Space-time is a reality

Discussion in 'Physics & Math' started by Fork, Sep 28, 2013.

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  1. chinglu Valued Senior Member

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    Within this "space" and I mean metric space, it is a necessary condition of the "metric" or distance formula that d(x,y) = 0 iff x=y.
    http://en.wikipedia.org/wiki/Metric_space

    However, on a light sphere, it is the case that the Minkowski metric claims all points satisfy d(x,y) = 0 and yet the 4D vectors on that light sphere are not equal.

    You need to step from metric space to topological space to manifold and yet this cannot be done unless you have a valid metric space.

    It is therefore the case that space-time is not a valid metric space.
     
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  3. rpenner Fully Wired Valued Senior Member

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    First of all, nowhere in this whole thread do I see the word "metric" used before your chinglu's most recent post. "Metric" is most often used in general relativity work where it is short-hand for "the indefinite bilinear form used to raise and lower indices when doing tensor calculus on a pseudo-Riemannian manifold." Since I have restricted the majority of my discussion to special relativity, I found no need to use the term. In fact, I went out of my way to avoid it, specifically calling out that what is analogous to distance and dot product in Minkowski space is not exactly the same thing as in Euclidean space.
    (Emphasis added.)

    So not only is chinglu committing a etymological fallacy where he attempts to substitute one particular definition of the word "metric" for the definition used by physicists everywhere, but I haven't relied upon the term so he invents an objection out of whole cloth. We have too many people skilled at logical reasoning, math or physics for this baseless argument to gain any ground.

    I mean, if chinglu want to be SR's equivalent of a grammar Nazi then the burden is on him to actual master formal reasoning and mathematics, topics which have given him much problems in the recent past. And to object to actual use or misuse of language instead of inventing objections.

    Obviously, you are using the wrong definition of metric. Wikipedia, like a dictionary, is descriptive and not prescriptive, and here they choose to describe a definition of metric which applies to Euclidean geometry and Riemannian manifolds. In Wikipedia terms, Minkowski space is a pseudo-Euclidean space just like GR uses pseudo-Riemannian manifolds. It's a more advanced mathematical concept, 20th century math rather than 18th or 19th century math.

    But the analogy between pseudo-metrics and metrics is a close one and in Minkowski space, given a time-like vector you can induce a 1-D metric space in the direction of the vector and a 3-D metric space in the spatial directions orthogonal to the time-like vector. Wikipedia agrees that these are close analogies.


    Your reliance on one particular math is not justified by experiment. My embrace of a different math is justified by experiment since 1859. You cannot win this argument with a dictionary or reliance on particular mathematical axioms -- you have to use the axioms and definitions I use or you aren't even in the game.

    Words have more than one meaning. Here I follow the convention of physicists talking to physicists and refer to the bilinear form of flat space-time, \(\eta\), as a metric even though if I was talking to a mathematician inexperienced in SR or GR I might make the concession that I am talking about a pseudo-metric. As you, chinglu, have taken upon yourself to oppose my mathematics, the burden is still on you to demonstrate their physical invalidity or internal inconsistency. As you, chinglu, have done this for quite some time you aren't allowed now to object to a trivial matter of jargon which I nowhere used in this thread. Nor given your extensive post history can you be easily forgiven for failing to grasp the sense in which other people use the terms. Lots of textbooks spell out that they use "metric" for "pseudo-metric" to save on the cost of ink and the wear and tear on readers' eyeballs.
     
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  5. chinglu Valued Senior Member

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    Let me correct you.

    I said the logic over and over that space-time is not a space.

    Now, in mathematics, what is a space?

    Primarily, it is a metric space. In particular, Euclidean geometry is a metric space. I just assumed you understand since any college freshman in math would know what that means.

    So, I am sorry that you have some invalid concept of "space" in mathematics and that concept is not a "metric space". Now, if you would like to present your pseudoscience concept of "space", please have at it.

    Now, let's get back on task.

    The burden is on you. I am presenting the mainstream mathematical definition of space. You are not.

    It is on you to demonstrate that coordinates on the same light sphere have the same distance 0 from the origin "using the Minkowski metric" in the space, yet they are different coordinates and this is also a valid mathematical space.

    Show me the mathematics.
     
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  7. arfa brane call me arf Valued Senior Member

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    Someone's wrong on the internet again.
     
  8. rpenner Fully Wired Valued Senior Member

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    I did. You are not arguing that Minkowski space is not a topological space, not a vector space, not an affine space or not a pseudo-Euclidean space. You are arguing that it isn't Euclidean, which is already covered by the concept of non-Euclidean geometry so you are arguing something that has been understood since 1908 when Minkowski a mathematician, demonstrated it was a perfectly fine vector space.

    You can't argue with a definition, which is what you are attempting to do by claiming that the definitions of space for "Vector Space", "Affine Space" or "pseudo-Euclidean space" or "Minkowski space" are "not good enough." Even if you say you are familiar with a definition of "length" that suggests to you that the universe is Euclidean and not Lorentzian, I can show that given any time-like vector you can induce a spatial metric on a 3-dimensional slice of the universe that is simultaneous to any given "now" and a temporal metric on displacement from that slice. But since different time-like vectors induce different 3-dimensional slices one expects these notion of "length" and "coordinate time" to not be fundamental and thus time dilation and length contraction are consequences of the essentially Lorentzian nature of our universe.

    So all your objections boil down to "You claim space-time is not Euclidean" which has been the experimental picture since 1859, the theoretical picture since 1905.
    Since then, we have the work of Poincaré (1905), Minkowski (1908), Einstein (1916), Pauli (1928, 1958) that build on the work of Felix Klein (1872) to show that Special Relativity is not only fine mathematics but mathematics about a Klein geometry. Your insistence on using the wrong axioms, as before, leads you into error.

    Even if this were a discussion of abstract math and not physics, you are wrong to criticize a body of work which uses different axioms than you would use. You don't get to pick which axioms I work with -- at best you choose to work with the same axioms to prove my system of axioms is inconsistent. When I talk about special relativity I am using the axioms of Minkowski space and therefore you don't have any right to try and force a different set of definitions or axioms.
     
  9. chinglu Valued Senior Member

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    This post indicates you do not understand the nature of a metric space. I am not talking about a Euclidean space, which is a special case, I am talking about a general metric space.

    Rn is an example of a valid vector space and a valid metric space.

    See, the whole point of a space is that a point represents a unique address in the space. For example, if two folks had the same phone number, then that is a failure of logic. Also, if two websites had the same IP address, then is also a failure.

    So, the whole point of a metric space is that given the axioms, each vector is unique in that space and therefore has a unique address.

    However, the Minkowski space does not satisfy these basic conditions of a metric space. An address or vector in that space maps to an infinite number of locations on any given light sphere.

    So, in the Minkowski non-space, given a vector, I can find an infinite number of possibilities of where that vector may be in the space and I cannot find that vector in that space.

    Now, to simplify this ever further, a function must map 1 value to another value. If you can map the same value to multiple locations, then it is non-functional under the rules of logic.

    Therefore, the Minkowski non-space is non-functional under the rules of logic.

    So, prove to me that the Minkowski non-space is a topological space and a manifold.

    Again, I am trained to use any set of axioms as long as they are functional. The failed Minkowski non-space cannot map an arbitrary vector into a unique position in the space. Hence, it is not a valid logical space.
     
  10. arfa brane call me arf Valued Senior Member

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    A 'space' equipped with a metric has a 'precise' definition of distance between points, regardless of their "address". The metric also means the object that defines or measures a distance.
    Yes it does, it's called the Minkowski metric, loosely the 'separation' between events.
    That's almost funny.
    . . .
     
  11. rpenner Fully Wired Valued Senior Member

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    I disagree and you have not given me a reason to agree. You can't even stick to one definition of space and so you don't make a clear point.
    I disagree. \(\mathbb{R}^n\) doesn't have a metric until you specify one. You are assuming the Euclidean metric \(\sqrt{\sum_{k=1}^{n} \left(\Delta x_k \right)^2 }\) but so far have failed to rule out the pseudo-metric \(\left(\Delta x_n \right)^2 - \sum_{k=1}^{n-1} \left(\Delta x_k \right)^2 \) which induces the Euclidean metric \(\sqrt{\sum_{k=1}^{n-1} \left(\Delta x_k \right)^2 }\) on any slice with fixed value of \(x_n\). The Euclidean metric is unsuitable for 4-dimensional space-time because it would allow treating time exactly the same as any spatial direction, which is not observed; however the pseudo-metric you object to without cause allows us to induce a Euclidean metric in any slice of simultaneity and to induce a 1-D Euclidean metric along any time-like direction.

    But \(\mathbb{R}^n\) is not a metric space until you specify a metric.

    That cannot be the issue since Minkowski space uses a 4-dimensional Cartesian coordinate system just like a Euclidean 4-dimensional space.

    No it doesn't. You are just advertising your ignorance and lack of experience with Minkowski space. You are also confusing if you are talking about an address or a vector which is modeled as a difference of addresses. This means we don't know if you are talking about a vector space (which has a well-defined origin) or an affine space (which has no concept of an origin).

    It's like you can't do geometry at all, but that is not a failure of Minkowski space but merely a statement of your inability.

    A function must map one element of its domain to exactly one element of its range. Famously, the Lorentz transforms and Euclidean rotation maps are functions -- indeed, they are one-to-one functions so their inverse mappings are functions. Specifically, there are unit vectors in Minkowski space such that any address turns into location via \(\mathcal{X}(x_1, x_2, x_3, x_4) = \sum_{k=1}^{4} x_k \hat{e}_k\) which is as functional as any function can be.

    //Edit:
    That's \(\mathcal{X}(x_1, x_2, x_3, x_4) = \mathcal{O} + \sum_{k=1}^{4} x_k \hat{e}_k\) for an address or \(\mathcal{X}(x_1, x_2, x_3, x_4) = \sum_{k=1}^{4} x_k \hat{e}_k\) for a vector but the distinction is blurred when \(\mathcal{O}\) takes on the properties of 0 as it does in a vector space.​

    I never claimed it was a manifold -- I claimed it was pseudo-Riemannian manifold.

    But this can be done if you consider first the hyperbolic neighborhood of a point a with "radius" r. Thus this is all points with \(\left( <x-a, x-a> \right)^2 < r^2\). Then we choice as our basis all the non-empty intersections of two or more of these hyperbolic neighborhoods where coordinate values don't run off to infinity. This it turns out results in a topological space on Minkowski space where the basis sets have the property of being Lorentz-invariant. Further, this is a topology we are well-versed in as it is the Euclidean topology.
    http://mathpages.com/rr/s9-01/9-01.htm

    And for a pseudo-Riemannian manifold, Minkowski space is the canonical example since the whole point of a pseudo-Riemannian manifold is to show that locally it approximates a Minkowski space and nothing approximates a Minkowski space like a Minkowski space.

    Lies! You don't have the math credentials to do multivariate calculus, so you can't follow logical axioms to theorems.
    Lies! (As demonstrated above.)
    That does not follow, so you have perverted the term "logic" by your use here.
     
    Last edited: Oct 12, 2013
  12. brucep Valued Senior Member

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    Great post. The good thing is: Not everybody have ears of stone.
     
  13. brucep Valued Senior Member

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    chinglu is the poster child for the intellectually dishonest. The only good thing about his troll is responses such as yours.
     
  14. chinglu Valued Senior Member

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    First, let's get some opinions from the mainstream.

    http://math.ucr.edu/home/baez/gr/metric.html

    So, as we can see, mathpages agrees the "distance" function for space-time as a metric space is the Minkowski metric.

    Also, to make sure you understand space-time claims to be a valid metric space, we have the following link which indicates the Minkowski metric is for space-time as is the normal 3-D metric is for 3-D Euclidean space.

    http://mathworld.wolfram.com/MinkowskiMetric.html


    Now, I have already proven that the Minkowski metric does not satisfy the rules of a metric space since for any point on the light sphere d(x,o) = 0 where x is on the light sphere and o is the origin. However, in a valid metric space, d(x,y) = 0 iff x = y. And obviously since x != o, with d(x.o) = 0, therefore we have an invalid metric space as I already explained.

    This is the part you have failed to handle.
     
  15. chinglu Valued Senior Member

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    I will give you more information since you quoted mathpages.

    http://www.mathpages.com/home/kmath413/kmath413.htm

    Make sure you understand what the above means before you post.
     
  16. paddoboy Valued Senior Member

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    Oh the Irony of it all!
     
  17. Trapped Banned Banned

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    I don't really know who Chinglu is, but the hypocrisy in this statement is astounding, considering you claim to know bullshit when you see it, but incapable of knowing the basics of redshift.

    Glass houses and all that.
     
  18. paddoboy Valued Senior Member

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    That one paragraph just about sums chinglu up and is so patently obvious, that it sticks out like the proverbial dogs balls.
     
  19. paddoboy Valued Senior Member

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    Giving a simple layman's view on space, time, and space/time, here is how I see it.....
    Space is what separates everything....If we had no space, everything would be together....It also has three dimensions, L B H
    Time separates events...If we had no time, [as of the exact instant of the BB,] everything would be together.......

    Space/time is the three dimensions of space, and the single dimension of time interwoven into a framework against which the calculations of GR are decyphered/worked out.
    It also exists in reality as is shown by GP-B and other probes, and the warping curving and twisting [Lense Thirring Effect] that has been shown to present itself in the presence of matter/energy.

    Would our online experts agree with those simplistic non mathematical definitions?...Any errors, alterations or corrections?
     
  20. rpenner Fully Wired Valued Senior Member

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    Both "metric" and "measures" are in scare-quotes to indicate non-standard usage. For example, on that same page, John Baez writes:
    So he is talking about a Lorentzian pseudo-metric. He also uses the term Minkowski space which shows my usage of the term is mainstream and your objections are baseless.

    Also I have restored the context to your equation and John Baez is not the author of mathpages, a different web site. Nowhere did John Baez use the definition of metric space in talking about Minkowski space but uses scare quotes around metric because as he says on the overview page he knows the properties of his generalization of an inner product has different properties than a Euclidean metric. Your page is linked to from this following overview.
    http://math.ucr.edu/home/baez/gr/outline2.html#metric

    This was your spurious objection. I have never confused the two.
    Yes, it is used in an analogous manner, but that doesn't mean the authors of that page are calling Minkowski space other than a pseudo-metric space.
    (emphasis added)​
    So far every source you have brought agrees with my usage and none yet supports your vague idea that mainstream physicists are deluding themselves that Minkowski space is a Euclidean metric space when it is not. But you aren't in a position to claim physics are using the wrong mathematics when you haven't even argued that point -- at best you have simply asserted that only Metric Spaces are valid, so you have failed the debate because you have not argued for your position and have instead brought in evidence that plenty of other mathematical objects are described as spaces thus weakening your claim before you start to support it.

    But Metric Space is nowhere the description of space-time. Every source of physics, including me, gets this straight and point out that the Minkowski space ignores some of the axioms of a metric space, and this relaxation is specifically called out every time they describe the space or the metric as a generalization of metric space or metric, respectively.

    Or rather the point you have missed in [post=3119113]#22[/post].

    All it said it that space-time is continuous and has an associated topology which is destroyed with a non-continuous transform. Since the Lorentz transform is linear your quote is off-topic.
     
  21. chinglu Valued Senior Member

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    1) John Baez claimed the Minkowski metric is a metric just like the Euclidean metric or distance formula.

    2) But, what is the point of a metric? It measures distances.

    3) As I have shown over and over, the Minkowski metric shows all points on the light sphere are equivalent to the origin.

    4) I then gave the mathpages logic which proves the Minkowski metric does not induce a metric space. To wit.

    On the other hand, even the Lorentz transformation leads to a non-Hausdorf topology if we took the invariant spacetime interval as the measure of proximity, because of the singularity of this measure along light rays.

    A non-Hausdorf topology is not a metric space and the statement above says everything I have been saying, you cannot tell the difference of any point on the light sphere from the origin. Since any light sphere will eventually traverse any point in the space, then topologically, all points in the space look like the origin. That is the mainstream.

    5) I also claimed the Minkowski metric does not induce a manifold. Again, let's quote mathpages.

    Of course, unlike the Lorentz transformations, a permutation of light cones does not preserve the identity of all worldlines, it preserves only the emitting wordline and those parallel to it. In general, the usual objection to discontinuous transformations of the spacetime coordinates is that they violate the topology of the manifold, in the sense that the open sets and neighborhoods of the manifold are not preserved.

    What this is saying, if you attempt to map an interval of time from one frame of the light cone to another frame, you do not preserve the light cone in the mapping. The result is not a light cone in the other frame, therefore, you have violated the topology of the manifold since you do not end of with concentric light spheres in the other frame.

    In other words, if you take some neighborhoods of the mapping, they will not look like a neighborhood that satisfies the light postulate in the other frame.


    So, let's summarize.

    Space-time is not a metric space as I have said and further, it is not a Hausdorf topology which any reasonable person would agree reality must be. More specifically, space-time claims any coordinate in the space looks like the null vector.

    Next, mapping the light cone violates the topology of the manifold. By considering all points that are acquired by the light sphere during a time interval, the resulting mapping of all those coordinates by LT does not look anything light concentric light spheres. So, local local neighborhoods in the mapping do not look anything like the original manifold. More specifically, some local neighborhoods do not satisfy the light postulate of a light pulse. Thus the manifold is not preserved.

    You are left with nothing.
     
  22. rpenner Fully Wired Valued Senior Member

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    It is your burden to prove this as I have only so far found places where he was making an analogy.
    In a metric space it is used to compute distance. In a pseudo-metric space it is used to compute the interval. This is an analogy not a synonym.
    No, while the metric of a metric sub-space can be used as an equivalence relation of points projected into that subspace because d(P(A)-P(B)) = 0 implies d(P(C)-P(A)) = d(P(C)-P(B)), the Minkowski "metric", being a pseudo-metric does not have that property. Thus <B-A,B-A> = 0 and <C-A,C-A> = 0 does not mean <C-B,C-B> = 0 so the pseudo-metric does not imply an equivalence relation exists between the points of a light cone.
    No one, not me, not John Baez, not any mainstream physicist, not any mainstream mathematician says it does. Most of us go out of our way to say it doesn't, so you get no credit for raising this point.
    Kevin Brown is clearly saying that any discontinuous transformation of any manifold destroys properties associated with the topology of that manifold. He is also saying that these sorts of discontinuous permutations are not the type of transform demonstrated by the Lorentz transform. From: http://www.mathpages.com/home/kmath413/kmath413.htm
    I'm quite certain that isn't close to what Kevin Brown was writing in that he was distinguishing a continuous light-cone preserving transform like the Lorentz transform or a rotation from a discontinuous permutation that manages to preserve the light cone.


    • You haven't raised any math objections internal to the mathematics of special relativity, you have simply asserted that so-and-so isn't good enough.
      • You assert that Minkowski space doesn't have a natural Hausdorf topology but fail to state a reason why it must.
      • You assert that Minkowski space violates the axioms of a metric space but fail to identify a single source on special relativity relying on the axioms of a metric space.
      • You claim that the space-time interval is an equivalence relation between points which is not true -- the space-time interval is preserved by the Lorentz transform but that fact alone is not enough to say that equivalence exists between the apex of the light cone and a point distinct from the apex. For example, the Lorentz transform preserves orthochronosity of events within the light cone, and the interval does not.
    • You haven't raised any physics objections at all.

    Ah, but that was a mapping distinct from that of the Lorentz transform.
    Yes, a light cone transforms as a light cone and a slice of a light cone transforms as a slice of a light cone, but that doesn't mean that a hyperplane of simultaneity transforms to a hyperplane of simultaneous events, so you fail in comprehension of the Lorentz transform.
    This you have not demonstrated and thus your claims are like the witless chirping of a bird.
    Except a reputation, the ability to read English, the ability to do math and the ability to do physics. While largely intangible, I submit that they amount to more than nothing.
     
  23. chinglu Valued Senior Member

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    You keep running on the burden thing.

    So, let's get specific. What is your definition of continuity along a light ray in the Minkowski space. That is the reason why the Minkowski space must be Hausdorf. Once you fail to produce a definition, tyhen perhaps you will understand any reasonable person would demand a definition of continuity in a "natural space".

    Next, a time slice of concentric light spheres in the rest frame does not produce a set of concentric light spheres in the moving frame. You claim the relativity of simultaneity justifies this.

    However, the LT's are supposed to mirror the actual physical reality of the moving frame. Yet, the LT's do not produce a set of concentric light spheres in the moving frame and therefore, LT does not mirror the physical reality of the moving frame given any time interval in the rest frame.

    Now, of course, you can parrot the relativity of simultaneity all you want, but the moving frame only views concentric light spheres and does not see the relativity of simultaneity.

    So you parroting is worthless.

    This proves SR does not correctly produce the physical reality of the moving frame and is therefore a false theory.
     
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