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Is curvature possible for a 2D metric?

  1. Aug 31, 2014 #1
    I was recently trying to test something out with the Riemann tensor. I used only 2 dimensions for simplicity sake. As I was deriving the Riemann tensor, I noticed that it looked as if all of the elements were going to come out to be 0 (which they all did). Therefore, this coordinate system is flat space.

    Then I realized something. I was working in 2D. The very definition of 2D alone essentially means flat planes. Therefore it would make sense for a 2D coordinate system to be flat space.

    Just to make sure however, I must ask this: Is it possible for a 2 dimensional Riemann tensor to ever have a non zero element, or does the fact that it is 2D mean that there can never be any non zero elements (and therefore no curvature)?
     
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  3. Aug 31, 2014 #2

    WannabeNewton

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    No this is incorrect. A 2-sphere is 2-dimensional and has non-zero Riemann curvature. If every 2-manifold was trivially a plane then the extremely elegant classification theorem of compact 2-manifolds would be a waste :wink:
     
  4. Aug 31, 2014 #3

    Orodruin

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    Yes, it is perfectly possible for a 2-dimensional manifold to have a non-vanishing curvature tensor. The prime example would be a sphere embedded in three euclidean dimensions with the induced metric.
     
  5. Aug 31, 2014 #4

    A.T.

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  6. Aug 31, 2014 #5

    Matterwave

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    You gotta go one more dimension down. A 1-D Riemann tensor is always 0. Any curve can only have extrinsic curvature and no intrinsic curvature.
     
  7. Aug 31, 2014 #6

    pervect

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    Not so, for a counterexample, the previously mentioned surface of a sphere. But if you start with a plane, you'll get zero curvature, that much is true.
    There are 16 components of the Riemann in 2d

    ##R_{ijkl}## is zero whenever i=j or k=l. Since i,j,k and l can take only the values 0 and 1, that leaves only four non-zero components for the Riemann in 2d.

    These four components will all have the same value in an orthonormal basis, there is really only one degree of freedom for (intrinsic) curvature in 2d.
     
  8. Aug 31, 2014 #7

    Matterwave

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    Adding on to pervect's reply a little, in general the number of independent components of the Riemann tensor is ##n^2(n^2-1)/12## where n is the dimension of the manifold. You can immediately see from this equation that for n=1 the Riemann has actually no independent components, it must vanish, and for n=2 the answer is 1.
     
  9. Sep 1, 2014 #8

    stevendaryl

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    Wikipedia gives an expression for the simplification of [itex]R_{ijkl}[/itex] in the 2D case:

    [itex]R_{abcd} = K (g_{ac} g_{db} - g_{ad}g_{cb})[/itex]

    where [itex]K[/itex] is a real number that varies from point to point and is equal to the Gaussian curvature.
     
  10. Sep 1, 2014 #9
    If you define the 2D space to have the geometry of the surface of a sphere would that be flat?
     
  11. Sep 1, 2014 #10

    robphy

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    No. On a sphere, ...
    -the sum of the interior angles of a triangle won't add up to 180 degrees (you can make a triangle with three right angles).
    -a triangle with one right angle won't satisfy the Pythagorean theorem.
    -the circumference/radius ratio of a circle (generated by a fixed length string tied to a post) won't be 2*pi.
     
  12. Sep 1, 2014 #11
    OK so a sphere gives us a 2D space that isn't flat.

    Extending that into one dimension (e.g. the great circle of a sphere) don't we get a 1D space that isn't flat?
     
  13. Sep 1, 2014 #12

    Nugatory

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    No. Mark two points on a piece of thread, and no matter how the thread is laid out the distance between the two points is the same - and in one dimension that's all the geometry there is.
     
  14. Sep 1, 2014 #13

    ChrisVer

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    or in other words there is no difference between a circle of radius R and a line of length 2πR
    you can cut the circle, extend it and make it a line. That's actually not the case for a sphere - you can't cut it and make a flat plane. That's because the sphere has curvature.
     
  15. Sep 1, 2014 #14
    Imagine a 3D sphere entering a 2D flatland space at constant speed. The inhabitants of flatland would see a circle - growing and then diminishing (this is in Abbot's book Flatland if memory serves...)

    Now imagine the edge of a circle entering (i) my curved 1D space (ii) a straight 1D space. Assuming the circle moves into the 1D spaces at the same speed the "shadow line" would grow at different rates in the different 1D spaces.

    Fun animation showing flat case:

    http://www.rmcybernetics.com/science/physics/dimensions_1_dimensional_space.htm
     
    Last edited: Sep 1, 2014
  16. Sep 1, 2014 #15

    Nugatory

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    That's true, but that's not how intrinsic curvature (the kind described by the Riemann tensor and that matters in general relativity) is defined. Instead, it is defined by the geometrical relationships between points in the manifold itself - look at robphy's post above for examples of such relationships.

    Another example worth considering is the two-dimensional surface of a cylinder. It doesn't appear at first glance to be flat, but it is - rolling a sheet of paper up into a cylinder doesn't stretch the paper or distort any of the geometrical relationships between points on the sheet of paper.
     
  17. Sep 1, 2014 #16

    ChrisVer

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    Since we are talking about the sphere, I have one question... Is it because of the sphere's curvature that the Rieamman sphere projection on [itex]\textbf{C}^{2}[/itex] has the pole of the sphere appearing at "[itex] \pm ∞ [/itex]"?
     
  18. Sep 1, 2014 #17
  19. Sep 1, 2014 #18

    A.T.

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    A 1D curve can have extrinsic curvature, which depends on its embedding in higher dimensional space. For actual intrinsic curvature, that exists independently of any embedding you need at least 2D.
     
    Last edited: Sep 1, 2014
  20. Sep 1, 2014 #19
    I don't quite understand that, I find it difficult to visualise a space without its embedding in a higher dimension. Wolfram says:

    A 2D intrinsic curvature, "is detectable to the inhabitants of a surface and not just outside observers. An extrinsic curvature ... is not detectable to someone who can't study the three-dimensional space surrounding the surface on which he resides."
    http://mathworld.wolfram.com/IntrinsicCurvature.html

    Fair enough, but in the 1D case, surely you need to add there must be no observable intrusion from the embedding 2D space? (thinking of the circle passing through...)

    Also, if an inhabitant of circular 1D space went for a walk and found himself back where he started couldn't he say the space was curved? It certainly isn't flat or he'd be walking to infinity. So shouldn't we also disallow "long walks" in the definition of extrinsic curvature?
     
  21. Sep 1, 2014 #20

    ChrisVer

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    You are mixing the Riemannian curvature with the Gaussian...

    He could say that his space is closed, it doesn't have to be curved to be closed...
     
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