel?? I`m essentially asking, if I had a wire in a circuit with a point a distance X away from a voltage source, and turned on the curcuit (i.e. applied voltage), how long would it take for X to know that the circuit was switched on??

Thanks,

PS

2. Jun 2, 2005

### Dr.Brain

If it is vacuum , it would take the velocity of light.
Otherwise depends on the medium /air around.

3. Jun 2, 2005

### primal schemer

If current is the flow of electrons, and electrons have mass, how can it travel at the speed of light???

What am I missing??

4. Jun 2, 2005

### robphy

You need to distinguish the "speed of the electrons in an electric current" [which is surprisingly very slow!] from the "speed of the electromagnetic interaction" [the speed of light].

5. Jun 2, 2005

### Staff: Mentor

The electrons themselves move extremely slowly; it's the signal that moves at an appreciable fraction of light speed.

6. Jun 2, 2005

### Meir Achuz

It's like falling dominoes. Each electron moves slowly, but the effect in a copper wire moves at the speed of light. This was proven by Gustav Kirchoff in 1857.

7. Jun 2, 2005

### primal schemer

What is this 'effect'?? (or 'signal' as Doc Al called it)

Is it that one electron moves, then the electron immediately beside it realises that the first electron moves and moves itself, which is turn sets off the electron beside it to move etc?? So it is the time it takes for the electrons a great distance away to 'realise' that the first one has moved that travels at the speed of light??

I always assumed that it was the flow of electrons that powered things in a circuit. Is this incorrect so??

PS

8. Jun 2, 2005

### HallsofIvy

Staff Emeritus
If you are talking about direct current (wire attached to a battery) then it is the flow of the electrons that is the current. However, in alternating current, electrons just move back and forth a tiny distance. The "signal" is the wave caused by such motion.

9. Jun 2, 2005

### pervect

Staff Emeritus
If you have a simple straight piece of wire, it would actually be very difficult to determine how fast the signal propagated down it. (You asked "what is the signal" - the signal arrives when, in your original quesiton, point "X" "knows" that the circuit is switched on.

The problem is that the signal transmission depends not only on the wire, but the entire room that the wire is in, the ground return wire, and a bunch of other factors. The high frequencies are especially distorted - and those are just the ones you need to get an accurate measurement of the "leading edge" of the signal.

But there is a practical way around this. When you use a "coaxial cable" aka coax (this is what's used to wire cable TV outlets), the circuit does not depend on the geometry of the room, only on the geometry of the cable - which includes the return signal path as part of the "wire".

When you measure the speed of transmission of a signal in a coaxial cable of the sort commonly used is slower than light, about 70% slower. It is slower because of the dielectric material (the insulation) used in the cable.

This may help illustrate why the fields that propagate the signal are distinct from the electrons that carry the current. The material that the wire is made of doesn't affect the speed much if at all, but the material that separates the inner wire from the outer jacket, the insulation, does.

An even more extreme example along the same lines is the waveguide, which is essentially a coaxial cable with the center conductor removed. A waveguide actually does a perfectly fine job of transmitting high frequency signals, though it cannot carry a direct current (DC) or low frequences below the "cutoff" of the waveguides.

Perhaps this is too much detail, but I thought it might illuminate your question a bit.

10. Jun 2, 2005

### Gokul43201

Staff Emeritus
No, it is not this way. The electrons move "virtually" independent of each other. By throwing a switch on a power source, you are changing the chrge distribution on this source. Associated with (or resulting from) a charge distribution is an electrostatic field. Changing the charge distribution changes the electric field. The leading edge of this field is what travels at a speed clode to c. An electron in a electric field feels a force proportional to the strength of the field. When a circuit is powered on, the field changes from zero to some non-zero number, virtually instantaneously throughout the circuit. Every electron in the circuit feels the force from this new field "almost" simultaneously, starting them all moving in the direction of the force.

Can time travel at the speed of light ? It is the feeling of the influence (known as the field) of the applied change in charge distribution that travels at nearly c.

It is the flow of electrons that dissipates power in a circuit.

Last edited: Jun 2, 2005
11. Jun 3, 2005

### primal schemer

12. Jun 5, 2005

### erickalle

1 No, it is not this way. The electrons move "virtually" independent of each other.

2 It is the flow of electrons that dissipates power in a circuit.

Now consider an established steady dc current, the electric field is not changing anymore. Electrons "feel" the field from the power source but also "feel" a back EMF by the moving electrons in front. If electrons would only be responding to the field they would accelerate continuously.

I do agree with point 2, but it must be true that a certain amount of pushing is going on which is responsible for the transport of power in a dc circuit. In my view this pushing is can be compared with the movement of fluids down a pipe, where atoms deal only with their immediate (and not so immediate) neighbour atoms.

eric

13. Jun 7, 2005

### Gokul43201

Staff Emeritus
No sir ! This isn't right. The primary scattering mechanism is electron-phonon scattering, or scattering of the eletrons by the ions in the conductor, and not by other electrons. Yes there is a contribution to the resistance from electron-electron scattering, but this is orders of magnitude smaller.

Here's a simple way to semi-understand why this effect is small. Imagine electron 1 feeling an "back EMF" from electron 2, (which is in front of it) causing it to slow down. Then, by Newton's Third Law, electron 2 should feel a "forward EMF" making it speed up. As a result, one could argue that the average speed of the electrons is barely affected, and this is both predicted by more sophisticated calculations as well as verified by experiment.

So, what prevents the electrons from accelerating continuously is mostly collisions with the atoms/ions in the conductor.

As explained before, the pushing is provided by the electric field set up by the battery, not by other electrons. Electron-electron interactions are known to be very small compared to the electric field response. This is why theories based on non-interacting electrons (such as the early Drude-Sommerfeld theory, all the way up to Band theory) work very well for things like electrical conductivity in regular metals.

No, the two can not be compared. They are different, and there is no reason that they should be analogous. The first is a system of nearly non-interacting particles, while the second is a system of interacting particles (hence you have viscous flow).

At best, you could think of it as similar to an inviscid fluid in a vertical pipe. The liquid flows down the pipe because all the molecules respond to an external field - gravity.

Last edited: Jun 7, 2005
14. Jun 7, 2005

### erickalle

Ok, when I wrote this post I sort of wanted some kind of good response (which I got). By the way this comparison between fluids and currents is often used.

Whats your view (or anyone's) of the next situation regarding electron-electron interaction:

say 10 capacitors are connected in series. I now charge each one by 10 Volts, so that we end up with a voltage between first and last one of 100 Volts.
I could have charged all at once with a 100 Volts source but I want to make the next point: all 10 charges must be interacting or "feeling" each other so as to make the total 100 Volts. If we now connect a resistor across the 2 ends ie across the 100 volts, a current starts to flow. In my view it must be true that all 10 charges are still (although slowly losing energy) interacting. Of course you can make the point that the electrons are still responding to a field but this field exist because of the electrons.

My next step is to extend this principle to millions of tiny capacitors with tiny charges, these charges being single electrons with an typical energy of the voltages encountered in a (atomic distance) potential difference of a conductor. The field again being produced by charges of a real big capacitor.

I'm sure you follow my drift, although I gotto go :zzz: back up at 04:55

eric

15. Jun 7, 2005

### brewnog

Can I just say thanks to Gokul for his last explanation? It's posts like that which convince me he deserves those green badges!

16. Jun 8, 2005

### Gokul43201

Staff Emeritus
I do not deny that a useful comparison between fluid flow and current flow exists at the macroscopic level. I often use circuit analysis to solve hydraulics problems. It is at the microscopic level that the analogy is no longer valid.

When using an analogy, it is useful to ask yourself "what drives the analogy" ?

In this case, the reason is that the fluid flow equation $P = QZ$ (where P is pressure, Q is the flow rate and Z is the impedance of the pipe) is of the same form as Ohm's Law. In both cases, the work done by the pump/battery is given by the product of the "same" terms. The analogy works at the macroscopic level. But a knowledge of the extents of feasibility of an analogy are crucial to its application.

At the microscopic level, there are several deviations between the two cases. For instance, the impedance of a pipe goes inversely like the third or fourth power of the radius of the pipe (depending on the Reynolds Number for the flow), unlike the resistance of a conductor, which goes inversely with the second power of the radius. These deviations result from differences at the microscopic level.

17. Jun 8, 2005

### ZapperZ

Staff Emeritus
You need to make sure you understand the terminology involved here. As Gokul has pointed out, when you measure a quantity called "resistance" or "resistivity", the PREDOMINANT mechanism that dictates such quantity is the electron-phonon interaction. We can verify this very easily because if you apply the same thing to a bunch of free electrons in vacuum, you will get virtually almost no resistivity, certainly many orders of magnitude lower than what you would get in a typical conductor.

Secondly, you also need to understand under what set of assumptions are the standard formulation of popular laws were obtained. Things such as Ohm's Law, basic circuit theories, etc. were obtained under one of two models: the degenerate free-electron gas, or the Landau's Fermi Liquid theory. Either one will produce practically all the standard charge transport equations that are popularly used. However, let us examine what is involved in those two models.

The free-electron gas is nothing more than an ideal gas! The ONLY interactions involved here is when electrons bump into each other and the ions elastically. That's all. There are no other potential involved, not even electron-electron coulomb repulsion. They are all treated as if they are point particles without any electric fields. Using this statistics, you end up with the Drude model. You can literally derive Ohm's Law using this Drude model. It is rather amazing to see how many descriptions that are popularly used in electrical transport that are actually based on such primitive set of assumptions.

Now, what if the electrons do "feel" an interaction with one another? What if they now start sensing the coulomb repulsion among themselves? This has now become a many-body problem. Landau showed that if such interactions are "weak", one can renormalize all these interactions and lump them into the particle's mass and get back the "free" electron case - the Fermi Liquid theory. So what he did was to change a single many-body problem (difficult to solve) into a many one-body problem (easier to solve). You now end up with particles with no quite the same mass (effective mass) as their free particle cousins. Since you again obtain a system consisting of "free" particles, you also can get all the results that the Drude model got. So a system of "weakly-interacting" electrons can give the identical result as free-electron gas.

However, once the electron-electron interactions goes beyond "weak", then all hell breaks loose. An important part in the study of condensed matter physics is the strongly-correlated electron system (an area that I used to study extensively). This is where the assumption of Fermi Liquid theory breaks down, that these electrons cannot be accurately described by renormalizing all their electron-electron interactons and converting them to one-body system. A number of exotic phenomena can occur once the electrons strongly interact with each other. You do not get Ohm's Law, you sometime get "fractionalization", and it may even violate several known transport laws that we know of.

The point I'm trying to get across is that when you say "electron-electron" interaction, you have to be VERY careful with that terminology because at the microscopic level, the term has a specific meaning especially in condensed matter physics. In practically all cases, the macroscopic scenario that you observe are described WITHOUT the use of such electron-electron interactions. Such interactions are so weak that they do not play a role in the description of such a system. When they DO play a role, I can tell you that you will no longer recognize the behavior of that system. It will not have the same characteristics that you know and love.

Zz.

Last edited: Jun 8, 2005
18. Jun 8, 2005

### pervect

Staff Emeritus
A good URL that does this is

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html

I thought I mentioned this link earlier in the thread, but I don't see it. Anyway, this link might clear up a number of questions that people have asked.

In this model the electrons are zipping around inside the wire at a high rate of speed, with only a very tiny amount of their velocity being due to the "drift" velocity imposed by the electric field.

The average drift velocity component is given by multiplying the acceleration due to the applied field by the mean time between collisions. One can basically assume that the average velocity of an electron after a collsion is "reset to zero" - the actual velocity after any specific collision may be anything, but the average over a large number of collisions is zero.

19. Jun 9, 2005

### erickalle

I've just read Zapper's answer and I'm very impressed. Its too late (again) to go deep into all his stuff (I will at a later stage) but a big thanks for all the time spend. I'm sure you are doing many people a big favour as it is simply too time consuming and too far above most heads. I will pick up this thread since no doubt more questions will pop up. Perhaps in a new thread since it doesn't really belong in this one.

eric