"Veritasium" (YT) - "The Big Misconception About Electricity" ?

[TimFox]

Mapping (imaging) electron density has been done for a long time using x-ray diffraction (crystallography) and different methods of electron microscopy.
Strictly, this is a map of the probability spatial distribution of electrons in, say, a large molecule.

[aetherist]

Will you please explain how "new electricity" leads to the conclusion that a capacitor discharging into a transmission line leads to a step-like voltage?

The main thing here is that old (electron) electricity can't explain the discharge, ie the half voltage for double the time.

The second thing here is that it might be possible to come up with lots of theories about what electricity is or isn’t, & for most of these to satisfactorily explain the discharge (ie to tick that box). I could come up with other theories that tick that box, but would they tick all of the boxes. My new (electon) electricity i think ticks all of the boxes (so far).

But old (electron) electricity duznt tick the capacitor discharge box, as far as i can tell. But u or someone else might indeed be able to explain a way that drifting electrons tick the box. If so then i would not be able to use that box to falsify old (electron) electricity. But that would then leave me with all of the other boxes that falsify old (electron) electricity. And i only need one box. One strike & old (electron) electricity is out.

Now to answer your question. New (electon) electricity says that in a charged capacitor the negative plate is full of (covered by)(saturated with) electons propagating in every direction, as is a short wire connected to the negative plate. On the short wire half of the electons are going one way & half are going the other way. When electons get to the end of the short wire they do a u-turn (in reality they go straight ahead as usual)(it is the surface of the wire that does a u-turn). When the short wire is connected to a new non-charged long wire the electons on the short wire that are already heading towards the long wire will instead of doing a u-turn will enter onto the long wire. Some of the electons going the "wrong way" on the short wire will have to go all the way to the end of that circuit, ie to the furthest end of the negative plate, & do a u-turn, & head back & enter the long wire.

So, the time taken for the last electon to leave the capacitor & enter the long wire is double the average time. The average time is the time taken for an electon to travel from the farthest point on the negative plate to the nearest point on the long wire.

That is the simple version of the electon discharge from the negative plate of a capacitor.
The positive plate is different. I think that it has no electons. It has an induced positive charge, ie due to the repulsion of electrons due to the negative charge on the negative plate (ie due to the negatively charged electons on the negative plate). I need to have a think about how the positive plate might discharge, & how that would affect current in the long wire. The discharge involves the flow of electrons on the surface, & this will be very slow, ie much slower than the speed of light. This will produce i think a long slow weak discharge, in addition to the almost instantaneous electon discharge mentioned above.  I need to have a think.

I might be overplaying the importance of the short wire. If the capacitor holds say 100 times the number of electons on the short wire then the length of the short wire might not make much difference to the time of the (main) discharge, ie the geometry of the capacitor itself would be paramount.

Here is a link to what Harry Ricker says. Harry has written lots of good articles.
https://beyondmainstream.org/the-wakefield-experiments-background-and-motivation/

[aetherist]

That Demjanov paper was not retracted by Demjanov, it was removed by the Journal. Hence it was not retracted, & they lied. So, u are supporting censorship, & a lie. And cheering it on.

Yet I was able to download it, and read it, so not censored. They merely offer their apologies for letting it slip through, without which we would not be discussing it. The journal did aetheriests (or whatever they are called) a service, you complain about the taking away of a part of something given in error, where's the lie?

I see your rational core bubbling up and causing these confusing surface vacillations in your logic. You sought evidence, it's your choice and I am merely observing an evolution towards imaginative rationalism. I don't want or expect you to abandon your ideas, but that does not preclude your journey to evidence-based thinking you clearly seek. Far be it from me to judge, but bravo.

Leapfrogging electons were the first electons that i thought of in Dec 2021. Shortly after, i realized that simple hugging must be the answer, no hopping. If the screw-thread X does not show the extra delay due to the simple extra distance then that would falsify my electons. And i don’t see how roo-tons could come to the rescue. Roo-tons would fail just as Beaty's silly leapfrogging em radiation must fail to rescue old (electron) electricity from the elephant in the room.

Ok but leapfrogging is due to someone else you said. If the "X" fails to support your electon theory then on one-strike rules you will be forced to invoke your roo-tons, you said. Why? If roo-tons would also likely fail as you now claim then that is effectively a prehumous admission of failure for your revised theory. You set up a false dilemma, by denying any possibility to revise your theory, by speculatively revising your theory into a form that would also fail. But that doesn't prevent you from seizing an opportunity. You have previously used the device which I expect you would again invoke upon failure of the experiment (which your rational core might have determined is quite likely): "I don’t agree that roo-tons explain this null result. But perhaps they could. Blah blah de blah ...". You have actively sought out this situation which has doomed you to fail then come to your own transparently ridiculous rescue, yet you chose a forum where you know you can get called up on this issue after all these years. Welcome.

Forum members around here seem to be unaware that it is almost impossible to prove something, especially a subatomic something. But it is of course possible to disprove something. Anyhow, it is easy for me to say that there is no proof for electrons & photons, because there will always be good alternative theories that fit the facts. Many scientists don’t believe in electrons & photons. Or, putting it another way, if u designed a page full of yes/no questions re electrons (or photons), the chances are that no 2 scientists in the whole world would tick the same boxes exactly.

It is impossible to disprove anything. It is merely a quicker path to the same false certainty one gets from 'proof' through absence of evidence - tipping the balance of probability quicker. There is no certainty, only belief. A composite of conscious hope and subconscious fear. Any belief I have in electrons and photons is therefore optional. I am not against alternative theories. Your comment about the quiz is probably true.

Forum members seem to be unaware that Einstein contradicted Einstein. His ideas changed right up to his death. Einstein would disagree with much of modern (supposedly Einsteinian) science. And modern science disagrees with much of Einstein. Einstein would be thrilled by my electons.

That may all be true, or at least not wildly untrue. But I was thrilled by your roo-tons. Does any of this particularly matter?

We have facts & we have hot air.  I came here & i have tried to point the way to replace hot air with facts.

And it's working. Refer to your rational core.

Can any members here use old electricity to explain the traces for the AlphaPhoenix X pt1 & (later) pt2?

Once again, this explanation is in this thread way back - the answer is yes. The Maxwell simulation (or even all of them) replicates the features seen in the measurement I think better than expected given the problems with 'X' technique. The only thing I found 'interesting' is the "subtle lift", in both. I didn't quite go to town on the scope screenshot to the degree you have, but I did pore over it for a time not to treat it as some kind of smorgasbord of  Dunning-Krugeresque intrigue but because I use scopes and know what to look for. You are ignoring the fact pointed out in one of my first replies to you that the result of the measurement matches the Maxwellian simulator's output, confirming the theory for that particular case, which is what you question, resulting in the answer "yes" which is a simple word with a stable meaning and unlikely to be confusing unlike this unnecessarily long sentence which you have no problem understanding. Ask your rational core, it asked the question.

I am still not happy with lumped element TL models. And i admit that they can replicate the initial 0.2 V that AlphaPhoenix (Brian) got in his white trace for V across his bulb.

But i should have made it clear that i was referring to his green trace for the voltage across the resistor near his positive terminal.
And re his X pt2 i should have made it clear that here i was referring to his (missing) X pt1 trace for the voltage across the resistor near his negative terminal.
Brian did not show us that trace, but he said that it was different to the green trace (they should be identical i think)(according to old (electron) electricity), & he said it was so different that his head nearly melted, & he said that he would show us that trace when he got around to showing us his X pt2 (but still no new update re if he has done X pt2 or ever will)(its coming up to 2 months since he did X pt1).

I reckon that i can explain the green trace, & the missing trace (after he divulges it). But old (electron) electricity will not be able to.

Actually i will try to finish my new (electon) electricity explanation for the green trace today, & i will put it on this thread.

When someone does the screw-thread X, will old (electron) electricity explain that?

Yes. But you already suspect it might - that's why you are here and asking the question.

My new (electon) electricity might explain (we will see).

It won't. You will sidestep it as described above, which you know full well because you have it planned.
But that's not the point. Nor are your reasons for being here, really.
Given that it is impossible to disprove anything, what if despite all your pushing and tests which (say) leave your theory in tatters, it turns out to be correct in large part ~100 years from now? We just didn't test it right. All this would undoubtedly have happened, leaving a mark on history weirder than Tesla's, but how could it in any way affect the validity of a theory years from now?

Can u or Howardlong or someone around here to the test?
Howardlong has a 20 GHz scope & might need only  say 3 ft of threaded rod to get a half decent measurement.
A 100 MHz scope can see down to say 10 ns which is 10 ft at the speed of light, so it might need say 50 ft of threaded rod (costing say $50) to detect the extra delay, & better still say 100 ft of rod ($100) to get a half decent accuracy of measurement.

U could repeat the X after painting the rods, to get a double dose of delay, but painting would ruin the rods.
Better to simply paint with oil, which can be washed off later.

I am willing to bet that my new (electon) electricity wins. Loser pays for the rods.
But if u are correct that old (electron) electricity can just as easily explain the delay then there would be little point in doing the X.
But how could old (electron) electricity explain?

[adx]

Will you please explain how "new electricity" leads to the conclusion that a capacitor discharging into a transmission line leads to a step-like voltage?

The main thing here is that old (electron) electricity can't explain the discharge, ie the half voltage for double the time.

I've already explained a page back how the test is of a transmission line not a capacitor, the source material calls it a capacitor but that is wrong. The test is not of a capacitor. Is there any way you could accept it's not a capacitor even if the source calls it one?

Your descriptions of photon-like things reflecting back and forth is barking up the right tree, and is well-known in conventional electricity and is a point made many times in this thread prior to your arrival. You have conflated electrons with photons, which might or might not have merit, I am currently not interested which one it is.

Your thoughts on experimental philosophy are just barking, and although I'm sure some people would think the same of me, I had better steer away from this too.

[TimFox]

In the current political environment, I have grown sick and tired of complaints that scientists change their mind when new evidence becomes available.
Einstein lived a long life, and his thoughts evolved.  His personal life might not qualify him for sainthood, but that is a different question not relevant here.
Rigidity of thought and refusal to change ones mind regardless of evidence, is common amongst ideologues.

[SandyCox]

I looked at the "Electronics World" paper by Ivor Catt and applied old-school transmission-line theory.

The setup
An 18 m long 75 Ohm transmission line. The reed switch is located at z=0m and the open end of the transmission line is at z=-18m. The transmission line is pre-charged to 8V. The reed switch is open-ended at t=0s. I chose a propagation speed of c=2.55e8 m/s.

What transmission line theory tells us

• The voltage wave is the sum of a wave travelling in the positive z direction (indicated by v+) and a wave travelling in the negative z direction (indicated by v-).
• The reflection coefficient at the open end is equal to 1.  The reflection coefficient at the reed switch is 0.
• Prior to t=0s, the voltage across the entire length (z=-18m to z=0m) of the transmission line is equal to 8V. There is no current in the transmission line. We can use this information to calculate v+ and v- for t<0s. It turns out that both are constant and equal to 4V.
• Closing the reed switch at t=0s initiates the transient. At z=0, v+ enters the 75 Ohm resistor, where it is terminated. v- is reflected from the open end of the transmission line and becomes part of v+

I attach figures for comparison with the results in Catt's paper. You do the comparison. A video showing the waves is available at



The error in Catt's paper is that he views the transmission line as a capacitor, which it most certainly is not.

A more detailed explain and the mathematical derivations will follow.

There is nothing strange about this setup. I remember doing experiments like this when I was an undergraduate student and using transmission line theory to explain the measurements.

Note that Catt's paper was published in the April 2013 issue.

[SandyCox]

Will you please explain how "new electricity" leads to the conclusion that a capacitor discharging into a transmission line leads to a step-like voltage?

The main thing here is that old (electron) electricity can't explain the discharge, ie the half voltage for double the time.

The second thing here is that it might be possible to come up with lots of theories about what electricity is or isn’t, & for most of these to satisfactorily explain the discharge (ie to tick that box). I could come up with other theories that tick that box, but would they tick all of the boxes. My new (electon) electricity i think ticks all of the boxes (so far).

But old (electron) electricity duznt tick the capacitor discharge box, as far as i can tell. But u or someone else might indeed be able to explain a way that drifting electrons tick the box. If so then i would not be able to use that box to falsify old (electron) electricity. But that would then leave me with all of the other boxes that falsify old (electron) electricity. And i only need one box. One strike & old (electron) electricity is out.

Now to answer your question. New (electon) electricity says that in a charged capacitor the negative plate is full of (covered by)(saturated with) electons propagating in every direction, as is a short wire connected to the negative plate. On the short wire half of the electons are going one way & half are going the other way. When electons get to the end of the short wire they do a u-turn (in reality they go straight ahead as usual)(it is the surface of the wire that does a u-turn). When the short wire is connected to a new non-charged long wire the electons on the short wire that are already heading towards the long wire will instead of doing a u-turn will enter onto the long wire. Some of the electons going the "wrong way" on the short wire will have to go all the way to the end of that circuit, ie to the furthest end of the negative plate, & do a u-turn, & head back & enter the long wire.

So, the time taken for the last electon to leave the capacitor & enter the long wire is double the average time. The average time is the time taken for an electon to travel from the farthest point on the negative plate to the nearest point on the long wire.

That is the simple version of the electon discharge from the negative plate of a capacitor.
The positive plate is different. I think that it has no electons. It has an induced positive charge, ie due to the repulsion of electrons due to the negative charge on the negative plate (ie due to the negatively charged electons on the negative plate). I need to have a think about how the positive plate might discharge, & how that would affect current in the long wire. The discharge involves the flow of electrons on the surface, & this will be very slow, ie much slower than the speed of light. This will produce i think a long slow weak discharge, in addition to the almost instantaneous electon discharge mentioned above.  I need to have a think.

I might be overplaying the importance of the short wire. If the capacitor holds say 100 times the number of electons on the short wire then the length of the short wire might not make much difference to the time of the (main) discharge, ie the geometry of the capacitor itself would be paramount.

Here is a link to what Harry Ricker says. Harry has written lots of good articles.
https://beyondmainstream.org/the-wakefield-experiments-background-and-motivation/

Can new electricity answer the following question? What is the equation for the voltage across the capacitor, as a function of time, as it discharges through the resistor?

[SiliconWizard]

In the end, the real underlying misconception is that electricity, as we usually define it, is not energy.

[adx]

In the end, the real underlying misconception is that electricity, as we usually define it, is not energy.

Yes, electrons (with the r) become a mental crutch for something weirder.

[adx]

Mapping (imaging) electron density has been done for a long time using x-ray diffraction (crystallography) and different methods of electron microscopy.
Strictly, this is a map of the probability spatial distribution of electrons in, say, a large molecule.

"Yes but" x-ray crystallography is more of a reconstruction, from a periodic structure and guesses of phase. AFM is more direct but also harder to argue it's "a picture" because it is done by feel even if it looks much the same.

An ordinary camera photo is a map of the probability distribution of photons over a known integration time, complete with shot noise. The lens resolves and combines the phase onto the spatial array. It's hardly more intuitive, but is the sense of sight, so comes with instinctive understanding (clouds, motion blur etc).

As far as I can tell, electron cloud densitometry produces an image directly and really is worthy of the usual meaning of "photo". The language is what struck me as useful - being able to point to a picture and say "this is the electron cloud".

[aetherist]

I looked at the "Electronics World" paper by Ivor Catt and applied old-school transmission-line theory.
The setup
An 18 m long 75 Ohm transmission line. The reed switch is located at z=0m and the open end of the transmission line is at z=-18m. The transmission line is pre-charged to 8V. The reed switch is open-ended at t=0s. I chose a propagation speed of c=2.55e8 m/s.

What transmission line theory tells us

• The voltage wave is the sum of a wave travelling in the positive z direction (indicated by v+) and a wave travelling in the negative z direction (indicated by v-).
• The reflection coefficient at the open end is equal to 1.  The reflection coefficient at the reed switch is 0.
• Prior to t=0s, the voltage across the entire length (z=-18m to z=0m) of the transmission line is equal to 8V. There is no current in the transmission line. We can use this information to calculate v+ and v- for t<0s. It turns out that both are constant and equal to 4V.
• Closing the reed switch at t=0s initiates the transient. At z=0, v+ enters the 75 Ohm resistor, where it is terminated. v- is reflected from the open end of the transmission line and becomes part of v+

I attach figures for comparison with the results in Catt's paper. You do the comparison. A video showing the waves is available at



The error in Catt's paper is that he views the transmission line as a capacitor, which it most certainly is not.

A more detailed explain and the mathematical derivations will follow.

There is nothing strange about this setup. I remember doing experiments like this when I was an undergraduate student and using transmission line theory to explain the measurements.

Note that Catt's paper was published in the April 2013 issue.

(1) I don’t see how u can split the electricity into 2 half currents, ie 2 half voltages.  I can see that on one side of the switch we have +4V & on the other side of the switch we have -4V. But when the switch is closed we immediately have 8V.
Looking simply at drifting electrons producing a wave, there will be an 8V wave going right (to the terminating resistor), this will be a depletion wave, ie that wire (which includes the outer sheath of the coax) is electron rich, & the conduction electrons will begin to spread out.
And, there will be an 8V wave going left (along the core wire of the 18m long coax), this will be an enrichment wave, ie the core wire is electron poor, & the electrons will begin flow into it & begin to bunch up.
If the core wire was neutral, ie with 00V, then in that case there would be a 4V wave going right & a 4V wave going left. But it aint neutral, it has +4V.

(2) Nextly i could explain why it is an impossibility for an electron wave to reflect at a dead end of a wire. Or, putting it another way, why such a reflexion would not add to the gradual addition of electrons into that wire, ie it would not add to the gradual bunching up of electrons in that wire. But not today, i might explain later.

(3) Here is my main objection. U chose the propagation speed of light, but in a part of your explanation u invoke an infinite speed of light for one of your  4V waves. U have this wave starting to wave at t=0 s, at the end of the coax, which is 18m from the switch.

(4) I noticed a few things, as an interesting aside, not necessarily a criticism of your explanation. And u might address some of these things in your detailed explanation later. I noticed that u did not mention drifting electrons (& what parts they played), surface charge (ie surface electrons)(& surface charge distribution), Poynting, & whether electrical energy is in or on or near the wires.

(5) Re a capacitor not being a transmission line. Perhaps so, in a way. But a transmission line is a capacitor. Perhaps it depends on whether ends are open or shorted.  Perhaps it depends on the kind of source producing the electricity (ie the charge)(eg a lead acid battery). Anyhow i don’t see how arguing about that stuff would help us today.

[aetherist]

Can new electricity answer the following question? What is the equation for the voltage across the capacitor, as a function of time, as it discharges through the resistor?

I don’t know what the old (electron) electricity equation(s) is for discharge of a capacitor. But the new (electon) electricity equation(s) would be almost identical, except that it would have to show the correct steady half voltage for the correct double the distance, ie for double the time (at least it would for the case of zero reflexions).

And it would need an additional equation for the additional voltage from electons leaving the length of the wire from the capacitor to the resistor. This extra voltage would be for a doubled time, ie electons have to go the wrong way along that wire & later return along that wire, hence a doubled distance & a doubled time. For a giant capacitor & a short or thin wire this voltage might be insignificant.

And it would need an additional equation for the very small extra voltage happening for a very long time due to surface electrons gradually slowly entering the positive plate. But this voltage might be insignificant.

[aetherist]

I just noticed today. Erik Margan did a 2m long coax version of Wakefield's 18m X. And Margan confirmed Wakefield's half voltage for double time. But, Margan erred in his own explanation (page 11). He said that photons had to go up & back in his 2m test piece.  No. The photons (my electons actually) came from his 12.8m intermediate coax. Yes, they had to propagate up & back in the 2m test coax, & hence he got his desired trace for the pulse, but his pulse came from his negative wire, ie the coax sheath of the 12.8m coax, not from the positive core wire of his 2m test piece. Hence i think that the sign of his voltage for his pulse might be wrong too (perhaps he fudged his graphs)(perhaps non-intentionally)(i might have a think about that some other day)(my brain hurts).

Anyhow, his mistake shows that u have to follow the electons. Electons live on the negative wire-plate-terminal. Margan (& everybody else but me) think that photons (or whatever) live on both plates, positive plate & negative plate. Or, they think that photons live in the space tween plates, & that the plates themselves dont matter much. No. The energy is on the surface of the negative plate-wire-terminal, in the electons (& in their radiation).

U will notice that both Wakefield & Margan both connected the negative terminal of their lead acid battery to the outer sheath of their coax. They thought that this was a non-critical detail, outer sheath, inner core, who cares. But, i care, it makes (or can make) lots of difference (depending on the problem in question). U have to follow the electon. Its not just a matter of drawing silly Poynting lines on a bit of paper.

http://www.ivorcatt.co.uk/x37p.htm
Erik Margan repeated Wakefield's X.
http://www.ivorcatt.co.uk/x726.pdf

[adx]

From an abandoned post a few days ago: The voltage on a capacitor you might use (ie, a real one you can buy) rises linearly when a constant current (coulombs per second) is injected into it, and falls when that charge is removed. The rate at which the voltage rises is predicted by the capacitance value, it can vary for non-quality capacitors (a well known set of problems) but is generally stable and almost exact for good ones. Not 100% out. There isn't a "charging time" independent of what I just described.

So there is no stair step change in voltage. There is no steady half voltage. There is no distance. There is no double the time. There are no "reflexions". This is all complete nonsense.

There also no polarity effects except electrolytic capacitors which have been formed to a particular polarity - many can be reformed (carefully) and used in reverse. There are no differences between positive and negative beyond the sign.

There is no "very small extra voltage happening for a very long time" in a vacuum capacitor and the small amount that occurs in usual capacitors is due to dielectric absorption, ie the insulator taking a 'set'. That goes away if the insulating material is removed.

[SandyCox]

(1) I don’t see how u can split the electricity into 2 half currents, ie 2 half voltages.  I can see that on one side of the switch we have +4V & on the other side of the switch we have -4V. But when the switch is closed we immediately have 8V.
Looking simply at drifting electrons producing a wave, there will be an 8V wave going right (to the terminating resistor), this will be a depletion wave, ie that wire (which includes the outer sheath of the coax) is electron rich, & the conduction electrons will begin to spread out.
And, there will be an 8V wave going left (along the core wire of the 18m long coax), this will be an enrichment wave, ie the core wire is electron poor, & the electrons will begin flow into it & begin to bunch up.
If the core wire was neutral, ie with 00V, then in that case there would be a 4V wave going right & a 4V wave going left. But it aint neutral, it has +4V.

(2) Nextly i could explain why it is an impossibility for an electron wave to reflect at a dead end of a wire. Or, putting it another way, why such a reflexion would not add to the gradual addition of electrons into that wire, ie it would not add to the gradual bunching up of electrons in that wire. But not today, i might explain later.

(3) Here is my main objection. U chose the propagation speed of light, but in a part of your explanation u invoke an infinite speed of light for one of your  4V waves. U have this wave starting to wave at t=0 s, at the end of the coax, which is 18m from the switch.

(4) I noticed a few things, as an interesting aside, not necessarily a criticism of your explanation. And u might address some of these things in your detailed explanation later. I noticed that u did not mention drifting electrons (& what parts they played), surface charge (ie surface electrons)(& surface charge distribution), Poynting, & whether electrical energy is in or on or near the wires.

(5) Re a capacitor not being a transmission line. Perhaps so, in a way. But a transmission line is a capacitor. Perhaps it depends on whether ends are open or shorted.  Perhaps it depends on the kind of source producing the electricity (ie the charge)(eg a lead acid battery). Anyhow i don’t see how arguing about that stuff would help us today.

I will attempt to address your misconceptions in the note I am writing.

In the meantime, let me set a challenge: Let's change the termination resistor in Catt's paper from 75Ohm to 47Ohm. How, according to "new electricity", will the measured pulses look if we do that?
 

[adx]

<big snip>

Can any members here use old electricity to explain the traces for the AlphaPhoenix X pt1 & (later) pt2?

Once again, this explanation is in this thread way back - the answer is yes. The Maxwell simulation (or even all of them) replicates the features seen in the measurement I think better than expected given the problems with 'X' technique. The only thing I found 'interesting' is the "subtle lift", in both. I didn't quite go to town on the scope screenshot to the degree you have, but I did pore over it for a time not to treat it as some kind of smorgasbord of  Dunning-Krugeresque intrigue but because I use scopes and know what to look for. You are ignoring the fact pointed out in one of my first replies to you that the result of the measurement matches the Maxwellian simulator's output, confirming the theory for that particular case, which is what you question, resulting in the answer "yes" which is a simple word with a stable meaning and unlikely to be confusing unlike this unnecessarily long sentence which you have no problem understanding. Ask your rational core, it asked the question.

I am still not happy with lumped element TL models. And i admit that they can replicate the initial 0.2 V that AlphaPhoenix (Brian) got in his white trace for V across his bulb.

As I posted in my first reply on page 42, lumped TL models don't replicate the 0.2V, they predict higher, because they are an incomplete subset of conventional theory not intended for antennas. You are right to not like them in this particular application - they are not intended to be accurate for the job.

What does match experiment quite consistently including shape of the pulses, is the collection of field solver simulations based on Maxwell's theory. You ran through them in your second post, and they show conventional electricity theory matching measurement for the white trace.

But i should have made it clear that i was referring to his green trace for the voltage across the resistor near his positive terminal.

<snipperoo>

Yes, that is what I meant too. I can see why you have a problem with it now, and needed more detail...

If the scope were truly isolated (or ground lifted, depending on where EMC caps go) then the green trace should rise sharply more like the yellow.

and

I explained some of the problems with AlphaPhoenix's result many pages back, one of the main ones which distorts the send waveform I think is common mode coupling. I explained what I think it should look like, if it is measured with a better technique. Others did too, and went into quite some detail.

At 7:27 in the video is a diagram of the setup. The probe "reference GND" is the ground clip of the scope. This is tying one side of the pulse generator to Earth, loosely via extension cords and perhaps an inverter from the cars (described in discussions here at the time). The green probe, which is on the other side of the resistor, can thus not see the step directly from the step generator, because it is shorted to ground at the send end (by the ground clip). In essence it can only see voltage due to current getting around the circuit the long way, and a slow change of the GND voltage (which we can't directly see, because there is no probe measuring the voltage between this scope's GND and Earth under the desk).

This is not the way it's meant to be, but surprisingly the experiment still works. It's not necessarily an error if the person doing the test knows that taking this shortcut will still work. Again, I agree the green trace is "wrong", and this does represent the current in that resistor, and hence the current sent into that leg of the 'apparatus'. The other leg should be taking the balance, so it should be seeing nearly all the initial pulse missing from the green side (because that is shorted to ground).

From the clean white trace I can infer that the differential send current is probably fairly rectangular. But subtract the green trace and add the generator step, and that's going to make for a pretty messy voltage on the far side of the unprobed resistor, possibly best not to think about because it is guaranteed to confuse.

The situation would be the same but inverted traces (voltages) if the polarity of the generator is changed - other than that there is no difference and I likely would not test to confirm if I were doing the experiment.

BTW all this isn't so much from theory, as from experience. It just helps explain what is seen. The result is in the white trace, and matches simulation as you already know and I can now see you never really had a problem with. I rarely think about theory when doing engineering stuff, but I sometimes calculate things, and sometimes put it in a circuit simulator if I have really turned my brain inside out. By "Trevor's theorem" I try not to think through tricky situations to arrive at an answer, especially if it is about something that is inverted a number of times - he says you're most likely to get that out by one so would be better off flipping a coin, so save yourself the bother and guess, test, and then swap it round if it's wrong (sort of thing). The important thing is that anyone can be wrong at any time so don't put too much trust in thoughts. Or scopes. In either case trying to bulldoze through a problem with your mind is asking for trouble. It's not about intelligence, but experience in the biz.

Also small apology that I didn't say "subtle lift" originally, or if I did I edited that to "some sort of frequency dependent tilt". This is the white trace before the first reflection arrives. I'm not very interested in why, just noticed it (it appears in the simulations too).

[aetherist]

From an abandoned post a few days ago: The voltage on a capacitor you might use (ie, a real one you can buy) rises linearly when a constant current (coulombs per second) is injected into it, and falls when that charge is removed. The rate at which the voltage rises is predicted by the capacitance value, it can vary for non-quality capacitors (a well known set of problems) but is generally stable and almost exact for good ones. Not 100% out. There isn't a "charging time" independent of what I just described.

So there is no stair step change in voltage. There is no steady half voltage. There is no distance. There is no double the time. There are no "reflexions". This is all complete nonsense.

There also no polarity effects except electrolytic capacitors which have been formed to a particular polarity - many can be reformed (carefully) and used in reverse. There are no differences between positive and negative beyond the sign.

There is no "very small extra voltage happening for a very long time" in a vacuum capacitor and the small amount that occurs in usual capacitors is due to dielectric absorption, ie the insulator taking a 'set'. That goes away if the insulating material is removed.

Margan explains the stepped charging of a capacitor on page 14 of his paper. I think i am happy with his explanation of steps.
Erik Margan repeated Wakefield's X.
http://www.ivorcatt.co.uk/x726.pdf

However, Margan makes a mistake on page 15.  He says that the battery continues to supply photons (ie energy) after the capacitor is full.
As the line is filled, the battery supplies the current (less after each reflection at V1), until several µs later the current has dropped practically to zero, and the voltage is practically equal to the battery voltage. But the important question to ask is this: Did the battery stop supplying energy to the circuit?
The correct answer is: no! Photons cannot stand still! The battery continues to supply the energy to the system, but the system now reflects all the energy back to the battery.
The effective current is zero, so no power is dissipated, but the energy continues to flow back and forth throughout the system. Another point to make is the following: if we put back the 50 k resistor as H VB and charge again the cable, we will not see any stairsteps. This is because the impedance mismatch is very high, 50 to 50 k , so the steps are less than 9 mV, too H H small to see (at the input sensitivity of 2V division). Because of the much smaller Î steps a much larger number of reflections is needed to charge up the line, leading to the classical capacitor charging equation e . It is important to…..

I don’t agree with Margan. I reckon that once the system is fully saturated with electons then the negative lead plate of the negative terminal of the lead acid battery can't feed any more electons out of the battery fluid onto the lead plate (or onto the lead strap or onto the lead terminal). The chemical process stops. Margan reckons in effect that the chemical process continues, but in effect he reckons that electons are both coming out of the fluid & going into the fluid. No.

U mentioned that i reckoned that the discharge of a capacitor must have a very weak additional long term discharge current due to the redistribution of the induced surface charge (electrons) on the positive plate.
And u said that i said that i reckoned that there will be a corresponding very weak additional long term charge current, ie a mirror image of the discharge.
Yes & No.
No i never said that.
And yes, i do believe that there is a mirror image effect during charging.
If the speed of surface electrons is say c/10,000 then their charge/discharge will be very weak compared to the c/1 speed of electons. But i am not sure how far the surface electrons need to move. If they move from the surface of the wire or plate to just under the surface then that distance might be only 1.0 nm. Or they might have to move 1.0 nm along the surface. I don’t think that any electrons have to go all of the way to the battery. Still thinking.

[aetherist]

I will attempt to address your misconceptions in the note I am writing.
In the meantime, let me set a challenge: Let's change the termination resistor in Catt's paper from 75 Ohm to 47 Ohm. How, according to "new electricity", will the measured pulses look if we do that?

There will be stepped reflexions of current & voltage. I have never worked on that kind of stuff.
But i see that Wakefield already has some traces for 75 Ohm coax with a 40 Ohm termination.
Shown on page 42 & 43 of Forrest Bishop's paper re Reforming Electromagnetic Units…
http://www.naturalphilosophy.org//pdf//abstracts/abstracts_6554.pdf


One thing i can explain, that i think no-one else could explain.
See how the bottom trace jumps up a few V too high, & then falls down to the correct 4V.
That jump is due to the 40 Ohm resistor being saturated with my electons. When the switch is closed electons in that short wire near the switch already heading for the switch will instead of doing their usual u-turn at the open switch they will propagate through the closed switch & onto the core wire of the coax, followed by electons from the resistor. The short wire was saturated with electons too, but the resistor holds a lot more electons per m length than the wire. Hence the brief spike of over-voltage.
U can see the same spike of over-voltage in some of the other traces, but there it is a negative spike of under-voltage i suppose u could call it.

But i have to have a think about what happens to electons inside resistors. Are they annihilated. Do they looz energy. Do they convert to infrared photons.
Anyhow, if there are lots of voids or porous bits or interface surfaces then there must be lots of electons on thems surfaces, & the electons would be doing lots of u-turns.
And they would be jostling lots of surface electrons, which would jostle atoms, & produce heating & resistance & would we know produce a voltage drop.
Its the voltage drop that has me worried. I am glad that SandyCox did not ask me re how exactly do electons produce a voltage drop across a resistor. Still thinking.

[bsfeechannel]

Shown on page 42 & 43 of Forrest Bishop's paper re Reforming Electromagnetic Units…
http://www.naturalphilosophy.org//pdf//abstracts/abstracts_6554.pdf

https://naturalphilosophy.org/about/

The John Chappell Natural Philosophy Society (CNPS) provides an open forum for the study, debate, and presentation of serious scientific ideas, theories, philosophies, and experiments that are not commonly accepted in mainstream science.

Yeah. Right.

[adx]

You misquote me, emphasis mine:

...
U mentioned that i reckoned that the discharge of a capacitor must have a very weak additional long term discharge current due to the redistribution of the induced surface charge (electrons) on the positive plate.
And u said that i said that i reckoned that there will be a corresponding very weak additional long term charge current, ie a mirror image of the discharge.
Yes & No.
No i never said that.
And yes, i do believe that there is a mirror image effect during charging.
If the speed of surface electrons is say c/10,000 then their charge/discharge will be very weak compared to the c/1 speed of electons. But i am not sure how far the surface electrons need to move. If they move from the surface of the wire or plate to just under the surface then that distance might be only 1.0 nm. Or they might have to move 1.0 nm along the surface. I don’t think that any electrons have to go all of the way to the battery. Still thinking.

You said (in full for context, again emphasis mine):

Can new electricity answer the following question? What is the equation for the voltage across the capacitor, as a function of time, as it discharges through the resistor?

I don’t know what the old (electron) electricity equation(s) is for discharge of a capacitor. But the new (electon) electricity equation(s) would be almost identical, except that it would have to show the correct steady half voltage for the correct double the distance, ie for double the time (at least it would for the case of zero reflexions).

And it would need an additional equation for the additional voltage from electons leaving the length of the wire from the capacitor to the resistor. This extra voltage would be for a doubled time, ie electons have to go the wrong way along that wire & later return along that wire, hence a doubled distance & a doubled time. For a giant capacitor & a short or thin wire this voltage might be insignificant.

And it would need an additional equation for the very small extra voltage happening for a very long time due to surface electrons gradually slowly entering the positive plate. But this voltage might be insignificant.

I then said (also in full for context, now emphasising):

From an abandoned post a few days ago: The voltage on a capacitor you might use (ie, a real one you can buy) rises linearly when a constant current (coulombs per second) is injected into it, and falls when that charge is removed. The rate at which the voltage rises is predicted by the capacitance value, it can vary for non-quality capacitors (a well known set of problems) but is generally stable and almost exact for good ones. Not 100% out. There isn't a "charging time" independent of what I just described.

So there is no stair step change in voltage. There is no steady half voltage. There is no distance. There is no double the time. There are no "reflexions". This is all complete nonsense.

There also no polarity effects except electrolytic capacitors which have been formed to a particular polarity - many can be reformed (carefully) and used in reverse. There are no differences between positive and negative beyond the sign.

There is no "very small extra voltage happening for a very long time" in a vacuum capacitor and the small amount that occurs in usual capacitors is due to dielectric absorption, ie the insulator taking a 'set'. That goes away if the insulating material is removed.

You'll see I didn't say you reckoned anything. I merely quoted your words to show you how it is totally wrong in the context you used it. Your thoughts are irrelevant to the facts in this instance (where you are wrong). I didn't comment on what you think, only your wrong claims, using some of your exact words for clarity. I took no mental leap of assumption (like I did with your roo-tons, claiming that they were part of a planned deception scheme on your part - is that what you're upset with?).

You introduced the concept of a slow charge effect. We were both talking about a long-term charge retention and discharge mechanism, where the capacitor maintains a small voltage for longer than expected while discharging.

You are now starting to believe I have said things I didn't say. It seems to me you might be both externally and internally grasping at straws to ignore your "rational core". How can I know how you think if I am merely under your skin? To use your own device, I'm not saying I am your rational core, but what if I were?

Of course if you want to define capacitors to be transmission lines in either practical or theoretical effect (or both), then by your new definition then of course you go right ahead and show stair step changes in voltage, distance and that whole first line of nonsense. But I (or is it you?) think you should use the term "capacitor (transmission line)" to avoid ambiguity over your definition.

[aetherist]

You misquote me, emphasis mine:

...
U mentioned that i reckoned that the discharge of a capacitor must have a very weak additional long term discharge current due to the redistribution of the induced surface charge (electrons) on the positive plate.
And u said that i said that i reckoned that there will be a corresponding very weak additional long term charge current, ie a mirror image of the discharge.
Yes & No.
No i never said that.
And yes, i do believe that there is a mirror image effect during charging.
If the speed of surface electrons is say c/10,000 then their charge/discharge will be very weak compared to the c/1 speed of electons. But i am not sure how far the surface electrons need to move. If they move from the surface of the wire or plate to just under the surface then that distance might be only 1.0 nm. Or they might have to move 1.0 nm along the surface. I don’t think that any electrons have to go all of the way to the battery. Still thinking.

You said (in full for context, again emphasis mine):

Can new electricity answer the following question? What is the equation for the voltage across the capacitor, as a function of time, as it discharges through the resistor?

I don’t know what the old (electron) electricity equation(s) is for discharge of a capacitor. But the new (electon) electricity equation(s) would be almost identical, except that it would have to show the correct steady half voltage for the correct double the distance, ie for double the time (at least it would for the case of zero reflexions).

And it would need an additional equation for the additional voltage from electons leaving the length of the wire from the capacitor to the resistor. This extra voltage would be for a doubled time, ie electons have to go the wrong way along that wire & later return along that wire, hence a doubled distance & a doubled time. For a giant capacitor & a short or thin wire this voltage might be insignificant.

And it would need an additional equation for the very small extra voltage happening for a very long time due to surface electrons gradually slowly entering the positive plate. But this voltage might be insignificant.

I then said (also in full for context, now emphasising):

From an abandoned post a few days ago: The voltage on a capacitor you might use (ie, a real one you can buy) rises linearly when a constant current (coulombs per second) is injected into it, and falls when that charge is removed. The rate at which the voltage rises is predicted by the capacitance value, it can vary for non-quality capacitors (a well known set of problems) but is generally stable and almost exact for good ones. Not 100% out. There isn't a "charging time" independent of what I just described.

So there is no stair step change in voltage. There is no steady half voltage. There is no distance. There is no double the time. There are no "reflexions". This is all complete nonsense.

There also no polarity effects except electrolytic capacitors which have been formed to a particular polarity - many can be reformed (carefully) and used in reverse. There are no differences between positive and negative beyond the sign.

There is no "very small extra voltage happening for a very long time" in a vacuum capacitor and the small amount that occurs in usual capacitors is due to dielectric absorption, ie the insulator taking a 'set'. That goes away if the insulating material is removed.

You'll see I didn't say you reckoned anything. I merely quoted your words to show you how it is totally wrong in the context you used it. Your thoughts are irrelevant to the facts in this instance (where you are wrong). I didn't comment on what you think, only your wrong claims, using some of your exact words for clarity. I took no mental leap of assumption (like I did with your roo-tons, claiming that they were part of a planned deception scheme on your part - is that what you're upset with?).

You introduced the concept of a slow charge effect. We were both talking about a long-term charge retention and discharge mechanism, where the capacitor maintains a small voltage for longer than expected while discharging.

You are now starting to believe I have said things I didn't say. It seems to me you might be both externally and internally grasping at straws to ignore your "rational core". How can I know how you think if I am merely under your skin? To use your own device, I'm not saying I am your rational core, but what if I were?

Of course if you want to define capacitors to be transmission lines in either practical or theoretical effect (or both), then by your new definition then of course you go right ahead and show stair step changes in voltage, distance and that whole first line of nonsense. But I (or is it you?) think you should use the term "capacitor (transmission line)" to avoid ambiguity over your definition.

Aha, i see the problem. In red -- when i said surface electrons slowly entering the positive plate, this entering was a part of the overall system discharging, not charging.
The positive plate has positive charge, hence it makes a half of the capacitance. And it loozes all of that when electrons enter onto that plate & eventually snuff out the positive charge.
And i havent given it much thort but charging of the system would have a mirror image process i think.

[aetherist]

<big snip>

Can any members here use old electricity to explain the traces for the AlphaPhoenix X pt1 & (later) pt2?

Once again, this explanation is in this thread way back - the answer is yes. The Maxwell simulation (or even all of them) replicates the features seen in the measurement I think better than expected given the problems with 'X' technique. The only thing I found 'interesting' is the "subtle lift", in both. I didn't quite go to town on the scope screenshot to the degree you have, but I did pore over it for a time not to treat it as some kind of smorgasbord of  Dunning-Krugeresque intrigue but because I use scopes and know what to look for. You are ignoring the fact pointed out in one of my first replies to you that the result of the measurement matches the Maxwellian simulator's output, confirming the theory for that particular case, which is what you question, resulting in the answer "yes" which is a simple word with a stable meaning and unlikely to be confusing unlike this unnecessarily long sentence which you have no problem understanding. Ask your rational core, it asked the question.

I am still not happy with lumped element TL models. And i admit that they can replicate the initial 0.2 V that AlphaPhoenix (Brian) got in his white trace for V across his bulb.

As I posted in my first reply on page 42, lumped TL models don't replicate the 0.2V, they predict higher, because they are an incomplete subset of conventional theory not intended for antennas. You are right to not like them in this particular application - they are not intended to be accurate for the job.

What does match experiment quite consistently including shape of the pulses, is the collection of field solver simulations based on Maxwell's theory. You ran through them in your second post, and they show conventional electricity theory matching measurement for the white trace.

But i should have made it clear that i was referring to his green trace for the voltage across the resistor near his positive terminal. <snipperoo>

Yes, that is what I meant too. I can see why you have a problem with it now, and needed more detail...

If the scope were truly isolated (or ground lifted, depending on where EMC caps go) then the green trace should rise sharply more like the yellow.

and

I explained some of the problems with AlphaPhoenix's result many pages back, one of the main ones which distorts the send waveform I think is common mode coupling. I explained what I think it should look like, if it is measured with a better technique. Others did too, and went into quite some detail.

At 7:27 in the video is a diagram of the setup. The probe "reference GND" is the ground clip of the scope. This is tying one side of the pulse generator to Earth, loosely via extension cords and perhaps an inverter from the cars (described in discussions here at the time). The green probe, which is on the other side of the resistor, can thus not see the step directly from the step generator, because it is shorted to ground at the send end (by the ground clip). In essence it can only see voltage due to current getting around the circuit the long way, and a slow change of the GND voltage (which we can't directly see, because there is no probe measuring the voltage between this scope's GND and Earth under the desk).


This is not the way it's meant to be, but surprisingly the experiment still works. It's not necessarily an error if the person doing the test knows that taking this shortcut will still work. Again, I agree the green trace is "wrong", and this does represent the current in that resistor, and hence the current sent into that leg of the 'apparatus'. The other leg should be taking the balance, so it should be seeing nearly all the initial pulse missing from the green side (because that is shorted to ground).

From the clean white trace I can infer that the differential send current is probably fairly rectangular. But subtract the green trace and add the generator step, and that's going to make for a pretty messy voltage on the far side of the unprobed resistor, possibly best not to think about because it is guaranteed to confuse.

The situation would be the same but inverted traces (voltages) if the polarity of the generator is changed - other than that there is no difference and I likely would not test to confirm if I were doing the experiment.

BTW all this isn't so much from theory, as from experience. It just helps explain what is seen. The result is in the white trace, and matches simulation as you already know and I can now see you never really had a problem with. I rarely think about theory when doing engineering stuff, but I sometimes calculate things, and sometimes put it in a circuit simulator if I have really turned my brain inside out. By "Trevor's theorem" I try not to think through tricky situations to arrive at an answer, especially if it is about something that is inverted a number of times - he says you're most likely to get that out by one so would be better off flipping a coin, so save yourself the bother and guess, test, and then swap it round if it's wrong (sort of thing). The important thing is that anyone can be wrong at any time so don't put too much trust in thoughts. Or scopes. In either case trying to bulldoze through a problem with your mind is asking for trouble. It's not about intelligence, but experience in the biz.

Also small apology that I didn't say "subtle lift" originally, or if I did I edited that to "some sort of frequency dependent tilt". This is the white trace before the first reflection arrives. I'm not very interested in why, just noticed it (it appears in the simulations too).

Interesting. Here is my opinion re the AlphaPhoenix X pt1. Keeping in mind that i don’t know what a scope smells like. And i am allergic to electrons.  In my opinion Brian has 1 hit & say 7 strikes.
Hit 1. Brian has shown that there is an early significant current in the bulb. White trace (black in my drawing).
Strike 1. Brian fails to show whether the 1/c answer (ie the optional answer (d) ie 3.3 ns) is correct, because his scope can only see down to about 10 ns, & he needs to see down to 1 ns or 0.1 ns if he wants to confirm the 3.3 ns. However Howardlong has confirmed that (d) is correct (his 20 GHz scope can see down to 0.05 ns).
Strike 2. Brian fails to tell us the exact length of his two loops of wires. Hence we can't check to see the (time delay) effect of the heavy enamel on his wires. Oh, & he fails to tell us whether there is enamel on the wires.
Strike 3. Brian's wires etc on his table are all over the place. They should be symmetrical or rectangular or something. And there is lots of hardware in the joints & leads & clips & buckles & bows, but i don’t know whether that pile of krapp can be cleaned up a bit, perhaps he could have used some solder.
Strike 4. Brian's table has more wt of Fe in the frame under the plastic table top (i think it is plastic) than there is Cu wire in his circuit. If u look u will see that there is say 3 mm tween the Fe & the Cu near his bulb, & likewise near his source. In his X pt2 he could show us a trace for a probe on his Fe. Not good. The plastic top might make it worse, it would act as a capacitance multiplier (ie a dielectric).
Strike 5. His source is a 5V DC charger. I would prefer a lead acid battery, like Veritasium had. In fact i reckon that a lead acid battery is essential.
Strike 6. His say 1000 m of Cu is overkill. However i might make this his Hit 2, koz i think the extreme length is going to (accidentally) help me with my explanation for the green trace (which will follow hereunder)(we will see)(a lucky punch perhaps).
Strike 7. Brian failed to show us the trace for the resistor near the switch. This missing trace would for sure help us to explain the green trace (the main topic today).

I noticed the slight rise in the white trace (black in my drawing), but i haven’t thort about it, i doubt that any explanation (& there would be hundreds ovem) would be of much interest. And there are lots of interesting things in the traces. Ok, i have had a think about it (the slight rising grade), i might explain it tomorrow.

Anyhow Hit 1 was a big shock to me. I thort that there might be a brief weak spike, at 3.3 ns, due to radio crosstalk, but instead we see a strong capacitive inductance crosstalk, at 3.3 ns we think. I was shocked (pun alert). However, it lead me to my new (electon) electricity. This feel-good story will someday be folklore. Hell, i might get a Nobel medallion. bsfeechannel might nominate me (& a murmuration of pigs will darken the Sun). But now my genius is needed to explain the Green Trace. Green, my favourite colour – NO, ITS GREY (python joke alert). Ok, i had a think about the green trace, i will explain it in a new reply later today or early tomorrow.

Re any adverse comments that i might have made about TL models for the white trace, i think that these were re the initial part of the transient, ie what i called stage-1 of the transient, ie the part that AlphaPhoenix didn’t & couldn’t measure with his mickey mouse 100 MHz scope. I said that a TL model might be ok for the stage-2 transient, ie AlphaPhoenix's  0.2V bit of the white trace, but that a TL model was almost certainly not ok for the stage-1 transient, but might be with lots of new clever tweeking.

[adx]

Hell, i might get a Nobel medallion.

That's what thingverse is for.

[SandyCox]

I will attempt to address your misconceptions in the note I am writing.
In the meantime, let me set a challenge: Let's change the termination resistor in Catt's paper from 75 Ohm to 47 Ohm. How, according to "new electricity", will the measured pulses look if we do that?

There will be stepped reflexions of current & voltage. I have never worked on that kind of stuff.
But i see that Wakefield already has some traces for 75 Ohm coax with a 40 Ohm termination.
Shown on page 42 & 43 of Forrest Bishop's paper re Reforming Electromagnetic Units…
http://www.naturalphilosophy.org//pdf//abstracts/abstracts_6554.pdf


One thing i can explain, that i think no-one else could explain.
See how the bottom trace jumps up a few V too high, & then falls down to the correct 4V.
That jump is due to the 40 Ohm resistor being saturated with my electons. When the switch is closed electons in that short wire near the switch already heading for the switch will instead of doing their usual u-turn at the open switch they will propagate through the closed switch & onto the core wire of the coax, followed by electons from the resistor. The short wire was saturated with electons too, but the resistor holds a lot more electons per m length than the wire. Hence the brief spike of over-voltage.
U can see the same spike of over-voltage in some of the other traces, but there it is a negative spike of under-voltage i suppose u could call it.

But i have to have a think about what happens to electons inside resistors. Are they annihilated. Do they looz energy. Do they convert to infrared photons.
Anyhow, if there are lots of voids or porous bits or interface surfaces then there must be lots of electons on thems surfaces, & the electons would be doing lots of u-turns.
And they would be jostling lots of surface electrons, which would jostle atoms, & produce heating & resistance & would we know produce a voltage drop.
Its the voltage drop that has me worried. I am glad that SandyCox did not ask me re how exactly do electons produce a voltage drop across a resistor. Still thinking.

Your theory is useless unless it can predict values of voltage and current.

[aetherist]

Your theory is useless unless it can predict values of voltage and current.

My electons explain what happens in the various stages of transients.
And if fully developed my theory will give numbers for transients.
Old electricity can't even give good numbers for steady state, eg the half voltage double time for discharge of a capacitor (which my electons explain in the simplest possible way).