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Lorentzian metrics w/ timelike closed curves on cylinder that is a 2-manifold

  1. Oct 22, 2011 #1
    I defined a covering map ℝ^2 → S^1 x ℝ in order to work with the manifold.

    1) How can I find lorentzian metrics (=metric tensors) on S^1 x ℝ (2-dimensional manifold)?
    I know that the diagonal matrix (2x2 matrix) of such a lorentzian metric must have signature 1. and there are some famous metrics like g= -dx^2 + dy^2. but how can i find/ define lorentzian metrics myself?

    Approach: I know that a lorentzian metric has signature 1. This means the matrix elements of the normal form are g11= +1, g22= -1, g12= g21= 0.
    But how can I find DIFFERENT lorentzian metrics? And since they all have the same normal form with signature 1, how can I distinguish them?

    2) How does a timelike closed curve looks like on S^1 x ℝ ?
    Let's take the metric above g= -dx^2 + dy^2.
    How can I figure out if a curve on the cylinder with this metric is closed (and timelike; see examples below)?

    I defined some curves on the cylinder (= I define them in the ℝ^2 since the cylinder is a 2-manifold). What does "closed curve" mean in this case? Is a curve c: [0, 2∏ ] → ℝ^2 closed when c(0) =c(2∏)?
    Even if in the ℝ^2 they do not look closed? but they are closed on the cylinder? Is this right?
    Are the following ideas correct?


    c(x) = (x, sinx) and c'(x)= ( 1, -cosx)
    => <c'(x), c'(x)> = -1 + cos^2 (x) ≤ 0
    but this means the curve is not timelike and not spacelike? How is this possible?

    d(x) = (0,x) and d'(x)= (0,1)
    => <d'(x), d'(x)> = 1 >0
    this means this curve is timelike


    e(x)= (x,x)
    but this curve is not closed? Right?


    f(x)= (cosx, sinx) and f'(x)= (sinx, -cosx)
    => <f'(x), f'(x)> = -sin^2(x) + cos^2(x)
    what does this mean? is f(x) not a closed curve?


    It would be very helpful if somebody could just show me all steps using a simple example.
    Last edited: Oct 22, 2011
  2. jcsd
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