Accelerating Charges: Classical Theory and Light Radiation

[Cheman]

According to source i read, in classical theory it its believed that charged particles radiate when the accelarate - but do they have to be accelerating or can they just be moving? So if we had an electron gas, each time "collision occured" light would be radiated as the particles decellerated, but not inbetween when they are traveling at constant speed?

Thanks in advance.

 

[dextercioby]

Check out the Liénard-Wiechert potentials and fields in any serious electrodynamics book,J.D.Jackson being the first to turn to.

Daniel.

 

[shyboy]

yes, there is certainly some radiation called Bremsstrahlung, produced by electrons colliding with ions. I am not so sure about electron-electron scaterring.

 

[pmb_phy]

No. Charges moving at constant velocity do not radiate. The charges must be accelerating to radiate.

Pete

 

[dextercioby]

How do you prove the charges moving at constant velocity do not radiate...?I have a hunch they do radiate.

Daniel.

 

[pmb_phy]

How do you prove the charges moving at constant velocity do not radiate...?I have a hunch they do radiate.

Daniel.

Take my word for it when I tell you that this kind of derivation is not something I can remember. With me physics is a "Use it or loose it" situation.

The power radiated by a charged is proportional to the magnitude of the acceleration. The derivation can be found in Jackson. A buddy of mine is an expert in this area. He did a lot of work on this subject. Unfortunately he's on an extended vacation. Basically he says that classical radiation is defined as when the EM field "breaks away" from the field of the charge.

However that does not mean that there is no energy flux when the charge is moving at constant velocity. The Poynting vector is not zero. In fact anytime you have a non-zero Poynting vector there is a flow of energy - as surprizing as it may seem. This fact is well worth your attention. I'll dig out the references it you'd like? One of them is in Shadowitz's text on radiation. There is also 1 or two articles in the American Journal of Physics on this as I recall. I think I have them in fact.

Pete

 

[Andrew Mason]

No. Charges moving at constant velocity do not radiate. The charges must be accelerating to radiate.

Pete

But do they radiate because they accelerate? Or do they accelerate because they radiate? After all, no one has been able to detect radiation from a charge accelerating due to gravity.

According to Feynman, the radiation of a charge depends on the third time derivative of position (ie non-uniform acceleration). So constant acceleration does not produce radiation. This is still somewhat controversial. see: Does A Uniformly Accelerating Charge Radiate?

AM

 

[dextercioby]

Alright,Pëte.Computing the Poyting vector for a Liénard-Wiechert field generated by a constant velocity moving electric charge,one gets a nonzero result.Ergo,radiation is being emitted.

I'm aware people always ponder about Larmor's formula,but this case must not be excluded.

Daniel.

 

[dextercioby]

But do they radiate because they accelerate? Or do they accelerate because they radiate? After all, no one has been able to detect radiation from a charge accelerating due to gravity.

According to Feynman, the radiation of a charge depends on the third time derivative of position (ie non-uniform acceleration). So constant acceleration does not produce radiation. This is still somewhat controversial. see: Does A Uniformly Accelerating Charge Radiate?

AM

It is a thorny subject indeed.But I'm sure i was not referring to a generally covariant theory of classical electrodynamics,but only about a Lorentz invariant theory.

Daniel.

 

[Cheman]

So when we say an accelerating charge "radiates" what do we actually mean? I mean, say we have an electron moving around at constant velocity - this is clearly exerting an electric/ megnetic field on any charged particles near it. How come as soon as it accelerates this field suddenly becomes light? How come a charge accelerating would cause an oscillating "light field"? How does this field become different to be classed as light rather than just an electric field? How is classical light absorbed? eg - people would have said that things were green because they reflected green light and absorbed everything else - what would they say it would have been absorbed by? Would it be electrons, as in quantum theory, and if so how is this "force" absorbed> ( ie - if light is just a force field oscillating as the theory suggests, then how could it be "absorbed"? I mean, afteral, if you had an electron and put a proton near it and then another one behind that, the second protons energy is not altered due to force/ energy being absorbed by the proton in front and thus weakening the force - the only thing that makes the force from the electron slightly weaker for the 2nd proton is that this proton is further away? (if we say that the protons do not hypothetically repel each other.) Forces at a distance can't have their energy absorbed by objects the act on - so why does this happen with classical light?

I always feel that the classical model of light is always completey over looked in my studies - we are told "its wave - the end" and then quantum begins to be mentioned - but no one actual explains what people before quantum believed light to be like, and what form it took. I would be very interested if anyone could explain to me from scratch what classical light is like, how its acts, is formed, and was believed it would act with particles. (obviously, this was wrong as later proved by photoelectric effect,etc.)

Thanks in advance.

 

[McQueen]

cheman
According to source i read, in classical theory it its believed that charged particles radiate when the accelarate - but do they have to be accelerating or can they just be moving?

One explanation that is given is that charged particles moving with uniform velocity , give rise to magnetic fields while accelerating charges give rise to electromagnetic fields. For instance in a DC circuit as soon as the current is established it ceases to effect conductors in its vicinity although it will still deflect a compass needle , attract or repulse another wire carrying a current and so on. An AC circuit would induce a current in conductors in its vicinity..

 

[dextercioby]

cheman
According to source i read, in classical theory it its believed that charged particles radiate when the accelarate - but do they have to be accelerating or can they just be moving?

One explanation that is given is that charged particles moving with uniform velocity , give rise to magnetic fields while accelerating charges give rise to electromagnetic fields. For instance in a DC circuit as soon as the current is established it ceases to effect conductors in its vicinity although it will still deflect a compass needle , attract or repulse another wire carrying a current and so on. An AC circuit would induce a current in conductors in its vicinity..

That's not true.Do some reading,okay and stop posting inaccurate info.

Daniel.

 

[McQueen]

Dextercioby
That's not true.Do some reading,okay and stop posting inaccurate info.
Check out this site , I think the author is saying more or less what I said in my post , although he uses the word a changing field as opposed to accelerating charges.
http://www.ch.ic.ac.uk/local/physical/mi_5.html

 

[dextercioby]

I give you this formula

\vec{E}_{\mbox{Lienard-Wiechert}} =\frac{q}{4\pi\epsilon_{0}}\frac{1-\frac{v^{2}}{c^{2}}}{\left(R-\frac{\vec{R}\cdot\vec{v}}{c}}\right)^{3}} \left(\vec{R}-R\frac{\vec{v}}{c}\right)\left|_{t'}\right +\frac{\mu_{0}}{4\pi}\frac{q}{\left(R-\frac{\vec{R}\cdot\vec{v}}{c}}\right)^{3}} \left\{\vec{R}\times\left[\left(\vec{R}-R\frac{\vec{v}}{c}}\right)\times\vec{a}\right]\right\}\left|_{t'}\right

,where t' is the retarded time,\vec{v} is the velocity of the charge "q",\vec{a} is the acceleration of the charge and

\vec{R}=\vec{r}-\vec{r}_{0}(t')

,where \vec{r}_{0}(t') is the position of the charge at retarded time.

Daniel.

P.S.The magnetic field is obtainable from the expession of the electric one.

 

[dextercioby]

One sees that there's a nonzero time dependent component of the electric field (and of the magnetic one of course),even in the absence of acceleration (case in which the second term of sum in the electric field's expression vanishes).

Daniel.

 

[Cheman]

So suppose I had a highly charged rod - if i moved it across at constant velocity from one side to another we would not class its field as light. Yet if we wave it side to side it IS electromagnetic radiation. Why? Surely the fields are exactly the same and both do exactly the same things?! (ie - impose force on other charged objects.)

(NB - if anyone can answer the questions on my previous post that would also be useful. )

Thanks in advance.

 

[Cheman]

Also, how does classical wave theory of light explain why light is absorbed? Eg - if we put a green filter infront of our eyes all other light is absorbed - but how according to classical theory? It obviously can't use the resonace idea ( like people tried to use on the photoelectric effect, even though attempts at getting this to fit failed. ) since that would not cause the wave to be absorbed as it is, in essence, simply a long distance force. However does this force field, except for the force field of green, in this example with a green filter, manage to be absorbed and not poass through the filter according to classical wave theory? Was it even related to the atomic structure of matter at all, or to something elese?

Thanks in advance.

 

[Cheman]

Or would it be (now that I've had a think about it. ) anything to do with if the light frequency casued "particles" (electrons or whatever ) to resonate, then they will do it in anti-phase to the light - therefore setting up their own light which destructively interferes with it? I think I'm confussing myself now... lol.

Any help appreciated!

Thanks.

 

[Cheman]

Erm... could anybody please add some input and help?

Thanks in advance.

 

[Crosson]

One explanation that is given is that charged particles moving with uniform velocity , give rise to magnetic fields while accelerating charges give rise to electromagnetic fields

The reason that this is wrong is that there is a very important distinction between charged particles moving at a constant velocity and a "constant current" ala magnetostatics.

 

[pmb_phy]

According to Feynman, the radiation of a charge depends on the third time derivative of position (ie non-uniform acceleration).

Feynman is talking about the radiation reaction. Not the radiation.

So when we say an accelerating charge "radiates" what do we actually mean?

Loosley speaking we mean that the EM field "breaks away" from the charge. And yes. They must be accelerating to radiate. A charge moving at constant velocity will not radiate even though it will produce both electric and magnetic fields. The presence of E and M fields does not dictate whether radiation is present.

Pete

 

[dextercioby]

Creating an electromagnetic field with nonzero Poynting vector means emitting radiation...An electric charge moving with constant velocity emits radiation.

The end.

Daniel.

 

[Cheman]

So if the field only "breaks way" when a wave is formed, does that mean that electrical fields do not usually take the time for light to travel to cause an effect? ie - are they instantaneous?

 

[dextercioby]

Nothing is instantaneous.Electromagnetism is a relativistic theory.

Daniel.

 

[pmb_phy]

Creating an electromagnetic field with nonzero Poynting vector means emitting radiation...An electric charge moving with constant velocity emits radiation.

The end.

Daniel.

That is quite wrong. Charge a stationary magnet. Wait for the charges to come to rest on the surface of the magnet. The field around the magnet will have a non-zero Poynting vector. There will be no radiation though. It appears to me that you're confusing the flow of EM energy with the flow of EM radiation.

You really should read the article
Physical Significance of the Poynting Vector in Static Fields, Emerson Pugh and George Pugh, Am. J. Phys. 35, page 153-156, 1967.

Or better yet read section 11-3 on the Poynting Vector in
The Electromagnetic Field, Albert Shadowitz, Dover Pub

Its a fascinating text and a fascinating section.

 

[pmb_phy]

So if the field only "breaks way" when a wave is formed, does that mean that electrical fields do not usually take the time for light to travel to cause an effect? ie - are they instantaneous?

No. It means that as you accelerate a charged particle part of the field continuosly separates from the charge and becomes radiation. That's a very fuzzy way to phrase this but I know of no precise definition of the term "radiation."

Pete

 

[Adrian Baker]

What about relativity here? Depending on the frame of reference, a charge can be at rest, with constant velocity, or accelerating. Would observers in different frames of reference see the charge doing different things? I'm pretty sure that the answer to this is yes, but how can this be so in energy terms??

Unfortunately I don't have the maths to work this out - can it be explained easily without maths or not, or is my original assumption about frames of reference wrong?

 

[dextercioby]

Don't involve general relativity and electromagnetic radiation.Hilbert-Einstein-Maxwell action gives good results for charged BH,but the issue of accelerating electric charges emitting radiation is still open.

Daniel.

 

[pmb_phy]

What about relativity here? Depending on the frame of reference, a charge can be at rest, with constant velocity, or accelerating. Would observers in different frames of reference see the charge doing different things? I'm pretty sure that the answer to this is yes, but how can this be so in energy terms??

If the charge is accelerating in one inertial frame then its accelerating in all inertial frames. Therefore if radiation is detectable in one inertial frame then its detectable in all inertial frames. If the charge is moving uniformly then it its rest frame there is only an E-field. According to relativity if you now transform to another inertial frame there will be both an E field and a B field thereby giving a flow of energy as defined by the Poynting vector defined as E cross B.

Pete

 

[reilly]

By no means does a non-zero Poynting Vector imply radiation. Rather, it means there is a local energy/momentum flow. For a steady linear current, the Poynting vector is constant in time, and points in the direction of the wire. But there's no radiation. The momenta carried by the field and by the current are constant. And, if the system is turned on adiabatically, starting with everything at rest, the total momentum should be zero.

Radiation means energy flux through the surface at infinity, so the radiation fields must go as 1/r. In the L-W potential, given in one of dextercioby's prior posts,the 1/r term is proportional to the acceleration. No acceleration, no radiation.

Why? Maxwell says it's so. That is, to understand why charges emit and absorb radiation one must understand why maxwell's eq.s work. That's a very big order. But, the assumption that radiation/photons are emitted and absorbed classically and quantum mechanically is extraordinarily powerful in explaining how much of the world works.

In fact, the integral of the Poynting Vector over all space for a non-accellerating charge will be constant in time, even though vector itself can be locally variable, as common sense suggests must be the case.

The "origins" of radiation along the lines pmb phy mentioned do give a good picture of the process. As my E&M prof put it, when a charge accelerates, its Coloumb field is constantly changing, and the emited radiation between t and t+dt is necessary to compensate for the change in that field. Like, pull out a few photons here, and put a few more there. This can be more-or-less demonstrated with a bit of algebra applied to the L-W fields.

Why charge? Why fields? Who knows.

Regards,
Reilly Atkinson

 

[pmb_phy]

Hi Reilly

Thanks for stopping by. Always great to see you here!

Radiation means energy flux through the surface at infinity, so the radiation fields must go as 1/r. In the L-W potential, given in one of dextercioby's prior posts,the 1/r term is proportional to the acceleration. No acceleration, no radiation.

Now see? This is something I don't get. A plane wave is a solution of Maxwell's equation and it does not decrease with distance traveled. Let me guess - It doesn't appear in nature so we don't talk about it?

Pete

 

[reilly]

Pete -- Not really. The basic idea is that, locally, a spherical wave front far from its source will look like a plane wave. And, macroscopic light bouncing off mirrors comes pretty much in plane waves. In such a case, the plane waves are superpositions of lots of sources. There's lots of math to go with all of this.
Plane waves, spherical waves are all alive and well. But, indeed plane E&M waves are a bit of an abstraction, but a useful one at that, particularly in QED.

Thanks & regards,
Reilly

 

[pmb_phy]

But, indeed plane E&M waves are a bit of an abstraction, but a useful one at that, particularly in QED.

Yep. That's what I was referring to by not occurring in nature. Thank God I don't have to teach this point in a class. I think the students would razz me.

Pete

 

[pmb_phy]

It just occurred to me how simple the answer is to the question regarding the Poynting vector. Simply provide a simple example of the Poynting vector and then see if the wave equation is satisfied. E.g. Let an uncharged current carrying wire sit on the x-axis. Here there is only a B-field. Now transform to a frame moving parallel to the wire. Now there is both an E field and a B field. These fields are static and therefore do not satisfy the wave equation and therefore there is no radiation in the new frame. This can be placed in the covariant form

\Box A^{\alpha} = 0

which is the wave equation in the absense of charges and current as is the case above outside the wire. So no radiation in one inertial frame -> no radiation in all inertial frames. If you're wondering about the case A^{\alpha} = constant it represents a wave with infinite wavelength. At least this is what sounds right. I'm not 100% sure. Hopefully Reilly or Rob will chime in and make sure.

Actually "to radiate" is different than "there is radiation" but it should be convincing about the absolute existence of detecting radiation in an inertial frame. But it does address the question in an indirect way.

Pete

 

[Borogoves]

Creating an electromagnetic field with nonzero Poynting vector means emitting radiation...An electric charge moving with constant velocity emits radiation.

The end.

Daniel.

Oh don't be so arrogant young man. The Poynting vector is locally invariant.

_________________________________________________________________

 

[Meir Achuz]

Creating an electromagnetic field with nonzero Poynting vector means emitting radiation...An electric charge moving with constant velocity emits radiation.

The end.

Daniel.

I just came across this thread. Each statement above is completely wrong.
It is contradicted in several posts in this thread, and by all textbooks.
Radiation must fall off no faster than 1/r^2, and this only occurs if there is acceleration.

 

[Pwantar]

Alright,Pëte.Computing the Poyting vector for a Liénard-Wiechert field generated by a constant velocity moving electric charge,one gets a nonzero result.Ergo,radiation is being emitted.

I'm aware people always ponder about Larmor's formula,but this case must not be excluded.

Daniel.

Yes, but did you integrate the Poynting vector over a sphere containing the charge? This gives you the power. A non-zero result there means it radiates... a non-zero Poynting vector does not. That is what it means for something to radiate... To carry energy out to infinity.

What we are looking for, is a pointing vector whose radial component is inversely proportionate to r^2, so that the limit as r approaches infinity of the Poynting vector integrated over the sphere containing the point charge is not equal to 0. (Note, the area element in the radial direction is given by r^2*sin(t)*dt*dp, where t=theta, p=phi... it makes sense that an inversely proportionate r^2 dependence would give a non-zero result.)

And if you look closely, the Poynting vector given by the E and B fields given by the Lienard-Weichart formula has an inverse dependence on r^4... Thus, the Poynting vector integrated over a sphere containing the charge tends to 0 as r approaches infinity. Thus, it does not radiate.

As for the original question of the thread, maybe it's an if and only if thing. We already know the answer in one direction.. Can we show it the other? I'm unsure, and will certainly ponder over it.

By the way, everyone interested in this stuff, check out chapter 10 of Griffith's Introduction to Electrodynamics... it's great for a high-level introduction. I'm actually studying for an exam on this stuff now... well, back to stuff that will actually be on my exam.

-Tina the Bunny

 

[PhyDude]

Hi Everyone

Is it possible to determine the force that a positive test charge experiences when it is released near a stationery/fixed positive charge by simply using Coulomb's law? The positive test charge obviously accelerates non-uniformly but can Coulombs law really be used to determine the force that the stationery point charge exerts on the positive test charge?

Any help would be appreciated and I know that this is kinda a homework question and not really a radiation question but its is related to the topic of this thread.

 

[jason12345]

Check out the Liénard-Wiechert potentials and fields in any serious electrodynamics book,J.D.Jackson being the first to turn to.

Daniel.

It's far too complicated in Jackson. Giffiths' Introduction to Electrodynamics gives a far easier to understand derivation