Moving electrical charges and Maxwell's equations

[closet mathemetician]

I'm confused about Maxwell's Equations.

1) does an electrical charge (say, an electron) traveling with a constant velocity (say, in the x-direction) travel as an electromagnetic wave?

I'm thinking of an analogy with flowing mass. Suppose you have massive particles, evenly distributed through some volume, traveling with constant velocity, say the x-direction. The flux across the (y,z) surface of a volume is calculated as the integral of the current density, vector \vec{J} over the (y,z) surface area.

where

\vec{J}=\rho\cdot\vec{v}

and
\int\int_A\vec{J}\cdot dA

The result is the flow of mass across the (y,z) surface area per unit time, with units of kg/s. There is no curl in this vector field.

Now, imagine that instead of flowing mass, we have flowing electric charge in the x-direction across the (y,z) surface of a volume. Assume the electrons are not being accelerated as they flow across the (y,z) surface.

With this substitution the flux of mass across the surface becomes the flux of electrons across the surface with units of q/s, which is electrical current.

Does each electron rotatate as it travels in a straight line, producing an "infintesimal" rotation, which, when integrated over the surface area (y,z), produces an overall rotation of the vector field (Stokes theorem).

Does any of this cause the electrons to radiate? I don't see any light radiating when current travels through a metal wire.

Maybe I'm confusing electromagnetic waves traveling along the electric field lines with electrons traveling through space.

[uzair_ha91]

I'm confused about Maxwell's Equations.

1) does an electrical charge (say, an electron) traveling with a constant velocity (say, in the x-direction) travel as an electromagnetic wave?

MY knowledge (which still remains untested) is that an accelerating electric charge produces a changing electric field. This changing electric field then produces a changing magnetic field which again sets up a changing electric field. This continuous process is known as an electromagnetic wave. Thus electromagnetic wave is actually caused by oscillations of electric and magnetic fields which arise due to an accelerating charge not a charge moving with constant velocity.
A charge moving with constant velocity only creates a constant magnetic field.

[kcdodd]

I'm confused about Maxwell's Equations.

1) does an electrical charge (say, an electron) traveling with a constant velocity (say, in the x-direction) travel as an electromagnetic wave?


The simple answer is no. If a particle is moving at a constant (non-zero) velocity, that means there exists an inertial frame in which the particle has velocity of zero. Clearly there would be no EM waves in this frame since you simply have E = kq/r^2 (EM waves cannot be "gotten rid of" in this way). The other way to know it is not a EM wave is that EM waves travel at the speed of light (in free space). However, this "wave" simply travels at the speed of the particle.

As a side note, if you know electric field in this "rest frame", then you can perform a certain type of transformation (called a Lorentz transformation) to know the electric and magnetic fields in the original frame.

[closet mathemetician]

Thanks for the responses. I guess the key is that to radiate the charge must be accelerating, but must the acceleration always be in the form of an oscillation? What if you accelerated a charge in a straight line? Or, maybe if you tried that the charge would always oscillate anyway?


Maybe what you are saying is that the very occurrence of the mutual induction of E and B fields causes the oscillation, so any time you accelerate a charge it will oscillate?

And yes, kcdodd, I know about Lorentz transformations (got lots of questions about that too, but I'll save for another thread).

[elect_eng]

... key is that to radiate the charge must be accelerating, but must the acceleration always be in the form of an oscillation? What if you accelerated a charge in a straight line? Or, maybe if you tried that the charge would always oscillate anyway?


An oscillating charge will create a steady wave with a fixed frequency. The oscillation can be charge flowing around a loop, or a straight line back and forth motion.

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

[kcdodd]

If I get too technical just tell me. The definition of radiation has to do with integrating the flux of power, or pointing vector (S = ExB), over a closed surface around the particle. For a single particle, the flux across this surface is proportional to the acceleration of the particle. 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.

In fact, a particle which is rotating is the one which gives bursts of radiation (which, at least at one time, was the explanation for pulsars). Since the perpendicular direction to the path also rotates (except along the axis of rotation), you get an outward spiral of radiation. You actually call the "bursts" as waves, but depending on the exact parameters they may not look like waves (aka, broad spectrum of waves as the poster mentioned). So in short, radiation is not necessarily the same thing as waves.

[closet mathemetician]

you're good, kcdodd, I'm getting it, little, by little.

[elect_eng]

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.

.

So, according to you, a charge accellerates by you and continues traveling on out to an infinite distance. When the the charge is 100 million light years away, you will still be experiencing a constant radiation field. Also, you will be seeing photons with infinite frequency (implying infinite energy) since you say they have no period (I assume no period means 0 period, hence f=1/T=infinity).

[elect_eng]

In fact, a particle which is rotating is the one which gives bursts of radiation (which, at least at one time, was the explanation for pulsars).

I didn't say rotating, I said traveling in a loop. A charge traveling in a circle can be viewed as a sinusoidal oscillation in both the x and y directions. Photons are generated at the frequency of the charge circulation (cyclic frequency). If the angular velocity is constant and this exists for a long time, then you have a steady state solution, which is what I mean by a steady wave with fixed frequency.

[kcdodd]

By no period, I mean it is not periodic. And I also meant travelling in a loop.

[bjacoby]

I'm confused about Maxwell's Equations.

1) does an electrical charge (say, an electron) traveling with a constant velocity (say, in the x-direction) travel as an electromagnetic wave?


As pointed out the answer is "no". But that has to do with waves that are electromagnetic as you are thinking of. In that case acceleration is needed to create propagating waves.

But there is more to this than that. Apparently waves of some type do exist even for electrons traveling at constant velocity. And the evidence for that is that electrons, even single electrons exhibit wave behavior such as diffraction at a slit. So if one has an electron beam traveling at constant velocity through space one has to wonder just what it is that causes the statistical paths of electrons to be on average solutions to a wave equation? It's not clear at all why such at thing is true. But it is. But these waves are apparently not electromagnetic in any sense.

[elect_eng]

By no period, I mean it is not periodic. And I also meant travelling in a loop.

If we are talking about outward radiation of real power (not reactive power in the near field), then this energy transmission must take the form of photons. Each photon will have a quantized energy and a corresponding frequency. How can photons have no period (or frequency)? If we restrict the interpretation to classical EM theory, how can a traveling electromagnetic wave not be periodic? Doesn't it need to be a solution of the wave equation?

[Born2bwire]

If we are talking about outward radiation of real power (not reactive power in the near field), then this energy transmission must take the form of photons. Each photon will have a quantized energy and a corresponding frequency. How can photons have no period (or frequency)? If we restrict the interpretation to classical EM theory, how can a traveling electromagnetic wave not be periodic? Doesn't it need to be a solution of the wave equation?

I too am at a lost on kcdodd's views here. Charges that have a regular oscillation that elect_eng described previously give rise to electromagnetic waves. All electromagnetic radiation are waves, I am not sure why kcdodd would state to the contrary. The only exception I would give is if we are talking about alpha or beta radiation which involves particles. Synchrotron radiation is just an example of highly directed electromagnetic wave excitation.

[kcdodd]

I am simply stating a definition of radiation as the Poynting vector. If you integrate this flux around a closed surface, and is non-zero, then energy is moving out from whatever is inside the surface. I have said nothing about waves. The poynting vector of a classical em plane wave is periodic in magnitude, but always in the direction of motion indicating the movement of energy. If you accelerate a charge with constant acceleration, then you must be doing work on EM field at a rate proportional to v*a (which is proportional to the total flux of energy out). Nowhere here is a period.

How this relates to photons I cannot answer you.

[elect_eng]

Nowhere here is a period.

How this relates to photons I cannot answer you.

I disagree with you. There can indeed be a period. Also, any explanation must be related to photons. This is what I tried to do in my first post, yet you chose to shoot it down even though you can't correlate your explanation with simple physics.

I'm OK with your explanation of the Poynting vector, but I disagree with your statement that my description of a radiation pulse is not correct. Imagine that you are a stationary observer and you see a charge accellerating by you. What will you observe from the "Poynting vector"? You will see a radiation pulse as the charge accellerates by you. The charge is initially far away and you see a very small intensity, however when the charge is going by near you, maximum intensity is observed. Then as the charge fades in the distance you see small intensity again. Now think of Fourier analysis. How can a pulse have a period? Well, it can't have one period, but it can be an integration of a spectrum of sinusoidal waves. This is how the solution obeys the wave equation. Superposition applies and the summation of an infinite number of sine waves forms a pulse in the time domain. This is the classical explanation in terms of Maxwells equations. However, in the quantum world, this radiation is interpreted as photons. As an observer, you will see the emission of photons with a broad energy spectrum.

At least that is my opinion. This is the only way I can make sense of your description of the Poynting vector radiating power away (which I agree with), and the well known fact that the energy must disappear as energy carried away by photons (which you should agree with).

[kcdodd]

Firstly, photons are not "simple physics". If you cannot explain classical events with classical physics, then adding photons and quantum mechanics is not going to help. How do you "simply" explain the electric force on a charged particle with photons? That is out of the scope of this forum.

Second, you do not see radiation of a particle coming directly toward you, only the perpendicular component.(thinking now about this it is hard to believe, but look at a dipole antenna radiation pattern). Granted, if a particle passes you it's distance will go something like tan(angle), and you will get a magnitude change due to the distance effect, but that is not radiation. You would see that even from a particle with a constant velocity that is just passing by.

I suppose you might call that the induction definition of radiation. If a charged particle passes by, you will see a changing E and B, and you can use a "probe" to extract energy, and therefore conclude that radiation exists. However, it is really the probe which is doing work on the field, and so the flux through a surface surrounding the probe will be non-zero. But the flux through the surface around the particle of interest still is zero. So this is not a good definition of radiation.

The magnitude of the poynting vector, like I said already, is proportional to the rate of work done on the particle, which is also proportional to v*a. If acceleration is constant, you will get something like a^2*t (where i just subbed in the velocity as a function of time). If you wish to do Fourier analysis on this function that's fine, but I just don't think it illuminates very much.

edit: I realized there is something wrong with the power relation I have stated here. see my later post.

[elect_eng]

Firstly, photons are not "simple physics". If you cannot explain classical events with classical physics, then adding photons and quantum mechanics is not going to help. How do you "simply" explain the electric force on a charged particle with photons? That is out of the scope of this forum.

I did give a conceptual explanation in terms of classical physics. By bringing in the notion of photons, I'm just trying to show an additional point of view. I could avoid any further discussion of photons if it is viewed as out of the scope of this forum, or this thread. It's not a critical part of my view on what is happening here. Still, why ignore the well-known wave-particle duality? It is a useful concept.

In any event, you do make a good point about "explaining classical events with classical physics". I should present a mathematical derivation in terms of Maxwells equations in the classical context. If I don't do this, then I'm just giving an opinion without proof. The weekend is coming up, so I can spare an hour or two to write something up.

If you are inclined, you can also present a more detailed explanations with equations and derivations. It's hard to follow some of your points without more details. Some things are confusing. For example you said "The magnitude of the poynting vector is proportional to the rate of work done on the particle", but some of the work done is used to generate kinetic energy, since any charged particle has mass and velocity is continually increasing. This may just be a wording issue, but equations will make it more clear and leave no ambiguity. There are other examples of confusion I have in interpreting, but I don't want to be nit-picky. -I'm just saying that sometimes equations leave less wiggle-room for misinterpretation.

[kcdodd]

That sounds good.

By work done on the particle I did mean by the em field. It would probably be better to say the work done on the field by the particle, since the field energy is going up.

That leads to a misunderstanding I realized I have when I posted earlier, and that is the acceleration dependence of radiation power. Apparently it is the second power of a (acceleration), not the first. I only had an intuitive idea that the radiation reaction force on the particle was simply an induction like effect, resisting the change of velocity. Basically, an induced electric field in the opposite direction as acceleration. And so you get work done against this efield as you accelerate putting energy into the field.

In math terms

\nabla \cdot \vec{S}= - \vec{J}\cdot\vec{E}

where, now we have for a particle;

\nabla \cdot \vec{S} = - q\vec{v}\cdot\vec{E}

at the position of the particle, and zero everywhere else.

The electric field on the right hand side, in my mind, was simply proportional to the first time derivative of velocity in the way I already mentioned (so that v and E oppose each other). I am missing where the second power of acceleration comes from. If you go to the power of radiation formula you also loose the velocity dependence, which doesn't seem logical because that seems to indicate you could get radiation power out with no mechanical power in (when v = 0). So, I need to do some homework on where the disconnect is.

[elect_eng]

The electric field on the right hand side, in my mind, was simply proportional to the first time derivative of velocity in the way I already mentioned (so that v and E oppose each other). I am missing where the second power of acceleration comes from. If you go to the power of radiation formula you also loose the velocity dependence, which doesn't seem logical because that seems to indicate you could get radiation power out with no mechanical power in (when v = 0). So, I need to do some homework on where the disconnect is.

I ran across this article which seems interesting and relevant to what you are saying here.

http://www.mathpages.com/home/kmath528/kmath528.htm

It seems there is some controversy about the case of a uniformly accelerating charge. One of the paradoxes seems tied to the equivalence principle. If a uniformly accelerating charge radiates, then a stationary charge in a gravitational field should radiate also. These types of questions are beyond my expertise. For this reason, anything I say should apply to a charge accelerating in one direction, but not necessarily a constant acceleration. The above link makes reference to the first and third derivatives of position. It's also an interesting discussion.

I'm still putting together a derivation and explanation for why an observer would experience a radiation burst with a broad spectrum if a charge accelerates while traveling by in one direction. I've mapped out the approach and it is straightforward and based on the standard derivation of radiation from a short dipole. I'm trying to keep the explanation on the lowest possible level, and also make the explanation clear. Hence, it takes some time to put this together. The approach I'll show allows someone familiar with basic antenna theory to follow.

My basic approach considers that the radiation seen at any particular time is the result of the current from the accelerating charge at one particular section of it's path. This small section can be viewed as a short antenna with a current that is an impulse function. Fourier analysis can be used to represent this current pulse as a broad spectrum (that is, an integral of sinewaves over a frequency band). The full effect of what an observer sees over time is then the integration over the entire path of the charged particle. The overall effect is a pulse of radiation due to distance effect (low radiation intensity when particle is far away), combined with the broad spectrum generated by a current pulse in each short section of the particle's path.

[kcdodd]

I ran across this article which seems interesting and relevant to what you are saying here.

http://www.mathpages.com/home/kmath528/kmath528.htm

It seems there is some controversy about the case of a uniformly accelerating charge. One of the paradoxes seems tied to the equivalence principle. If a uniformly accelerating charge radiates, then a stationary charge in a gravitational field should radiate also. These types of questions are beyond my expertise. For this reason, anything I say should apply to a charge accelerating in one direction, but not necessarily a constant acceleration. The above link makes reference to the first and third derivatives of position. It's also an interesting discussion.


Yes I encountered that "paradox" a while back, but it is not really the issue here. The short answer is uniform acceleration does radiate. But if acceleration is constant, then magnitude of poynting flux is constant (assuming literature is correct), which has a spectrum of a single frequency of zero. I am not really sure how this frequency domain relates to photon frequency domain. Interested to see what you get with a different approach.

[Phrak]

A charge undergoing constant linear acceleration over a time interval emits a pulse of radiation. It's a single pulse; not periodic. But the frequency components are periodic. The probability of measuring a particular frequency for an emitted photon is related to the energy amplitude (or rather, the energy per unit of fequency) of the spectrum at that frequency.

This is the particle view in talking about an emitted photon, which I'm not particularly fond of. You might better consider evolving fields which are only quantized upon measurement.

[kcdodd]

If you restrict it to a finite time interval, then it is not undergoing constant linear acceleration. It's undergoing a step function linear acceleration.

[Phrak]

By the way, kcdodd. You can obtain some unphysical results integrating the Poynting vector over a surface. In a periodic electromagnetic wave, the Poynting vector is also periodic, passing through zero. The flux energy2 for a monochromatic wave, for instance, is proportional to the time averaged Poynting vector.

[kcdodd]

How is that unphysical?

[elect_eng]

Yes I encountered that "paradox" a while back, but it is not really the issue here. The short answer is uniform acceleration does radiate.

Yes, I agree with that. I just thought this is interesting and relevant to the thread, even if not the real issue. Also, since resolution of this type of issue is more in the realm of a theoretical physicist, rather than an engineer, I don't want my answers to encroach into this area. However, it does seem well established that a uniform charge radiates, so my comments should apply to the constant acceleration case too.

I also ran across another article which claims there is no real paradox, but states that the equivalence principle does not hold for the charged particle. Again, I can't confirm or deny these ideas, I'm just adding them in for the OP to consider.

A PDF of the article can be downloaded here.

http://arxiv.org/abs/gr-qc/9303025

[kcdodd]

When I looked into this issue I came to the conclusion that the equivalence principle holds just fine. I believe the author there assumed that a stationary charged particle in a gravitational field "weighs" the same as an uncharged particle of the same mass. It does not. The em field associated with the charged particle is not supported by whatever is supporting the charge, and so it creates a stress on the charge, adding an extra downward force making it seem heavier. In the free fall frame this extra force IS the radiation reaction force. The conclusion therefore is that radiation is not present in accelerated frame, but is present in the free-fall frame, which is interesting all by itself.

[elect_eng]

If you restrict it to a finite time interval, then it is not undergoing constant linear acceleration. It's undergoing a step function linear acceleration.

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? Do you acknowledge that the nonuniform case does create a pulse and a broad frequency spectrum? If this is your position, then we are left with the conclusion that you are saying that the effect of the entire particle path over all time is able to effect what an observer sees at one particular instant in time. This is not logical. Obviously, the effect of the future can not be felt in the present, and the effects from the past have already propagated past the observer (at the speed of light).

As I described above, what the observer sees at any one instant is the effect of one small portion of the particle's path. This is because the current is localized both in space and in time. The current by its very nature is transient even in the constant acceleration case. If you take any point on the path, the current is zero, for all time accept one instant as the charge passes by. There will of course be a delay between the current pulse at that point and the observer (retardation effect) due to the finite speed of light. However, there is a one to one correspondence (or mapping) between a location on the particle's path and the time that the observer sees the effect of the charge as it passes that location.

It's still not clear to me how you are viewing the situation. Perhaps you are considering a frame of reference that moves with the particle. This would be a non-inertial (accelerating) frame of reference and might not be appropriate to consider. Or, at least great care would be needed to make interpretations in this frame.

[Phrak]

I hope everyone realizes this all very speculative.

Some of your equations need some help. You used the current density J and substituted qv. J is current density. It is not charge in motion. The units don't match-up either. To obtain J for your Poynting vector equation begin with a charge density rho, and Lorentz boost to obtain a current component.

If you use a classical point particle, both rho and J go to infinity, so it might be better to consider some small charged body.

How did you conclude that a charged particle couples to the vacuum in proportion to av?

[kcdodd]

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? Do you acknowledge that the nonuniform case does create a pulse and a broad frequency spectrum? If this is your position, then we are left with the conclusion that you are saying that the effect of the entire particle path over all time is able to effect what an observer sees at one particular instant in time. This is not logical. Obviously, the effect of the future can not be felt in the present, and the effects from the past have already propagated past the observer (at the speed of light).


Yes, of course I agree that a pulse of acceleration will give a pulse of radiation. Just like periodic acceleration gives periodic radiation, etc.

I find your argument of future effect interesting. However, your example of periodic motion of a particle has the same downfall. The periodic motion must be specified for all time to give you a single frequency. You must specify the entire motion of the particle to do a Fourier transform on its trajectory, whether it be periodic acceleration, a burst of acceleration, or uniform acceleration.


It's still not clear to me how you are viewing the situation. Perhaps you are considering a frame of reference that moves with the particle. This would be a non-inertial (accelerating) frame of reference and might not be appropriate to consider. Or, at least great care would be needed to make interpretations in this frame.

I am considering an inertial frame of reference like a good boy. For the case of constant/uniform acceleration, I am sort of imagining an infinity long cylinder as my "closed surface", where the particle accelerates along the axis of symmetry, and ignoring the ends of the cylinder at infinity since the integral over the end should go to zero.

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 realized the relation a*v is incorrect. It came from the idea that E would be proportional to dJ/dt (or q*a), and the force would be proportional to E, and so the power into the field, which is F*v, would be proportional to a*v. However, what I was missing is that the self field of a particle is non-linear, and so it can't be solved in that manner. I think the correct relation for this situation is simply a^2, as I already mentioned.

[elect_eng]

Yes, of course I agree that a pulse of acceleration will give a pulse of radiation. Just like periodic acceleration gives periodic radiation, etc.

OK, it's good to understand that you are only talking about uniform accelleration.


I find your argument of future effect interesting. However, your example of periodic motion of a particle has the same downfall. The periodic motion must be specified for all time to give you a single frequency. You must specify the entire motion of the particle to do a Fourier transform on its trajectory, whether it be periodic acceleration, a burst of acceleration, or uniform acceleration.

That sounds like a valid point to me.




I am considering an inertial frame of reference like a good boy. For the case of constant/uniform acceleration, I am sort of imagining an infinity long cylinder as my "closed surface", where the particle accelerates along the axis of symmetry, and ignoring the ends of the cylinder at infinity since the integral over the end should go to zero.



OK, so isn's this a fundamentally different thing than what I'm doing. I am considering a fixed observer at a point, much the same way an antenna analysis would be done. We are often interested in the received radiation at a point due to excitation of a distant current source. If you integrate over an infinite cylinder, this is a completely different observation, isn't it?

I still say you need to have radiation with frequencies and a spectrum, but if you integrate over an infinite surface, perhaps you may not see a pulse if you talk about total integrated power. I'm not saying you will or won't - just that the answer is not clear to me and I would do a calculation before giving an answer about power. Have you actually done this calculation? If so, it would be interesting to see it.

[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.

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

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 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 accross 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, Az), where you might notice that Phi and Az transform to Phi' and Az' 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=Ao34iFuNZgIC&pg=PA72&lpg=PA72&dq=coaxial+waveguide+modes&source=bl&ots=dsBl23u4ft&sig=zDicjQIBA6uuKG6d1fWBc1uXP_Y&hl=en&ei=Ghq_S9PxKsugkQX-3rTrBQ&sa=X&oi=book_result&ct=result&resnum=2&ved=0CBEQ6AEwAQ#v=onepage&q=coaxial%20waveguide%20modes&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%E2%80%93Savart_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%27s_law


Now, 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 fundemental 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.

[kcdodd]

Energy and momentum won't be invariant. Just think about the same quantities of a particle. However, you can write it in covariant form. It is interesting that the lagrangian has the form of energy and yet is invariant. But then it must be or it wouldn't work.

This is handwavy, but E^2+B^2 and ExB is a way to look at the energy and momentum of the field itself. If you start with a pure electric field, say in the x direction, and then go to a frame moving in the -z direction, you will see an ExB in the +z direction. You might say the field now has momentum in the +z direction, as well as increase in total energy. In analogy to a particle, it might be called kinetic energy. That is why you can consider ExB as energy flow, when technically energy is only a scalar. With particles, the kinetic energy can be related to the momentum. If momentum is zero, then the total energy is just the rest energy.

[Phrak]

Energy and momentum won't be invariant. Just think about the same quantities of a particle. However, you can write it in covariant form.

Yes, I realize all that, which is why I used the toy equation that I did. Unfortunately it works for energy and momentum, but not energy density and momentum density.

So, with your inspiration to this thread, I got out the crayons and have been looking for a manifestly covariant energy transport equation in terms of electromagnetic fields. The direct (and skew) product of F and G,

F_{[\mu \nu} G_{\sigma \rho]}

has units of energy. F is the covariant Maxwell tensor, G is the Maxwell tensor multiplied by the mixed index 4 dimensional Levi-Civita symbol in the Minkowski metric. The seemingly imbalanced brackets are deliberate and denote skew multipliction.

There are 24 nonzero terms in the product. The elements with evenly permutated indeces equal -2E2. The odd permutations have entries of 2B2.

BTW, I don't know that this is going in the right direction. I'm still hunting. And if all this sounds like complete jibberish, you're not alone. I'm just hoping some of it is understandable.

[Dunnis]

The radial electric field strength of a coaxial cable operating in the fundemental 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.

What fields and what distance are you talking about exactly, the same ones I draw on my diagrams, and for which I provided equations and links to Wikipedia? -- So, you are just saying everything I said is wrong, including my reference, and what you say is correct, but please, if you mean to disagree, provide some link, just show me some article, paper or some other web-site where I can see what in the world you are talking about, ok?


- "coaxial cable" or not, has nothing to do with this, wires are electrically neutral at any macroscopic distance away from them, which means "around them".

- "operating mode", whatever that is supposed to mean, makes no difference to the magnitude of electric field potential around any cables.

- "alternating currents" has nothing to do with 'radial distance and electric field potential around any cables'.

Wires would NOT be SAFE if they were electrically charged, i.e. if they had much more positive or negative charges per any given point over length, but cables are uniformly made both of positive and negative charges, regardless of any changes in the direction or magnitude of their drift velocity along that length.


You may disagree, and if so please be specific this time, but most importantly I just want to see that information from the "1st hand", some valid source, as you seem to be misinterpreting something.

[Phrak]

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.

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.

In reply to myself, you said:

You may disagree, and if so please be specific this time, but most importantly I just want to see that information from the "1st hand", some valid source, as you seem to be misinterpreting something.

Let's start from the beginning, with your first critical statement. You might read very carefully the conditions set up by elect_eng again.

Being both insistent and wrong about the most elementary physics doesn't play well on this forum. Take this as a word to the wise.

[kcdodd]

Phrak, is that not the em field lagrangian?

Dunnis, I am not even sure what you are taking issue with. If you have a difference in voltage, then there is an electric field. A point's field drops off as 1/r^2, an infinite line as 1/r, and a infinite plane as a constant. This goes for both electric and magnetic sources.

[Dunnis]

Being both insistent and wrong about the most elementary physics doesn't play

I'm not interested in your opinions and advices given out ignorance. Basic physics are Coulomb's and Biot-Savart law, and you are the one here that wishes to refute them, so I'm asking for the 3rd time now:

1.) Show me Wikipedia article where I can see those "other" formulas for E and B field.

2.) Google whatever other web-site, paper or article where I can see those two equations.


I'm not insistent any more than you're refusing to support your claims. Whatever your status on this forum is, this is, I hope, still scientific discussion, so provide EVIDENCE and don't expect me to believe a complete stranger that uses threats and intimidation instead of valid sources and references.

[Phrak]

I'm not interested in your opinions and advices given out ignorance. Basic physics are Coulomb's and Biot-Savart law, and you are the one here that wishes to refute them, so I'm asking for the 3rd time now:...

I'm not interested in your discourse.

[Dunnis]

I'm not interested in your discourse.

Ok, I just thought you would like to know those equations are incorrect, and also that pretty much anything you attempted to disagree with me is wrong too, but I do not mean to be bothersome, so I'm sorry, and please, go on, as you were...

[Dunnis]

Dunnis, I am not even sure what you are taking issue with.


These two equations are wrong:


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



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



No mater if "inside" or "outside" the cable, the Coulomb's law and Biot-Savart law should hold. Now, I may be mistaken of course, and all I'm asking is to see some reference where I can see exactly where and how did those formulas come to be.



If you have a difference in voltage, then there is an electric field.


You mean if you change voltage over time? Otherwise it is called "difference in electric potential" OR "voltage". It is not "difference in voltage", because the "voltage" IS "difference in electric potential". There is no such thing as "difference in voltage" when talking about a single wire, but only if you change voltage during some time period.


What "electric field" are you talking about?

ELECTRIC FIELD: E(r)=Ke*(-Q+Q)/r^2 = zero
|
| MAGNETIC FIELD: B(r)=Km*I*L*sin(90)/r^2
| r |
| | r
| |
MORE+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+->>LESS
ELECTRONS-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+->>ELECTRONS
LESS PROTONS+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+--+-+-+-+>>MORE PROTONS
| |
|<-------------POTENTIAL DIFFERENCE (VOLTAGE)---------->|


Is there anything here you disagree with?

Do you think that E(r>0) can be anything else but zero?



A point's field drops off as 1/r^2, an infinite line as 1/r, and a infinite plane as a constant. This goes for both electric and magnetic sources.

No, that is impossible. Please show me where do you draw your conclusions from. Can you please Google some link or show me Wikipedia article so I can see for myself what is it you are talking about?

[Born2bwire]

No, that is impossible. Please show me where do you draw your conclusions from. Can you please Google some link or show me Wikipedia article so I can see for myself what is it you are talking about?

Coulomb's law only deals with a single point charge. You need to do the appropriate integrals for multiple charges. Though with line and planar sources you can use Gauss' law to derive the appropriate electric fields.

[kcdodd]

For an infinite line, those equations are correct. You can do a google search for youself, and I suggest actually doing the calculus as an excersise to see where they come from.

Electric field is defined as the spatial derivative of voltage. In si it has units of volts/meter. So whe I say a change in voltage, that's literally what I mean. If you have two conductors at different potentials, you know there has to be an electric field between them.

[Dunnis]

Coulomb's law only deals with a single point charge. You need to do the appropriate integrals for multiple charges. Though with line and planar sources you can use Gauss' law to derive the appropriate electric fields.

Why is it so hard to Google some link where I can see what you telling me is not just your imagination, again? It is physically and mathematically impossible for a single electric field potential to change its distribution and gradient, perhaps in special relativity, but in any case, if what you say is true, then surely there must be some internet article to mention it. -- Show me some evidence that can support your opinion, some links and papers, please.

[Dunnis]

For an infinite line, those equations are correct. You can do a google search for youself, and I suggest actually doing the calculus as an excersise to see where they come from.

Electric field is defined as the spatial derivative of voltage. In si it has units of volts/meter. So whe I say a change in voltage, that's literally what I mean. If you have two conductors at different potentials, you know there has to be an electric field between them.

No, that is nonsense and is impossible, there is nothing on the internet that agrees with your assumptions. I do this calculus every day, it's my job, and if you want to support your assertions than you should be able to Google at least one of those links.

Why I can not I find anything about it in Wikipedia, and would it not be easier for you to just copy/paste the damn link? -- Seriously now, CAN YOU provide any reference to support YOUR CLAIMS, or not?

[kcdodd]

Don't say I never did anything for you. Second link in google.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

[elect_eng]

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.

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:

...

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

Can you show me your equations anywhere in Wikipedia?

...

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%27s_law


Now, can you provide some reference for this nonsense: E_r={{q}\over{2\pi \epsilon r}}




Good God man. What is wrong with you?

I did say "inside". You can check my post and read it for yourself. Unfortunately, because you have not bothered to read the thread carefully, you don't even know what "inside" means in the context we are talking about. I'm talking about a coaxial cable that has a central thin wire and a cylindrical shell outer conductive shield. "Inside" means the volume with the dielectric between the two conductors. Outside the shield, the fields are zero. Inside the shield the fields are what I said. One of a thousand references I could site is as follows.

"Fields and Waves in Communications Electronics", by Ramo, Whinnery and van Duzer. Pages 76 and 77 show the magnetic field.

On page 10, there is example 1.4b entitled "Field about a line Charge, or between coaxial cylinders". The solution is:

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

Since we are talking about two conductors and a coaxial transmission line, the concept of capacitance and charge applies here. A coax line can act as a capacitor. If you are not aware of this, don't criticize me. Go back and study what you should know. This can be found under chapter 8 of the above reference. The chapter title is "Waveguides with cylindrical conducting boundaries".

By the way, you keep talking about Wikipedia. This is not a reliable source of information. You can pick up ANY electromagnetics book and find this all worked out. Besides, don't you even know how to apply Maxwell's equations to a basic problem like this?

From Maxwell's expression of Ampere's Law in the static case \ointop \vec H \cdot d\vec l = 2\pi r H_{\phi}=I

therefore, H_{\phi}={{I}\over{2\pi r}}}

Was that so hard?

EDIT: OOPS, I forgot that I need to provide a reference for Maxwell's Equations, otherwise I will be viewed as an alien from another planet. Please see the following. "A Treatise on Electricity and Magnetism" by James Clerk Maxwell, 1873.

And, lets not forget the Wikipedia link: http://en.wikipedia.org/wiki/Maxwell's_equations
http://en.wikipedia.org/wiki/Maxwell's_equations

[elect_eng]

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%27s_law


Now, can you provide some reference for this nonsense: E_r={{q}\over{2\pi \epsilon r}}



In the previous post, I gave a reference for what you call "nonsense", but your stubbornness convinces me this will not be enough for you. Let's ignore your large number of misconceptions and focus on this one area. This is an easy one to handle because you claim to know about Coulomb's law from the first chapter of your EM fields books.

So let's proceed to chapter two and apply Coulombs law to a line of charges. I've attached a derivation as a jpg file. So, now you have a reference and a clear derivation in terms of a law you can understand and a basic application of calculus. Contrary to two of your statements above. There is something to derive, and the final equation is not nonsense.

I'd like to state for the record that I don't appreciate someone who is unwilling to read past the first chapter of his book saying I'm "dreaming", "stating nonsense" and "from another planet".

[Dunnis]

I did say "inside". You can check my post and read it for yourself. Unfortunately, because you have not bothered to read the thread carefully, you don't even know what "inside" means in the context we are talking about. I'm talking about a coaxial cable that has a central thin wire and a cylindrical shell outer conductive shield. "Inside" means the volume with the dielectric between the two conductors. Outside the shield, the fields are zero. Inside the shield the fields are what I said.


You "shield" magnetic fields, electric fields you "NEUTRALIZE" with superposition, simply by having conductors be made of regular material, all of them are electrically neutral, there is nothing to "shield" with electric fields, but only to prevent "sparking" or direct contact, i.e. current flow. There are no any electric fields around the conductor at any macroscopic distance, as I said. The dielectric between the two conductors is to prevent microscopic contact and keep the distance, but you are wrong to refer to it as INSIDE, as that is just another layer of OUTSIDE, the only difference is that it is not air between those two but some plastic.



E(r2),B(r2)
|
|
| r2
AIR OUTSIDE 2 |
|
========================================RUBBER==== ==========
+-+-+-SHIELD INSIDE-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+- |
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ | C
. . . . . . . . . . . . . | O
. . . . . . . . . . . . . E(r1),B(r1) | A
. . PLASTIC OUTSIDE 1 . . | | X
. . . . . . . . . . . . . | r1 | I
. . . . . . . . . . . . . | | A
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+- ->> | L
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+- ->> |
+-+-CORE INSIDE-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+- ->> | C
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+- ->> | A
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+- ->> | B
. . . . . . . . . . . . . . . . . . . . . . . . . . | L
. . . . . . . . . . . . . . . . . . . . . . . . . . | E
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ |
================================================== ==========


Any objection to the diagram? We are talking about the same fields and different materials do not make difference to what is "outside" and what the "inside" of the CONDUCTOR ITSELF, which means - ONLY THAT WHICH IS MADE OF METAL, any point that is not located in this substance is OUTSIDE of that substance, ok? There ere TWO conductors here and that case simplifies to the case of two parallel wires, and then it boil down to the line integration against a single charge - coaxial cable or not. Having a complex combination of insulators and conductors changes nothing as the problem need to be split to its basic components to be solved, and only then to be combined (integrated) and superimposed macroscopically. -- Nothing is wrong with me, but with your equations, understanding and mathematics, as I will demonstrate it step by step, hold on...

[Dunnis]

Don't say I never did anything for you. Second link in google.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imgel2/elin2.gif



\lambda= 1C/1m = ?

E = \frac{\lambda}{2\pi\epsilon_0*r} => \frac{1C/1m}{2\pi\epsilon_0*1m} = \frac{1C}{2\pi\epsilon_0*1m^2} \ N/C


Coulomb's law: E = \frac{Q}{4\pi\epsilon_0*r^2} => \frac{1C}{4\pi\epsilon_0*1m^2} \ N/C


So, all that mathematics to change 1/4Pi to 1/2Pi? -- That is not right, it's some kind of butchering and symbolic derivation, while that integral was never supposed to be derived, but numerically integrated at the time of application, with given numerical input and REAL NUMBERS, not symbols. Symbols and derivation can not generalize arbitrary scenarios, and so that equation must stay in its integral form so it actually can be applied to wires of different and arbitrary lengths in relation to different and arbitrary geometry, where given input is defined by the REAL-WORLD numerical values.

E = \int_a^b \frac{Q*dl}{4\pi\epsilon_0*r^2} => \int_a^b \frac{1C*1m}{4\pi\epsilon_0*1m^2} \ Nm/C = 1 Volt = N/(m/C)





================== LET'S DO SOME REAL WORD PHYSICS, shall we? ==================

CASE A: CASE B: CASE C:


E(r),B(r)=? E(r),B(r)=? E(r),B(r)=?
| | |
|r=2m |r=1m | r=0.5m
| | |
|===================| |==========| + ==================infinite >> -
| I1= 1A | | I1= 1A | I1= 1A
|<------- 9m ------>| |<-- 1m -->|
+ - + -


E = \frac{\lambda}{2\pi\epsilon_0*r} \ N/C


\lambda= 1C/1m = ? How do you get lambda, and what result your equation predicts for the above three scenarios: with 9 meters, 1 meter and infinite wire?

[elect_eng]

Any objection to the diagram?

I object to your diagram.

I object to every word you have said in this thread.

I object to you having no consideration to try to understand an existing thread.

I object to you interrupting this thread with nonsense.

I object to your harassing tone.

I object to anyone who doesn't know that Coulomb's law and Biot Savart's Law apply to differential elements and integration is required to get the answer for real objects, yet purports himself to an expert "who does this for a living".

I object to needing to waste my time trying to educate an ignoramus who will not admit when he is wrong even in the face of overwhelming evidence.

Mostly, I object to all of the stupid comments you will be spouting in response to this post.

I won't be responding to any more of your nonsense. However, I will be reporting your posts as inappropriate.

[Dunnis]

What is wrong with you?
...
One of a thousand references I could site is as follows.

"Fields and Waves in Communications Electronics", by Ramo, Whinnery and van Duzer. Pages 76 and 77 show the magnetic field. On page 10, there is example 1.4b entitled "Field about a line Charge, or between coaxial cylinders". The solution is:
E_r ={{q}\over{2 \pi \epsilon r}}


That equation is wrong and demonstrates nicely what you know about it.

What do you imagine 'q' stands for?


E(r)
|
| r=2.5m
|
-----------------I= 23A -------------------->


q= ??! How do you solve this problem?


q= 1.602176487(40)×10^-19 coulombs

Do you realize that equation predicts the same result regardless of any wire length? What that equation does is to evaluate electric field of a SINGLE electron, regardless of any voltage, amperes or oscillation or wire length. It is the same thing as Coulomb's law for POINT CHARGES, only it has 1/2pi*r, where it should be 1/4pi*r^2. -- It is not "q", but it should be Q(net), which is zero to start with, since wires are made of atoms and atoms are electrically neutral when measured at any macroscopic distance, either from a single atom or volume of atoms, and even a line of atoms. -- Why is it surprising all the wires in your house are electrically neutral? Is that some news?

Q(net) = sum(-q,+q) --- made of atoms ---> electrically neutral (mostly and usually)


THREE DIMENSION, three integrals: C/m, Cm^2, C/m^3

1.) Q(net Line) ; 2.) Q(net Surface) ; 3.) Q(net Volume)

http://en.wikipedia.org/wiki/Charge_density


You can not "derive" integrals, integral equations must stay in their integral form so you can INTEGRATE over given distance segments and certain geometry defined by the NUMERICAL VALUES given by the specific given problem - i.e. integrals and not "derived", but numerically integrated. And to realize this you only need to actually try to USE YOUR EQUATION on some REAL WORLD SCENARIO, where the actuality is different than the one in Wonderland.



From Maxwell's expression of Ampere's Law in the static case \ointop \vec H \cdot d\vec l = 2\pi r H_{\phi}=I

therefore, H_{\phi}={{I}\over{2\pi r}}}

And, lets not forget the Wikipedia link: http://en.wikipedia.org/wiki/Maxwell's_equations


What the... ?!? There is no such equation in that Wikipedia article or on the whole internet. Nowhere in Wikipedia there is any mention of any of your equations, is that why you're angry(?) -- This is what Wikipedia says:

Free charge and current:
\oint_{\partial S} \mathbf{H} \cdot \mathrm{d}\mathbf{l} = I_{f,S} + \frac {\partial \Phi_{D,S}}{\partial t}

Total charge and current:
\oint_{\partial S} \mathbf{B} \cdot \mathrm{d}\mathbf{l} = \mu_0 I_S + \mu_0 \varepsilon_0 \frac {\partial \Phi_{E,S}}{\partial t}



So let's proceed to chapter two and apply Coulombs law to a line of charges. I've attached a derivation as a jpg file. So, now you have a reference and a clear derivation in terms of a law you can understand and a basic application of calculus. Contrary to two of your statements above. There is something to derive, and the final equation is not nonsense.

I'd like to state for the record that I don't appreciate someone who is unwilling to read past the first chapter of his book saying I'm "dreaming", "stating nonsense" and "from another planet".
Attached Images File Type: jpg Efield.jpg


You mean this below? Is that supposed to be the same?
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imgel2/elin2.gif
-----------

You got it wrong it seems, can you please decide which one:

E = \frac{\lambda}{2\pi\epsilon_0*r}

-OR-

E = \frac{q}{2\pi\epsilon_0*r}



...since your equation always give the same result regardless of any specific input, I'm really curios how do you solve for DIFFERENT scenarios we just happen to have in REAL WORLD, so can you show me how do you apply that equation to the following setups and what result do you get:

CASE A: CASE B: CASE C:


E(r),B(r)=? E(r),B(r)=? E(r),B(r)=?
| | |
|r=2m |r=1m | r=0.5m
| | |
|===================| |==========| + ==================infinite >> -
| I1= 1A | | I1= 1A | I1= 1A
|<------- 9m ------>| |<-- 1m -->|
+ - + -


E = \frac{q = ?}{2\pi\epsilon_0*r}


E = \frac{\lambda = 1C/1m = ?}{2\pi\epsilon_0*r}

[kcdodd]

Can we please get a moderator?

[Dunnis]

Can we please get a moderator?

Huh? What for? What Wikipedia articles and what equation do you believe is wrong? What kind of discussion is this where I provide all the reference and get called "stupid", yet those who can not support their claims just scream around and wave hands without actually saying anything? Just say it already, what exactly are you so nervous about?

[elect_eng]

Can we please get a moderator?

I've reported his crazy posts. I recommend others do the same. He can't even tell the difference between charge and charge density, yet he wants to claim every EM book ever written is wrong. This is yet another clear example of how a little bit of knowledge can be a dangerous thing.

[Dunnis]

I've reported his crazy posts.

He can't even tell the difference between charge and charge density..

Interesting reaction, and I only wanted to see some evidence for your assertions.


YOUR FALSE EQUATION: E = \frac{q}{2\pi\epsilon_0*r}

It is you who has a single charge of a single electron in your equation, and moderators will tell you that, whatever the reason you want to call them. You may as well call your mum too, that will not change the reality and what has come to past. Next time be careful about the equations and try to use them before you make your conclusion, so to not embarrass yourself like this. Ok? -- At least you realized there is a TOTAL amount of charges here, now learn about it:

http://en.wikipedia.org/wiki/Coulomb

http://en.wikipedia.org/wiki/Charge_density

http://en.wikipedia.org/wiki/Elementary_charge

[elect_eng]

It is you who has a single charge of a single electron in your equation, and moderators will tell you that, whatever the reason you want to call them. You may as well call your mum too, that will not change the reality and what has come to past. Next time be careful about the equations and try to use them before you make your conclusion, so to not embarrass yourself like this. Ok? -- At least you realized there is a TOTAL amount of charges here, now learn about it:



No, it is you that has ignored all of the explanations given to you. I provided you a derivation in an attached jpg file. Did you even look at it after you asked for references and explanations? No you did not. I quoted a book. Did you bother to consult it? No you did not. The variable q is linear charge density with units of Coulombs per meter. It is not the single electron charge as you say. Anybody can go back into this thread and read proof that I explained this. Besides, it needs no explanation. This is such an obvious fact from the context of the equations. You are asking for explanations that are like asking what 2+2 is. Who do you think you are kidding? You are the one who is embarrassing himself.

[elect_eng]

YOUR FALSE EQUATION: E = \frac{q}{2\pi\epsilon_0*r}



Why do you call this "my false equation"? This equation can be found in every electromagnetics book. EVERY SINGLE ONE! I gave reference to one book above. The well known book by Krauss is another, and Jackson can be consulted too. I gave a jpg file with my own derivation and kcdoddd gave a link to another derivation. You keep asking for references and we give them, yet you ignore them. What is your problem? Are you just doing this as a prank? Are you just unwilling to admit when you are wrong? Whatever the issue is you'd better come to terms with it. By the way, at one point you mentioned you do this for a living. I dare you to show this thread to your boss, or your customers if you are self-employed.