Electrons - when they move from Atom to Atom - where do they end up?

[mivey]

work in joules (J)
thus
current is proportional to work per volt-second. NOT energy per volt-second.

The change in energy equals the work and they have the same units. Does not change the fact that your units did not agree.

J = kg*m*m/s/s
V = kg*m*m/C/s/s

V = kg*m*m/C/s/s = J / C
so
C = J / V
which is what I had in the prior post.

Coulomb is the base unit for the magnitude of energy not joules. Joules and Volts are derived measurements of behaviour.

Coulomb is not an energy unit. Your units will not agree (see prior post).

 

[mivey]

The interaction is typically the most dense in the metal because that's where the field center is located. But it doesn't have to interact inside the metal or at least the interaction can be balanced within the metal incurring the same result. Sort of like the beads on an abacus you don't have to push on the center to make it move down the wire but it's still constrained to the wire, the center still rests in the wire, and the field density is greatest in the wire.

Consider the simple, steady-state DC case with negligible resistance in the wire (approximately a zero axial electric field component). The net radial electric field is zero inside the wire because the wire has become polarized at equilibrium. This polarization produces an internal electric field that balances the applied electric field from the battery.

The remaining electric field is radially directed between the surfaces of the two conductors and is where the traveling energy resides.

With a current in a resistive wire, we never reach equilibrium in the axial direction because we have a charge pump (battery) that keeps the charges circulating (current).

 

[pfalcon]

Not so. It's joules.

Coulomb is not an energy unit. Your units will not agree (see prior post).

No. You won't find a definition of Joule that doesn't refer to it as work. Work is a limited characteristic of energy referenced to units of time and distance. If you neglect the static forces of a charge then the remaining forces are proportional to work. Which is exactly what most current equations do - they neglect the static forces. Coulomb is the base unit. Joule is the coulomb in motion.

CV = J

(Energy)*(Work potential per Unit of Energy) = Work

Meaning, as you have, when you only deal with Work equations if follows since energy is proportional to work those equations can treat energy and work as though they were identical.

 

[pfalcon]

Consider the simple, steady-state DC case with negligible resistance in the wire (approximately a zero axial electric field component).

Negligible resistance in a wire is akin to a hockey puck on ice. It's axial component is independant of the resistance. Your equations are not going to work unless there is real resistance to create an equilibrium state. Otherwise any axial component will create acceleration.

The net radial electric field is zero inside the wire because the wire has become polarized at equilibrium. This polarization produces an internal electric field that balances the applied electric field from the battery.

Polarized?!
This is more again to pushing an object across a rough surface where friction negates the acceleration from a force. The friction balances the force to create a net zero and therefore constant velocity. The forces in the wire reach an equilibrium. But forces in equilibrium are not the same as without energy. That the terms in equilibrium are crossed out when solving most of your equations is not the same as their not existing.

The remaining electric field is radially directed between the surfaces of the two conductors and is where the traveling energy resides.

Which immediately fails the test of a capacitor since that requires only one conductor to create a current. There is no second conductor for the field to travel between. Think lightning bolt if you must. Leyden jar if you will. Electricity requires an energy source and an energy sink. It does NOT require two conductors. It does NOT require a return path. Those are artifacts of electrical generators and practical application of electricity. Important artifacts that you won't run your AC without - important for our usage of electrical energy not important to the transmission of electrical energy.

With a current in a resistive wire, we never reach equilibrium in the axial direction because we have a charge pump (battery) that keeps the charges circulating (current).

We always reach equilibrium in a resistive wire. The voltage (potential for work) that pushes the current is balanced by the resistance (frictional work) to create a constant current (velocity). Otherwise the current would go exponential. Or are you proposing that Newton's second law doesn't apply?

 

[ggunn]

We always reach equilibrium in a resistive wire. The voltage (potential for work) that pushes the current is balanced by the resistance (frictional work) to create a constant current (velocity).

I do not agree with current as velocity. Current is like the number of cars that pass a point on a highway per unit of time; it doesn't tell you how fast they are moving. If there are more lanes (lower resistance), they don't need to travel as fast. Or water in a hose; to get the same number of gallons per hour (current) through a smaller hose, the water must move faster.

 

[mivey]

No. You won't find a definition of Joule that doesn't refer to it as work. Work is a limited characteristic of energy referenced to units of time and distance. If you neglect the static forces of a charge then the remaining forces are proportional to work. Which is exactly what most current equations do - they neglect the static forces. Coulomb is the base unit. Joule is the coulomb in motion.

CV = J

(Energy)*(Work potential per Unit of Energy) = Work

Meaning, as you have, when you only deal with Work equations if follows since energy is proportional to work those equations can treat energy and work as though they were identical.

Can you provide some reference material? I'm all for exploring new things, but if it is outside standard accepted electrical theory I would like some background material for a frame of reference.

Thanks.

 

[mivey]

Negligible resistance in a wire is akin to a hockey puck on ice. It's axial component is independant of the resistance.

The axial component of the electric field is not independent of the resistance. It is directly related.

Your equations are not going to work unless there is real resistance to create an equilibrium state. Otherwise any axial component will create acceleration.

We've covered that. If there is an axial component, there is resistance and thus a voltage drop along the conductor. Basic circuit theory stuff.

Polarized?!

Yes. That is what happens when you apply an electric field across a metal conductor. Basic physics.

This is more again to pushing an object across a rough surface where friction negates the acceleration from a force. The friction balances the force to create a net zero and therefore constant velocity. The forces in the wire reach an equilibrium. But forces in equilibrium are not the same as without energy. That the terms in equilibrium are crossed out when solving most of your equations is not the same as their not existing.

The net electric field is zero. The field from the polarized metal offsets the applied field from the battery. Basic physics again.

Which immediately fails the test of a capacitor since that requires only one conductor to create a current. There is no second conductor for the field to travel between. Think lightning bolt if you must. Leyden jar if you will. Electricity requires an energy source and an energy sink. It does NOT require two conductors. It does NOT require a return path. Those are artifacts of electrical generators and practical application of electricity. Important artifacts that you won't run your AC without - important for our usage of electrical energy not important to the transmission of electrical energy.

Electrical energy transmission actually does not require any conductor...not even one. I've already covered that.

A capacitor requires the presence of both an electric field and magnetic field to store or release energy. Besides, please recall that we were discussing conductors and the energy traveling to the load.

We always reach equilibrium in a resistive wire.

No we do not. More basic physics. The current in a wire keeps the metal from reaching equilibrium.

The voltage (potential for work) that pushes the current is balanced by the resistance (frictional work) to create a constant current (velocity). Otherwise the current would go exponential. Or are you proposing that Newton's second law doesn't apply?

I said nothing of the sort. One of my points, and the point made in the paper I referenced, was that the axial component of the electric field pushed the current along against the resistance of the wire. The component of the Poynting Vector that is directed radially into the wire equals the wire losses. That was covered in detail in the paper.

 

[pfalcon]

I do not agree with current as velocity. Current is like the number of cars that pass a point on a highway per unit of time; it doesn't tell you how fast they are moving. If there are more lanes (lower resistance), they don't need to travel as fast. Or water in a hose; to get the same number of gallons per hour (current) through a smaller hose, the water must move faster.

Yes, ggunn, it's closer to a flow rate than a velocity. Adjust the discussion accordingly.

 

[pfalcon]

Can you provide some reference material? I'm all for exploring new things, but if it is outside standard accepted electrical theory I would like some background material for a frame of reference.

Thanks.

http://physics.nist.gov/cuu/Units/units.html
energy, work, quantity of heat joule J N?m m2?kg?s-2
electric charge, quantity of electricity coulomb C - s?A

According to NIST which I assume is a sufficiently reliable source for you, the base SI units for the Joule are Newton-Meters. That's force applied over a distance which is exactly the definition of Work aka Work energy, Energy of work.

When people use work equations they typically stop using the word work as it becomes inconvenient and repetitious. In work equations other forms of energy are cancelled out. Balanced work energy is typically cancelled out. That doesn't mean they cease to exist. It just means they have no impact on that specific equation. Net zero NOT equal zero.

 

[Besoeker]

No. You won't find a definition of Joule that doesn't refer to it as work. Work is a limited characteristic of energy referenced to units of time and distance. If you neglect the static forces of a charge then the remaining forces are proportional to work. Which is exactly what most current equations do - they neglect the static forces. Coulomb is the base unit. Joule is the coulomb in motion.

CV = J

(Energy)*(Work potential per Unit of Energy) = Work

Meaning, as you have, when you only deal with Work equations if follows since energy is proportional to work those equations can treat energy and work as though they were identical.

Hey ho...........
A coulomb is a measure of electric charge.
Not energy.

A joule is energy.
One volt times one amp times one second.

It's an impracticably small unit of energy for billing purposes so you pay for electrical energy in kWh as a rule.

 

[pfalcon]

Mivey, I'd include the post quotes but they're get disjointed.

The electrical charge ALWAYS has an axial component. In an ideal conductor it is net zero NOT equal zero. The axial component is independant of resistance. The axial component of Work energy is proportional to resistance. The field effects are still there. They don't disappear. They just have nothing to latch onto.

Further in a normal conductor - the axial component of work energy is balanced by the equal and opposite reactive resistive work energy. If it didn't balance then the current would go infinite. The axial component doesn't create the resistance - it performs work on the existing resistance.

My apologies but this may sound harsher than I mean Mivey: You constantly argue that the energy travels between conductors then wave the keyboard and announce it doesn't require any conductors to travel. That's contradictory. Which is it? You can't sustain the first argument and simply declare the second can occur without explaining how it bypasses all your other requirements for transmission.

 

[mivey]

According to NIST which I assume is a sufficiently reliable source for you, the base SI units for the Joule are Newton-Meters. That's force applied over a distance which is exactly the definition of Work aka Work energy, Energy of work.

Yeah NIST works fine for me. Work and energy have the same units. NIST agrees.

But don't lose focus that your contention is that Coulombs are the base unit for energy. The NIST site does not seem to agree with you so perhaps you can explain?

When people use work equations they typically stop using the word work as it becomes inconvenient and repetitious. In work equations other forms of energy are cancelled out. Balanced work energy is typically cancelled out. That doesn't mean they cease to exist. It just means they have no impact on that specific equation. Net zero NOT equal zero.

What does that have to do with Coulombs being the "base unit of energy"?

For the metal conductor with the net radial electric field at equilibrium, the net electric field is zero. The axial electric field does not reach equilibrium because the current loop prevents an equilibrium state. The current against the resistance creates losses. These losses are equal to the radial component of the Poynting Vector (PV). This inward-pointing component of the PV is the energy delivered to the conductor at that point, not the energy flowing to the load at the end. The axial component of the electric field is directly proportional to the energy delivered to the conductor and thus directly proportional to the conductor resistance.

 

[mivey]

The electrical charge ALWAYS has an axial component.

The electric field has an axial component along the wire for a normal wire. We have already covered that.

In an ideal conductor it is net zero NOT equal zero. The axial component is independant of resistance.

The axial component of the electric field is not independent of resistance but is directly proportional to it. If it were not so, we would have a conflict with Ohm's Law.

For the steady-state DC case where we reach equilibrium in the radial direction, the radial component of the applied electric field is offset by the electric field from the polarized metal and the net radial electric field is zero.

The axial component of Work energy is proportional to resistance. The field effects are still there. They don't disappear. They just have nothing to latch onto.

The axial component of the electric field is proportional to the resistance. The radial component of energy is proportional to the resistance. The axial component of energy is proportional to the down-line load.

Don't forget that the energy vector is the cross product of the electric field vector and magnetic field vector and thus the energy vector is perpendicular to both.

You constantly argue that the energy travels between conductors then wave the keyboard and announce it doesn't require any conductors to travel. That's contradictory. Which is it?

Energy to the load travels outside the conductor. Without the conductors, we do not have as much control over the energy direction. The conductors act like a waveguide for the energy.

You can't sustain the first argument and simply declare the second can occur without explaining how it bypasses all your other requirements for transmission.

I covered it earlier when I talked about antennas and waveguides.

The IEEE Press "Power Definitions and the Physical Mechanism of Power Flow" by Emanuel is primarily concerned with the in-depth analysis of metering electrical energy and the analysis of the energy flow. It has a very detailed coverage of the application of the Poynting Vector (PV) across many load and conductor types. It is an excellent text covering the details of how and where energy flows plus how we measure it. From the text:

The last result emphasizes the fact that the flow of electric energy toward the load takes place within the dielectric that surrounds the transmission line conductors. One may figure the conductors as a wave-guide for the electromagnetic wave. Equation (1.24) shows that the density of the energy increases as one nears the conductors. In the vicinity of a superconductor [the PV] is perfectly parallel to the conductor; however, in the vicinity of a lossy conductor the Poynting vector streamlines bend slightly toward the conductor due to a small component perpendicular to the conductor surface. This transversal component of [PV] transfers to the conductors the power that sustains the Joule and eddy-current losses dissipated in the conductor.

 

[pfalcon]

The electric field has an axial component along the wire for a normal wire. We have already covered that.

The axial component of the electric field is not independent of resistance but is directly proportional to it. If it were not so, we would have a conflict with Ohm's Law.

And there is a conflict with Ohm's law. In a super-conductor the resistance is zero but the current and voltage are finite values. Ohm's law FAILS in an ideal conductor.

This is just not sensible mathematics. Sensible mathematics involves neglecting a quantity when it turns out to be small - not neglecting it just because it is infinitely great and you do not want it!

Yet that's exactly what Ohm's law does in a superconductor. It predicts an infinitely great quantity in a superconductor and we ignore it. We ignore it because Ohm's law is like Newton's gravity. Unlike Einstein we're unlikely to see the conditions under which the equations fail. The axial component IS present in a superconductor without the presence of resistance. Therefore the sensible conclusion is NOT that resistance creates the field component but that the field acts upon resistance.

For the steady-state DC case where we reach equilibrium in the radial direction, the radial component of the applied electric field is offset by the electric field from the polarized metal and the net radial electric field is zero.

The voltage generated by the charge field is offset by the reactive force from the resistance to flow. CV ~ IR.

The axial component of the electric field is proportional to the resistance. The radial component of energy is proportional to the resistance. The axial component of energy is proportional to the down-line load.

1 No. 2 Almost. 3 Why?

1 The electric field is constant not proportional to anything but the charge.
2 The radial WORK ENERGY is proportional to the resistance.
3 Load = Resistance. Why repeat the first statement?

 

[pfalcon]

Yeah NIST works fine for me. Work and energy have the same units. NIST agrees. ...

Wow. Wish I'd read this reply from you before the other post. If you can't separate system energy from work energy then you're not gonna grasp this. So I see no point in going further with this thread.

It was fun, but I'm out. Thanks Mivey.

 

[Torxtimer]

Please don't stop you guys, I have learned new things "THEORY" from this thread since it was started.
Having gotten my start some 42 years ago playing with the "Elusive Electron" and getting so far ahead in the past couple of days has been a real learning experience for me and I am sure many others that have viewed it.

 

[ggunn]

3 Load = Resistance. Why repeat the first statement?

Load = 1/Resistance. No load = open circuit.

 

[mivey]

And there is a conflict with Ohm's law. In a super-conductor the resistance is zero but the current and voltage are finite values. Ohm's law FAILS in an ideal conductor.

Actually, the axial voltage and resistance both go to zero as the temperature drops. The current thus can flow without the voltage drop we have in the normal conductor.

Yet that's exactly what Ohm's law does in a superconductor. It predicts an infinitely great quantity in a superconductor and we ignore it.

No infinity, just that resistance and the axial voltage go to zero.

We ignore it because Ohm's law is like Newton's gravity. Unlike Einstein we're unlikely to see the conditions under which the equations fail. The axial component IS present in a superconductor without the presence of resistance.

The resistance and axial voltage go to zero.

Therefore the sensible conclusion is NOT that resistance creates the field component but that the field acts upon resistance.

Not sensible since that contradicts accepted physics.

The voltage generated by the charge field is offset by the reactive force from the resistance to flow. CV ~ IR.

Without some reference material to frame this line of thought, it is just gobbledygook.

The axial component of the electric field is proportional to the resistance.

1 No.

Yes. Also known as the voltage drop along the conductor (since we are considering just the resistive component). Common knowledge for almost everybody with electrical experience.

1 The electric field is constant not proportional to anything but the charge.

The voltage at the source is not the same as the voltage at the load because of the voltage drop along the conductor. The conductor impedance causes the drop.

The radial component of energy is proportional to the resistance

2 Almost.
2 The radial WORK ENERGY is proportional to the resistance.

Gobbledygook.

The axial component of energy is proportional to the down-line load.

3 Why?

Because the energy is being delivered to the load.

3 Load = Resistance. Why repeat the first statement?

The 1st was about the axial electric field (see previous info about the voltage drop). The 3rd was about the energy to the load. The energy to the load is axial in the conductor while the energy to the conductor is radially inwards towards the conductor (which is a small parasitic load along the path).

The load is a resistor also and once the energy gets there, it becomes radially (perpendicular) directed into the load. On the way to the load, the load energy is axial (parallel) to the conductor. Part of the traveling energy gets bent inwards towards the conductor as the conductor consumes energy through losses. See the IEEE book "Power Definitions and the Physical Mechanism of Power Flow" by Emanuel or some of the other prior references for in-depth discussions.

 

[mivey]

Wow. Wish I'd read this reply from you before the other post. If you can't separate system energy from work energy then you're not gonna grasp this.

If you want to discuss physics and electrical theories that are no main-stream, I would require reference material to grasp the framework.

So I see no point in going further with this thread.

I think we have about covered it also.

It was fun, but I'm out. Thanks Mivey.

I had fun as well but I'm sure there will be more fun to be had on other topics. Later.

 

[mivey]

Please don't stop you guys, I have learned new things "THEORY" from this thread since it was started.
Having gotten my start some 42 years ago playing with the "Elusive Electron" and getting so far ahead in the past couple of days has been a real learning experience for me and I am sure many others that have viewed it.

Any points of clarification you would like to discuss?