# Electrical conduction

1. May 16, 2006

### erickalle

Hi all,

If we state that low (drift) speed free electrons are responsible for the transport of power than it follows we need big forces. Regardless the explanation of what happens inside a wire i.e. classical / semi / qm etc. as far as I’m concerned this wire is a box in which I push small marbles with speed and force.

For example say we generate a power of 2 Watt in a piece of wire area 2 mm2 and length 1 m. Normally electronic drift speed caused by voltage in good conductors is ~0.001 m/s, therefore it follows that this force needs to be ~2000 N !
The pressure becomes even more impressive!

One way we could check whether any such forces are generated is by sticking a couple of probes of a power supply horizontally in some liquid mercury. Since we have roughley as many +ve ions as free electrons there should be a drift of liquid towards the –ve probe. I’ve got a feeling there will be no shift of liquid.

Back to the drawing board. Is our original statement correct? Under the rules and regulations of this good forum I can state the following. When we determine the size of a capacitor we look at the important features such as plate area, length and material in between. We are not interested in the free electronic surface density of the conducting plates.

eric

2. May 16, 2006

### Staff: Mentor

Well, it isn't. Electrical energy is not the same as kinetic energy, and it can't be expressed that way.

3. May 16, 2006

### pervect

Staff Emeritus
then we'd be wrong :-(.

The power is actually being transferred by the fields, not the electrons.

Imagine a small stream that flows over a cliff. The stream does not have a high pressure, yet much power can be genreated at the waterfall where the stream flows over the cliff.

4. May 17, 2006

### erickalle

You will have to show me a website etc. with which we can get to grips.
Personally (as a start) I like Feynmann’s idea in his lectures vol 2 / 27-5 halfway page 27-8.
Sorry gotto go. I’m in a bit of a hurry.

5. May 22, 2006

### erickalle

You are correct…but
If you work out drift velocity and power you will find that your numbers and mine match, almost perfect.
So the problem is’nt so much the way in which power is transferred as the fact that the forces involved are so high. The reason for this is the enormous amount of charge we use to calculate the drift velocity.

6. Jun 16, 2006

### erickalle

In the above mentioned lecture Feynman points out that energy enters the wire with the Poynting vector at a right angle to the wire. This vector represents force x velocity per unit area, which means that instead of a length wise force, the force acts radially.
If Feynman’s theory is correct then most first chapters of solid state / condensed matter books on electrical conductivity need to be rewritten. Not only that, but I think that his theory is also of big value to people exploring super conductivity or any kind of electrical conductivity.

7. Jun 16, 2006

### ZapperZ

Staff Emeritus
Aren't you forgetting something?

The classical picture done in the solid state text WORKS, at least in producing the Drude model.

And think, where do you think the "B" or "H" came from for you to find the Poynting vector when you apply a static E field across a conducting wire? I also suggest you go beyond just the "first chapters of solid state/condensed matter books", because in Chap. 3 of Ashcroft and Mermin, you'll find a whole chapter on the failure of the free electron model. People working in superconductivity (like I did) do not deal with such classical model anymore. And that classical model doesn't produce something as obvious as a semiconductor.

Zz.

8. Jun 17, 2006

### erickalle

The question is can we apply a static electrical field across a conducting wire? We can across a capacitor but such a field would disappear extremely quickly across a wire. So perhaps we can apply a dynamic electric field to a wire, in which case Feynman’s theory makes more sense.

That chapter (and more comments in that book) are even more reasons for me to query electrical conduction theory’s.

9. Jun 17, 2006

### uart

Erik you have to remember that the power is not typically being lost in the wire itself (where the drift velocity is low) but in some type of load, a resistance for example.

The drift speed of electrons can vary dramatically depending on what material they are flowing in and at what current density. For example the drift speed you quote (0.001m/s = 0.1 cm/s) is actually very high for say copper wire as it would actually correspond to the slightly unreasonable current density of about 25 to 30 Amps per square mm. For the 2 mm^2 wire in your example at a current of 0.5 Amps the drift velocity in copper wire would be more like 0.001 cm/sec (0.00001 m/s).

The reason for the low drift velocity is simply that there are so many charge carriers taking part. In the case of copper for example it's around 1.7 x 10^23 per cm^2. So with the drift velocity of only 0.001cm/sec (0.00001 m/s) the total charge transport is 1.7E23 * 0.001 * 1.6E-19 which gives about 27 Amps per square cm, or about 0.54 amps total in the 2 mm^2 wire example.

In contrast electron velocity can be very fast indeed for materials of higher resistivity where the charge carrier density is much lower. Typical useful semiconductor charge carrier densities range from about 10^14 per cm^3 up to about 10^19 per cm^3. As you can see these are many orders of magnitude lower than that of copper and therefore drift velocities are correspondingly much higher. Drift velocities in silicon semiconductor devices can be tens, hundreds or even thousands of meters per second.

Take the case of say a minority carrier device like a MOSFET. If it is designed for reasonably high voltages then the carrier density must be limited to avoid breakdown, something like to 10^15 per cm^2 would not be unreasonable. But power devices like this also can carry a high current density, so a figure like 100 Amps/cm^2 is reasonable. If you take these two figures (100A/cm^2 and 10^15 carriers/cm^3) you end up with a mean drift velocity of 625000 cm/sec (6250 m/sec).

If you did wish to insist that the 2 Watts of you example was dissipated in the copper wire itself then given the values of 2 mm^2 wire and say 0.5 Amps then you would need a length of nearly a kilometer of wire to drop the required 4 volts to get 2W dissipation. In this length of wire there would be approximately 3 x 10^26 electrons taking part in that 2W dissipation. I hope this puts your example into perspective.

Last edited: Jun 17, 2006
10. Jun 17, 2006

### ZapperZ

Staff Emeritus
Have you never done any experiment?

When you connect a battery across a conductor, what exactly are you doing here? Are you saying that the potential difference across the conductor is varying, and thus, you can never get a static E-field across a conductor? Please explain the discrepancy between what you are saying, and what you are measuring.

But we ALL know that there's a limit to the free-electron model. However, this limit is NOT based on your faulty scenario, but rather based on invocation of quantum mechanical effects. You have not shown a single empirical evidence where deviation from the free electron model is DUE to your assertion. I can, however, point to deviation from the free electron model due to neglecting periodic boundary conditions, neglecting the electron-electron interactions, etc. None of these matches what you are claiming.

Zz.

11. Jun 18, 2006

### erickalle

I did’nt want to include things like loads because it only diverts from my main point. No matter what is connected in a circuit I can always work out pd and power loss across a wire.

This is true. This density should ideally not exceed 10. In an experiment I can get round that by connecting the power for a short time only.

I’m not sure where this value of 0.5 comes from. Current I = J * A = 25 * 2 = 50 amps. No doubt this big different result is going to have an effect on the rest of your calculations.

12. Jun 18, 2006

### erickalle

I’ve used batteries in the past and noticed that their electrical field are not all that static.
Say I connect a good length of wire across a battery so as not to drain it too quickly. There’s then a H field of same magnitude and direction around wire and battery. The changing E fields internal and external are in different directions and different in magnitude. Internally inside the battery, the Poynting vector points outwards at a right angle into space and externally across the wire it points inwards. (By the way this is not my theory!).
Another example. I can replace this battery with a big charged plate capacitor. Field theory tells us that energy stored is proportional to E^2 * volume. With other words all the energy is in between the plates, stored in empty space. Upon discharge, Poynting tells us that this energy flows outwards into space. It does not flow towards the plates it flows away radially into space. How do you think the connecting wire is going to receive this energy? Again Poynting tells us through space!
Of course now you are going to say that in case of eg a regulated power supply there will not be such loss in voltage / field. My answer to that is that we still have moving charges which could give rise to a dynamic E field although I cannot prove that. What’s rather crucial here is that there cannot be 2 different ways of transport. Either power comes through the wire length ways or not. In my op I stated that I felt a reason to doubt the length wise theory. As it happens Feynman and Poynting show me a different approach.
If those forces are directed radially they are spread out over a much lager area.

This post is placed here because I don’t want to be drawn into a discussion of what happens when we zoom into the wire but want to do the opposite. I want to see whether it is possible to first look at macroscopic field theory and see if that can solve my (possibly faulty) idea of force and velocity. If I’m wrong I’m sorry for corrupting impressionable young minds, if not then I can only hope a similar young mind takes this further, becomes famous and sends me a postcard from Sweden.

13. Jun 18, 2006

### ZapperZ

Staff Emeritus
Place a resistor in a series circuit with a battery. Now look at the current going through the resistor. If you see the current varying over a reasonable period of time (say 5 minutes), then I would like you to write this up carefully and send it in to, let's say, AJP to have it published, because you have discovered something utterly weird. After you've done that, we'll talk. Till then, you will understand that I will not believe what you have said here, because I can show you TONS of observation that contradicts what you have claimed.

But you are forgetting one thing: how do you account for your "scenario" with experimental evidence? Again, tell me how do you calculate the poynting vector in such a case? You require the existence of an E field, AND a B field. Now for a static E field, where do you think this B field came from? You'll see that you HAVE to make an explicit assumption of the direction of the current! Look at this direction! Now look at what happen if the charges are NOT moving along the length of the wire, but instead radially. Where do you think is the direction of the poynting vector NOW?

In Harrison's text on Solid State Physics, there is an exhaustive chapter on the classical transport problem in solids using the Boltzmann Transport equation using a distribution function. Such description describes ALL the transport scenario based on classical model. This includes electrical and heat transport, and even allows one to derive the Wiederman-Franz relation. So this is extremely well-verified. This model clearly contradicts what you have described here.

Zz.

14. Jun 19, 2006

### uart

That was just an example based on copper with a carrier density of 1.7x10^23 and a mean drift velocity of 0.001 cm/sec (0.00001 m/s).

But power is the product of voltage times current, if you only consider the wire and not the load then from where does the voltage component arise?

As I pointed out above (in the 0.5 Amp example), you need nearly 1000 meters of 2 mm^2 copper wire to dissipate 2Watts at 0.5 amps. That's a lot of interactions with a lot of atoms (more than 10^26 of them) to produce that net retarding force on the electrons that you are considering.

15. Jun 19, 2006

### erickalle

Voltage arises from the power supply. In a way this wire is the load. In my circuit wire and load are the same thing.

Ok if you are unhappy about my figures in the op lets have a look at yours. Your current density is J= I / A = 0.5 / 2 = 0.25 A/mm^2 (2.5E5 A/m^2) which is low but perfectly legal. Lets take resistivity of cupper at room temperature ρ=~1.7E-8 Ωm. Then field E = J * ρ = 2.5E5 * 1.7E-8 = 4.25E-3 V/m and U = E * L = 4.25E-3 * 1000 =4.25 Volts. Which is close enough if we work it out from U = P / I = 2 / 0.5 = 4 Volts (your ρ is a bit lower then mine). Your velocity is v=1E-5 m/s. Now net retarding force F = P / v = 2 / 1E-5 = 2E5 N. I know it’s boring but let’s work out the rest. Volume wire V= L * A = 1E3 * 2E-6 = 2E-3 m^3. Your Density D = 1.7E23 cm^-3 (=1.7E29 m^-3). Number of electrons (atoms) N = D * V = 1.7E29 * 2E-3 = 3.4E26 which tallies again with you number. Therefore force per electron Fe = F / N = 2E5 / 3.4E26 = 5.9E-22. Of course this force can also be worked out from Fe = E * e = 4.25E-3 * e = 6.8E-22. Again close enough.
The point I’m going to make is that force F must be applied in the same direction as the drift velocity ie length wise. In which case there must exist a pressure of Pr = F / A = 2E5 / 2E-6 = 1E11 N/m^2. To bring this number in prospective Young’s modulus of cupper = ~1.2E11 N/m^2. Of course in lab conditions we can increase your current density many times and why not melt this wire in a channel? As long as we are not boiling it away it’s perfectly ok. For a short period of time we really can now chase some amps through. We are now in a similar experiment as the mercury one in my op. Why is this metal not moving length ways?

16. Jun 19, 2006

### erickalle

For your very good Duracell products the same answer applies as the one of the regulated power supply.

I never said charges are moving radially. Energy is moving radially! For a diagram of the E and B fields and Poynting vector involved look at the lecture I mentioned before.

I’ve not got Harrison’s text but I’ll order a copy (if its affordable).

17. Jun 19, 2006

### ZapperZ

Staff Emeritus
So you are saying that in that setup, the current doesn't stay the same? Please publish this result of yours.

But why would this matter to the charge transport? The poynting vector simply tells you the direction of energy being transfered by the fields - the static E and the static B due to the current. Why is this even relevant to the charge transport in that conductor? You are nowhere close to doing any self-energy effects in here.

This is very puzzling.

Zz.

18. Jun 20, 2006

### erickalle

ZapperZ thanks for still talking to me.
I’m going to be busy for a couple of days but I’ll be back. By the way you better let me win this argument otherwise I’ll start a thread on airplanes and conveyor belts!
regards,
eric
Ps. perhaps by that time there will be some more idea's of other people.

19. Jun 23, 2006

### erickalle

Sorry for the delay but I’m on a 12 hours shift system and to reply properly I need some time to ponder.

At this stage I do not want to talk about charge. As shown I’ve got reason to doubt the theory which says that power is supplied length ways. If I can show that the power is supplied by an energy field entering the wire radially then as far as I’m concerned I can show a much improved picture regarding the huge forces I’ve calculated above. Clearly where as before the energy enters through a small surface area of 2 mm^2 it now enters (uarts example) through an area of A =~ 2 mm * 1E3m =~ 2 m^2 and the surface of entering is therefore ~1E6 times bigger. Also the direction of the force is radially inwards which is a far better idea, because it would explain why the wire is not moving.
But before I go on, are you content with the idea that energy is transported radially instead of axially (length wise)?

20. Jun 23, 2006

### ZapperZ

Staff Emeritus
Energy of WHAT?

Again, you are confusing the energy gained by the conducting electrons from the static E-field, versus the energy of the FIELD generated by the magnetic field DUE to the moving charges. You have mixed them up into the same thing.

Can you explain why the Drude model works?

Zz.