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I What does the Einstein tensor actually tell you?

  1. Jul 26, 2016 #1
    I recently calculated the Einstein tensor for the exterior Schwarzchild solution. Here it is:

    G00 = 0
    G11 = [-2GM/(r3c2 - 2GMr2)] - G2M2/[r2(rc2 - 2GM)2] - [-G2M2/(r4c4 - 2GMr3c2)] - (-2GM)/(r3c2) - (-2G3M3)/[r3c2(rc2 - 2GM)2]

    G22 = (2G2M2rc2 - 2G3M3) / (r3c6 - 2GMr2c4)

    G33 = G22sin2(θ)

    Every other element is 0. Just so you know, I used the (+ - - -) signature, so here is the line element:
    ds2 = [1 - 2GM/(rc2)]c2dt2 - dr2/[1 - 2GM/(rc2)] - r2(dθ2 + sin2(θ)d∅2)

    Now what exactly does this Einstein tensor (or any Einstein tensor for that matter) tell you about the curvature of space-time? I have some speculation about this question based on some dimensional analysis that I did.

    After calculating this Einstein tensor (while keeping in all factors of c and G) I did some dimensional analysis on this Einstein tensor using SI units, and I found that in SI units, the radial element G11 of this tensor has the units:

    1/m2 (or just simply m-2). I noticed that this is a common unit for things involving curvature. On an interesting note however, I found the angular elements G22 and G33 to be dimensionless. I wonder if this is because the Schwarzchild metric is isometric, and doesn't have a function in front of the angular section of the line element.

    Any way, based on the dimensions that I found for the radial coordinate, I speculated that G11 gives you the amount of space-time curvature in the radial direction. Would this speculation be correct? Also, what would the angular components tell you, since they are dimensionless in this case? If my speculation is not correct then what exactly would this tensor tell me?
     
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  3. Jul 26, 2016 #2

    PeterDonis

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    It's probably best to start by understanding what the Ricci tensor, from which the Einstein tensor is derived, tells you about the curvature of spacetime. A good brief description is in this article on the meaning of the Einstein Field Equation:

    http://math.ucr.edu/home/baez/einstein/

    The Einstein tensor is basically an alternate version of the Ricci tensor which has the useful property that its covariant divergence is zero. The best geometrical interpretation of this fact that I have found is in Misner, Thorne, and Wheeler, where they describe it as the property that "the boundary of a boundary is zero".

    As a simple example of this property, consider the boundary of a cube, which is six square faces, each with an orientation--for example, we can assume that the normal vectors to each face are pointing outward. Now consider the boundary of this boundary, which will be six squares, i.e., six sets of four line segments each, where each line segment has a particular direction, corresponding to the direction in which the segment is traversed when going around the square counterclockwise--the direction that corresponds to the normal pointing outward by the right-hand rule. If you then consider all six squares together, you will see that each line segment is part of two squares, in which it is traversed in opposite directions. So adding up all the line segments including their directions gives a boundary of a boundary which is zero.

    The reasont this property is useful is that the Einstein Field Equation relates the Einstein tensor to the stress-energy tensor, i.e., to the presence of matter and energy (and pressure and stress and other things). The covariant divergence being zero equates to the stress-energy tensor being locally conserved, i.e., stress-energy can't be created or destroyed, it can only be converted from one form to another. In other words, this is GR's version of local conservation of energy.
     
  4. Jul 26, 2016 #3

    George Jones

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    Since the exterior Schwarzschild solution is a vacuum, by Einstein's equation, all components (in all coordinate systems) of the Einstein tensor should be zero.
     
  5. Jul 26, 2016 #4
    I understand your logic, but I also find something kind of odd about this statement. The Schwarzchild metric has non-zero Ricci tensor elements as well as a non-zero curvature scalar. When plugging the metric into the formula Gμν = Rμν - (1/2)gμνR, not all terms canceled out and came out to be 0. Some of the terms came out to be actual expressions (which could be set equal to 0, but that would mean that the elements were only 0 given certain values of r and M).

    Now this could just be arithmetic error on my part. I can check my math again. I think it will also help if I show you this:

    http://web.stanford.edu/~oas/SI/SRGR/notes/SchwarzschildSolution.pdf

    If you start on page 4 and read down through page 9, you'll see where they give you the Ricci tensor elements as well as other terms such as the Christoffel symbols and the curvature scalar in terms of U, V and derivatives of U and V. It later concludes that U = 1 - 2GM/(rc2) and V = 1 / [1 - 2GM/(rc2)].
    I simply plugged these terms into the formulas given for the elements of the Christoffel symbols and the Ricci tensor, and then I used Rμν - (1/2)gμνR to get my Einstein tensor that way.
     
  6. Jul 26, 2016 #5

    haushofer

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    Well, all the components of the Einstein tensor should be zero. That's how you solve for the Ansatz in the first place.
     
  7. Jul 26, 2016 #6

    PeterDonis

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    Check your computations. It is straightforward to show as a general theorem that if the Einstein tensor is identically zero, the Ricci tensor must be as well. We simply raise an index in the definition of the Einstein tensor and then take the trace (using the fact that raising an index on the metric gives the Kronecker delta):

    $$
    R^a{}_b - \frac{1}{2} g^a{}_b R = 0
    $$

    $$
    R^a{}_b = \frac{1}{2} \delta^a{}_b R
    $$

    $$
    R^a{}_a = R = \frac{1}{2} \delta^a{}_a R = \frac{1}{2} 4 R = 2 R
    $$

    This can only be satisfied if ##R = 0##. But if ##R = 0## and the Einstein tensor is zero, then we must have ##R_{ab} = 0##.
     
  8. Jul 26, 2016 #7

    robphy

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    Equation 8 gives you an expression for a component of Einstein for arbitrary U(r) and V(r) as introduced in the static and spherically form of the metric given in Equation 1. This is just tensor calculus... given the metric, find the curvature tensors... no use of the Field Equations yet.
    For arbitrary U(r) and V(r), these components are not all identically zero.
    (Possibly useful: http://pages.pomona.edu/~tmoore/grw/Resources/DiagMetricb.pdf )

    As this next section says "2.1 Solving the Einstein equation"...
    For the metric to satisfy the Einstein Equations in vacuum (so the left-hand side of Eq 8 is zero), the expression on the right involving V,V', and r must equal zero, which provides a differential equation (Eq. 11 that is obtained from Eq 8) that determines what V(r) must be [for the vacuum case].
    So that if you use that V(r) back in Eq 8, that component of Ricci is zero. And so on for the rest.

    EDIT:
    Possibly useful thread:
    https://www.physicsforums.com/threads/geometric-interpretation-of-einstein-tensor.865834/
     
    Last edited: Jul 26, 2016
  9. Jul 28, 2016 #8
    The Einstein tensor is a linear machine with two slots to pass an "input" to. If you choose a time-like unit vector ##t^{\mu}## and insert it into each of these slots, you find that

    [tex]\displaystyle{G_{\mu \nu}t^{\mu}t^{\nu}=-\frac{1}{2}R_{(D-1)}}[/tex]

    So, if you choose a "time direction" on your manifold, then the Einstein tensor will tell you the average scalar curvature across the remaining spatial directions. In vacuum this is zero, which does not imply that you are in a flat space, it just means that world lines will diverge in some directions, and converge in others, so that the sum of those effects balances out to zero.
     
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