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Eigenvalues of curvature tensors as curvature scalars?

  1. Nov 30, 2015 #1

    bcrowell

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    I've been playing around with the Carminati-McLenaghan invariants https://en.wikipedia.org/wiki/Carminati–McLenaghan_invariants , which are a set of curvature scalars based on the Riemann tensor (not depending on its derivatives). In general, we want curvature scalars to be scalars that are continuous functions of the Riemann tensor (i.e., continuous functions of its components). As I've fiddled around and tried to gain insight into the CM invariants, I've found that the four invariants R, R1, R2, and R3 really just seem to be encoding information about the four eigenvalues of the Ricci tensor. I think R is basically the mean of the eigenvalues, and R1 their variance.

    It seems that this is a nice general way to think about curvature scalars. Is it true in some sense that we can get all the curvature scalars we need as eigenvalues of certain curvature tensors?

    The roots of a polynomial are continuous functions of the coefficients (all of this being over the complex numbers). Therefore the eigenvalues of the Ricci tensor, which are roots of its characteristic polynomial, are continuous functions of the Riemann tensor.

    Can one do something similar with the Weyl tensor? The Weyl tensor has 4 indices, but can we characterize the curvature scalars that depend on the Weyl tensor by writing the Weyl tensor in terms of two bivector indices and then taking its eigenvalues?

    Computationally, it can in principle be extremely expensive to compute curvature polynomials. If the Riemann tensor is known as an expression involving n terms, and we want to compute a curvature polynomial of order k, then the amount of computation would probably go like n^k. This can be extremely big, e.g., for the Kerr-Newman metric it looks like n is on the order of 10^2 or 10^3 when I compute it by brute force, and k goes up to 5 for the CM invariants. I wonder if it could be more practical computationally to compute eigenvalues.
     
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  3. Dec 3, 2015 #2

    jedishrfu

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  4. Dec 3, 2015 #3

    bcrowell

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    Aha-- that's very helpful, thanks!
     
  5. Dec 3, 2015 #4

    bcrowell

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    Good to have some confirmation from @jedishrfu that this was a reasonable thing to pursue. I think if you look at the Ricci decomposition, it's clear that the parts built out of the Ricci tensor have 4 pieces of information built into them, corresponding to the 4 eigenvalues of the Ricci tensor. This makes sense when you compare with the CM invariants: there are four of them that use only the Ricci tensor.

    [The following is wrong, see #12.]

    I'm slightly less clear on the Weyl part of the decomposition. If you consider the Weyl tensor as a matrix with two bivector indices, it's a real antisymmetric matrix of dimension 6, so it should have three pairs of imaginary eigenvalues, each pair being of the form ##\pm k i##, where ##k## is real. This seems like three degrees of freedom.

    Now if it's a vacuum solution, the Weyl part is the only part there is. In this situation, I would expect that there would be three independent curvature invariants of order 0 (i.e., not involving any derivatives of the Riemann tensor). For a vacuum solution, the only CM invariants that don't vanish are ##W_1## and ##W_2##, which are both complex, so that seems to make 4 degrees of freedom rather than 3. But maybe there is some constraint on ##W_1## relative to ##W_2## in the vacuum case...?
     
    Last edited: Dec 4, 2015
  6. Dec 3, 2015 #5

    bcrowell

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    Hm...WP says this:

    https://en.wikipedia.org/wiki/Petrov_classification#The_classification_theorem

    I don't understand this. For example, suppose the Weyl tensor is zero. Then any 6 linearly independent bivectors are a counterexample to the above...? They're all eigenvectors with eigenvalues of zero.
     
  7. Dec 3, 2015 #6

    PeterDonis

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    Are there 6 linearly independent bivectors once the symmetries of the Weyl tensor are taken into account?
     
  8. Dec 3, 2015 #7

    bcrowell

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    There are 6 linearly independent bivectors just because the space of bivectors is a six-dimensional space...right?
     
  9. Dec 4, 2015 #8

    martinbn

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    That seems strange. Perhaps it's just poorly phrased. May be what was intended was to say that the eigenvalue problem for the Weyl tensor reduces, because of symmetries, to an equivalent (but different) eigenvalue problem in 4 dimensions.
     
    Last edited: Dec 4, 2015
  10. Dec 4, 2015 #9

    PeterDonis

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    But the possible eigenbivectors of the Weyl tensor, given the symmetries of that tensor, might not span the entire 6-dimensional space; they might span only a 4-dimensional subspace. I don't have the mathematical tools to verify this; I'm just suggesting it as a possible way to reconcile the discrepancy you mentioned.
     
  11. Dec 4, 2015 #10

    bcrowell

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    Right, but then wouldn't the case where the Weyl tensor vanishes be a counterexample? I think I'll take a simple example such as the Schwarzschild metric and compute the eigenvalues and eigenvectors, and see what I get.
     
  12. Dec 4, 2015 #11

    PeterDonis

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    If you want an example where the Weyl tensor vanishes, wouldn't FRW spacetime be a better choice?
     
  13. Dec 4, 2015 #12

    bcrowell

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    Minkowski space is the simplest example where the Weyl tensor vanishes. My motivation for trying the Schwarzschild metric was that I thought maybe the WP article's statement was true in nontrivial cases, but false in the trivial case where the Weyl tensor vanishes completely.

    What I realized after calculating the Weyl tensor explicitly in bivector form, for the Schwarzschild spacetime, is that it is *not* antisymmetric. For example, the Weyl tensor for that spacetime has ##C_{tr}{}^{tr}=-2m/r^3##. If I let ##[tr]## be bivector index 1, then in bivector form, we have ##C_{11}=-2m/r^3##. The bivector form of the Weyl tensor for this spacetime is actually diagonal when expressed in a bivector basis corresponding to the Schwarzschild coordinates. The following is my Maxima code, using my package https://github.com/bcrowell/cm_invariants :

    Code (Text):

    load(ctensor);
    ct_coords:[t,r,theta,phi];
    assume(m>0);
    lg:matrix([(1-2*m/r),0,0,0],
      [0,-1/(1-2*m/r),0,0],
      [0,0,-r^2,0],
      [0,0,0,-r^2*sin(theta)^2])$
    load("cm_invariants.mac");
    cm_init();
    cm_c_bivector;
     
    The result is ##C_{AB}=(m/r^3)diag(-2,1,1,1,1,-2)##. Since it's diagonal, obviously all six of the eigenvectors are linearly independent. This leaves me even more confused by the WP article's statement.

    I believe the Weyl tensor has the same symmetries as the Riemann tensor, so the bivector form of the Weyl tensor is always *symmetric*.

    What might be true is that the number of distinct eigenvalues of ##C_{AB}## is always 4 or less. This would be different from what WP says, but maybe they just made a mistake in writing it. It would make sense because for a vacuum solution, the information in the CM invariants is only the two complex invariants W1 and W2, which makes a total of 4 d.f.
     
    Last edited: Dec 4, 2015
  14. Dec 4, 2015 #13

    Ben Niehoff

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    The invariants of a matrix are exactly its eigenvalues. Another basis of invariants are the traces of nth powers of the matrix. Another basis of invariants are the coefficients of the characteristic polynomial (modulo rescaling; otherwise you would have one too many invariants).

    For tensors of higher rank, the idea is essentially the same. So yes, eigenvalues are one way to look at it.
     
  15. Dec 4, 2015 #14

    PeterDonis

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    Yes, but it also has zero curvature period, so I thought it might be too simple. :wink: Your interpretation of the number 4 as referring to the maximum number of distinct eigenvalues makes sense to me; it certainly wouldn't be the first time that a Wikipedia article used less than precise terminology. :wink:
     
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