OK, let's hear it from some of you that can give me a good explanation.

The principle of equivalence (upon which General Relativity is based) states that a uniformly accelerated frame is equivalent to a gravitational field of the same magnitude, and it is impossible to devise an experiment to detect the difference.

Now consider this: If I accelerate a charged particle it obviously radiates EM radiation. But suppose I place the same charged particle at rest (or nearly so) in a gravitational field. It does not radiate. So there is no equivalence; They are quite distinguishable. Have I found a hole in GR?

Or can somebody give an explanation that will preserve Einstein from destruction?

Don't worry. Einstein is quite safe. It was actually something like this that got him started on special relativity in the first place. Only that had to do with the question of how moving charges make magnetic fields, and how there can still be measureable magnetic fields if you move to a frame in which a given charge is not moving.

But that's not what you asked about. The solution to your question is right in what you wrote: if the charged particle is at rest, it is not accelerating, and hence there is no radiation. No problem.

If it were in free fall in a gravitational field, it would radiate, just as if it were being accelerated through some other method: hance, equivalence.

Yours,

Don Smith

I think the original question is the opposite though: If you take an isolated particle and accelerate it at 9.8m/s^2, it will radiate (how much?), and the field it will experience will be indistinguishable from gravity by the equivalence principle. Then, if you set the same particle on the surface of the earth, it will again experience the same field. Is its behavior different? Is that a way to distinguish between a gravity field, and one produced by acceleration?

This was discussed OB (old board), right?

It won't experience acceleration if it's sitting on the surface of the Earth, only if it's falling.

Except for the small acceleration due to the Earth's rotation. Now, would that be expected to produce any radiation?

A further befuddlement:

Say we accelerate a particle, and (from our non-accelerating reference frame) detect it radiating. All well and good.

But what if we accelerate along with the particle? It's not accelerating, from our perspective. Should we expect it to radiate? If not, how can the particle be seen simultaneously to radiate (from the non-accelerating frame) and not to radiate (from the co-accelerating frame)?

Is this one of those issues that depends on the presence of the rest of the cosmos, like the twin paradox?

Eh? Twins paradox.

[quote]
On 2002-01-22 09:38, GrapesOfWrath wrote:
[quote]
On 2002-01-22 08:37, Donnie B. wrote:
9 IX 12 MUAN ? hmm? now lemme see if i can find "MY"
Frames string, B4 i have to go Places?
8:23 A.M. searching : 8:25 A.M. searching
ok found it / in dec maybe not the newesst
:|({["1,2,3"r]g}s)c|p/t: and you speak to 1 frame in 1 other {he he}
5:tempral: 4Polar/ 3|Cylindrical| 2(Spherical) 1{Gravitational} 0[Rectangular]

Try this "thought experiment" on for size:

Situation 1) Mile tall building. With a very sensitive spring scale, we weigh an object on the ground floor and on the top floor. Since the difference in distance from the center of the Earth is ~ 1 part in 4000, the diference in weight will be ~ 1 in 16 million.

Situation 2) Mile long space ship accelerating @ 1 gee. If we weigh the object at the bottom and top of the space ship, we will find no weight difference. In fact, if we take the mass of the ship into account, the object will weigh more at the top than at the bottom, since it will be gravitationally attracted to the center of the ship.

__________________
Any day you wake up on "the right side of the dirt" is a good day.

T. Anderson

Thank you Grapes of Wrath, Doctor Don , and Donnie B.
Yes, I was originally referring to a particle at rest in a gravity field, as Grapes clarified.
Nevertheless, yes, the 'free fall' of the charge will make it behave equivalently to acceleration, Don, BUT (as Donnie said) only as observed from the rest frame of the gravitational source, no?

From the co-accelerating frame of the free falling charge, can radiation be observed? Has there been any exp. test of such?
And what is this strange stuff we call EM radiation if its existence is dependent upon the frame of the observer?!

G^2


<font size=-1>[ This Message was edited by: Gsquare on 2002-01-22 10:41 ]</font>

<font size=-1>[ This Message was edited by: Gsquare on 2002-01-22 11:17 ]</font>

I'm sorry for being so thick, GOW, but I don't see how that page explains away the paradox.

In the first case, A stays fixed and B accelerates away, then accelerates retrograde and returns. A is older than B.

In the second case, we take B as fixed. A accelerates away, then accelerates retrograde and returns. Yet A is still older than B.

The difference between the two cases is only one thing: B felt the acceleration in both cases. Yet, why should this be? With B taken as fixed, it was A who accelerated in the second case.

The only difference between them is that A remained fixed with respect to the greater universe. In the second case, B felt the acceleration because the whole cosmos went soaring past him. Good thing he had his rocket motors turned on, or he'd have been dragged along!

I'm not trying to be argumentative, but I'd be interested in getting this further clarified.

They aren't equivalent situations, since the gravitational force for the earth is r^-2, but the acceleration of the spaceship is constant.

Don Smith

I understand that, but wasn't the point of Einstein's original thought experiment that an observer in a closed room could not determine (experimentally) whether he was under gravitational or inertial acceleration?

_________________
TANSTAAFL!

<font size=-1>[ This Message was edited by: Kaptain K on 2002-01-22 11:47 ]</font>

Again, good question, Donnie B.
Does anyone have an answer?
Does anyone know of any radiation detection experiments with detectors being accelerated along side of accelerated charges?

My only question here, Donnie:
In reality, from the perspective of B, the whole cosmos does not 'go soaring past him'. He only sees very local objects soaring past him; (the distant stuff shows so little movement that it is essentially 'fixed'). Would this not more logically indicate that he 'feels' acceleration as a result of some 'local' phenomena?
G^2


<font size=-1>[ This Message was edited by: Gsquare on 2002-01-22 12:22 ]</font>

Donnie, acceleration is not relative. Motion is.

The "relativity" part of Einstein's theories only says that any inertial frame can be considered at rest. Consider this:

A and B are both at rest. B accelerates so he is moving "that way.". B decelerates so he is again at rest. B accelerates so he is moving "this way". When he gets back to A, he again decelerates so he is at rest.

Now: A and B are both moving "this way". B decelerates so he is at rest, and A continues to move away from him. B accelerates so he is moving "this way" faster than originally. When he catches up to A, he decelerates to the original velocity.

And: A and B are both moving "that way". B accelerates so he is moving "that way" even faster. B decelerates so he is at rest. Once A catches up to him, he accelerates to the original velocity again.

In those three cases, we have three different "rest frames", but in all three cases it is B that changes frames -- while the frames can only be described relative to other frames, changing frames is an absolute act.

In your thought experiment, if B remained in the original inertial frame while A and the whole cosmos accelerated away from him and then returned, B would be older than A. If A (and the rest of the cosmos) remain in the same inertial frame while B accelerates away and returns, B is younger than A.

So, the question is, how does the "universe" know whether it was B that accelerated or the rest of the cosmos, right? Newton's Law of Inertia stated that a body at rest will remain at rest and a body in motion will remain in motion unless a force acts upon it. Although in Relativity, there's no absolute rest or absolute motion, that Law still holds: a body cannot change inertial frames arbitrarily. Whichever body ("B" or "A and the rest of the cosmos") has the requisite force act on it is the one which ultimately accelerates.

Does this help any?

__________________
SeanF

"Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher

The contents of this post are ©2008 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF

In practice, it only works in a small room... Arthur C. Clarke made much of this, saying that it only worked for points. Well, that's not true...

1) I can easily arrange masses around the room in such a way as to make the gravitational vector always downward (Clarke noted that, on the earth, it would point toward the center of the earth, and thus only exactly straight downward from points above the center of the room's floor...)

2) Harder... I *think* I could arrange masses around the room so that the force of gravity doesn't vary from the elevator's floor to its ceiling, but I'm not sure...

Anyway, these are artifacts: Einstein was talking about "force" in the perfect abstract, whereas on a planet or in a spaceship, we are acted upon by the sum of a large number of forces.

Silas

In the derivation for the radiation from a moving point charge, one finds that the fields are proportional to the time derivative of velocity as measured by an observer. If the observer is accelerating with the point charge, the derivative of the velocity will be zero. Hence no radiation.

If math does not scare you, take a look at Glenn Smith's book, "Classical Electromagnetic Radiation". It does a very good job showing the fields from a moving charge. It has pictures that show how the fields change when the charge accelerates, and by extension why a co-accelerating observer would not notice this change.

Thanks Wiley for the reference. I am familiar with the standard derivation of radiation from accelerated charge but haven't considered it for a co-accelerating frame. Apparently, you have given the correct answer.

And no, the math doesn't scare me; what is disconcerting are the implications. As I said in my second post, what is EM radiation that it can be made to disappear and reappear depending on the frame of the observer?
We now have a situation where someone co-moving (accelerating) in free fall with a charged particle cannot observe any radiation, but the stationary observer does see radiation.

So now we appparently have as great an unresolved paradox as before, no?

G^2









<font size=-1>[ This Message was edited by: Gsquare on 2002-01-22 20:50 ]</font>

Dang. No arguments? This could get to be a boring place.

I did move the discussion to a new thread.

It's important to remember that while observer may or may not see any radiation, (s)he will always see the fields. If the source is moving with a constant velocity with respect to the observer, the fields will decay at 1/R^2, but if the source is accelerating with respect to the observer, the fields will decay at 1/R. (R is the distance between source and observer at the retarded time.)

Interestingly, both vector and scalar potentials decay at 1/R regardless of the observer's frame. (Exception: if source and observer are at rest with respect to each other, the vector potential vanishes, leaving only the scalar (static) potential.)


<font size=-1>[ This Message was edited by: Wiley on 2002-01-24 17:25 because he zigged when he should've zagged ]</font>

<font size=-1>[ This Message was edited by: Wiley on 2002-01-24 17:26 ]</font>

Thanks Wiley; very interesting.
You are really challenging me on this one. Manipulating vector & scalar potentials are not exactly my strengths.
Nevertheless; a couple of questions:

1). I can see it intuitively, but in the co-accelerating frame, how do we establish vanishing vector potential? Simply from the absence of a magnetic field?
And...

2). how do you arrive at 1/R dependence of the potentials in both frames.
and...

3). More importantly, how do you arrive at the E field (in the co-accelerating frame) as having 1/R dependence when we know E = - gradient of the scalar potential?

I assumed you worked this out and I've got a feeling I shouldn't have asked; since it will probably involve lots of div's, grad's and Laplacians.

G^2

Only if the accelerations were the same in both cases. You can tell the difference between a r^-2 acceleration and a constant acceleration, just not whether either of them is inertial or gravitational.

Don Smith