Explaining Bohmian Mechanics: Wavefunction, Pilot Wave & Particles

[Dmitry67]

The interaction between the (Minkowski) vacuum and the accelerated measuring apparatus (made up of Minkowski particles) creates new Minkowski particles in very special (actually, squeezed) states.

is it compatible with

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 ρ0, the thermal boundary condition for the field theory at ρ0 is the temperature of the wall. By making the positive ρ side of the wall colder, the extension of the wall's state to ρ > ρ0 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

 

[Demystifier]

Lets talk about the predicting power.
If you dint know about the Unruh effect, based on the BM you would probably say: the number of REAL particles is an objective reality, it is the same in all inertial and all accelerating frames. Hence, if vacuum does not contain real particles then there is neither Unruh effect nor Hawking radiation. BM can be equivalent to QM in inertial frames but lead to different results (so it is testable) when we include gravity. Also,as I understand, BM should deny the existence of gravitons and their existence is frame-dependent.

You misunderstood something about BM.
Anyway, irrespective of BM, see this:
http://xxx.lanl.gov/abs/0904.3412

 

[Ilja]

Lets talk about the predicting power.
If you dint know about the Unruh effect, based on the BM you would probably say: the number of REAL particles is an objective reality, it is the same in all inertial and all accelerating frames. Hence, if vacuum does not contain real particles then there is neither Unruh effect nor Hawking radiation. BM can be equivalent to QM in inertial frames but lead to different results (so it is testable) when we include gravity. Also,as I understand, BM should deny the existence of gravitons and their existence is frame-dependent.

Particle approaches to QFT use and have to use variable particle numbers.

Then, pilot wave theory has a preferred frame. Thus, there will be different theories related with different preferred frames. But the empirical predictions will show the same observable symmetries. At least I see absolutely no reason to expect that the equivalence proof does not work in QFT or quantum gravity.

If some version of BM will deny the existence of something observable, like gravitons, it is reasonable to expect that it will deny their existence in a similar way as it "denies" the existence of spin in Bell's version of particle theory with spin. (A theory I don't like, but it is useful to illustrate that one does not need much, one can even deny the existence of spin, but nonetheless prove the empirical equivalence to QM.)

 

[Demystifier]

More seriously, I would have preferred a published paper, but I will certainly take a look at your reference.

If someone is interested, now the final version accepted for publication in Int. J. Mod. Phys. A is available:
http://xxx.lanl.gov/abs/0904.2287

 

[Mentz114]

Thanks. I look forward to reading it.

 

[Dmitry67]

From another thread:

The point is that the number of created particles does NOT depend on the observer. Instead, it depends on whether a particle detector exists and on how does it move. In the case of accelerated particle detector, the created particles exist for any observer, not only for the observer comoving with the detector.
Does it make sense to you? Or will you ask me the same question again?

Wait, I don't understand.

In Unruh effect particles comes not from the detector, but from apparent cosmological horizons.

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

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 ρ0, the thermal boundary condition for the field theory at ρ0 is the temperature of the wall. By making the positive ρ side of the wall colder, the extension of the wall's state to ρ > ρ0 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.

How your theory can explain that?

 

[Demystifier]

As I said, there are two inequivalent versions of the Unruh effect. The one you mention is not measurable and the Bohmian theory cannot explain it. The Bohmian theory can only explain the other, measurable, version of the Unruh effect.

Let me briefly explain the two versions of the Unruh effect. The first version is defined with respect to various choice of spacetime coordinates and does not rest on properties of the measuring apparatus. It is based on Rindler quantization. In the second version there is only one definition of particles, which are standard Minkowski particles, while the Unruh effect manifests as clicks in an accelerated particle detectors. In the case of uniform acceleration, both approaches predict the thermal distribution of particles with the same temperature. Yet, the two approaches are not completely equivalent.

And of course, don't trust everything you find on wikipedia.

By the way, if you accept that the existence of many worlds in MWI only makes sense in the context of decoherence, then MWI also cannot explain the first version of the Unruh effect. The first version of the Unruh effect only makes sense in some variants of the Copenhagen interpretation (and in the Tegmark mathematical-universe interpretation, but this is much more than MWI because it asserts that any mathematical structure exists). As you probably know, decoherence makes sense only if you have some objective environment (measuring apparatus), so a vague notion of an "observer" not described by wave functions cannot explain decoherence, which is why decoherence-based MWI cannot explain Unruh effect based on such a vague notion of "observers".

 

[Dmitry67]

1 Why it is not measurable? Say, if I accelerate a ball, its front (and rear?) side will melt from the Unruh effect (theoretically). If you accelerate a ball in the box, then box melts but the ball is unaffected, correct?

2 What is a problem with MWI and the first type of Unruh effect? Why do we need a measurement (CI)?

 

[Demystifier]

1 Why it is not measurable? Say, if I accelerate a ball, its front (and rear?) side will melt from the Unruh effect (theoretically). If you accelerate a ball in the box, then box melts but the ball is unaffected, correct?

2 What is a problem with MWI and the first type of Unruh effect? Why do we need a measurement (CI)?

1. Well, nothing is measurable without a measuring apparatus. If you include a QUANTUM description of the measuring apparatus in your theory, then fine. But nobody has done it yet for the first version.

2. I assume that MWI needs decoherence. (Do you agree?) On the other hand, the first version does not include it.

Even worst, the first version does NOT even assert that the total state is a superposition of a state without particles and a state with particles. Therefore, MWI is not applicable. The first version asserts that THE SAME STATE has two different interpretations (vacuum interpretation and thermal bath interpretation), depending on motion of the classical observer NOT DESCRIBED BY QUANTUM MECHANICS. It is unacceptable in MWI because MWI asserts that everything is described by quantum mechanics. There are no classical observers in MWI, only quantum states in the Hilbert space. Therefore, I think you should be among the first who reject the first version of the Unruh effect.

All this reinforces what we already know, that MWI and BI are almost the same. :-)

 

[Dmitry67]

'Ball melted' is macroscopic event. Why do we talk about the measurement here while the effects of the Unruh radiation are macroscopic?
Lets talk only about the macroscopically observable version.

So, what happens to to ball and to the ball in the box? I provided you my version of what is going to happen. What do you think?

 

[Count Iblis]

You can treat your melting ball experiment in an inertial frame. In that frame the ball accelerates and there are no Unruh particles. The effect is then related to the Casimir effect. It is not the same as the "dynamic Casimir effect", but essentially it is just a matter of the acceleating object imposing certain boundary conditions on the fields.

 

[Dmitry67]

What? If we agree on the macroscopic realism, then if ball melted in one frame, it must melt in all others. If ball is accelerating then it can not be in inertial frame

Well, technically, in GR you can use any frames, but in any frame co-moving with a ball there will be rindler horizons.

 

[Dmitry67]

BTW, how particle can be accelerated? What we see in "accelerators" is a sum-over-histories of charged particles, exchanging virtual photons with magnets, increasing the momentum after each interactions. But between interactions the momentum is conserved.

 

[Count Iblis]

What I mean is that you an describe the accelerating ball in an intertial frame and do the calculations in tat frame. Then there is no Unruh radiation, but the ball will still melt.

See here for an example worked out from the two points of views:

http://arxiv.org/abs/gr-qc/0104030

 

[Demystifier]

'Ball melted' is macroscopic event. Why do we talk about the measurement here while the effects of the Unruh radiation are macroscopic?
Lets talk only about the macroscopically observable version.

So, what happens to to ball and to the ball in the box? I provided you my version of what is going to happen. What do you think?

First, I don't understand the word "macroscopic" except in terms of decoherence.
Second, I think that accelerated ball will melt and that nothing will happen with the box. This is because the forces act on the ball and not on the box.

 

[Demystifier]

See here for an example worked out from the two points of views:

http://arxiv.org/abs/gr-qc/0104030

I partially disagree with this paper, i.e., I do not think that the Unruh effect is really necessary to describe this process in the accelerated frame. I have actually submitted a comment on this paper, but PRL has rejected it.