Electrical Conduction: Exploring Drift Speed and Force Requirements

[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

 

[russ_watters]

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.

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

 

[pervect]

Hi all,

If we state that low (drift) speed free electrons are responsible for the transport of power

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.

 

[erickalle]

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

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.

 

[erickalle]

then we'd be wrong :-(.

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

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.

 

[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.

 

[ZapperZ]

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.

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.

 

[erickalle]

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?

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.

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.

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

 

[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.

 

[ZapperZ]

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.

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.

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

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.

 

[erickalle]

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.

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.

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.

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.

For the 2 mm^2 wire in your example at a current of 0.5 Amps

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.

 

[erickalle]

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.[

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.

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.

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.

 

[ZapperZ]

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!).

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.

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.

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.

 

[uart]

I’m not sure where this value of 0.5 comes from

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).

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.

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.

 

[erickalle]

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?

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

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.

Ok if you are unhappy about my figures in the op let's 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. Let's 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?

 

[erickalle]

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.

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

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.

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.

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.

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

 

[ZapperZ]

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

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

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.

But why would this matter to the charge transport? The poynting vector simply tells you the direction of energy being transferred 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.

 

[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 ideas of other people.

 

[erickalle]

But why would this matter to the charge transport? The poynting vector simply tells you the direction of energy being transferred 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.
Zz.

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)?

 

[ZapperZ]

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)?

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.

 

[uart]

Eric you're really confused there. There is no mystery about this. Yes the total force on the electrons is huge but so is the total force on positive ions in the metal latice equally huge and of course opposite. That's why there is no net force.

 

[erickalle]

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.

At this stage you have to make a choice and tell me at which of the next sequence of events you want to stop me.
1 Energy is stored in the empty space in between the plates of a capacitor.
2 Upon discharge this energy is flowing radially outwards.
3 This energy is therefore not entering the plates axially.
4 The wire responsible for the discharge is therefore not receiving energy axially.
5 This wire is receiving the energy radially.

From here it’s not too difficult to imagine some circuitry involving several capacitors fed by a power supply to demonstrate a nearly steady dc current in this wire, generated from energy flowing entirely radially into this wire.

 

[erickalle]

Eric you're really confused there. There is no mystery about this. Yes the total force on the electrons is huge but so is the total force on positive ions in the metal latice equally huge and of course opposite. That's why there is no net force.

As I suggested before, increase your current (in a lab) more then 100x so that the wire melts. The drift velocity is then > 1 mm/s. Any +ve ion see as much –ve / +ve ions in front as behind so that there’s no net coulomb force. The force on each +ve ion due to the applied field is > 7E-20 N. The force of gravity on each +ve ion is >10,000x less. Why are the +ve ions not flowing to the –ve terminal?

 

[ZapperZ]

At this stage you have to make a choice and tell me at which of the next sequence of events you want to stop me.
1 Energy is stored in the empty space in between the plates of a capacitor.
2 Upon discharge this energy is flowing radially outwards.
3 This energy is therefore not entering the plates axially.
4 The wire responsible for the discharge is therefore not receiving energy axially.
5 This wire is receiving the energy radially.

From here it’s not too difficult to imagine some circuitry involving several capacitors fed by a power supply to demonstrate a nearly steady dc current in this wire, generated from energy flowing entirely radially into this wire.

You seem to want me to answer your question, but you never answered mine!

1. Show me, under the standard metallica case, why the Drude model works!

2. Energy of WHAT? Energy in the FIELDS, or energy TRANSFERED to the charges? If it is energy of the FIELDS, what fields? When you "discharge", are you shorting out a capacitor? How? How are you able to figure out where the "field" is when such a thing is going on because you have a TIME VARYING fields for E?

3. Why is this capacitor being discharged similar to charge transport in a conductor under a constant potential? Why are we even going to this capacitor analogy?

4. Do you now see that you have garbled several DIFFERENT things and confusing them to be the SAME thing, something that I've mentioned way back when?

The Sommerfeld-Drude model of charge transport in simple enough without needing one. So why don't you use your axially transported energy and derive Ohm's Law with it the way we can with the Drude model? I'd pay more attention to what you are claiming if you can do that.

Zz.

 

[ZapperZ]

As I suggested before, increase your current (in a lab) more then 100x so that the wire melts. The drift velocity is then > 1 mm/s. Any +ve ion see as much –ve / +ve ions in front as behind so that there’s no net coulomb force. The force on each +ve ion due to the applied field is > 7E-20 N. The force of gravity on each +ve ion is >10,000x less. Why are the +ve ions not flowing to the –ve terminal?

Er... I'm sorry to say this, but this is getting sillier by the minute.

If you have MOBILE ions, they WILL move. What do you think occurs in an electrolysis?

But you are forgetting that in a solid, you have bonds that causes the ions to be in a rigid lattice. Gravity plays zero part in this. If you want to know what causes these various types of bonding, you have to open a solid state/chemistry text.

Zz.

 

[Farsight]

Interesting thread.

 

[erickalle]

Er... I'm sorry to say this, but this is getting sillier by the minute.

If you have MOBILE ions, they WILL move. What do you think occurs in an electrolysis?

But you are forgetting that in a solid, you have bonds that causes the ions to be in a rigid lattice. Gravity plays zero part in this. If you want to know what causes these various types of bonding, you have to open a solid state/chemistry text.

Zz.

And you forget that I’m now talking about a liquid, not a solid. Why are the electrons allowed to move but the +ve ions have to stay put in a metallic liquid? Right back in my op I mentioned liquid mercury so as to get rid of all the solid bonds. This is the second time you misquoted me!

 

[ZapperZ]

And you forget that I’m now talking about a liquid, not a solid. Why are the electrons allowed to move but the +ve ions have to stay put in a metallic liquid? Right back in my op I mentioned liquid mercury so as to get rid of all the solid bonds. This is the second time you misquoted me!

And what makes you think the mercury ions don't move IF you can get it to remain electrically charged? How do you think we do electroplating?

However, under normal conditions, the electrons are MORE mobile than the ions. It means that if you apply any reasonable potential across mercury, the electrons are zipping in and out of the mercury so fast, the mercury ions, on average, NEVER become electrically charged! So these so-called "ions" are never ions in the first place!

But why are we talking about this? Since when is applying normal potential causes a wire to melt? And I still want to know how you would explain why the Drude model works so well, which clearly contradicts what you are suggesting in this thread. You have ignored that question several times.

Zz.

 

[erickalle]

And what makes you think the mercury ions don't move IF you can get it to remain electrically charged? How do you think we do electroplating?

Electroplating is a completely different process from the normal conduction current we are discussing here.

However, under normal conditions, the electrons are MORE mobile than the ions. It means that if you apply any reasonable potential across mercury, the electrons are zipping in and out of the mercury so fast, the mercury ions, on average, NEVER become electrically charged!

If that would be true the total mass of mercury would have an enormous net –ve charge! That statement contradicts the fact that macroscopically wires are neutral.

But why are we talking about this? Since when is applying normal potential causes a wire to melt?

If you would have followed the conversation between Uart and me then you would have seen that I’ve made a point of making this potential abnormally high so as to melt the wire so as to get rid of the solid bonds between atoms.

And I still want to know how you would explain why the Drude model works so well, which clearly contradicts what you are suggesting in this thread. You have ignored that question several times.

I said before that I don't want to zoom into the wire which I would be doing if we talk about Drude. Having said that at the moment we are slowly drifting that way!

 

[ZapperZ]

Electroplating is a completely different process from the normal conduction current we are discussing here.


If that would be true the total mass of mercury would have an enormous net –ve charge! That statement contradicts the fact that macroscopically wires are neutral.

How does what I explained causes this? All I said what that the electrons are highly mobile and are very efficient at moving. I didn't say there's any accumulation of charges of any kind! In fact, I even said that on average, the mercury ions "... NEVER become electrically charged..." How did you arrive as what you said above based on what I said? You'd think I would know enough about electrical conduction to not make that silly conclusion.

If you would have followed the conversation between Uart and me then you would have seen that I’ve made a point of making this potential abnormally high so as to melt the wire so as to get rid of the solid bonds between atoms.

I said before that I don't want to zoom into the wire which I would be doing if we talk about Drude. Having said that at the moment we are slowly drifting that way!

But you ARE! Your premises started from the erroneous assertion that you would have electrons in a SOLID CONDUCTOR gaining such a high speed that it can attain ridiculously high temperature! Would you like me to point out where you said such a thing, or did you forget? This is what I am countering with the Drude model, and the fact that in most application, a wire simply doesn't melt that easily. So I am attacking your starting premise, which if show to be false, makes the rest of your scenario utterly moot.

You may not believe it, but I happen to be working on a similar problem. I'm trying to investigate breakdown in copper and other metallic structure under very high field gradients, especially when there are "sharp point" or field enhancement regions. And we're working with RF fields up to 90 MV/m! The initiation of breakdown effects is thought of to be preceeded by the melting of a localized region of Cu due to either heating effects of the surface currents that these field enhancement regions, or due to the field-emitted current though through very small surface area and thus generating huge current density and Ohmic heating. But these things do not occur THAT easily, and it certainly does not occur in non-field enhanced region! It isn't THAT easily to melt copper under such a condition using charge transport in itself.

Zz.

 

[Gokul43201]

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!

Why the exclamation? The average force on a single electron due a small field of say 1000V/m is F = eE ~ 10^{-16} N. In a wire with 10^{20} electrons, that's a total force of about 10,000N from the field. Naturally, the net average force on the electrons is zero, since they are not accelerating. There's nothing bizarre about this.

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.

So did you actually do the experiment before going back to the drawing board?

 

[erickalle]

How does what I explained causes this? All I said what that the electrons are highly mobile and are very efficient at moving. I didn't say there's any accumulation of charges of any kind! In fact, I even said that on average, the mercury ions "... NEVER become electrically charged..." How did you arrive as what you said above based on what I said? You'd think I would know enough about electrical conduction to not make that silly conclusion.Zz.

If the mercury ions never become charged then we are left with the charge of the –ve electrons. So the total sum is: zero charge on the ion and 1 negative electronic charge per atom. This makes for a lot of –ve charge on the whole wire.

But you ARE! Your premises started from the erroneous assertion that you would have electrons in a SOLID CONDUCTOR gaining such a high speed that it can attain ridiculously high temperature! Would you like me to point out where you said such a thing, or did you forget? This is what I am countering with the Drude model, and the fact that in most application, a wire simply doesn't melt that easily. So I am attacking your starting premise, which if show to be false, makes the rest of your scenario utterly moot.

If you got objections to me melting the wire with a high current I can always melt it in another way and then send a high current through. My point here was that the ions are not moving length wise even in the liquid state.

You may not believe it, but I happen to be working on a similar problem. I'm trying to investigate breakdown in copper and other metallic structure under very high field gradients, especially when there are "sharp point" or field enhancement regions. And we're working with RF fields up to 90 MV/m! The initiation of breakdown effects is thought of to be preceeded by the melting of a localized region of Cu due to either heating effects of the surface currents that these field enhancement regions, or due to the field-emitted current though through very small surface area and thus generating huge current density and Ohmic heating. But these things do not occur THAT easily, and it certainly does not occur in non-field enhanced region! It isn't THAT easily to melt copper under such a condition using charge transport in itself.

You’re very privileged.

 

[erickalle]

Why the exclamation? The average force on a single electron due a small field of say 1000V/m is F = eE ~ 10^{-16} N. In a wire with 10^{20} electrons, that's a total force of about 10,000N from the field. Naturally, the net average force on the electrons is zero, since they are not accelerating. There's nothing bizarre about this.

Hi Gokul.
1000V/m is a huge field for the conductor we are considering!
What do you think is providing a counter force of 10,000 N per atom so that there’s no net acceleration? What is the total pressure exerted by this counter force?

edit: 10,000 N in total. Not per atom

So did you actually do the experiment before going back to the drawing board?

I do this experiment each time I connect a mercury switch in a circuit. If the liquid was moving, it and the current would soon start to oscillate.

 

[ZapperZ]

If the mercury ions never become charged then we are left with the charge of the –ve electrons. So the total sum is: zero charge on the ion and 1 negative electronic charge per atom. This makes for a lot of –ve charge on the whole wire.

Er... say what?

A "liquid metal" is still a metal. Stick your electrodes in liquid mercury and tell me if you have a charge liquid after 5 minutes.

If you got objections to me melting the wire with a high current I can always melt it in another way and then send a high current through. My point here was that the ions are not moving length wise even in the liquid state.

And my question is, why should they? The whole lump of metallic liquid is neutral. A liquid metal is STILL a metal!

You’re very privileged.

And I don't just talk the talk without having to get to DO it also, which means I don't just make it up as I go along.

Zz.

 

[erickalle]

Stick your electrodes in liquid mercury and tell me if you have a charge liquid after 5 minutes.

It’s probably me but I don’t understand this sentence.

You have stated (#28) that on average the mercury ions are neutral. That still leaves a non neutral net charge per atom because of the –ve electron. Are you saying that the electrons are on average neutral as well?