Energy flux direction in a conducting wire?

[fluidistic]

On the Internet, I have read that the energy doesn't flow in the wire, for example in a very simple electric circuit made of a battery and a closed loop. When one computes the Poynting vector $\vec S \propto \vec E \times \vec B$, one gets that its direction is towards the center of the wire. The closer to the center of the wire, the smaller the magnitude of $\vec S$, and it vanishes right at the center. So far so good.

Now, from a thermodynamics point of view, there exist a relation between the internal energy $U$ and the electrochemical potential $\overline{\mu}$. This relation implies that the internal energy flux $\vec J_Q=\overline{\mu}\vec J$ where $\vec J$ is the current density (I am ignoring thermoelectric effects for simplicity here). However this means that the energy flux's direction is along the wire, not perpendicular to it, i.e. the direction is perpendicular to that of $\vec S$.

I am thus left confused. What is the direction of the energy flux created by the battery? Why do I get 2 different directions? Are these different energies? That's very confusing. I can derive Joule effect starting from any of the 2 energies mentioned above... so... shouldn't they be the same? If so, why do they have a different direction? What is going on?

 

[Baluncore]

It cannot be understood by considering only one wire. The analysis must consider the two wires that make up the complete circuit.

The electric field is guided by the two conductors from the generator to the load. Between those two conductors, the electric field is proportional to voltage, while the magnetic field is the sum of the fields generated by the currents in the two conductors, (or rather the currents are proportional to the magnetic field being guided by the surface currents on the wires). The cross product of the E and M fields is then the Poynting vector, the energy directed from the generator to the load.

The voltages and currents measured on conductors, are proxies for the electric and magnetic fields that direct the energy through the space between the wires.

 

[fluidistic]

It cannot be understood by considering only one wire. The analysis must consider the two wires that make up the complete circuit.

The electric field is guided by the two conductors from the generator to the load. Between those two conductors, the electric field is proportional to voltage, while the magnetic field is the sum of the fields generated by the currents in the two conductors, (or rather the currents are proportional to the magnetic field being guided by the surface currents on the wires). The cross product of the E and M fields is then the Poynting vector, the energy directed from the generator to the load.

The voltages and currents measured on conductors, are proxies for the electric and magnetic fields that direct the energy through the space between the wires.

I must say I am a bit confused about the setup. Hmm, magnetic field created by surface currents? I thought the problem assumed a constant and homogeneous $\vec J$ inside the wire. The magnetic field outside the wire is due to the current density of the whole wire, shouldn't it be that way, rather than surface currents?

Also, I do not really see why we need the 2 wires. I can just consider a battery attached to a loop of a single material, right? I mean, I understand that the Poynting vector exists outside the wire, and that it does penetrate the wire too. What I don't understand is why isn't the internal energy flux inside the wire pointing in the same direction than the Poynting vector.

 

[bob012345]

It cannot be understood by considering only one wire. The analysis must consider the two wires that make up the complete circuit.

The electric field is guided by the two conductors from the generator to the load. Between those two conductors, the electric field is proportional to voltage, while the magnetic field is the sum of the fields generated by the currents in the two conductors, (or rather the currents are proportional to the magnetic field being guided by the surface currents on the wires). The cross product of the E and M fields is then the Poynting vector, the energy directed from the generator to the load.

The voltages and currents measured on conductors, are proxies for the electric and magnetic fields that direct the energy through the space between the wires.

Are you referring to a sort of wave guide?

 

[Baluncore]

What I don't understand is why isn't the internal energy flux inside the wire pointing in the same direction than the Poynting vector.

The Poynting vector component into the wire represents the resistive I²R losses in the wire.

The surface of a conductor makes a very good reflector that keeps most of the energy outside the conductor. Energy that enters the wire will be lost as heat. Better conductivity makes a better mirror surface.

Are you referring to a sort of wave guide?

That is one way to consider the model.
The magnetic fields of the two conductors sum between the wires, but cancel away from the wires, so the circuit is really a two wire transmission line.

 

[bob012345]

The Poynting vector component into the wire represents the resistive I2R losses in the wire.

The surface of a conductor makes a very good reflector that keeps most of the energy outside the conductor. Energy that enters the wire will be lost as heat. Better conductivity makes a better mirror surface.That is one way to consider the model.
The magnetic fields of the two conductors sum between the wires, but cancel away from the wires, so the circuit is really a two wire transmission line.

Are arguing the waveguide is along each wire as an inherent property of the wire but you need to complete the circuit or are you arguing the fields go between two wires?

Also, what happens if you use wires that are shielded so there can be no field outside them?

 

[Dale]

Now, from a thermodynamics point of view, there exist a relation between the internal energy U and the electrochemical potential μ―. This relation implies that the internal energy flux J→Q=μ―J→ where J→ is the current density (I am ignoring thermoelectric effects for simplicity here). However this means that the energy flux's direction is along the wire, not perpendicular to it, i.e. the direction is perpendicular to that of S→.

Do you have a source for this? I am skeptical of it.

As you say the electromagnetic energy flux with the Poynting vector is clear and is directed radially inward inside the wire and roughly parallel to the wire outside the wire. This energy flux is derived directly from Maxwell’s equations, so it applies any time Maxwell’s equations apply.

I am not sure where this other flux comes from nor what assumptions are made in its derivation.

 

[Dale]

Are you referring to a sort of wave guide?

This is how a simple DC circuit behaves. No need to consider waves.

 

[bob012345]

This is how a simple DC circuit behaves. No need to consider waves.

Are these static fields then?

 

[Baluncore]

Are arguing the waveguide is along each wire as an inherent property of the wire but you need to complete the circuit or are you arguing the fields go between two wires?

The waveguide (or transmission line) is between the two wires carrying the equal and opposite (surface) currents.

The fields that propagate the energy are mostly through the insulation and space between the conductors. Only for DC or very low frequencies does the skin effect allow the time needed for the current and magnetic field to diffuse into, and flow deep inside the conductors.

Also, what happens if you use wires that are shielded so there can be no field outside them?

The fields remain inside the shield, between the conductor and the shield which completes the circuit.

 

[fluidistic]

Do you have a source for this? I am skeptical of it.

As you say the electromagnetic energy flux with the Poynting vector is clear and is directed radially inward inside the wire and roughly parallel to the wire outside the wire. This energy flux is derived directly from Maxwell’s equations, so it applies any time Maxwell’s equations apply.

I am not sure where this other flux comes from nor what assumptions are made in its derivation.

Yes I do have a reference. Non equilibrium thermodynamics by de Groot and Mazur. Essentially it's just the usual expression of dU in terms of mu dn that becomes a heat flux in terms of a particle flux, which is nothing but the usual current density.

 

[bob012345]

The waveguide (or transmission line) is between the two wires carrying the equal and opposite (surface) currents.

The fields that propagate the energy are mostly through the insulation and space between the conductors. Only for DC or very low frequencies does the skin effect allow the time needed for the current and magnetic field to diffuse into, and flow deep inside the conductors.The fields remain inside the shield, between the conductor and the shield which completes the circuit.

Thanks. I don't mean to sound obtuse but I don't have a clear picture in my mind of what you are saying. I would love to see a picture or diagram of how the energy flows around a simple DC circuit with a battery and a resistor.

Also confused how if there was a shield around the wire that that would complete the circuit?

 

[Dale]

Essentially it's just the usual expression of dU in terms of mu dn that becomes a heat flux in terms of a particle flux, which is nothing but the usual current density.

That doesn’t sound like an electromagnetic energy flux.

 

[Dale]

Are these static fields then?

They are any EM fields, static or not

 

[fluidistic]

That doesn’t sound like an electromagnetic energy flux.

Sorry, I didn't mean a heat flux. I meant an energy flux, of course.

As I wrote in my OP, I can derive Joule heat from it, just like with the poynting vector theorem. That does sound like an electeomagnetic flux, therefore?

 

[Dale]

Sorry, I didn't mean a heat flux. I meant an energy flux, of course.

Yes, but I don’t think it is an electromagnetic energy flux. Different forms of energy need not have parallel fluxes

As I wrote in my OP, I can derive Joule heat from it, just like with the poynting vector theorem. That does sound like an electeomagnetic flux, therefore?

No, why do you think it does? I don’t have that reference, so you will have to evaluate it. But it doesn’t sound like it was derived from Maxwell’s equations.

 

[Baluncore]

Are these static fields then?

The static fields of DC obey exactly the same mathematics as high frequency fields. DC is really just very low frequency AC, you build it and turn it on, then someone disconnects it, or the battery is exhausted later.

Also confused how if there was a shield around the wire that that would complete the circuit?

The inside of the shield is conductive, so it is a mirror. An equal and opposite current will be seen in the mirror, which completes the virtual circuit.
Why is a conductor a mirror? Because the incident magnetic field induces a perpendicular current in the conductive surface, which in turn generates an almost equal and opposite magnetic field that cancels into the mirror but is reflected back. In effect, turning left twice sends you back the way you came; i² = -1.

 

[bob012345]

The static fields of DC obey exactly the same mathematics as high frequency fields. DC is really just very low frequency AC, you build it and turn it on, then someone disconnects it, or the battery is exhausted later.The inside of the shield is conductive, so it is a mirror. An equal and opposite current will be seen in the mirror, which completes the virtual circuit.
Why is a conductor a mirror? Because the incident magnetic field induces a perpendicular current in the conductive surface, which in turn generates an almost equal and opposite magnetic field that cancels into the mirror but is reflected back. In effect, turning left twice sends you back the way you came; i² = -1.

Practically, in a simple DC circuit with say a 1 mm diameter wire, the energy that flows along the wire but outside it, what kind of additional radius essentially contain the fields?

 

[Baluncore]

... what kind of additional radius essentially contain the fields?

The radius is virtually infinite, that is how wireless signals radiate.

Both the E and M fields away from the two wires tend to cancel, so are very small. The magnetic fields between the wires reinforce, so the vast majority of energy is propagated between the wires. That is why two wire lines make such good transmission lines, and such poor antennas.

Any conductive material nearby will restrict the effective radius by shielding, so reflecting the small radiating fields back. Placing a two wire cable in a conductive conduit will minimise the radiation radius.

 

[hutchphd]

I am thus left confused. What is the direction of the energy flux created by the battery? Why do I get 2 different directions? Are these different energies? That's very confusing. I can derive Joule effect starting from any of the 2 energies mentioned above... so... shouldn't they be the same? If so, why do they have a different direction? What is going on?

Consider a battery, a length of twin conductor (flat untwisted) cable, and a resistor (>> battery resistance) at the far end. Estimate the fields between the wires. Calculate the Poynting Vector. Energy flows in the fields.
Questions?

 

[Dale]

what kind of additional radius essentially contain the fields?

The radius is virtually infinite, that is how wireless signals radiate.

There may be some energy far away from the wire, but the majority of the energy is quite close to the wire for DC circuits. I have seen this calculation before, but don’t recall the exact result. I will try to find it.

 

[Baluncore]

There may be some energy far away from the wire, but the majority of the energy is quite close to the wire for DC circuits.

I agree. There is almost no energy out there, but the radius is still theoretically infinite.

 

[fluidistic]

Consider a battery, a length of twin conductor (flat untwisted) cable, and a resistor (>> battery resistance) at the far end. Estimate the fields between the wires. Calculate the Poynting Vector. Energy flows in the fields.
Questions?

Yes. First, it's not as simple as this. In the original problem the current density is uniform in the wire. This means that a charge density is built up on the surface of the wire, and in principle should be computed so that it makes J constant throughout the wire. If you do the math, you should find that the charge builds up linearly from a terminal of the battery to the other. Then I hope the wire is circular, else it might become a nightmare to compute the field you ask. But that's irrelevant, I mean, I already buy that the Poynting vector points inward, in the wire. This is not.the problem.

The problem is that from a thermodynamics point of view, to a non vanishing particle number passing through a cross surface is associated an energy flux. And that this flux has the direction of the current, whereas the Poynting vector, which should be the energy flux, points in a direction perpendicular to this. That's the problem.

 

[Baluncore]

There is little electric field inside the conductor, so there is little energy propagating along the inside of the conductor.

Some of the energy does not reach the resistive load. The resistance of the conductor results in a small voltage drop along the conductor. Some of the energy therefore turns into the conductor and is converted to heat by I²R.

 

[vanhees71]

There is a very nice discussion of these issues for a DC in a coaxial cable in Sommerfeld, Lectures on Theoretical Physics, vol. 3.

 

[fluidistic]

There is little electric field inside the conductor, so there is little energy propagating along the inside of the conductor.

Some of the energy does not reach the resistive load. The resistance of the conductor results in a small voltage drop along the conductor. Some of the energy therefore turns into the conductor and is converted to heat by I²R.

I know this, this doesn't change the problem.
J equals sigma E (sorry on the phone, hard to write latex). Where E equals minus grad phi. The electrochemical potential is worth the chemical potencial plus e grad phi. Imposing the condition that the div of the energy flux vanishes in the steady state, one finds the heat equatiob in the wire, and we see that therr is indeed a joule term.

The problem that the direction of the energy flux does not match the direction of the poynting vector rrmains.

 

[Baluncore]

There is a very nice discussion of these issues for a DC in a coaxial cable in Sommerfeld, Lectures on Theoretical Physics, vol. 3.

Perhaps you can give the article number or the relevant figure number.

 

[vanhees71]

That's a textbook. You can of course find it online (though not legally ;-)).

You find a writeup about it also here (but in German):

https://itp.uni-frankfurt.de/~hees/faq-pdf/coax.pdf

For the relativistic part of the infinitely long cylindrical wire but without the discussion of the Poynting vector, see (in English):

https://itp.uni-frankfurt.de/~hees/pf-faq/relativistic-dc.pdf

 

[Baluncore]

That's a textbook.

I have a copy here of the 1952 English translation, but could not find the relevant section. I would have thought the section numbers and figure numbering would stay the same.

 

[fluidistic]

I am making progress. I thibk I figured out that JU and S are different. Because in steady state, div JU equals 0, and this lead to the heat equation of the wire, it contains the conduxtion term as well as a Joule term. Whereas in steady state, div S equals Joule heat. So the Poynting vector is a different energy flux than the interbal ebergy flux from thermodynamics. And so, maybe, all the youtubers claimjng that energy doesn't flow in the wires are wrong.

I wiol come back to you later, but it's as if S equals mu J, except that.the direction do not match.

 

[vanhees71]

I have a copy here of the 1952 English translation, but could not find the relevant section. I would have thought the section numbers and figure numbering would stay the same.

In this edition it's Paragraph 17 (p. 125ff). A nice diagram is Fig. 23 on p. 129 showing the electric field lines and the those of the Poynting vector.

 

[Dale]

The problem is that from a thermodynamics point of view, to a non vanishing particle number passing through a cross surface is associated an energy flux. And that this flux has the direction of the current, whereas the Poynting vector, which should be the energy flux, points in a direction perpendicular to this. That's the problem.

Why is that a problem? There is no reason to expect that the thermodynamic flux should be the same direction as the EM energy flux.

 

[Dale]

There may be some energy far away from the wire, but the majority of the energy is quite close to the wire for DC circuits. I have seen this calculation before, but don’t recall the exact result. I will try to find it.

I found the calculation for a coaxial cable in section 3.5.3 of this text courtesy of @vanhees71

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

It looks like the key point is that the energy density falls off as $1/R^2$

 

[fluidistic]

Why is that a problem? There is no reason to expect that the thermodynamic flux should be the same direction as the EM energy flux.

I expected that the internal energy flux would be equal to the EM ebergy flux in the isothermal case where grad T vanished.

I mathematically reached that div S equals div (mu times J). However I cannot conclude that S equals mu times J, whixh is pretty clear in that they do not share the same direction.

I just want to understabd what's going on regarding the energy flux inside the wire in such conditions, and in particular whether it is true or not that the energy doesn't travel inside the wire, as is claimed by famous youtubers.

 

[Dale]

I just want to understabd what's going on regarding the energy flux inside the wire in such conditions, and in particular whether it is true or not that the energy doesn't travel inside the wire, as is claimed by famous youtubers.

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