Solving the Mystery of Guided Particles in a Black Hole

[Dmitry67]

And for this reason I doubt that what you say above--that Bohmian mechanics can be viewed as equivalent to MWI + decoherence, just with one outcome "tagged" as real--could really be backed up in any precise mathematical sense. Can you point to any papers or textbooks that justify your argument rigorously?

I am not dBB proponent :)
It is not an argument, it is my understanding.

Based on MY understanding, only one, tagged observer is consicous. It is not derived from anything: it is dBB axiom. This axiom is not physical: it is not a formula or any kind a physical law: it is pure handwaving, it is expressed in words, and can't be put into mathematical framework.

Every time you listen to dBb proponent you get some inconsistent picture:
- Is wavefunction real? Yes, it is
- Are particles real? Yes, but they are undetectable
- Are 'empty' worlds real?
- No
- But how? They are described by wavefunction, and as you say, wavefunction is real.
- Hmm... Well, to be real, it must have particles inside
- So wavefunction is real, but it is not enough to be effectively real?
- Yes, non-empty branch is much more real :)

To avoid confusion, when I discussed it with Demystifier, we agreed to call branches 'tagged' to avoid using different 'flavors' of being real.

 

[Demystifier]

Every time you listen to dBb proponent you get some inconsistent picture:
...
- Are particles real? Yes, but they are undetectable

No, this is not what dBB proponents say. In fact, according to dBB, particle positions are THE ONLY THING that is directly detectable.

However, the source of confusion is the fact that, even though the particle positions are detectable, the particle trajectories predicted by dBB theory are not. But how that can be? This is because, in order to measure the trajectory, you actually need to measure the position MANY TIMES. And whenever you do that, you disturb the wave function in a way which you cannot control (i.e., decoherence takes place), which disturbs the trajectory as well, so that the measured trajectory is not the same as the one predicted by dBB in the absence of measurements. In fact, due to decoherence, the measured trajectory looks stochastic to an experimentalist, even though it is deterministic at the fundamental microscopic level.

If one could do a measurement without decoherence, then one could actually observe the Bohmian trajectories (or prove that the Bohmian theory is wrong). Unfortunately, nobody knows how to make a measurement without decoherence. Decoherence seems to wash out all the differences between various interpretations of QM.

 

[Dmitry67]

Good example

particle positions are THE ONLY THING that is directly detectable.

To clarify your statement

particle positions are THE ONLY THING that is directly detectable by a tagged observer

Tagged observers observe tagged things, empty observers observe empty things... So you must accept the handwavy axiom to get rid of the addition I highlighted with bold

 

[JesseM]

I am not dBB proponent :)

Well, I didn't say you were, and neither am I.

It is not an argument, it is my understanding.

But your understanding here is nonmathematical, correct? Do you agree that because decoherence never causes interference terms to go to zero exactly, the MWI cannot quite be viewed as an ordinary statistical ensemble of different "worlds" or "branches", so that it cannot be exactly correct to say that the dBB is simply taking a "branch" of the MWI and tagging it as conscious? Or are you claiming that dBB itself contains different "branches" and one is tagged in this way? If the latter can you point to some reference that shows that the pilot wave is mathematically equivalent to some sort of sum of different branches?

 

[JustinLevy]

Your "s'=s=0 at particle one and s'=0,s=1 at particle 2" specifies two points in spacetime. But it has nothing to do with foliation, because two points do not define a hypersurface.

Demystifier, please be reasonable.
As I said,
The Lorentz violating factor is the fixed background telling what pairs of spacetime points in the worlds lines X_1 and X_2 correspond to a "simultaneous time". This is provided by the foliation given by s. The evolution equation is only invariant to foliations that don't change this pairing

So yes, what I specified was not a unique foliation since it only gave two points. It specified a class of foliations. The point still remains that the predicted evolution changed with the choice of foliation.

But there is no FIXED background that tells it.

How can you deny this? You are not making sense.

Look at the initial value problem. If you are given a slice of spacetime with the wavefunction, its derivatives, and particle positions, can one use your evolution equation to answer whether this is enough to determine the derivatives of the particle positions? The answer is no. You need ADDITIONAL INFO. You need to know whether that slice matches with the pairings of simultaneous time defined by the structure you give in the scalar s.

There is information IN ADDITION to the wavefunction and particle positions that is needed. You must provide a FIXED background.

Furthermore, to get the full evolution (not just the particle derivative at one slice) it is not even enough to give an INITIAL pairing (ie. given where in spacetime s=0 for particle 1 and particle 2), you need to specify ALL the pairings, because changing the foliation so that s'=s at particle one and at particle two s'=2s will change the pairing and therefore effect the evolution, even though the INITIAL pairing is the same.

If you want to claim a dBB state now include the following elements:
- wavefunction
- particle positions
- s structure on spacetime

then fine, you can call those 'initial conditions' in a limited sense (limited, because you can predict the evolution of the first two state pieces, but not s). But you are claiming instead that the evolution in invariant to choice in foliation. That is clearly false.

Arguing your theory is then Lorentz invariant is the same as attempting to argue that modern Aether theories are lorentz invariant, because one can claim the background field is merely specifying the vacuum state's initial condition. While I can see what you mean by that, it is not what people mean by Lorentz invariance. You cannot change the meaning of Lorentz invariance to suit your needs. Your theory, undeniably, requires a foliation structure on spacetime which violates lorentz invariance.

 

[JesseM]

Your way of thinking is very close to mine.

I would like to see what do you think about my recent alternative to Bohmian mechanics
http://xxx.lanl.gov/abs/1102.1539 ?
Would you find it less contrived and more appealing than BM?
(In short, the equations for particle trajectories are local, but the physical time is a non-classical time non-locally related to the classical one, which provides a form of nonlocality simpler and more elegant than that in BM.)

Thanks, but the problem is that as I mentioned to JustinLevy I have only a layman-level understanding of Bohmian mechanics, I haven't actually studied the math! So I would need more study to understand your paper, but I'll save a copy and hopefully come back to it sometime when I get around to reading up on the details of dBB.

 

[Dmitry67]

Do you agree that because decoherence never causes interference terms to go to zero exactly, the MWI cannot quite be viewed as an ordinary statistical ensemble of different "worlds" or "branches", so that it cannot be exactly correct to say that the dBB is simply taking a "branch" of the MWI and tagging it as conscious? Or are you claiming that dBB itself contains different "branches" and one is tagged in this way? If the latter can you point to some reference that shows that the pilot wave is mathematically equivalent to some sort of sum of different branches?

If there are some non-zero interferation terms of wavefunction (weird combinations of an observer observing 2 fuzzy images of both cats) then in dBB particles have non-zero probability of 'leaking' into these areas. As a result, there is (a very low) chance of observing these things. So in that sense there I absolutely no difference between dBB and MWI. Otherwise one would be able to tell what theory is right using the 'partial' measurements.

 

[JesseM]

If there are some non-zero interferation terms of wavefunction (weird combinations of an observer observing 2 fuzzy images of both cats)

What do interference terms have to do with "weird combinations of an observer observing 2 fuzzy images of both cats"? I don't think that's what they'd mean in orthodox QM, I suspect they would just mean that the probabilities of different outcomes on measurement of the cat cannot be calculated in terms of a weighted sum of different classical histories, and isn't MWI intended to duplicate the predictions of orthodox QM?

then in dBB particles have non-zero probability of 'leaking' into these areas.

What do you mean by "areas"?

 

[Dmitry67]

Lets begin from the first part.
In "Ortodox" (Copenhagen) QM measurement is 100% final and irreversible, hence the following

decoherence never causes interference terms to go to zero exactly

does not happen. So modern (decoherence) based QM is slightly different from the "Ortodox" one, even the difference (for macrocopic measurement devices) is far beyond any experimental ability.

It does not include interaction-free measurements (I wonder how that types of measurements were explained in CI era)
(is my understanding correct)?

 

[JesseM]

does not happen. So modern (decoherence) based QM is slightly different from the "Ortodox" one, even the difference (for macrocopic measurement devices) is far beyond any experimental ability.

But decoherence can be modeled in the orthodox version, and indeed I think this is the way it's usually done, it's not like decoherence is characteristically associated with MWI or non-collapse interpretations in general (though certainly it strengthens the case for dropping the idea of collapse from QM). To model it using orthodox QM, the idea as I understand it is that you use the standard QM rules to model a system composed of both some subsystem S and the remainder of the system which constitutes its "environment"...then if S interacts with the environment at time T0, and if you later measure S at time T1 and model this as "collapsing the wavefunction", you can show that the probabilities of finding different outcomes for S at T1 will be nearly identical to what you'd predict if you had modeled the earlier interaction with the environment as causing another "collapse" at T0, even though in fact you modeled it using the standard deterministic rules of wavefunction evolution where the interaction of S with its environment simply resulted in their becoming entangled. Someone correct me if I'm wrong, but I think this is one valid way of understanding the meaning of decoherence and its relevance to the measurement problem.

 

[Demystifier]

Look at the initial value problem.

Yes, that is the right way of looking at it.

If you are given a slice of spacetime with the wavefunction, its derivatives, and particle positions, can one use your evolution equation to answer whether this is enough to determine the derivatives of the particle positions? The answer is no. You need ADDITIONAL INFO. You need to know whether that slice matches with the pairings of simultaneous time defined by the structure you give in the scalar s.

My claim is the following. If I know the particle positions $$X^{\mu}_a(s=0)$$ and if I know the wave function and its derivatives at these positions, then my evolution equation is sufficient to determine the derivatives (with respect to s) of the particle positions at $$s=0$$. Indeed, this is exactly what Eq. (23) in my
http://xxx.lanl.gov/abs/1002.3226
says. Don't you agree with that?

 

[JustinLevy]

Demystifier,
You seem to be ignoring the points. This is math, there should be no confusion. The s-structure is needed to evaluate your equations, in contrast to what you were claiming before.

Don't you agree with that?

There, you specified an intial foliation as well. So yes, you can obtain the derivative at s=0 then. The point is you need to specify this foliation, since the solution depends on the foliation.

Nothing you said is disproving my point. In fact you seem to have just ignored it completely.

It is not clear from your response if you are still incorrectly claiming that your formulation is independent of the foliation (since you again, require one to specify an initial pairing for "s simultaneity"). So are you still claiming your formulation is independent of the s structure or not?

Another thing you avoided from the previous post.
Are you claiming a dBB state now includes the following elements?
- wavefunction
- particle positions
- s structure (a "folliation" on spacetime)

To make this even more clear to you, consider a slice of spacetime with the wavefunction, its derivatives, and particle positions. If we choose this slice as s=0, and solve for the evolution, you now have the wavefunction and particle paths in a chunk of spacetime. Now it is possible to take a different slice in this solution, and decide THAT as you initial conditions (with s=0 on this slice), you will get a DIFFERENT solution. Because, as already noting several times, the solution depends on the s structure.

Either your equations demand a preferred foliation, or they are inconsistent mathematically.

 

[Demystifier]

There, you specified an intial foliation as well. So yes, you can obtain the derivative at s=0 then. The point is you need to specify this foliation, since the solution depends on the foliation.

It seems that we don't understand each other. To repeat, I specified:
a) The spacetime positions at s=0
b) The wave function and its derivatives at these positions.
That is all that I specified, nothing more and nothing less. So are you claiming that a) also specifies a foliation?

Assuming that your answer is - yes, then my response is:
1. Although it is not strictly a foliation that a) specifies, for 2 particles it does specify a particular Lorentz frame. So you are more right than wrong about that. However,
2. This specification of the Lorentz frame is encoded in the specification of the INITIAL CONDITION. The EQUATIONS OF MOTION themselves do not specify a Lorentz frame. Consequently, the theory is still covariant, because covariance is a property of the equations of motion, not of the initial conditions.

In other words, even if one can say (by using an imprecise language) that the theory specifies a foliation, it is simply totally wrong to say that this specification of the foliation ruins the covariance. One who says that this ruins the covariance does not understand what covariance is. Covariance is a property of the equations of motion (which do not specify a foliation or anything like that). Covariance is not a property of initial conditions (which do specify something like that).

Or if you still don't get it, let me also quote from wikipedia
http://en.wikipedia.org/wiki/Bohmian_quantum_mechanics#Relativity
which says the same, but in different words:
"The relation between nonlocality and preferred foliation can be better understood as follows. In de Broglie–Bohm theory, nonlocality manifests as the fact that the velocity and acceleration of one particle depends on the instantaneous positions of all other particles. On the other hand, in the theory of relativity the concept of instantaneousness does not have an invariant meaning. Thus, to define particle trajectories, one needs an additional rule that defines which space-time points should be considered instantaneous. The simplest way to achieve this is to introduce a preferred foliation of space-time by hand, such that each hypersurface of the foliation defines a hypersurface of equal time. However, this way (which explicitly breaks the relativistic covariance) is not the only way. It is also possible that a rule which defines instantaneousness is contingent, by emerging dynamically from relativistic covariant laws combined with particular initial conditions. In this way, the need for a preferred foliation can be avoided and relativistic covariance can be saved."

Yet another way to explain that is to exploit the analogy with spontaneous breaking of symmetry (which many physicists are familiar with). The equations of motion for such a theory still obey the symmetry. Yet, a solution (more precisely - the vacuum solution - but this is not important here) does not obey the symmetry. With an abuse of terminology, one could say that relativistic Bohmian mechanics is a Lorentz invariant theory with a spontaneous breaking of Lorentz invariance.

 

[Demystifier]

So are you still claiming your formulation is independent of the s structure or not?

Actually yes, because, as long as the only goal is to calculate the trajectories in spacetime, the parameter s can be eliminated from the equations. In this sense, s is only an auxiliary parameter. If you don't see how s can be eliminated, see
http://xxx.lanl.gov/abs/quant-ph/0512065
Eq. (30).

 

[JustinLevy]

It seems that we don't understand each other. To repeat, I specified:
a) The spacetime positions at s=0
b) The wave function and its derivatives at these positions.
That is all that I specified, nothing more and nothing less. So are you claiming that a) also specifies a foliation?

Yes, more precisely, you specified:
1] the particle positions
2] the wavefunction and its derivatives (at the particle positions)

(the typical bohmian mechanics state, at least at the particle positions)
AND

3] the "simultaneity pairing structure" / foliation / s structure (at least for s=0)

If you only gave the state of the Bohmian mechanics system on a slice of spacetime (ie. not specifying 3), you could NOT solve for the dynamics.

You can ONLY consider specifying the "pairing structure" as an initial condition, if you decide that you have now added an additional structure to the state of bohmian mechanics: wavefunction, particles, and pairing-structure. You have stated otherwise, so your further statements on initial conditions are wrong, and pointless to debate until we can agree on this simple math.

I'll ignore your issues with Lorentz invariance (since they are based on misunderstanding the intial conditions situation) until we can at least agree on the math.

So are you still claiming your formulation is independent of the s structure or not?

Actually yes, because, as long as the only goal is to calculate the trajectories in spacetime, the parameter s can be eliminated from the equations. In this sense, s is only an auxiliary parameter. If you don't see how s can be eliminated, see
http://xxx.lanl.gov/abs/quant-ph/0512065
Eq. (30).

Please stop introducing new ways to write the same dynamics. You are making a similar error in that paper. We are having enough trouble agreeing on the math, that I don't want to branch out into discussing other errors at this time. Let's focus on one math error at a time.Defining the s-structure as well as the wavefunction and particle positions is clearly necessary as shown back in post 24. Unless it is always the case that
$$\partial^\mu_1 \partial^\nu_2 \mbox{Arg}\left[ \psi(X_1,X_2) \right] = 0$$
then clearly changing the foliation will change the evolution.

Put another way, and as I explained earlier:
Consider a slice of spacetime with the wavefunction, its derivatives, and particle positions. If we choose this slice as s=0, and solve for the evolution, you now have the wavefunction and particle paths in a chunk of spacetime. Now it is possible to take a different slice in this solution, and decide THAT as you initial conditions (with s=0 on this slice), you will get a DIFFERENT solution.

Either your equations demand a background s-structure to be specified, or they are inconsistent mathematically.

 

[Demystifier]

Yes, more precisely, you specified:
1] the particle positions
2] the wavefunction and its derivatives (at the particle positions)

(the typical bohmian mechanics state, at least at the particle positions)
AND

3] the "simultaneity pairing structure" / foliation / s structure (at least for s=0)

No, I have not specified 3] as something additional. I simply don't need it, 1] and 2] are enough. What you fail to realize is that the "simultaneity pairing structure" is already contained in 1], because 1] includes not only the 3-space positions at s=0, but also the TIME positions at s=0. Therefore, it would be redundant to specify 3] as something additional.

 

[JustinLevy]

No, I have not specified 3] as something additional.

Yes you have.
For example, if I just randomly picked a point on each particle's path and gave you the wavefunction and derivatives evaluated at those points, that would give you
#1
and
#2

however, this deoes not mean you could then solve for the correct dynamics (derivative of particle positions) given that information. You could only calculate the derivatives if you were ALSO told #3 (that those particle points happened to all be paired with the same s simultaneity).

I would appreciate if you could answer two direct questions for me.

Q1] If I just randomly picked a point on each particle's path and gave you the wavefunction and derivatives evaluated at those points, are you claiming you could calculate the derivatives of the particle positions given that information?


If yes, then your theory is inconsistent, since different foliations will predict different dynamics. If no, then for consistency you have to finally admit that the s-structure is indeed necessary.

The math is so clear here; I don't understand what the issue is. I am worried that you have something you really want to claim, and you will ignore any math that shows otherwise. You should become immediately suspicious if someone claims that by merely rewriting an equation using different notation, that the symmetry properties have changed. All you are doing is making it harder to see, but the properties have not changed, nor could they, since it is the same mathematical sentence just written with different notation.

Q2] Do you disagree with the following? (and please show some math if you do, as I have shown math supporting this)
Consider a slice of spacetime with the wavefunction, its derivatives, and particle positions. If we choose this slice as s=0, and solve for the evolution, you now have the wavefunction and particle paths in a chunk of spacetime. Now it is possible to take a different slice in this solution, and decide THAT as your initial conditions (with s=0 on this slice) to then get a DIFFERENT solution. So the solution depends on more than the wavefunction and particles positions: it also depends on the choice of s-structure.

 

[Demystifier]

For example, if I just randomly picked a point on each particle's path and gave you the wavefunction and derivatives evaluated at those points, that would give you
#1
and
#2

however, this deoes not mean you could then solve for the correct dynamics (derivative of particle positions) given that information. You could only calculate the derivatives if you were ALSO told #3 (that those particle points happened to all be paired with the same s simultaneity).

You are absolutely right about that. However, your #3 is already included in my initial conditions by the very fact that "initial" means "at s=0". Namely, since s=0 for ALL these initial points (otherwise it would not be called - INITIAL condition), this condition already contains the information that the positions are simultaneous.

 

[JustinLevy]

Before you claimed the s-structure wasn't even needed. Now you agree the structure needs to be specified with the initial conditions (since you seem to be using the s-structure to define what initial means). You keep using the s-structure in your intial conditions, yet keep claiming your formulation is independent of the s-structure. You are contradicting the math and yourself.

I would very much appreciate it if you answered my two direct questions above. It is not helpful to ignore the underlying math issues here. And I feel answers to those questions would best help us figure out the root issue of our disagreement here on the math.

Please answer the two direct questions in my previous post.

 

[Demystifier]

Q1] If I just randomly picked a point on each particle's path and gave you the wavefunction and derivatives evaluated at those points, are you claiming you could calculate the derivatives of the particle positions given that information?

No, I do not claim that.

Q2] Do you disagree with the following? (and please show some math if you do, as I have shown math supporting this)
Consider a slice of spacetime with the wavefunction, its derivatives, and particle positions. If we choose this slice as s=0, and solve for the evolution, you now have the wavefunction and particle paths in a chunk of spacetime. Now it is possible to take a different slice in this solution, and decide THAT as your initial conditions (with s=0 on this slice) to then get a DIFFERENT solution. So the solution depends on more than the wavefunction and particles positions:

I disagree. If these two different slices coincide on the initial particle positions (but not elsewhere), then these two solutions are THE SAME.

Can you agree with the following?:
If I know the wave function everywhere, and if I choose n points in spacetime and say that they all have the same $s$, then I can solve my n-particle equations uniquely. Yet, these n points do not define a unique slice. They define an infinite class of different slices, where each of the slices crosses these n points. It doesn't matter which of these slices I choose, because my solution does not depend on it.

 

[JustinLevy]

No, I do not claim that.

Good, so we are in agreement that specifying the wavefunction (and its derivatives) and particle positions on a slice of spacetime is not necessarily sufficient to calculate the evolution.

Put more succinctly: Given only the dBB state, and not the s-structure, one cannot determine the evolution.

It's nice to find a nugget we can agree on. But it is still very confusing, because if you agree the dBB state is not enough, how can you not take that tiny little step to the remainder of what the math says and agree that the missing piece is the s-structure. The s-structure is required.

If you maintain the s-structure isn't needed, then if I give you a dBB state on a slice of spacetime, you agreed above we need more information ... what do YOU call that information? Since it is separate from the debroglie bohm state, can't we agree it is additional information?

I disagree. If these two different slices coincide on the initial particle positions (but not elsewhere), then these two solutions are THE SAME.

Oh come on. That is disingenuous. That is not what I asked, and you know that is not what I asked since you already tried this earlier.

I said:
Now it is possible to take a different slice in this solution, and decide THAT as your initial conditions (with s=0 on this slice) to then get a DIFFERENT solution.

You instead restricted your answer to slices that preserve the simultaneity / pairing structure / s-structure between particles. My point is that it is possible to take a different slice and get a different solution. Your response that you can change the slice everywhere but at the particle locations and still get the same solution is clearly moot.

Can you agree with the following?:
If I know the wave function everywhere, and if I choose n points in spacetime and say that they all have the same $s$, then I can solve my n-particle equations uniquely. Yet, these n points do not define a unique slice. They define an infinite class of different slices, where each of the slices crosses these n points. It doesn't matter which of these slices I choose, because my solution does not depend on it.

Yes, I agree it defines a class of slices. Let me quote myself when I started this:
The evolution equation is only invariant to foliations that don't change this pairing

Yet you complained, as you are doing now, that defining s at each particle is not a hypersurface and I said:
So yes, what I specified was not a unique foliation since it only gave two points. It specified a class of foliations. The point still remains that the predicted evolution [can change] with the choice of foliation.

In summary:
The only part of the "foliation" that the evolution depends on is the s-structure pairing of "simultaneous points" on the particle paths. If you change this pairing the predicted evolution changes. So clearly, the evolution depends on this s-structure.


Since you side-stepped my question, could you please go back and answer it this time?

 

[Demystifier]

Your response that you can change the slice everywhere but at the particle locations and still get the same solution is clearly moot.

It seems to be the source of our disagreement. Unfortunately, right now I don't have time to discuss it in more detail.

 

[JustinLevy]

It seems to be the source of our disagreement. Unfortunately, right now I don't have time to discuss it in more detail.

Please please don't focus on trying to give more detail on that moot point.

The root issue is your claim that your formulation is independent of the s-structure. I have shown that the evolution prediction can change with a change in s-structure, which clearly demonstrates that your claim is false. Repeatedly explaining that there exists a class of foliations that don't change the evolution does not change this fact. It's like you are claiming x + y = x is independent of y, and I show that it is not true if you consider y=1, but you respond 'no! consider y=0, so x + y = x is independent of y'. If a counter example for a proposition is given, you cannot remove the counter example and prove the proposition by giving an example case. This is basic logic. Your point is clearly mathematically moot.

We agree that specifying the dBB state is not enough. And we agree that if we specify the s-structure and give the dBB state for simultaneity paired particle positions, that it is enough. I have even shown you that the evolution can give a different prediction if one changes the pairing s-structure (although you keep avoiding this point). So it is NOT an issue of whether there exist some foliations which your theory is invariant to, it is an issue of whether your theory is really, as you claim, completely independent of the choice of foliation. The math clearly shows it is not.

So instead of trying to find time to discuss a completely moot subset of the issue (ie. focusing on only a restricted class of foliation changes), instead I would very much appreciate it if you answer my direct question you avoided. This should take much less time.The question again for reference is:
Question] Do you disagree with the following? (and please show some math if you do, as I have shown math supporting this)
Consider a slice of spacetime with the wavefunction, its derivatives, and particle positions. If we choose this slice as s=0, and solve for the evolution, you now have the wavefunction and particle paths in a chunk of spacetime. Now it is possible to take a different slice in this solution, and decide THAT as your initial conditions (with s=0 on this slice) to then get a DIFFERENT solution. So the solution depends on more than the wavefunction and particles positions: it also depends on the choice of s-structure.

 

[Demystifier]

The question again for reference is:
Question] Do you disagree with the following? (and please show some math if you do, as I have shown math supporting this)
Consider a slice of spacetime with the wavefunction, its derivatives, and particle positions. If we choose this slice as s=0, and solve for the evolution, you now have the wavefunction and particle paths in a chunk of spacetime. Now it is possible to take a different slice in this solution, and decide THAT as your initial conditions (with s=0 on this slice) to then get a DIFFERENT solution. So the solution depends on more than the wavefunction and particles positions: it also depends on the choice of s-structure.

I agree with this.

Now it is my turn to ask a question. Do you agree with the following?

Assume
a) That the n-particle wave function is known everywhere in the 4n-dimensional configuration space.
b) That the quantities $$X^{\mu}_a(s=0)$$ are known. (These quantities define n points at s=0.)
c) That a shape of a whole slice that crosses these n points is NOT specified.

Then
1) From this knowledge one can calculate the functions $$X^{\mu}_a(s)$$ for all s.
2) These functions fully determine the spacetime trajectories of particles.

And please, don't elaborate whether you find this question relevant or not. Just say whether you agree or not, because it is crucial for MY way of looking at it.

 

[JustinLevy]

I agree with this.

Thank you for answering.

Now it is my turn to ask a question. Do you agree with the following?

Assume
a) That the n-particle wave function is known everywhere in the 4n-dimensional configuration space.
b) That the quantities $$X^{\mu}_a(s=0)$$ are known. (These quantities define n points at s=0.)
c) That a shape of a whole slice that crosses these n points is NOT specified.

Then
1) From this knowledge one can calculate the functions $$X^{\mu}_a(s)$$ for all s.
2) These functions fully determine the spacetime trajectories of particles.

Yes, I agree with that.
I feel this was already asked and answered. So I apologize if my answers were not clear before. I'll try to answer straight up yes/no when possible from now on to avoid this in the future. Or if you were just repeating this in more mathematical detail to remove any possible confusion, that is fine too.To make this more clear, we agree:
Given:
#1] That the n-particle wave function is known everywhere in the 4n-dimensional configuration space.
#2] That the quantities $$X^{\mu}_a$$ are known on some spacetime slice (These quantities define n points on the slice.)

we are NOT able to solve for the dynamics. However if we add information about the simultaneity-pairing / s-structure, such as

#3] specify that the spacetime slice is s=0

NOW we can
1) From this knowledge one can calculate the functions $$X^{\mu}_a(s)$$ for all s.
2) These functions fully determine the spacetime trajectories of particles.
I'm glad we can finally agree on this. But since you need MORE information than the dBB state, and in particular we agree the s-structure satisfies this need, and furthermore that changing the s-structure can change the solutions, then it seems obvious that your formulation depends on the s-structure. If you want to continue to argue this is part of the initial conditions, that would be tantamount to claiming the s-structure is now a required additional part to the dBB state: wavefunction, particle positions, and the "simultaneity-pairing"/s-structure.

So please please don't start down that path until we finish this discussion of the math, since despite agreeing to above, I have the feeling we still are not agreeing completely on the root points. If we don't agree on the root points, it is useless to try to progress forward.We agree that without the s-structure specified we can't solve for the dynamics, and we agree that with the s-structure we can solve the dynamics, and furthermore we agree that the dynamics can change if we change the s-structure.
Question]: So do you now agree that your formulation of the dynamics is NOT independent of the "simultaneity-pairing"/s-structure?

 

[Demystifier]

Question]: So do you now agree that your formulation of the dynamics is NOT independent of the "simultaneity-pairing"/s-structure?

Yes I do.

Do you agree that this extra information (not present in nonrelativistic BM) is encoded in the initial conditions?
And if you do, do you agree that this means that equations of motion are relativistic covariant?

 

[JustinLevy]

I don't think we can jump to that yet, as there are still some math issues to discuss.

Do you agree that this extra information (not present in nonrelativistic BM) is encoded in the initial conditions?
And if you do, do you agree that this means that equations of motion are relativistic covariant?

Unfortunately I cannot answer these with yes/no, as I feel we don't even agree on what those questions mean currently.

First, this extra information was present in the "nonrelativistic BM" as well, in the form of a preferred inertial frame. Also, these questions are starting to get away from the math and get into terminology issues: you use terms differently from the way mainstream physics does. That being said, here's a first attempt at answering your questions fully.

your Q1] If we take the dBB state to be the wavefunction and particle positions, then no. If we take the dBB state to be the wavefunction and particles positions AND the simultaneity pairing structure, then in this bizarre sense yes. I have strong reservations on this though, as it is strikingly similar to choosing a preferred frame and then adding some scalar fields with values set to the coordinate labels, and then claiming this is not a choice of frame but initial conditions for a scalar field. These 'set once' features are best described as parameters or external structure required by the theory, and not an initial conditions. Like a constant lorentz violating vector in some Lorentz violating theories in literature. These are coupling parameters or an external structure, not initial conditions.

your Q2] The easiest way to answer this is just no.

I'm not sure what to consider "s" really as a geometric entity. I once referred to it as a scalar field, but this isn't really true since the "simultaneity pairing" structure cannot always be reduced to such (for example if particle paths cross). However, it doesn't seem unreasonable to claim this s-structure can be interpreted as a geometric object of some kind (however, as noted above, even coordinate systems can be interpreted as such if taken as values on spacetime). So by introducing that s-structure, and if you constrain the metric, it seems okay to consider your tensor notation like equations as coordinate system independent. If that is all you meant (some kind of 'coordinate system independent' way of writing the equations), then yes.

However, you said "relativistically covariant", which I think you were trying to ask: does this have Lorentz invariance? There is absolutely no debate here, the answer is no.

Question for you:
Do you agree that writing an evolution equation in tensor notation can make it coordinate system independent but does not automatically imbue it with Lorentz symmetry?

For consideration, note that even Newtonian gravity can be written in tensor notation (Newton-Cartan). And if we start allowing promoting of coordinate systems to scalar fields, we can bastardize the notation to make anything written in one coordinate system to be in tensor notation.

Even if I hold my nose and say the s-structure is part of the state and therefore specified in the initial conditions, the answer is still no. Because the very existence of the s-structure break Lorentz invariance. It's not like the theory has Lorentz invariance, and then choosing a particular initial condition breaks it ... the very existence of the s-structure in the theory breaks the Lorentz invariance.

Similarly, in an aether theory, one could claim the choice of rest frame of the aether is an initial condition. The use of the term "Lorentz symmetry" by the mainstream does NOT consider such theories as having Lorentz symmetry.

For an example of mainstream use of the phrase, I searched for vector field lorentz violations and picked one with lots of citations:
Spacetime-varying couplings and Lorentz violation
http://arxiv.org/abs/astro-ph/0212003
cite: 97 times

If the lorentz violating terms are non-zero the theory is called Lorentz violating. If the theory is only an effective theory, it is possible that these violating terms arise from a more fundamental theory which has a lorentz invariant action, in which case the Lorentz violating low-energy effective theory is said to have dynamically broken lorentz symmetry. In your case we are not generating a lower energy effective theory from your theory ... so all we need to worry about is simple Lorentz symmetry yes or no. The existence of the lorentz violating s-structure makes this a simple no. It is not an issue of initial conditions.

Yes I do.

Good.

So I assume you understand now why your other claims about removing s-dependence using similar reformulations are wrong as well, such as:

Actually yes, because, as long as the only goal is to calculate the trajectories in spacetime, the parameter s can be eliminated from the equations. In this sense, s is only an auxiliary parameter. If you don't see how s can be eliminated, see
http://xxx.lanl.gov/abs/quant-ph/0512065
Eq. (30).

That formulation still requires a simultaneity pairing structure.

I'm also worried about other issues. For example it is well known that the Klein-Gordon equation has serious issues interpreting the wavefunction as a probability distribution, but you went right ahead and did so anyway.