[yor_on]

Awh are you going to take away the magic Graham.
Don't do that, to me an electron is very strange
Nice explanation, I mean, even I could understand it  )

you treat it like a light wave here, but with a magnetic field?
Or am using the wrong analogy?
==
I'm using the wrong analogy, it was when it accelerated that fooled me for a minute

[Ron Hughes (OP)]

I agree, I think it fits perfectly with GR.

[yor_on]

"It is the change of the field (in fact the second derivative if I remember correctly) that produces radiation. If an electron is moving at constant speed relative to you, then you see a steadily changing field but not em radiation. If it accelerates relative to you, the field strength and direction (vector) propagates towards you at speed "c".

Why would it 'fire photons' at me (distant observer) at 'c' when falling down a gravity well Graham?
I'm not sure how to see that.

[graham.d]

I am not sure it would fire photons, yor_on. I don't think I can reason this without doing some hard sums :-) But how about this... an electron orbiting a neutron star (say) is in free fall and it is just following a geodesic. Provided there are no "frozen" magnetic fields associated with the star, I think it will continue to orbit and not lose energy, so can't be radiating. Perhaps the common statement that an accelerating charge emits radiation is too simplistic. I think my earlier statement regarding a charged particle falling in a gravity well may be incorrect.

[Ron Hughes (OP)]

Imagine a gravity well at 12 O'clock. Some distance away at 6 O'clock you have a falling charged particle (accelerating with respect to the gravity well). To the right of the charged particle some distance away is an observer in a rocket ship that can remain stationary with respect to the gravity well. The charged particle is accelerating with respect to the observer and will emit radiation at the observer.  http://www.cv.nrao.edu/course/astr534/LarmorRad.html

An observer stationary with respect to the neutron star will see the electron emit radiation. An observer in the same orbit as the electron would not see the electron emitting radiation.

[graham.d]

That's what I thought at first, Ron. But then the electron is losing energy and its orbit will decay wouldn't it? If you were in orbit with an electron you would not see it emit em radiation but its orbit will magically decay and yours will not. I ahve confused myself now. I thought I understood this.

[Ron Hughes (OP)]

As far as the star is concerned the electron is not losing energy. The emission of radiation is strictly relative to an observer who sees the particle accelerating with respect to the observer. The particle is not accelerating with respect to the star.

[yor_on]

Ron, both you and Graham makes sense
But we seem to be discussing a principle here.
About energy.

And the main question seems then to fall down to how we define energy. In the fall towards the EV we seem to have two possibility's, depending on observer. In one there must be energy spent, otherwise there can be no radiation observed, in the other (being at rest relative the electron) there is no such 'energy' spent, no radiation to observe.

If we say that entropy is 'work transforming' why would we from one frame be able to see 'work done' (radiation) and from the frame at rest with the electron see it as nothing happens at all? If we want to see entropy as correct 'energy' should be lost in both circumstances (frames) it seems to me?
==

And yes, I agree that there can be no 'uniform acceleration' around the neutron star, as long as we are talking about a stable 'orbit' by the electrons 'orbital' orbiting  (Couldn't let that one pass:)

[Ron Hughes (OP)]

If I drop a one kg block an observer standing on the ground can actually measure the gain in energy of that block as it falls. An observer falling along side the block will see the block's energy as always staying the same as they fall.

[yor_on]

I'm not discussing potential energy here. I'm saying that from one frame you will see radiation, defined as photons as far as I know? And from the other frame we observe nothing at all.
==
---Quote---

The motion of a free electron (i.e unbound to an atom) may produce x-rays if the electron is undergoing any one of these motions:

    * accelerated past a charged particle,
    * moving in a magnetic field,
    * accelerated by another photon.

---End of quote-- NASA:s view on it

[graham.d]

But if there is em energy emitted from the accelerated electron it's orbit must decay, mustn't it? It must be losing kinetic energy.

[yor_on]

You are thinking of the neutron star now Graham? But the framework is different there as you don't have an uniformly accelerating orbit, as long as it is stable. But if it would radiate then it should behave like you think, considering that it is a particle of restmass. At least as I see it (and then the same weirdness should apply there too, as it now will accelerate, I guess it will, at least?
==
I'm wrong there ain't I, it's orbit should become wider, shouldn't it. As there is less 'attraction' between it and the neutron star as they fall into each other gravity wells?
==

And in the first example depicting a BH? If the electron from one frame is seen as emitting radiation, shouldn't it follow that there have to be a equivalent loss of energy in the other frame too?

[graham.d]

An electron in orbit around a gravitating body is accelerating. In Classical terms, inwards toward the centre of the body at a rate of (v^2)/r. From my distant position I see that electron accelerating due to the force of gravity exerted by the body. If it is accelerating will it emit radiation? If so will it's orbit decay? If it's orbit decays then if you are orbiting with the electron I will see the electron orbit decay but you stay in your orbit. Now, the fact that we are different observers cannot account for two different events. From your co-orbit perspective why would the electron move away from you as you are both in free fall and following the same geodesic?

[yor_on]

Assuming that it is as you present here we will then have the exact same situation Graham, just as you stated. Then an electron won't have a stable orbit if I read you right? Or is there something else I'm missing here?

As for the rest of your scenario I agree and the explanation would then be that it's  acting the same in both frames. But there is no explanation to why you can't measure the loss of energy in the 'near situation', only in the 'distant' do you see a causality chain.
==
Thinking of it, do stable orbits exist
Naaah..

[Ron Hughes (OP)]

GR covers this situation. Whether or not a charged particle emits radiation depends on the frame of reference from which you make the measurement. The particle must be accelerating with respect the observer for the observer to see radiation.

[graham.d]

Well I tend to believe this is true, but I'm having a job to see how to reconcile this with an apparant paradox. Will an orbiting electron's orbit decay or not?

[yor_on]

It's just not reasonable. The only way it could change its orbit under those circumstances, without us being able to measure any such loss in the near situation seems to be some sort of 'action at a distance'? The idea of 'locality' is already open for discussion, but the idea that information always will obey 'c' is not, if that one would be questioned Einsteins universe would have to be revised totally. But here there have to be information exchanged presuming your 'system' would change?

Or is there a better description of it.
One could be that the 'energy loss' happens in both frames and are measurable?
Anyone?

==

The way I thought here was to reason from the causality chain as that is the one I see as clearly following times arrow. The 'near situation' if we can't measure why the electron orbits break with the near observer is not following any causality chain. And therefore I give a 'higher importance' to the distant observer as it fits the universe we see normally. And from there follows my conclusion that information have to be exchanged.
==

But? with what, if so. )
Awh sh*

[yor_on]

SpaceTime is a 'whole'. No matter how many 'observers' we will introduce under what circumstances SpaceTime will still act as a whole to those observing each other. So even if there is an action that we can see taking place under different causality chains, as seen from different frames of reference, there will still be a gold standard hidden behind our different observations when compared. And a 'causality logic', but here I fail to see the reason why the 'near case' wouldn't observe any energy loss (equivalent to the radiation observed from the 'distant case'?

So there has to be a measurable one, if the original idea is correct that is. Not only goes this idea against the principle of equivalence but it seems also to introduce a very strange sort of 'action at a distance' to me, strange because both observations depict the same object under 'the same time' if you see how I think here

Not easy I got to admit )

[yor_on]

This is pretty weird reasoning, ain't it. But if we are treating it as frames, and that we are, the same sort of twisted logic is applicable on the electron falling down to the EV. In both cases you will have two frames observing, one near, one distant. According to the idea the near frame won't measure a thing as there in fact is nothing to measure, the distant one will measure a radiation. To do so there has to be photons involved. What you could argue here is that there is a acceleration involved and that this acceleration is a relation only noticeable from the distant observer being at rest with the Black Hole/neutron star.

But the 'acceleration' is in both falls a 'free fall', well, as I see it? Following an 'easiest/shortest path' through SpaceTimes geodesics. In none of the cases do we expend any energy to get to this acceleration. You have to understand that. If you think that they do expend energy you disagree with the whole idea of Einsteins SpaceTime and its geodesics. So, without expending energy they 'accelerate' uniformly. If we define uniformly as always being at the same G factor then both examples fall short I expect, as the gravitation will become steeper the closer their falls will take them though?

But if assuming that we have two laboratories, one in space accelerating (K'), and this time I expect, expending energy to keep to a constant 1G, and the other one 'at rest' (K) on Earth will the reasoning become different?;

---Quote---

It is easy to make sure, that with the help of charges it is possible to differ system K from system K′. In fact, let us place the charges in both laboratories. The charge in the laboratory K′ has to radiate, so it is moving with acceleration. The observer who is in this laboratory can register this radiation having placed, for example, a charge into the water. Then a part of the radiating energy will be absorbed by the water and will warm it up. On measuring the temperature of the water the observer will be to register the radiation. We won’t discuss the technical details how to carry out such an experiment. It is enough that it is a principal possibility to find out the radiation in immediate proximity to the charge. In laboratory K a motionless charge will not radiate. Thus, the observer in every laboratory very easily can define the character of his motion. Therefore the principle of equivalence is violated.
---End of quote---

  Why electrons don't radiate in Rutherford's atom

But if you look at this case the radiation measured will be at the same frame, it seems that the distant observer isn't involved at all here? "Thus, the observer in every laboratory very easily can define the character of his motion. Therefore the principle of equivalence is violated."

And that is contrary to what we have been discussing, isn't it?
But the discussion we have had is to be found on a lot of other places too. take a look at this f.ex.

---Quote---

A charge at rest in a gravitational field is accelerated (assume uniformly) yet does not radiate. Therefore (by equivalence) a charge at rest in a uniformly accelerating reference frame does not radiate *in that frame*. Thus if you suppose you are next to a charge in an elevator that is undergoing uniform 1 G acceleration, it will not appear to emit radiation to *you*.

But an observer in a nearby non-accelerated frame will measure the presence of both electric and magnetic fields changing as a function of time. Time-changing fields (in free space) will result in radiation. There IS radiation coming from the accelerating charge which can be observed in other frames. The energy for this radiation comes from the mechanical source which is accelerating the charge, it's prime mover.


The observer in the elevator sees no radiation, but *does* measure an anisotropic field in the elevator *and through all of space*. That is, the static electric Coulomb field at the top of the elevator is different than at the bottom. There is a time-static spatial potential energy variation in this Coulomb field that has an equivalent mass which takes work by the elevator's prime mover to accelerate. If you transform this time-static spatial variation of the Coulomb field back into the uniformly moving reference frame, you will recover the radiation fields. . .

Think of it this way. The static field is in an accelerating frame, so viewed from any other inertial frame, it's a time-changing field. It's precisely because it's not a time changing field in the accelerating frame that it is not seen as radiation in that frame. Since we're having fun with this, let me confound the issue even more. An interesting consequence of this is that the radiation of a uniformly accelerating charge does not appear as radiation in ANY uniformly accelerating frame so long as the rate of acceleration is the same in
both frames.

Elevator 1 with a charge starts falling at time T=0. We all see the radiation except for the observer on the elevator. At time T=15, a second not-colinear elevator starts dropping. The radiation field in the second elevator must disappear and appear to be the field of charge moving uniformly with velocity V. This is because with the same rate of acceleration, two falling frames will have a constant velocity of separation.

I reach my own limit of knowledge on the subject when the two elevators reach relativistic speeds. I don't know if the velocity of separation would still be a constant as observed from the second elevator. It probably is.

---End of quote-- By Antiphon .Look here

But now we have two scenarios. The PDF presents one where no distant observer is involved at all, and where the difference will be seen by the near observer in both cases. And so presumably by an distant observer too. So far it seems consequential to me.

The case we, and others have been discussing though?
I can't accept it as possible?

In the Pdf's case?
Well, there it will be K' (uniformly accelerating at one G) that will radiate from the near observers point of view, and if we introduced a distant observer I see no reason why he too shouldn't observe the same?

The only thing similar in both our scenarios is the idea of being at rest. But in our own we defined being 'at rest' as traveling inside the same 'frame of reference' being a 'free fall'. In the Pdf the 'frame of rest' is Earth, uniformly moving, invariantly massing at one G.

So? does this mean that we need a gravitational field? Yes, but we have have a gravitational field acting on both scenarios, don't we? Where it seems to fall when I look at it is where we equal an constantly accelerating electron with one being exposed to exact same accelerating force (one G).

And furthermore, where those conclusions are drawn. In our original case, an electron either accelerated at 12 G in a 'G-accelerator'. Or accelerating as I thought towards the EV, or orbiting a neutron star. The last two we can count out if we define the G-force as needed to be constant for it to be called 'uniformly accelerating', and that I think we need to do here.

Then we have Ron's example left, and the one I referred too above by Antiphon. To me they seem similar. But the conclusions made seem contrary to the PDF, so, which one is possible?

That you by observing in the same frame being at rest with what you observe, will see an radiation, due to you and the charge being in a artificially created frame of acceleration at a constant one G?

Or that you, as in Ron's and Antiphons cases, won't?

---Pick your choice---

Ah, there can be a third too.
That both scenarios are wrong

[yor_on]

Now, that doesn't really seem to exclude my or Grahams example.
But I choose Ron's and Antiphons because of their similarity, as I saw It.

If you look at Does Free-falling Electron Radiate we have a similar idea.

The question is as follows..

---Quote--
I have a question. According to Einstein, gravity and acceleration are indistinguishable. . .

How about the following experiment. If you have a free-falling on Earth electron, it should radiate e/m waves according to e/dynamics (as it is moving with acceleration g). On the other hand, if this electron is simply at rest in the gravitationla field g, it does not radiate. So this way you can tell apart acceleration and gravity. Or can't you?

---End of quote---

And reading Bruce.

-Quote---

Interesting question which I don't have an answer for. The local shell observer observes that the electron freely falling by his shell radiates. The observer in freefall and at rest with respect to the free falling electron notes that it does not radiate. Kip Thorne in his 'Black holes and Time Warps' discusses the discovery by William Unruh and Paul Davies that accelerated observers just above (and at rest with respect to) the event horizon see the sea of virtual particles as real particles emmiting radiation of a black body spectrum while freely falling observers note these same particles are only virtual. He goes on to say "This starling discovery revealed that the concept of a particle is relative, not absolute; that is, it depends on ones's reference frame". The far away observer (GR's bookkeeper) would note that the electron radiates all its energy into the gravitational field while never reaching the event horizon because the bookkeeper perceives the distance to the horizon to be infinite. E/m=(1-2M/r)dt/dt so when r->2M E/m->0. It must all have to do with the information each frame receives about the event.

---End of quote--

So Grahams and mine question seems to be an added complication then
But let's start with Ron's and Antiphons.

And if we 'ever' reach a conclusion there we can look at the last ones too.
He*, we can do it, I'm positively positive we can. Oh yes, a piece of cake 

*Runs away to hide under the table, shouting Glooriaa*