How can the signs change with an acceleration? Unruh radiation.

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Offline yor_on

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Anyone that know how to see this one?

In a uniform motion two observers in relative motion will see different frequencies for the same source, but they can not change the sign if you go by Lorentz transformations, meaning that particles anti-particles are recognizably 'the same' for both observers. That is not true in a acceleration though, there, if i got it right, the sign can be interchangeable and you will observe radiation that the rocket 'coasting' won't. And that you can't see it 'coasting' has nothing to do with the velocity/speed you have uniformly moving.

And that's the Unruh radiation. I've seen it explained as your engine becoming your 'detector', but I have difficulties turning my head around this 'sign switch' possible here?

To me that seems like a reversal of 'flow'?
Should I look at it as a reversible effect of time?
How do the signs know when to change.
And why.

What do you think?
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Offline yor_on

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You really should try to answer this one, if you can :)

People love to make statements that are not correct it seems, and we all do it from time to to time. One of the worst is to assign plus and minus to a so called 'gravitational acceleration' like throwing a ball in the air. Worthless and superfluous to me, and only confounding the reader (me:).

But what I'm wondering of here is not that. As far as I understand the 'sign switch' is expected to be 'real' and a part of the explanation to why we see a Unruh radiation. so how is it possible? What would we see.
==

Or is it the same?
I've seen some putting Hawking radiation and Unruh radiation as one and the same?
And then, if so, I will ignore the statement of reversing signs as that seems to be a mathematical joke with no relation to the factual effect.
==

And then we should have an extreme pair production in a Unruh radiation which if the idea was right should cancel itself out as the particles and 'anti particles' react with each other.
But there are some weird effects to it.

1. We shouldn't see it. Not if it cancels itself out, if it doesn't though you can expect radiation to escape, hitting your retina.
2. it will leave a positive 'rest product' whatever that is thought to be. And so, once more, we meet that scarlet pimpernel 'Energy'.

Hmm?

To see the inherent weirdness of 2. You need to remember that Unruh radiation doesn't exist for our uniformly moving observer. So he won't observe any such 'effect'. But what about that 'positive rest product' then?
« Last Edit: 03/01/2011 15:36:54 by yor_on »
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Offline yor_on

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There is a clear difference between Hawking radiation and Unruh radiation.

In a Hawking radiation you will have the so called 'negative' particle-half inside a event horizon negating the 'positive' particles, and even if leaving a positive rest product not being inside our SpaceTime. That will also slow down the Black hole relative our universe and so have an uneven equivalence.
==

I forgot to mention that here you will see a radiative effect, with all right too as we will have a surplus.
But in a Unruh radiation I do not expect you too, that is, if it's expected to be the same?
==

In a Unruh radiation it all happens in 'SpaceTime'. and if you want to assume that we all share the same, then the rest product will 'exist' for us all. Even though the radiation didn't :)
« Last Edit: 03/01/2011 15:58:50 by yor_on »
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Offline graham.d

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Yor-on, this is related to the discussion we had some time ago about radiation of accelerated charges. This paper may help:

http://www.physics.princeton.edu/~mcdonald/accel/unruhrad.pdf

I don't find much of this easy and I don't think all the theory has been resolved.

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Offline yor_on

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Thanks Graham :)

I will read it and see what I can understand ::))
Not much huh :)
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Offline yor_on

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You know, I don't doubt the effect in itself. It should be there as I see it, it's the explanations I doubt :)
==
Also, depending on what we expect that 'rest product' to be we should possibly be able to test the prediction at CERN, if we can measure that 'positive rest product'?

It should be a radiation too, shouldn't it. I don't know? 'Energy' huh :)
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Offline graham.d

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I think the conclusion is that firstly, a uniformly accelerated charge radiates, as outlined in the paper by Schott et al. So if we assume a sea of virtual charged particles, then we can assume that by accelerating through them there will be radiation from them but that at low values for the acceleration this radiation is negligibly small.  I think, and this is where I am rather unsure, if the acceleration is large enough then there will develop an asymmetry in the radiation similar to that caused by the gravitational event horizon so that there will then be a significant net radiation of photons as observed by the accelerating body. I will have to follow references to get a better understanding. Regrettably I have to go to work tomorrow after the seasonal break so time is at a premium :-(

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Offline Ron Hughes

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Virtual particles, there in lies the problem. I seriously doubt the existence of Hawking or Unruh radiation.
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Offline yor_on

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I like it Graham, I find your idea perfectly understandable, as well as simpler :)
Which either mean that it's a 'superior one', or that we both are bicycling here.

I think I will open a new thread soon, debating whether gravity is a 'force' or not.
You can look at it this way, Earth is accelerating at one gravity as this is the equivalence to an uniform acceleration. Now, the really interesting thing here, to me.

If it is true, and i mean true true here :)
Where is it accelerating.

Normally we use a vector right? Defining it as a motion. But in this case we have a invariant object, staying inside a same coordinate system as defined by the solar system. But if you believe it to be an exact equivalence?

I don't know why, but that one been on my mind for some days?
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Offline yor_on

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Yeah Ron, that might be a problem. Did you see that definition of (orange) virtual light "The vacuum is defined to have zero energy. Therefore yes, the virtual particles violate the law of energy preservation. But the particles disappear again directly after (e. g. eighty attoseconds for a photon of orange light) and give the energy back to the vacuum."

?

I really would like to know how you, they I mean, define that?
As we can measure down to twelve?
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Offline Ron Hughes

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How can the signs change with an acceleration? Unruh radiation.
« Reply #10 on: 04/01/2011 15:30:15 »
The electron and proton have an electric field. That field has been propagating away at c since their creation. Therefore space can never be energyless as long as matter exists. Virtual photons, electrons or protons can only pop into existence if there is a disturbance of sufficient energy in that field to create them. They can and do occur somewhat like giant waves in the ocean where small waves can add up to a giant. To say that space is a mush of virtual particles seems to me an excuse to justify some theory.
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Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #11 on: 04/01/2011 17:01:47 »
Nonetheless Ron, it is the most commonly held view that there is a sea of virtual particles. The theory seems to fit with measurements of the Casimir effect, although there could be other explanations for this. The theory is consistent with Heisenberg's uncertainty priciple which should be valid in a vacuum as much as anywhere else. I am not saying that you are wrong, but for the sake of other readers, saying that your view (as I'm sure you will agree) is not the conventional one.

There is a long way to go to get this concept to agree with cosmological theories of vacuum (dark) energy though - like an error of a factor of 10^120 !!

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Offline yor_on

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How can the signs change with an acceleration? Unruh radiation.
« Reply #12 on: 04/01/2011 18:07:11 »
I'm also a believer :)
But I get a big headache reading people 'timing them'.
That's a outrageous statement, where we suddenly have virtual photons existing longer than what we call 'real photons' :)

Let us assume that what steer SpaceTime is the 'arrow of time' a 'device' pointing in one direction. Let us also assume that real photons are of a shorter duration than 'virtual' :) Anyone more than me that gets a headache?

Maybe we could assume that 'virtual photons' then goes backward in time :)
Then the duration for orange light is eighty negative attoseconds ::))
Kind of brilliant statement isn't it :) Entropy and all that..

But if you make that assumption you are actually declaring that you can measure those seconds macroscopically in 'reversed (ah, positive) time'. That's your own personal frame of reference you did that measurement in? And now I'm flabbergasted :) If you did..
==

And the only way of making sense of that is to assume that radiation 'doesn't move'. Then you can have such possibilities open possibly as we now are talking about something 'unmoving' with the only thing creating motion being the 'changes' we observe. Such a universe will in fact be 'magic' to us and if it was, then we would need see why it isn't?

As I said, I'm weird :)
« Last Edit: 04/01/2011 18:12:53 by yor_on »
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Offline yor_on

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How can the signs change with an acceleration? Unruh radiation.
« Reply #13 on: 04/01/2011 19:22:02 »
Here's an alternative. To Unruh radiation and Rindler observers.

"The Unruh effect tells us that an accelerated observer will detect particles in the Minkowski vacuum state. An inertial observer, of course, would describe the same state as being completely empty; indeed, the expectation value of the energy-momentum tensor would be ** . But if there is no energy-momentum, how can the Rindler observers detect particles? This is a subtle issue, but by no means a contradiction. If the Rindler observer is to detect background particles, she must carry a detector - some sort of apparatus coupled to the particles being detected. But if a detector is being maintained at constant acceleration, energy is not conserved; we need to do work constantly on the detector to keep it accelerating. From the point of view of the Minkowski observer, the Rindler detector emits as well as absorbs particles; once the coupling is introduced, the possibility of emission is unavoidable. When the detector registers a particle, the inertial observer would say that it had emitted a particle and felt a radiation-reaction force in response. Ultimately, then, the energy needed to excite the Rindler detector does not come from the background energy-momentum tensor, but from the energy we put into the detector to keep it accelerating."

That was the 'detector idea' I referred to earlier
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Offline yor_on

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How can the signs change with an acceleration? Unruh radiation.
« Reply #14 on: 05/01/2011 02:47:17 »
There is one thing to ponder here, and that depending on how far one should draw an equivalence? A uniform constant acceleration is equivalent to a gravity right?

"But if a detector is being maintained at constant acceleration, energy is not conserved; we need to do work constantly on the detector to keep it accelerating."

Remember me wondering about what I should see Earths 'acceleration' as above. I think I just realized why I was thinking of that one :) A black hole would the have a Hawking radiation, much for the same reasons as Unruh radiation, but as earths invariant mass is to small for that 'acceleration' we won't see it.

Maybe :)
==

So what happen with non-uniform accelerations then?

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Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #15 on: 05/01/2011 10:25:49 »
"So what happen with non-uniform accelerations then?"

This all gives me a headache too. It is definitely a related topic to the subject of whether a uniformly accelerated charge radiates; we had a lengthy debate about this on this forum a while back. I found a few papers discussing Schott energy which seemed to solve the paradox of whether a uniformly accelerated charge radiates, but at the expense of having to accept that a charge stores Schott energy (seemingly without limit) in its near-field only (if I remember correctly). The paradox comes about from examining what happens to a free-falling charge in a gravity field which, to a observer falling in the same field, should behave the same (by the equivalence principle) to a similar observer with a nearby charge in free space. The question being that if it's radiating it should be losing energy but the charge (by equivalence) must be acclerating the same as the uncharged observer - so where is the energy coming from? I think this was solved by Schott though I can't say I fully understood it all and it seemed quite counter-intuitive, but then lots of this stuff is that way.

All that you have to do is to tie these concepts together with the maths :-)

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Offline Soul Surfer

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How can the signs change with an acceleration? Unruh radiation.
« Reply #16 on: 05/01/2011 10:39:09 »
A uniformly accelerated charge radiates and this radiation is used a great deal as a powerful source .  Remember anything moving in a circular orbit is accelerating continuously towards the centre and synchrotrons to this with magnetic fields requiring a continuous input of energy to keep the particles in orbit.  This sort of radiation is usually called "synchrotron radiation"  QV.  Most of the radiation appears in a narrow beam in the direction the particle is travelling.
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Offline jartza

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How can the signs change with an acceleration? Unruh radiation.
« Reply #17 on: 05/01/2011 11:11:40 »
When there is an accelerating electric field and an electron floating in that electric field, there is no reason for the electron to radiate. It is staying in the same place in the field, you see.

Only when the electric field jerks, does radiation happen.
« Last Edit: 05/01/2011 11:23:36 by jartza »

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Offline Soul Surfer

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How can the signs change with an acceleration? Unruh radiation.
« Reply #18 on: 05/01/2011 11:38:49 »
What are you trying to say jartza?  Is it that synchrotron radiation sources don't work?   When they plainly do because they are in use every day! or 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 in just the same way that stopping high energy electrons with a target produces x-rays.
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Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #19 on: 05/01/2011 11:40:12 »
SS, this is a lot more complicated than you think and it is still a debated subject. Jartza may be right, and was also proposed by Feynman, that it is the change in acceleration in a charge that produces radiation. I am not sure (and have not the time right now) how to refer back to our previous debate on this topic - maybe Yor_on can do it. An orbiting charge is indeed continually accelerating inwardly but if you imagine a ring of such charges orbiting a gravitating body then you will simply establish a static magnetic field. There is no loss of energy once this field is established. You should look up papers on Schott energy.

The issue is similar when comparing a free falling charge. Where is the radiating energy coming from? If you, as an observer, are falling with the charge does the charge fall at a slower rate than you because it loses kinetic energy. But you would not see it radiating because, as a comoving observer in freefall and by the principle of equivalence, neither of you are accelerating with respect to each other.


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Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #20 on: 05/01/2011 11:50:26 »
IT may be worth looking at this site to see that solutions to this question are not obvious:

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

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Offline jartza

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How can the signs change with an acceleration? Unruh radiation.
« Reply #21 on: 05/01/2011 12:12:17 »
What are you trying to say jartza?  Is it that synchrotron radiation sources don't work?   When they plainly do because they are in use every day! or 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 in just the same way that stopping high energy electrons with a target produces x-rays.


Surprisingly circular motion is not uniformly accelerating motion [:)]

There is a constant jerk. (change of acceleration)
« Last Edit: 05/01/2011 12:39:25 by jartza »

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Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #22 on: 05/01/2011 12:46:39 »
How so, Jartza?

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Offline graham.d

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How can the signs change with an acceleration? Unruh radiation.
« Reply #23 on: 05/01/2011 14:25:40 »
The previous thread on this subject is:

http://www.thenakedscientists.com/forum/index.php?topic=28617.0

I worked out how to use "search" :-)

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Offline imatfaal

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How can the signs change with an acceleration? Unruh radiation.
« Reply #24 on: 05/01/2011 14:46:20 »
I guess Jartza is stating that the third derivative of the position is not zero.  If we assume that the position is given by an equation similar to x=k.e^iwt or x=coswt - isinwt then it is clear that no time derivative will be a constant nor zero.  

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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 »
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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 »

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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.

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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 »
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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 »
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Offline Ron Hughes

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« Reply #30 on: 05/01/2011 19:44:41 »
a. would not see radiation
b. would
c. would
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Offline yor_on

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« 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 »
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Offline jartza

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« 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

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« Reply #33 on: 06/01/2011 04:43:18 »
Interesting Jartza.
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Offline jartza

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« 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.

 

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Offline graham.d

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« 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!!

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Offline Ron Hughes

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« 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.
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Offline graham.d

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« 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. 

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Offline Ron Hughes

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« 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,
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Offline yor_on

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« 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 »
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Offline graham.d

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« 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 »

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Offline jartza

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« 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.

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Offline graham.d

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« 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.

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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.
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« 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.

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Offline yor_on

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« Reply #45 on: 10/01/2011 23:11:39 »
Graham, you might like this one. The Rindler Horizon.
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Offline jartza

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« 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 »

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Offline graham.d

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« 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?

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Offline jartza

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« 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? 


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Offline graham.d

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« 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 :-)