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

Independent components of the curvature tenso

  1. Sep 16, 2011 #1

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    A formula I know for the number of functionally independent components of the curvature tensor is: (n^2)(n^2 -1)/12. It gives 1 for n=2, 6 for n=2, 20 for n=4.

    However, for a metric space (with symmetric metric), the curvature tensor is completely specified by the metric tensor. For n=4, there are only 10 different components of the metric, and one can argue there are only 6 functions of the manifold needed to specify the geometry, as all metric representations connected by diffeomorphism represent the same geometry.

    How does one square this with 20 functionally independent components of the curvature tensor?
     
  2. jcsd
  3. Sep 16, 2011 #2

    atyy

    User Avatar
    Science Advisor

    http://books.google.com/books/about/A_panoramic_view_of_Riemannian_geometry.html?id=d_SsagQckaQC, p215:

    "one cannot recover from the Rijkh, which form a total of d2(d2 − 1)/12 numbers, all of the second derivatives ∂ijgkh, which form a total of (d(d + 1)/2)2 numbers. ... is there enough room between (d(d+1)/2)2 and d2(d2 − 1)/12 to find a nonisometric map between two metrics which still preserves the whole curvature tensor? The subject was initiated quite recently: we know now many examples of nonisometric Riemannian manifolds admitting diffeomorphisms preserving their respective curvature tensors."
     
  4. Sep 16, 2011 #3

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    Interesting, but sort of the opposite direction I'm coming from. Roughly, they seem to talk about the possibility of different manifolds having the same curvature tensor. I'm thinking there are too many curvature tensors for the number of distinct manifolds (when the latter are limited to symmetric metric, with 'metric compatible' connection). I'm sure there is a resolution, but I couldn't find it either in my books or my own searches of the internet. I'm thinking that the generic (n^2)(n^2-1)/12 formula does not account for hidden dependencies in connection values due to the fact that a metric compatible connection is completely determined by metric components modulo diffeomorphism.

    Thanks for anything you can find.
     
    Last edited by a moderator: Apr 26, 2017
  5. Sep 17, 2011 #4
    You need only ten because you are thinking of the vacuum solution with the ten components of Ricci tensor curvature vanishing , therefore only has the ten nonzero Weyl components.
    There are no known solutions in general relativity with particles stress-energy.
     
    Last edited: Sep 17, 2011
  6. Sep 17, 2011 #5

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    I don't see vaccuum as having anything to do with it - symmetry says the metric has only 10 different components. Then coordinate conditions (which amount to modulo diffeomorphism) say the number of distinct geometries would be specified by 6 functions. Whether exact solutions are known is also completely unrelated - this is an implicit function argument.
     
  7. Sep 17, 2011 #6

    WannabeNewton

    User Avatar
    Science Advisor

    What conditions are you talking about? [itex]\triangledown _{\nu }G^{\mu \nu } = 0[/itex]?
     
  8. Sep 17, 2011 #7

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    No, on top of the field equations, you can specify 4 additional, rather arbitrary conditions without loss of generality. These are called coordinate conditions. They arise because any given metric 'expression' is connected to family of geometrically equivalent metrics by coordinate transformation, defined by 4 continuous functions on the manifold. This implies 6 independent functions are sufficient to specify all distinct geometries defined by symmetric metric tensor (in 4-d).
     
  9. Sep 17, 2011 #8
    Sorry I might not have read your question correctly, I assumed you were talking about GR.
    The metric tensor is not the curvature tensor, you need additional info, the connection, to obtain the curvature tensor.
     
  10. Sep 17, 2011 #9

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    I am talking about GR. In GR, the choice is made to use a metric compatible connection. This is fancy talk for the connection is completely specified by the metric with the standard Christoffel formula. Thus there cannot be more geometrically distinct connections than metric tensors. Then, the curvature tensor is completely determined by these connections. Thus 6 continuous functions should be enough to generate all geometrically distinct curvature tensors. I'm looking for how to coordinate this fact with the curvature tensor is normally said to have 20 functionally independent components (not 4^4 = 256) in dim=4.
     
  11. Sep 17, 2011 #10

    WannabeNewton

    User Avatar
    Science Advisor

    Right if you specify initial conditions for the metric you can still make arbitrary coordinate transformations in the future so you end up with 4 arbitrary functional degrees of freedom. Isn't this made explicit by [itex]\triangledown _{\nu }G^{\mu \nu } = 0[/itex]?
     
  12. Sep 17, 2011 #11

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    I'm not sure. This is four equations, but my understanding is that this divergence equation is true by construction of G(mu,nu) from the full curvature tensor. Maybe it is equivalent to the ability specify 4 coordinate conditions. I'll have to think on that. Post any further thoughts of yours on this, as well.
     
  13. Sep 17, 2011 #12
    PAllen, if you carefully thought about it I think you'd remember the connection used in GR is not only metric compatible but also symmetric. That 's all you need to go from 20 to 10 components IMO.
     
  14. Sep 17, 2011 #13

    atyy

    User Avatar
    Science Advisor

    Hmmm, Berger says that 20 curvature conditions are too few to determine the metric, whereas you say those are too many (ie. the 20 cannot all be independent)?

    Naively, I'd think it'd be too few, maybe like there are more component Maxwell equations than E and B field components, and yet the fields are underspecified, since one has to add in boundary conditions.

    Berger does comment that if we go to Riemann normal coordinates, then at the origin the curvature components do determine all second partials of the metric. However, this is because going to normal coordinates brings in additional information beyond the curvature (I'm not sure what exactly this additional information is).
     
  15. Sep 17, 2011 #14

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    Right, I thought about that later, but didn't mention it. So then there is only the discrepance from 10 to 6.

    [Edit: wait, I'm not sure it decreases it by that much. You may be right. Can you post an argument or reference that symmetric metric reduces independent curvature components that much?]
     
    Last edited: Sep 17, 2011
  16. Sep 17, 2011 #15

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    The whole point of my question is how to square arguments like this with the fact that (in GR at least), the metric completely determines the curvature tensor and has only 6 functional degrees of freedom.
     
  17. Sep 17, 2011 #16

    atyy

    User Avatar
    Science Advisor

    Hmm, let's see: the metric defines the curvature, but the curvature does not define the metric.

    I thought you were asking about the second part of that statement, but you are asking about the first?
     
  18. Sep 17, 2011 #17
    I don't have a reference, but you certainly save all the torsion tensor components of the curvature, it is kind of obvious.
     
  19. Sep 17, 2011 #18
    From ten to six usually the argument is used that four of the ten would be fixed by the test particle coordinates, but this would only work for vacuum solutions, as I explained above no other kind of solutions are known for test particle stress-energy so no one cares much about that discrepancy (two body problem is still not solved in GR).
     
  20. Sep 17, 2011 #19

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    I'ts obvious to me you save some. It's not obvious to me that you get from 20 to 10 just by assuming symmetric connection.
     
  21. Sep 17, 2011 #20

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    I don't see the 10 to 6 argument as having anything to do with vaccuum/non-vaccuum. It is derived assuming nothing other than that the metric is symmetric, and you have diffeomorphism invariance.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook