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

[T3sl4co1l]

It's a good prompt, because it encourages thinking about how the fields work.

They aren't just some free-willed inscrutable thing, after all.  The fields depend upon motion of charges, and boundary conditions.  Which is to say, more motion of charges, but we can treat shields, dielectrics, etc. as boundaries to media with characteristic properties, no need to do a full particle simulation every time you throw the light switch and all.

Say you have a near-plane wave propagating along the wire.  (I suppose it won't be a perfect plane wave, as it's dragged inward a little bit by the wire.)  The wire ducks into a hole in a metal sheet.  Currents flowing on the wire, reflect image currents in the sheet, and so the field effectively scrunches down as the wire approaches the sheet.  If we're talking high frequencies, most of the energy reflects off the plane, and yes, it effectively blocks radiation -- the wire has a fairly small cross-section (locally it acts like a whip antenna, albeit of somewhat indeterminate length since, well, it's long -- particularly if we're assuming an infinite wire passing though a small (not touching) hole in a perpendicular sheet of infinite size).  Some remainder couples through, is re-radiated in the same way on the other side, and there you are.

Well, inside and near the hole: you have current on the wire all the way through, some potential with respect to ground, and thus some electric field.  Outside the hole, the field will be fairly mild (making roughly 1/4-circular arcs from wire to plane, up in the near field of course i.e. we can locally ignore wave propagation), rising in intensity as you approach the edge of the hole (the arcs get shorter and shorter).  Inside the hole, the E-field lines are largely radial, approximating a coaxial cylinder structure.  Likewise the currents are largely radial outside the hole (converging on it like a black hole), then longitudinal inside.

The field is, and isn't, blocked.  For sure, there's no abrupt brick-wall stoppage, it's not like it's a...cloud of tennis balls or something.   Specifically, we can consider the superposition of free propagating (and subsequently reflected), and wire-guided (and subsequently transmitted), waves, and realize that one is blocked, the other not.  There is some coupling (the wire picks up some propagating field on both sides) so it's not completely independent, and, I'm not entirely sure how you'd set up a problem around an infinite wire (if it's resistive, it'll drag down all the field around it, until nothing's left -- depending on which axis you consider "infinite-er"); maybe you still need the excitation source at some finite distance, so you're really modeling waves inside a large box, rather than infinite space.  And then you get standing waves, severity depending on how the other boundaries are modeled (an impedance-matched source wall would be good to keep the bandwidth stable).  Anyway, it'll be something -- boot up the simulator yourself, I'm sure I've missed something here.

Tim

[Bud]

Yes but there should be at least a measurable difference in power delivered to the load with the plate vs no plate. And I bet my dollar to a donut that there will be None at DC.

[adx]

Oh no he didn't!:

Now the question is, after I close this switch, how long would it take for the bulb to light up?

...

...(converging on it like a black hole) ...

What exactly is it that got us all here! I've got the T-shirt, handy visualisation aid for situations like this.

Meanwhile, earlier in the day:

...It's not even a chicken/egg dilemma, the charges can be moving for any reason you like.  It can be a bag of protons on a hamster wheel.

That last sentence made me hungry for some reason, for a very specific thing that doesn't even make sense (bag of hamster flavoured protons).

I watched the ElectroBOOM vid - I found it rather thought provoking and entertaining in a "what's up slappers" kind of vibe. Don't know if that's new. But new to me, was that I had not directly noticed the semantic contradiction in Veritasium's video between movement of electrons creating the magnetic field, and his statement that energy is only in the field (or whatever it was). Potentially 3 "Ow my balls" moments if you include the diagram? Ah, what could have been. Follow your dreams, kids.

I watched much of Veritasium's video again and I think completely this time, in 1440p.  I got the battery polarity reversed in my simulation, I knew that would happen if I guessed.

I'm currently stumped on where the energy flows, without any pun intended because current can be located to one place, the pressure difference can't. I'd like to see a good argument for why a magnetic field is a better position for the current than the current itself. Where would a photon go?

Oops I had more, getting late. Hang on, new post:

@Bud: The magnetic field at DC will go straight through a copper plate. The electric field AKA voltage difference will crowd into the gap. From this time of night I can't see how that makes you wrong. The bulk of the magnetic field will be in a place with no electric field.

Time for bed.

[bdunham7]

So if energy flows not in wires but near the surface, then if we take a fairly large sheet of copper, ground it, drill a hole in the center with the hole diameter being very close to the wire diameter,  and run the wire thru the hole, then we will block the flow of energy (assuming the hole walls are so close to the wire surface but not touching it ) along the wire surface , and therefore when we close the switch the load will not get any energy in DC steady state because no electromagnetic field can break through the shield . I do not think anyone here would believe that at DC no energy will flow into the load in this experiment. Therefore the claim that energy flows not in wires is bogus.

You've beat me to it.  My version of this was going to be to take two blocks of material, machine cavities in one to hold the battery, wires and the base of the lamp, then put them together so that the entire circuit was encased in the block.  If you make various blocks out of acrylic (non-conductive, non-magnetic), aluminum (conductive, paramagnetic) and iron (conductive, ferromagnetic) you can then see that the fields and Poynting vectors in each case will be radically different, but I think we'd all agree that in all of the steady state DC cases, after say 5 seconds of stabilization time, the actual current, voltage and 'energy flow' will be indistinguishable.  However, any change you make to the actual wires--different material, different size, hollowing out the center, etc--will be readily distinguishable.  So you can say that there is an S-field and that is the 'energy flux', but that in the DC cases all of the different S-fields always amount to the same thing. 

What does that get you?  I'm not going to try and reconcile Poynting's theorem with this example--Mr. Veritaseum should do that.  And, he should use the correct depiction of the right-hand rule because even though the thumbs are all pointing the right way, I'm not sure how they got that way using curled fingers.

[SandyCox]

Hi Guys,

Here's my take on this problem.

The most important point is that each of the two transmission lines looks like resistors with a resitance of Zo before the first reflection is recevied back from the end of the lines, i.e before 1 second. So if the resistance of the bulb is 2Zo then the voltage across the bulb will be equal to 1/2 of the battery voltage between 0 and 1 seconds and will be equal to the battery voltage for the rest of eternity.

Sandy

[bdunham7]

Based on the title, I'd argue that's he's still "right".
The video is fundamentally about the misconception that energy flows in wires, and he did a fairly decent job of explaining why that's not the case.

Well, if you put "right" in quotes, I suppose you can get away with anything! 

His explanations are full of glossed-over holes and errors that you miss if you glide through it, although it is hard to be 'wrong' if you aren't precise or complete.  But if you go through and really nitpick it there are fatal errors in multiple locations and Electroboom picked up on some of those--the wood sawing example--but there are others.  For example, in the DC-case Poynting vector explanation he uses the wrong right-hand depiction (although you can make it work--it's still a right hand I suppose) and doesn't explain how a uniform surface charge would cause electron drift one way or another. 

Now obviously the circuit works as advertised and I'm not saying Poynting's theorem is wrong or doesn't apply.  However, as I pointed out in my reply to Bud, you could encase the circuit in different materials which would result in radically different external fields and Poynting vectors, but the resulting current, voltage and power would always be the same.  I think the key is in understanding the difference between the curled magnetic field around moving charges vs magnetic flux induced by charges moving in a closed loop.  That and some math.

[Sredni]

So if energy flows not in wires but near the surface, then if we take a fairly large sheet of copper, ground it, drill a hole in the center with the hole diameter being very close to the wire diameter,  and run the wire thru the hole, then we will block the flow of energy (assuming the hole walls are so close to the wire surface but not touching it ) along the wire surface , and therefore when we close the switch the load will not get any energy in DC steady state because no electromagnetic field can break through the shield . I do not think anyone here would believe that at DC no energy will flow into the load in this experiment. Therefore the claim that energy flows not in wires is bogus.

Why do you think it will block the 'flow' of energy? The energy is in the fields, and most notably in the fields generated by the surface charge. Not that of the electrons 'flowing' inside the conductor. When you create a hole in your shield, you let the conductors through it, without touching the shield. This means that the surface charge (albeit a tad disturbed by the capacitance of the shield, but let's neglect that) will go through as as well. And with the surface charge, so do the fields. And the energy.

The wires are there to conduct the fields. And the fields most relevant to the transfer of energy are those in the space between the conductors. Far from the other wires, the magnetic field, in the hypothesis of uniform current density inside the wire, grows linearly with the radius of the conductor and then decreases as 1/r - it is due to the current inside the wire that comes from the electrons subject to the field E_copper resulting from the effects of the surface charge.
Inside the wire the electric field is negligibly small (E_copper = j / sigma_copper) and the material is such that the electrons do not lose any significant amount of energy (they do not get hot); inside the bulb the electric field is strong (E_bulb = j / sigma_bulb) and the material is such that electrons lose a significant amount of energy (the bulb get so hot that it starts to glow). But the energy they lose in the bulb was not carried by the electrons in the wires. It's the potential energy that comes from the large potential difference across the bulb resistor. A large potential difference associated with the strong electric field produced by the surface/interface charge at the resistor-conductor boundary.

This is reflected by the direction of the Poynting vector: nearly parallel to the wires, irradiating from the battery, plunging into the bulb resistor.

Edit: added part on magnetic field.

[bdunham7]

The energy is in the fields, and most notably in the fields generated by the surface charge. Not that of the electrons 'flowing' inside the conductor.

Why 'fields' (plural) without naming them?  Since the Poynting vector is the cross product of the E and B fields, why do you say that the E-field generated by the surface charge is somehow more important than the curled magnetic field, which is indeed due to the electrons flowing inside the conductor?

[iMo]

The E and B fields in the Veritasum's experiment do not follow the 2x 300k km distance, imho.
The E and B fields follow the "path of least resistance/effort" (a fundamental law of physics, afaik) - that means directly from the battery to the bulb 1 meter apart.
Whatever shape and length of the wires you choose, the bulb X meters apart from the battery will always lit in X meters/c secs.

[Sredni]

The energy is in the fields, and most notably in the fields generated by the surface charge. Not that of the electrons 'flowing' inside the conductor.

Why 'fields' (plural) without naming them?  Since the Poynting vector is the cross product of the E and B fields, why do you say that the E-field generated by the surface charge is somehow more important than the curled magnetic field, which is indeed due to the electrons flowing inside the conductor?

You probably posted this before I edited my post. I edited before reading yours because my crystal ball told me this would come up as an objection. Now, re-read my edit and consider the case of a perfect conductor that would make E = 0 inside the conductor.
Compute the Poynting vector inside the conductor with E = 0. What do you get? Zero.
There is no energy being carried inside the perfect conductor. It's all in the space between conductors.

[Bud]

So if energy flows not in wires but near the surface, then if we take a fairly large sheet of copper, ground it, drill a hole in the center with the hole diameter being very close to the wire diameter,  and run the wire thru the hole, then we will block the flow of energy (assuming the hole walls are so close to the wire surface but not touching it ) along the wire surface , and therefore when we close the switch the load will not get any energy in DC steady state because no electromagnetic field can break through the shield . I do not think anyone here would believe that at DC no energy will flow into the load in this experiment. Therefore the claim that energy flows not in wires is bogus.

Why do you think it will block the 'flow' of energy? The energy is in the fields, and most notably in the fields generated by the surface charge. Not that of the electrons 'flowing' inside the conductor. When you create a hole in your shield, you let the conductors through it, without touching the shield. This means that the surface charge (albeit a tad disturbed by the capacitance of the shield, but let's neglect that) will go through as as well. And with the surface charge, so do the fields. And the energy.

Because E and B fields do not crawl *On* the wire surface in their entiety, they also exist at a distance from the surface albeit at a deminishing amplitude. Still, you'd need to integrate them thruoghout the entire space in order to arrive to a total. Therefore if we only leave a tiny hole very close to the wire surface , only the field portion that is closest to the surface will squeeze in and the rest will be blocked by the shield. That is why i said "there should be measureable difference in delivered power with the shield vs no shield".

[bdunham7]

Compute the Poynting vector inside the conductor with E = 0. What do you get? Zero.
There is no energy being carried inside the perfect conductor. It's all in the space between conductors.

Perhaps we don't disagree, I never implied that the Poynting vector in the wire was anything other than zero or close to it.  Rather I'm pointing out that the existence of one of the two required fields is directly due to the actual movement of electrons. And the other is due to where they end up.  If you'll amend 'between' to 'around' I would be even more agreeable.  'Between' is confusing and technically wrong in most cases.  Anyway, I regard this as mostly a semantic argument and would point out that the S-field is not a tangible phenomenon, even it if its is a useful model in some cases. 

In the DC steady-state example, the electric fields produced are solely the result of charge distributions, except perhaps internally within the battery at a microscopic level where things are arguable more complex. The light bulb lights because the system produces a charge distribution at its terminals that results in an electric field that moves electrons through it and they do work.  How those electrons are moved so as to create that charge distribution that creates the electric field is a matter of magnetic fields, electric fields and math.  The S-field and Poynting vectors are a way of representing the results of that math, no more and no less. 

[Sredni]

So if energy flows not in wires but near the surface, then if we take a fairly large sheet of copper, ground it, drill a hole in the center with the hole diameter being very close to the wire diameter,  and run the wire thru the hole, then we will block the flow of energy (assuming the hole walls are so close to the wire surface but not touching it ) along the wire surface , and therefore when we close the switch the load will not get any energy in DC steady state because no electromagnetic field can break through the shield . I do not think anyone here would believe that at DC no energy will flow into the load in this experiment. Therefore the claim that energy flows not in wires is bogus.

Why do you think it will block the 'flow' of energy? The energy is in the fields, and most notably in the fields generated by the surface charge. Not that of the electrons 'flowing' inside the conductor. When you create a hole in your shield, you let the conductors through it, without touching the shield. This means that the surface charge (albeit a tad disturbed by the capacitance of the shield, but let's neglect that) will go through as as well. And with the surface charge, so do the fields. And the energy.

Because E and B fields do not crawl *On* the wire surface in their entiety, they also exist at a distance from the surface albeit at a deminishing amplitude. Still, you'd need to integrate them thruoghout the entire space in order to arrive to a total. Therefore if we only leave a tiny hole very close to the wire surface , only the field portion that is closest to the surface will squeeze in and the rest will be blocked by the shield. That is why i said "there should be measureable difference in delivered power with the shield vs no shield".

You seem to think that fields are 'material arrows' connected along lines that must pass inside the hole. No, they are manifestation of charges and currents. If the surface charge passes through the hole, they will create an electric field between conductors on the other side of the hole. They will also create the conditions for the field inside the conductor to comply with Ohm's law, and that will make a current flow, with the corresponding magnetic field on the other side of the hole.
Again, in the case of a perfect conductor, E inside is ZERO. So the vector product with B, whatever that may be will be ZERO. Ergo, no energy is being carried inside the wires.
The wires carry the surface charge that creates the E field (inside and outside) and the current that creates the B field (inside and outside). But, like it or not, only outside the wires the contribute of the fields combined gives a non-negligible transfer of energy.

The wires are necessary, but they do not carry the energy, they just guide it.

[Sredni]

Compute the Poynting vector inside the conductor with E = 0. What do you get? Zero.
There is no energy being carried inside the perfect conductor. It's all in the space between conductors.

Perhaps we don't disagree, I never implied that the Poynting vector in the wire was anything other than zero or close to it.  Rather I'm pointing out that the existence of one of the two required fields is directly due to the actual movement of electrons. And the other is due to where they end up.  If you'll amend 'between' to 'around' I would be even more agreeable. 

[/quote]

Ok, 'around' is a better description. But most of the energy transfer happens between the cables. Please read that paper by Jackson I referenced a few times already. It's "Surface charges on circuit wires and resistors play three different roles" published on the American Journal of Physics 64 (7), July 1996.

[HuronKing]

So if energy flows not in wires but near the surface, then if we take a fairly large sheet of copper, ground it, drill a hole in the center with the hole diameter being very close to the wire diameter,  and run the wire thru the hole, then we will block the flow of energy (assuming the hole walls are so close to the wire surface but not touching it ) along the wire surface , and therefore when we close the switch the load will not get any energy in DC steady state because no electromagnetic field can break through the shield . I do not think anyone here would believe that at DC no energy will flow into the load in this experiment. Therefore the claim that energy flows not in wires is bogus.

Others have answered and given good descriptions but I think it's worth pointing out that what you've described is explicitly used as an example of waveguide properties in Kraus Chapter 10:
http://amasci.com/graphics/kraus_poynt.gif

The accompanying text is easy to find online. I highly recommend the chapter. And it says it there explicitly - all wires are waveguides. If you go looking for the energy in the central conductor of a (ideal) coaxial cable and not in the dielectric between the inner conductor and the shield, you'll never find it. That was Oliver Heaviside's great insight in inventing the coaxial cable in the first place (and independently discovering the Poynting Vector, but the pun is too good to ignore to have named it after Heaviside)!

This is actually one of the frustrating parts of Veritasium's video - not using the coaxial cable or giving proper credit to Heaviside as a VERY explicit example of where the energy is located in/around current carrying wires.

[Someone]

lol, still going strong. Yet only one mention of Litz!

The "argument" is framed around such a nonsense assumption/premise, its intentionally trying to break peoples heads by equating radically different modes of propagation.

Back in the real world, with the sort of power people actually use to power lighting devices (I'll send you a $1 if you actually have an installed and in use microwave power system) is so low in frequency that all these "clever" ways of moving power are irrelevant and lost in the noise.

Does power travel in the conductor/wire, yes it does, otherwise hollow shell conductors (just the shield of an empty coax) would be as effective or better than a solid of the same outside dimensions and material. You're going to revolutionise the world with that material saving, send me just 1% of the sales as royalties . Litz wire gives the practical example that people can play with at reasonable frequencies and see the effects, at some frequency the energy does concentrate to the surface/outside, and below that it appears the same as a solid of the same cross section...

Now redo the same experiment with the litz wire exploded/disassembled to have larger gaps between the conductors...

What point does that become significant?

[Sredni]

lol, still going strong. Yet only one mention of Litz!

The "argument" is framed around such a nonsense assumption/premise, its intentionally trying to break peoples heads by equating radically different modes of propagation.

Back in the real world, with the sort of power people actually use to power lighting devices (I'll send you a $1 if you actually have an installed and in use microwave power system) is so low in frequency that all these "clever" ways of moving power are irrelevant and lost in the noise.

You seem to think, like Dave, that the Poynting vector is directed in that way - or worse, has only sense - at high frequencies. (By the way, by not correcting his video, at least in the description, Dave has joined the long list of Youtube crackpots that are not capable to correct their biggest blunders.)
The fact that the Poynting vector is usually introduced in later chapters in introductory physics and EM book does not mean you cannot consider it at DC.

Does power travel in the conductor/wire, yes it does,

That's not what math and classical electrodynamics say.

otherwise hollow shell conductors (just the shield of an empty coax) would be as effective or better than a solid of the same outside dimensions and material. You're going to revolutionise the world with that material saving, send me just 1% of the sales as royalties .

Seriously? It's the resistivity that makes the difference. If you use perfect conductors then the resistance in the wire is zero, no matter how thin the wires are. Zero electric field inside means zero Poynting vector inside. There is zero energy carried inside the wires.

Back to the real world where copper has a finite conductivity. Yes, now size matters. Still, the power is not carried inside the wires. Why size matters? Because for a given value of current I, the bigger the section of the cable the smaller is its resistance. The less power is lost in the cable. In terms of Poynting vector: the smaller the conductivity the biggest will be the electric field inside the conductor for the same current density j; with a small nonzero electric field inside the wire, the Poynting vector will be directed slightly toward the wire. Part of the battery power ends up dissipated in the cable.

You need bigger cables because for big currents - when your cable has nonzero resistance - you want to minimize I²R.

Litz wire gives the practical example that people can play with at reasonable frequencies and see the effects, at some frequency the energy does concentrate to the surface/outside, and below that it appears the same as a solid of the same cross section...

Litz wire and the skin effect has NOTHING to do with the transfer of power at steady state in DC.
One might argue that the proximity effect can change a hair here or there but that is again missing the point.
But I am sure that a lot of people who did not study EM and just heard Dave - uncorrected - talking about the Poynting vector pointing the other way around in DC (LOL) and the supposed role of the  skin effect (in what is not clear), well they will think you have a point.
This is how disinformation works on Youtube. It's self-feeding.

Edit: syntax

[bdunham7]

The fact that the Poynting vector is usually introduced in later chapters in introductory physics and EM book does not mean you cannot consider it at DC.

You can consider it, but IMO it is poyntless.  For a given DC circuit made up of specific components and wires of a given resistance, no matter how you physically rearrange it all solutions will degenerate to the exact same result.  You could encase the circuit in iron--which would certainly change the overall arrangement of the fields--but the end result will be identical.  That's not true for an AC circuit, where something more interesting is happening with the magnetic fields...

The concept of power 'travelling' is interesting in itself.  If you have a motor transmitting rotational power (torque x speed) through a shaft to a load in a completely constant manner, so that the energy of the shaft--which consists of the angular momentum and flex torsion--is absolutely constant, how does the power 'travel'?  No energy is going in or out of the shaft itself, yet power is applied at one end and applied to the load at the other.  Likewise, the static fields of a DC system store some energy, but since they are static and their energy never increases or decreases, can you say that energy flows 'through' them even though it doesn't flow in or out?  Even if you do (a semantic issue IMO), you'd have to concede that this is at least nominally distinguishable from the case where energy is put into a field, it's energy measurably increases and then at a later time and perhaps at a different location, that energy is dissipated in a load and the field's energy decreases.

[iMo]

..
The concept of power 'travelling' is interesting in itself.  If you have a motor transmitting rotational power (torque x speed) through a shaft to a load in a completely constant manner, so that the energy of the shaft--which consists of the angular momentum and flex torsion--is absolutely constant, how does the power 'travel'?  No energy is going in or out of the shaft itself, yet power is applied at one end and applied to the load at the other.
..

Every energy/force/power/momentum etc. in our Universe is transferred via fields. All matter around you consist of atoms/particles which are small and far away from each other. From microscopic view the shaft of a motor is an almost empty space. The motor's torque is transferred to the load via EM fields in the shaft therefore. At one side of the shaft you put in the energy which moves the fields between the particles in a specific direction - the fields propagate with the speed of light - and on the other side of the shaft the particles follow that movement because the fields affect them. The power transfer through the shaft happens in almost empty space (of course not "fully" empty - there is the energy of vacuum, black matter, black energy, quantum foam, strings, etc. as well  )..

[Sredni]

The fact that the Poynting vector is usually introduced in later chapters in introductory physics and EM book does not mean you cannot consider it at DC.

You can consider it, but IMO it is poyntless.  For a given DC circuit made up of specific components and wires of a given resistance, no matter how you physically rearrange it all solutions will degenerate to the exact same result. 

What. Are. You. Talking. About. ?

Wh---
No, don't answer. Please read the following before answering:

Understanding Electricity and Circuits: What the Text Books Don’t Tell You
Ian M. Sefton (School of Physics, The University of Sydney)
Science Teachers’ Workshop 2002

Energy flow from a battery to other circuit elements: Role of surface charges
Manoj K. Harbola
2010 American Association of Physics Teachers.
DOI: 10.1119/1.3456567

Energy transfer in electrical circuits: A qualitative account
Igal Galilia and Elisabetta Goihbarg
Am. J. Phys. 73 (2), February 2005
DOI: 10.1119/1.1819932

and, of course

John D. Jackson
Surface charges on circuit wires and resistors play three different roles
American Journal of Physics 64 (7), July 1996

While you're at it, try to get a copy of Sommerfeld's Lectures on Theoretical Physics (it's 6 Volumes but you need only the third volume about Electrodynamics). There you will find an exercise about an infinitely long resistive wire.
It's on page 125, "Detailed treatment of the field of a straight wire and a coil"
Back in 1942 the role of surface charge and the direction of Poynting vector for a DC circuits was no mystery at all. Keep in mind that Sommerfeld is considering a very long resistor, but I copy here the conclusion:

[bdunham7]

What. Are. You. Talking. About. ?

Mock me if you like, but what I said was crystal clear--and correct.  It was also pretty basic and simple.  If you don't understand that (and your replies are non-responsive to the point) I'll be happy to try and explain with examples.

[HuronKing]

"The Poynting Vector is pointless."

Please, kindly, toss all your coax cables out the window then.

[bdunham7]

"The Poynting Vector is pointless."

Please, kindly, toss all your coax cables out the window then.

Apparently you missed the context.  I was referring to steady-state DC circuits exclusively, which ironically I often use coax cables for--but that's a different issue.  And I didn't say that they (Poynting vectors) can't be calculated, drawn or that they somehow don't apply--I said that they are poyntless in that they serve no purpose other than to give you the smug satisfaction that you are above all the other heathens because you somehow "know what is really going on". 

If I have the basic circuit with a completely defined battery, two uniform wires of known resistance and a completely defined light bulb, I can model the circuit for voltage, current, power, fields, heating, etc.  I can also draw some nice Poynting vectors--but only after I use my model to determine the E and B fields at each point in space.  Now if I physically rearrange my circuit--put the wires in a circle, twist them, wrap the circuit in tin foil, encase it in an iron block--each of those instances will have a different set of fields and a different S-field and Poynting vectors at each point in space.  However, no matter how I do that in the DC case, my power transfer to the light bulb will be exactly the same.  That's what I mean by 'degenerate'--you have infinitely many Poynting vector solutions corresponding to infinitely many physical configurations that will always work out to the same result, the result that I got by just considering the characteristics of the battery, bulb and resistance of the wires.

[Sredni]

I said that they are poyntless in that they serve no purpose other than to give you the smug satisfaction that you are above all the other heathens because you somehow "know what is really going on". 

The discussion - the context - was about how energy is transferred in a DC circuit. And energy is transferred by means of the fields in the insulator around the wires (and mostly between the wires). How is the Poynting vector useless when it is the answer, the solution to what was being discussed?
To know what is 'really' going on (and 'really' here means in the context of classical electrodynamics, at least for me) is the whole point of the discussion.

If I have the basic circuit with a completely defined battery, two uniform wires of known resistance and a completely defined light bulb, I can model the circuit for voltage, current, power, fields, heating, etc.  I can also draw some nice Poynting vectors--but only after I use my model to determine the E and B fields at each point in space.  Now if I physically rearrange my circuit--put the wires in a circle, twist them, wrap the circuit in tin foil, encase it in an iron block--each of those instances will have a different set of fields and a different S-field and Poynting vectors at each point in space.  However, no matter how I do that in the DC case, my power transfer to the light bulb will be exactly the same.

So what? I can say the same about the electric field and the current density inside the wires and resistor. No matter how you twist and place the wires, they will always be the same. Ironically, it's thanks to the surface and interface charges. Does that make the electric field and the current density pointless?

That's what I mean by 'degenerate'--you have infinitely many Poynting vector solutions corresponding to infinitely many physical configurations that will always work out to the same result, the result that I got by just considering the characteristics of the battery, bulb and resistance of the wires.

But that solution does not give you any insight on how the energy is actually transferred. As a matter of fact it can lead you to think that energy travels through the wire, when it is not so.
I gave you a paper by Jackson and a book by Sommerfeld (plus a few papers here and there) to support this 'pointless' point. I mean, have you seen the pictures of Sommerfeld and Jackson? I wouldn't argue with those two. I will probably have to sleep with my lights on for a week after seeing how unforgiving they look...

[adx]

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You seem to think that fields are 'material arrows' connected along lines that must pass inside the hole. No, they are manifestation of charges and currents. ...

... So the vector product with B, whatever that may be will be ZERO. Ergo, no energy is being carried inside the wires.
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The wires are necessary, but they do not carry the energy, they just guide it.

That's a contradiction. If the (external) fields are (merely) a manifestation of the physical, then it's entirely invalid to use their (non-)existence to define location in this way. It sounds like belief in maths over physics to me.

I ask again: I'd like to see a good argument for why a magnetic field is a better position for the current than the current itself.

And also (was intended to be a separate question): Where would a photon go?

Having slept on it for a night, I notice for the first question that the magnetic field disappears from the picture (both inside and outside the wire) from the perspective of the moving charge carriers. Unsurprising from an energy (conceptual) point of view, because we define power as a product of pressure and flow, and we can safely ignore it from the integral. But it does break, in the way I described a page or so ago. Change the frame of reference by rotating the circuit (after making it round, and equal cross section components). At certain very similar speeds, there is no magnetic field at all (differences arise from the compressability of charge).

For the second question, I've come to the conclusion (wrong word) that it is all down to its time. And all that implies.

BTW for those curious about my black hole shirt (especially in case it came a cross as some crude dig!), pic below. Mine's black, looks like they no longer make it. Lifted from cafepress where apparently it came from (I don't buy my own clothes, the process confuses me  ). Assuming this is fair use, if not go buy one from them and I hope I don't get in trouble here. I get strange looks sometimes because it looks a bit like a gang patch, but colourful. Not as bad as the caffeine molecule one without the word "caffeine" (which they seem to have now), people seem to assume a whole different skill set.