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Help me make this mathematical connection in general relativity

  1. Nov 11, 2014 #1
    As some may know, I have been studying the Morris-Thorne wormhole metric for quite some time now.

    ds2= -c2dt2 + dl2 + (b2 + l2)(dθ2 + sin2(θ)d∅2)

    Now, from this space-time interval, it is easy to see how I would deduce the following metric tensor:

    g00= -1
    g11 = 1
    g22= (b2 + l2)
    g33= (b2 + l2)sin2(θ)

    where
    x0 = ct
    x1= l
    x2= θ
    x3= ∅

    Now with this metric tensor, the Christoffel symbols yield:
    Γ122= - L
    Γ133= - Lsin2(θ)
    Γ212 and its counterpart with the lower indices reversed (Γ221) = L/(b2 + l2)

    Γ233 = - sin(θ)cos(θ)

    Γ313 and its counterpart = L/(b2 + l2)
    Γ323 and its counterpart= cot(θ)

    If you plug these Christoffel symbols into the Riemann tensor formula:

    Rabmv = (∂Γavb/ ∂xm) - (∂Γamb/ ∂xv) + ΓamcΓcvb - ΓavcΓcmb

    you will see that the following Riemann tensor elements equal as follows (I am doing these specific elements for a certain reason):

    R2323 = (b2sin2(θ))/(b2 + l2)
    R0323 , R1323 and R3323 = 0

    Now, I did these specific elements for the purpose of calculating the purely covariant version of the Riemann tensor element R2323 using the formula:

    R2323 = g2fRf323

    Doing this yields the following result:

    R2323= b2sin2(θ)

    The following sources however, (http://www.spacetimetravel.org/wurmlochflug/wurmlochflug.html) (http://www.physics.uofl.edu/wkomp/teaching/spring2006/589/final/wormholes.pdf [Broken]) (pg. 4 on the PDF) said that Rθ∅θ∅ (which is in fact R2323) = b2/ (b2 + l2)2

    This is the part that I do not understand. I have shown you clearly with the use of formulas how I derived

    R2323= b2sin2(θ)

    They however, did not show any work on how they got
    R2323 = b2/ (b2 + l2)2

    Can anyone please tell me how they got that (starting with the metric tensor in matrix form and going forward)?

    I know that they worked in an orthonormal basis. My problem with that is this: I know what a basis is and I know what an orthonormal basis is. I know what basis vectors are and I know (or at least I think I know) how to derive basis vectors using a metric tensor. What I do not know, is what to do with said basis vectors after I have derived them or how to take a tensor product (despite the fact that I know what one is).

    Essentially, I don't know how you convert from a coordinate basis to orthonormal with regards to these tensors (like the Riemann).

    That is why I ask if someone can please tell me what they did differently in the beginning of their calculations to get:
    R2323 = b2/ (b2 + l2)2

    instead of

    R2323= b2sin2(θ)
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Nov 11, 2014 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    But you are not using one; you are using a coordinate basis, as you appear to be aware. That is, your basis vectors are ##\partial / \partial t##, ##\partial / \partial l##, ##\partial / \partial \theta##, and ##\partial / \partial \phi##. These are orthogonal (i.e., the inner product of any two different ones is zero), but not normalized (i.e., the inner product of each one with itself is not 1).

    Since they are already orthogonal, normalizing them is easy. I'll just write down the normalized vectors, and you should be able to see how I obtained them. The normalized vectors are ##\partial / \partial t##, ##\partial / \partial l##, ##\left( 1 / \sqrt{b^2 + l^2} \right) \partial / \partial \theta##, and ##\left( 1 / \sin \theta \sqrt{b^2 + l^2} \right)\partial / \partial \phi##.

    We can express the transformation between these two sets of basis vectors by a matrix, ##L##, such that ##\hat{e}_a L^{\mu}{}_a e_{\mu}##, where the "hatted" vectors (on the LHS) are the normalized ones, and the non-hatted ones (on the RHS) are the coordinate ones. Obviously, in this case, the matrix ##L## is diagonal, with components ##\left( 1, 1, 1 / \sqrt{b^2 + l^2}, 1 / \sin \theta \sqrt{b^2 + l^2} \right)##.

    With the matrix ##L##, you can transform any tensor from the coordinate basis to the orthonormal basis; you just use one "copy" of ##L## for each index of the tensor. For example, ##\hat{R}_{2323} = L^{\mu}{}_2 L^{\nu}{}_3 L^{\rho}{}_2 L^{\sigma}{}_3 R_{\mu \nu \rho \sigma}##. Since ##L## is diagonal, the only coordinate basis component that appears on the RHS is ##R_{2323}##, so we have

    $$
    \hat{R}_{2323} = \frac{1}{\sqrt{b^2 + l^2}} \frac{1}{\sin \theta \sqrt{b^2 + l^2}} \frac{1}{\sqrt{b^2 + l^2}} \frac{1}{\sin \theta \sqrt{b^2 + l^2}} b^2 \sin^2 \theta = \frac{b^2}{\left(b^2 + l^2\right)^2}
    $$

    which matches the references you gave.
     
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