Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Einstein’s Religion

  1. May 23, 2003 #1
    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:

    Last edited: May 25, 2003
  2. jcsd
  3. May 26, 2003 #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.
  4. May 27, 2003 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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. May 27, 2003 #4
  6. May 27, 2003 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
  7. May 27, 2003 #6
    I thought alis had explained that to you.


    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?

    Same question.

    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.

    Do you not believe in controlled gedanken experiments?

    It’s impossible to prove global theorems about time order with an insufficient array of synchronized clocks.

    Every manifold is locally flat. SR applies locally at every point in GR according to EEP.
    Last edited: May 27, 2003
  8. May 27, 2003 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
  9. May 27, 2003 #8

    Do you have some empirical data supporting this claim?
  10. May 27, 2003 #9
  11. May 27, 2003 #10
    Re: Re: Einstein’s Religion

    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.

  12. May 27, 2003 #11


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
  13. May 27, 2003 #12
    The primary focus of this thread

    The first post states:

    The referenced link states my argument with even greater clearly:

    The first post states explicitly

    Last edited: May 27, 2003
  14. May 27, 2003 #13
    According to Eistein, his greatest ever blunder in science was the Cosmological Constant.
  15. May 27, 2003 #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.
  16. May 27, 2003 #15
    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.

    You are right. So let’s move on and debate Einstein’s second greatest blunder.

    Perhaps in the future you’ll believe in God.
  17. May 27, 2003 #16
    Heck, you provided a religious text reference! I guess that makes it true, eh?

    Perhaps in future you will give up on organised superstitions.
  18. May 27, 2003 #17

    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?

    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?

  19. May 28, 2003 #18


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.

    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 travelling photon and the counterclockwise travelling photon are precisely two revolutions apart at the great illumination, so one of them has been travelling 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook