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## Main Question or Discussion Point

Could someone show me a derivation of the Weyl tensor or link me to a good site?

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- #1

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Could someone show me a derivation of the Weyl tensor or link me to a good site?

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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 [Broken]

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.)

[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 [Broken]

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.)

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- #4

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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].

- #5

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- #6

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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.

[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.

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- #7

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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.)

[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.)

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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?

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Thanx

D

- #10

Chris Hillman

Science Advisor

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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.

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

Thanx

- #12

Chris Hillman

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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

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

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 [Broken]

However, for this topic an even better source is the superb review article by Bonnor, "Physical interpretation of vacuum solutions",

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- #13

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

Thank you again

Dara

- #14

Chris Hillman

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- #15

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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.

- #16

Chris Hillman

Science Advisor

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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

- #18

Chris Hillman

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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.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.

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.

- #19

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Salaam ChrisHi, darashayda,

I am not sure I understand the problem, 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.

here at PF.

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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