Spacetime is doomed.

[mistermack]

14 hours ago, Schmelzer said:

Unfortunately, length measurements are also distorted, and there is also no undistorted way to identify contemporaneity.  All these distortions taken together make your proposal impossible. 

What happens when you have a pair of quantum entangle particles? What I read is that it's looking likely that any change of state happens instantaneously to both particles, so maybe there will one day be a way to establish what real contemporaneity looks like. 

[dimreepr]

17 minutes ago, mistermack said:

What happens when you have a pair of quantum entangle particles? What I read is that it's looking likely that any change of state happens instantaneously to both particles, so maybe there will one day be a way to establish what real contemporaneity looks like. 

What happens when you have a pair of shoes?

[mistermack]

43 minutes ago, dimreepr said:

What happens when you have a pair of shoes?

I like to take things one step at a time.

[swansont]

16 hours ago, mistermack said:

What happens when you have a pair of quantum entangle particles? What I read is that it's looking likely that any change of state happens instantaneously to both particles,

"Change of state" implies the particles were in a particular state to begin with, and that's incorrect. They are in superposition, and the state is unknown. The determination of which state they end up in is simultaneous, since measuring one particle instantly tells you the state of the other.

[Schmelzer]

23 hours ago, studiot said:

No, just because some of the entries are zero, does not mean they don't exist.
Can you demonstrate a rectangular matrix with only 10 entries that is also a tensor?

I did not say that some entries will be zero, I said the tensor is symmetric.  That means, [math]g_{\mu\nu}(x)=g_{\nu\mu}(x)[/math].  This symmetry remains, a symmetric tensor will remain symmetric in other systems of coordinates.  

Quote

2) So are you telling me that you can add a single real number to a tensor or are you telling me that when you multiply all the elements of a tensor by a single real number you don't get another tensor?
If you want to use a 'cosmological' constant you have to apply it everywhere, which means you multiply g_υν by it.

If I want to take a look at the Einstein equations, Wiki gives me quite immediately the formula 

[math]G_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8 \pi G}{c^4} T_{\mu\nu}[/math]

for copypasting. The term there is the cosmological constant [math]\Lambda[/math] multiplied by the metric tensor.  

Quote

3. SR is only global in the absence of all mass and energy in the system.

SR is always global.  If the masses and energies in the system are too big, SR is no longer a valid approximation, and GR has to be used.  This does not change the fact that SR is global.  

Quote

4. Thank you for that confirmation. So can you reference a clock that can directly read time units in this system, and what does it read? (perhaps swansont will help here - this is his field not mine.)
I would also respectfully suggest that adopting this practice whilst saying nothing is baffling for many members here who may be unfamiliar with natural units in general and geometrised ones in particular.

It is always difficult to decide how much information has to be given.  Those really interested in formulas will have to follow anyway a particular textbook, or whatever replaces it like Wiki.  If one cares about c factors, then the formula I have given for proper time is assumed to have the unit of a distance, thus, you have to divide it by c to get the unit of seconds.

Edited November 15, 2019 by Schmelzer

[studiot]

7 minutes ago, Schmelzer said:

I did not say that some entries will be zero, I said the tensor is symmetric.  That means, gμν(x)=gνμ(x) .  This symmetry remains, a symmetric tensor will remain symmetric in other systems of coordinates.  

If I want to take a look at the Einstein equations, Wiki gives me quite immediately the formula 

Gμν+Λgμν=8πGc4Tμν

for copypasting. The term there is the cosmological constant Λ multiplied by the metric tensor.  

SR is always global.  If the masses and energies in the system are too big, SR is no longer a valid approximation, and GR has to be used.  This does not change the fact that SR is global.  

It is always difficult to decide how much information has to be given.  Those really interested in formulas will have to follow anyway a particular textbook, or whatever replaces it like Wiki.  If one cares about c factors, then the formula I have given for proper time is assumed to have the unit of a distance, thus, you have to divide it by c to get the unit of seconds.

Thank you for half a discussion.

 

You were correct I misremembered the equation, but you did not answer my comments.

So I can take it you are avoiding saying that  when you multiply a tensor by a constant you get another tensor.

Edited November 15, 2019 by studiot

[Schmelzer]

In science, it is often very useful to be pendantic. 

22 hours ago, mistermack said:

What happens when you have a pair of quantum entangle particles? What I read is that it's looking likely that any change of state happens instantaneously to both particles, so maybe there will one day be a way to establish what real contemporaneity looks like. 

Maybe one day.  But this would require knowledge of some theory beyond quantum theory.  Of course, if there is real contemporaneity, there is only one candidate in reality for it, defined by the CMBR frame. But without a more fundamental theory, this is not more than a guess. 

Edited November 15, 2019 by Schmelzer

[Markus Hanke]

On 6/23/2019 at 10:31 PM, koti said:

Why is there such a dissonance between us knowing the underlying laws governing satellite orbits or alleles in cells and not knowing seemingly simple stuff like cigarette smoke and other fluid turbulance?

Cigarette smoke rising in air is an example of an open system, the dynamics of which are chaotic. That means we understand the underlying laws very well, and the system is completely deterministic and classic, yet it is still not predictable into the indefinite future. That is because any future state of the system is highly dependent on initial and boundary conditions, and initially small inaccuracies in determining those conditions compound very quickly, making final outcomes increasingly unpredictable the further you go into the future.

A satellite orbit is different, since such a system is not chaotic in nature; the orbit of course still depends on initial conditions, but once it has been entered, it will remain stable and not evolve in unpredictable ways. Scale is also a factor - any inaccuracies in initial condition will generally be negligibly small when compared with the total orbital radius of a satellite, whereas even the tiniest movement of air will have a large impact on cigarette smoke.

To make a long story short, the differences are not in our understanding of the underlying laws, but in the dynamics of how a system evolves, specifically in the way those dynamics depend on initial and boundary conditions.

[Strange]

1 hour ago, Markus Hanke said:

A satellite orbit is different, since such a system is not chaotic in nature

Unless there are three bodies involved!

[Markus Hanke]

22 hours ago, Strange said:

Unless there are three bodies involved!

Indeed, even though not all 3-body systems evolve chaotically. There are special cases when you get regular, periodic solutions; also, if the mass of one of the bodies is very much smaller than that of the other two, the setup effectively reduces to a 2-body problem. So it depends on the circumstances. But you are right, most gravitational 3-body systems evolve chaotically.

[swansont]

!

Moderator Note

Discussion of Boltzmann brains has been split https://www.scienceforums.net/topic/121007-boltzmann-brains-split-from-spacetime-is-doomed/

 

 

[Schmelzer]

On 11/13/2019 at 7:07 PM, swansont said:

No, I disagree. If you do this, you have to invent a new notion of time, and you need new physics to go along with it.  

Given that new physics would be speculation, I'm forbidden to present them here. Fortunately, nothing new is necessary here. I can take good old Newton's definitions.  He already distinguished absolute time which flows independent of any human action and apparent time, the result of the attempts of humans to measure time. Technically, absolute time survives in GR as coordinate time - if there would be some absolute time, it would be defined by some time coordinate as the corresponding coordinate time. But GR has no equations for absolute time, and it does not appear in the spacetime interpretation.  

Quote

What is the reference for this happening?

Standard twin experiment.  Use two identical clocks moving from a common starting event to a common end event on different trajectories.  The result will be that the clocks show different time.  In an interpretation based on the Newtonian concept of time, this would not be possible for absolute time, only for apparent time, and would prove the existence of some distortion of the clocks. 

Quote

I think if you expect different answers you need a different model. What I am not getting is why you think there is a different interpretation to be had here.

I think having different interpretations for the same equations is always useful.  It helps very much to distinguish what follows from experiments and observation (these things would be the same in different interpretations) from metaphysical assumptions (which can be different in different interpretations).   

Quote

Yeah, actually it does say it's a curved four-dimensional spacetime, and not a physical distortion of a device (see the equivalence principle). And if you think it's a physical effect you would need to explain kinematic dilation. The burden of proof here is yours.

No problem.  Except that I'm not allowed to deliver such things here. 

Except, I hope, for the simple formula \(\tau = \int \sqrt{g_{\mu\nu}(x,t) \frac{d\gamma^\mu}{dt}\frac{d\gamma^\nu}{dt}} dt\)  which quite explicitly described that (and how) the gravitational field (described by \(g_{\mu\nu}(x,t)\))  distorts the time measurement for a clock moving along the trajectory \(\gamma^\mu(t)\).  

[swansont]

16 minutes ago, Schmelzer said:

Given that new physics would be speculation, I'm forbidden to present them here. Fortunately, nothing new is necessary here. I can take good old Newton's definitions.  He already distinguished absolute time which flows independent of any human action and apparent time, the result of the attempts of humans to measure time. Technically, absolute time survives in GR as coordinate time - if there would be some absolute time, it would be defined by some time coordinate as the corresponding coordinate time. But GR has no equations for absolute time, and it does not appear in the spacetime interpretation. 

Is my coordinate time the same as your coordinate time? Does my physics work just as well as yours?

 

16 minutes ago, Schmelzer said:

 

Standard twin experiment.  Use two identical clocks moving from a common starting event to a common end event on different trajectories.  The result will be that the clocks show different time.  In an interpretation based on the Newtonian concept of time, this would not be possible for absolute time, only for apparent time, and would prove the existence of some distortion of the clocks. 

Since we know Newtonian time is wrong, what's the point of invoking it? And I was hoping for an actual experiment, not a thought experiment (which is not actual evidence)

 

16 minutes ago, Schmelzer said:

I think having different interpretations for the same equations is always useful.  It helps very much to distinguish what follows from experiments and observation (these things would be the same in different interpretations) from metaphysical assumptions (which can be different in different interpretations).   

No problem.  Except that I'm not allowed to deliver such things here. 

Not here in this thread, but nothing is preventing you from opening a thread in speculations (or putting this in an existing thread of yours in speculations, as appropriate) so this is hardly an excuse.

 

 

[Schmelzer]

1 minute ago, swansont said:

Is my coordinate time the same as your coordinate time? Does my physics work just as well as yours?

In GR, the physical predictions do not depend on the choice of coordinates, thus, the physical predictions will be the same, even if you choose another coordinate time. 

Quote

Since we know Newtonian time is wrong, what's the point of invoking it? And I was hoping for an actual experiment, not a thought experiment (which is not actual evidence)

We don't know this. We know that the Newtonian theory of gravity is wrong. We don't know that the Newtonian definitions for time are wrong.  

We talk here about GR, and interpretations of the GR equations which differ from the spacetime interpretation.  If one talks about different interpretations, even thought experiments distinguishing them cannot be expected.  Even less actual experiments.  To discuss alternative theories of gravity, which could provide differences in actual experiments, would be forbidden here as speculation. 

Quote

Not here in this thread, but nothing is preventing you from opening a thread in speculations (or putting this in an existing thread of yours in speculations, as appropriate) so this is hardly an excuse.

I do not need excuses.  Take a look at these threads.  If you miss something, ask where it is not forbidden to answer. 

[Mordred]

So let me get this straight we have experimental evidence of time dilation at a distance of 1 metre apart. Yet you feel time is absolute.

 In all reference frames in Newtonian physics time is absolute. That is not true for reference frames under SR/GR. So how can you claim that Newton isn't wrong in regards to time ?

More importantly why do you feel GR must be incorrect in regards to time with your knowledge of GR ?

Which from your equation above and your last few posts seem to describe that the equivalence principle with regards to inertial mass and gravitational mass doesn't apply.

 Yes I do specify your equation 

 

 

Edited February 13, 2020 by Mordred

[swansont]

16 hours ago, Schmelzer said:

In GR, the physical predictions do not depend on the choice of coordinates, thus, the physical predictions will be the same, even if you choose another coordinate time. 

You didn't answer my question. 

Is my coordinate time the same as yours?

 

 

[Schmelzer]

1 hour ago, Mordred said:

Ok I have a question according to the above paper you have a Lorentz style eather that is non interacting with regular matter fields. This eather frame according to your paper allows for FTL. So how does it mediate entangled particles to allow supposed FTL signaling between entangled particles in violation of Einstein causality ? Without considering that a form of interaction ? Is there some arbitrary mediator boson as per regular matter fields ? Secondly what property of this eather allows FTL when massless non coupling particles propogate at c in regards to our observable matter and force fields ? Ie via {ct}.

No. There is absolute space and absolute time, as described by the preferred coordinates.  Then there is an ether.  Density, velocity and stress tensor of the ether are described by the gravitational field, all other properties of the ether described by matter fields. So, all the degrees of freedom of the ether interact with each other.  

The frame of absolute space and time is simply what is used to write down the equations of quantum theory.  And these equations describe, in a very explicit way, FTL causation (not signalling). You know, from the Schroedinger equation follows a continuity equation for \(\rho(q)=|\psi(q)|^2\), and the velocity in this continuity equation, which defines how the probability distribution changes in time, depends on the whole actual configuration, in the whole universe.

1 hour ago, Markus Hanke said:

These aren’t basic principles of science, they are basic attributes of classicality only. There is no good reason to believe they are scale-independent.

That the only formula we need from realism to prove Bell's inequality is part of the logic of plausible reasoning (the objective Bayesian interpretation of probability) I have shown in  Schmelzer, I. (2017). EPR-Bell realism as a part of logic, arxiv:1712.04334.  That without Reichenbach's principle of common cause we we have to give up key methodological principles which distinguish science from astrology I have already argued.   Correlations require explanation, and no observation can force me to give up this principle because this is part of the scientific method. 

[Strange]

1 minute ago, Schmelzer said:

Schmelzer, I. (2017). EPR-Bell realism as a part of logic, arxiv:1712.04334

That style of citation make it look like it is a published paper. Maybe just provide the link: https://arxiv.org/abs/1712.04334, it's a bit less pretentious.

6 minutes ago, Schmelzer said:

no observation can force me to give up this principle

You are not doing science then.

[Schmelzer]

2 hours ago, swansont said:

You didn't answer my question. 

Is my coordinate time the same as yours?

In the context of GR, probably not, because GR does not fix a time coordinate as preferred.  In a theory with preferred absolute time, if you choose absolute time as the time coordinate, yes, but in principle even in such a theory you are allowed to use other time coordinates.  

2 minutes ago, Strange said:

That style of citation make it look like it is a published paper. Maybe just provide the link: https://arxiv.org/abs/1712.04334, it's a bit less pretentious.

You are not doing science then.

I prefer to give also the title, so that people know at least something about the paper.  

I'm doing science.  That's why I don't give up principles of scientific methodology.  That correlations require causal explanations is such a principle.  If you think you can do science without any principles, your choice. Without some basics like logic, you will end nowhere.  

[Bmpbmp1975]

Going through this thread there are a few things I understand and a few I don’t, my question is if we still see the universe from the past and continue to see it as present how is SpaceTime considered doomed? It’s still there is it not?

Edited February 14, 2020 by Bmpbmp1975

[iNow]

As you're part of the universe, then if you're there to "see" it, then clearly it's still there. Your question is also (unsurprisingly given your posting behavior at this site) a few layers separated from the actual topic.

[Strange]

2 hours ago, Schmelzer said:

I prefer to give also the title, so that people know at least something about the paper. 

It might surprise you to know that you can do both. 

2 hours ago, Schmelzer said:

I'm doing science.  That's why I don't give up principles of scientific methodology.

One of those is that you cannot ignore observational evidence (therefore "no observation can force me to give up this principle" is an anti-scientific stance).

 

1 hour ago, Bmpbmp1975 said:

Going through this thread there are a few things I understand and a few I don’t, my question is if we still see the universe from the past and continue to see it as present how is SpaceTime considered doomed? It’s still there is it not?

It was a statement about the theory that is based on the mathematical model of spacetime: that this theory is incomplete and will, eventually, need to be replaced (i.e. the theory is "doomed"). 

[Markus Hanke]

2 hours ago, Schmelzer said:

Correlations require explanation

Standard quantum physics explains entanglement very well, without the need for hypothetical constructs such as superluminality. Also, as you know, correlation does not imply causation.

[Bmpbmp1975]

32 minutes ago, Strange said:

It might surprise you to know that you can do both. 

One of those is that you cannot ignore observational evidence (therefore "no observation can force me to give up this principle" is an anti-scientific stance).

 

It was a statement about the theory that is based on the mathematical model of spacetime: that this theory is incomplete and will, eventually, need to be replaced (i.e. the theory is "doomed"). 

Thank you 

[Schmelzer]

33 minutes ago, Strange said:

It might surprise you to know that you can do both. 

Which is what I have done.

Quote

One of those is that you cannot ignore observational evidence (therefore "no observation can force me to give up this principle" is an anti-scientific stance).

Once the principle is a methodological principle (correlations require causal explanations), the stance is not anti-scientific.  And I don't ignore any observational evidence.  I work until I find an explanation of the observations which is compatible with the principles of scientific methodology. 

8 minutes ago, Markus Hanke said:

Standard quantum physics explains entanglement very well, without the need for hypothetical constructs such as superluminality. Also, as you know, correlation does not imply causation.

No. Quantum theory allows you to compute the numbers without giving any explanation for them.  Note also that what is required is causal explanation.  

Of course, correlation does not imply causation. This is what makes the search for causal explanations a serious scientific problem.  And Reichenbach's common cause principle defines precisely what counts as a causal explanation and what does not count.