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Weyl Tensor

  1. Jan 30, 2007 #1
    Could someone show me a derivation of the Weyl tensor or link me to a good site?
     
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  3. Jan 30, 2007 #2

    robphy

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    One way to obtain Weyl is to decompose Riemann into its trace and trace-free part. Weyl is that trace-free part.

    [tex]
    C_{abc}{}^d=
    R_{abc}{}^d
    - \frac{2!}{n-2}g^{de}\left(\left(g_{a[c}R_{e]b}-
    g_{b[c}R_{e]a}\right)
    - \frac{1}{n-1}Rg_{a[c}g_{e]b}\right)
    [/tex]

    Another approach is to apply a conformal transformation to Riemann and somehow isolate that part of Riemann which is invariant under the transformation. Weyl is that conformally-invariant part. (This calculation is more involved that the algebraic calculation above.)

    For an online site with some of these ideas, try
    http://io.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/WeylTensor.htm

    For a text, try General Relativity by R.M. Wald or Relativity on Curved Manifolds by F. de Felice and C. J. S. Clarke or The Large Scale Structure of Space-Time by Stephen Hawking and G. F. R. Ellis.

    (There is also something called the "Weyl Projective Curvature Tensor", but I don't think this is what you are looking for.)
     
  4. Jan 30, 2007 #3
    Yes i know its the trace free part. So starting with the full riemann curvature tensor how would we subtract out the correct components to make it trace free?
     
  5. Jan 31, 2007 #4

    robphy

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    This might not be the best motivation... If you find a better one, post it here.

    It's convenient to work with the lowered-index tensor [tex]C_{abcd}=C_{abc}{}^e g_{ed}[/tex].

    Start by writing
    [tex]
    C_{abcd}
    = R_{abcd} +\lambda(g_{a[c}L_{d]b} - g_{b[c}L_{d]a})
    [/tex]
    where [tex]\lambda[/tex] and [tex]L_{ab}[/tex] are to be determined by requiring that the left-hand-side is totally traceless. This form is chosen because the combination on the right has the same index symmetries as the lowered-index Riemann.

    To determine [tex]\lambda[/tex] and [tex]L_{ab}[/tex], start by contracting with [tex]g^{bd}[/tex], then with [tex]g^{ac}[/tex].
     
  6. Feb 1, 2007 #5
    What does it mean to be traceless for a 4 tensor? You mean 1111,2222,3333,4444 are sum to zero right? Also waht is the notation with the sqaure brackets in the subscript?
     
  7. Feb 1, 2007 #6

    robphy

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    Traceless for this tensor means that
    [tex]C_{abcd}g^{ab}=0[/tex]
    [tex]C_{abcd}g^{ac}=0[/tex]
    [tex]C_{abcd}g^{ad}=0[/tex]
    ...etc.
    [tex]C_{abcd}g^{cd}=0[/tex]
    ...Contraction with the [inverse] metric with any pair of indices is zero [more correctly, the zero tensor of type (0,2)]. Of course, some of the above are obviously zero because of the symmetries of the curvature tensor.

    The square-brackets mean "antisymmetric part".
    Given a tensor [tex]Q_{ab}[/tex], its antisymmetric part is
    [tex]Q_{[ab]}\equiv \frac{1}{2!} (Q_{ab}-Q_{ba})[/tex].
    So, any tensor can be written as the sum of its "symmetric and antisymmetric parts".
    [tex]\begin{align*}
    Q_{ab}
    &= \frac{1}{2!} (Q_{ab}+Q_{ba})+\frac{1}{2!} (Q_{ab}-Q_{ba})\\
    &= Q_{(ab)} +Q_{[ab]}
    \end{align*}
    [/tex]


    Given a tensor [tex]Q_{abc}[/tex], its [totally] antisymmetric part is
    [tex]Q_{[abc]}\equiv \frac{1}{3!} (Q_{abc}+Q_{bca}+Q_{cab}-Q_{bac}-Q_{acb}-Q_{cba})[/tex].
    ...etc.
     
    Last edited: Feb 1, 2007
  8. Feb 7, 2007 #7

    robphy

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    By request, here is the first step:
    [tex]
    \begin{align*}
    C_{abcd}g^{bd}
    &= R_{abcd}g^{bd} +\frac{\lambda}{2}g^{bd}\left(
    (g_{ac}L_{db}-g_{ad}L_{cb})
    - (g_{bc}L_{da}-g_{bd}L_{ca})\right)\\
    0 &= R_{ac} +\frac{\lambda}{2}\left(
    (g_{ac}L-L_{ca})
    - (L_{ca}-nL_{ca})\right)\\
    &= R_{ac} +\frac{\lambda}{2}\left(
    (g_{ac}L-L_{ca})
    - (1-n)L_{ca})\right)\\
    &= R_{ac} +\frac{\lambda}{2}\left(
    (g_{ac}L+(n-2)L_{ac}\right)\\
    \end{align*}
    [/tex]
    So, by transvection with [itex]g^{ac}[/itex], we have
    [tex]
    0=R+\frac{\lambda}{2}\left(2(n-1)L\right)
    [/tex],
    or equivalently,
    [tex]\lambda L = -\frac{1}{n-1}R[/tex],
    which could be used above to find [tex]L_{ab}[/tex] in terms of [tex]R_{ab}[/tex].

    (One comment missing from above: [tex]L_{ab}=L_{(ab)}[/tex]... it's a symmetric tensor.)
     
    Last edited: Feb 7, 2007
  9. Feb 13, 2007 #8

    rdx

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    Great stuff!!

    I really like the presentation here. here's my question. To sort out the meaning of traceless in a qualitative way, I think that the trace represents radial terms, expansion/contraction related, whereas the traceless terms are more related to rotational terms, analogous to divergence and curl. So, is this a reasonable interpretation or am I way off base?
     
  10. Apr 26, 2007 #9
    Is it possible if you could give me a link on the Weyl Geometry, I have one of this papers and nothing else to learn from.

    Thanx

    D
     
  11. Apr 27, 2007 #10

    Chris Hillman

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    Caution: "Weyl geometry" can mean many things

    Caution: don't confuse "Weyl geometry", which could mean any of a dozen things, even in the context of gravitation physics, much less pure mathematics generally, with the "Weyl tensor". Weyl was a prolific researcher who made many fundamental contributions in diverse areas of mathematics, so his name is attached to many important notions, including notions used in physics. If you read the literature on classical gravitation theory, one thing "Weyl geometry" might mean is the geometry of the family of all static axisymmetric vacuum solutions, which were found by Weyl about 1918. Another is the geometry involved in Weyl's early attempt at a "unified field theory", also from about 1918, which nowadays is best treated as a certain Cartanian manifold with the role of SO(1,n) replaced by a larger group including a dilation. Both of these notions are distinct from the Weyl or conformal curvature tensor.
     
  12. Apr 27, 2007 #11
    Hello Chris

    And thank you. I have and am studying the 1918 paper, where can I find the 'axisymmetric vacuum solutions'

    Thanx
     
  13. Apr 28, 2007 #12

    Chris Hillman

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    The wondrous Weyl vacuums

    The Weyl vacuums comprise the family of all static axisymmetric vacuum solutions of the EFE. This family includes such familiar examples as the Schwarzschild vacuum solution, as well as the exact solutions corresponding to other isolated objects, such as a uniform density disk, ring, or rod. (These terms require, as it turns out, considerable qualification!) They were found by Weyl in about 1918, and can be expressed in the following memorable form:
    [tex]
    ds^2 = -\exp(2 \, u) \, dt^2 + \exp(-2 \, u) \;
    \left( \exp(2 \, v) \; (dz^2 + dr^2) + r^2 \, d\phi^2 \right), [/tex]
    [tex] -\infty < t, \, z < \infty, \; \; 0 < r < \infty, \; \; -\pi < \phi < \pi [/tex]
    where [itex]u, \, v[/itex] are axisymmetric (functions of z,r only) and satisfy the vacuum field equations, which reduce to the following simple "triangularized" system:
    [tex]
    u_{zz} + u_{rr} + \frac{u_r}{r} = 0, \; \;
    v_r = r \, \left( u_r^2 - u_z^2 \right), \; \; v_z = 2 \, r \, u_z \, u_r
    [/tex]
    Here, the first equation involves only u and it is the Laplace equation, written in cylindrical coordinates and applied to an axisymmetric function.
    That is, to obtain a Weyl vacuum solution, you can choose an arbitrary axisymmetric harmonic function and then obtain v by quadrature (partial integration). Needless to say, this is delightful because it means we can rather easily obtain a great variety of examples of exact axisymmetric static vacuum solutions.

    Even better, if we expand everything to first order in u, we find the weak-field Weyl vacuums are controlled by axisymmetric harmonic functions, which correspond to the Newtonian gravitostatic potential:
    [tex]
    ds^2 = -(1+2\, u) \, dt^2 + (1-2 \, u) \;
    \left( dz^2 + dr^2 + r^2 \, d\phi^2 \right), [/tex]
    [tex] -\infty < t, \, z < \infty, \; \; 0 < r < \infty, \; \; -\pi < \phi < \pi [/tex]
    where
    [tex]
    u_{zz} + u_{rr} + \frac{u_r}{r} = 0
    [/tex]
    Thus, we have a clear and immediate interpretation of the master metric function u in the weak-field approximation.

    However, things are not as simple as this summary would have you believe! To begin to see the difficulties, note that in the Weyl canonical chart (the cylindrical-like chart written above), the radial coordinate r does not possess the "Schwarzschild property", according to which the area of the sphere [itex]t=t_0, \, r=r_0[/itex] should be [itex]4 \, \pi \, r_0^2[/itex]. Nor does it have the "arc length property", according to which [itex]r_2-r_1[/itex] should be the length of a radial segment with [itex]\phi=\phi_0, \; t=t_0, \; z=z_0, \; r_1 < r < r_2[/itex]. Thus, it is more challenging to give a geometric interpretation of these coordinates than it is to interpret the Schwarzschild coordinates. That is, we want to give a coordinate-free characterization of the radial coordinate r we are using, as we know how to do for the Schwarzschild chart for the Schwarzschild vacuum, but it is not so obvious how to do that here.

    Why should anyone care about such quibbles? One reason to see why this issue is important is to ask how to express the Schwarzschild vacuum using the Weyl canonical chart. Since the Schwarzschild vacuum is the unique spherically symmetric static vacuum, and since we seemingly obtained above a simple correspondence between Newtonian potentials and Weyl vacuum solutions, we naturally expect that this should be given by starting with the axisymmetric harmonic function [itex]u(z,r) = -m/\sqrt{z^2+r^2}[/itex], which is of course the spherically symmetric Newtonian potential (independent of time). But in fact this gives a vacuum solution called the Chazy-Curzon vacuum, which is not in fact a spherically symmetric spacetime at all! It turns out that to obtain the Schwarzschild vacuum, we need to start with the Newtonian potential of a finite thin rod with just the right "effective mass density"! In a similar vein, I note that the Newtonian potential of a thin uniform density ray (i.e., limiting case of a rod with one endpoint drawn off to infinity) gives a Weyl vacuum which is isometric to the Rindler wedge, i.e. is locally flat.

    There is much, much more to say, but I'll only hint at a few more points. First, the Weyl vacuums can be generalized to the Ernst vacuums, the family of all stationary axisymmetric vacuum solutions of the EFE, which includes the Kerr vacuum, and which involves a system of coupled second order PDEs in two variables (the simplest formulation does not involve the metric functions directly but certain derived variables), plus a quadrature for a third metric function. Second, the Ernst and Weyl families are closely related to several other families of exact solutions. Perhaps the most interesting of these are the colliding plane wave models (CPW), and then close formal analogies led Chandrasekhar to the discovery that the Kerr and Schwarzschild vacuums can be realized (locally!) as CPW models. Third, if we compute the point symmetry group of the Weyl or Ernst system, we can find explicit solutions obeying symmetry conditions (since the Ernst system, unlike the Weyl system, is a nonlinear system of coupled PDEs, the general solution is not easy to write down!), but we also find that "mass parameters" as in the Schwarzschild vacuum are not so straightforward as Newtonian intuition suggests. Ultimately, this is because the very notion of "parameterization of solutions" is fraught with difficulties when we try to apply this notion to the solution space of a nonlinear PDE!

    A good place to start reading about any topic involving exact solutions is the monograph on exact solutions of the EFE, by Kramer et al., which is cited in full at http://www.math.ucr.edu/home/baez/RelWWW/HTML/reading.html#advanced
    However, for this topic an even better source is the superb review article by Bonnor, "Physical interpretation of vacuum solutions", Gen. Rel. Grav. 24 (1992): 551-573. For the connection with CPW models, try part II of this paper plus the monograph by Griffiths, Colliding Plane Waves in General Relativity, Oxford University Press, 1991. Various arXiv eprints discuss specific Weyl vacuum solutions, some of are "named", such as the Bach-Weyl dumbell, the Chazy-Curzon asymmetric particle, and so on. There is in fact a substantial literature on the Weyl vacuums; this is one of the best studied classes of vacuum solutions, yet in many respects it remains curiously misunderstood.
     
    Last edited: Apr 28, 2007
  14. Apr 28, 2007 #13
    As we say in Farsi Dasteh Shoma raa boosidam, thank you for such wonderful help.

    I will study and get back to you, give me a day or so

    Thank you again

    Dara
     
  15. Apr 28, 2007 #14

    Chris Hillman

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    Sure thing. The Weyl vacuums are one of my fav exact solution topics, so you can probably inveigle me into talking some more about them.
     
  16. May 2, 2007 #15
    Salaam Chris

    Toiling over the math to get all the concepts right prior to running my mouth again. got some of the arXiv papers to help out.

    A bit more time.
     
  17. May 2, 2007 #16

    Chris Hillman

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    An important point about the Weyl vacuums which generalizes to many other important "tractable" families of exact solutions of the EFE is that the integrability condition for the two first order equations giving v in terms of u via quadrature is precisely the equation for u (which happens to be the Laplace equation, for an axisymmetric function). That is we have a formula for [itex]v_z[/itex] and another formula for [itex]v_r[/itex]. But partial derivatives commute, so the partial wrt z of the rhs of the first formula must agree with the partial wrt r of the rhs of the second formula! Writing this out gives precisely the equation for u. It turns out that this phenomenon arises in other important areas of mathematics, including the theory of solitons.
     
  18. May 5, 2007 #17
    Salaam Chris

    I am ready to type in some equations and ask you questions, however no matter where I get the equations i.e. from which editor or even your own, I cannot cut and paste into this text box.

    I know I am doing something stupid, is there a trick into pasting equations here?

    Sorry for dumb question.

    D
     
  19. May 16, 2007 #18

    Chris Hillman

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    Obtaining formatted mathematical expressions in PF

    Hi, darashayda,

    I am not sure I understand the problem, but I am guessing you wanted to quote some LaTeX formatted expression from another PF post and tried to cut and paste. You'll probably need to do something more indirect: left click on the equation and paste the code you'll see in the popup window into your clipboard. Then return to this thread, hit "reply" and paste the code into your message in the edit window. Unfortunately due to an unfixed goof, you can't "preview" latex formatted expressions, so hit "save" and if any errors appear, try editing your post.

    Hope this makes sense. If not you should repost your question, describing as precisely as possible what you want to do, in the "Forum Feedback" board here at PF.
     
  20. May 16, 2007 #19
    Salaam Chris

    Yes I tried to cut & paste some latex and I did not see any rendering of the symbols and thought I was doing something wrong.

    I will start a new thread shortly with your write up and ask some questions.

    Sorry about my slowness around her

    and thanx for the resposne

    D
     
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