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Weyl tensor for the Godel metric interpretation

  1. Jul 21, 2015 #1
    I have recently derived both the purely covariant Riemann tensor as well as the purely covariant Weyl tensor for the Gödel solution to Einstein's field equations. Here is a wiki for the Gödel metric if you need it:


    There you can see the line element I was working with.

    Now I will give you my Ricci tensor:

    R00 = 1
    R03 and R30 = ex
    R33 = e2x

    Every other Ricci tensor element is 0.

    Now for my Riemann tensor Rabcd:

    R0110 & R1001 = -1/(4ω2)
    R1010 & R0101 = 1/(4ω2)
    R0330 & R3003 = -e2x/(8ω2)
    R3030 & R0303 = e2x/(8ω2)
    R0113 & R1301 & R1031 & R3110 = -ex/(4ω2)

    R1013 & R0131 & R3101 & R1310 = ex/(4ω2)

    R1331 & R3113 = -3e2x/(8ω2)
    R3131 & R1313 = 3e2x/(8ω2)

    Every other element was 0.

    Now for the Weyl tensor Cabcd:

    C0110 & C1001 = -1/(12ω2)
    C1010 & C0101 = 1/(12ω2)
    C0220 & C2002 = 1/(6ω2)
    C2020 & C0202 = -1/(6ω2)
    C0330 & C3003 = -e2x/(24ω2)
    C3030 & C0303 = e2x/(24ω2)
    C0113 & C1301 & C1031 & C3110 = -ex/(12ω2)

    C1013 & C0131 & C3101 & C1310 = ex/(12ω2)

    C0223 & C2302 & C2032 & C3220 = ex/(6ω2)

    C2023 & C0232 & C3202 & C2320 = -ex/(6ω2)

    C1221 & C2112 = 1/(12ω2)
    C2121 & C1212 = -1/(12ω2)

    C1331 & C3113 = -e2x/(6ω2)
    C3131 & C1313 = e2x/(6ω2)

    C2332 & C3223= 5e2x/(24ω2)
    C3232 & C2323= -5e2x/(24ω2)

    If you'd like to verify my math, feel free to put this in some software, because I don't have any.

    Anyway, I learned that the R00 element of the Ricci tensor tells you how the volume of objects traveling along geodesics change, while the Weyl tensor tells you how the shape of objects change as they travel along geodesics due to space-time curvature.

    Using my Gödel metric as an example, would someone please explain how I get the info about how objects change shape using this Weyl tensor? In other words, what info does the value of C0110 for example give me (as well as the other elements)?

    Thank you.
  2. jcsd
  3. Jul 21, 2015 #2


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    Please calculate the Einstein tensor for the local frame given there and compare with the value on the wiki page.

    My frame field calculation agrees with the wiki Einstein tensor, and with your Ricci components when transformed.
    Last edited: Jul 21, 2015
  4. Jul 21, 2015 #3


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    I calculated the tidal tensor ##T_{ab}=C_{a\mu b\nu}u^\mu u^\nu## ( ##a,b## are spatial indexes) in the coordinate basis using Maxima. ##u^\mu = \partial_t / \sqrt{g_{00}}##. This gives the three components
    ##T_{xx}=-1/6,\ T_{yy}=1/3 \ T_{zz}= -e^{2x}/12 ##.

    This is not a very meaningful form but converting to the mixed components gets
    ##{T^{x}}_{x}=-w^2 /3,\ {T^{y}}_{y}=2w^2/3, \ {T^{z}}_{z}= -w^2 /3 ##. (This is identical to the frame field result so I have high confidence in it).

    which is traceless as required. The tidal components suggest equal shrinking in the x and z directions with expansion in the y-direction.
    Last edited: Jul 21, 2015
  5. Jul 21, 2015 #4
    I see, so the components of this tidal tensor tells how much an object would expand or shrink in the direction of the corresponding component? Is this correct? Also, is it correct to say that mixed tensor forms are the forms that actually give you information, while covariant forms are just used for conversions and mathematical simplicity?

    Is the vector uμ 4-velocity?

    Next, you say a and b are spatial indices, so they only span 1 through 3 correct?

    Finally, where do gravitational waves and closed timelike curves come into play with these tensors?
  6. Jul 21, 2015 #5


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    It is true in the case of the tidal tensor.

    In the case of a spatially projected tensor it is a cheap trick to mimic the frame field calculation. To be sure of getting local physics do all calculations in a frame field.

    Yes. It is the velocity of the 'fiducial' frame, which is at rest wrt the source.

    Yes. ##T## is a spatial tensor in the local Minkowski coordinates.

    I cannot answer that from memory. Petrov classification is type 1 (I) which is non-radiating I think.

    I'm glad you posted this because the importance and distinction of the Weyl tensor is something to remember.
    Last edited: Jul 21, 2015
  7. Jul 21, 2015 #6


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    In general, gravitational waves involve fluctuations in the Weyl tensor. However, there are many spacetimes which have variations in the Weyl tensor that are not gravitational waves. As Mentz114 notes, Gödel spacetime is one of them.

    I don't know of any simple relationship between the presence or absence of closed timelike curves and the Weyl tensor (or any of the others you mention).
  8. Jul 22, 2015 #7
    I forgot to ask about this:

    What are you differentiating with respect to time here? Are you saying that you are differentiating the square root of g00 with respect to time, or are you saying that the formula is the derivative of something with respect to time divided by the square root of g00?
  9. Jul 22, 2015 #8


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    I presume you are referring to the vector ##u^\mu=\partial_t /\sqrt{g_{00}}##. This is an algebraic way of writing a vector in terms of basis vectors.
    A tangent space vector ( or vector field) can be written ##T\vec{\partial_t} +X\vec{\partial_x} +Y\vec{\partial_y} +Z\vec{\partial_z} ## where ##T,X,Y,Z## are numbers or functions of the coords.

    If you're not familiar with this, think of ##u^\mu## as ##(1/\sqrt{g_{00}},0,0,0)##
    Last edited: Jul 22, 2015
  10. Jul 22, 2015 #9


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    Maxima and ctensor are free and open-source.
  11. Jul 23, 2015 #10
    Hey I noticed a problem about uμ being ##(1/\sqrt{g_{00}},0,0,0)## .

    g00 = -1/(2ω2) in this case, so the square root of g00 would be imaginary.

    How should I resolve this issue? Should I just take the absolute value of g00 first and then square root it?
  12. Jul 23, 2015 #11


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    There's nothing to resolve. ##g_{00}## is always greater than zero. The negative sign comes from the signature (-1,1,1,1) or the sign of ##dt^2##.
  13. Jul 23, 2015 #12


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    That's not quite true.

    If you use the convention of a signature (−1,1,1,1), then ##g_{00} < 0## and the 4-velocity is ##(1/\sqrt{-g_{00}},0,0,0)##.

    If you use the convention of a signature (1,−1,−1,−1), then ##g_{00} > 0## and the 4-velocity is ##(1/\sqrt{g_{00}},0,0,0)##.

    Or write it as ##(1/\sqrt{|g_{00}|},0,0,0)## which works either way.
  14. Jul 23, 2015 #13


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    I've been expecting this post. Thank you. My way also works for both signatures without changing the sign in the radical. But it is not conventional.
  15. Jul 23, 2015 #14


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    True enough.
    And that will only cause confusion if you redefine symbols to mean something different than how everyone else defines them.
  16. Jul 23, 2015 #15
    Hey, I just calculated those 3 components of the Tidal tensor tha you did by hand just to make sure that what I got matched what you got. It did match except Txx and Tzz were positive for me where as Tyy was negative for me (your results are vice versa).

    This is just because of sign convention correct? I used the (- + + +) signature in my calculations. I assume you used (+ - - -). Is this correct?
  17. Jul 23, 2015 #16


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    I used -+++ but I think your signs on ##T_{aa}## may be correct. I'm glad we have agreement (up to a sign) in any case.

    I'll check if there is something like swapped anti-symmetric indexes in my calculation.
    Last edited: Jul 23, 2015
  18. Jul 26, 2015 #17
    Did you figure out why our signs turned out different by any chance?
  19. Jul 26, 2015 #18


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    I think the signs on the Weyl components Maxima gets are reversed from yours. It's difficult to check because Maxima alters the index ordering.

  20. Jul 27, 2015 #19
    By the way, here is a question:

    We know that the tidal tensor Tab tells how objects in a space-time expand and contract due to tidal forces. Well we have both seen in our calculations examples of tidal tensor elements, like Txx = ω2/3

    Now here is where the question comes in. According to the Godel solution wiki (http://en.wikipedia.org/wiki/Gödel_metric) the term ω refers to the angular velocity of the body around the y-axis. Now angular velocity has units of radians per second (or some other angular unit per unit time). Therefore ω2 would have units rad2/s2

    This would in turn be the units of the tidal tensor elements (in the case if this metric). Shouldn't tidal forces however, have units of Newtons? Are there perhaps any hidden terms (like any c terms) that you set equal to 1 when you derived uμ that would change the units of the tidal tensor elements to the appropriate units for tidal forces had you included said terms?
  21. Jul 27, 2015 #20


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    That's a good question. Our calculations have used geometric units where ##c=G=1## and everything has dimensions ## [T]^n## which is equivalent to ##[L]^n## !

    The good news is that Dr Wiki has some help here https://en.wikipedia.org/wiki/Geometrized_unit_system
    Last edited: Jul 27, 2015
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