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

1. 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.

2. Originally Posted by pfalcon
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?

Originally Posted by pfalcon
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.

3. Originally Posted by pfalcon
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.

Originally Posted by pfalcon
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.

Originally Posted by pfalcon
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.

Originally Posted by pfalcon
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.

Originally Posted by pfalcon
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.

4. Originally Posted by mivey
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.
Originally Posted by Paul Dirac
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.

Originally Posted by mivey
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.

Originally Posted by mivey
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?

5. Originally Posted by mivey
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.

6. Junior Member
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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.

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Originally Posted by pfalcon
3 Load = Resistance. Why repeat the first statement?
Load = 1/Resistance. No load = open circuit.

8. Originally Posted by pfalcon
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.

Originally Posted by pfalcon
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.

Originally Posted by pfalcon
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.

Originally Posted by pfalcon
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.

Originally Posted by pfalcon
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.

Originally Posted by pfalcon
Originally Posted by mivey
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.

Originally Posted by pfalcon
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.

Originally Posted by pfalcon
Originally Posted by mivey
The radial component of energy is proportional to the resistance
2 Almost.
2 The radial WORK ENERGY is proportional to the resistance.
Gobbledygook.

Originally Posted by pfalcon
Originally Posted by mivey
The axial component of energy is proportional to the down-line load.
3 Why?
Because the energy is being delivered to the load.

Originally Posted by pfalcon
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.

9. Originally Posted by pfalcon
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.

Originally Posted by pfalcon
So I see no point in going further with this thread.
I think we have about covered it also.

Originally Posted by pfalcon
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.

10. Originally Posted by 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.
Any points of clarification you would like to discuss?