Moving electrical charges and Maxwell's equations

[kcdodd]

Now you are saying that radiation is observer dependant? How can an observer influence what a particle is radiating? As long as you stick with a single reference frame, the radiation power from the particle will not depend on where you stick an antenna. I thought I had already addressed that point.

The surface integral is a way to tell you what the radiation power is without getting into the mechanics of the particle itself. If I were to describe the radiation in terms of the source itself would you still have a complaint? According to Jackson's EM, the radiation power is a function of acceleration only. I don't know how correct that is, but it's the best answer I have seen so far. I have already addressed the issue of doing a Fourier transform on this. If you have a specific issue with what I have said I will try to address it. Energy is going from one place to another, and you can see that from poyntings theorem. If you don't call that radiation, what do you call it.

 

[Phrak]

Now you are saying that radiation is observer dependant?

Certainly this is true. More to the point, radiation is dependent upon the relative motion of the observer.

 

[kcdodd]

My bad. I think I see your point elect_eng. For circular motion you have constant total radiation. But through any particular solid angle you have a periodic amplitude. Likewise for proper acceleration, you will have constant total power, but get varying amplitude in particular locations, though I think it would instead be defined through a particular section of my aforementioned cylinder.

I'm not 100% sure of the math here, but I have simply amended some of Jacksons equations. A more accurate approach would be to start from first principles, but I don't have much time to devote to this.

For constant proper acceleration in x direction, with alpha as acceleration:

$$x =\frac{c^2}{\alpha}(\sqrt{1+(\alpha t/c)^2} - 1)$$

and power flux through cylinder for electron (it seems there should be at least a factor of gamma, but anyway):

$$dP/dA = \frac{e^2}{4\pi c^3 R^2}\alpha^2$$

$$R = \sqrt{(x_0 - x)^2 + r^2}$$

where r is the radius of the cylinder. so you get something like

$$dP/dA = \frac{e^2 \alpha^2}{4\pi c^3 ((x_0 - \frac{c^2}{\alpha}(\sqrt{1+(\alpha t/c)^2} - 1))^2 + r^2)}$$

This is very strange to me since the "pulse" width seems to flattens out the further away the cylinder is, due to radius dependence. Surly wavelength would not depend on distance? A good exercise to actually work out sometime I suppose.

 

[elect_eng]

I see your latest response, but just to clarify some of the previous comments.

Now you are saying that radiation is observer dependant? How can an observer influence what a particle is radiating? As long as you stick with a single reference frame, the radiation power from the particle will not depend on where you stick an antenna.

The observer need not influence the particle, and this assumption is generally made in classical theory. I am certainly not claiming that the observer affects the solution.

The radiation field is position and time dependent. Received power is observer dependent because an observer never receives all of the transmitted power. The observer only receives the portion of the fields that reache that point (power density), and only what their receiver antenna, or power meter captures (integration of power density).

If you are talking about total transmitted power, that does not bother me, but your terminology was "radiation field" originally. The radiation field can not be constant since it moves with the particle like a zipper being unzipped. If you really mean (as seems clear now) total transmitted power, I have less cause to complain if you say this is constant. I still say an observer actually measures a frequency spectrum of EM waves (or photons with a spectrum of energies)? If I install a radio dish receiver antenna I would measure a broadband (white noise) spectrum in the form of a pulse in time as the particle goes by.

Energy is going from one place to another, and you can see that from poyntings theorem. If you don't call that radiation, what do you call it.

I call it radiation of power in the form of electromagnetic waves, or photons. I don't call it a constant radiation field in this case, but I have no complaint if you say total transmitted power may be constant, at least under some assumptions.

 

[kcdodd]

I see. But the poynting flux need not necessarily be varying in time. I think one of the counter examples is a coaxial cable with dc voltage. The poynting vector does indeed show the direction of energy through the em field, and yet it does not vary. You might say this means frequency of zero, and so any photon would have zero energy. I honestly don't know the connection here. Are there infinite number of photons with zero energy passing through the cable? Or no photons at all, and it is a completely different effect? If it was ac, would you then say there are photons with specific energies? What, then, is the difference in the mechanism of energy transfer?

 

[elect_eng]

... the poynting flux need not necessarily be varying in time. I think one of the counter examples is a coaxial cable with dc voltage. The poynting vector does indeed show the direction of energy through the em field, and yet it does not vary. You might say this means frequency of zero, and so any photon would have zero energy. I honestly don't know the connection here. Are there infinite number of photons with zero energy passing through the cable? Or no photons at all, and it is a completely different effect? If it was ac, would you then say there are photons with specific energies? What, then, is the difference in the mechanism of energy transfer?

That is an interesting question. I would normally take more time to answer this type of question with certainty, so take my comments as food for thought and not as a definitive answer.

Clearly, this is a static case and a wave solution or radiation solution is not relevant.

The statement of the problem is a little ambiguous in this case because we have to ask how long the coax cable is. Is it infinite in length, or finite? If it is finite, how are the ends terminated? Are they open, shorted, or terminated with impedance (matched or otherwise)?

The solution inside a long open coax cable (perhaps with a voltage source at one end) is essentially that of a capacitor with fixed charge generating the electric field. A DC field is present only if there is a charge on the capacitor. The field is in the radial direction and proportional to the stored charge, and inversely proportional to radial distance.

$$E_r={{q}\over{2\pi \epsilon r}}$$

The solution inside a long shorted coax cable (perhaps with a current source at one end) is essentially that of an inductor. The DC magnetic field is in the angular direction and proportional to the current and inversely proportional to radial distance.

$$H_{\phi}={{I}\over{2\pi r}}$$

If we take these two independent solutions, then they are orthogonal and there appears to be a Poynting vector and power flow. However, this can't be done because a shorted cable has no charge and an open cable has no current, hence power is zero in either case.

The more interesting case is putting a voltage or current source at one end and a resistor at the other end. Here we have power generation at one end and power dissipation at the other end. And, it seems to be true that fields (E and H) are orthogonal inside the coax, at least in the middle far from the ends.

This is a distributed case and Poynting's theorem seems more appropriate than simple circuit equations, even though the DC case is valid with circuit equations. Poynting's theorem is simplified since all time derivatives are zero, as follows.

$$\int_V (\vec E \cdot \vec J) \;dV=-\ointop_{\partial S} (\vec E \times \vec H) \cdot dS$$

So presumably, you would want to take a surface that captures one end of the cable and cuts through the middle of the cable. This would show power flow in the middle and power dissipation at the end. So, no problems here.

The question of the photons is a little tricky and I can't say I have a definitive answer. However, generally when I try to equate a traveling EM wave with photon flux, I only do it when the boundary dimensions are much larger than the wavelength of the wave. This works well for light in everyday objects such as lenses. It also works for radio waves traveling in free space. However, in this case we are talking about zero frequency, which implies infinite wavelength. Hence, a photon interpretation is more difficult, at least for me.

 

[Phrak]

These are several points of misunderstaning presented in this thread. So many that I surely missed most.

First, the Poynting vector, integrated over a surface is NOT proportial to the energy transported across the surface. Consider a planar wave. At some surface both the electric field and the magnetic field are zero valued. Their cross product is zero valued. The Poynting vector is everywhere zero on this surface. Equating the Poynting vector with energy times any constant of proportionality predicts that no energy is transmitted across the boundry. I think we can agree this is false.

Second, I provided an equation for the energy spectrum due to the proportion of the world line of a charge undergoing constant acceleration. Did any of you notice; the frequency is exclusively zero Hertz?

 

[kcdodd]

No energy is transported at that instant, but as the wave moves ExB will become nonzero and energy will be transported across the surface. The time average will still work out, so I don't see a problem. And the zero frequency thing is a current question I think.

 

[Phrak]

Third, no momentum is exchanged, one to another, for comoving charges. This is independent of whether the two charges are in relative motion with respect to one another or not. This is something of a simplification subject to relative phase:-

Consider an oscillating charge. At some distance is another charge which we see bathed in alternating electic and magnetic fields. There exists an oscillitory state of motion of this particle in which the Lorentz transform of the magnetic field due to the motion of the particle cancels at least some of the the electric field so that the particle is subjected to less electric field in its frame of reference. Clearly, less work is done on the particle.

3a) The influence of one particle upon another is depentent upon their relative states of motion.

3b) The propagating fields influencing a charged particle is a function of that particle's state of motion.​

---------------------------------------------------------------

What is the 4-vector field for a planar wave?

 

[kcdodd]

By four vector, I'm guessing you mean four potential. You can probably guess it from EM wave solution:

$$E_x = Re(E_0e^{i(k z-\omega t)})$$
$$B_y = Re(B_0e^{i(k z-\omega t)})$$

And if you work out the solutions for

$$E = -\nabla \phi - \partial_t A$$
$$B = \nabla \times A$$

you can see that phi must be zero, and A has the form

$$A_x = -iA_0e^{i(k z-\omega t)}$$

Which has a divergence of zero

where now you have

$$E_0 = \omega A_0$$
$$B_0 = k A_0$$

 

[Phrak]

And if you work out the solutions for

$$E = -\nabla \phi - \partial_t A$$
$$B = \nabla \times A$$

you can see that phi must be zero, and A has the form

$$A_x = -iA_0e^{i(k z-\omega t)}$$

Which has a divergence of zero

where now you have

$$E_0 = \omega A_0$$
$$B_0 = k A_0$$

Phi would be zero in the Coulomb gauge, but in general, otherwise. It should be sufficient to consider for the purposes of futher argument the magnetic potential to be a Fourier spectrum of sinewaves in the z direction, normal to the direction of propagation x, and parallel with E.

In this form it's fairly easy to Lorentz transform A = (Phi, 0, 0, A_z), where you might notice that Phi and A_z transform to Phi' and A_z' in the manner of time and space coordinates.

 

[kcdodd]

Well, phi can be up to a constant for this solution to be valid. If you want phi to vary then you must come up with a different solution.

 

[Phrak]

Well, phi can be up to a constant for this solution to be valid. If you want phi to vary then you must come up with a different solution.

OK, on second thought, the Coulomb gauge should be sufficent. If for a test charge we change to the frame of reference of the particle, some of the magnetic potential will become time-like and the 4-potential will change to one with a nonzero electric potential. I don't expect this is could be a problem.

 

[kcdodd]

Also, something that should have been obvious to me earlier, but slipped by, is that the frequency at which the poynting vector changes is not the same as the frequency of the wave. For a plane wave, it is exactly twice the frequency. So, I wonder if there is a general rule of connecting poynting frequency spectrum to wave spectrum. Does it shift everything down by a factor of two? Then, what would be the meaning of 0/2?

 

[elect_eng]

... Are there infinite number of photons with zero energy passing through the cable? Or no photons at all, and it is a completely different effect? If it was ac, would you then say there are photons with specific energies? What, then, is the difference in the mechanism of energy transfer?

So, with more time to think, I can add a few more comments to my previous thoughts about this.

It is an experimentally known fact that an accellerating charge emits photons. Granted, this fact does not fit well with a classical field description with Maxwell's equations, but it is true none the less. The DC case you are talking about can be viewed as a case in which charges are not accellerating, and it is known that charges with constant velocity don't emit photons. Hence, I would conclude that there are no photons at all, and this is a different effect.

Is the above a circular argument? Yes, but I like the answer I get. Can I prove that it is not an infinite number of photons (with no energy) that are responsible for the energy transfer from one place to another. No, but that idea seems strange and unverifyable to me.

So what is the mechanism for energy transfer, from an intuitive point of view in terms of Maxwell? Well in reality, the coax cable has resistance, even when that resistance is small compared to a load resistance. Hence, there is a distributed voltage drop around the entire conduction path. The fields inside the dielectric of the coax represent force that could be placed on a hypothetical charge in the dielectric (stationary for electric field, and moving charge or current for magnetic field), but there are no charges inside, ideally. Hence, the interior fields are doing no work and not actually transfering power. There are charges in the conductor, and the voltage due to the electric field is the electromotive force driving the charges at constant velocity around the conduction loop. The magnetic field that results from the moving charges is just a stored energy that came from whatever transient event happened long ago to get the system into a constant DC (steady state) condition. That field does no work in the DC case. The electric field around the loop is doing work because it represents a force acting on charges in the direction they are moving.

This is analogous to a locomotive (moving at constant velocity) dragging a bunch of cars with low friction, but with the last car with it's wheels locked (hence high friction). The force is linked from the generation of energy to the dissipation of energy in both cases. The beauty of Maxwell's equations is that the fields inside the dielectric (on a cross section) give us information about the power flow, even though the fields at those points are not actually transfering power (doing work) in and of themselves.

In a nutshell, the electric field does the work in this case, just as an electric field would do work if it pulled a charged sphere at constant velocity through a liquid at its terminal velocity. The "accidental" generation of a magnetic field is irrelevant because a moving charge (which is current generating a magnetic field) does not place a Lorentz force on itself. However, the generated magnetic field does represent information about the rest of the system, and this is one of the amazing qualities of Maxwell's field theory. This is why the simplified version of Poynting's Theorem, in the DC case, works even if the surface of integration slices the middle of the coax cable, where there is no power generation and no power dissipation.

I'm not always in favor of stating these kinds of intuitive descriptions and interpretations because they are open to criticism. However, you asked the question, and I feel compelled to stick my neck out. This is the best answer I can come up with.

 

[Phrak]

elect_eng: Which state of a coaxial cable are you discussing? I've lost track of the thread on this.

Either a simple resistor or an imperfect conductor with a DC current has a radial Poynting vector. The Poynting vector is constant. E and B are constant. The usual treatment is to say that the surface integral of S round a portion of the conductor is equal to the energy lost to heat in that portion. The heat is the result of lattice collisions. I don't see any mystery.

The Poynting vector is directed inward.
http://www.cheniere.org/images/EMfndns1/LorentzInt%20sm.jpg"

Maybe I'm being a bit dense. So I considered one electron or charged object at a time. You accelerate some charged body in an electric field and crash it into something. Repeating this process should give you an average inwardly pointing Poynting vector. It would even happen if the charge carrier were subject to a continuous resistant to acceleration such as friction.

To make it look more like a bunch of charge carriers in a wire, let the charge carrier be a uniformly charged rod of material subject to sliding friction which is immersed in a uniform co parallel electric field.

If you are inclined to presume the mathematics is correct, then at this point it may seem uninteresting. But why should the local actions of a charged body resisting acceleration by an electric field appear as the cross product of the applied electric field and the magnetic field induced by the body?

 

[elect_eng]

elect_eng: Which state of a coaxial cable are you discussing? I've lost track of the thread on this.

...

The Poynting vector is directed inward.

Perhaps the issue is the definition of a coax cable. A coax cable has a wire conductor on the axis and a cylindrical shell conductor as a shield. This is a very effective waveguide, and a good example of one is cable TV coax that transmits hundreds of TV channels with low loss over a wide bandwidth range in the RF spectrum. I showed the E and H fields for the coax in a previous thread. The E field is in the radial direction and the H field is in the angular direction (using cylindrical coordinates, of course). You can see that they are orthogonal and the Poynting vector is in the axial z-direction, not inward.

I agree there is no real mystery with the coax. I was just trying to give a physical interpretation directed at a specific question from kcdodd.

If you are inclined to presume the mathematics is correct, then at this point it may seem uninteresting. But why should the local actions of a charged body resisting acceleration by an electric field appear as the cross product of the applied electric field and the magnetic field induced by the body?

I'm not sure I understand this question, but I interpret you to mean, "why did kcdodd even want to ask the question". I think it is because one can apply Poynting's theorem over a closed surface that encloses the power dissipation end of the coax cable and the middle of the coax cable, but excludes the power generation end of the coax cable. One then sees power being dissipated without a generator. The only interpretation left is to note that the cross product of E and H looks like a power flow into that volume over a portion of the closed surface. Whatever interpretation we want to give to this, I don't think it should include photons.

 

[kcdodd]

Phrak is talking about a resistive coax. Since power is being dissipated throughout the cable, you don't get as much poynting out one side as goes in the other. The only way that can happen is if the divergence of the poynting vector field is negative within the cable, which is caused by the energy flowing into heat. In a perfect coax you don't get this, and so the vector points along the axis all the way through. My point is the energy is "going from one place to another". This is a classical problem, and so doesn't really require photons to begin with. I am sure (quantum) field theory has a more correct explanation, I just don't know it.

 

[Dunnis]

The solution inside a long open coax cable (perhaps with a voltage source at one end) is essentially that of a capacitor with fixed charge generating the electric field. A DC field is present only if there is a charge on the capacitor. The field is in the radial direction and proportional to the stored charge, and inversely proportional to radial distance.

$$E_r={{q}\over{2\pi \epsilon r}}$$

Where did you get that information and formula from? cables are electrically neutral, there will be no measurable net electric field outside the cable. And, in the case cable happens to have some charge, then the magnitude of electric field potential will be inversely proportional to radial distance squared. Why are you comparing a wire with a capacitor? For electric fields it is irrelevant whether the wire is finite or not.

The solution inside a long shorted coax cable (perhaps with a current source at one end) is essentially that of an inductor. The DC magnetic field is in the angular direction and proportional to the current and inversely proportional to radial distance.

$$H_{\phi}={{I}\over{2\pi r}}$$

Magnetic field potential drops radially from the wire with "1/r^2" as well.

Where did you get your equations from?

 

[Phrak]

I think we've may have swapped positions on this mystery thing about coaxial cables. Signs of a good discussion. I have a thought experiment.

Take a nice room where all the walls are turned into a coaxial shield with perfectly conductive mesh. Run a perfectly conductive wire from one wall the opposite wall. At one end of the room insert a voltage source in the wire. At the other end of the room insert a resistor into the wire. Change the shape of the room into a cylinder surrounding the wire at it's center. Now we're inside this coaxial cable arrangement that was a room.

The magnetic fields form concentric hoops, uniform from one end of the room to the other because the current is uniform through the wire, battery and resistor. The radial electric field is the same as well, whether about the battery, reististor or conductor. The battery and resistor have a voltage across them so the axial electric field within the volume of each is nonzero and opposite. The axial electric field within the wire is zero valued.

What does the Poynting field look like in the cylinder?

 

[elect_eng]

Where did you get that information and formula from? cables are electrically neutral, there will be no measurable net electric field outside the cable. And, in the case cable happens to have some charge, then the magnitude of electric field potential will be inversely proportional to radial distance squared. Why are you comparing a wire with a capacitor? For electric fields it is irrelevant whether the wire is finite or not.

Magnetic field potential drops radially from the wire with "1/r^2" as well.

Where did you get your equations from?

Wow, talk about misunderstandings. Clarifications as follows ...

1. Where did I get the information from?

I got it from ... Memory, ... Derived from Maxwells Equations, ... Verified in every EM book I own. Take your pick of the answer.

2. Cables are electrically nuetral?

A coax cable with open ends is also a capacitor, it can be charged up with no problem. Every coax cable as a capacitance per unit length.

3. There will be no measureable electric field outside the cable?

Did you notice that I said "INSIDE" the cable? Yes, the shield has the effect of making the magnetic and electric field zero outside the cable.

4. The electric field will be inversely proportional to radius squared?

No, it's inversely proportional to radius. You can look up the solution or derive it if you don't believe me.

5. Why am I comparing a wire with a capacitor?

I'm not, I'm comparing a waveguide with a capacitor. So it's actually two wires. However, even a wire by itself has capacitance and one could make the comparison if it served a purpose.

6. For electric fields length is irrelevent?

Not sure why you mentioned this, but it's not true. A finite length object has end effects and a distorted electric field when compared to the solution of a similar, but infinitely long object.

7. Magnetic field drops as 1/r^2 as well?

Nope. Look it up. Solution for a coax is proportional to 1/r, and again I specified this is inside the coax, not outside.

Clearly, you have not read the post and the rest of thread carefully (or at all). It's important to do this before you jump to conclusions too hastily. It wastes time and distracts from the thread. I dont' mind some misuderstandings because I know I don't explain things in a perfectly clear way, but you need to make at least the smallest of efforts.

 

[elect_eng]

I think we've may have swapped positions on this mystery thing about coaxial cables. Signs of a good discussion. I have a thought experiment.

Take a nice room where all the walls are turned into a coaxial shield with perfectly conductive mesh. Run a perfectly conductive wire from one wall the opposite wall. At one end of the room insert a voltage source in the wire. At the other end of the room insert a resistor into the wire. Change the shape of the room into a cylinder surrounding the wire at it's center. Now we're inside this coaxial cable arrangement that was a room.

The magnetic fields form concentric hoops, uniform from one end of the room to the other because the current is uniform through the wire, battery and resistor. The radial electric field is the same as well, whether about the battery, reististor or conductor. The battery and resistor have a voltage across them so the axial electric field within the volume of each is nonzero and opposite. The axial electric field within the wire is zero valued.

What does the Poynting field look like in the cylinder?

I'm confused by what you are describing. You said to make the walls ( I assume this means only wall and not ceiling and floor) a conductor. But, then you said to run a wire from one wall to the other. This would short the wire, but then it seems you say to insert a resistor and battery. I guess this is in series so that the wire is no longer shorted. It's not clear to me how the magnetic field forms concentric hoops uniformly. This strikes me as a very asymmetrical geometry that would need a numerical solution.

Please clarify if I'm completely misinterpreting. Maybe a diagram would help.

 

[Born2bwire]

I think we've may have swapped positions on this mystery thing about coaxial cables. Signs of a good discussion. I have a thought experiment.

Take a nice room where all the walls are turned into a coaxial shield with perfectly conductive mesh. Run a perfectly conductive wire from one wall the opposite wall. At one end of the room insert a voltage source in the wire. At the other end of the room insert a resistor into the wire. Change the shape of the room into a cylinder surrounding the wire at it's center. Now we're inside this coaxial cable arrangement that was a room.

The magnetic fields form concentric hoops, uniform from one end of the room to the other because the current is uniform through the wire, battery and resistor. The radial electric field is the same as well, whether about the battery, reististor or conductor. The battery and resistor have a voltage across them so the axial electric field within the volume of each is nonzero and opposite. The axial electric field within the wire is zero valued.

What does the Poynting field look like in the cylinder?

I don't know this off-hand and I'm just jumping into the end of this discussion but I'm pretty sure I could find the answer in Balanis' electromagnetics text as he does analyze the coaxial cable transmission line. Suffice to say though, it would probably depend on your mode and frequency specifically. My gut intuition is that the Poynting vector will be normal to the \phi direction and will "bounce" back and forth between the shield and conductor. That is, the there is a cylindrical wavefront the is a standing wave in the radial direction and propagating along the axial direction.

I could be misinterpreting what you are asking, but I assume you are just treating the voltage and resistor as discrete elements simply providing an excitation and load respectively. Now there will also be reflection off of the load and the source and so forth but I assume you are familiar with the transient behavior of transmission lines.

But you are asking about in the DC case though right?

With an electromagnetic wave, the power is propagated by the waves being guided between the shield and conductor. But in the DC case... Hmmm... I'd have to think about it. If we naively ignore the return path on the shield, then the Poynting vector points along the surface of the conductor only. But with a coax we now have a static electric field applied between the shield and conductor. Question then would be what would be the magnetic field from the return current on the shield I guess. I'll take a look at Balanis tomorrow morning. The first mode of the coax problem extends down to DC and so I might be able to gleam what the answer is from that solution.

EDIT: elect_eng has mentioned some stuff I was also wondering about. I am just assuming we have a cylindrical conducting shield, a wire conductor and at one end of the coax we put a resistor between the shield and conductor and at the other end we place a voltage source across it.

 

[Phrak]

But you are asking about in the DC case though right?

That's correct. The underlying question that I'm asking is: "Does the Poynting vector correspond to energy flux in a physically meaningful way?"

With an electromagnetic wave, the power is propagated by the waves being guided between the shield and conductor. But in the DC case... Hmmm... I'd have to think about it. If we naively ignore the return path on the shield, then the Poynting vector points along the surface of the conductor only. But with a coax we now have a static electric field applied between the shield and conductor. Question then would be what would be the magnetic field from the return current on the shield I guess. I'll take a look at Balanis tomorrow morning. The first mode of the coax problem extends down to DC and so I might be able to gleam what the answer is from that solution.

EDIT: elect_eng has mentioned some stuff I was also wondering about. I am just assuming we have a cylindrical conducting shield, a wire conductor and at one end of the coax we put a resistor between the shield and conductor and at the other end we place a voltage source across it.

I made a mistake in my little DC thought experiment. There cannot be a radial component of electric field in any length of the conductor or out of the components without a net electric charge present.

Is it necessary to have a charge imbalance to have a current flow?

We are allowed to have both positive and negative charge carriers at once, so than any net charge might be canceled. If no imbalance, there is the peculiar circumstance that the Poynting vector points outward in a disk about the battery and inward in a disk about the resistor with both terminating on the shield. So there would be no interconnecting Poynting vector field to transport energy from battery to resistor.

 

[Born2bwire]

That's correct. The underlying question that I'm asking is: "Does the Poynting vector correspond to energy flux in a physically meaningful way?"

With an electromagnetic wave, the power is propagated by the waves being guided between the shield and conductor. But in the DC case... Hmmm... I'd have to think about it. If we naively ignore the return path on the shield, then the Poynting vector points along the surface of the conductor only. But with a coax we now have a static electric field applied between the shield and conductor. Question then would be what would be the magnetic field from the return current on the shield I guess. I'll take a look at Balanis tomorrow morning. The first mode of the coax problem extends down to DC and so I might be able to gleam what the answer is from that solution.

EDIT: elect_eng has mentioned some stuff I was also wondering about. I am just assuming we have a cylindrical conducting shield, a wire conductor and at one end of the coax we put a resistor between the shield and conductor and at the other end we place a voltage source across it.

I made a mistake in my little DC thought experiment. There cannot be a radial component of electric field in any length of the conductor or out of the components without a net electric charge present.

Is it necessary to have a charge imbalance to have a current flow?

We are allowed to have both positive and negative charge carriers at once, so than any net charge might be canceled. If no imbalance, there is the peculiar circumstance that the Poynting vector points outward in a disk about the battery and inward in a disk about the resistor with both terminating on the shield. So there would be no interconnecting Poynting vector field to transport energy from battery to resistor.

Bah, wouldn't you know it, Balanis doesn't have the coaxial waveguide, he does do circular cross-section though. But here we go: http://books.google.com/books?id=Ao...Q#v=onepage&q=coaxial waveguide modes&f=false

That text gives the fields for the DC mode and we can see that the Poynting vector is in the z-direction, which would point from your source to your load.

 

[elect_eng]

The underlying question that I'm asking is: "Does the Poynting vector correspond to energy flux in a physically meaningful way?"

This was exactly the question that I took kcdodd to be asking - whether of not the power flow was photons (or waves) in the DC case. My long winded answer to his question was looking exactly at the case of a long coax with a voltage source at one end and a resistor at the other end. My opinion was that the proper choice of a closed surface (one that slices a cross section in the middle of the coax dielectric far from the ends) would force us to make an interpretation that the Pointing vector is representing power flow even in the DC case. It was also my opinion that this is not physically meaningful, but that the DC fields in the dielectric of the coax represent "encoded information" about physically meaningful mechanisms of power flow in other parts of the system. Note that if one uses a closed surface that encloses the entire coax including both ends and the resistor and voltage source, then there will be no Poynting vector on the surface and only power dissipation and power generation will show up as canceling (net zero) power in the volume integral portion of Poynting's Theorem for DC case.

 

[kcdodd]

If you look at the resistor, the electrons are dropping in potential, and depositing that energy into the resistor. It is the electric field which is doing the work on the electrons, not some other mechanism. If the electric field is doing work, then the energy in the field would drop. That is poyntings theorem.

$$\nabla \cdot \vec{S} + \partial_t u + \vec{J}\cdot\vec{E} = 0$$

If you balance mechanical work and poynting flow

$$\nabla \cdot \vec{S} + \vec{J}\cdot\vec{E} = 0$$

then you reach steady field strength:

$$\partial_t u = 0$$

Same for the generator side. So, to me, without poyting's theorem, you do not construct a self-consistent scenario. That energy must be getting to the field somehow. So, how is it not physically meaningful? How else does the energy get there?

 

[elect_eng]

So, to me, without poyting's theorem, you do not construct a self-consistent scenario. That energy must be getting to the field somehow. So, how is it not physically meaningful? How else does the energy get there?

Keep in mind that we are talking about interpretations which are like opinions in that they are hard to prove. However, the way I'm thinking about it is that a DC electric field and a DC magnetic field each represent a stored energy, not an energy flow or transfer. That stored energy got there during a transient event that happened long ago in the past before the DC steady state conditions were created. Once we try to interpret the DC state, the physical meaning of power flow in the dielectric over a portion of a surface due to a Poynting vector becomes questionable.

If you choose to interpret the DC field as being an energy flow in and an energy flow out with perfect balance, I won't try to say you are wrong. It's just an interpretation that does not appeal to me personally. I like the idea of the fields representing information about power flow, rather than the power flow itself.

 

[kcdodd]

In your interpretation, how is the energy getting from the generator to the resistor?

 

[elect_eng]

In your interpretation, how is the energy getting from the generator to the resistor?

I prefer to think of the energy as force pushing charges over a distance. One can view this as an analogy with a train. Here that last car of the train gets its energy through a force linkage through all of the intermediate cars. Essentially a tension force through the chain is the linkage. Similarly the electric field around the conduction path is a force per unit charge. This force pushes charges over distance. The analogy is not perfect but is helpful. A water pipe analogy might be preferred, with a pump pushing water through a wide hose with a small outlet at the end. But, the idea of a linkage appeals to me. In mathematical terms this is similar to applying Poynting's theorem over the entire volume of the system, rather than slicing through the center. Here the integral of E.J is the only part of the equation that matters since the Poynting vector is zero everywhere on the surface. Basically, I prefer to think of the E.J as more physically meaningful than the ExH in the DC case.

Again, this is just an interpretations to help in understanding. I can't prove one interpretation is better than another. It is interesting that Poynting's theorem works for any surface. This tells us that more than one interpretation is possible. Clearly there are surfaces that show a power flow with ExH. Are these fields actually doing anything at those points in space? I think not, since they represent the ability to apply a force and there is nothing to apply a force on in the dielectric. Hence, I prefer to think of those fields as information about what is happening in other places in the system. These other places actually have charges that are moving under the influence of force pushing through a resistance.

 

[Phrak]

I'm confused by what you are describing. You said to make the walls ( I assume this means only wall and not ceiling and floor) a conductor. But, then you said to run a wire from one wall to the other. This would short the wire, but then it seems you say to insert a resistor and battery. I guess this is in series so that the wire is no longer shorted. It's not clear to me how the magnetic field forms concentric hoops uniformly. This strikes me as a very asymmetrical geometry that would need a numerical solution.

Please clarify if I'm completely misinterpreting. Maybe a diagram would help.

Sorry. Take a large conducting cylindrical surface with conducting disks at each end. Run the wire, resistor and voltage source down the axis of the cylinder. The symmetry makes the analysis easier. To make things even better, take the radius and length to infinity.

I don't see anything taking energy from the voltage source to the resistor.

-------------------------------------------------------------------------------

I'd like to see the Lorentz invariant energy continuity equation which should apparently include the Poynting vector as three out of 6 components. Maybe kcdodd already posted it. I can't tell.

 

[kcdodd]

I think you're looking for the stress tensor. This will give both energy and momentum conservation between field and matter.

 

[Dunnis]

Wow, talk about misunderstandings. Clarifications as follows ...

7. Magnetic field drops as 1/r^2 as well?

Nope. Look it up. Solution for a coax is proportional to 1/r, and again I specified this is inside the coax, not outside.

Look it up where, in your dreams? $$H_{\phi}={{I}\over{2\pi r}}$$

No, you did not specify "inside", you said: -"inversely proportional to radial distance", which now tells us that you do not even understand the meaning of what you said yourself previously. Look at the diagram below, that's what radial distance is. There, now even you know what was that which you were talking about.

                  B(r) = Km* I*L*sin(90)/r^2
                   |
                   |
                   | r
                   | 
          alpha=90 |
------------------------------------WIRE--------->>

We do not measure any magnetic fields INSIDE wires, nor electric fields for that matter - what we measure inside wires is called voltage and el. potential difference. Your equations talk about OUTSIDE too, it's just that they are not completely correct. This is correct relation of magnetic field, current and radial (and even non-radial) distance:

http://en.wikipedia.org/wiki/Biot–Savart_law -The Biot–Savart law is used to compute the magnetic field generated by a steady current, for example through a wire... The equation in SI units is:

$$B = \int \frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2}$$

1. Where did I get the information from?

I got it from ... Memory, ... Derived from Maxwells Equations, ... Verified in every EM book I own. Take your pick of the answer.

Then wake up, or sober up, you are dreaming.

Can you show me your equations anywhere in Wikipedia?

2. Cables are electrically nuetral?

A coax cable with open ends is also a capacitor, it can be charged up with no problem. Every coax cable as a capacitance per unit length.

What kind of questions are those? Have you not any cables around your house to confirm for yourself? Yes, in relation to radial distance cables are electrically neutral, of course. Wires are made of atoms and atoms are made of about the same number of positive and negative charges, so due to principle of superposition these fields combine to neutral net charge around the wire, regardless of any drift velocity, voltage or whatever is going on inside. This is macroscopic approximation so we do need some insulator to prevent direct contact and this distance does have a minimum other than zero, which is where "outside" becomes the "inside".

3. There will be no measureable electric field outside the cable?

Did you notice that I said "INSIDE" the cable? Yes, the shield has the effect of making the magnetic and electric field zero outside the cable.

Your equations and you said: "inversely proportional to RADIAL DISTANCE". What do you imagine is this "radial distance" you were talking about? Yes, as explained above, there will be no measurable electric field with any macroscopic radial distance from the cable, i.e. outside the cable.

4. The electric field will be inversely proportional to radius squared?

No, it's inversely proportional to radius. You can look up the solution or derive it if you don't believe me.

What planet are you from? I don't think you should be allowed to make any more statements without providing some source to back it up. There is nothing to derive, it is experimentally established relation and the 1st chapter in electromagnetism on this planet, it's called Coulomb's law:

$$E = {1 \over 4\pi\varepsilon_0}\frac{q}{r^2}$$

http://en.wikipedia.org/wiki/Coulomb's_lawNow, can you provide some reference for this nonsense: $$E_r={{q}\over{2\pi \epsilon r}}$$

5. Why am I comparing a wire with a capacitor?

I'm not, I'm comparing a waveguide with a capacitor. So it's actually two wires. However, even a wire by itself has capacitance and one could make the comparison if it served a purpose.

And I'm asking you about this "purpose". Why make comparison, what is the purpose and where are you getting at with it? What is the application, what is conclusion or prediction? What is it you are talking about and what is the point you are trying to make?

6. For electric fields length is irrelevent?

Not sure why you mentioned this, but it's not true. A finite length object has end effects and a distorted electric field when compared to the solution of a similar, but infinitely long object.

I mentioned it because I saw your confusion. You need to decide what kind of "objects" you are talking about. There is not many objects in this world that are electrically charged, and even batteries and el. conductors do not have any net electric charge at any macroscopic distances away from them. You need objects made exclusively, or largely, of one type of charges for that object to have some eclectic potential AROUND it, say electron beams are electrically charged even AROUND them.

+ positive charge (nucleus)
- negative charge (electron)                  E(r) = Ke* (-Q+Q)/r^2 = ~zero 
                   |
                   |
                   | r
                   | 
                   |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+WIRE ZOOM+-+-+-+-+->>
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+->> 
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+->>
                  E(r) = Ke* -Q/r^2 > 0
                   |
                   |
                   | r
                   | 
                   |
- - - - - - - - - - - - - - - - - ELECTRON BEAM ZOOM- - - - - ->>
 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ->> 
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ->>

You are mixing "electric field" with electric current and voltage. The equations you gave and the term "radial distance" very specifically describe field potentials in relation to their sources and/or center of their density distribution, which means "OUTSIDE". - Now, do you really think electric field potentials around cables, at some radial distance, depend on the length of the wire?

 

[Phrak]

I think you're looking for the stress tensor. This will give both energy and momentum conservation between field and matter.

You have gotten me curious about this. I'm looking for something similar to the charge continuity equation, that might look something like

$$\frac{\partial W}{\partial t} = -\nabla \cdot p$$

where W and p are energy and momentum densities in terms of the fields.

 

[Phrak]

Look it up where, in your dreams?

The radial electric field strength of a coaxial cable operating in the fundamental mode has 1/r dependence. Note this is about alternating currents.

In the limt of DC current, it is also 1/r dependent, but the radial component is zero valued; 0*(1/r)=0.

The currents in different phases along the length of the cable are not equal. Charge continuity requires that the charge be not always zero valued. It is not much different than a wave guide, where there are oscillations in all of B, E, phi and J.