Einstein's Religion: Mistakes in Physics & Time Order

  • Thread starter Eugene Shubert
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In summary: There's a problem with that assumption, too. If there's only one reference frame that admits the space-time decomposition, then that frame must be the only one in the universe. If it's the only frame, then it must be stationary, and the laws of physics in that frame must be the same for all inertial frames. But that's absurd! In summary, Einstein was wrong about true religion. This paper discusses outrageously religious ideas in the philosophy of physics. Einstein's greatest blunder was his stubborn, unrealistic faith in a deterministic universe, and his second greatest scientific blunder was his fallacy of no absolute time order for all events in the universe. These errors led to his belief that we may not conceptual
  • #1
Eugene Shubert
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You all know that Albert Einstein was deeply religious in his pantheistic veneration of the physics of nature. No scientist today should be surprised to learn that Einstein was completely wrong about true religion. This paper is about outrageously religious ideas in the philosophy of physics.

Einstein’s greatest blunder in science was his stubborn, unrealistic faith in a deterministic universe. His belief in a mechanistic interpretation for all natural law is widely recognized as a direct denial of quantum physics and the Hebrew Bible. Einstein would express his faith by saying, “Gott wurfelt nicht!” (God does not play dice!) Of course God plays dice with light and matter. God not only plays dice with the universe, —He cheats. (I don’t mean to review the philosophical/religious underpinnings of quantum mechanics in this paper).

Einstein’s second greatest scientific blunder, which he never repudiated, was his fallacy of no absolute time order for all events in the universe and that we may not conceptualize time being divided into an absolute past, present and future.

“For us believing physicists, the distinction between past, present, and future is only an illusion, even if a stubborn one.” —Albert Einstein.

Einstein’s sophistry about time order being relative is clever and compelling but it’s inconsistent with Einstein’s favorite cosmological model. All of Einstein’s watchful, guarded reasoning in the famous train and embankment gedanken experiment derails itself in a spatially closed and bounded universe. I will demonstrate how an absolute time order follows from the laws of physics in Einstein’s universe. The argument is trivial. Here are the key ideas:

If a law is a true law of physics, then it’s true everywhere, for all time. There is a universal law of light propagation. It’s impossible to prove global theorems about time order with an insufficient array of synchronized clocks. A consistent, global view of synchronization and spacetime, based on a universal law of light propagation, outranks all local, partial and limited views of the universe.

The mathematical details continue in this link:

http://www.everythingimportant.org/relativity/simultaneity.htm
 
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  • #2
That's funny I always interpreted "God does not play dice with the universe" to mean there are causes to things not to mean that chances and probabilities didn't exist in physics, randomness.
 
  • #3
I don't see what makes your scheme any different from all the other schemes of decreeing a particular reference frame of SR/GR as absolute.
 
  • #5
I would have to go with decree.

Your particular frame is interesting for your particular topology because it's the unique flat "reference frame" that admits the space-time decomposition S*R.

But it is by no stretch of the imagination the only "reference frame" that admits an S*R decomposition, and there are certainly other frames that don't admit the decomposition.


Anyways, the notion of a universal reference frame in a periodic universe has its problems. Not only do two different spatial coordinates refer to the same place in the universe, but for any particular photon you received, you don't know how many times it wrapped around the universe before reaching you. That's why differential geometry works only with local coordinate charts, and thus general relativity only with local reference frames.


And the main problem with any universal synchronization scheme is that it adds an unnecessary layer of computation. Why bother converting my local time into universal time to do analysis when I can just do the calculations in my local reference frame? Universal synchronization is good for giving directions and scheduling apponitments, but not much else.




I have another objection to your website, but it's late and I'm tired and may not have read it through fully... when you use the formula:

ds = ds0 * sqrt(1 - (v/c)2)

you justify it via the EEP.. but the EEP says that the laws of special relativity hold in locally flat reference frames. Since the Schwartzchild metric is flat only at infinity, I don't think you're justified in using this formula.
 
  • #6
Originally posted by Hurkyl
I don't see what makes your scheme any different from all the other schemes of decreeing a particular reference frame of SR/GR as absolute.
I thought alis had explained that to you.

http://www.everythingimportant.org/relativity/Soleimani.htm

Originally posted by Hurkyl
Your particular frame is interesting for your particular topology because it's the unique flat "reference frame" that admits the space-time decomposition S*R.
What do you mean by the asterisk? What is a flat reference frame? Which inertial frame in SxR might have an observer that wouldn’t agree to the space-time decomposition SxR?

Originally posted by Hurkyl
But it is by no stretch of the imagination the only "reference frame" that admits an S*R decomposition, and there are certainly other frames that don't admit the decomposition.
Same question.

Originally posted by Hurkyl
Anyways, the notion of a universal reference frame in a periodic universe has its problems. Not only do two different spatial coordinates refer to the same place in the universe
The patchwork nature of a manifold is just the nature of mathematics, over which I have no control. However, there is a well-defined global law of light propagation that works perfectly well. Consequently, a global time order exists.

Originally posted by Hurkyl
but for any particular photon you received, you don't know how many times it wrapped around the universe before reaching you.
Do you not believe in controlled gedanken experiments?

Originally posted by Hurkyl
Why bother converting my local time into universal time to do analysis when I can just do the calculations in my local reference frame?
It’s impossible to prove global theorems about time order with an insufficient array of synchronized clocks.

Originally posted by Hurkyl
I have another objection to your website, but it's late and I'm tired and may not have read it through fully... when you use the formula:

ds = ds0 * sqrt(1 - (v/c)2)

you justify it via the EEP.. but the EEP says that the laws of special relativity hold in locally flat reference frames. Since the Schwartzchild metric is flat only at infinity, I don't think you're justified in using this formula.
Every manifold is locally flat. SR applies locally at every point in GR according to EEP.
 
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  • #7
Yah, I was mixing up my terms. That'll teach me to post on differential geometry 4 hours past by bedtime! :smile:


Everywhere I said "flat", I meant "diagonal metric".

For any point on a manifold there is a reference frame where the metric is locally diagonal, but in general it's not so. The formulae of special relativity only hold where the metric is diagonal, that's why I think you're not allowed to use that one formula.
 
  • #8
Originally posted by Eugene Shubert
Of course God plays dice with light and matter. God not only plays dice with the universe, —He cheats.


Do you have some empirical data supporting this claim?
 
  • #10


Originally posted by RSM1000
Do you have some empirical data supporting this claim?
During all my communications with G_d I never thought to ask Him for any empirical data. And He never brought up the subject. I suppose it’s not important.

http://www.everythingimportant.org
 
  • #11
Yah, that's a big fat duh when I look at the equation. Since I'm obviously misremembering something, I'll hold off on the black hole discussion until I figure out why I was thinking what I did.



Anyways, back to SxR.

I'm not sure if you noticed, but you assumed that the SxR decomposition holds in both Σ and Σ'. Allow me to demonstrate how that assumption fails and how it resolves the paradox of the great illumination.

(I hope my directions for drawing pictures is up to par)


Start off by drawing a long rectangular vertical strip on a sheet of paper, maybe 2 inches wide. We shall topologically identify the pairs of points on the left and right sides that are on the same horizontal line so that our strip is a representation of SxR. (at least it would be if the strip is infinitely long. The length of the paper is sufficiently infinite for our problem)

Draw a horizontal line across the cylinder somewhere near the middle or possibly near the bottom) of the paper. This is the t=0 line for Σ.

For directions, north on the page is up and east is right.

Label the intersection of the t=0 line with the left side 'A'. (Of course, the intersection on the right side must also be 'A') From 'A', draw a north-by-northeast line that makes a 15 degree angle with the side of the rectangle. 15 degrees isn't magical, it just has to be less than 45 degrees... my picture was drawn at 15 degrees and it seemed to work out nicely both for ease of drawing and for keeping the later steps clear.

This line will represent the trajectory of the primed observer. (In other words, the x'=0 line)

Now, draw an east-by-northeast line from A that makes a 15 degree angle with the horizontal line. This is the special relativistic line of simultaneity in the primed coordinates. (In other words, the t'=0 line). "Wrap around" this line so it intersects with the x'=0 line. (I presume you know what I mean by wrap around, but for the sake of completeness I shall explain it. Because the left and right sides are identified, the intersection of the t'=0 line with the right side is the same point as the one on the left side displaced horizontally. Continue the 15 degree line from there until it intersects the x'0 line)


So now we've established the axes of two different intertial frames in our universe. Now, let's release the two photons.

Draw two 45 degree lines from point A one going northeast and one going northwest. (Of course, the northeast one starts at the left representation of A and the northwest one starts at the right representation of A) These lines are the paths of the two photons. They should both meet back up at the x=0 line at the same point (the two identified points, of course). We have now identified the event of the great illumination. Draw lines parallel to the t'=0 line from the great illumination so we can get the line of simultaneity in the primed coordinates.


Now, let's draw the picture aligned with the primed coordinates on another part of the paper. You have drawn a parallelogram that wraps around the cylinder whose sides are those 15 degree lines. That should be a rectangle in the primed coordinates. Draw that rectangle. Mark A's left hand representation at the lower left corner of that rectangle. Note, however, that the lower right corner is not identified with A. (This is the failure of the SxR decomposition assumption) Let's find out where the right hand representation of A should be. If you follow the t'=0 line around the cylinder, you eventually hit the x'=0 line. That point is the same as the lower right corner of our rectangle. Mark it as B. Since B lies on the x'=0 line, you should be able to (roughly) place where it should go on the rectangle on the left hand side. The vertical displacement between A and B should be the same on both sides, so now we know where the right hand representation of A should go.

Now, let's draw the photons' trajectories again. Draw a 45 degree northeast line from the left hand A. It should intersect the top of the rectangle roughly 30% of the way from the top right corner. Now, draw the 45 degree northwest line from the right hand representation of A. It will intersect the left hand side someplace, depending on how you've drawn your figure. Now, it has to wrap around with the same offset as we've seen with B and A. Continue the 45 degree line until it meets the other photon at the great illumination on the top of the rectangle.


It should be clear now why there's no paradox. The reason we thought there was a paradox was because the northwest photon crosses the edge of the cylinder once or twice, and we didn't realize that wrapping around the cylinder induced a time displacement. Now that we've recognized the time displacement, it's easy to see that the northwest photon had more coordinate time in which to travel, easily explaining why it could cover a greater coordinate distance.



To summarize:

In Σ, (0, t) = (1 revolution, t)
In Σ', (0, t) [x=] (1 revolution, t)...
instead, (0, t) = (1 revolution, t + δt) for some δt


Another way to see the paradox resolved is to resist the urge to clip the coordinates just because we happen to be living on a circle.

In Σ, assuming the perimeter of the universe is 1 and c is 1, the great illumination happens when the two photons coincide at:

(-1, 1) and (1, 1)

Applying the Special Relativistic Lorentz transformations to yield the points of coincidence in &Sigma' gives:

(-γ*(v + 1), γ*(v + 1)) and (γ*(1 - v), γ*(1-v))

Both permitting the same value for the speed of light.
 
  • #12
The primary focus of this thread

The first post states:

I will demonstrate how an absolute time order follows from the laws of physics in Einstein’s universe. …

The mathematical details continue in this link:

http://www.everythingimportant.org/relativity/simultaneity.htm
The referenced link states my argument with even greater clearly:

The issue I’m raising is that of time order.
In special relativity, it’s meaningful to say that there are events E1 and E2 in spacetime such that for observer1, event1 comes before event2 but for observer2 event1 comes after event2. (E1=event1, E2=event2). However—and don’t let this alarm you—spacetime is sufficiently different in a spatially closed and bounded universe. In this instance, as my global theorem reveals, there is an absolute past, present and future.
The first post states explicitly

Einstein’s greatest blunder in science was his stubborn, unrealistic faith in a deterministic universe. His belief in a mechanistic interpretation for all natural law is widely recognized as a direct denial of quantum physics and the Hebrew Bible. Einstein would express his faith by saying, “Gott wurfelt nicht!” (God does not play dice!) Of course God plays dice with light and matter. God not only plays dice with the universe, —He cheats. (I don’t mean to review the philosophical/religious underpinnings of quantum mechanics in this paper).
 
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  • #13
Originally posted by Eugene Shubert

Einstein’s greatest blunder in science was his stubborn, unrealistic faith in a deterministic universe.
According to Eistein, his greatest ever blunder in science was the Cosmological Constant.
 
  • #14
Eugene - Who are you to decide Einsteins "greatest blunder"?

He proposed a deterministic universe. Now we have proof otherwise. Perhaps in the future we'll obtain more fundamental proof it is indeed deterministic.

Who are you to say the facts of now are the facts of always?

That's a very anti-scientific comment.
 
  • #15
Originally posted by RSM1000
Eugene - Who are you to decide Einstein’s "greatest blunder"?
Who are you to say the facts of now are the facts of always?
I’m a spiritual man. The spiritual man makes judgments about all things, but he himself is not subject to any man’s judgment. 1 Corinthians 2:15.

Originally posted by RSM1000
He proposed a deterministic universe. Now we have proof otherwise.
You are right. So let’s move on and debate Einstein’s second greatest blunder.

Originally posted by RSM1000
Perhaps in the future we'll obtain more fundamental proof it is indeed deterministic.
Perhaps in the future you’ll believe in God.
 
  • #16
Originally posted by Eugene Shubert
I’m a spiritual man. The spiritual man makes judgments about all things, but he himself is not subject to any man’s judgment. 1 Corinthians 2:15.
Heck, you provided a religious text reference! I guess that makes it true, eh?

Originally posted by Eugene Shubert

Perhaps in the future you’ll believe in God.
Perhaps in future you will give up on organised superstitions.
 
  • #17
Originally posted by Hurkyl
I'm not sure if you noticed, but you assumed that the SxR decomposition holds in both Σ and Σ'. Allow me to demonstrate how that assumption fails and how it resolves the paradox of the great illumination.
Hurkyl,

I appreciate you calling my riddle “the paradox of the great illumination.” If there’s a chance that I can figure out what you’re attempting to say mathematically, why not just explain in simple language what happens when my gedanken experiment is performed. How complicated must the descriptive language be in a universe that’s just a circle?

Originally posted by Hurkyl
It should be clear now why there's no paradox. The reason we thought there was a paradox was because the northwest photon crosses the edge of the cylinder once or twice, and we didn't realize that wrapping around the cylinder induced a time displacement. Now that we've recognized the time displacement, it's easy to see that the northwest photon had more coordinate time in which to travel, easily explaining why it could cover a greater coordinate distance.
Permit me to ask a question. Each inertial observer should think of his frame of reference as stationary. In these stationary frames, each observer should be able to measure the distance around the universe. In a stationary frame, for any line segment, isn’t the distance from right to left the same as left to right?

http://www.everythingimportant.org/relativity/simultaneity.htm
 
  • #18
Pictures are worth a thousand words! (Evidenced by my requireing a thousand words to try and describe how I drew the picture!) I was trying to describe how to draw the space-time diagram for the problem.


In a stationary frame, for any line segment, isn’t the distance from right to left the same as left to right?

Yes. But in Σ', the line segment for clockwise transversal is a different segment than counterclockwise transversal. (Still the same length though).

In Σ', if you transverse the universe clockwise and then shift your x-coordinate back to 0, you have to simultaneously shift your t-coordinate forward δt. If you transverse the universe counterclockwise and shift back to 0, you simultaneously shift your t-coordiante backward δt. The clockwise traveling photon and the counterclockwise traveling photon are precisely two revolutions apart at the great illumination, so one of them has been traveling 2δt longer than the other one, allowing it to transverse a greater distance.

I shall try to demonstrate this mathematically.


Let's start by describing &Sigma's frame.

Let us suppose that the universe's perimeter is 1, so that the point (0, t) = (1, t) for all t. (the essense of SxR)

There are 4 objects in question that are all simultaneuosly at the point (0, 0). They are the unprimed observer, the primed observer, the eastward photon, and the westward photon. The trajectories for 3 of them are:

unprimed observer: (0, ζ)
eastward photon: (ζ, ζ)
westward photon: (-ζ, ζ)

Let us select a particular trajectory for the primed observer. How about:

primed observer: (ζ/2, ζ)
The primed observer's trajectory is, by definition, the line for which x'=0

It is a theorem of Special Relativity that when drawing space-time diagrams, the lines of simultaneity (trajectories with constant time-coordinate) for any (inertial) observer are simply the reflection of his trajectory across a photon trajectory. In particular, this means that the equation for the line of simultaneity for our primed observer that containes (0, 0) must be:

t'=0 equation: (ζ, ζ/2)

Now, let us find where the t'=0 line meets back up with the x'=0 line after one revolution around the universe. The equation is:

(ζ/2 + 1, ζ) = (ξ, ξ/2)

The solution is ζ = 2/3, ξ = 4/3, leading to the intersection point (4/3, 2/3), which is identified with the point (1/3, 2/3) on the x'=0 trajectory by the nature of the SxR decomposition of Σ.

From this we can find the distance around the universe in Σ the length of the line in Σ from (0, 0) to (4/3, 2/3) is 2/sqrt(3), so this must be the x-displacement observed in Σ' for a trip around the universe.

However, in the primed coordinates this trip around the universe did not take you back to (0, 0) (in Σ), it took you back to the point (1/3, 2/3) (in Σ). We can compute the length of that line too, and come to the following conclusion:

In Σ', the point (2 / sqrt(3), 0) is the same as the point (0, 1 / sqrt(3)), so to bring yourself back into sync after circumnavigating the universe to the right, you must add the vector:
(-2 / sqrt(3), 1 / sqrt(3))
to you coordinates.

That is, unlike Σ (SxR) where if you take a trip around the universe you just correct your x coordinate to resynchronize, in Σ' you have to adjust both your x coordinate and your t coordinate to resynchronize. (Failure of SxR decomposition)


The trajectories of the two photons in primed coordinates must, of course, still be:

Eastward: (ζ, ζ)
Westward: (-ζ ζ)

And to compute the event of the great illumination, we add the vector displacements to adjust for the circumnavigation of the universe to bring them back into sync with the primed observer:

Eastward: (ζ - 2 / sqrt(3), ζ + 1 / sqrt(3))
Westward: (-ξ + 2 / sqrt(3), ξ - 1 / sqrt(3))

Solving gives &zeta = 1 / sqrt(3), so the great illumination occurs, in Σ', at (-1 / sqrt(3), 2 / sqrt(3))


However, if we don't clip the photon's paths, we see that the Eastward photon went:

from (0, 0) to (1 / sqrt(3), 1 / sqrt(3))

And the Westward photon went

from (0, 0) to (-3 / sqrt(3), 3 / sqrt(3))


The coordinate displacement of the westward photon was greater, but it traveled further through coordinate time as well. We thought there was a paradox because we didn't anticipate that Σ' required both a temporal correction and a spatial correction to synchronize frames after a circumnavigation, but once we recognize that, everything's fine again.
 

1. What is the main argument of "Einstein's Religion: Mistakes in Physics & Time Order"?

The main argument of "Einstein's Religion: Mistakes in Physics & Time Order" is that Albert Einstein made mistakes in his theories of physics, specifically in his concept of time order, which has led to misconceptions and misunderstandings in the scientific community.

2. How does "Einstein's Religion" challenge traditional beliefs in physics?

"Einstein's Religion" challenges traditional beliefs in physics by questioning the validity of Einstein's theories and proposing alternative explanations for phenomena such as time order and gravity.

3. What evidence does "Einstein's Religion" present to support its argument?

"Einstein's Religion" presents evidence from Einstein's own writings and equations, as well as from other scientists and experiments, to support its argument that Einstein made mistakes in his theories of physics.

4. Does "Einstein's Religion" suggest that Einstein's contributions to science are invalid?

No, "Einstein's Religion" does not suggest that Einstein's contributions to science are invalid. Rather, it argues that his theories should be re-examined and potentially revised in light of new evidence and alternative explanations.

5. What impact could "Einstein's Religion" have on the scientific community?

"Einstein's Religion" could potentially have a significant impact on the scientific community by challenging long-held beliefs and theories and encouraging further exploration and experimentation in the field of physics.

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