Unruh Effect: Virtual Particles Become Real in Accelerated Frames

[pervect]

Would it be fair to describe the Unruh effect by saying that from the perspective of an accelerated observer, some virtual particles in an inertial frame become real particles in the accelerated frame?

Wald talks about how the appropriate transformations between inertial and accelerated frames (bogoliubov transformation) map positive frequencies to negative frequencies (page 414).

I think it is correct to say that the negative frequency particles would have negative energy and hence be virtual particles, leading to my description above, but I'm not terribly confident about QFT.

 

[Mentz114]

Yes, I get that impression. The full calculation is very heavy but I found this much more accessable work some time ago. The authors quantize a scalar field in 2D Minkowski spacetime and then transform to Rindler coords. They show that the vacuum expectation is different from the unaccelerated state. Re negative frequency states they say

Hence the Rindler modes cannot be a linear superposition of pure positive frequency Minkowski modes, but must also contain negative frequencies.

I've attached the paper because I don't remember where I found it ( I think it's course notes from U. Texas).

 

[MiljenkoM]

I don't think this is correct. Unruh effect is about describing same physics from different perspectives. One is inertial perspective, and other is non-inertial accelerating perspective (offten called Rindler). Main point is how you define what is particle. In inertial case you solve wave equation, and separate solutions into positive and negative frequency solution with respect to inertial time, and define vacuum and particles with respect to them (lets call them Minkowski particles and vacuum).
Now you do the same thing in Rindler coordinates (coordinates adopted to accelerating observer), solve the wave equation in Rindler coordinates, and separate solutions into positive and negative frequencies with respect to proper time (proper time is what accelerating observer is using), and define Rindler particles and vacuum with respect to these mode functions (or solutions to wave eq.).
Now you want to know how is Minkowski vacuum described using Rindler perspective, and you get that Minkowski vacuum is infect described as thermal bath of Rindler particles. This is telling you that concept of a particle is observer dependent, and unless you specify state of a detector being used to measure particles, not very useful.
I found that some author's don't distinguish Rindler from Minkowski particles in a sense that they equate thermal bath of Minkowski particles with thermal bath of Rindler particles that is seen by acceleration observer. This is not apriori true because mode functions for Rindler particles are different from mode functions for Minkowski particles, or another way of saying this is that thermal radiation in inertial motion is measurably different from thermal radiation in acceleration motion.

Wald talks about how the appropriate transformations between inertial and accelerated frames (bogoliubov transformation) map positive frequencies to negative frequencies (page 414).

Bogoliubov transformation used here map positive frequency modes with respect to inertial time to positive and negative frequency modes with respect to proper time, and vice versa. If there where no mixing between the two, Minkowski and Rindler perspective would have same vacuum state, and hence would agree on particle content of a theory.

 

[MiljenkoM]

I think it is correct to say that the negative frequency particles would have negative energy and hence be virtual particles, leading to my description above.

I don't think negative energy particles are virtual particles, but antiparticles.

 

[Mentz114]

I don't think negative energy particles are virtual particles, but antiparticles.

True. So is the affect of acceleration to break a vacuum symmetry ? If the (effective) zero point is moved, then either particles become anti-particles or vice-versa depending on whether the zero-point is raised or lowered. I have the Dirac 'sea' in mind.

In your first post you don't seem to be contradicting what Pervect and the paper I attached are saying.

 

[MiljenkoM]

In your first post you don't seem to be contradicting what Pervect and the paper I attached are saying.

Oh.. you are right, I didn't say what exactly I was disagreeing in OP. I'm not disagreeing with paper you attached.

I'm disagreeing on a way that OP is defining a particle. Positive and negative modes functions are not particles. So when you use Bogoliubov transformations you are not turning one particle into another.
Historically it was not immediately realized that relativistic particles can be described only by quantized fields. At first, fields were regarded as wave functions of point particles.

I never tough of Unruh effect in terms of Dirac sea, that is interesting observation. But I don't think it is accurate description of processes. I find Dirac see a historical curiosity, ingenious solution to problem of negative energy particles. But I don't consider this sea real and taking this sea to seriously can lead to faulty conclusions.

 

[Mentz114]

I'm disagreeing on a way that OP is defining a particle. Positive and negative modes functions are not particles. So when you use Bogoliubov transformations you are not turning one particle into another.
Historically it was not immediately realized that relativistic particles can be described only by quantized fields. At first, fields were regarded as wave functions of point particles.

I understand.

I never tough of Unruh effect in terms of Dirac sea, that is interesting observation. But I don't think it is accurate description of processes. I find Dirac sea a historical curiosity, ingenious solution to problem of negative energy particles. But I don't consider this see real and taking this sea to seriously can lead to faulty conclusions.

Yes. I'm near novice level in QFT. Thanks for the clarifications.

 

[Naty1]

Would it be fair to describe the Unruh effect by saying that from the perspective of an accelerated observer, some virtual particles in an inertial frame become real particles in the accelerated frame?

Yes.
You have just [properly] described hovering outside a black hole versus free falling toward it...that's the Unruh effect...called Hawking radiation for black holes.

This from the paper in post #2 is consistent with

John Baez's description of Hawking Radiation:

"An observer at rest has his own definition of a vacuum: it is the state in which he sees no particles. An accelerated observer also has his own vacuum, using the same definition.
We will show that these two vacuums are not the same, so that an accelerated observer actually sees particles in the inertial observer's vacuum."

http://www.obscure.org/physics-faq/Relativity/BlackHoles/hawking.html

The usual computation involves Bogoliubov transformations. The idea is that when you quantize (say) the electromagnetic field you take solutions of the classical equations (Maxwell's equations) and write them as a linear combination of positive-frequency and negative-frequency parts. Roughly speaking, one gives you particles and the other gives you antiparticles. More subtly, this splitting is implicit in the very definition of the vacuum of the quantum version of the theory! In other words, if you do the splitting one way, and I do the splitting another way, our notion of which state is the vacuum may disagree!

This should not be utterly shocking, just pretty darn shocking. The vacuum, after all, can be thought of as the state of least energy. If we are using really different co-ordinate systems, we'll have really different notions of time, hence really different notions of energy—since energy is defined in quantum theory to be the operator H such that time evolution is given by exp(-itH). So on the one hand, it's quite conceivable that we'll have different notions of positive and negative frequency solutions in classical field theory—a solution that's a linear combination of those with time dependence exp(-iωt) is called positive or negative frequency depending on the sign of ω—but of course this depends on a choice of time co-ordinate t. And on the other hand, it's quite conceivable that we'll have different notions of the lowest-energy state.

 

[Mentz114]

Quoted by Naty1 above ( can't work out where it's from )

We will show that these two vacuums are not the same, so that an accelerated observer actually sees particles in the inertial observer's vacuum.

Hmmm... can the accelerated observer ever 'see' the inertial observers vacuum ? Or am I misunderstanding ?

 

[MiljenkoM]

I'll like to point out that this view is not completely accurate. What autor, Naty1 is quoting, is describing is Unruh effect in extended Schwarzschild spacetime. The analog of Rindler vacuum state in extended Schwarzschild spacetime is known as Boulware vacuum and analog of Minkowski vacuum is known as Hartle-Hawking vacuum.

I'll paraphrase R.M.Wald from his book Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics - page 129.

"This thermal property of Hartle-Hawking vacuum state often is presented as though it were a derivation of effect of particle creation by a Schwartzschild black hole (Hawking radiation). This viewpoint is not correct."

Than he explains why is Hartle-Hawking state highly unphysical, which is somewhat technical, that's why I'm not reproducing it now.

"The difference in nature between the "Unruh effect" and the "Hawking effect" of particle creation by black holes is dramatically illustrated by considering the case of the Kerr metric. As already noted above, it has been proven that no analog of Hartle-Hawking vacuum state exists in Kerr spacetime. Thus, there is no analog of "Unruh effect" in Kerr spacetime. However, there is no corresponding difficulty with the derivation of particle creation in the case where gravitation collapse produces a Kerr black hole."

 

[MiljenkoM]

http://www.obscure.org/physics-faq/Relativity/BlackHoles/hawking.html

Let me correct my self.. Actually this link gives pretty good description of Hawking radiation.

Naty1 by saying

hovering outside a black hole versus free falling toward it

gave what author in the link calls watered-down version, and this version is usually given because it simplifies math considerably. That "watered-down" version is what I was opposing in my previous post.

 

[Naty1]

Hi MiljenkoM, I don't really understand all of your post #10...

Do you agree this is or is not accurate:

Would it be fair to describe the Unruh effect by saying that from the perspective of an accelerated observer, some virtual particles in an inertial frame become real particles in the accelerated frame?

I think its fine...its one explanation; but there could well be other intepretations.


MY limited understanding is that the Rindler horizon associated with the Unruh effect and a Schwarzschild horizon, for example, associated with a black hole are pretty much equivalent descriptions regarding the appearance of 'thermal radiation' ...
but I do not know all the math so am happy to learn more.

...can the accelerated observer ever 'see' the inertial observers vacuum ? Or am I misunderstanding ?

Not as I understand things. In each case (Unruh, black hole) acceleration introduces a horizon and with with it a different view of the energy of the vacuum state. So as in other aspects of relativity, different observers make different observations. In the Black Holes case for example, a free falling observer sees no horizon and passes that theoretical boundary without incident...and is unable to make any detection of it. An accelerating observer, however, in this case meaning a stationary observer near the horizon will be fried
by thermal radiation in an instant.

This description SUGGESTS something different between the Unruh and Black Hole situation, but other than the fact that in the Unruh effect the horizon is way distant [maybe infinity, I'm not positive] and the Black Holes horizon is right nearby, I'm unsure what else may be different.

ok, found this...Here is what Wikipedia offers:

http://en.wikipedia.org/wiki/Unruh_effect#Vacuum_interpretation

...The energy states of any quantized field are defined by the Hamiltonian, based on local conditions, including the time coordinate. According to special relativity, two observers moving relative to each other must use different time coordinates. If those observers are accelerating, there may be no shared coordinate system. Hence, the observers will see different quantum states and thus different vacua.

In some cases, the vacuum of one observer is not even in the space of quantum states of the other. In technical terms, this comes about because the two vacua lead to unitarily inequivalent representations of the quantum field canonical commutation relations. This is because two mutually accelerating observers may not be able to find a globally defined coordinate transformation relating their coordinate choices.

An accelerating observer will perceive an apparent event horizon forming (see Rindler spacetime). The existence of Unruh radiation could be linked to this apparent event horizon, putting it in the same conceptual framework as Hawking radiation# On the other hand, the theory of the Unruh effect explains that the definition of what constitutes a "particle" depends on the state of motion of the observer...

# I think what they mean is this:
Elsewhere I have read that just as an acelerating [stationary]black hole observer sees the positive energy photons appear on the observer's side of the horizon, so an accelerating Unuh observer sees a similar radiation on the observer side of the Unruh horizon.
On the other hand, as has been noted in other discussions in these forums, nothing in the math reportedly discusses the 'appearance' of virtual particles...this description was an 'intuitive' description which Hawking ,for one, used and said was not specifically related to his own mathematics.

 

[Naty1]

Mentz:
Quoted by Naty1 above ( can't work out where it's from )

We will show that these two vacuums are not the same, so that an accelerated observer actually sees particles in the inertial observer's vacuum.

This quote is from the introduction to the paper you reference in Post #2.

That statement is perfectly consistent with other sources I have read.

 

[Mentz114]

Naty1, I wasn't doubting it. I just don't know what it means. How can one see particles in another observers vacuum ?

As I understand Beckenstein, the horizon changes the entropy so the accelerated observer detects particles in their vacuum.

 

[MiljenkoM]

Hi MiljenkoM, I don't really understand all of your post #10..

Hi, Naty1... I'll try to make my points more clear in this post, and if you have any questions after, please ask.

Do you agree this is or is not accurate:

Would it be fair to describe the Unruh effect by saying that from the perspective of an accelerated observer, some virtual particles in an inertial frame become real particles in the accelerated frame?

I think its fine...its one explanation; but there could well be other intepretations.

Well, there is parse nothing particularly wrong with putting it like that, but I don't think it is very accurate description of effect. It raises a lot of questions if you put it like that, for example: why only some virtual particles, how do you decide which? Now, the main problem with this is that you are using classical concept of particles that involves poin-like objects moving along trajectories. Now we know that this concept does not actually apply on subatomic scales. As you know, one needs QFT in which basic object are not particles but quantum fields. Quantum states of the fields are interpreted in terms of corresponding particles. Experiments are than described as computing probabilities for specific field configuration and so on...
As I sad in my previous post, I got a feeling that OP was thinking that mode functions are wave functions of particles, and that Bogoliubov transformations are turning one particle into another...
Since Unruh is quantum field effect, I don't see any point in making it "particle" effect.

MY limited understanding is that the Rindler horizon associated with the Unruh effect and a Schwarzschild horizon, for example, associated with a black hole are pretty much equivalent descriptions regarding the appearance of 'thermal radiation' ...
but I do not know all the math so am happy to learn more.

Agreed, horizon is crucial in derivation of Hawking radiation. Consider for example neutron star, whose radious may be close to Schwarzschild radius but now there is no horizont, and neutron stars as far as I know do not emit Hawking radiation. But with Unruh effect horizon is only crucial for thermal spectrum of detected particles. You can set for example mind experiment with model detector that is accelerating only for finite time, than there is no horizon, all the photons will eventually catch up with detector. And you'll find out that transition probability for detector ≠0, meaning, it is detecting particles but with different spectrum than thermal spectrum [arXiv: gr-qc/0306022v2]. BTW how do you set a link in this forum?

This description SUGGESTS something different between the Unruh and Black Hole situation, but other than the fact that in the Unruh effect the horizon is way distant [maybe infinity, I'm not positive] and the Black Holes horizon is right nearby,
I'm unsure what else may be different.

Horizon for accelerating observer is 1/a away from him, where a is acceleration, but I don't think this is important.
Main difference is that Unruh effect is flat space effect, and Hawking is curved space.
Now you can draw analogy between this two affects. First you need to find analog to Minkowski vacuum state, state that is regular throughout Schwarzschild spacetime, and this state is called Hartle-Hawking vacuum. State that is analog to Rindler vacuum (vacuum defined by accelerating observer) in case of black hole is called Boulware vacuum. And now by analogy with Unruh effect you find that Hartle-Hawking vacuum is thermal state, same as you will find that Minkowski vacuum is thermal state in Unruh effect.

Now, my point in post #10 was that Unruh and Hawking effect are not infect analog, where one in flat spacetime, other in curved. And way to illustrate this is to look at Kerr black hole, where there is no Hartle-Hawking vacuum, and you can't draw analogy to Unruh effect, but nevertheless there is emission of particles from black hole (Hawking effect).

 

[MiljenkoM]

Naty1, I wasn't doubting it. I just don't know what it means. How can one see particles in another observers vacuum ?

You can't if you are by yourself :) . But let say that you have a friend named Peter, and you fly by him in accelerating rocket.

You: Dude, what do you see?
Peter: Dude, I don't see anything, this must be vacuum.
You: Strange, my thermometer is showing temperature.

 

[Mentz114]

You can't if you are by yourself :) . But let say that you have a friend named Peter, and you fly by him in accelerating rocket.

You: Dude, what do you see?
Peter: Dude, I don't see anything, this must be vacuum.
You: Strange, my thermometer is showing temperature.

But that means the particles are in my frame, not in his. Direct contradiction to

We will show that these two vacuums are not the same, so that an accelerated observer actually sees particles in the inertial observer's vacuum.

 

[MiljenkoM]

Hmm... I think you are interchanging vacuum with frame. There is no contradiction, or am I missing something :/ .

 

[Mentz114]

Hmm... I think you are interchanging vacuum with frame. There is no contradiction, or am I missing something :/ .

No, I'm not. In your scenario it is I, the accelerated observer who sees particles, thus they are in my vacuum. That contradicts this

... so that an accelerated observer actually sees particles in the inertial observer's vacuum.

(my bold)

I may not know QFT but I know English and that is a contradiction. Either the particles are in both observers vacua, or one of those statements is wrong. According to Beckenstein, the one I quote above is wrong. I don't think it is even as good as wrong - it is devoid of meaning.

 

[harrylin]

No, I'm not. In your scenario it is I, the accelerated observer who sees particles, thus they are in my vacuum. That contradicts this
(my bold)

I may not know QFT but I know English and that is a contradiction. Either the particles are in both observers vacua, or one of those statements is wrong. According to Beckenstein, the one I quote above is wrong. I don't think it is even as good as wrong - it is devoid of meaning.

I know nothing at all about this but I find it fascinating.
Also - and despite my complete lack of knowledge of this topic - I dare to expand on that line of argument: if indeed this is about field theory, then I guess that acceleration could imply radiation. And some people may refer to the prediction of possible photon detections as "particles", even if nothing material is implied. Am I on the right track?

 

[MiljenkoM]

I think I see your point Mentz114, correct me please if I misunderstood something...
So, my definition of vacuum is this: State coresponding to intuitive notion of "the absence of anything", or "empty space", without any particles, state with lowest possible energy. Now you say this:

No, I'm not. In your scenario it is I, the accelerated observer who sees particles, thus they are in my vacuum.

This makes no sense. If I see particles, I'm not in vacuum state.

I may not know QFT but I know English and that is a contradiction. Either the particles are in both observers vacua, or one of those statements is wrong.

This also makes no sense. Either you have particles or you have vacuum (no particles). You can't have particles in vacuum.
So my conclusion was that you define vacuum as space or frame or something like that. To get our terminology straight - What do you mean by vacuum? What do you mean when you say that you have particles in your vacuum?

In QFT you first define vacuum state, that is your "reference", and particle states are then build upon it. So when you say you "see" particles, you are observing exited state. So definition of your vacuum is crutial, becouse you need it to define what particle means to you.
Now, inertial and accelerating observer have different vacuum states, they define particles differently. So when you ask accelerating observer to observe quantum state that inercial observer is calling his vacuum, he reports that that state is thermal state.

Booth observers are observing same state. Inertial is calling this state vacuum, and accelerating observer is not agreeing, because they have different notion of what is vacuum, and what is particle.

 

[MiljenkoM]

I know nothing at all about this but I find it fascinating.
Also - and despite my complete lack of knowledge of this topic - I dare to expand on that line of argument: if indeed this is about field theory, then I guess that acceleration could imply radiation. And some people may refer to the prediction of possible photon detections as "particles", even if nothing material is implied. Am I on the right track?

You are on the right track. There are many examples of this apparent disagreeing in interpretations of same process. I found this paper particularly interesting - arXiv:gr-qc/9901008v1

Authors discuss weak decay of noninertial protons. Althought inertial protons are stable according to Standard model, noninertial protons are not, because of accelerating agent provides the required extra energy. It is interesting to see that what inertial observers interpret as being p^{+}\rightarrow n^{0}e^{+}\nu, accelerating observer is interpreting as beeing combination of following channels:
p^{+}e^{-}\rightarrow n^{0}\nu\:,\qquad p^{+}\bar{\nu}\rightarrow n^{0}e^{+}\:,\qquad p^{+}e^{-}\bar{\nu}\rightarrow n^{0}\:
Where election and antineutrino are absorbed from Unruh thermal bath.
But although two observers interpret same process very differently, there is no paradox, because mean lifetime is same is both frame, and they agree on the outcome of the experiment.

 

[Mentz114]

... So when you ask accelerating observer to observe quantum state that inertial observer is calling his vacuum, he reports that that state is thermal state.

Booth observers are observing same state. Inertial is calling this state vacuum, and accelerating observer is not agreeing, because they have different notion of what is vacuum, and what is particle.

Thanks for trying clear this up. But we have gone full circle. How can the accelerating observer 'see' the inertial observers vacuum state ? His thermometer is showing a higher temperature, but that tells him nothing about what the same thermometer will read in a non-accelerating state. He has to rely on his friend for that information.

I think my confusion is deep-seated and I should at least check how the authors of the paper I attached come to their ( strange ?) conclusion.

 

[MiljenkoM]

I think my confusion is deep-seated and I should at least check how the authors of the paper I attached come to their ( strange ?) conclusion.

If you have access to fantastic book by Sean Carroll - Spacetime and Geometry, you can find best explanation of Unruh effect in chapter 9, he assumes no knowledge of QFT, and derives all you need to know from QFT to understand Unruh effect step by step. I have read a lot on Unruh effect for my graduation thesis, and found this chapter very good introduction.

But we have gone full circle. How can the accelerating observer 'see' the inertial observers vacuum state ? His thermometer is showing a higher temperature, but that tells him nothing about what the same thermometer will read in a non-accelerating state. He has to rely on his friend for that information.

I think you have answered you question, you are right, you need two observers. Same as you need two observers to observe Lorentz contraction.

 

[Mentz114]

I think you have answer you question...

No, I have not. I don't think we are talking about the same question.

The problem I had is caused by this

An accelerated observer also has his own vacuum, using the same denition. We will show that these two vacuums are not the same, so that an accelerated observer actually sees particles in the inertial observer's vacuum. In other words, \vacuum" is a relative concept that depends on the observer.

They refer 'two vacuums', but as you say, there is only one, which they see differently. I think that sorts it out.

[Edit]I've now read and fully understood the paper, and I can't fault the logic.

 

[MiljenkoM]

They refer 'two vacuums', but as you say, there is only one, which they see differently. I think that sorts it out.

No, its not that there is one vacuum. There is one quantum field, inertial observer is reporting that this field is in its "ground" state, or vacuum. And accelerating observer is saying it is not a vacuum state, it is thermal state. They disagree because they have different notion of vacuum state.
Let me put this in a perspective, this is not what is actually happening:
Let's say you have quantum harmonic oscillator. You "define vacuum" as ground state of the oscillator. Exited states are interpreted as describing particles. (n-th level is interpreted as describing n particles). And let's say someone else has different "definition of vacuum state", let's say that he calls n=1 level vacuum. Than from perspective of first observer (n=0 vacuum) what second observer (n=1 vacuum) is calling his vacuum contains particles, precisely one particle.

 

[Mentz114]

Than from perspective of first observer (n=0 vacuum) what second observer (n=1 vacuum) is calling his vacuum contains particles, precisely one particle.

So there are two vacuums

I understand what is happening - my problem is only with the terminology.

This is Fock space, isn't it ? So we can do this without mentioning the word 'vacuum' by invoking the changed ground state of the oscillator(s) measured by the observers.

I think your example above is a simplification. The state space basis in the unaccelerated field is completely different from the basis in the accelerated frame. Neither basis is bounded below by 0 because the negative frequencies are included. The difference is that the accelerated definition of the ground state must include some -ve frequency modes while the unaccelerated basis does not. But they are different Fock spaces, so how can we compare them ? N=0 in one space may not be the same as n+anything in the other.

[Edit]It looks as if the accelerated Fock space is a kind of rotated Minkowski Fock space. The 'rotation' has just moved the ground state. This probably nullifies my objections above.

 

[MiljenkoM]

So there are two vacuums

I understand what is happening - my problem is only with the terminology.

This is Fock space, isn't it ? So we can do this without mentioning the word 'vacuum' by invoking the changed ground state of the oscillator(s) measured by the observers.

Yep, I figured it was question of terminology.

I think your example above is a simplification.

My example above was just for illustrative purposes.

But they are different Fock spaces, so how can we compare them ?

Hilbert space for the theory is the same in either representation, its interpretation as a Fock space is different.

 

[Naty1]

Mentz posts:

I understand what is happening - my problem is only with the terminology.

well, yes, and I agree, but the statement in the paper which you question IS in fact contradictory. It could have been stated more clearly...and such poor phrasing happens to also be evident in the quotes below.This is likely the source of my earlier claim regarding the close association of Unruh and Hawking radiation.


I thought others might find this except from Kip Thornes BLACK HOLES AND TIME WARPS, Chapter 12, of interest.

Regarding Hawking's initial calculations on black hole radiation:

...The worlds dozen top experts on the partial marriage of general relativity with quantum theory were quite sure Hawking had made a mistake. [but] Gradually one expert after another came to agree with Hawking...and in the process firmed up the partial marriage producing a new set of physical laws. The new laws are called the laws of quantum fields in curved spacetime...where the black hole is regarded as a non-quantum mechanical general relativisitic spacetime object, while the gravitational waves, electromagnetic waves, and other types of radiation are regarded as quantum fields.

Later in that chapter Thorne goes on to say:

...William Unruh...discovered (using the laws of quantum fields in curved spacetime)
that accelerated observers just above a black hole's horizon must see vacuum fluctuations there not as virtual pairs of particles but rather as an atmosphere of real particles, an atmosphere Unruh called 'acceleration radiation'. This startling discovery revealed that the concept of a real particle is relative...it depends on one's reference frame.

[One 'funny' aspect here: One cannot 'see' virtual particle pairs, and Thorne repeats this misconception in the following paragraph. Who 'edits' this stuff??]

...Observers in freely falling frames...see no real particles outside the horizon, only virtual ones. Observers in accelerated frames who, by their acceleration always remain above the horizon, see a plethora or real particles...

This sure SEEMS like a different view than presented by Wald in post #10.

 

[Naty1]

I was just reading the rest of Unruh effect in Wikipedia:


http://en.wikipedia.org/wiki/Unruh_effect#Vacuum_interpretation

..The Rindler spacetime has a horizon, and locally any non-extremal black hole horizon is Rindler. So the Rindler spacetime gives the local properties of black holes and cosmological horizons. The Unruh effect would then be the near-horizon form of the Hawking radiation.

Anyone care to offer further insights?? This sure seems to closely link Hawking and Unruh radiation.

MiljenkoN: How should I interpret the quote above with this
from earlier:
"Now, my point in post #10 was that Unruh and Hawking effect are not infect [in effect?] analog[ous], where one [is] in flat spacetime, other in curved. And [the] way to illustrate this is to look at Kerr black hole, where there is no Hartle-Hawking vacuum, and you can't draw analogy to Unruh effect, but nevertheless there is emission of particles from black hole (Hawking effect).

 

[timmdeeg]

[One 'funny' aspect here: One cannot 'see' virtual particle pairs, and Thorne repeats this misconception in the following paragraph. Who 'edits' this stuff??]

I wouldn't take this too serious. Of course one cannot "see" virtual particles, but can think of the static Casimir-effect. Whereas the dynamical Casimir-effect predicts real particles.

 

[Mentz114]

My example above was just for illustrative purposes.

Sure. It was apposite.

Hilbert space for the theory is the same in either representation, its interpretation as a Fock space is different.

Thanks.

 

[Darwin123]

Yes, this is fair. However, this explanation requires quantum mechanics.
Electromagnetic radiation can only be described as particles (i.e., photons)
if one accepts quantum mechanics. Since there is a famous discrepancy
between quantum mechanics and general relativity, I would prefer an
explanation that avoids quantum mechanics.
Therefore, I will state an explanation given in terms of stochastic
electrodynamics. This is a model where there are fundamentally no photons.
However, there is a Lorentz invariant distribution of electromagnetic waves
in the universe that provides a background called zero point radiation.
Zero point radiation is usually undetectable in our universe since the
distribution is Lorentz invariant. If two inertial observers are moving
at a constant velocity relative to each other, they see the same
distribution of electromagnetic radiation because it is Lorentz invariant.
The blue and red shifts cancel each other out.
Suppose one observer is accelerating relative to an inertial frame. He will see
some of those electromagnetic modes red or blue shifted in such a way that
the distribution follows a black body distribution instead of a Lorent invariant
distribution.
Thus, one doesn't have to visualize particles at all. This is a quasiclassical
way of looking at it. Basically, one is using special relativity. One isn't using
quantum mechanics. One is also using only a small part of general relativity.

 

[yuiop]

What if an Unruh particle observed by the accelerating observer collides with a real particle that is visible to both the accelerated observer and to an inertial observer. Does this cause a paradox?

Does the inertial observer see the real particle suddenly accelerate for no apparent reason because he does not see the Unruh particle that collided with it?

Does the accelerated observer see the Unruh particle collide with the real particle (that they both see) and have no effect upon it?

Does the accelerated observer see the real particle that they both see accelerate as a result of the collision with the Unruh particle, while the inertial observer does not see any change of momentum of the real particle (that they both see) as a result of the collision (that only the accelerated observer sees)? Does the real particle involved in the collision with the Unruh particle have two different realities?

Does the universe arrange things so that "real" Unruh particles are never on a collision course with really real particles to avoid such paradoxical situations?

 

[PeterDonis]

What if an Unruh particle observed by the accelerating observer collides with a real particle that is visible to both the accelerated observer and to an inertial observer.

In Wald's book Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics, he discusses a similar question; he analyzes the case of a particle detector on an accelerating worldline, where the quantum field is in a state that, according to an inertial observer, is vacuum. There is a nonzero probability of a state transition, which is described as follows (I think I've got this right--I don't have my copy of the book handy to check):

* From the viewpoint of an observer who is accelerated with the detector, the detector absorbs a particle and registers a corresponding increase in energy, and changes its motion slightly as a result of the absorbed particle's momentum.

* From the viewpoint of an inertial observer, the detector *emits* a particle, and registers a corresponding *decrease* in energy and a change in momentum due to "radiation reaction".

The underlying quantum field viewpoint is that there is a state transition from the "inertial" vacuum state to the "accelerated" vacuum state (which the inertial observer does *not* see as a vacuum state), and a corresponding state transition in the detector, as required by the appropriate conservation laws. It all hangs together, but I admit it is certainly not intuitive.

 

[yuiop]

* From the viewpoint of an observer who is accelerated with the detector, the detector absorbs a particle and registers a corresponding increase in energy, and changes its motion slightly as a result of the absorbed particle's momentum.

* From the viewpoint of an inertial observer, the detector *emits* a particle, and registers a corresponding *decrease* in energy and a change in momentum due to "radiation reaction".

OK, this sort of makes sense. It suggests to me that the Unruh particles only interact with accelerating objects and the accelerating observer only observes the Unruh particle from the effects on his accelerating detectors and does not see them in transition.

Next question. As I understand it, the Unruh particles are only observed coming from behind the accelerating observer from the location of the Rindler horizon. Any intervening cold wall between the Rindler horizon and the accelerating observer casts a shadow that seems to block the Unruh radiation. Now let's say that we have one accelerating object that has an intervening wall between it and its Rindler horizon and another accelerating object that does not have an intervening wall between it and its Rindler horizon. Would the inertial observer now see a radiation reaction and particles being emitted from the accelerating observer without an intervening wall and no radiation reaction or particles emitted from the accelerating object with no intervening wall behind it? Is not radiation reaction an intrinsic part of accelerating particles and shouldn't it be independent of any wall behind the accelerating object?

 

[stevendaryl]

I don't think negative energy particles are virtual particles, but antiparticles.

That's not exactly correct. A particle's charge doesn't change with energy, so a negative energy electron is still an electron, and not a positron. In the old-fashioned "Dirac Sea" interpretation of quantum field theory, it was postulated that all the negative-energy electron states were filled. An energetic photon could boost a negative energy electron to make it positive energy. This would produce a new, positive-energy negatively-charged particle (an ordinary electron) and also would leave a "hole" in the negative-energy states, which means a net positive charge among negative energy states. This hole could be reinterpreted as a positron.

Anyway, according to this (as I said, old-fashioned) way of looking at things, a positron is not a negative energy electron, it is the absence of a negative energy electron.

By the way, this way of viewing a hole in an otherwise filled shell of electron states as a positively-charged particle is no longer used in quantum field theory, but is still commonly used in solid-state physics, especially semiconductors.

 

[Demystifier]

I don't think negative energy particles are virtual particles, but antiparticles.

But in the case of real (hermitian) field, such as electromagnetic field, antiparticles do not exist. And yet, Unruh effect exists for such fields as well. Thus, you are right that negative-energy particles are not virtual, but they are not antiparticles either. Negative-energy particles simple do not exist in the physical Hilbert space, essentially because such states would have negative norm.

In fact, the Rindler vacuum is a superposition of various Minkowski-particle states all of which have a POSITIVE energy. Technically, this superposition of positive energy states is known as a squeezed state.

Minkowski particles and Rindler particles are states in the same Hilbert space. Two definitions of particles correspond to two definitions of particle-number operator which do not commute. The fact that the same state |0> may be either vacuum or not-vacuum with two different definitions of the particle-number operator is analogous to the more familiar fact that the same spin-1/2 state |down> may be either down or not-down state with two different definitions of the z-axis. More physically, just like rotation of the Stern-Gerlach apparatus (which measures spin) may turn down-state into a not-down state, acceleration of a particle detector may turn vacuum state into a not-vacuum state.

 

[PeterDonis]

OK, this sort of makes sense. It suggests to me that the Unruh particles only interact with accelerating objects and the accelerating observer only observes the Unruh particle from the effects on his accelerating detectors and does not see them in transition.

This is true of any kind of particle; we only observe them from their effects on detectors, we don't observe them "in transition". The "Unruh particles" are bona fide particles just like any others. As Demystifier said, the fundamental physics here is that the "inertial" and "accelerated" particle number operators (physically, the particle number operator is basically what a "particle detector" realizes) do not commute; they have different sets of eigenstates, so a state that is an eigenstate of the inertial operator (such as the vacuum state, with eigenvalue zero, so there is zero probability of detecting a particle) is *not* an eigenstate of the accelerated operator (so there is a nonzero probability for it to detect a particle).

Next question. As I understand it, the Unruh particles are only observed coming from behind the accelerating observer from the location of the Rindler horizon.

I'm not sure this is true; can you give a reference that leads you to this conclusion? The "cold wall" scenario you're describing doesn't match up with what I know about how the Unruh effect works (which isn't a lot).

 

[yuiop]

I'm not sure this is true; can you give a reference that leads you to this conclusion? The "cold wall" scenario you're describing doesn't match up with what I know about how the Unruh effect works (which isn't a lot).

I read about this a long time ago but I cannot recall where. A quick Google turned up this mention in a Wikipedia article http://en.wikipedia.org/wiki/Unruh_effect#Calculations

The Unruh effect could only be seen when the Rindler horizon is visible. If a refrigerated accelerating wall is placed between the particle and the horizon, at fixed Rindler coordinate , the thermal boundary condition for the field theory at is the temperature of the wall. By making the positive side of the wall colder, the extension of the wall's state to is also cold. In particular, there is no thermal radiation from the acceleration of the surface of the Earth, nor for a detector accelerating in a circle[citation needed], because under these circumstances there is no Rindler horizon in the field of view.

 

[PeterDonis]

A quick Google turned up this mention in a Wikipedia article http://en.wikipedia.org/wiki/Unruh_effect#Calculations

Ah, I see; basically the idea is that the Unruh effect only occurs if the Rindler horizon is "visible", as it would be for linear acceleration in free space, but is not for the two counterexamples they give (but note the "citation needed"):

* For acceleration in a circular path, there is no Rindler horizon (I think that's right) and no Unruh effect (I think that's also right but I'm not sure).

* For the "acceleration" of objects at rest on the Earth's surface, the Rindler horizon is not "visible" because the spacetime is curved--the Earth is only 4000 miles in radius but the Rindler horizon would be a light-year away for a 1 g acceleration. So there is no Unruh effect here either (again, I think that's right for the Earth--but note that for a black hole, there is Hawking radiation, the mathematics of which is very similar to that of the Unruh effect).

 

[Physics Monkey]

It seems physically reasonable to me that uniform circular motion will be accompanied by something like the Unruh effect, although the details are more complicated.

For example, Bell and Leinaas have a series of papers, 25+ years old now, where they try to interpret the known partial spin polarization of electrons orbiting in a storage ring as experimental evidence in favor the unruh effect. The original paper is I think http://www.sciencedirect.com/science/article/pii/0550321387900472 while an accesible discussion is at http://arxiv.org/pdf/hep-th/0101054.pdf

Another discussion of radiation in the circulating case can be found at http://arxiv.org/abs/gr-qc/9903054

I'm not sure about this "cold wall" situation, mostly because I'm not sure what that even means, but I doubt that such a thing could seriously interfere with unruh radiation.

 

[Physics Monkey]

In Wald's book Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics, he discusses a similar question; he analyzes the case of a particle detector on an accelerating worldline, where the quantum field is in a state that, according to an inertial observer, is vacuum. There is a nonzero probability of a state transition, which is described as follows (I think I've got this right--I don't have my copy of the book handy to check):

* From the viewpoint of an observer who is accelerated with the detector, the detector absorbs a particle and registers a corresponding increase in energy, and changes its motion slightly as a result of the absorbed particle's momentum.

* From the viewpoint of an inertial observer, the detector *emits* a particle, and registers a corresponding *decrease* in energy and a change in momentum due to "radiation reaction".

The underlying quantum field viewpoint is that there is a state transition from the "inertial" vacuum state to the "accelerated" vacuum state (which the inertial observer does *not* see as a vacuum state), and a corresponding state transition in the detector, as required by the appropriate conservation laws. It all hangs together, but I admit it is certainly not intuitive.

There is a simple and neat explanation for part of this story, namely that both observers must observe an increase in the Rindler energy. The Minkowski observer sees a particle emitted which can reasonably be expected to increase the energy. On the other hand, the Rindler observer sees a particle absorbed which sounds contradictory. However, one may think of the observation of the particle at the detector by the rindler observer as a partial measurement of the state of the field. In particular, more highly excited states are more likely to lead to detection and thus, conditioned on a detection, they become more likely. Thus the average energy can actually increase as a result of a particle being absorbed. As a simple extreme example, think of a state of the radiation like \rho = \frac{N-1}{N} |0 \rangle \langle 0 | + \frac{1}{N} |N \rangle \langle N | labeled by energy quanta. If a quanta is absorbed then the field had to be in the N state because you can't absorb from 0. The initial average field energy is 1 but the final average field energy is N-1 (because the state had to be N and you lost one unit of field energy).

This is discussed in a paper by Wald and Unruh although I'm forgetting the reference right now.

 

[Physics Monkey]

That was easy, found it: http://prd.aps.org/abstract/PRD/v29/i6/p1047_1

 

[PeterDonis]

That was easy, found it: http://prd.aps.org/abstract/PRD/v29/i6/p1047_1

Ah, good, I was remembering correctly:

"It is shown in detail for the simple case of a two-level detector how absorption of a Rindler particle corresponds to emission of a Minkowski particle."

 

[yuiop]

"It is shown in detail for the simple case of a two-level detector how absorption of a Rindler particle corresponds to emission of a Minkowski particle."

OK, that seems reasonable as in both instances, the velocity of the accelerating Rindler observer is effectively increased and momentum is presumably conserved. I also assume that while the Minkowski observer does not see the Rindler particle, that the Rindler observer does not see the emitted Minkowski particle. Now I have to ask, what if the emitted Minkowski particle (that the Rindler observer does not see) collides with a pre-existing normal particle (that both observer's see), how does the Rindler observer explain the sudden change in momentum of the normal particle after the collision?

P.S. Yet another question. By the equivalence principle, the Unruh radiation effect implies that an observer that is stationary observer outside a black hole sees Hawking radiation and observes the black hole evaporate (possibly to nothing in a mini explosion) while a free falling observer would not be able to see the Hawking radiation and would either see no change in the mass of the black hole or would be unable to explain the loss of mass of the black hole. Which position is more accurate?

 

[PeterDonis]

OK, that seems reasonable as in both instances, the velocity of the accelerating Rindler observer is effectively increased and momentum is presumably conserved. I also assume that while the Minkowski observer does not see the Rindler particle, that the Rindler observer does not see the emitted Minkowski particle.

All correct as far as I can see.

Now I have to ask, what if the emitted Minkowski particle (that the Rindler observer does not see) collides with a pre-existing normal particle (that both observer's see), how does the Rindler observer explain the sudden change in momentum of the normal particle after the collision?

If there is a "pre-existing normal particle that both observers see", then the overall state of the quantum field is not a vacuum state in either sense--it's neither the Rindler vacuum nor the Minkowski vacuum. I'm not sure if there is an analysis out there of how this affects things; all the analyses I have seen of the Unruh effect assume that the field state transition is from one vacuum to the other (meaning no other "pre-existing" particles).

Edit: To speculate a bit, I would guess that in the case where a "pre-existing" particle is present, it won't look the same to the inertial observer as it does to the accelerated observer, for reasons similar to why the vacuum state for one is not the vacuum state for the other. Also, I would expect that, once the emitted Minkowski particle collides with the pre-existing particle, the Minkowski particle will no longer be in the state that the Rindler observer thinks is "vacuum"; i.e., it will now be visible to that observer. So I would guess that the Rindler observer's interpretation of events would be that his detector absorbs a particle (so the field is now in the vacuum state); then the pre-existing particle causes a particle-antiparticle pair to pop out of the vacuum (what the Rindler observer "thinks" is the vacuum), one of the pair flies off, while the other is absorbed by the pre-existing particle and changes its state. That's just a guess, though; as I said, I haven't seen any mathematical treatment of this case.

 

[yuiop]

If there is a "pre-existing normal particle that both observers see", then the overall state of the quantum field is not a vacuum state in either sense--it's neither the Rindler vacuum nor the Minkowski vacuum. I'm not sure if there is an analysis out there of how this affects things; all the analyses I have seen of the Unruh effect assume that the field state transition is from one vacuum to the other (meaning no other "pre-existing" particles)...

I think perhaps you are applying a too rigid interpretation of "vacuum" here. For example Physics Monkey mentions a study of the Unruh effect in an electron storage ring. Presumably the experiment was done in an evacuated chamber approximating a vacuum, but it is well known that it is impossible to obtain a true vacuum in a laboratory because at extreme low pressure the walls of the chamber starts to evaporate and there will always be stray particles present in an experimental chamber. Another example is the Schwarzschild metric which is a vacuum solution. This does not prevent us assuming test particles and observers (with none zero rest mass) when carrying out an analysis as long as masses are insignificant compared to the mass of the gravitational body and the average vacuum density reasonably approximates a true vacuum.

 

[PeterDonis]

I think perhaps you are applying a too rigid interpretation of "vacuum" here.

When I say "vacuum" I'm talking about the quantum state used in the theoretical treatment. Every theoretical treatment I have seen uses only two states, both of which are vacuum states: the Minkowski vacuum state and the Rindler vacuum state. The notion of "vacuum state" has a very precise meaning theoretically, and that's what I was talking about. I wasn't intending to say that an actual experiment would have to be done in a perfect vacuum. I was only saying that, theoretically, if there is a "pre-existing particle", then the quantum state of the field is *neither* of the states that I have seen used in theoretical treatments of the Unruh effect.

 

[Demystifier]

Every theoretical treatment I have seen uses only two states, both of which are vacuum states: the Minkowski vacuum state and the Rindler vacuum state.

There are also more general theoretical notions of the "vacuum" and "particle":
http://xxx.lanl.gov/abs/gr-qc/0409054