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

[rs20]

Yet he doesn't mention transmission lines at all does he? (I still haven't watch the entire thing from end to end)

Correct, I just searched the transcript and the word "transmission" isn't used at all. And it makes sense that he doesn't, because mentioning transmission lines and how useful they are as a model would weaken his "everyone's doing it wrong" strawman.

[Kalvin]

On the undersea cable example, if it would be correct physics to say that the transmission of a signal through the cable happens purely by electrical impulses inside the metal of the conductor, then distortion would not be expected. It is only when you bring in the physical understanding that the signal propagation depends on the surrounding environment outside the cable that you can explain the distortion.

Sound like they didn't know how to make proper control impedance transmission lines...

It was stated in the video that there were two views in 1800s for how the energy is being transmitted in a cable.

The simpler model was this (wrong) "water hose"-model in which electrons are moving through cable.

The other model (correct one) was using EM-fields and transmission line model to describe how the energy was transmitted through cable.

Edit: Probably they used Occam's razor for selecting the model they were using 

[EEVblog]

Given that we are talking about ENERGY here, that means time.
And with the example of a DC battery and the switch right next to it, yes you'll get a brief initial almost instant surge of power to the light bulb due the nearby line capacitance (effectively just a 1m loop at t=0), and then transmission line propagation effects after that, half way to the moon each side. But once you hit steady state DC, is the ENERGY transported in the electromagnetic field?
No, it's not.
If you think it is, please explain how with the mm/s slow electron drift at DC.

Engineering has transmission line theory, transient anlysis, and steady state theory seperate for a reason.

[rs20]

But once you hit steady state DC, is the ENERGY transported in the electromagnetic field?

I'd say that's more of a philosophical question than a strictly scientific one.

If you choose to use Poynting vectors to answer that question, you'd find that yes, even in DC, the interior of the (superconducting) wire is devoid of an E gradient so it is not carrying power, yet the space around the wire has both E and H (even in steady state), and therefore Poynting vectors. So from the perspective of maxwell's equations, even in DC, the EM field around the wires is carrying the power in some sense.

If you choose to use the much more pragmatic and useful circuit theory, then P=IV at the light bulb, *current* is carried in the wires, voltage is "across" the wires (whatever that means, might be worth thinking about this before leaping to the conclusion that wires must obviously be carrying the power), and meh, P=IV and kirchoff and all that. Point is, the theory has perfect predictive power for simple circuits, which is why it's taught.

Oh, and I'm ignoring your distinction between power and energy. One is just the other integrated over time, that doesn't make any meaningful difference if we're talking about a system that's in steady state anyway.

[Kalvin]

Given that we are talking about ENERGY here, that means time.
And with the example of a DC battery and the switch right next to it, yes you'll get a brief initial almost instant surge of power to the light bulb due the nearby line capacitance (effectively just a 1m loop at t=0), and then transmission line propagation effects after that, half way to the moon each side. But once you hit steady state DC, is the ENERGY transported in the electromagnetic field?
No, it's not.
If you think it is, please explain how with the mm/s slow electron drift at DC.

The cable is carrying energy from A to B. There is also a time aspect in it: Transmitting morse symbols from one cable end to another is carrying/transmitting/conducting energy from one cable end to another. Similarily, the cable is carrying/transmitting/conducting energy from the battery to the lamp.

I would not use capacitance-model only as a way to explain why there is energy flowing in the circuit after the switch is closed. The more appropriate model is to use transmission line model, which will also explain accurately why the load doesn't get the full power after the switch is closed. Transmission line model will introduce impedance, which will limit the power to the load after the switch is closed. Using only capacitance-model, the power into the load would be maximum right after the switch is closed.

When the system has reached its steady state again (ie. constant current is flowing in the circuit), the energy will flow in the wire as usual: The EM-model and transmission line model will still apply, although the L and C in the transmission line model can now be ignored (it is constant current after all). However, it is the EM-field which is still carrying energy from the battery to the load, and Maxwell's equations will still apply.

[sandalcandal]

Given that we are talking about ENERGY here, that means time.
And with the example of a DC battery and the switch right next to it, yes you'll get a brief initial almost instant surge of power to the light bulb due the nearby line capacitance (effectively just a 1m loop at t=0), and then transmission line propagation effects after that, half way to the moon each side. But once you hit steady state DC, is the ENERGY transported in the electromagnetic field?
No, it's not.
If you think it is, please explain how with the mm/s slow electron drift at DC.

Engineering has transmission line theory, transient anlysis, and steady state theory seperate for a reason.

You seem to be asking for someone to play devil's avocado here so...

In the DC ON steady-state, the electric field exists outside the wires. Specifically the electric field exists "outside" the wire in the free space directly between the battery and the bulb. The "energy" comes from a charge flux through this electric field i.e. not electrons/charge carriers "flowing through the length of the wire". Indeed as you note, the electron drift velocity in a typical electrical wire is in the mm/s. No electrons are flowing all the way down the wire and back a light-second in distance.

The "energy" is conveyed via this "external" electric field. If we were to impose something to block this field, then the flow of energy would stop.

Catch here is that "blocking the field" would mean applying an equal/opposite battery or simply disconnecting the wire.

Following this more things start to stray towards the Walter Lewin vs Electroboom "what is a voltage/potenital" stuff. There are probably other ways to approach it to try define the "energy transfer" as still happening "outside the wires".

[Ditto the above two comments]

[Trader]

Dave?

A lone image floating in space with no context... we aren't psychic.

Context: "Make a Video About This"

[EEVblog]

But once you hit steady state DC, is the ENERGY transported in the electromagnetic field?

I'd say that's more of a philosophical question than a strictly scientific one.

If you choose to use Poynting vectors to answer that question, you'd find that yes, even in DC, the interior of the (superconducting) wire is devoid of an E gradient so it is not carrying power, yet the space around the wire has both E and H (even in steady state), and therefore Poynting vectors. So from the perspective of maxwell's equations, even in DC, the EM field around the wires is carrying the power in some sense.

But energy is flowing, how does it flow? What's doing the moving at DC?

Oh, and I'm ignoring your distinction between power and energy. One is just the other integrated over time, that doesn't make any meaningful difference if we're talking about a system that's in steady state anyway.

But power is being continuously delivered from the source to the load under steady state conditions. How if the magnetic field is not moving?

[aneevuser]

The difficulty  in understanding whats going on is that "Veritasium" has a clever setup of having the loop a parallel ultra long wires, spaced apart by 1 meter all along the way.  With a degree of BS assumptions, like the cable and switch having 0 resistance, the battery having millions of amps of current with 0 ohm impedance, and the lamp being infinitely high in impedance.  What you end up with is the equivalent of 1 gigantic dipole antenna on the battery and switch side making a massive VLF transmitter at switch on, while on the lamp side, you have the parallel receiving dipole antenna.

This is precisely the way that I thought about the problem, once I had understood what claim was actually being made in the rather clickbaity video. I'm guessing that there's no disagreement here that there can in principle be energy transfer in 1/c s via dipole-to-dipole transmission?

However, the in-my-opinion more interesting case of steady state operation looks like a mess. If we claim that the energy is transmitted "by the fields" outside of the wires, then we clearly need both a non-zero E and B field outside of the wires, else |S| = |E x B| = 0. But AFAICS, there is not even universal agreement about what exactly *is* the E field outside of a conducting wire.

For example, here: https://www.feynmanlectures.caltech.edu/II_27.html, in section 27.5, Feynman briefly talks about the field outside of a *resistive* wire, says that it is non-zero, and shows an inward directing Poynting vector based on that claim. However, Veritasium's claim is about an ideal wire, where the internal E field, at least, would be 0, so this argument doesn't seem to be appropriate (and would also seem to argue for a zero inward-directed Poynting vector in the case of an ideal wire - where does the energy come from then?)

On the other hand here: https://physics.stackexchange.com/questions/61884/does-a-current-carrying-wire-produce-electric-field-outside is a StackExchange thread with one reply claiming that an E field exists, and another that it doesn't. One reply states that Jefimenko has experimentally observed an E field outside a conducting wire.

For another point of view, here: https://physics.stackexchange.com/questions/623858/surface-charge-on-a-current-carrying-conductor-is-impossible?rq=1, we have a reply to a question which seems to demonstrate that there is, in fact, a *radial* E field inside a current carrying wire, which arises as the electrons drift to the surface of the wire as the experience a radial v x B Lorentz force, due to their own B field. (This seems intriguing, since if true, we have both non-zero E and B fields *inside* the wire, and S = E x B points along the interior of the wire in the direction of v, unless my 3D geometry is way off)

Anyway, I've spent enough fruitless hours looking at this to realise that there seems to be a variety of conflicting arguments, and it probably needs a good physicist (which I'm not) to sort the wheat from the chaff.

[EEVblog]

Given that we are talking about ENERGY here, that means time.
And with the example of a DC battery and the switch right next to it, yes you'll get a brief initial almost instant surge of power to the light bulb due the nearby line capacitance (effectively just a 1m loop at t=0), and then transmission line propagation effects after that, half way to the moon each side. But once you hit steady state DC, is the ENERGY transported in the electromagnetic field?
No, it's not.
If you think it is, please explain how with the mm/s slow electron drift at DC.

Engineering has transmission line theory, transient anlysis, and steady state theory seperate for a reason.

In the DC ON steady-state, the electric field exists outside the wires. Specifically the electric field exists "outside" the wire in the free space directly between the battery and the bulb. The "energy" comes from a charge flux through this electric field i.e. not electrons/charge carriers "flowing through the length of the wire". Indeed as you note, the electron drift velocity in a typical electrical wire is in the mm/s. No electrons are flowing all the way down the wire and back a light-second in distance.

The "energy" is conveyed via this "external" electric field. If we were to impose something to block this field, then the flow of energy would stop.

Yes, there is a magnetic field around the wire, but it's not moving. The electrons however are moving, albeit slowly.
So how is the energy being transfered in the non-moving magnetic field?
Is the magnetic field just a byproduct of the current beign conducted in the wire in the DC case?

We all know that magnetic fields can store and transfer energy, it's the basis for transformers and motors and inductors, all very basic theory. But it is how power (and hence energy) is transferred at DC? If so, how?
Yes, I'm postulating that the power/energy transfer via electromagnetic field theory is not valid at DC. 

[EEVblog]

The difficulty  in understanding whats going on is that "Veritasium" has a clever setup of having the loop a parallel ultra long wires, spaced apart by 1 meter all along the way.  With a degree of BS assumptions, like the cable and switch having 0 resistance, the battery having millions of amps of current with 0 ohm impedance, and the lamp being infinitely high in impedance.  What you end up with is the equivalent of 1 gigantic dipole antenna on the battery and switch side making a massive VLF transmitter at switch on, while on the lamp side, you have the parallel receiving dipole antenna.

This is precisely the way that I thought about the problem, once I had understood what claim was actually being made in the rather clickbaity video. I'm guessing that there's no disagreement here that there can in principle be energy transfer in 1/c s via dipole-to-dipole transmission?

However, the in-my-opinion more interesting case of steady state operation looks like a mess.

That's my huge problem with this. And effectively also with LF 50/60Hz power transmission for example.

For another point of view, here: https://physics.stackexchange.com/questions/623858/surface-charge-on-a-current-carrying-conductor-is-impossible?rq=1, we have a reply to a question which seems to demonstrate that there is, in fact, a *radial* E field inside a current carrying wire, which arises as the electrons drift to the surface of the wire as the experience a radial v x B Lorentz force, due to their own B field. (This seems intriguing, since if true, we have both non-zero E and B fields *inside* the wire, and S = E x B points along the interior of the wire in the direction of v, unless my 3D geometry is way off)

If so then that implies skin effect at DC.

[Kalvin]

It may be possible to explain the current flow in DC steady state as the following simplified explanation: There is a longitudinal electrical potential gradient in the conductor, and the electrons are passing charge from one electron to another, like passing buckets of water from a person to another when putting out a fire: Although the people do not move, the water gets passed through. Only full buckets will be passed through the chain (as a charge is quantized). In a wire, the charge is passed from one electron to another at the speed determined by the wire's electrical properties. Electromagnetic field-theory will provide more accurate explanation and model for the current flow in conductor(s) in DC steady state as well.

[EEVblog]

It may be possible to explain the current flow in DC steady state as the following simplified explanation: There is a longitudinal electrical potential gradient in the conductor, and the electrons are passing charge from one electron to another, like passing buckets of water from a person to another when putting out a fire: Although the people do not move, the water gets passed through. Only full buckets will be passed through the chain (as a charge is quantized). In a wire, the charge is passed from one electron to another at the speed determined by the wire's electrical properties. Electromagnetic field-theory will provide more accurate explanation and model for the current flow in conductor(s) in DC steady state as well.

Yep, that's called charge conduction, and it happens inside the wire.

[han]

Hi Dave and all.


I'm not a native English speaker so I will put my explanation in picture.
Many things in the video I disagree, especially about the electricity.
But first things i want to address about the electron / electric field propagation.
(see picture)
and for the switch since it assume ideal switch. the waveform will be like very steep from 0 to 1. Is it like a half impulse function (broadband),
so I assume the 300KM / 2 cable length is not important since the wave is very broadband,

And because the Side of the battery always connected to the cable, so I assume whole circuit beside negative side battery and half of switch is in Positive.



Sorry for long time inactive...

[sandalcandal]

In the DC ON steady-state, the electric field exists outside the wires. Specifically the electric field exists "outside" the wire in the free space directly between the battery and the bulb. The "energy" comes from a charge flux through this electric field i.e. not electrons/charge carriers "flowing through the length of the wire". Indeed as you note, the electron drift velocity in a typical electrical wire is in the mm/s. No electrons are flowing all the way down the wire and back a light-second in distance.

The "energy" is conveyed via this "external" electric field. If we were to impose something to block this field, then the flow of energy would stop.

Yes, there is a magnetic field around the wire, but it's not moving. The electrons however are moving, albeit slowly.
So how is the energy being transfered in the non-moving magnetic field?
Is the magnetic field just a byproduct of the current beign conducted in the wire in the DC case?

We all know that magnetic fields can store and transfer energy, it's the basis for transformers and motors and inductors, all very basic theory. But it is how power (and hence energy) is transferred at DC? If so, how?
Yes, I'm postulating that the power/energy transfer via electromagnetic field theory is not valid at DC. 

Let me try paint a more "complete" picture of the DC system with fields.

The battery can be thought of as an "electric dipole" where there is a positive side/terminal with slightly lower electron density and a negative side/terminal with slightly higher electron density and as a result there exists an electric field (and a voltage potential) between the positive and negative terminals within the battery as well as the surrounding space. [This imbalance is setup by the electrochemical reactions within the battery][This can also be though of conversely, the battery has a electric field which causes a charge imbalance] When the battery is connected to a load, charge carriers (typically anions) move against the electric field (against the potential) thus there is generation/sourcing of electrical energy/power.

When the load (let's just assume a simple resistive load for simplicity) is connected, the electrons being "pushed" in see a resistance and bunch up on one side while spreading out on the other. Again there is an imbalance of charge and an electric field (and voltage potential) develops across the load. For the load however, the charge moving "with" the electric field so there is loss/sinking of electrical energy/power. In the ideal case where the load and the source are connected with an ideal (lossless) wire, the electric dipoles of the source and the load will perfectly match, in other words the voltage across the source and load will be equal but opposite relative the direction of electrical current.

Now we ask "how is the energy being conveyed?" There is no electric field in the (ideal) wires, any flow of charge through the wires has no generation/loss of electrical energy by definition of the wires being ideal wires in a DC system. In fact, in our theoretical system, the perfectly matched source and load dipoles effectively (but not completely apart from within the wire) cancel each other out with only substantial fields in a parallel region between the dipoles. Again, where is the energy "coming and going"? the creation and loss of electrical energy (from and into other forms) happens only "inside" the source and load.

But what if we disconnected the wire? Without an exit for electrons to flow out of the source negative and without a supply of electrons to flow into the load negative, the flow will stop. The imbalance across the battery will remain but the continuous imbalance across the load resistance will actually disappear as electrons spread themselves out within the load resistor till there is zero electric field within the resistor. [The resistive load will actually being an opposite polarity dipole in order to cancel out the net electric field within is self, no net current flow so no energy]. Now we can see the purpose of the wire, it supplies/takes charge carriers from the source and load terminals in order to continue the ON state equilibrium: On the negative side wire, some electrons leave the battery negative and enter the wire and some other electrons exit the wire enter the load negative. On the positive side wire, some electrons leave the load and enter the wire while some other electrons leave the wire and enter the battery. You don't have electrons travelling the full length of the wire (mostly, statically). You have some jumping on, some jumping off. There is a net zero charge change in the ideal wire itself.

Thus we see the "true" purpose of the wire is just charge conduit to maintain an certain equilibrium where there is a flow of electrons through both the source and the load but no net change in the total charge of the system, charge is cycled in a circular manner between the source and load, a "circuit" if you will

The actual "energy" is coupled via the electric field.

Edit: Attached a rough sketch. Might be much prettier if someone can to do a complete 2D field sim instead... I wonder if FEMM can do it, otherwise maybe something in Python...

[han]

My respond to the video:

The Frist analogy of the electricity using chain and hose is wrong according to Derek(IMHO chain analogy can be used)
(video time 2:30)
1. He said the electron is not continuously came to your house (yes)
(video time 2:34)
2. There is a Transformer (yes) ,but the transformer is change the Current in the wire into magnetically field in Primary winding
and because change in magnetically field is happen too in secondary winding, so the energy is transferred from one side to other side
(video time 2:34)
3. He ask why electron don't carry the energy back to the power station.
ans: if he using the chain analogy then the answered is same like motorcycle engine to the tire.
because the one do the work (push(+) and Pull(-)) electron is the Battery(DC) or Generator (AC)

[EEVblog]

Now we ask "how is the energy being conveyed?" There is no electric field in the (ideal) wires, any flow of charge through the wires has no generation/loss of electrical energy by definition of the wires being ideal wires in a DC system. In fact, in our theoretical system, the perfectly matched source and load dipoles effectively (but not completely apart from within the wire) cancel each other out with only substantial fields in a parallel region between the dipoles. Again, where is the energy "coming and going"? the creation and loss of electrical energy (from and into other forms) happens only "inside" the source and load.

*snip*

Thus we see the "true" purpose of the wire is just charge conduit to maintain an certain equilibrium where there is a flow of electrons through both the source and the load but no net change in the total charge of the system, charge is cycled in a circular manner between the source and load, a "circuit" if you will

The actual "energy" is coupled via the electric field.

Yup, inside the wire.

I'll say it explicity in case it's not clear what I'm getting at. The Veritasium video is all about the power/energy flowing outside the wire.
At DC that doesn't happen, it's inside the wire. Whatever physics mechanism you want to use the describe that doesn't matter, because whilst there is an external magnetic field around the wire at DC, it's not moving. And therefore by definition the mechanism of power transfer must hence be inside the wire.

[han]

At low frequency (DC) the Electro-magnetic field the effect is very small.
At high frequency the electron flow inside of wire is decrease in effect, as the wave became more dominants force.

[aneevuser]

If so then that implies skin effect at DC.

I'm far from convinced by the argument in the StackExchange post - to feel the Lorentz force, an electron would have to feel the B field generated by the other electrons in the wire - but the other electrons are stationary in the rest frame of any given electron, so should feel no B field due to them, as far as I can see.

[vk6zgo]

Not Electron flow again.   

No, his argument would work as well if he was talking about "pretend positive charge carriers".
I think everybody knows that energy flows through the interaction of Electrical & Magnetic fields----I just don't think it is quite as simple as he makes it.

For a start, his 300 million metre long cable is really a bunch of series inductors & parallel capacitors, so any practical lighting of his lamp will take longer than with an ideal cable.

[aneevuser]

I'll say it explicity in case it's not clear what I'm getting at. The Veritasium video is all about the power/energy flowing outside the wire.
At DC that doesn't happen, it's inside the wire. Whatever physics mechanism you want to use the describe that doesn't matter, because whilst there is an external magnetic field around the wire at DC, it's not moving. And therefore by definition the mechanism of power transfer must hence be inside the wire.

By magnetic field "not moving", I guess you mean that it's time-independent? If so, this seems to be irrelevant to Poynting vector based energy transfer arguments - take a look at the final example in section 27.5 of the Feynman Lectures, where he talks about energy circulation in a system comprising a stationary bar magnet and stationary charge:

https://www.feynmanlectures.caltech.edu/II_27.html

[EEVblog]

I'll say it explicity in case it's not clear what I'm getting at. The Veritasium video is all about the power/energy flowing outside the wire.
At DC that doesn't happen, it's inside the wire. Whatever physics mechanism you want to use the describe that doesn't matter, because whilst there is an external magnetic field around the wire at DC, it's not moving. And therefore by definition the mechanism of power transfer must hence be inside the wire.

By magnetic field "not moving", I guess you mean that it's time-independent? If so, this seems to be irrelevant to Poynting vector based energy transfer arguments - take a look at the final example in section 27.5 of the Feynman Lectures, where he talks about energy circulation in a system comprising a stationary bar magnet and stationary charge:

https://www.feynmanlectures.caltech.edu/II_27.html

That does not really explain how the power/energy flows through a DC circuit. And certainly not how it flows outside the cable.

[rs20]

But energy is flowing, how does it flow? What's doing the moving at DC?

The electric field is just static, I don't think there's any debate there. The magnetic field arises from the drift velocity of the electrons in the wire (and before you say "but that's only millimeters per hour" or whatever, see my site for a proof/justification that even a tiny drift velocity + relativity leads to accurately calculated magnetic effects. So yeah, the magnetic field arising from the drift of the electrons X the electric field between the wires leads to this apparent flow of power/energy.

But power is being continuously delivered from the source to the load under steady state conditions. How if the magnetic field is not moving?

The Poynting vector is E x H, not E x dH/dt or something like that. So not sure why you're asserting that the magnetic field must be moving or changing over time for the Poynting vector to be nonzero?

(This raises an interesting question: if you have a charged capacitor and an orthogonal permanent magnet, does that not mean the dielectric of the capacitor sees an E gradient and an orthogonal H gradient, leading to a non-zero Poynting vector implying a flow of power, even though I just described a completely static situation? Answer is at https://en.wikipedia.org/wiki/Poynting_vector#Static_fields if you want to dive in. It gets pretty wild, a "circular flow of electromagnetic energy" ensues, which doesn't actually carry any energy anywhere and somehow is even necessary for angular momentum to be conserved (?).)

Key point: Here's the thing, if you compute the Poynting vector across an entire plane slicing through a battery/lamp circuit, you'll find the Poynting vector is zero within most/all the cross section of the wires, and the integrated total of the Poynting vectors in the free space outside the wires will perfectly match the P=IV calculations you expect from the circuit. All despite the magnetic field being static/non-moving/time-independent. I think that makes the viewpoint that the energy/power is flowing outside the wires justifiable enough -- no more justifiable than normal circuit theory or anything, but just in the sense that you can have perfectly self-consistent yet very different-looking explanations of things. (E.g. gravity as a force vs gravity as curvature in spacetime: both views hold a lot of predictive power and value, but they look totally incompatible to each other in a philosophical sense)

I'm far from convinced by the argument in the StackExchange post - to feel the Lorentz force, an electron would have to feel the B field generated by the other electrons in the wire - but the other electrons are stationary in the rest frame of any given electron, so should feel no B field due to them, as far as I can see.

Be careful with moving rest frames; you might be neglecting the positively charged nuclei which are now appearing to move backwards with respect to your moving electrons.

[sandalcandal]

Now we ask "how is the energy being conveyed?" There is no electric field in the (ideal) wires, any flow of charge through the wires has no generation/loss of electrical energy by definition of the wires being ideal wires in a DC system. In fact, in our theoretical system, the perfectly matched source and load dipoles effectively (but not completely apart from within the wire) cancel each other out with only substantial fields in a parallel region between the dipoles. Again, where is the energy "coming and going"? the creation and loss of electrical energy (from and into other forms) happens only "inside" the source and load.

*snip*

Thus we see the "true" purpose of the wire is just charge conduit to maintain an certain equilibrium where there is a flow of electrons through both the source and the load but no net change in the total charge of the system, charge is cycled in a circular manner between the source and load, a "circuit" if you will

The actual "energy" is coupled via the electric field.

Yup, inside the wire.

I'll say it explicity in case it's not clear what I'm getting at. The Veritasium video is all about the power/energy flowing outside the wire.
At DC that doesn't happen, it's inside the wire. Whatever physics mechanism you want to use the describe that doesn't matter, because whilst there is an external magnetic field around the wire at DC, it's not moving. And therefore by definition the mechanism of power transfer must hence be inside the wire.

This is why I think (and probably others) that this is getting a bit philosophical, akin to the Lewin vs Electroboom. How you you define where energy is "flowing"?

There is no electric field in the (ideal) wires, any flow of charge through the wires has no generation/loss of electrical energy by definition of the wires being ideal wires in a DC system.

What is "flowing"? The electrons are "flowing" but the flow of electrons has no associated energy interactions while inside the wire so how is the energy flowing "through" the wire? The energy exchange only happens inside the source/load and it only happens there due to the electric field present there.


Consider, instead of a "wire" we had some magical device we could attach to the terminals of the load which allowed free sinking and sourcing of charge (like an earth but allowed to be different potentials?) Then we brought in the battery or otherwise had the system setup with the same electric field as with the battery connected. Again, no wire connecting the battery and the load. The load would still see an electric field and a steady flow of charge hence energy/power from the electric field.

Then consider the opposite, connect the source and load with a wire but remove the electric field (and still have a current flowing in the wire even). There will be no energy.

So what's more important, the wire or the electric field?

On the other hand, the electric field is "static" in the DC state though so how can that be "flowing" energy?

[Ultimately the different models are self-consistent and make equivalent predictions]

For the first hypothetical, the magical stubs end up replacing the potential "from the wires" i.e. you effectively just connected the load to another battery. For the second hypothetical, you have to make the load resistance zero and you just have current source connected to an ideal wire loop. For the DC system to work as described/expected, you need both an electric field/voltage potential present and circular transfer of charge.

Perhaps there is again some linguistic/definition difference where for something to be considered a part of an energy system in typical academic physics it must have an associated loss/gain of energy. I think in this case if one want to be really "strict" this sort if thing gets resolved by some (perhaps virtual) particle transfer between the source and load in standard model quantum physics? [rs20's description with the proper Poynting vector description works.]

I also edited in a sketch to my previous post. https://www.eevblog.com/forum/chat/veritasium-(yt)-the-big-misconception-about-electricity/msg3829100/#msg3829100

[rs20]

One question for you Dave (because I'm going to go out on a limb and guess that your next objection is me conflating "power flowing in a particular place" with "magnitude of Poynting vector", i.e., just accepting Veritasium's video at face value):

Veritasium (and many physicists) propose the Poynting vector, S = E x H as the best way to answer the question "how much electromagnetic power is flowing in this particular point in space". If you don't like this, what do *you* propose as an alternative? IMHO P = IV isn't really up to the task of answering "how much power is flowing in this particular point in space", because I and V aren't fields in space (or at least, V isn't well defined until you choose a ground point).

Or, do you agree that the Poynting vector is the right formula to use, but disagreeing with Veritasium about what that field actually looks like in the case of a simple steady state battery+lamp circuit?