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General relativity and differential geometry

  1. Sep 28, 2015 #1
    Dear all
    I am studying general relativity and i have a question as follow. We have the 2- sphere can be scanned totally by a coordinate system (theta, phi) with the metric tensor written in terms of theta and phi. Now i want to divide the 2-sphere into charts 4 charts then each will have its own coordinate system. Then what will happen for the metric? initially we were having the same formula for the metric for the whole sphere but now shall i say that each chart will have its own metric formula?
    Thank you
  2. jcsd
  3. Sep 28, 2015 #2


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    I think that, although the spherical coordinate system covers the whole sphere, the coordinate representation of the metric tensor is discontinuous in that coordinate system at the Greenwich Meridian, where the longitude flips from 0 to 360 degrees. So in fact we don't initially have the same formula for the metric tensor everywhere on the sphere, because there will always be a half-great-circle for which it (the formula, not the tensor itself) is undefined. I think you need at least two charts to cover a sphere, and they will have different coordinate representations of the metric tensor.
  4. Sep 28, 2015 #3


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    If you have a chart with coordinates [itex](X,Y)[/itex], then you will have metric components [itex]g_{XX}, g_{XY}, g_{YX}, g_{YY}[/itex]. These will be related to the metric components for [itex](\theta, \phi)[/itex] through:

    [itex]g_{IJ} = \sum_{ij} g_{ij} \dfrac{\partial x^i}{\partial X^I} \dfrac{\partial x^j}{\partial X^J}[/itex]

    where [itex]I, J, i, j[/itex] take values [itex]1[/itex] and [itex]2[/itex], and where [itex]X^1 = X, X^2 = Y, x^1 = \theta, x^2 = \phi[/itex].
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