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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:

    http://en.wikipedia.org/wiki/Gödel_metric

    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

    Mentz114

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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

    Mentz114

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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

    Mentz114

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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

    PeterDonis

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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

    Mentz114

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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

    bcrowell

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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

    Mentz114

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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

    DrGreg

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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

    Mentz114

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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

    DrGreg

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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

    Mentz114

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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

    Mentz114

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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.

    Inconclusive.
     
  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

    Mentz114

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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
  22. Jul 27, 2015 #21

    Mentz114

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    Yes. I worked out the symmetries of the indexes of ##C_{\mu\nu\alpha\beta}## as presented by Maxima - swapped them to match in the code and now I agree with your tidal tensor.

    I always favoured your result because of the oblateness. The vorticity vector shows that the matter (which is evenly distributed ) is rotating around the y-axis and so we would expect it to shrink along the y-direction and expand in the x and z-directions. Although how one shrinks a universe I don't know.
     
  23. Jul 27, 2015 #22

    PeterDonis

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    Remember that the tidal tensor describes the behavior of geodesics, not "the universe". In other words, if you have a cloud of freely falling test particles that are initially at rest relative to each other, over time the cloud will shrink along the y direction and expand in the x and z directions.
     
  24. Jul 28, 2015 #23

    Mentz114

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    I think the tidal tensor in the local frame is specific to a congruence which does not have to be geodesic. However this does not invalidate your point.

    I suppose because this is Weyl curvature we can introduce our ball of particles which does not interact with the rotating dust. The comoving ball will be rotating with the dust (which comes out in the vorticity of the congruence) and shrinking along the y-axis while expanding along x and z. It makes sense.
     
  25. Jul 29, 2015 #24
    Thanks for the help. I now have one more thing I want to ask about:

    When you derived ∂t earlier (in the process of deriving the 4-velocity vector), here is my hypothesis as to how you did it:

    When I first learned tensor analysis, the playlist taught me how to derive basis vectors from coordinate systems (such as spherical coordinates) that would then be used to derive metric tensor elements. According to what I learned then, the first thing you need is a vector field of transformation properties R ("vector field of transformation properties" is not an official name for the vector field R, but I don't know what else to call it. You'll see why I call it that in a minute.)

    In 3D spherical coordinates for example, the vector field R is as follows:

    R= ( rsin(Θ)cos(∅) , rsin(Θ)sin(∅) , rcos(Θ) )

    [Notice that the components of this vector are how you transform from spherical coordinates to Cartesian coordinates, which is why I call this the vector of transformation properties]

    After you have your R, you can then derive your basis vectors by doing ∂R / ∂xi. The notation that the playlist used for the basis vectors was ei. Here is an example of a basis vector calculation:

    er = ( sin(Θ)cos(∅) , sin(Θ)sin(∅) , cos(Θ) ) [As you can see, I just differentiated R with respect to r]

    Once you had all of your basis vectors derived, you could just dot product them together to derive the elements of your covariant metric tensor.

    Now it is my speculation that er is the same as ∂r just written in a different way.

    Is this speculation correct?

    Potential evidence to back up my speculation is the fact that vectors can be written in the form V = Viei, and you yourself also mentioned that vectors could be written as T∂t + X∂x + Y∂y + Z∂z. Furthermore, you called this a tangent space, while the video playlist I watched sometimes called ei tangential vectors. That is why I have this speculation.

    Now if this speculation is correct, then how exactly would I derive R and the basis vectors ∂μ for a metric such as the Godel metric where I am only given the line element? Even if it is not correct, still how do I derive said basis vectors?
     
  26. Jul 29, 2015 #25

    Mentz114

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    So R contains the transformation of coordinates from a Cartesian basis to the spherical polar basis, for instance
    ##x=r \, \sin\theta \, \cos\varphi##

    To convert this to a transformation of differentials we differentiate
    ##dx= \partial_r(r \, \sin\theta \, \cos\varphi)dr##. The ##\partial_r## here is the differentiating operator, not ##\vec{\partial_r}## which is a basis vector. One usually can work it out from the context.

    Now substituting for ##dx^2## and the other differentials in the metric gives the transformed metric.
    This is the process of finding the covariant line element after the transformation of contravariant vectors.

    I don't think the 'tangent' is means the same in the two contexts.

    We don't need an R because we already have the final form of the metric.

    What we are trying to do is find the coordinate ( also called holonomic) basis vector ##e^\mu## or cobasis vectors ## e_\mu## of the spacetime defined by the metric.
    The line element is ##ds^2 = - e_t\cdot e_t + e_x\cdot e_x + e_y\cdot e_y + e_z\cdot e_z## ( for -+++ signature).

    It is clear that ##e_x=\sqrt{g_{xx}}dx## and ##e_y=\sqrt{g_{yy}}dy## are correct, but there will be both ##dt## and ##dz## components in ##e_z## and ##e_t##.

    I would just put in variables ##a,b,c,d## so ##e_t=adt+bdz## and ##e_z=cdt+ddz## and solve algebraically. There are probably more elegant ways to understand (explain) this but I generally just calculate. People who have practice at this and can think fast are able to read the components from the line element.

    (This business of basis changing is handled better by the tetrad formalism which I have avoided here, at some cost)
     
    Last edited: Jul 29, 2015
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