?Accelerating a charged particle

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Offline Ron Hughes

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?Accelerating a charged particle
« on: 08/02/2010 15:52:22 »
An accelerated charged particle emits radiation with respect to a stationary observer but if the observer is accelerated along with the particle does the observer still detect the particle emitting radiation?. Suppose we put a detector in the G machine at NASA and get it up to say 12G's would it detect radiation coming from the matter it is mounted on? I don't think so and if it doesn't that means radiating particles obey the laws of general relativity which I find extremely interesting.

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Offline Bored chemist

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« Reply #1 on: 08/02/2010 19:15:21 »
12G isn't a lot of acceleration so the simple answer is no any effects would be far too small to observe.
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Offline lightarrow

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« Reply #2 on: 08/02/2010 19:28:34 »
That's true, it wouldn't radiate (significantly) even respect to the stationary observer.

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Offline Ron Hughes

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« Reply #3 on: 08/02/2010 19:40:24 »
Ok, how about say a million G's? That should produce some detectable radiation. BTW, our G machine is in outer space millions of kilometers from any matter. Now if we have a detector just outside the arc of the machine that transmits it's readings to us it will detect radiation coming from the machine but I don't believe the detector on the machine will detect radiation except from our nearby detector which would appear to be accelerating to the detector on the machine.
« Last Edit: 08/02/2010 19:55:18 by Ron Hughes »
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Offline yor_on

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« Reply #4 on: 08/02/2010 22:05:13 »
"An accelerated charged particle emits radiation with respect to a stationary observer"

By a G machine I assume you are thinking of an electromagnetic accelerator?
And as that EM field interact with your particle (electron f.ex) sure, you will have an radiation. What you're not asking, if so, is if the electron would radiate if being in a 'free fall' toward f.ex a Black Hole. Or was it this you meant after all?

How exactly would that electron free falling towards the BH be able to radiate?
Tell me Ron because I don't know?
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You will have an acceleration there too, as seen from an observer.
« Last Edit: 08/02/2010 22:08:03 by yor_on »
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Offline Ron Hughes

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« Reply #5 on: 09/02/2010 00:48:39 »
The G machine at NASA used to test Astronauts in high G conditions. My point is that an accelerated charged particle emits radiation relative to an observer. If the observer is accelerated along with the particle then the observer does not detect radiation being emitted by the particle. If that is true, and I have no way knowing if it is true, it should be telling us something important.
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Offline yor_on

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« Reply #6 on: 09/02/2010 00:57:35 »
You mean that the radiation only will be noticeable for observers outside that frame perhaps? Which would make a weird statement if so?

Well I'm finding this subject to be jungle of differing views.
So? What about it, do you have a view on my scenario?

It's a free and, as I understands it, uniformly accelerating fall for that electron towards the EV of the BH. Does it radiate?
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Offline Ron Hughes

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« Reply #7 on: 09/02/2010 01:05:00 »
Yes, an observer outside the frame would see both observer and particle radiate.

To an observer stationary with respect to the BH I would expect the electron to radiate.
« Last Edit: 09/02/2010 01:07:54 by Ron Hughes »
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Offline Soul Surfer

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?Accelerating a charged particle
« Reply #8 on: 09/02/2010 20:52:48 »
There is no need for normal rotating machines which are far too slow and weak to produce significant radiation.

We observe radiation from accelerated particles all the time.  The most common form is in the form of heat it is the accelerations of the particles as they collide that produces the heat radiation with which we are all familiar. particle velocities at room temperature are typically thousands of miles an hour and interaction times to reverse this in a collision about the size of one atom extremely short  so accelerations are truly enormous.

Radiation can be produced in a more controlled way using particles confined in orbits by magnetic fields. This is called synchrotron radiation and some of the most powerful x ray sources use this sort of radiation and there are several research facilities all over the world.

synchrotron radiation can also be observed in the earth's ionosphere and in outer space at all sorts do frequencies from radio to light.

a gravitational field could also provide the orbital confinement and it is probably a significant part of quasar radiation.
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Offline yor_on

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« Reply #9 on: 09/02/2010 22:27:14 »
Radiation due to interaction I have no problems with. It's radiation of the electron when falling towards the EV I'm questioning here. Where is the interaction? And how does it radiate?

---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


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Offline Ron Hughes

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« Reply #10 on: 10/02/2010 02:06:11 »
yor, any charged particle that is accelerated emits radiation perpendicular to the direction of acceleration. I'll find you some sites.
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Offline Soul Surfer

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« Reply #11 on: 10/02/2010 20:43:09 »
An accelerating charged particle in a gravitational field should radiate but in general this radiation will not be observed because of the way in which the particles will in general be arranged and interact with the fields.

A lot of space is filled with charged particles moving under gravity and electromagnetism.  This can be seen very well in observations of the sun.

A plasma contains equal quantities of positive (usually protons or other nuclei)  and negatively charged (usually electrons) particles  now the motions and resonances of protons and electrons in a magnetic field are very different because they have different masses.  they therefore behave differently and their effects are different.  However the motion of particles under gravity does not depend on their mass at all heavy and light particle tend to fall at the same rate so the radiation effects are equal and opposite on plasma falling freely or in orbit under gravity and nothing else.

The critical difference comes when turbulence and viscosity sets in and this allows the generation of magnetic effects
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Offline Ron Hughes

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Offline JP

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« Reply #13 on: 11/02/2010 06:02:45 »
Larmor radiation is for a charged particle accelerating relative to an observer who isn't accelerating. 

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Offline yor_on

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« Reply #14 on: 11/02/2010 12:08:44 »
Soul Surfer are you saying that an electron accelerated by a gravitational field will radiate or were you thinking of some combination? I'm not discussing interactions with a EM field here, just gravitation and an electrons 'free uniformly accelerating fall' towards a EV.

Why, or should it be how:)?

And looking at Larmor radiation I only see it used in interacting EM fields?
And in those you will have an radiation.

Larmor's Formula

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JP do you mean that you can use this formula if you were observing the electron, whilst yourself standing still relative that Black Hole?

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Anybody up for a headache:)

Found this one too.
Hey, don't blame me ::))
I'm soon gonna be out of them too, ah, headache pills I mean..
Does Free-falling Electron Radiate

(taking them by the dozens now)

So, to summarize that one it seems to say that if the acceleration is equivalent to gravity then the electron will radiate to the the observer standing still relative the BH, but if free falling beside it you would tell me that it doesn't radiate.

It's all about frames then :)

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The problem is that you won't get a clearcut answer it seems?
Well, if you're not going to use the concept of frames to describe how an electrical engine works. I would love that one :) Also. Why doesn't it violate the equivalence principle?
Or does it violate it?
« Last Edit: 11/02/2010 15:03:20 by yor_on »
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Offline lightarrow

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« Reply #15 on: 11/02/2010 13:18:39 »
The G machine at NASA used to test Astronauts in high G conditions. My point is that an accelerated charged particle emits radiation relative to an observer. If the observer is accelerated along with the particle then the observer does not detect radiation being emitted by the particle. If that is true, and I have no way knowing if it is true, it should be telling us something important.
As I wrote in another thread, the answer to this question is not simple. Probably the co-moving observer would still measure a radiation, if its distance from the accelerating source is (significantly) greater than the radiation's wavelength.

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Offline yor_on

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« Reply #16 on: 11/02/2010 13:54:56 »
The principle of equivalence which is the base of the general theory of relativity reflects the close connection between inertial coordinate system K with uniform gravitational field and noninertial coordinate system K′ moving with uniform acceleration in empty space. According to this principle any physical process proceeds absolutely identically in the both systems. In other words, if we imagine the system K as a closed laboratory, which is at rest on the Earth, and the system K′ as an identical laboratory, moving with acceleration g in a distance of gravitating masses, and the sizes of the laboratories are such chosen, that it will be ossible to neglect the nonuniformity of the gravitational field in system K, then the observer, who is in one of these laboratories and who does not have any connection with the external world, i. e. without having any possibility to look out of its limits, could not make any experiment inside the laboratory in order to find out in which of them he is, i. e. he could not define the character of his motion.


That can be explained by the fact, that in system K′ during the uniform accelerating the field of inertial forces appears. The action of this field does not differ from the action of the gravitational field. It is easy to make sure, that the mechanical processes will proceed identically in both laboratories. In fact, free bodies in every laboratory will move with acceleration g, identical pendulums will oscillate with equal periods, etc. Thus, systems K and K′ are equivalent in respect of mechanical processes. Einstein spread the equivalence of systems K and K′ on all physical processes without exceptions, having formulated the principle of equivalence, according to which, not only mechanical, but also any physical processes have to proceed identically in systems K and K′. But we can point to the process which violates the principle of equivalence. It is the radiation of charges.

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 wont 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.

Now we will consider uniformly rotating coordinate systems. We can imagine one of these systems (we mark it as S′) as a disk, revolving with constant angular velocity on its axis. In accordance with the principle of equivalence the noninertial coordinate system S′, in which a field of centrifugal forces of inertia exists, can be considered as an inertial system K with an uniform gravitational field. Let us consider two charges: one of them is on the disk, i. e. rotates with this disk. The second one is at rest at system K. In the first case the charge has to radiate and in the second one it does not. It has been already told how to differ a radiating charge from a charge, which does not radiate. And what is more, a braking force of radiation friction f must act on uniformly moving round a circle charge, which action one can observe at an as short as we want distance from the charge. Therefore, in this case the principle of equivalence is violated. So far as we have mentioned the radiation force of friction, it is necessary to point out another problem, which was first remarked by M. Born. By motion of the charge with uniform acceleration, force f turns into zero. That leads to the violation of energy balance. 

If the motion firmly accelerates, r=const and  r=0. an uniformly accelerating charge radiates without any losses of energy. That contradicts the law of conservation of energy. We will return to laboratories K and K′ again. We will change the character of their motion. Let laboratory K′, which is in empty space far from gravitating masses, move uniformly and straightforward and let K freely move in a gravitational field. There will be state of weightlessness in both laboratories. According to the principle of equivalence, in this case systems K and K′ are also equivalent. I. e. as in the last instance, the observer, who is in one of the laboratories and does not have any connection with the external world, could not make any experiment inside the laboratory in order to find out the character of his motion. But we will place a charge in each of these laboratories again. In laboratory K charge must radiate, for it moves with acceleration under the action of the gravitational field. The observer in this laboratory will be able to register the radiation. In laboratory K′ the charge will not radiate. Therefore, in this case the principle of equivalence is not fulfilled.

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Sorry, for the math you better read this one Why electrons don't radiate in Rutherford's atom
===

Einstein combined the equivalence principle with special relativity to predict that clocks run at different rates in a gravitational potential, and light rays bend in a gravitational field, even before he developed the concept of curved spacetime.

So the original equivalence principle, as described by Einstein, concluded that free-fall and inertial motion were physically equivalent. This form of the equivalence principle can be stated as follows. An observer in a windowless room cannot distinguish between being on the surface of the Earth, and being in a spaceship in deep space accelerating at 1g. This is not strictly true, because massive bodies give rise to tidal effects (caused by variations in the strength and direction of the gravitational field) which are absent from an accelerating spaceship in deep space.

Although the equivalence principle guided the development of general relativity, it is not a founding principle of relativity but rather a simple consequence of the geometrical nature of the theory. In general relativity, objects in free-fall follow geodesics of spacetime, and what we perceive as the force of gravity is instead a result of our being unable to follow those geodesics of spacetime, because the mechanical resistance of matter prevents us from doing so.
Equivalence_principle
« Last Edit: 11/02/2010 14:06:17 by yor_on »
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Offline graham.d

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« Reply #17 on: 11/02/2010 14:20:24 »
I think if you are in free fall next to a charged particle it will not be perceived to radiate. Assuming GR is correct, being in free fall would be equivalent to being next to (as long as you were sufficiently close, to avoid tidal effects) the charged particle in free space so it cannot radiate.

If a charged particle is free falling in a gravitational field and you, as an observer, are some distance away then the charged particle will be seen to emit Bremsstrahlung radiation. Similarly if the particle is accerated by other means, it will also emit radiation.

There is some controversy about this because it is said that because an accelerating charge emits radiation, a static charge in a gravitational field (say on the earth's surface, should also emit radiation, which it clearly does not. This being because the gravity field should be the same (as far as the particle is concerned) as being accelerated. Some people say this points to there being a difference between inertial and gravitational forces. However, the reason, I think, that this is not so is because what you observe as radiation is the changes in the electric field. It is not something that necessarily affects the particle itself. So a particle in a static environment, whether in a gravity field or not, will not emit radiation, but a particle whose state changes, relative to an observer, from a static or linearly moving state (i.e. accelerates) will significantly disrupt the field that is radiating out from it. It is this, second order, change that is responsible for em radiation.
« Last Edit: 11/02/2010 14:22:15 by graham.d »

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Offline yor_on

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« Reply #18 on: 11/02/2010 14:42:11 »
Very good Graham.
I liked that, We have both frames to consider and 'relations'.
That is how I see it, Frames of reference will describe the relations, but then we also have a dichotomy between matter and radiation, even when discussing a single electron, depending on frame. You have a very conceptual way of describing things :)
==

"from a static or linearly moving state (i.e. accelerates) will significantly disrupt the field that is radiating out from it. It is this, second order, change that is responsible for em radiation."

There is something I can't express here, but I think there is a difference between what I name 'relations' and 'frames of reference' :) Doesn't make sense does it? The relations follows the frames, but in this case there is a difference. I saw someone saying something of the EM field interfering with itself due to motion. Would that be what you're thinking of?
==

For me the thing is, if we have a thing of 'matter' like our electron. Is it then an object?
And if it is a a defined object, geometrically existing inside SpaceTime, free
falling, why will it radiate in one case and not in the other. What is radiation?

So assuming that it is an object it have a property that we call radiation that is frame-dependant, but radiation is no ephemeral thing. It's energy, the same energy that we use to our batteries to my computer. Why will you from one frame see energy but from another not?

Would it f.ex be possible to 'tap' that radiation from one frame at the same time as you when you measure it free- falling beside your 'object' measure no radiation at all? What would that distant observer see as our free falling fellow made his measurement of radiation, assume a 'cosmic voltmeter' :)
==
Or should it had been an electronmeter?
Ah well..
« Last Edit: 11/02/2010 16:03:52 by yor_on »
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Offline graham.d

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« Reply #19 on: 11/02/2010 17:14:17 »
If you are accelerating with the electron then field lines emanating from the electron will be observed by you to be unchanging relative to you. 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". A common example is an oscillating charge in a piece of metal (i.e. a radio aerial).

I think it is more straightforward than you first think and I don't think there is any paradox created by the concepts of GR here. But some people do think that.

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Offline yor_on

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« Reply #20 on: 11/02/2010 17:54:35 »
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 :)
« Last Edit: 11/02/2010 17:59:46 by yor_on »
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Offline Ron Hughes

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« Reply #21 on: 11/02/2010 19:21:21 »
I agree, I think it fits perfectly with GR.
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Offline yor_on

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« Reply #22 on: 12/02/2010 15:03:19 »
"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.
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Offline graham.d

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« Reply #23 on: 12/02/2010 15:57:10 »
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.

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Offline Ron Hughes

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« Reply #24 on: 12/02/2010 18:50:18 »
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.
« Last Edit: 12/02/2010 18:56:27 by Ron Hughes »
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Offline graham.d

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« Reply #25 on: 12/02/2010 20:20:20 »
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.

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Offline Ron Hughes

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« Reply #26 on: 12/02/2010 21:37:44 »
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.
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Offline yor_on

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« Reply #27 on: 12/02/2010 21:40:16 »
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:)

« Last Edit: 12/02/2010 21:45:22 by yor_on »
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Offline Ron Hughes

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« Reply #28 on: 12/02/2010 21:58:05 »
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.
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« Reply #29 on: 12/02/2010 22:12:50 »
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



« Last Edit: 12/02/2010 22:16:49 by yor_on »
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Offline graham.d

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« Reply #30 on: 12/02/2010 22:16:29 »
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.

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« Reply #31 on: 12/02/2010 22:22:28 »
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?
« Last Edit: 12/02/2010 22:41:30 by yor_on »
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Offline graham.d

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« Reply #32 on: 12/02/2010 22:30:57 »
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?

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« Reply #33 on: 12/02/2010 22:49:47 »
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..
« Last Edit: 12/02/2010 23:15:53 by yor_on »
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Offline Ron Hughes

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« Reply #34 on: 12/02/2010 23:17:08 »
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.
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« Reply #35 on: 12/02/2010 23:21:18 »
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?

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« Reply #36 on: 12/02/2010 23:29:38 »
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*
« Last Edit: 12/02/2010 23:44:10 by yor_on »
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« Reply #37 on: 12/02/2010 23:58:33 »
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 ::))
« Last Edit: 13/02/2010 06:27:27 by yor_on »
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« Reply #38 on: 13/02/2010 05:57:15 »
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 wont 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 :)
« Last Edit: 13/02/2010 07:32:29 by yor_on »
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« Reply #39 on: 13/02/2010 07:23:38 »
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  :)

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« Last Edit: 13/02/2010 07:27:06 by yor_on »
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« Reply #40 on: 13/02/2010 09:23:10 »
Yor_on, I don't think there is a problem regarding the difference in observation of em radiation between the comoving observer and a distant one (in the orbiting electron case). The issue is whether the electron is losing energy and its orbit decaying. This is a distinct measurable difference in how the electron will behave depending on the point of observation and would be a paradox. I am sure there is a fault in my reasoning somewhere, but I need to see, and be convinced, where it is.

I will read some of the references you cite later.

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« Reply #41 on: 13/02/2010 15:00:02 »
"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."

I disagree with the writer's assumption. He is assuming that the electron will radiate in the accelerating laboratory (K) but to the observer in (K) it will not radiate. It will appear to radiate to the stationary observer in (K'). Conversely if the observer (K) can look at (K') then he will see the electron radiate. The equivalence principal is satisfied.
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Offline Ron Hughes

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« Reply #42 on: 13/02/2010 16:17:28 »
Have I missed something Graham? Why is it assumed that the orbiting electron's orbit will decay? Unless some force acts on it the orbit will be stable. As an aside, shortly I will post an explanation of why we see the photon as having both a magnetic and electric component at ninety degrees to each other. I will be busy for the rest of the weekend and will post Monday in the thread "The Photon".
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« Reply #43 on: 13/02/2010 16:51:50 »
Ron, in order for em energy to flow away fron the star and orbiting electron, the energy has to come from somewhere. The energy from an accelerating charge is emitted as Bremsstrahlung radiation and is normally considered to be (literally) from the braking effect on the charge. If there is such radiation the electron orbit must decay I think. And, as far as I undestand, there has to be em radiation from an orbitting electron.

I should have said Larmor radiation rather than Bremsstrahlung.
« Last Edit: 15/02/2010 13:11:28 by graham.d »

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« Reply #44 on: 13/02/2010 18:25:40 »
Yor_on, I don't think there is a problem regarding the difference in observation of em radiation between the comoving observer and a distant one (in the orbiting electron case). The issue is whether the electron is losing energy and its orbit decaying. This is a distinct measurable difference in how the electron will behave depending on the point of observation and would be a paradox. I am sure there is a fault in my reasoning somewhere, but I need to see, and be convinced, where it is.

I will read some of the references you cite later.

Well Graham, that is why I say I don't think it's possible. I can't find fault in your logic, presuming that the electron will accelerate the scenario is the same as mine I think? the far observer should both observe radiation and the electron 'breaking orbit' as I see it. But for the near observer, presuming that we can't observe the energy disappearing it will be as 'magic' if it is so. So the scenario seems too weird for me :)

Now, that's a thing to remember ::))
« Last Edit: 13/02/2010 19:59:36 by yor_on »
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« Reply #45 on: 14/02/2010 20:30:46 »
   Graham, you post is is basically the same as an orbiting electron in the hydrogen atom and produces a possible error in my own logic about how radiation is produced. For my idea to be correct then that electron must emit radiation with respect to all observers outside that system. I have not as yet looked for conformation of such radiation. I will endeavor to do so now.
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« Reply #46 on: 15/02/2010 02:46:17 »
So far everything I've read suggests that the electron is a wave. That being said it would mean the electron around the proton would not emit radiation and my original premise holds.

The electron orbiting a star does not lose energy with respect to the star. As far as the star is concerned the electron does not emit radiation. The electron only emits radiation with respect to an observer outside the system as suggested by GR.
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« Reply #47 on: 15/02/2010 08:47:11 »
So far everything I've read suggests that the electron is a wave. That being said it would mean the electron around the proton would not emit radiation and my original premise holds.

The electron orbiting a star does not lose energy with respect to the star. As far as the star is concerned the electron does not emit radiation. The electron only emits radiation with respect to an observer outside the system as suggested by GR.

I am not sure about the star. It may depend on whether an observer is on the surface of the star and moving with respect to the orbiting charge. I think we established that we think a comoving observer in free fall with the charge would not see any radiation but that someone external to the system would see radiation. The question still is whether or not the orbit would decay. If energy is coming out as a result of the orbit then the orbit must decay - GR or not - a co-moving observer should also see this decay. That is the paradox. What are we missing here?

I am happy to consider electrons in atoms as a QM problem. You can replace the electron with a large charged body to avoid QM issues.

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« Reply #48 on: 15/02/2010 12:49:14 »
An accelerated charged particle emits radiation with respect to a stationary observer but if the observer is accelerated along with the particle does the observer still detect the particle emitting radiation?
Yes. SoulSurfer mentioned a synchrotron. Here in the UK we have the "Diamond Light Source", see http://www.diamond.ac.uk/ which I've visited. They use dipole magnets to deflect the electrons, which as a result emit X-rays. If you were racing around the ring along with the electrons being similarly deflected, you'd still get a faceful of X-rays. You'd be emitting X-rays too and it wouldn't be too comfortable in there, but those X-rays don't go away. 
« Last Edit: 15/02/2010 12:54:19 by Farsight »

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« Reply #49 on: 15/02/2010 13:10:11 »
I have done a little reading on whether there is a paradox regarding: a co-moving observer with a charged particle in orbit (who sees no radiation because he in the same free falling frame as the charge) vs a distant stationary observer who sees radiation from the accelerating charged particle. The paradox is that the distant observer should see the orbit decay and the comoving observer should not; events on which they must agree.

It seems this is an unresolved issue. Feynman reasoned that the equation for radiation needs to involve the third derivative (changing acceleration) but this is disputed. A good description of the problem, though not exactly as I state above, though related is:

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