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Author Topic: How can the signs change with an acceleration? Unruh radiation.  (Read 15103 times)

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?


 

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 »
 

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 »
 

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.
 

Offline yor_on

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Thanks Graham :)

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

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 :)
 

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 :-(
 

Offline Ron Hughes

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

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?
 

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?
 

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.
 

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 !!
 

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 »
 

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
 

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?

 

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 :-)
 

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.
 

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 »
 

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.
 

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.

 

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
 

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 »
 

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?
 

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" :-)
 

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