Energy flux direction in a conducting wire?

[fluidistic]

I don’t think that the YouTube discussion you reference was talking about any thermal flux. They were specifically describing the electromagnetic flux.

I know. I am also talking about an energy flux, the internal enerfy flux, whixh I expected to equal the poynting flux. The energy flux can be decomposed into a thermal flux (that I can neglect in isotgermal conditions), and an energy flux due to the current, worth mu times J. By neglecting.the thermal flux, only the em flux should remain, and this should match the poynting flux, though I don't get this mathematically. I get that their div must match, even though their directions are different. Both are able to yield.the joulean dissipation in the material.

 

[fluidistic]

Ok, I think I fibally figured the mystery out! Skipping all mathematical steps, insode the material, S is equal to kappa grad T (note that the direction makes sense, and the fact that they vanish at the center too). There is a non zero poynting vector inside only if the conductivity is finite. If thr thermal gradient ie neglected then it implies that.the electrical conductivity is infinite. Everythibg makes srnse. And claiming thst the ebergy doesn't flow in wires cannot make sense as soon as the electric conductivity is finite. And when it is infinite I am not sure yet, but there is no poynting vector inside the material for sure, no thermal gradient. I am not sure if there is an ebergy associatee to the current, i suppose yes but not a hindrede percebt sure.

Edit 2: when i take the limit of large sigma, i get that grad T and therefore S become small inside.the wire. However, mu times J becomws big, and so does.the energy flux in the current's direction. Taking the limit rho the resistivity equals 0 shows that the energy's direction is pureley along the wire, and non null. The videos are wrong in their claim, i believe.

 

[vanhees71]

Sure, microscopically the resistance/electrical conductivity is due to friction and indeed the energy is converted to heat. The total flux of energy per unit time indeed is $R I^2$, as demonstrated in Sect. 3.5.3 (with the bad clash of notation using $R$ all of a sudden also for the resistance, which I'll correct in a moment).

 

[fluidistic]

I will write up my answer when i get the time in front of a computer instead of my phone. I have everything figured out. The interesring questions that remain are: 1)what would happen in the case of a superconduxtor, how does it differ from a conduxtor with rho equals 0? 2) what fraction of the total energy of tge power source ends up in the wire, as rho is increased from 0 up to. a finite value? So that we'd know if the conduxtivity impacts on the quantity of the total energy being stored in the em fields outide the wire vs inside of it. I already know that for an ideal conduxtor, there is an energy flux flowing along the wire inside every point in the wire, contradixting the claim that energy doesn't flow in the wires.

Sure, there is a joule heat when rho isn't zero, there was no problem with that part. Note that it's not a flux, it's a volume heat generated, it doesn't have a direction if you prefer.

 

[fluidistic]

The long awaited promise has been fullfilled. I join a pdf explaining what is going on. There are still a few things to figure out, but what is crystal clear is that.there is an energy flux along the direction of the wire, regardless of the value of.the resistivity. This goes against.the claim of many youtubers, some of which are experts in electronics, or scientific popularizers.
Have fun!

 

[Baluncore]

My argument is that electrical energy does not propagate from the generator to the load along the inside of the conductor. You seem to be saying that the joule heating is propagated into the wire, which I agree with. But I don't believe joule heating propagates useful electrical energy from the generator to the load.

I have trouble knowing when the energy you are considering is thermal waste energy, or useful electrical energy. It is important to distinguish between the thermal energy wasted and the transmitted electrical energy.

How do you say electrical energy is propagating from the generator to the load along the inside of the conductor, without it being wasted as heat.

 

[fluidistic]

My argument is that electrical energy does not propagate from the generator to the load along the inside of the conductor. You seem to be saying that the joule heating is propagated into the wire, which I agree with. But I don't believe joule heating propagates useful electrical energy from the generator to the load.

I have trouble knowing when the energy you are considering is thermal waste energy, or useful electrical energy. It is important to distinguish between the thermal energy wasted and the transmitted electrical energy.

How do you say electrical energy is propagating from the generator to the load along the inside of the conductor, without it being wasted as heat.

I should wait to go home before replying hastily, but here it goes.
I say that joule heat is produced everywhere in the wire, at every point if you prefer, it is homogrneously produced when no material properties depend on temperature, which I assumed it holds to keep things simple. In steady state, the joule hest generated in any volume element must be conducted out of that element. In particular the heat flux is radial in this homogeneous case (stuff can get much messier if the thermal conductivity is anisotropic and thermoelectric effets are taken into account).

I say that.there is an energy flux that is proportional to the electric current density, i.e. it has the same direction. If we take a superconductor where no power is lost by dissipation, I guess I am saying that the fact that there's a non vanishing electric current implies that there is an energy flux (even though Poynting vector vanishes inside the SC wire) going through the wire. If I integrate this energy flux with respect to position along the wire, I should get a non zero energy. I think this energy is due to the fact that the electrochemical potential does not vanish, and neither does the electric current. (Is this true also in the case of Cooper pairs? I guess so.)

I have just found 2 refs of papers using the same terms albeit with a different notation, regarding the total energy flux. Paper by Callen called The application of Onsager's reciprocal relations to thermoelectric, thermomagnetic, and galvanomagnetic effets''. Paper by Domenicali ''Irreversible thermodynamics of thermoelectricity''. I think they address your question regarding ''wasted'' energy. Callen says entropy increases because of heat conduction and because of the degradation of electrochemical potential (essentially this means that.there's a potential drop). Both vanish when the resistivity vanishes, so in that case no entropy increase, no wasted heat.

 

[vanhees71]

The thermodynamical approach to electrodynamics is rarely treated in the literature afaik. A nice EJP paper is here:

https://iopscience.iop.org/article/10.1088/1361-6404/aa9caf

 

[fluidistic]

Some more thoughts. Maybe the "extra" energy flux term $\overline{\mu}\vec{J_e}$ (note that there's a slight units mistake, the charge of the electrons should hang around, but this doesn't change the discussion) accounts for an energy the power source needs to "use" initially, but needs not to maintain, in the steady state for example.
The thing is, an ideal wire (zero resistivity/resistance) with a current has more energy than the same wire with zero current. This also means that when the current goes away, this energy has to go somewhere (it cannot go away by Joule heat if the resistivity is null), maybe it is radiated away, I do not know. But I do know there is an energy due to the current (comes from integrating $\overline{\mu}\vec{J_e}$ over the volume of the wire).

What I could do soon, is to make a sketch and draw $\vec{J_U}$, $\kappa \nabla T$ (or the Poynting vector if you prefer), and $\mu \vec J_e$, to get the full mental picture of how the energy is flowing inside the wire. When the resistivity is non zero, $\mu$ grows/decrease (I'd have to check the signs) when going along the direction of the current density $\vec J_e$ is constant throughout the volume, so this means that this energy term the youtubers have missed either grows or decrease along the wire, i.e. it's not constant. It is constant radially, for a given position along the wire. The Poynting vector follows what the termal gradient does, i.e. vanishes at the center of the wire and increases as we move away from the center radially. It does not depend on the position along the wire, only depends on the radial coordinate. The total energy flux (which correspond to the internal energy flux) is the sum of these 2 energy fluxes.

In the case of zero resistivity, the first energy flux doesn't increase/decrease anymore along the wire ($\mu$ is constant), and it is non zero.

Therefore, in every single possible cases imaginable, there is an energy flux that goes along the wire, and energy "flows" inside the wire, no matter what happens to the Poynting vector (could vanish, or not, doesn't matter).

 

[fluidistic]

The thermodynamical approach to electrodynamics is rarely treated in the literature afaik. A nice EJP paper is here:

https://iopscience.iop.org/article/10.1088/1361-6404/aa9caf

Thanks for the ref. Lots of insights in there. Eq. 21 is pretty similar to what I wrote in my doc (similar notation too). From a quick glance at the paper, he refers to "ionization energy" for what I have as $\overline{\mu}\vec{J_e}$ but there is more to it, as he splits an electronic contribution to a lattice one...

 

[bob012345]

It seems confusing to me to talk about energy flowing through or around a wire instead of some form of potential energy. Consider a simple circuit consisting of a battery and a resistor and wires completing the circuit in a loop. If energy flows out of the battery and then is virtually all consumed in the resistor as heat, what flows out of the resistor and through or around the wire back to the battery? The current still flows the same in both paths into and out of the resistor. Aren't the Poynting vectors the same? I'm confused. No, I'm really confused.

 

[vanhees71]

I think @fluidistic 's manuscript is pretty clear though I've not checked the details carefully yet. It's clear that here you have a current in a conductor due to electrons, which are driven by the electric field and subject to friction through scatterings with the ion lattice of the metal, which leads to dissipation and thus the production of heat. The manuscript treats this as an energy-transport process in the usual way of hydrodynamical transport equations. The Poynting vector is of course the em. part of the energy flux needed in the energy balance, including in addition the heat flow.

 

[fluidistic]

Thanks Vanhees71. I promise a new version soon (I'm losing my job tomorrow, I may have more free time soon). I have reorganized the doc. a little bit, and there is more juice to squeeze in order to obtain an accurate sketch of the energy flux. For.example, the fact that the potential drop is linear w.r.t. z (because the current density is uniform), etc.

 

[fluidistic]

Here's the new doc.

 

[Baluncore]

Here's the new doc.

"I investigate the claim that energy doesn’t flow in wires, but in the space around them. The claim comes from focusing on the Poynting vector and by assuming that the conducting wire of a circuit has no resistance. However, it turns out that there is energy ”flowing” in the wire in all cases, including when the resistivity vanishes."

You are making a different claim. The web claim is that electrical energy from a generator, that reaches the load, does not travel inside the conductor(s). The web claim does NOT assume the conducting wires of the circuit have no resistance. We know real wires get hot, but that thermal energy does not reach the load as useful electrical energy.

 

[fluidistic]

You are making a different claim. The web claim is that electrical energy from a generator, that reaches the load, does not travel inside the conductor(s). The web claim does NOT assume the conducting wires of the circuit have no resistance. We know real wires get hot, but that thermal energy does not reach the load as useful electrical energy.

Well, they are making the claim that the energy doesn't travel in the wire (and that Poynting vector points radially inward the wire, this part is correct). Veritasium says that the energy flux flows one-way from the battery to the lightbulb when an AC is used. But this is wrong. He also says the energy that flows out of the battery ends up in the load and doesn't come back to the battery, whereas in reality there's an energy flux component that follows the current's direction, so that statement is also wrong. Also, I already wrote earlier that I focused on the claim ''Energy doesn't flow in wires'', which btw is the thumbnail of Veritasium's video. I found this shocking since I thought there was some energy flowing there (that.energy comes from the battery by the way).

Overall they focus only on the Poynting vector part, missing out an energy flux that goes like the current density.

 

[hutchphd]

I found this shocking since I thought there was some energy flowing there (that.energy comes from the battery by the way).

So your point can be summed up as :
"Of course the energy flows in wires, they get hot"

And your paper explains that there is heating that is quadratic in J. I don't see why this disputes the Veritasium picture at all.

 

[vanhees71]

You are making a different claim. The web claim is that electrical energy from a generator, that reaches the load, does not travel inside the conductor(s). The web claim does NOT assume the conducting wires of the circuit have no resistance. We know real wires get hot, but that thermal energy does not reach the load as useful electrical energy.

Of course not. Also in the writeup the standard Ohm's Law $\vec{j}=\sigma \vec{E}$ is assumed, which means you have dissipation. The energy flow is indeed not along the wire but through the em. field outside of it. Nevertheless the heat is generated due to the "friction" of the electrons inside the wire, and this is described in the manuscript by heat transport, making use of the standard thermodynamical approach, which is however rarely found in the textbook literature, which indeed is a pity.

"Zero resistance" is of course not described by simply making $\sigma \rightarrow \infty$. This doesn't make sense mathematically. If you want to describe a superconducting wire you have to use other constitutive equations like the London theory. A modern quantum-theoretical approach can be found in the Feynman lectures vol. III (it's a gem!).

 

[Baluncore]

Of course not.

Of course not what?
@vanhees71 I think we agree with each other.

To put it really simply.
The energy that reaches the load flows outside the wire in the direction of the Poynting vector.
Some Poynting vector turns into the resistive wire and becomes I²R heat.

 

[vanhees71]

Of course not what?
@vanhees71 I think we agree with each other.

Zero resistance. The naive assumption of a zero-resistance material, i.e., simply making $\sigma \rightarrow \infty$ makes no sense. I also think we agree with each other.

To put it really simply.
The energy that reaches the load flows outside the wire in the direction of the Poynting vector.
Some Poynting vector turns into the resistive wire and becomes I²R heat.

Exactly. Microscopically it's the friction of the conduction electrons.

 

[Baluncore]

Microscopically it's the friction of the conduction electrons.

If the energy flux inside the conductor travels with the electrons, does it start at the generator, flow through the load, then close the circuit by returning with electrons to the generator ?

 

[vanhees71]

The energy flux doesn't travel with the electrons. If this were the case it would take several minutes until you have light when switching it on. One should not forget that the drift velocities of the electrons making up house-hold currents are on the order of 1mm/s (millimeters per second!). The energy transport is indeed descrbied by the Poynting vector and that's why the signal propates with (nearly) the speed of light rather than via the crawling electrons making up the current. The heat is then generated locally at the place of the electrons through friction of these electrons with the ion lattice making up the wire.

 

[bob012345]

The energy flux doesn't travel with the electrons. If this were the case it would take several minutes until you have light when switching it on. One should not forget that the drift velocities of the electrons making up house-hold currents are on the order of 1mm/s (millimeters per second!). The energy transport is indeed descrbied by the Poynting vector and that's why the signal propates with (nearly) the speed of light rather than via the crawling electrons making up the current. The heat is then generated locally at the place of the electrons through friction of these electrons with the ion lattice making up the wire.

Yes the drift velocity is very slow but the signal transmitted by those electrons is very fast. An analogy is to move a one meter rod by one millimeter per second. The signal through the lattice defined here by the time it takes for the other end to begin to move is not instantaneous because nothing is perfectly rigid but it is extremely fast compared to the rod. I think the same is true for electron drift currents. The current itself moves slowly but it is set up around the whole circuit virtually instantaneously.

Also consider how the fields in the Poynting vector are set up virtually instantly if not through the electrons in the wire and all along the wire? Fields do not mysteriously jump out of wires at the generator source, flow along wires and suddenly jump back into the loads. Of course, I may be totally out to lunch.

Further consider, if the energy is carried outside the wire but used up in the load, what exactly is flowing outside the return wires if the current, fields and thus the Poynting vectors are the same? I think the Poynting vector must be different on the return path from the load.

 

[nasu]

@bob012345 Do you have a problem with light from a star reaching Earth without any electrons beeing involved in between? How are the fields jumping through space without wires or electrons pushing each other?

 

[bob012345]

@bob012345 Do you have a problem with light from a star reaching Earth without any electrons beeing involved in between? How are the fields jumping through space without wires or electrons pushing each other?

I think you are missing my points. The drift current may only 1 mm/s but it does not take several minutes for the drift current itself to exist throughout the whole circuit. That is set up virtually instantaneously. That was my first point. The fact that fields can exist in space without a medium of any kind is not the issue here. That is not a proof that the fields around a wire in a DC circuit are not intimately connected to the electrons in the wire which was my second point. My last point was merely an observation that if energy flows along outside the wire, it must be less after the load on the return path that before it or energy would not be conserved. Do you disagree?

 

[nasu]

@bob012345 This is what I was talking about, from your post:
"Fields do not mysteriously jump out of wires at the generator source, flow along wires and suddenly jump back into the loads."

 

[bob012345]

@bob012345 This is what I was talking about, from your post:
"Fields do not mysteriously jump out of wires at the generator source, flow along wires and suddenly jump back into the loads."

Well of course you can make them do just that with antennas but I meant in the context of this discussion. The generator, such as a battery in a DC circuit, is not an antenna that send out fields into space, independent of the wires as I understand it. Again, I might be completely wrong in my understanding.

 

[vanhees71]

What's discussed in my above quoted writeup or in Sommerfeld, Lectures on Theoretical Physics, Vol. 3 was the DC case for a coaxial cable (an example chosen, because it's pretty easy to calculate), i.e., that's valid only for the situation after a sufficiently long time the circuit is "switched" on and you are in the stationary state ("magneto statics"). Then the energy transport from the source ("battery") along the cable is clearly due to the electric field in the free space between the coaxial conductors, because that's the only place, where the Poynting vector has a component along the direction of the wire (the $z$-axis in the calculation). This energy is dissipated into heat along the wire ("Ohmic loss"). You find the complete discussion also in my E&M manuscript, Sect. 3.5 (in German only):

https://itp.uni-frankfurt.de/~hees/publ/theo2-l3.pdf

To understand the transient state after switching on the circuit, you have to solve the wave equation (or "telegrapher's equation"). I guess you can also find this calculation somewhere, but I've no reference at hand at the moment.

 

[bob012345]

What's discussed in my above quoted writeup or in Sommerfeld, Lectures on Theoretical Physics, Vol. 3 was the DC case for a coaxial cable (an example chosen, because it's pretty easy to calculate), i.e., that's valid only for the situation after a sufficiently long time the circuit is "switched" on and you are in the stationary state ("magneto statics"). Then the energy transport from the source ("battery") along the cable is clearly due to the electric field in the free space between the coaxial conductors, because that's the only place, where the Poynting vector has a component along the direction of the wire (the $z$-axis in the calculation). This energy is dissipated into heat along the wire ("Ohmic loss"). You find the complete discussion also in my E&M manuscript, Sect. 3.5 (in German only):

https://itp.uni-frankfurt.de/~hees/publ/theo2-l3.pdf

To understand the transient state after switching on the circuit, you have to solve the wave equation (or "telegrapher's equation"). I guess you can also find this calculation somewhere, but I've no reference at hand at the moment.

Yes but originally this thread and my recent posts was about a conducting wire not a coaxial cable? Reading back through it it seems the discussion is confusing with different trains of thoughts intermingling. No pictures to help either...

 

[hutchphd]

If energy flows out of the battery and then is virtually all consumed in the resistor as heat, what flows out of the resistor and through or around the wire back to the battery?

If energy flows out of the battery and then is virtually all consumed in the resistor as heat, what flows out of the resistor and through or around the wire back to the battery? The current still flows the same in both paths into and out of the resistor. Aren't the Poynting vectors the same? I'm confused. No, I'm really confused.

Yes perhaps I can point out why.

Further consider, if the energy is carried outside the wire but used up in the load, what exactly is flowing outside the return wires if the current, fields and thus the Poynting vectors are the same? I think the Poynting vector must be different on the return path from the load.

The Poynting Vector shows the energy flow in the fields. There is no "return" path except by our naming convention

My last point was merely an observation that if energy flows along outside the wire, it must be less after the load on the return path that before it or energy would not be conserved. Do you disagree?

Their is no return path! I think you need to look at the Veritasium video again carefully. The Poynting vector always comes out of the battery. Everywhere. On one "leg" it is parallel to the current and on the other it is antiparallel. At the resistor it goes into the resistor. The pertinant image is at 8:00:


I hope that helps

 

[bob012345]

I'll look at it again but if there were a medium to completely dis-allow fields completely surrounding the wire do you say no energy would get to the resistor?

 

[Lord Jestocost]

Energy transfer in electrical circuits: A qualitative account
Igal Galili and Elisabetta Goihbarg
Am. J. Phys. 73, 141 (2005); doi: 10.1119/1.1819932

Understanding Electricity and Circuits: What the Text Books Don’t Tell You
Ian M. Sefton
Science Teachers’ Workshop 2002

Energy flow from a battery to other circuit elements: Role of surface charges
Manoj K. Harbola
American Journal of Physics 78, 1203 (2010); doi: 10.1119/1.3456567

 

[vanhees71]

Yes but originally this thread and my recent posts was about a conducting wire not a coaxial cable? Reading back through it it seems the discussion is confusing with different trains of thoughts intermingling. No pictures to help either...

The single wire is not different from the coax cable. It's only even less "realistic" ;-).

 

[bob012345]

I forgot it's just a shielded single wire.

 

[bob012345]

Energy transfer in electrical circuits: A qualitative account
Igal Galili and Elisabetta Goihbarg
Am. J. Phys. 73, 141 (2005); doi: 10.1119/1.1819932

Understanding Electricity and Circuits: What the Text Books Don’t Tell You
Ian M. Sefton
Science Teachers’ Workshop 2002

Energy flow from a battery to other circuit elements: Role of surface charges
Manoj K. Harbola
American Journal of Physics 78, 1203 (2010); doi: 10.1119/1.3456567

This is very useful! Thanks.