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

Information content of the metric, Riemannian and Ricci tensors

  1. Aug 25, 2009 #1
    I'm getting myself up to speed on GR to try to understand a book by John Moffat called "Reinventing Gravity....". So far I've been using Sean Carroll's sort of classic course notes and a fair bit of Wikipediea (sp?). It may be a naive question, but the point I wouldn't mind comments on is the following: Often in writeups on GR, some sort of comment is made the the metric tensors describes the geometry of the GR (semi-Riemannian) manifold. Then, however, the Riemann tensor is derived and it indicated as containing ALL the information about the geometry of the manifold. However the contraction of the Riemannian tensor, the Ricci tensor, is actually used in the Einstein GR equation.

    The metric is symmetric hence, in component form, it has ten independent components. The Riemann tensor components are constructed from, after all is said and done, combinations of products of the metric tensor components and their partial derivatives. It has 20 independent components, i.e., we have extracted additional information from the metric by the operations performed on it to get the Riemann tensor. We then contract the Riemann tensor to get the Ricci tensor, symmetric again, hence ten independent components. So we gain some information, then lose or transform some of it in going from the metric to the Ricci tensor. One specific question is, what information is no longer available upon the contraction, and why is it (whatever it may be) not important?

    I have a few other musings about how much information content can, in some sense, be attributed (contained) in an operator or function (like these tensors) for the domains that they are relevant for, and (when I finally read Einstein's paper), why he contracted the Riemann tensor, supposedly the ne plus ultra. (One suspects that the stuff he had to work with (the stress-energy tensor) was a (0,2) tensor and he was stucK.
     
    Last edited: Aug 25, 2009
  2. jcsd
  3. Aug 25, 2009 #2

    atyy

    User Avatar
    Science Advisor

  4. Aug 26, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, the metric tensor contains all (local) information about the space. No, the Riemann tensor does not contain all information the metric tensor does but it does contain curvature information. The Ricci tensor contains information about specific kinds of curvature that are related to gravity.

    (And it is NOT true that just because one tensor has more components than another, it must contain more information. Any tensor derived from the metric tensor alone cannot contain more information than the metric tensor, though it may contain less.)
     
  5. Aug 26, 2009 #4
    Looks like an interesting discussion, and I thank you both for your input. Having been (way back in the day) a particle experimentalist, I usually need to visualize and internalize a concept before I can proceed farther with a line of reasoning. Quoting John Baez (a kinda flippant and fun guy, but with a good visual way of presenting things (and a faculty member at my old alma mater); "In 4 dimensions, it takes 20 numbers to specify the curvature at each point. 10 of these numbers are captured by the "Ricci tensor", while the remaining 10 are captured by the "Weyl tensor". Now the Riemann tensor has 20 components ("numbers", at any point). The Weyl tensor is constructed from the Riemann tensor and the metric. In fact, I think I have seen (don't remember where), the derivation of the metric tensor from the Riemann tensor. Sort of chicken and egg situation.

    HallsofIvy said: "Any tensor derived from the metric tensor alone cannot contain more information than the metric tensor, though it may contain less".
    This sort of jibes with my thoughts, but is there, anywhere, a "Fundamental theorem of information content" ?

    Lastly, just a little numerology ;-{) : The Einstein tensor, the curvature part of the basic GR equation, has from the Ricci tensor, 10 independent (from each other) components, as does the metric tensor, so, 20. Any significance, or just numerology?
     
    Last edited: Aug 26, 2009
  6. Aug 26, 2009 #5

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Last edited by a moderator: May 4, 2017
  7. Aug 26, 2009 #6
    Thank you, robphy,for the references. A fast scan of the first looks to be very informative, and not too daunting other than the usual mind-numbing mathematical terminology.
     
  8. Aug 26, 2009 #7

    atyy

    User Avatar
    Science Advisor

    Trying to read up on HallsofIvy's and robphy's comments, googling gave me section 4.5 of
    http://books.google.com/books?id=d_...al+curvatures+dimension&source=gbs_navlinks_s
    http://www.math.umn.edu/~xuxxx225/docs/A%20Panoramic%20View%20of%20Riemannian%20Geometry.asp [Broken]

    "one cannot recover from the Rijkh, which form a total of d2(d2 − 1)/12 numbers, all of the second derivatives ∂i∂jgkh, which form a total of (d(d + 1)/2)2 numbers .......... we know now many examples of nonisometric Riemannian manifolds admitting diffeomorphisms preserving their respective curvature tensors."
     
    Last edited by a moderator: May 4, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Information content of the metric, Riemannian and Ricci tensors
Loading...