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Visualization of metric tensor

  1. Sep 15, 2012 #1
    Barbour writes:
    the metric tensor g. Being symmetric (g_uv = g_vu) it has ten independent components,
    corresponding to the four values the indices u and v can each take: 0 (for the
    time direction) and 1; 2; 3 for the three spatial directions. Of the ten components,
    four merely reflect how the coordinate system has been chosen; only six
    count. One of them determines the four-dimensional volume, or scale, of the
    piece of spacetime, the others the angles between curves that meet in it. These
    are angles between directions in space and also between the time direction and
    a spatial direction.

    I please to write more visually and more precisely.
    I suppose that the first four values are on diagonal?....
     
  2. jcsd
  3. Sep 15, 2012 #2

    Bill_K

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    Science Advisor

    Nope. At a single point, ALL TEN components of the metric tensor reflect how the coordinate system has been chosen. You can always choose the coordinates so that the metric tensor at the origin is equal to the Minkowski metric. The only things that count are the derivatives of the metric tensor.
     
  4. Sep 15, 2012 #3
    As I understand, it is not necessary that axes are rectangular. So he measure their angles? I think that it is possible here to ignore next derivatives?
     
  5. Sep 15, 2012 #4
    Let us imagine Spacetime around Schwarchild black hole, which is written in the last equation:
    http://en.wikipedia.org/wiki/Metric_tensor
    It is described with four numbers. Thus, I suppose all angles between axes are 90°? (In one point it can be also described with metric of Minkowski, and it really is, in infinity.)

    Barbour:
    One of them determines the four-dimensional volume, or scale, of the piece of spacetime,
    Thus, I suppose that this information is hidden in Sch. metric?
    Probably he has his own idea
    four merely reflect how the coordinate system has been chosen.
    What is his idea?
     
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