So what exactly is the use or the validity ...

Well, we already all agree that it is not so useful, at least not for engineers dealing with transmission lines. However, if you use Maxwell's equations at optical frequencies and try to calculate the physics of a laser system, you are going to bump into the concept of photons. The EM fields will need to be converted to equivalent photon flux and the interactions with atoms can be calculated from that perspective.

The validity is very important because Maxwell's equations are a classical limit of relatavistic quantum mechanics. It's nice to know the foundations of all theories and to know the limits of validity of the theory you are using.

 

If I were the OP here I would feel confused since even the experts can't agree.

Unfortunately his question has not been answered and the discussion is about a different one.

Let us be clear.

The velocity of light is the speed of propagation of a disturbance such as switching the power on or off, through the wires.

This is not the speed of propagation of electric power flow which is the speed of charge flow in the wires.

Charge does not flow at the speed of light or anywhere near it.

This speed is very much lower and it is this speed that should be used in the equation

λ = v/f

This then leads to more reasonable wavelengths for the ac mains asked about.

 

If I were the OP here I would feel confused since even the experts can't agree.

Unfortunately his question has not been answered and the discussion is about a different one.

I agree the discussion has diverged and has resulted in three different opinions that are likely to confuse, but I disagree that we did not answer his original question. I did that immediately in the second post (Papabravo's answer in post 3 is basically the same answer), but he never responded to let us know if this is sufficient.

 

With respect, Steve, the impression I received reading through the thread was that 'very long wavelengths as acceptable' were mentioned several times and it seemed to me that there was a tacit acceptance of the OP's use of c in the equation.

The actual speed of charge flow depends greatly upon conditions but may be only centimetres per hour, which point I don't think has really been brought out before.

 

This is not the speed of propagation of electric power flow which is the speed of charge flow in the wires.

This is the exact point I was disagreeing with Bill about. There is very little energy actually carried by charge. The main power flow is electromagnetic and is described by Poynting's Theorem.

I know that both you and Bill are aware of how slow the drift velocity is of electrons in a wire. Do you really want to wait that long for your light bulb to turn on?

I maintain that my answer in #2 and Papabravo's answer in #3 are the correct answers for the OPs question.

 

Here is a good reference that discusses this issue of power flow, which is always electromagnetic in nature. Pay particular attention to the section 8.3.1 which talks about 50 Hz power transmission. Also, note the section before that that descibes power flow to a resistor in a simple circuit. Even there, the point is made that no significant power flows through the wires, and it is the fields that carry the power (from source to load).

http://www.guspepper.net/electro/Primer%20semestre/Poynting%20vector.pdf

 

The velocity of light is the speed of propagation of a disturbance such as switching the power on or off, through the wires.

The disturbance is not the same as the effective displacement of charge.

The whole point of any form of wave action is that the disturbance moves (much) faster and farther than the motion (parameter variation) of the individual carrier particles.

 

The disturbance is not the same as the effective displacement of charge.

The whole point of any form of wave action is that the disturbance moves (much) faster and farther than the motion (parameter variation) of the individual carrier particles.

Exactly, and the disturbance in this case is an electromagnetic disturbance propagating at the speed of light. And, that is where the energy flow is.

 

But the charge is already at the other end of the wire.
The disturbance is merely the 'signal' that tells the charge to execute its parameter variation when, for instance we switch on.
Or more precisely the disturbance at the switch contacts tells the next element of charge along the wire, which tells the next element, which tells the next element and so on. It is this chain of action which propagates at the speed of light and results in near instantaneous activity from our light bulb.

 

But the charge is already at the other end of the wire.
The disturbance is merely the 'signal' that tells the charge to execute its parameter variation when, for instance we switch on.
Or more precisely the disturbance at the switch contacts tells the next element of charge along the wire, which tells the next element, which tells the next element and so on. It is this chain of action which propagates at the speed of light and results in near instantaneous activity from our light bulb.

Yes, i understand that, but that fact is not in conflict with anything I've said and is not in conflict with the reference I provided. Do you disagree with the reference I provided? If so, why? Can you provide a reference that supports your view? It seems to me that we now have 4 "experts" adding confusion. Let's step back and use outside sources to clear up the muddy water.

To help clarify what I'm saying I'll stress again that energy has to flow from one point to another. Charge traveling over a length of wire does not transport significant energy itself. The energy flow must be in the form of fields transporting energy, as shown with Poynting's Theorem, or it needs to be thought of as photons (i.e. actual particles with energy hv) that physically move from one point to another.

Consider the question posed and the answer given (about a source connected to a resistive load) in the reference I provide.

Question: How much power flows through the wires to the load?

Answer: For perfect wires, σ → ∞, and E → 0 within the wires. Thus P = E × H → 0, within the wires, and we conclude that NO power flows through the wires. From a field theory perspective, the power flows through the air. The power (i.e. the capacity to do work) is carried in the EM field, which travels from source to load. Why then do we need wires? The pair of wires serve to guide the wave from source to load. Transmission line theory is used to predict the behaviour of such guided waves when the length of the transmission line is comparable to the wavelength.

This is all consistent with what I've said throughout this thread. If the above reference is not viewed as suitable by the members here, I can start scanning pages from my EM field theory books. I only used this PDF reference because it is conveniently available. The author of the reference (http://www.guspepper.net/electro/Primer semestre/Poynting vector.pdf ) is (see the webpage of) Dr. Gustavo Perez M, Prof. of Physics, http://www.guspepper.net/

EDIT: I'll just add a quote from "Electromagnetics", by John D. Kraus, McGraw-Hill 1984, Section 10-19

"...Thus, integrating the Poynting vector over the resistor yields the same power VI, given by circuit theory. However, from the field point of view the power entering the load is negative power since S is inward, or negative, through the surface enclosing the load. ... It is evident that the power flow is through the empty space surrounding the circuit, the conductors of the circuit acting as guiding elements. From the circuit point of view we usually think of the power as flowing through the wires, but this is an oversimplification and does not represent the actual situation"

As my own note, I'd like to remind any reader that circuit theory is an approximation that is derived from field theory. Hence, any conflict in interpretation between the two must default to the higher theory, which is in this case Maxwell's field equations and the Poynting vector point of view.

The OP has asked a good and proper question, and a good and correct answer has been immediately provided. Unfortunately, he did not stick around to appreciate it.

 

I don't think we are talking conflict here.



However, doesn't the wavelength and velocity alter dramatically in conductors?

after Plonus.

A 60 hz wave in free space has a wavelength of λ = 5x10E6 metres and a velocity v =3 x 10E8 metres/sec. Upon penetrating copper (σ = 5.8 x 10E7 S/m) it becomes a wave of λ = 5.36 cm and velocity v = 3.2 m/s

Note the frequency does not alter in the quoted situation, ie it remains at 60 hz.

 

However, doesn't the wavelength and velocity alter dramatically in conductors?

Yes, it does, but waves don't typically propagate well in conductors due to losses, and we typically see the skin effect, which is an exponential decay into the conductor near the surface.

This means that wave propagation is in the dielectric space around the conductor and the speed is typically dictated by that medium. Hence, an open transmission line design would have most of the wave in air/freespace, with some of the wave in the wire insulation. The net result is a speed less than, but near the speed of light in vacuum.

A coaxial transmission line confines the wave completely to the internal dielectric and the wave speed might be one half or 2/3 the speed of light in vacuum.

I would expect that the fact that the field does penetrate into the conductor has some affect on speed, and it certainly impacts the wave attenuation.

 

Now let ω tend to zero in the equations.

As a matter of interest the Poynting vector has the units of Energy per area-second. (The vector bit comes from the fact that the area is a directed vector area.)
If you divide this by the energy density at any point you get another vector with the units of velocity. I have seen this called the energy velocity.

 

Now let ω tend to zero in the equations.

... while letting the circuit size increase proportionally with wavelength, ... or while maintaining the circuit size constant?

One has to define the limiting process more carefully. The OP asked about a circuit that is 6000 km long and at a frequency of 50 Hz.

 

I think someone said that the OP has long gone and we are now discussing round the subject and the theory need to work down to zero frequency.

Incidentally I don't think he was specifying a particular size of circuit, just worried by the apparent wavelength as calculated from velocity-frequency realation.

But surely the reverse mechanism to your proposal (and my figures) actually applies?

1) Low frequency voltage is applied directly to the wire.
2) This develops the EM field by the skin effect along the length of the wire

The velocity and wavelength in (1) will be the smaller figures, the velocity and wavelength in (2) will be the larger.

 

But surely the reverse mechanism to your proposal (and my figures) actually applies?

1) Low frequency voltage is applied directly to the wire.
2) This develops the EM field by the skin effect along the length of the wire

The velocity and wavelength in (1) will be the smaller figures, the velocity and wavelength in (2) will be the larger.

OK, I agree to forget about the OP.

I'm confused about what you are asking here though. I'm not sure what you mean by reverse mechanism.

In (1) low frequency implies larger wavelength, not smaller? Also, I'm not sure why velocity would be lower. I agree skin depth would be greater and would eventually completely penetrate the conductor, but the fields are still extensively in the free space around the wire.

In (2), I don't understand what you are saying at all.

Maybe you can reword what you are asking me?

 

OK, I agree to forget about the OP.

I'm confused about what you are asking here though. I'm not sure what you mean by reverse mechanism.

In (2), I don't understand what you are saying at all.

Maybe you can reword what you are asking me?

I think (2) is backwards, the skin effect is caused by the flow of EM energy in the free space around the wire, not the other way around.

http://amasci.com/elect/poynt/poynt.html

 

There is nothing electromagnetic about plugging a plug into a mains socket.

It is pure metal to metal contact and whatever interaction there is must therefore take place within the metal to start with.

My post #31 envisioned what would happen if a TEM impinged upon a wire not currently carrying current. In fact it only described the transmitted component.

You need to think outside the box (sorry textbook).

 

There is nothing electromagnetic about plugging a plug into a mains socket.

It is pure metal to metal contact and whatever interaction there is must therefore take place within the metal to start with.

If electrical energy is being moved electromagnetic fields are doing it even at DC. They might not radiate into free space as RF and be bound close to the transmission medium but it's all the same thing. The Metal just creates a point in space that routes the energy in a very low impedance zone around it. The basic thing is the current flow is not electrical energy, current flow is a property seen in matter in the presence of energy.

 

There is nothing electromagnetic about plugging a plug into a mains socket.

It is pure metal to metal contact and whatever interaction there is must therefore take place within the metal to start with.

My post #31 envisioned what would happen if a TEM impinged upon a wire not currently carrying current. In fact it only described the transmitted component.

You need to think outside the box (sorry textbook).

I'm still not following your logic and I don't understand what specific question you are asking me.

All circuit theory equations are dervied from electromagnetic equations, so it seems to me that you can't say that there is nothing electromagnetic about plugging into a mains. If you want to say that it is OK to approximate it that way, then I'll agree with you all day long.

The argument that "metal to metal contact implies that all interactions MUST be in the wire to start with" seems reasonable since current and charge are the sources of fields, and hence that is where they start. But, I don't think it considers the whole picture of the process after the start. The fields propagate out from the source. This propagation is an electromagentic effect during any transients. Certainly the transient is electromagnetic. Also, if it is AC, the steady state is strictly electromagnetic. Again, if you want to say the fields behave quasistatically at low frequency, I'll agree with you all day long.

Why do I need to think outside the box? Maxwell's equations define a pretty big box. The photon viewpoint is outside this box, and I've already commented on that. The circuit theory approximation is inside this box, and is something I'm well aware of. I ask for you to provide references for your viewpoints. Or, is it studiot's box that exists outside of Maxwell's box.