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Locally flat coordinates on the Poincaré half plane

  1. Feb 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the locally flat coordinates on the Poincaré half plane.
    Problem I.6.4 by A. Zee
    2. Relevant equations

    Poincaré Metric: ##ds^2 = \frac{dx^2 + dy^2}{y^2}##

    3. The attempt at a solution
    First, I'm having problems with the explanation in Zee's book. He said that we can always choose our neighborhood to be locally flat for any space of any dimension D.
    "Look at how the metric transforms when you go to a new set of coordinates:

    ##g'_{λσ}(x') = g_{μν}(x) \frac{∂x^μ}{∂x'^λ} \frac{∂x^ν}{∂x'^σ} ##

    Within reason, you could choose any x you want, and for each choice, you get a new form for the metric. You have a lot of freedom to massage the metric into the form you want. The proof simply amounts to counting how much freedom you have on hand. So, look at our space around a point P. First, for writing convenience, shift our coordinates so that the point P is labeled by x = 0. Expand the given metric around P out to second order:

    ##g_{μν}(x) = g_{μν}(0) + A_{μν,λ}x^λ + B_{μν,λσ}x^λx^σ + ... ##

    As always, if you get confused, you should simply refer to the sphere. Thus, let the coordinates of the point P be ##(θ_∗, ϕ_∗)##, so that ##x^1 = (θ − θ_∗)## , ##x^2 = (ϕ − ϕ_∗)##. (Of course, in this simple case, nothing depends on ##ϕ_∗##.) What we just wrote down is then simply, for example, ##g_{ϕϕ} = \sin^2θ = sin^2θ_∗ + 2sinθ_∗cosθ_∗x^1 + . . . ##, so that ##A_{ϕϕ},1 = 2sinθ_∗cosθ_∗## and ##A_{ϕϕ},2 = 0##. Nothing profound at all.
    Change coordinates according to ##x^μ = K^μ_ν x'^ν + L^μ_{νλ} x'^νx'^λ + M^μ_{νλσ}x'^νx'^λx'^σ + . . .## . Again, nothing profound: K , L, M, . . . are just a bunch of coefficients to be determined. At the point P, the new metric is given by

    ##g'_{λσ}(0) = g_{μν}(0) K^μ_λK^ν_σ ##

    Regard this as a matrix equation ##g = K^TgK##, where T denotes transpose. Since ##g_{μν}(0)## is symmetric and real, there always exists a matrix K that will diagonalize it. After ##g_{μν}(0)## becomes diagonal (with positive diagonal elements—we will take that as a definition of space), we could scale each coordinate, one by one, by an appropriate factor, so that the diagonal elements become 1. We end up with the Euclidean metric ##g_{μν}(0) = δ_{μν}##."


    Following the first part of his explanation,

    ##ds^2 = \frac{dx^2 + dy^2}{y^2}## → ##g_{xx} = g_{yy} = \frac{1}{y^2}##

    The metric is already diagonal so K is just the identity matrix, but how do I find L (for the second order term)? Generally, what is the exact way to find the locally flat coordinates. The explanation is kinda fuzzy. I can't point the exact method to find those.
     
  2. jcsd
  3. Feb 18, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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