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

[Terry Bites]

Gentlemen vs Spanners and all that.

So whats this really about? Electrons do not flow in conductors, current is a wavefront of charge is whizzing along. Get over it, sorry. Tube aficionados should probaly go for a long walk now.
Veritasium does not understand that electric fields can be converted to magnetic fields and vice versa. This is how his baffling power transformer works- or not.
Heavyside’s work was (among other important stuff like inventing phasors analysis, impedance operators, making sense of Maxwell, giving us the terms curl and div....) on transmission line theory and in particular characteristic impedance and the importance of line termination.  All electrical conductors are transmission lines, even at DC. It follows that energy flowing in is an EM traveling wave. It has a static field component as well. A battery creates a charge gradient not a bucket of balls to pour down a wire.
This is not lost on chip designers- the interconnects on the IC are high speed transmission lines not just some sploge of metalisation. Every bit is transmitted can be viewed as as a line transient. If they waited for electons to actually flow none of it would work. Mr Ivor Catt (heritic and martyr to EM flow in condcutors) had similar ideas and sorted out early ic fabrication issues!

My first comment on this got bumped- what was that about Ref?
Made you laugh though!

[bdunham7]

It's certainly a bit of a matter of semantics.  "Power flow" is not really an observable property.  The power produced by the battery and dissipated in the resistor/lamp are easily defined but the power flow requires a bit more care.

With that and most of the other points I agree.  I understand that when you analyze current flow in the wires, you can't derive power even if you look at both wires individually.  You need to know the potential between them, and Poynting is a way of modelling that. But the function of a wire is to transfer charge from one end to the other with minimal loss.  It does that via a process that is internal (mostly) to the wire and heavily dependent on the characteristics of the wire.  And it is also true that the energy or power is never actually in the wire, just like a rotating shaft may transmit a huge amount of power but only have a tiny amount of energy contained in the angular momentum of the shaft itself.  So perhaps 'power flow' is best left aside.

My issue Vertiaseum with isn't over Poynting, which I probably passed an exam on many decades ago and certainly don't dispute any more than I dispute Faraday, it is with his casual mixing of two different issues in a way that I have seen cause confusion already elsewhere.  He is conflating the Poynting model for DC/LF with things like transmission lines and EM radiation.  This leads to people thinking that their speaker cables and electric cables in their wall work as a transmission line, among other things.  They don't.  Despite his example which theoretically can transmit power over a meter of space, that effect has nothing to do with how power gets to your house.  Instead, to the extent that it occurs, it is a parasitic effect that results in losses.  Yes, for very long power lines this effect is part of the model, but not because it is helpful and has been deliberately incorporated, but because it is unavoidable.

[sectokia]

1- When the switch is flipped an electron will accelerate (the first one at the switch itself). This causes a changing magnetic field. This field is mediated by a gauged boson - which for electromagnetism is the photon. So it 'travels' at the speed of light. The changing magnetic field will induce a movement of charge at the light bulb after 1m/c seconds. Veritasium defined his light bulb as lighting from *any* amount of current. So the induced current, no matter how small, turns on his light globe 'wirelessly'.

So does the capacitance between the wires.
Everyone with even fairly basic knowledge of electronics can understand the capacitor model. No guaged bosons or magnetic field knowledge required.
But hey, that's not the cool physics explanation, engineers are boring 

I guess my point is... all you need for this experiment is: electron, photon, and the electromagnetic field. You don't need capacitors, or transmission lines, and its a round about way to arrive at the same conclusion. Why not break it down the most elementary minimum of components? His bulb lights lights on any current. Therefore it can be 1 electron. Another electron 1 meter away accelerating will light the light bulb after 1m/c seconds due to electromagnetism. There is really nothing more that needs explaining.

Sure, multiple ways to explain it. Some people know more about or are more familiar with capacitors than photons and electromagnetic fields and virce-versa. The best explanation is the one that works for the individual.

With respect, the simulation in the video and capactor/inductor model does not give the answer as 1m/c. It gives the answer as 0s.

To get the actual answer of 1m/c you would need to bring in the electro/magnetic scaling constants - as their mean inverse is the speed of light.

For example how would distance change be simulated? By changing the value of C? This will not change the time to the instant spike.

[EEVblog]

With respect, the simulation in the video and capactor/inductor model does not give the answer as 1m/c. It gives the answer as 0s.
To get the actual answer of 1m/c you would need to bring in the electro/magnetic scaling constants - as their mean inverse is the speed of light.
For example how would distance change be simulated? By changing the value of C? This will not change the time to the instant spike.

Of course, it's assuming that you know that all electric fields and current travel at the speed of light (or a bit slower with a dielectric), which is something that Derek mentioned in the video. I should have mentioned that in the video, but it's fairly obvious that you have a 1m wide loop, so it takes the 3ns or so to traverse, which is implied by the answer 1m/c. Capacitors don't magically violate the speed of light.

[besauk]

With respect, the simulation in the video and capactor/inductor model does not give the answer as 1m/c. It gives the answer as 0s.

To get the actual answer of 1m/c you would need to bring in the electro/magnetic scaling constants - as their mean inverse is the speed of light.

For example how would distance change be simulated? By changing the value of C? This will not change the time to the instant spike.

Not to be pedantic or anything, but the transmission line approximation doesn't seem to be fully correct here.   For the first 1/2 second or so (until the pulse hits the far end of the wires), this system is a dipole antenna transmitter and a dipole antenna receiver located 1 m away.  It isn't quite right to consider the bulb side a part of the transmission line since it is not being actively driven - it is a passive receiver.  A proper solution would be a rather complicated near field analysis of the transmitter / receiver pair.  With that said, I won't discount a solution that is close based on a transmission line approximation - nor would I be surprised if it varied significantly.  Doing that type of near field analysis is, how shall I put it, *difficult*.  I'm inclined to try actual measurements of a scaled down version - perhaps a few meters on each side, and varying the separation.  My limited equipment may not be fully up to the task, but it's worth a shot.  I'm not so much interested in measuring the time delay of seeing a signal at the "bulb" (I don't think there is any question it will arrive at 1m / c), but I would like to see how much current could actually be delivered with this setup.

[EEVblog]

I still don't quite get his explaination at 7:35
He shows the Poynting vector S coming out from the battery in a DC circuit. How? There is no EM radiation loss.

[emece67]

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[EEVblog]

With respect, the simulation in the video and capactor/inductor model does not give the answer as 1m/c. It gives the answer as 0s.
To get the actual answer of 1m/c you would need to bring in the electro/magnetic scaling constants - as their mean inverse is the speed of light.
For example how would distance change be simulated? By changing the value of C? This will not change the time to the instant spike.

Not to be pedantic or anything, but the transmission line approximation doesn't seem to be fully correct here.   For the first 1/2 second or so (until the pulse hits the far end of the wires), this system is a dipole antenna transmitter and a dipole antenna receiver located 1 m away.  It isn't quite right to consider the bulb side a part of the transmission line since it is not being actively driven - it is a passive receiver.  A proper solution would be a rather complicated near field analysis of the transmitter / receiver pair.  With that said, I won't discount a solution that is close based on a transmission line approximation - nor would I be surprised if it varied significantly.  Doing that type of near field analysis is, how shall I put it, *difficult*.  I'm inclined to try actual measurements of a scaled down version - perhaps a few meters on each side, and varying the separation.  My limited equipment may not be fully up to the task, but it's worth a shot.  I'm not so much interested in measuring the time delay of seeing a signal at the "bulb" (I don't think there is any question it will arrive at 1m / c), but I would like to see how much current could actually be delivered with this setup.

I can now see where this can be confusing. Yes, it's not the traditional transmission line simulation because the driving source is not across the line with the load at the end. It's effectively two transmission line stubs off to the sides as illustrated, with the source across the lower wire in each transmission line, and the load on the other terminal of both.
But physically that's what it is, two long transmission line stubs shorted at the end. How you actually simulate that is indeed more complex. But the fact of the nearby cable capacitance is what causes the initial 1m/c loop is not wrong.
As mentioned, there are other ways to look at it, like as an antenna, and you'll get the same result.

[EEVblog]

I'm inclined to try actual measurements of a scaled down version - perhaps a few meters on each side, and varying the separation.  My limited equipment may not be fully up to the task, but it's worth a shot.  I'm not so much interested in measuring the time delay of seeing a signal at the "bulb" (I don't think there is any question it will arrive at 1m / c), but I would like to see how much current could actually be delivered with this setup.

Yes, the more I think into this, the more convinced I feel that energy reaches the bulb 1 m/c after switching on, but not yet sure about the amount of energy or the time needed to achieve steady state. But, as you, my equipment is not suitable for the experiment.

I don't think it's really matters how long it takes to settle down and how the lamp lights up over time. It's just a basic question of how quickly does it initially light.
Certainly doesn't matter for the question as proposed.

Also, not sure now if transmission line models are applicable here (at least I realized now that I applied it in the wrong way some thread pages ago).

You certainly don't have the use the transmission line model, it's just one way to look at it to give you the 1m/c result.

[emece67]

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[EEVblog]

My issue Vertiaseum with isn't over Poynting, which I probably passed an exam on many decades ago and certainly don't dispute any more than I dispute Faraday, it is with his casual mixing of two different issues in a way that I have seen cause confusion already elsewhere.  He is conflating the Poynting model for DC/LF with things like transmission lines and EM radiation.  This leads to people thinking that their speaker cables and electric cables in their wall work as a transmission line, among other things.  They don't.  Despite his example which theoretically can transmit power over a meter of space, that effect has nothing to do with how power gets to your house.  Instead, to the extent that it occurs, it is a parasitic effect that results in losses.  Yes, for very long power lines this effect is part of the model, but not because it is helpful and has been deliberately incorporated, but because it is unavoidable.

Yes, I see it the same way.

[EEVblog]

I still don't quite get his explaination at 7:35
He shows the Poynting vector S coming out from the battery in a DC circuit. How? There is no EM radiation loss.

It is a DC circuit, but not a dead one. There is obviously a energy transfer from the battery to the load, so there must be Poynting vector "field lines" going from the battery to the bulb. The Poynting vector is neither an AC nor a wave related concept, but a general EM concept applicable even when there are no waves and fields are static.

Yes, so shouldn't the Poynting vector be along the wires instead?
i.e. visually, how does the vector coming out of the battery know where the battery is physically?

[ejeffrey]

My issue Vertiaseum with isn't over Poynting, which I probably passed an exam on many decades ago and certainly don't dispute any more than I dispute Faraday, it is with his casual mixing of two different issues in a way that I have seen cause confusion already elsewhere.  He is conflating the Poynting model for DC/LF with things like transmission lines and EM radiation.  This leads to people thinking that their speaker cables and electric cables in their wall work as a transmission line, among other things.  They don't.  Despite his example which theoretically can transmit power over a meter of space, that effect has nothing to do with how power gets to your house.  Instead, to the extent that it occurs, it is a parasitic effect that results in losses. 

I can see the point there, all but the longest power lines are not transmission lines in the sense that there is no appreciable propagation phase along them.  Although the veritasium video didn't actually talk about transmission lines or do any transmission line theory, even though the example was a classic transmission line geometry. But it does answer questions like why does the light come on immediately when I flip the switch even though the electron drift velocity is so slow?   Some people try to explain that with a hydrolic analogy of tightly packed electrons pushing on each other, but that falls apart if you look too closely: the question then is what is the equivalent of the speed of sound?  No amount of looking at just the conductor will answer that question it is almost entirely determined by the dielectric between the wires with a small modification for skin effect. In fact if you want to show that you don't have to treat your power line or audio cables as transmission lines you need to figure out the phase velocity and this is how you do it.

Likewise poynting vector analysis clearly answers questions such as how is power transmitted through capacitors and transformers when there is no path for electrons at all?  I think it's pretty cool that the poynting vector power flow will go straight through a transformer basically as if it wasn't there even when there is no electric current, not even a displacement current connecting the primary and secondary.  To me this is the killer reason to say that the poynting vector approach is the best interpretation of "where is power transmitted in an electric circuit" even at LF/DC despite the fact that it can give somewhat unintuitive answers they are always consistent and properly connect sources to loads.

[bdunham7]

I still don't quite get his explaination at 7:35
He shows the Poynting vector S coming out from the battery in a DC circuit. How? There is no EM radiation loss.

I think you are correct and I think people are making mistakes in analyzing DC/LF circuits and expecting them to look like the results they get with EM radiation.

In this case if you draw the Poynting vector at the conductor, where the E-field is always going to be perpendicular, you should have an arrow drawn straight up the wire.

Edit:  And you need the three-finger version of the right hand rule for the Poynting vector, as I realized when I looked it up.  At any point on the conductors you do get arrows right up or down the wire, but as shown in the drawing you do get an arrow away from the battery.  I'm not sure what that means since this a simplified drawing and we don't have magnitudes.  And I twisted my wrist in the process..

[ejeffrey]

I still don't quite get his explaination at 7:35
He shows the Poynting vector S coming out from the battery in a DC circuit. How? There is no EM radiation loss.

It is a DC circuit, but not a dead one. There is obviously a energy transfer from the battery to the load, so there must be Poynting vector "field lines" going from the battery to the bulb. The Poynting vector is neither an AC nor a wave related concept, but a general EM concept applicable even when there are no waves and fields are static.

Yes, so shouldn't the Poynting vector be along the wires instead?
i.e. visually, how does the vector coming out of the battery know where the battery is physically?

The current through the battery generates a magnetic field circling the battery.  The electric potential across the battery represents an electric field axial to the battery.  The cross product of these are vectors coming radially outward  from the battery.   Near the battery the poynting vector is radially outward in all directions but once you add in the fields created by the wires you get a power flow that is generally confined to the region between and near the wires, but it doesn't especially hug the individual wires.

[rs20]

I still don't quite get his explaination at 7:35
He shows the Poynting vector S coming out from the battery in a DC circuit. How? There is no EM radiation loss.

i.e. visually, how does the vector coming out of the battery know where the battery is physically?

There is current flowing through the battery + wires, which leads to magnetic flux/field/whatever (H) surrounding the battery and wire, as shown by the blue arrows.
There is also an electric field (E) formed by the potential difference across the two wires, and it takes the form shown by the red arrows (now, let's not get sidetracked by the fact that an AA battery has a case consisting almost entirely of its negative terminal so that E field [and poynting vectors] would actually look very different to what's shown in the picture, it's not actually an important difference for the sake of your question.)

And using the three-fingers rule to find S = E x H, the Poynting vector ends up pointing away from the battery. So the vector "knows" where the battery is by inferring it from the electric and magnetic fields surrounding the battery, even though those are static.



If you were to repeat the same exercise with a resistor dissipating power (positive voltage still and top but current now travelling down the page), then E would be unchanged but H would be reversed, so S would be reversed (-S = E x -H), and now the Poynting vector would be pointing towards the resistor, indicating that power is being dissipated there.

[ejeffrey]

I think you are correct and I think people are making mistakes in analyzing DC/LF circuits and expecting them to look like the results they get with EM radiation.

In this case if you draw the Poynting vector at the conductor, where the E-field is always going to be perpendicular, you should have an arrow drawn straight up the wire.

Yes.  The wires have no appreciable longitudinal electric field, so in their neighborhood the Poynting vector is mostly parallel to the wire.  The battery and load however have axial electric fields so the Poynting vectors go out or in of them respectively.

[emece67]

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[emece67]

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[bdunham7]

Some people try to explain that with a hydrolic analogy of tightly packed electrons pushing on each other, but that falls apart if you look too closely: the question then is what is the equivalent of the speed of sound?  No amount of looking at just the conductor will answer that question it is almost entirely determined by the dielectric between the wires with a small modification for skin effect. In fact if you want to show that you don't have to treat your power line or audio cables as transmission lines you need to figure out the phase velocity and this is how you do it.

Of course the hydraulic analogy falls apart, it doesn't even work properly for hydraulics if you don't factor in a lot additional factors.  The equivalent of the speed of sound in the wire is very nearly the speed of light unless you have a 'retardation' effect (as hypothesized by Thomson) from external sources. 

Likewise poynting vector analysis clearly answers questions such as how is power transmitted through capacitors and transformers when there is no path for electrons at all?  I think it's pretty cool that the poynting vector power flow will go straight through a transformer basically as if it wasn't there even when there is no electric current, not even a displacement current connecting the primary and secondary.  To me this is the killer reason to say that the poynting vector approach is the best interpretation of "where is power transmitted in an electric circuit" even at LF/DC despite the fact that it can give somewhat unintuitive answers they are always consistent and properly connect sources to loads.

I'm certainly not refuting Poynting.  I really haven't run into anyone before that uses that method on a regular basis and wasn't aware that it was a generally practical approach to most problems.

[bdunham7]

Another question. Suppose now that, in the original Veritasium circuit layout, the 1m long wire segments at the extremes are not present. Will the bulb light when the switch is closed? Will it reach maximum brightness? How much time, if any, will it be lit?

Relativity tells us that we won't know whether the ends are connected for at least 1 second after the switch is on, so to whatever extent the bulb lights initially, it does so in either case.  If the ends turn out not to be connected, the bulb will only flicker and then go out.

[penfold]

As if there's not already enough to nitpick about the video... if the bulb is 1m away, we close the switch (we are standing on the side with the switch), it is surely important that we consider the time it takes the light (let's assume that even the tiniest change in IR radiation is detectable) to reach the observer. There will therefore be only one measurable duration and that is the sum of the switch to bulb propagation delay and the light to observer propagation delay: from that sum (assuming only the experimental setup shown can be used) one cannot, without prior knowledge or assumption of the correct answer, deduce the correct answer.

The correct answer is, therefore: "None of the above, P.S. the measurement setup we originally proposed and for which grant money was provided is inadequate, could we request some more money in order to extend our research capabilities? What do you mean, no? Why not? Yes I know we ignored the engineers who designed saying it wouldn't work... ...fine then!"

[IanB]

Yes, so shouldn't the Poynting vector be along the wires instead?
i.e. visually, how does the vector coming out of the battery know where the battery is physically?

I don't think there is a single Poynting vector, I think there are an infinite number of them corresponding to the infinite number of points in space where you could calculate it. However, if you take the integral over all points in space (add up all the infinitesimal vectors), the resulting "total" vector will end up pointing from the battery to the lamp. (If I understand correctly--I have no expertise in electromagnetics beyond the most elementary level.)

[Sredni]

I still don't quite get his explaination at 7:35
He shows the Poynting vector S coming out from the battery in a DC circuit. How? There is no EM radiation loss.

That's what comments in your video tried to tell you.
You have a fundamental misconception about how power is transferred in DC. In your video you say that the direction of the Poynting vector is away from the source only in AC, and throw in for some reason the skin effect, while in DC it would be pointing toward the wires and towards the battery!
You also think that that figure on Feynman is that of a highly conductive wire, while Feynman himself very clearly states it's a resistive wire.
That figure is basically the resistor.

I tried to warn you here:
https://www.eevblog.com/forum/blog/eevblog-1439-analyzing-veritasiums-electricity-video/new/#new

You might want to read

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

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

(Yes, that Jackson)

[EEVblog]

Yes, so shouldn't the Poynting vector be along the wires instead?
i.e. visually, how does the vector coming out of the battery know where the battery is physically?

I don't think there is a single Poynting vector, I think there are an infinite number of them corresponding to the infinite number of points in space where you could calculate it. However, if you take the integral over all points in space (add up all the infinitesimal vectors), the resulting "total" vector will end up pointing from the battery to the lamp. (If I understand correctly--I have no expertise in electromagnetics beyond the most elementary level.)

My spidey sense tells me that too.