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 traveling 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 traveling 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 anyone 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 across the surface. The time average will still work out, so I don't see a problem. And the zero frequency thing is a current question I think.

 

[Phrak]

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

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

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

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

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

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

 

[kcdodd]

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

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

And if you work out the solutions for

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

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

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

Which has a divergence of zero

where now you have

E_0 = \omega A_0
B_0 = k A_0

 

[Phrak]

And if you work out the solutions for

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

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

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

Which has a divergence of zero

where now you have

E_0 = \omega A_0
B_0 = k A_0

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

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

 

[kcdodd]

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

 

[Phrak]

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

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

 

[kcdodd]

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

 

[elect_eng]

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

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

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

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

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

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

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

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