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Author Topic: How can the signs change with an acceleration? Unruh radiation.  (Read 15114 times)

Offline yor_on

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How can the signs change with an acceleration? Unruh radiation.
« Reply #25 on: 05/01/2011 17:10:51 »
Soul Surfer "is it that the concept that the acceleration of the high energy electron by deflecting it from its otherwise straight path creates the radiation is wrong because that's the way I understand the theory"

You have a very good point there. Normally when I think of a object breaking the geodesics I think of the object itself expending energy, but in this case we have the magnetic fields 'forcing it' expending the energy. So let us assume that an uniform acceleration can create this radiation, is that equivalent to a uniform motion in that synchrotron?

If we look at how Newton saw it, he considered any circular uniform motion a acceleration, just as  Jastra and Imfaatal seems to do. So I think we all agree on that one?

Then we comes to Grahams orbiting 'static' magnetic field? And now I don't know any more?  Graham could you explain how to think of that field? Is it the electrical charges themselves that creates the 'static field', each one of those charges interacting with another charge creating it?

would it if we only used one of those charges be possible to define a radiation, if it is it stands to reason that the field too could be seen as 'one' charge, isn't it so?

Then although the components of the field won't radiate any more the field in itself should? But if it also is following what we call a geodesic it's not expending any 'energy' rotating around the Earth and then if we treat it as one object it shouldn't expend any energy at all?

It's a weird one, isn't it? I'm not sure I'm seeing this one at all :)
==

And no, not Grahams idea, but the whole idea of Unruh and Hawking radiation.
« Last Edit: 05/01/2011 19:34:06 by yor_on »
 

Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #26 on: 05/01/2011 17:18:48 »
That is true, Matthew. In Cartesian coordinates, all the time derivatives of x and y would be sinusoids. I am unsure then whether it is then fair to say the acceleration is constant (which is what is normally done for circular motion). It is only the magnitude that is constant, and orthogonal to the motion. Nonetheless the particle is in freefall and should feel no net force upon it.

I rather gave up on this last time I looked into it. It is interesting to look at different states and see what we think would happen:

Would the following be seen to radiate to
(a) a comoving observer
(b) an stationary observer on the gravitating object
(c) a stationary observer at infinity (or a long way off)

1. A charged particle in circular orbit around a gravitating sphere (a, b and c)
2. A continuous circular line of charged particle in orbit around a gravitating sphere (a,b and c)
3. A charged particle in a straight line free fall toward the centre of a gravitating sphere (a, b and c)
4. A charged particle on the surface of the sphere (b and c)
5. A charged particle moving in free space and not in a gravity field (a and c)
6. A charged particle on an accelerating rocket, in free space, with a constant acceleration (a and c)

How would the charged particle behave in each observation compared to a non-charged particle in each case and does this stack up with the GR equivalence principle? Are the views all consistent with each other and what effect does the loss on energy (due to any radiation) have on each observer's view?

The a, b and c referenced are the questions applicable, not the anwers by the way. I realised belatedly that this was ambiguous.
« Last Edit: 05/01/2011 17:37:49 by graham.d »
 

Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #27 on: 05/01/2011 17:25:54 »
"Then we comes to Grahams orbiting 'static' magnetic field? And now I don't know any more?  Graham could you explain how to think of that field? Is it the electrical charges themselves that creates the 'static field', each one of those charges interacting with another charge creating it?"

I was just thinking that such an arrangement simply creates a very big bar magnet. A circulating electric current. I didn't really think it mattered how the in-line charges are held together and I'm sure that such a system is not naturally stable - it's just a thought experiment.
 

Offline yor_on

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How can the signs change with an acceleration? Unruh radiation.
« Reply #28 on: 05/01/2011 18:09:09 »
Found this Graham. But it's a 'pay for view'.

"We continue with our series of papers concerning a self-field approach to quantum electrodynamics that is not second quantized. We use the theory here to show that a detector with a uniform acceleration a will respond to its own self-field as if immersed in a thermal photon bath at temperature Ta=ħa/2πkc. This is the celebrated Unruh effect, and it is closely related to the emission of Hawking radiation from the event horizon of a black hole. Our approach is novel in that the radiation field is classical and not quantized; the vacuum field being identically zero with no zero-point energy. From our point of view, all radiative effects are accounted for when the self-field of the detector, and not the hypothetical zero-point field of the vacuum, acts back on the detector in a quantum-electrodynamic analog of the classical phenomenon of radiation reaction. When the detector is accelerating, its transformed self-field induces a different back reaction than when it is moving inertially. This process gives rise to the appearance of a photon bath, but the photons are not real in the sense that the space surrounding the accelerating detector is truly empty of radiation, a fact that is verified by the null response of an inertially moving detector in the same vicinity. The thermal photons are in this sense fictitious, and they have no independent existence outside the detector."
==

Thinking of it, I'm wrong in assuming that anything orbiting is in a geodesic, it can't be, it will always be acted on by gravity, making it into a spiral.

From Quantum electrodynamics based on self-fields. It seems similar?
« Last Edit: 05/01/2011 18:14:54 by yor_on »
 

Offline yor_on

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How can the signs change with an acceleration? Unruh radiation.
« Reply #29 on: 05/01/2011 19:13:24 »
Let's build it up from a QM level :) Highly unscientifically so. We know that electron doesn't 'radiate' away its energy as it makes that unspecified 'orbital' around the atoms nucleus. I found a very ingenious description of it from a Newtonian perspective. MAXWELL & THE REVOLVING ELECTRON This guy have a really clear view of the process, I got impressed reading him.
==

Better point out that the rest of the site is somewhat different, and not what I'm linking to, or defining as reality, although interesting. :) But the page referred too is as far as I understand perfectly correct.
==

So a 'single electron' refuse to radiate. How about putting several such atoms together. Would it radiate then? Nah, I don't think so. Radiation is about 'change' and 'energy expended' to me. Do you agree?
==

Maybe the next step should be to define what make something 'electro magnetic'?

As I understands it that come from the idea of charged particles "like 'negative' electrons and 'positive' protons. Without these charged particles, there can be no electric force fields and thus no electromagnetic waves." Here.

So what was a charge? There it becomes weird again, we have two definitions coexisting as I see it. One is what we call charges normally, observing them as 'whole' quantity's, the other also correct but there consisting of fractional quantity's of 'charge', also called 'quasi particles'. Here.

"In physics, a charge may refer to one of many different quantities, such as the electric charge in electromagnetism or the color charge in quantum chromodynamics. Charges are associated with conserved quantum numbers.

More abstractly, a charge is any generator of a continuous symmetry of the physical system under study.

When a physical system has a symmetry of some sort, Noether's theorem implies the existence of a conserved current. The thing that "flows" in the current is the "charge", the charge is the generator of the (local) symmetry group. This charge is sometimes called the Noether charge.

Thus, for example, the electric charge is the generator of the U(1) symmetry of electromagnetism. The conserved current is the electric current.

In the case of local, dynamical symmetries, associated with every charge is a gauge field; when quantized, the gauge field becomes a gauge boson. The charges of the theory "radiate" the gauge field. Thus, for example, the gauge field of electromagnetism is the electromagnetic field; and the gauge boson is the photon.

Sometimes, the word "charge" is used as a synonym for "generator" in referring to the generator of the symmetry. More precisely, when the symmetry group is a Lie group, then the charges are understood to correspond to the root system of the Lie group; the discreteness of the root system accounting for the quantization of the charge." Here.

So is there statements about what our electrical charge is?

"Charge is not energy. A set quantity of charge can have many different amounts of energy at the same time. Opposite charges moving along together are "mechanical", while opposite charges moving differently are "electrical." If the negative charge in an object should start moving while the object's positive charge stays at rest, then we call that motion an electric current. The words "electric current" mean the same as "charge flow. Charge is not invisible. Whenever light bounces off an object, it bounces off the outside of the atom, and the outside of an atom is made of negative charges. In other words, electric charge reflects light."

And "Charge is "poles"

When the positive and negative charges of matter are sorted out and pulled away from each other, "static electricity" is the result. When (+) is pulled away from (-), an invisible force field connects them and causes them to attract each other. This field is similar to magnetism in many ways, but it is not magnetism, it is called an Electrostatic Field, or "e-field." With magnetism, the lines of force spring from the north and south poles of magnets, and these lines seem to connect the opposite magnetic poles together. In Electrostatics, the electrical lines of force connect the (+) and (-) poles together. What is charge? It is the "pole" where the electrical lines of force come to an end. Follow the lines of a static "e-field" along, and eventually you'll arrive at a small bit of "charge." Electric charge is the glue which attaches the flux lines of e-field to the particles of matter. " From Here.
==

By the way, this guy above is one of my all time favorites :) when it comes to looking at things.. Here's something from his FAQ.   

"WHERE DID YOU GET THIS JUNK?

A: It's a secret. Here it is. Always tell the truth, and, more importantly, never lie. Even to yourself. What the heck does this have to do with anything? Well, once I realized that I was defending my ego by constantly telling myself a thousand subtle lies, I was able to stop. When I did, all this stuff started boiling up out of my unconscious and out onto my website. It must have been in there all along. It just wouldn't come out and play. Maybe it was embarassed about all the lying.

PS I strongly suspect that Richard Feynman accidentally stumbled across this same technique. It's a source of creativity like you wouldn't believe! It's a wellspring of amazing ideas which seem to arise fully formed, without you doing the work to assemble them. "

It's good.
« Last Edit: 05/01/2011 23:39:56 by yor_on »
 

Offline Ron Hughes

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How can the signs change with an acceleration? Unruh radiation.
« Reply #30 on: 05/01/2011 19:44:41 »
a. would not see radiation
b. would
c. would
 

Offline yor_on

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How can the signs change with an acceleration? Unruh radiation.
« Reply #31 on: 05/01/2011 21:34:41 »
So assume that this electrical particle (atom) is coasting along a geodesic in deep space, weightlessly sort of :) Will it show a radiation. I would say no, in none of the cases Graham suggest.

But when in-falling toward a object of invariant mass then? There I have two answers. The geodesics is no different from the one where I said no, they may be more and they may 'join' at some point into a big bump, but they are the same.

The other is to look at what an invariant mass is made from, other particles right? and they together create a electro magnetic field, don't they? So our particle is not only following the geodesics. It's also moving through a EM-field. And when a charge moves through a EM-field? Electricity is created, isn't it?

That is, if I'm thinking right here?
==

I know, I'm suddenly giving empty classical space 'EM-properties' :) but we can imagine it as dust perhaps, should be some close to a planet, shouldn't it? Thinking about it again it doesn't matter if there is one or not. We know that we can use EM-energy to communicate in space, so, no matter how that energy takes itself from 'A' to 'B' it will be there, when measured.. So it will from our particles frame of reference exist, am I right? That is, our particle becomes a 'detector'.
==

And thinking of it so it becomes the same strange phenomena as energy 'created' only in its interaction. You need to see that I differ between a 'electro magnetic' field existing on its own in 'space', and the fact that when 'existing' in that same space you become a focus for a lot of things that otherwise don't exist there. If that now make sense :)
« Last Edit: 05/01/2011 22:06:07 by yor_on »
 

Offline jartza

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How can the signs change with an acceleration? Unruh radiation.
« Reply #32 on: 06/01/2011 03:35:12 »
Jerk vector points backwards when you are moving in a circle. And you experience a braking force: jerk times your charge squared.
http://en.wikipedia.org/wiki/Abraham-Lorentz_force
 

Offline yor_on

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How can the signs change with an acceleration? Unruh radiation.
« Reply #33 on: 06/01/2011 04:43:18 »
Interesting Jartza.
 

Offline jartza

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How can the signs change with an acceleration? Unruh radiation.
« Reply #34 on: 06/01/2011 06:05:37 »
When you are driving in a very large circle, and your motor stalls periodically, then we have a situation where we can use the "force is charge squared times jerk" rule.

Here are the conditions of using the rule:
http://en.wikipedia.org/wiki/Radiation_reaction

Because jerk vector points sometimes forwards, sometimes backwards, you are radiating, but on the average radiation reaction force is not taking away energy from you.

 
 

Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #35 on: 06/01/2011 13:06:05 »
Lot's of comments here to catch up on

Ron, your answers don't make sense unless you intend the a, b and c answers to be true for all the cases 1 to 6 and I don't think you intend that.

Would the following be seen to radiate to
(a) a comoving observer
(b) an stationary observer on the gravitating object
(c) a stationary observer at infinity (or a long way off)

1. A charged particle in circular orbit around a gravitating sphere (a, b and c)

You say (a) no, but (b) and (c) yes.

But if the particle is radiating it is losing energy so its orbit will decay won't it? How would a comoving observer account for this loss of energy that he does not perceive and cannot measure? If its orbit does not decay where is the radiation energy (as seen by b and c) coming from?

2. A continuous circular line of charged particle in orbit around a gravitating sphere (a,b and c)

You say (a) no, but (b) and (c) yes.

The same as above. Plus, with a line of charge, how is this different from a large bar magnet. Sure this would just establish a static magnetic field.

3. A charged particle in a straight line free fall toward the centre of a gravitating sphere (a, b and c)

You say (a) no, but (b) and (c) yes.

This is good for compatibility with equivalence principle but still is complicated in showing from where the energy to radiate (in b and c) is coming from.

4. A charged particle on the surface of the sphere (b and c)

You say (a) no, but (b) and (c) yes.

I think everyone will agree that a charge sitting on your desk will not radiate (a and b are the same case here). But I think  a charge sitting on a (non-rotating) sphere would also not radiate to infinity either, so it should no to all 3 states I think.

5. A charged particle moving in free space and not in a gravity field (a and c)

You say (a) no, but (b) and (c) yes.

I think probably no to all of a, b and c. A charge moving at constant velocity in free space should not radiate.

6. A charged particle on an accelerating rocket, in free space, with a constant acceleration (a and c)

You say (a) no, but (b) and (c) yes.

The case b does not apply. By the equivalence principle I would also say no to (a) however, I am not so sure about c (I used to be!!). Generally I think it should radiate in case c also. Classically it should radiate but then where is the energy coming from? Part of the mass of the charged particle is the infinite electric field so I guess that the energy is provided by the rocket.

Do some of the anomalies here mean that there could be a difference between inertial and gravitational mass in the case of a charged object? This would be a novel concept!!
 

Offline Ron Hughes

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How can the signs change with an acceleration? Unruh radiation.
« Reply #36 on: 06/01/2011 16:57:44 »
graham said, "Would the following be seen to radiate to
(a) a comoving observer
(b) an stationary observer on the gravitating object
(c) a stationary observer at infinity (or a long way off)

1. A charged particle in circular orbit around a gravitating sphere (a, b and c)
2. A continuous circular line of charged particle in orbit around a gravitating sphere (a,b and c)
3. A charged particle in a straight line free fall toward the centre of a gravitating sphere (a, b and c)
4. A charged particle on the surface of the sphere (b and c)
5. A charged particle moving in free space and not in a gravity field (a and c)
6. A charged particle on an accelerating rocket, in free space, with a constant acceleration (a and c)"

Sorry, I miss understood your intent, I'll try again.

The co-moving observer always has the same frame of reference as the charged particle.
1.(a) The observer is also losing energy so both frames of reference are identical, he sees no radiation, (b) and (c) see radiation.
2. Same answer as above, for (a b and c).
3. Same
4. (a & b)no radiation, (c) could see minor radiation if the gravitating object is spinning.
5. (a & c) no radiation, (b) only if the object is spinning.
6.(a b & c) see radiation.
 

Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #37 on: 06/01/2011 17:55:20 »
Ron, for case 1, how is the comoving observer losing energy? I guess you maybe are assuming that because he is comoving he is being forced to deliberately lose energy to remain with the charged particle. This isn't what I meant really; I was meaning more that what he sees at some instant when they are matched in position and speed. They are both in freefall but you are saying (I think) that the charged particle will slow and its orbit will decay whereas the observer would have to take action to match this behaviour.

For case 2 can you explain why a rotating circular line of charge would produce anything but a static magnetic field. Once the charges had been set in motion I would expect the change in the em-field to continue out to infinity but not take any more energy once that had been set up. I'm not sure about this.

For case 3 would the comoving observer and charge continue to freefall at the same speed? I assume 'yes' as he does not perceive any energy loss. This may be puzzling to another observer though. This is where Schott energy needs to be invoked to explain it. I don't think these results are obvious.

For case 4 no radiation (body not spinning) - this one's clear

For case 5 no radiation - also clear

For case 6 why would a comoving observer see radiation? By equivalence if he was accelerating at the same rate as the charge it would be as if he were in the same gravity field (like case 4). I'm not 100% sure about whether the distant observer would see radiation either, though if you had asked me some time ago I would have said he would. 
 

Offline Ron Hughes

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How can the signs change with an acceleration? Unruh radiation.
« Reply #38 on: 06/01/2011 18:26:16 »
In case 2 the co-moving observer still has the same frame of reference as the orbiting charges so I would think no radiation. In the case c and b there is a difference in acceleration so I would think they see radiation.

In case 6 I was wrong, a does not see radiation,
 

Offline yor_on

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How can the signs change with an acceleration? Unruh radiation.
« Reply #39 on: 06/01/2011 19:14:26 »
What Jartza consider seem to be able to be expressed as a non-uniform acceleration as seen from a far observer, am I right? and in all accelerations you will have 'change' and as I presume 'energy expended'?

And from the inside of that charge the energy should be seen to fluctuate too, even if it 'evens out' statistically. What I don't get is the idea that it won't take any energy. As we're not discussing the charge being in a EM-field at all there, it's more of an action reaction, and also reaction action :) Now that hurts me head terribly..

Graham, what did you think of my idea of considering it as two 'causes'?
Splitting it in 'geodesics' and in a 'EM-field' when it comes to a charge spinning in towards a invariant mass (planet)?

Totally bicycling in the blue younder?
==

But then we have those 'self-fields' 
« Last Edit: 06/01/2011 19:16:33 by yor_on »
 

Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #40 on: 07/01/2011 13:59:50 »
Yor_on, I am not sure we need tbe concerned about geodesics or not. The paths followed by the particles are all undergoing hyperbolic motion (see definition), that is uniform acceleration. The ones in free fall in a straight line to the centre of a spherical gravitating body are, of course, following a geodesic, but orbital ones are not. I don't think this is a key issue, but I may be wrong.

In case other people are wondering why this is not as obvious as it first seems, please take a look at references to this in Google. An example would be searching for the word combination - hyperbolic motion rohrlich - the last word being the name of one of the main contributors to research in this area.

A paper, by Glass, that suggests that questions have been resolved can be found at:
www.arxiv.org/pdf/0801.1528
The maths is not trivial and assumes equations derived in other referenced papers.
« Last Edit: 07/01/2011 14:01:22 by graham.d »
 

Offline jartza

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How can the signs change with an acceleration? Unruh radiation.
« Reply #41 on: 08/01/2011 06:56:52 »
When you are driving in a very large circle, and your motor stalls periodically, then we have a situation where we can use the "force is charge squared times jerk" rule.

Here are the conditions of using the rule:
http://en.wikipedia.org/wiki/Radiation_reaction

Because jerk vector points sometimes forwards, sometimes backwards, you are radiating, but on the average radiation reaction force is not taking away energy from you.

 


Oh yes, when you start braking, radiation reaction force points backwards, and you move forwards, therefore you lose energy and momentum.

When you stop braking radiation reaction force points forwards, and you are not moving much, therefore you you gain the momentum back, but you don't gain the energy back.
 

Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #42 on: 08/01/2011 12:33:57 »
Jartza, I have read the link on this now. I had not heard the word "jerk" used to described the time derivative of acceleration, but I see that it is a sometimes used word for it. I understand the point you make and see that it refers to the Abraham-Lorentz force (or relativistically, the Abraham-Lorentz-Dirac force). This force applies only to periodic motion, which is an assumption required to reach the solution given. If this is the case then to maintain periodic motion this must imply that there needs to be external energy imparted to the charge in order to maintain its motion (if radiating) I think. I would assume, in the case of a cyclotron, this is exactly the case. In a charge orbiting in a radial gravity field this cannot provide an exact solution - if the charge is radiating there will be energy lost and the orbit lowered, so then not satisfying the criterion of periodic motion but only of "damped" periodic motion. However I expect this is close to what actually happens.

So what happens to a directly falling charge (following a geodesic radially in a gravitational field)? This is not covered by this theory but, I assume, will radiate to infinity (Larmor) but not to a comoving observer (equivalence principle). I still have problem with where the energy comes from except by using the slightly mysterious Schott energy theory.
 

Offline Soul Surfer

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How can the signs change with an acceleration? Unruh radiation.
« Reply #43 on: 09/01/2011 23:18:35 »
All Uniformly accelerating and orbiting charges do radiate both electromagnetically and gravitationally.  This can be observed in conditions where moving plasmas interact with magnetic fields like the Crab Nebula (M1)  the electrons orbit smoothly and uniformly around the magnetic fields (Cyclotron orbits) and emit radio waves.  This is also done in the magnetron in a microwave oven  the electrons orbit in a magnetic field and the orbital frequency produces the radiation.  In both of these cases there is very little higher order differentials in the electron velocity.

An electron orbiting the earth under gravitational attraction will radiate electromagnetically and lose energy but at a very low level and low frequency.    All orbiting gravitating bodies radiate gravitational energy as gravity waves but these are of such a low level and low frequency they are undetectable but the theory of the energy loss has been proved with great precision using a pair of orbiting pulsars.
 

Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #44 on: 10/01/2011 09:52:02 »
SS, I am gradually getting a better understanding of this though there are many aspects not wholly clear and much of the maths very daunting, often referring back to other derived equations. What is your opinion of what happens in the case of an orbiting, evenly distibuted line of charge? I know this is hypothetical and not stable, but nonetheless can be thought about. In one sense it should lose energy because it is an array of single charges, each of which will lose energy but, from another perspective it should just create a static magnetic field. And how does this relate to rotating spherically symmetric (charged) systems which do not radiate gravitationally or (I think??) electromagnetically.
 

Offline yor_on

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How can the signs change with an acceleration? Unruh radiation.
« Reply #45 on: 10/01/2011 23:11:39 »
Graham, you might like this one. The Rindler Horizon.
 

Offline jartza

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How can the signs change with an acceleration? Unruh radiation.
« Reply #46 on: 11/01/2011 07:52:43 »
You know why charge falling into dense and small sized massive object radiates?

Because part of electric field is hanging outside of the gravity field, so to say.

 
« Last Edit: 11/01/2011 07:56:33 by jartza »
 

Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #47 on: 11/01/2011 11:12:42 »
Yor_on, it may take me a while to read that paper. I had not even got on to black holes in this thread. Looks fun though.

I don't understand what you mean, Jartza. Are you talking about a charge falling into a black hole or any gravitating body? How is part of the electric field outside the gravity field?
 

Offline jartza

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How can the signs change with an acceleration? Unruh radiation.
« Reply #48 on: 11/01/2011 12:02:22 »
I guess I mean that tidal forces stretch the electric field.

Black holes, especially small ones, distort the electric field of vacuum, which produces Hawking radiation. Oh yes on the thread about Hawking radiation it was mentioned that tidal forces pull apart virtual particle pairs.

So I would guess there is radiation when electron's electric field gets mangled by gravity.

 
By the way:
What if we place a black hole between capacitor plates, and connect the capacitor to alternating voltage source? 

 

Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #49 on: 11/01/2011 13:55:04 »
I'm not sure about some of these issues. The observed effects will depend from where you are viewing and certainly your relative velocity and acceleration. A BH will suck in charge, as well as anything else, and the BH would end up as a charged BH. Because a non-rotating BH should not "have any hair" the charge's field lines should emanate perfectly radially and evenly distributed. I don't see why there should be any radiation except during the falling in of the charge which, from a distance, will take forever.

The capacitor plates idea is beyond my current understanding (no pun intended). I guess it is not possible to polarise the BH and I would assume then that the dielectric constant of the BH is zero - all the effective field lines from the plates would have to bend around the BH making the capacitance lower than would be the case if the BH wasn't there. I will build one in my garden and measure it :-)
 

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How can the signs change with an acceleration? Unruh radiation.
« Reply #49 on: 11/01/2011 13:55:04 »

 

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