Solving the Mystery of Guided Particles in a Black Hole

[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.