What if spacetime were a kind of fluid?

Great Debates
201

You are in a bowling alley filled with a supersolid. You roll the bowling ball. The bowling ball displaces the supersolid. The supersolid displaces the bowling ball as it fills in where the bowling ball had been.

The supersolid goes where the bowling ball isn’t.

As you bowl the bowling ball the bowling ball and the supersolid still occupy the same volume. What changes is where the bowling ball exists within the supersolid and that the supersolid fills-in where the bowling ball had been.

The total volume consisting of the supersolid and the bowling ball does not change as you roll the bowling ball. All which changes is where the bowling ball and supersolid exist in relation to each other.

202

My question regards the nomenclature of physics, not the actual equation of physics.

Is the theory that atoms “know when to decay” a “hidden variables” theory?

203

What does the “state” of the aether mean? Does the aether come in varieties, or levels of excitation, or gradations in density, or some other variation in “state?”

Can you define your terms, please?

204

Your terms contradict each other. If the bowling ball displaces the supersolid, then the supersolid is no longer occupying the same volume as the bowling ball.

There are some very serious problems with this theory.

205

In this supersolid, there is an atom in front of the ball. Where is that atom of supersolid “stuff” while the bowling ball moves through the place where the atom had been? The atom cannot go over the ball because the supersolid stuff that includes the atom cannot be compressed to allow the atom to move. It also cannot go under or around that ball for the same reason. In your world, where is the atom while it is being “displaced.”?

206

This part of mpc755’s theory isn’t necessarily incoherent, at least not by itself. When the bowling ball moves, the aether it displaces goes to the space the bowling ball leaves behind. The total volume of the bowling ball plus the aether doesn’t change.

However, this does come into conflict with other parts of his theory. If the aether is incompressible, and exists everywhere, then it can’t wave. It obviously can’t be a longitudinal wave, because that requires changes in density. It also can’t be transverse because that requires changes in density, or space that the aether doesn’t already occupy.

207

Not in the usual sense. A hidden variable typically denotes an observable quantity, such as position or momentum, whose exact value is perfectly determined at the microscopic level, but which can’t in general be perfectly known to us, because there exists a limit given by an uncertainty relation. I.e. it’s something whose value is definite, but which nature refuses to tell us.

If such things exist, we know they must exert nonlocal (faster-than-light) influences upon each other, due to Bell’s theorem.

Which is why what is known to us matters: if there are quantities that are always determinate, but not in generally perfectly known to us, then they must exert nonlocal influences upon one another in order to reproduce the observed phenomenology of entanglement (i.e. Bell inequality violation).

Then, the position and momentum of each of the pair must influence the others’ nonlocally.

If you’re meaning to say here that the entangled photons are simply in an unknown definite state, then your theory can’t reproduce observed phenomena. From what you’re saying, it sounds like each photon is prepared with a definite state, such that the other photon’s state is determined by conservation laws—i.e. say you have each photon in a box, know the total momentum is p, and you open one box and observe the photon’s momentum is p[sub]1[/sub], then you know the other photon’s momentum must be p - p[sub]1[/sub]—then this model doesn’t explain Bell inequality violations.

Bell inequalities are derived from making exactly this assumption—that is, if this assumption holds, Bell inequalities aren’t violated. But experimentally, they are violated—so this simple model for entanglement doesn’t work. You can build a working model, if you simply assume that, upon looking, the state of one photon is randomly determined, and the state of the other photon is instantaneously determined accordingly; then, you have a hidden variable theory with nonlocal influences.

208

Because the best way of explaining yourself is to continually parrot the same unfounded assertion without any variation, elaboration, or even definition:

Is the aether Beetlejuice? For that matter, could you please write me a sonnet on the subject of the Forth Bridge?

209

The state of the aether at every place determined by its connections with the matter and the state of the aether at neighboring places is the state of displacement of the aether.

“a “field” in physics may be envisioned as if space were filled with interconnected vibrating balls and springs, and the strength of the field can be visualized as the displacement of a ball from its rest position”

A “field” in physics is space filled with aether and the strength of the field is the displacement of the aether from its relativistic rest position.

The aether at the top of the mountain is less displaced than the aether at sea level. This allows the atomic clock to tick faster at the top of the mountain.

The faster the atomic clock moves with respect to the state of the aether in which it exists the greater the atomic clock displaces the aether the greater the force exerted by the displaced aether toward and throughout the atomic clock the slower the clock ticks.

210

The supersolid fills-in where the bowling ball had been.

211

The supersolid stuff is filling in where the bowling ball had been. This allows the supersolid in front of the bowling ball to be displaced by the bowling ball.

212

Dark matter is now known to fill ‘empty’ space. When galaxy clusters collide they create a ripple in the mass which fills ‘empty’ space. This ripple is a gravitational wave. This ripple is an aether displacement wave.

213

You think that due to your inability to understand entanglement is an exposed variable theory. Each of the photons can determine the position and momentum of their pair based upon their own due to conservation of momentum.

The photon pair are not physically or superluminally connected. They are entangled as they can determine each others state.

Bell’s inequality applies to hidden variable theories. Entanglement is not a hidden variable theory. Entanglement is an exposed variable theory. Therefore, Bell’s inequality does not apply to entangled pairs.

214

Even though the bowling ball is considered to be incompressible, it condenses slightly as it transitions from being at rest in the supersolid to moving through the supersolid. This condensation allows the supersolid to fill-in where the bowling ball had been. This allows for the supersolid in front of the direction the bowling ball is moving to be displaced.

The aether changes state at c and something similar is occurring as objects move through and displace the aether.

This condensation of objects due to their motion through the aether causes a pressure differential in the aether allowing it to wave.

215

This condensation of the object occurs during acceleration of the object through the aether. The aether pushing back is referred to as the ‘back reaction’ in the following article. This is the aether ‘displacing back’.

‘An Extended Dynamical Equation of Motion, Phase Dependency and Inertial Backreaction’

“We hypothesize that space itself resists such surges according to a kind of induction law (related to inertia); additionally, we provide further evidence of the “fluidic” nature of space itself. This “back-reaction” is quantified by the tendency of angular momentum flux threading across a surface.”

216

Define ‘exposed variable theory’. (Also, of course, saying things like ‘entanglement is a […] theory’ doesn’t help your case; entanglement is a characteristic of certain states in quantum theory, not in any sense a ‘theory’ unto itself.)

Then, they couldn’t violate Bell inequalities; but they do, so that assertion is false.

This doesn’t even make syntactic sense. Bell’s inequality applies to all theories in which the underlying state is always precisely defined, and there are no nonlocal influences. This is a strict logical inference: if both photons have a precisely defined position/momentum at all times, and they are not able to influence one another instantaneously, then they cannot violate a Bell inequality. Quantum mechanics, however, observationally violates Bell inequalities. Thus, photon pairs either cannot simultaneously have a well defined position and momentum (or any other suitable complementary observables), or they must be able to influence one another instantaneously.

If, in your theory, photon pairs always have a definite state, as they do in de Broglie’s double solution theory, then there must be nonlocal influences, again as in de Broglie’s theory.

I mean, the demonstration for this is very simple: consider two particles, A and B, and two observable properties of these particles—e.g. position and momentum—which can (for simplicity) only assume two different values (e.g. if the particle is detected in the left box, we mark a -1, if it’s in the right one, we mark a +1; if the momentum is oriented leftwards, we mark a -1, if it’s rightwards, we mark a +1—or we simply use spin measurements: spin is always found to be either +1 or -1, i.e. aligned with or aligned against the direction one measures).

So we have four observables: a[sub]1[/sub], a[sub]2[/sub], b[sub]1[/sub], and b[sub]2[/sub], where a[sub]i[/sub] refer to measurements carried out on particle A, and b[sub]i[/sub] refer to measurements on particle B, each of which can be either +1 or -1. Furthermore, because of the conservation of angular momentum (if we’re taking spin observables; otherwise, linear momentum), whenever we find particle A to have spin +1 along a particular direction, particle B will have spin -1 slong a particular direction—but that actually doesn’t strictly matter for now.

So, let’s make our measurements, that is, find out what value our four observables have. Then, consider the following quantity:
X = a[sub]1[/sub](b[sub]1[/sub] - b[sub]2[/sub]) + a[sub]2[/sub](b[sub]1[/sub] + b[sub]2[/sub])

It is not hard to see that if a[sub]1[/sub], a[sub]2[/sub], b[sub]1[/sub], and b[sub]2[/sub] have fixed values, then X can at most be two; it suffices to check these four cases:
[ul]
[li]b[sub]1[/sub] = b[sub]2[/sub] = +1: the first term vanishes, and the second one is equal to 2 if a[sub]2[/sub] = +1[/li][li]b[sub]1[/sub] = +1, b[sub]2[/sub] = -1: the second term vanishes, and the first one is equal to 2 if a[sub]1[/sub] = +1[/li][li]b[sub]1[/sub] = -1, b[sub]2[/sub] = +1: the second term vanishes, and the first one is equal to 2 if a[sub]1[/sub] = -1[/li][li]b[sub]1[/sub] = b[sub]2[/sub] = -1: the first term vanishes, and the second one is equal to 2 if a[sub]2[/sub] = -1[/li][/ul]

Slightly rearranged, we get the following:
a[sub]1[/sub]b[sub]1[/sub] + a[sub]2[/sub]b[sub]1[/sub] + a[sub]2[/sub]b[sub]2[/sub] - a[sub]1[/sub]b[sub]2[/sub] ≤ 2

This is the crucial bit: this inequality must ineluctably hold, if our observables have fixed values that cannot influence one another upon measurement; and it follows from nothing else but that. Therefore, if you say:

And:

What you are saying is that the above inequality must always hold; the statements are logically equivalent.

But if the above inequality always holds, then it also holds in the average; that is, if we denote by <a[sub]1[/sub]b[sub]1[/sub]> the result of the procedure ‘measure a[sub]1[/sub] on particle A and b[sub]1[/sub] on particle B many times, and average the result’, then the inequality
<a[sub]1[/sub]b[sub]1[/sub]> + <a[sub]2[/sub]b[sub]1[/sub]> + <a[sub]2[/sub]b[sub]2[/sub]> - <a[sub]1[/sub]b[sub]2[/sub]> ≤ 2
must likewise always hold.

But in quantum mechanics, it doesn’t! This is an experimental finding: for certain states of the two photons, the maximum observed value is 2√2 > 2. That means directly that

and

cannot both be true. If they were, we would never observe a value greater than 2. But we do.

If you disagree, please, with yes or no, answer the following questions:

[ol]
[li]In your theory, does each particle always have a definite position x = (some value) and a definite momentum p = (some value)?[/li][li]In your theory, does the position and/or momentum of one particle depend instantaneously on that of any other?[/li][li]Do you believe that in de Broglie’s double solution theory, every particle always has a definite position/momentum?[/li][li]Do you believe that in de Broglie’s souble solution theory, the position/momentum of a particle can be instantaneously influenced (via the nonlocal quantum potential) by those of other particles?[/li][/ol]

Even if you don’t see fit to answer any other part of this post, please, in the interest of honest debate, at least answer either yes or no (or perhaps, don’t know) to each of those questions.

217

An ‘exposed variable theory’ is where the position and momentum of one of the pair of downconverted photons can be determined by the other; the state of each of the pair is known to the other.

No, it doesn’t. Bell’s inequality applies to hidden variable theories.

" For a hidden variable theory, if Bell’s conditions are correct, the results that agree with quantum mechanical theory appear to indicate superluminal effects, in contradiction to the principle of locality."

Entanglement is not a hidden variable theory. Entanglement is an exposed variable theory as the state of each photon is knowable to its pair.

There is no reason to respond to the remainder of your post as you are in denial of understanding Bell’s Inequality applies to hidden variable theories which entanglement is not.

218

You’ll only “debate” those who already agree with your unfounded and/or undefined premises?

219

The rest of the post I did not respond to had to do with hidden variables and using the mathematics associated with hidden variables to make incorrect assumptions about de Broglie’s wave mechanics and double solution theory.

There are no such things as hidden variables, not in terms of having anything to do with being able to place constrictions on classical physics. You can’t place constrictions having to do with hidden variables on classical physics as entanglement has to do with exposed variable theories.

220

For that, of course, the state must be definite at all times, no? But since the theory must obey an uncertainty principle, that value can’t always be known, right?

‘Hidden variable theories’ and ‘theories in which the underlying state is always precisely defined’ are the same thing. A precisely defined—or, in technical language, ‘dispersion free’—state is one in which all observables have a definite value; these values are the hidden variables of the theory.

I asked some simple clarifying questions regarding your theory. It shows bad faith on your side to deny my polite request.

And regarding Bell inequalities, I explicitly showed you what kind of theories they apply to; entanglement on its own is neither a hidden variable theory nor a theory without hidden variables (well, and at any rate, it’s not a ‘theory’), but the observed phenomenology of entangled states can be reproduced either by a hidden variable theory with nonlocal influences, or by a non-realist theory, i.e. one without hidden variables. In both cases, you would observe a violation of Bell inequalities. ‘Entanglement is not a hidden variable theory’ is meaningless as such.

In any case, I would still be very much interested in your answers to these questions:
[ol]
[li]In your theory, does each particle always have a definite position x = (some value) and a definite momentum p = (some value)?[/li][li]In your theory, does the position and/or momentum of one particle depend instantaneously on that of any other?[/li][li]Do you believe that in de Broglie’s double solution theory, every particle always has a definite position/momentum?[/li][li]Do you believe that in de Broglie’s souble solution theory, the position/momentum of a particle can be instantaneously influenced (via the nonlocal quantum potential) by those of other particles?[/li][/ol]