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Locally Maximally Symmetric Spacetimes

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  1. Oct 18, 2013 #1
    Can one say that every general curved spacetime, locally is maximally symmetric?
    I know that one can say that every general curved spacetime is locally flat (and therefore maximally symmetric with R=0), but I'm talking about a very high curvature spacetime, where still we can consider nonzero curvatures.
     
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  3. Oct 18, 2013 #2

    PeterDonis

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    What do you mean by "maximally symmetric"? Most general curved spacetimes have no symmetries at all (i.e., no Killing vector fields).

    No, you can't say that. What you can say is that every general curved spacetime is locally Lorentz, meaning that at a given event, you can choose a coordinate chart that makes ##g_{ab} = \eta_{ab}## at that event, and makes all of the connection coefficients (first derivatives of ##g_{ab}##) vanish at that event. But that doesn't make ##R = 0##.
     
  4. Oct 18, 2013 #3
    Thanks for you replay Peter, but I'm still a little bit confused.

    I know it, but I meant "locally" not globally.


    What you mention here, looks like the Riemann normal coordinate, is it?

    Can we assume that R is almost constant as in maximally symmetric spacetimes and therefore [itex]R_{ab}\propto Rg_{ab}[/itex]?


    What I get from your answer, is we can not say that a general high curved spacetime, is "locally" maximally symmetric. Am I right?
     
  5. Oct 18, 2013 #4

    Bill_K

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    That's right. And you can strike the word "high". A general curved spacetime is not locally maximally symmetric.

    Since the Riemann tensor has dimension L-2, curvature only sets a length scale, and you can only say it's large or small in comparison to some other length.
     
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