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I Dimensional check of Einstein's Field Equations

  1. Mar 12, 2016 #1
    This is my first post in Physics Forums.

    I am trying to a dimensional check of Einstein's field equations. Unfortunately, most books consider c = 1 or sometimes even G = c = 1, when presenting the field equations. This makes it very difficult to do a dimensional check.

    In spite of this, I think that I have found the dimensions of the diagonal terms in the Ricci tensor. They all seem to be inverse length squared. The terms in the Ricci scalar too seem to have the same dimensions, inverse length squared.

    This means that the Robertson Walker metric tensor's diagonal elements should be dimensionless. But the last 2 diagonal terms in the metric tensor are a^2*r^2 and a^2*r^2*sin^2(theta). Clearly, they have the dimension squared length, whereas the first diagonal term -1 has no units and the second diagonal term a^2/(1-k*r^2) too has no units. (a is dimensionless, k has the dimension inverse length squared and r has the dimensions of length).

    Thanks in advance.

    Sundar
     
  2. jcsd
  3. Mar 12, 2016 #2

    phyzguy

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    The units of the metric tensor components ( and hence the Ricci tensor components) depend on the coordinate system. In the relation:
    [tex]ds^2 = g_{\mu \nu} dx^\mu dx^\nu [/tex]

    ds^2 has units of length squared. In a Cartesian coordinate system, dx^2 has units of length squared, so the metric tensor components are dimensionless. However, in spherical coordinates, dr^2 has units of length squared, so grr is dimensionless, but dθ^2 is dimensionless so gθθ has units of length squared.
     
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