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What dimension does space-time curve in?

  1. Jan 13, 2016 #1
    Is not space curvature the curving or projecting into a higher dimension? Like a curved sheet of paper perceived by a two dimensional creature? The mystery seems to reside in our ape brains being unable to perceive (but not conceptualize) higher dimensions than three or relativistic, quantized time.
     
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  3. Jan 13, 2016 #2

    Nugatory

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    It is not. The curvature produced by bending a two-dimensional sheet of paper through the third dimension is called extrinsic curvature, and it does not affect the two-dimensional geometry of the surface of the paper. For example, if you draw a triangle on the surface of the paper, the sum of the interior angles of the triangle will be 180 degrees. When you bend the paper or even roll it up into a cylinder nothing will change and the angles will still add to 180.

    The curvature of spacetime is intrinsic curvature and it does not need an extra dimension. The effect of intrinsic curvature is directly visible to hypothetical two-dimensional creatures living on the curved surface without any reference to the third dimension. For example, on an intrinsically curved surface the interior angles of a triangle will not add to 180 degrees.
     
    Last edited: Jan 13, 2016
  4. Jan 13, 2016 #3
    So to put this in simple terms, what is curving if other dimensions are not involved? How is it possible to perceive a straight line as curved and remain in three dimensions?
     
  5. Jan 13, 2016 #4

    PeterDonis

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    It isn't. A line cannot have intrinsic curvature. In order to have intrinsic curvature, you have to look at a manifold with at least two dimensions--for example, a 2-sphere. Then your question can be rephrased as: how is it possible to tell that a 2-sphere is curved, without making any use of an embedding of it into a space with more than 2 dimensions? The answer to that is, by looking at geodesic deviation, which can be measured purely within the surface.
     
  6. Jan 13, 2016 #5

    Orodruin

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    This might be a bit steep for a B level thread so let me try to put it in more accessible terms.

    Let us imagine you start at location A and travel in a straight line for a distance d. After doing this you turn 90 degrees to your right and travel a distance d in a straight line again. Repeating two more times, you will end up back at A if your space is flat. If you do not end up at A your space is therefore curved. To visualise this, I suggest thinking about how this would work on a sphere when d is one fourth of the circumference.
     
  7. Jan 13, 2016 #6

    Nugatory

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    Imagine that your two dimensional universe is a sheet of thin rubber stretched out on a flat table. Now I can tug on the corners and stretch it, distorting shapes on the surface, turning circles into ovals, making it so the distance between non-intersecting lines is not constant, and all the other effects of intrinsic curvature - and we're staying in two dimensions.
     
  8. Jan 13, 2016 #7
    Orodruin and Nugatory,

    It seems like all paths that stay on the surface of a sphere are curves; a straight path will leave the sphere's surface and doing so traversing a square path will place you back where you started (the straight line path comprising the square route in a flat plane intersecting the surface of the sphere with one of the route square's corners); travels that stay on the surface of a sphere are curved, not straight).

    It also seems like when a flat surface is distorted by displacement within the plane and what were circles become ellipses, etc., those ellipses are not circles anymore; they just need to be redrawn. Walking a square of straight lines will return one to there source point both before and after the distortion - what would throw off the path is if the distortion was ongoing during the walk...

    Both analogies seem ineffective at breaking out of referring to all curves with respect to flat space. How does one eliminate flat space as the basis from which curvature is a departure?
     
  9. Jan 13, 2016 #8

    PeterDonis

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    Here you are assuming a definition of "straight" that is only valid for a flat space. On the sphere, a "straight" path (i.e., a geodesic) is a great circle.

    By not using a definition of "straight" that privileges flat space. If you use the correct definition of "straight"--that a "straight" line is a geodesic--then flat space becomes just one option among many.
     
  10. Jan 13, 2016 #9

    Nugatory

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    When you say "a straight path will leave the sphere's surface" you have to mean a straight path in three-dimensional space, not a straight path on the two dimensional space that is the surface of the sphere - otherwise, it wouldn't be be able to leave the surface. That's fine if you're thinking of the points making up the curved surface of the sphere as a subset of the points making up a flat three-dimensional space, and using the three-dimensional space's definition of "straight" as a relationship between points in that three-dimensional space. But if you're doing that, you're already assuming the existence of points outside of the two-dimensional space so of course you conclude that there must be another dimension for those points to live in.

    However, that assumption is unnecessary. The mathematical object that is a set of points that collectively behave like the two-dimensional surface of a sphere is a perfectly good mathematical object in its own right, with its own perfectly good definition of "straight" (intuitively, the path of a string stretched tight between two points, called a "geodesic"). The third dimension is only needed because you and I live in three-dimensional space so if we're going to build a physical object with those mathematical properties we have to build a three-dimensional globe and then say "Hey - only pay attention to the points on the surface". If we were flatlanders (if you haven't read E.A. Abbott's classic book "Flatland", find a copy) we'd understand our world just fine without needing that third dimension; we'd just observe the geometrical behavior of our world, select a metric tensor that matched that behavior, and let differential geometry take it from there. Likewise, as creatures living in four-dimensional spacetime we can describe intrinsic curvature of that spacetime completely without ever needing a fifth dimension.

    You can't. There's flat spacetime in the absence of gravity, and if you add gravity you get curvature and departures from that flatness - no getting away from that. But you don't need an extra dimension to curve spacetime through to produce those departures. You just need a geometry in which the metric tensor, which describes the distance between adjacent points, is different from the flat-spacetime metric tensor.
     
    Last edited: Jan 14, 2016
  11. Jan 14, 2016 #10

    Dale

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    You are thinking about a 3D flat embedding space. A sphere is a 2D manifold and there is an intrinsic geometry entirely within that 2D surface. The fact that the path bends in the radial direction in 3D is utterly irrelevant to this intrinsic geometry because the radial direction does not exist in the 2D geometry.
     
  12. Jan 14, 2016 #11
    Well, I tried looking around in Wiki...

    "For any natural number n, an n-sphere of radius r is defined as the set of points in (n + 1)-dimensional Euclidean space which are at distance r from a central point, where the radius r may be any positive real number."

    "It is an n-dimensional manifold in Euclidean (n + 1)-space."

    "In particular:...
    a 2-sphere is the two-dimensional surface of a (three-dimensional) ball in three-dimensional space."
     
  13. Jan 14, 2016 #12

    Nugatory

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    And stuff like this is the reason why wikipedia is not, in general, an allowed source here. The text you've quoted is wrong (it's confusing "X is Y" with "Y is an example of X") in a way that doesn't always matter - but it matters here. You'll find some mention of the problem about two-thirds of the way down the "talk" page for the article on "n-spheres" where those quotes come from.
     
    Last edited: Jan 14, 2016
  14. Jan 14, 2016 #13
    Thanks for the thoughtful replies. But I am having a problem moving forward from the theoretical to the perceived reality of my (ageing) ape brain. If I travel in an airplane between two points in three dimensions, taking the shortest possible route (straight line) experiencing gravity the entire trip, where is the curvature in three dimensional space?
     
  15. Jan 14, 2016 #14

    jbriggs444

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    You are making that trip in four dimensional space-time. The path is not "straight" in four dimensional space time. That is, it is not a "geodesic path". You can tell because you are accelerated -- your seat is pushing up on you all the way through the trip. The problem is that you are using your pre-relativistic picture of three dimensional space with an independent one dimensional time. That picture is not appropriate for explaining gravity in terms of curved space time.
     
  16. Jan 14, 2016 #15

    Nugatory

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    Gravity is the result of intrinsic (no fifth dimension needed) curvature in four-dimensional spacetime, not three-dimensional space.

    Search this forum for member @A.T. 's excellent video showing how an apple falling freely to the ground is following a straight-line path through curved spacetime, while an apple hanging from the branch so not falling is being dragged away from that straight path by its stem which is connected to the branch.
     
  17. Jan 14, 2016 #16

    A.T.

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    Here the mentioned simpler examples, of hovering and falling radially in a gravitational field:






    Tajectories that require more than 1 spatial dimension, are very difficult to visualize though. See this post for explanation on the different meanings of "curvature":

     
    Last edited: Jan 14, 2016
  18. Jan 14, 2016 #17
  19. Jan 14, 2016 #18

    PeterDonis

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    If the airplane is traveling between two points on the Earth's surface, and it travels at a constant altitude, then the path it takes, viewed in the 3-dimensional space in which the Earth is embedded, is not straight; it's curved. But it's the shortest possible path between the two points given the constraints (both points are on the Earth's surface and the plane keeps a constant altitude above the Earth's surface); the constraints basically amount to restricting the plane's path to a 2-sphere, the Earth's surface, and within that restricted domain, the plane's path is "straight"--a geodesic.

    Now suppose that we're talking about a spaceship, not an airplane, and it's traveling from Earth to Mars, and it takes the shortest possible spatial path in the 3-dimensional space in which the Earth and Mars are embedded. This path will be (at least to a very good approximation--space in the solar system is actually not perfectly Euclidean because of the Sun's gravity, but the effect is too small to matter here) a Euclidean straight line, the sort of thing you're used to describing as a "straight" path.

    However, as others have mentioned, this path will be "straight" only in space. It will not be straight in spacetime. Why? Because the spaceship will not be in free fall; it will have to accelerate (fire rockets) in order to maintain this straight path. The free-fall path between Earth and Mars would be an elliptical orbit that, spatially, looks curved. But in spacetime, it is the free-fall elliptical orbit that is straight (a geodesic), and the accelerated path (the one that looks "straight" spatially) that is curved. This is a manifestation of the fact that, in the presence of gravity, spacetime itself is curved; the free-fall elliptical orbit is a straight (geodesic) path in curved spacetime.
     
  20. Jan 14, 2016 #19
    Thanks for the answers. I have a much better understanding of the concept as a model or map which helps explain and predict outcomes. I won't pretend I understand it beyond that any more than I can visualize four dimensional objects. I take my (small) hat off to those who can.
     
  21. Jan 14, 2016 #20
    Is it entirely proper to answer the original question,
    What dimension does space-time curve in?
    by saying:
    If you consider space and time as a four dimensional geometry, gravity IS the curvature of those dimensions.
     
  22. Jan 14, 2016 #21

    jbriggs444

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    While true, it is side-stepping the question. As has been stated in this thread, intrinsic curvature is perfectly meaningful and definable even without an extra dimension to curve through. A direct answer points this out: "no such dimension is needed".
     
  23. Jan 14, 2016 #22
    We do not need extra dimensions to explain curvature of our 1+3 dimension space.

    But sometimes hypothetical extra dimension in Euclid geometry helps us to imagine what is going on.

    For example our universe is regarded as 3D sphere in 4D Euclid space

    [tex]x_1^2+x_2^2+x_3^2+x_4^2= a^2[/tex] 

    ref. (107.5) of Landau-Lifshitz's Classical Theory of Field
     
  24. Jan 14, 2016 #23

    PeterDonis

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    This is not true for our current best-fit model of the universe; in that model, the universe is spatially flat and infinite in extent. The 3-sphere model is a model of a closed, spatially finite universe.
     
  25. Jan 14, 2016 #24
    Instead, 3D hyperbole in 4D fictional Euclid space
    [tex]x_1^2+x_2^2+x_3^2-x_4^2=a^2[/tex]
    works?
     
  26. Jan 15, 2016 #25

    pervect

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    Trying to visualize a 4d object is not the approach we are recommending at all.

    The first step is to understand curvature in 2 dimensions. Orodruin's definition in post #5 is one of the simplest tests for curvature, and you can use it to show that the surface of a globe is curved. I say a globe, rather than the Earth, because we can make our globe perfectly spherical, and don't have to deal with complicating issues of the Earth being not-quite a perfect sphere.

    You start off at some point on the surface of the globe, and draw a line on the surface of the globe (the line obviously does not leave the surface of the globe, it's painted on the surface). This is similar to the way we do not leave space-time, we're always in space-time, we never go "outside" space-time. But right now we are trying to describe curvature in only two dimensions, so you know what we mean by curvature, and how you can tell something like the surface of a globe is curved without ever imagining anything that is not on the surface of the globe. Later on, we can use the analogy to better understand space-time being curved.

    You draw your lines on the globe on the path that is basically the shortest distance between two points on the sphere. This is perhaps slightly oversimplified, but not by very much. This curve of shortest distance is a great circle. Sometimes this path is called a geodesic rather than a straight line. Because we are trying to keep things simple, the technical term "geodesic" wasn't used, rather the less formal and less precise term "straight line" was used. It appears the attempt at simplificaiton didn't work, so I thought I'd try introducing the concept of a geodesic, which at this point we can regard as being the shortest distance between two points that lies on the surface.

    So curvature may be slightly tricky, but the hopefully familiar example of the globe's surface being curved, and applying Odoruin's test of making 4 right angle turns and moving along geodesics (which can be regarded for our purposes as being the shortest path connecting points that lies entirely on the surface without ever leaving it) making 4 90 degree turns is sufficient to determine whether or not our 2d geometry is curved or not.


    OK - we've started out moving in some direction, for the sake of definiteness lets say we start heading north. Then we might a 90 degree turn (lets say to the right), and start to head East. But a circle of lattitude is not a geodesic path on the globe, because it isn't great circle, and we know that all geodesics on the surface of the globe are great circles. (You may have to look this up for yourself, or just trust us on this point, unless you have calculus of variations to prove it.)

    We make another 90 degree right turn, and we start to head south-west. We're not heading exactly south, because we veered from moving due east when we insisted in tracing out a goedisc path. It will be a mildly tricky exercise in spherical geometry to work out all the details, and follow the whole path, but basically when you follow Odoruin's description, you find that you do not wind up exactly at your starting point on the globe, but you do if you follow the description on a flat plane. Therefore Odoruin's procedure provides a test for curvature that doesn't require knowing anything about "higher dimensional spaces", you can do all your geometrical drawings and measurements on the surface of the globe.
     
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