[Ron Hughes (OP)]

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.

[Bored chemist]

12G isn't a lot of acceleration so the simple answer is no any effects would be far too small to observe.

[lightarrow]

That's true, it wouldn't radiate (significantly) even respect to the stationary observer.

[Ron Hughes (OP)]

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.

[yor_on]

"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?
==
You will have an acceleration there too, as seen from an observer.

[Ron Hughes (OP)]

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.

[yor_on]

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?

[Ron Hughes (OP)]

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.

[Soul Surfer]

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.

[yor_on]

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

[Ron Hughes (OP)]

yor, any charged particle that is accelerated emits radiation perpendicular to the direction of acceleration. I'll find you some sites.

[Soul Surfer]

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

[Ron Hughes (OP)]

http://www.cv.nrao.edu/course/astr534/LarmorRad.html

[JP]

Larmor radiation is for a charged particle accelerating relative to an observer who isn't accelerating. 

[yor_on]

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

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

==
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

*Throws away headache pills as he whistling lifts the light bulb and waits for it to light up*
==

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?

[lightarrow]

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.

[yor_on]

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 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.

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.

-----

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

[graham.d]

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.

[yor_on]

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
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"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..

[graham.d]

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.