Moving electrical charges and Maxwell's equations

[Phrak]

This statement seems to indicate that you agree with Phrak provided that the charge in not uniformly accelerating over all time. Are your statements limited to the case of uniform (constant) acceleration over all time?

Can you safely consider the fields in a region of flat spacetime only influenced by the charge while it is uniformally accelerating, so you don't have to worry about what happens when we either have to decelerate the charge or have its velocity exceed the speed of light?

At first I thought, yes. Now I'm not so sure.

 

[kcdodd]

Yes, it is different. A single observer will only detect a small fraction of the total radiated power from the particle. I suppose uniform acceleration is not really well defined to begin with. Once the particle reaches relativistic speeds acceleration cannot be uniform any more, and the angular distribution of the radiation even starts to change. So there is nothing uniform about it. Maybe proper acceleration is a better word. But I will have to look up relativistic radiation effects.

 

[Phrak]

And Phrak, yes I realize it is current density. There should be a delta function in there. That is why I specified that at the particle, and zero everywhere else.

I'm not sure what a Dirac delta function gets you, if that's what you are referring to, other than the infinities I referred to previously.

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Just for fun, I asked Mathematica for the Fourier transform of x=t, which seems to correspond to a constant acceleration:

-i \sqrt{2 \pi} \delta'(\omega)

 

[Phrak]

Yes, it is different. A single observer will only detect a small fraction of the total radiated power from the particle. I suppose uniform acceleration is not really well defined to begin with. Once the particle reaches relativistic speeds acceleration cannot be uniform any more, and the angular distribution of the radiation even starts to change. So there is nothing uniform about it. Maybe proper acceleration is a better word. But I will have to look up relativistic radiation effects.

In this case, the change in acceleration tends toward zero at each end, and we're back to a nonuniform acceleration pulse with smoother edges.

We might despense with the nicites of charge requiring mass, and the speed limit c, but now there is an event where the charge decelerates to less than c, and an event where it exceeds c, and perhaps a 'shockwave' from the past that dominates the resultant fields.

A charge sitting on the surface of a gravitating body seems the only viable landscape for a uniformily accelerating charge where we pretend the planet has been and will be there for all time.

This may not be as complicating as it seems, where you only have to redshift.

 

[elect_eng]

Yes, it is different. A single observer will only detect a small fraction of the total radiated power from the particle.

In that case, a point observer will see a pulse of energy.

OK, so now we have dispensed with two issues. First, my initial comment that acceleration in one direction generates a transient pulse was from the point of view of an observer at one point in space. Second, my comment was not restricted to the case of uniform acceleration. My exact words were as follows.

A straight line accelleration in one direction still creates radiation, but the wave is a transient pulse with a broad spectrum.

Your response to this was the following.

Contrary to the last post, a constant acceleration will create a constant radiation field in a direction perpendicular to the path of travel. Clearly it also has no period, but it is clearly not a burst.

Here you can see a few things. First you said it is a constant radiation field. Clearly the field is not constant. A constant radiation field is different than integrating power over a large surface and saying that is constant. Also, for a point observer, there can be a pulse (or burst as you say).

There now remains only one issue to dispense with. Your comment that there is no period still does not make sense. Maxwell's equations in vacuum is a linear theory. I've never heard anyone claim that you can't use Fourier analysis in a linear medium. Any time domain response has frequencies (and periods). There is either a period (or many of them), or the signal is constant. Are you saying that a constant electromagnetic field is able to propagate as electromagnetic radiation? Or, are you saying that the frequency is infinite? If it is not one of these extreme cases, then what does it mean to say that you have radiation with no period? Is this just a disconnect on terminology? Are you just saying that the signal is not periodic in the time domain?

 

[kcdodd]

Now you are saying that radiation is observer dependent? 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 dependent?

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 dependent? 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 boundary. 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.​

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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.