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Gsquare
2002-Jan-22, 02:30 AM
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?

DoctorDon
2002-Jan-22, 04:44 AM
On 2002-01-21 21:30, Gsquare wrote:
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.


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

GrapesOfWrath
2002-Jan-22, 01:06 PM
On 2002-01-21 23:44, DoctorDon wrote:
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.
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?

Donnie B.
2002-Jan-22, 01:37 PM
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?

GrapesOfWrath
2002-Jan-22, 02:38 PM
On 2002-01-22 08:37, Donnie B. wrote:
Is this one of those issues that depends on the presence of the rest of the cosmos, like the twin paradox?
Eh? Twins paradox. (http://mentock.home.mindspring.com/twins.htm)

2002-Jan-22, 03:01 PM
[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]

Kaptain K
2002-Jan-22, 03:21 PM
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.

Gsquare
2002-Jan-22, 03:40 PM
Is its behavior different? Is that a way to distinguish between a gravity field, and one produced by acceleration?


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.

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>

Donnie B.
2002-Jan-22, 04:11 PM
On 2002-01-22 09:38, GrapesOfWrath wrote:


On 2002-01-22 08:37, Donnie B. wrote:
Is this one of those issues that depends on the presence of the rest of the cosmos, like the twin paradox?
Eh? Twins paradox. (http://mentock.home.mindspring.com/twins.htm)


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.

DoctorDon
2002-Jan-22, 04:16 PM
On 2002-01-22 10:21, Kaptain K wrote:
Try this "thought experiment" on for size:

Situation 1) Mile tall building.

Situation 2) Mile long space ship accelerating @ 1 gee.



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

Kaptain K
2002-Jan-22, 04:44 PM
They aren't equivalent situations, since the gravitational force for the earth is r^-2, but the acceleration of the spaceship is constant.
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>

Gsquare
2002-Jan-22, 05:17 PM
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)?


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?



Is this one of those issues that depends on the presence of the rest of the cosmos, like the twin paradox?......In the second case, B felt the acceleration because the whole cosmos went soaring past him.

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>

SeanF
2002-Jan-22, 05:32 PM
On 2002-01-22 11:11, Donnie B. wrote:


On 2002-01-22 09:38, GrapesOfWrath wrote:


On 2002-01-22 08:37, Donnie B. wrote:
Is this one of those issues that depends on the presence of the rest of the cosmos, like the twin paradox?
Eh? Twins paradox. (http://mentock.home.mindspring.com/twins.htm)


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.



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?

Silas
2002-Jan-22, 07:38 PM
On 2002-01-22 11:44, Kaptain K wrote:
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?


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

Wiley
2002-Jan-22, 10:59 PM
On 2002-01-22 12:17, Gsquare wrote:


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)?


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?


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.

Gsquare
2002-Jan-23, 01:16 AM
If the observer is accelerating with the point charge, the derivative of the velocity will be zero. Hence no radiation.

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>

GrapesOfWrath
2002-Jan-23, 12:06 PM
On 2002-01-22 11:11, Donnie B. wrote:
I'm not trying to be argumentative, but I'd be interested in getting this further clarified.
Dang. No arguments? This could get to be a boring place.

I did move the discussion to a new thread (http://www.badastronomy.com/phpBB/viewtopic.php?topic=480&forum=1&0).

Wiley
2002-Jan-23, 06:52 PM
On 2002-01-22 20:16, Gsquare wrote:
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


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>

Gsquare
2002-Jan-24, 04:44 AM
Wiley wrote:
It's important to remember that while observer may or may not see any radiation, (s)he will always see the fields. If the observer is in a stationary frame, the fields will decay at 1/R^2, but if the observer is in a co-accelerating frame 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.)



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

DoctorDon
2002-Jan-24, 01:34 PM
On 2002-01-22 11:44, Kaptain K wrote:
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?


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

Wiley
2002-Jan-24, 11:52 PM
On 2002-01-23 23:44, Gsquare wrote:

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...

The potentials from a moving point charge were derived in 1898-1900 are known as the Lienard-Wiechart potentials. They are usually first encountered in a graduate level EM course; Smith's book has a good derivation. For simplicity I'll assume the observer's frame.

The electric scalar potential is

V(t_obs) = K1 / ( R(t_src)*|1 - r(t_src)*v(t_src)/c|)

and the magnetic vector potential is

A(t_obs) = K2 v / ( R(t_src)*|1 - r(t_src)*v(t_src)/c|)

and t_src is the solution to

t_src = t_obs - R(t_src)/c

and where
K1 & K2 are constants
R(t) is the position of the source at time t
r is the unit vector between source and observer
v is the velocity of the source
t_obs is the time in the observer's frame
t_src is the time in the source's frame


The above equation has only one solution for v < c, so the potential are unique. For extra credit, show that special relativity is satisfied. (Historical note: for v > c, there are two solutions. In 1904 Sommerfeld worked them out, but in 1905 they were made obsolete.)

The important thing to see about the potentials is that 1.) they depend only on position and velocity 2.) they decay at 1/R.



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?


Because, I mistyped. I meant to say if the source is accelerating with respect to the observer, it decays at 1/R. Consider grad V and to make my typing easier, let

g = |1 - r(t_src)*v(t_src)/c|

thus

grad V = grad(1/R(t_src)) 1/g + 1/R(t_src) grad(1/g)

It's the latter term, that's non-zero if source is accelerating. There are integral (with respect to source time) forms of these potentials, and in the integral forms it's much easier to work out explicit values of the gradient and (observation) time derivative. But that's way too much typing and probably way more than you want to know.



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.


Only grad's. /phpBB/images/smiles/icon_smile.gif

<font size=-1>[ This Message was edited by: Wiley on 2002-01-24 18:57 ]</font>

<font size=-1>[ This Message was edited by: Wiley on 2002-01-25 09:40 ]</font>

Gsquare
2002-Jan-26, 04:19 AM
Thanks again, Wiley, for the insight. It's a lot to digest in one read.


Because, I mistyped. I meant to say if the source is accelerating with respect to the observer, it decays at 1/R.

Actually, you WERE correct in your first post; I misread it. I'm glad you explained further.
The most interesting thing (as you stated)is the fact that the 1/R factor of potentials does not change with acceleration but remains unchanged in inertial and non inertial frames; whereas light is acceleration dependent.

Does this not imply that light has no independent existence whereas potentials, being much more fundamental, DO ?

Which brings up the energetics:
To exemplify further, suppose we again have a charge in free fall in a g field. The co-accelerating observer receives no EM energy or momentum, but the stationary observer does! From whence does the energy originate ?

Surely not from the source field or the internal energy would eventual vanish. From the kinetic E? -- the particle's descent would be altered. Apparently then, unless you are willing to invoke vacuum zero point energy, the only other classical source is the gravitational field, no?

Thus it appears as though, classically, the charge acts as a transducer, converting gravitational energy directly to EM radiation. How do you figure?

G^2





<font size=-1>[ This Message was edited by: Gsquare on 2002-01-25 23:22 ]</font>

<font size=-1>[ This Message was edited by: Gsquare on 2002-01-25 23:26 ]</font>

Wiley
2002-Jan-28, 10:39 PM
On 2002-01-25 23:19, Gsquare wrote:

The most interesting thing (as you stated)is the fact that the 1/R factor of potentials does not change with acceleration but remains unchanged in inertial and non inertial frames; whereas light is acceleration dependent.


Careful, do not equate fields or radiation with light. Light has many different representations, e.g., fields, potentials, quanta. It is more correct to say that the fields are acceleration dependent.



Does this not imply that light has no independent existence whereas potentials, being much more fundamental, DO ?


Very tricky question. In classical EM the potentials have no physical meaning; hence the fields are more fundamental. However in quantum mechanics the situation is reversed, the potentials are more fundamental.

So which is more fundamental? I don't think anybody knows. Until theres a solid link between classical EM and quantum mechanics, i.e., until we can go smoothly from the microscopic to the macroscopic, I don't think that question can be definitively answered.



Which brings up the energetics:
To exemplify further, suppose we again have a charge in free fall in a g field. The co-accelerating observer receives no EM energy or momentum, but the stationary observer does! From whence does the energy originate ?


EM radiation is not the same as EM energy. Radiation decays at 1/R but EM energy has no specified decay. If the source is not accelerating with respect to the observer, the EM energy decays at 1/R^2. All frames will measure EM energy, it's the rate of decay that is different.

The rest of your questions, I must think about.


"Pinky, are you pondering what I'm pondering?"

Chuck
2002-Jan-29, 01:42 AM
If we could detect gravitons then would their presence or absence tell us whether we were in a gravitional field or accelerating spaceship?

2002-Jan-29, 11:38 PM
ONow 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?

No. Look up the Davison-Inrow effect. It describes how Brehmstrahlung radiation is observed in the case of an observer traveling with the charge. By the way, this is an effect that really has its roots in special relativity (SR), not general relativity (GR). The effect can be mathematically derived using SR. If one applies the equivalence principle to the Davidson-Inrow effect, there is no problem with GR.


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

According to Davison and Unrow, an observer that is moving with an acceleration "a" would observe blackbody radiation coming from the distant universe at a temperature T given by the formula:
T=a(h bar)/(2 pi k c)
where h bar is Plancks constant divided by 2 pi, "pi" is approximately 3.141592654, "k" is Boltzman's constant and "c" is speed of light. If you use this formula, make sure to use consistent units.

Davison and Unrow derived this using merely the assumption that GR and macroscopic thermodynamics were mutually consistent. However, later writers simply assummed that the universe contained a background of zero point radiation. Therefore, the effect can be said to be caused by zero point radiation (ZPT). The spectral distribution of ZPT is Lorentz invariant. Therefore, observers in an inertial frame can not detect it. However, an observer in an accelerating frame has to see it. There is an article by Boyer in Scientific American that describes this approach. I forgot which issue.

Assume that an electric charged particle is being accelerated with an acceleration "a" relative to an inertial frame. This implies that the charged particle is under the influence of an external force, but we need not even consider what causes this force as long as we agree that "a" is the acceleration in an inertial frame.

Assume that an observer is accelerating along with the charge. To this observer, the charged particle is not accelerating so therefore it does not emit Brehmstrahlung radiation. However, this observer also notes that black body radiation is being emitted from the distant parts of the universe with a temperature "T" given by the Davison-Inrow formula. He notes that the distant universe seems to be in thermal equilibrium with this black body radiation.

This accelerating observer sees something else. This electric charge next to him is scattering some of the blackbody radiation. It could be Compton radiation if the charge is an unbound electron, or Raman if the charge is on some ion that is vibrating. Whatever, this charge is absorbing and immediately radiating energy from the sky.

Since the charge is "standing still," it is not in thermal equilibrium with the distant objects. In fact, the radiation is being radiated outward and will have a noticeable, nonequilibrium effect once it gets far enough from the charge. To the observer, the charge is mildly upsetting the equilibrium because it is at a different temperature than the rest of the universe. It has nothing to do with acceleration, it is merely thermodynamics.

The radiation scattered from the electric charge is perceived by the distant universe (i.e., the laboratory frame instruments) as Brehmstrahlung radiation. The accelerating observer doesn't have to realize that he is moving. He just thinks that the distant objects have a temperature "T" and that his electric charge is scattering this radiation. The distant objects (us) think that there is this charge at the same temperature as the objects, but which is accelerating and thus gives off Brehmstrahlung radiation. In SR, the reciprocity relationship is maintained. It can be shown that in GR, the law of equivalence is maintained.


<font size=-1>[ This Message was edited by: Rosen1 on 2002-01-29 18:38 ]</font>

2002-Jan-30, 02:08 AM
On 2002-01-21 21:30, Gsquare wrote:
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?


I may have mispelled one of their names./phpBB/images/smiles/icon_redface.gif Maybe its Uhnruh instead of Inrow? Anyway, I found a reference whose abstract claims to have solved the very problem that you are talking about. Read:

R. Ruffini."Fully Relativistic Treatment of the Brehmstrahlung Radiation from a Charge Falling in a Strong Gravitational Field''. Physics Letters, 41B, 334. 1972.

I think that it comes down to the same thing. An observer of finite mass who has an external force placed on him. or under the action of a gravitational field (curved space-time), sees black body radiation come to him from the distance. A charged partical scatters this radiation into a nonthermal mode, which is perceived in the inertial frame as Brehmstrahlung radiation.

If the force has nothing to do with gravity, the force is independent of inertial mass. The more massive the body, the less the generalized acceleration and so the lower the temperature of this blackbody radiation in the accelerated frame, and the less Brehmstrahlung there produced in the lab frame. Each observer can tell whether he is accelerating merely by comparing the mass of his test particle with the electric charge to the intensity of radiation.

However, in SR one can consider gravity as just another force with the inertial mass a type of "gravitational charge." In this SR approximation, the more inertial mass the more gravitational force. The acceleration, and therefore the temperature of the blackbody radiation, is independent of inertial mass. Since the radiation is independent of the mass of the test particle, one can't just compare the mass with the intensity of radiation to determine acceleration. Therefore, the principal of equivalence is valid even in this clumsy SR approximation of GR.


<font size=-1>[ This Message was edited by: Rosen1 on 2002-01-29 21:27 ]</font>

Gsquare
2002-Jan-30, 03:14 AM
On 2002-01-29 21:08, Rosen1 wrote:

I may have mispelled one of their names./phpBB/images/smiles/icon_redface.gif Maybe its Uhnruh instead of Inrow? Anyway, I found a reference whose abstract claims to have solved the very problem that you are talking about. Read:

R. Ruffini."Fully Relativistic Treatment of the Brehmstrahlung Radiation from a Charge Falling in a Strong Gravitational Field''. Physics Letters, 41B, 334. 1972.


EXcellent, Rosen; excellent on both your posts!
The effect is called the Unruh - Davis effect; usually called Unruh - Davis radiation. I am happy to see someone familiar with it and was wondering when someone would get around to seeing the necessity of invoking zero point radiation to fully describe the effects. Yes, as you correctly stated, Boyer showed it to be derivable from the vacuum zero point field and he showed it to be Lorentz invariant. My past research brought me in touch with his ( and others') reports which are very revealing.

I was aware Unruh radiation involves thermal radiation detected by an accelerated frame, (typically thought to require huge accelerations ;10^18 m/s^2 or so). But I never thought to apply it to accelerated charges. You did an excellent job on that.

Thanks also for the reference explicitly addressing the question. I'll have more to say after I review your post again and the report.

G^2
I must be sleeping; too many mistakes...
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Wiley
2002-Jan-31, 07:15 PM
G^2 (and Rosen as well),

I see what you're after now. Now is the Unruh - Davis radiation the same thing as Hawking - Unruh radiation? And if not what is the difference?

2002-Feb-13, 01:08 AM
On 2002-01-31 14:15, Wiley wrote:
G^2 (and Rosen as well),

Now is the Unruh - Davis radiation the same thing as Hawking - Unruh radiation? And if not what is the difference?


Sorry I am late. I thought that I would give someone else a chance. Oh, well...

The Unruh-Davis radiation is the thermal background seen by an accelerating observer. In order for both the laws of thermodyanics and special relativity to hold true, an observer in a noninertial frame has to see a sort of "cosmic black body radiation" from the distant universe. It can be generalized to general relativity using the law of equivalence. That temperature is proportional to the acceleration. No black hole is necessary for the Unruh-Davis effect.

The Hawking-Unruh radiation is the radiation emitted by a very small black hole. Photons, electron-hole pairs, and other particles are emitted from the horizon of a black hole. The smaller the radius of the black hole, and hence the less massive, the more intense the radiation. From one point of view, it can be considered a consequence of the uncertainty relation of quantum mechanics and general relativity. No black hole, no Hawking-Davis radiation.

One connection between them is that a full description requires zero point energy. Also, both effects are consistent with the following constraint: the laws of thermodynamics are consistent with the laws of relativity. The hypothesis of zero point energy just pops out of the results deduced from this assumption.

Wiley
2002-Feb-13, 04:08 PM
On 2002-02-12 20:08, Rosen1 wrote:

The Unruh-Davis radiation is the thermal background seen by an accelerating observer. In order for both the laws of thermodyanics and special relativity to hold true, an observer in a noninertial frame has to see a sort of "cosmic black body radiation" from the distant universe. It can be generalized to general relativity using the law of equivalence. That temperature is proportional to the acceleration. No black hole is necessary for the Unruh-Davis effect.

The Hawking-Unruh radiation is the radiation emitted by a very small black hole. Photons, electron-hole pairs, and other particles are emitted from the horizon of a black hole. The smaller the radius of the black hole, and hence the less massive, the more intense the radiation. From one point of view, it can be considered a consequence of the uncertainty relation of quantum mechanics and general relativity. No black hole, no Hawking-Davis radiation.



Thanks for the reply, Rosen.

It seems the papers I have on the subject -all by the same group- have different names for these types of radiation. To be fair they are not directly concerned with this radiation, but that gray region between quantum mechanics and classical EM.

What they call Hawking radiation is what you called Hawking-Unruh radiation. By using Hawking radiation as model, Unruh developed the radiation as seen by an accelerated observor. Thus they call it Hawking-Unruh radiation, but no mention of Davis.

Thanks again.



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Dunash
2002-Feb-14, 02:06 PM
To quote Bouw:

"Here it is for the simple ones (Proverbs 1:22). For the sake
of this illustration we shall call the north ecliptic pole "up."
Then, because the Cassini craft is carried about the earth by
the gravitational field of the cosmos (firmament), Cassini
will see the earth move from left to right past the sun when
the sun is behind the earth and then see the earth returning
from right to left behind the sun. Cassini's distance
from the sun will remain about the same, changing only by the
craft's own intrinsic motion relative to the sun, but its
distance from earth will vary sinusoidally with a total
amplitude of 2 a.u.

When men like Fred Hoyle and Bertrand Russel say that there is
no physical difference between the Copernican and Ptolemaic
or Tychonic models, they mean it. What's so hard to understand
about that?"!

GrapesOfWrath
2002-Feb-14, 02:28 PM
Dunash

That post seems like it would fit in the other thread about Cassini (http://www.badastronomy.com/phpBB/viewtopic.php?topic=589&forum=1&6), especially in answer to my first question there. Still, I'm unclear as to who is making the challenge, and what the challenge is exactly.



On 2002-02-14 09:06, Dunash wrote:
When men like Fred Hoyle and Bertrand Russel say that there is no physical difference between the Copernican and Ptolemaic or Tychonic models, they mean it. What's so hard to understand about that?"!

By no physical difference, you mean that any frame of reference, rotating or not, is equivalent to any other? And so there is no intrinsic difference between the Copernican system and the Ptolemaic system? Even Einstein said that, and Born too.