# What can be derived from the metric tensor?

• A
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I am working on a computational project about General Relativity. In this process, I want to code 'the stuff' that can be derivable from the metric tensor. So far, I have coded Riemann Tensor, Weyl Tensor, Einstein Tensors, Ricci Tensor, Ricci scalar. What are the other essential/needed quantities in the GR calculations that can be coded?

vanhees71

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I would say Christoffel symbols, and also you can try to implement something related to the geodesic equations.

Arman777, vanhees71 and Dale
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Kretschmann scalar, ##R^{abcd}R_{abcd}##

Arman777
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Thanks, I ll try those as well.

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Kretschmann scalar, ##R^{abcd}R_{abcd}##
What is the physical relevance of this for the simplest GR models, such as Schwarzschild and Kerr spacetimes?

e.g. in schwarzchild ##{R^{abcd}}_{abcd} \propto M^2/r^6## which diverges at ##r=0## any any coordinate chart, so you cannot analytically extend the metric through ##r=0##

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vanhees71 and Ibix
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What is the physical relevance of this for the simplest GR models, such as Schwarzschild and Kerr spacetimes?
It's a scalar measure of curvature. So is the Ricci scalar, but that's zero everywhere in vacuum because ##R^{ab}=0## so isn't that useful for Schwarzschild and Kerr. As etotheipi says it goes to infinity at the singularity, but not at the various horizons, and also it goes to zero as ##r## goes to infinity so you can see (or at least suspect) asymptotic flatness.

PeterDonis, dextercioby, vanhees71 and 1 other person
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Also, it's a scalar field dependent solely on the metric (at least if we assume the Levi-Civita connection). If you get a metric that you suspect is something familiar in funny coordinates ##(u,v,w,x)## then it may be useful to calculate the Kretschmann scalar ##K(u,v,w,x)## and see if you can see a relationship to ##K(t,r,\theta,\phi)##, or whatever the familiar coordinates are. (Ditto other scalar curvatures.)

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Kolmo, PeterDonis and vanhees71
Staff Emeritus
I am working on a computational project about General Relativity. In this process, I want to code 'the stuff' that can be derivable from the metric tensor. So far, I have coded Riemann Tensor, Weyl Tensor, Einstein Tensors, Ricci Tensor, Ricci scalar. What are the other essential/needed quantities in the GR calculations that can be coded?
It may be beyond the scope of what you want to do, but some one of the more useful functions is the ability to automate a change of basis, particularly to be able to use a set of coordinate basis vectors and orthonormal basis vectors.

GRTensor has one approach of doing this, and does a few other things such as kinematic decompositions, such as https://en.wikipedia.org/wiki/Bel_decomposition.

Ibix
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GRTensor has one approach of doing this, and does a few other things such as kinematic decompositions, such as https://en.wikipedia.org/wiki/Bel_decomposition.
Off topic dumb question - I recognise ##^\star R_{abcd}## as the Hodge dual of ##R_{abcd}##, but what's ##^\star R^\star{}_{abcd}##? The Sage code here (I guess, never having seen Sage code before) seems to imply that it's the Hodge dual of the Hodge dual - but if that's the case, what's wrong with notating it ##^{\star\star}R_{abcd}##?

romsofia
Off topic dumb question - I recognise ##^\star R_{abcd}## as the Hodge dual of ##R_{abcd}##, but what's ##^\star R^\star{}_{abcd}##? The Sage code here (I guess, never having seen Sage code before) seems to imply that it's the Hodge dual of the Hodge dual - but if that's the case, what's wrong with notating it ##^{\star\star}R_{abcd}##?
Don't see how that could be the case, ##\star \star x = x, \text{or} -x## so don't see how it would be useful.

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Don't see how that could be the case, ##\star \star x = x, \text{or} -x## so don't see how it would be useful.
Yeah. On closer reading of the code the ##\star## used in that code appears to be a four index tensor ##S^{ef}{}_{ab}##, which they apply twice: ##^\star R_{abcd}=S^{ef}{}_{ab}R_{efcd}## and ##^\star R^\star{}_{abcd}=S^{ef}{}_{ab}S^{gh}{}_{cd}R_{efgh}##. I think I need to find some more formal reading about Bel decomposition.

romsofia
Yeah. On closer reading of the code the ##\star## used in that code appears to be a four index tensor ##S^{ef}{}_{ab}##, which they apply twice: ##^\star R_{abcd}=S^{ef}{}_{ab}R_{efcd}## and ##^\star R^\star{}_{abcd}=S^{ef}{}_{ab}S^{gh}{}_{cd}R_{efgh}##. I think I need to find some more formal reading about Bel decomposition.
Personally, I've never even heard of this decomposition. So, if you figure out anything cool about it, please post about it!

Ibix
Staff Emeritus
Off topic dumb question - I recognise ##^\star R_{abcd}## as the Hodge dual of ##R_{abcd}##, but what's ##^\star R^\star{}_{abcd}##? The Sage code here (I guess, never having seen Sage code before) seems to imply that it's the Hodge dual of the Hodge dual - but if that's the case, what's wrong with notating it ##^{\star\star}R_{abcd}##?
The notation is weird. As I recall from memory, it's something like

$$*R_{klab} = \epsilon_{ijkl} R^{ij}{}_{ab}$$

##\epsilon## being the Levi-Civita symbol.

so one does not actually taking the dual of R to get the magnetic part of the Riemann, one divides the tensor into two parts, a "front half" and a "back half", tghe front half having the first two indices, the back half having the last two indicies, and one essentially takes the hodges dual of only the front part when computing the magnetic part of the Riemann.

*R* then does this twice, once for the front part and once for the back part.

There is a discussion of the "Electric part of the Riemann" and the "Magnetic Part of the Riemann" in Misner, Thorne, Wheeler's "Gravitation", on pg 360, exercise 14.14, which is not written for a general tensor basis, but is specific to an orthonormal basis. This is implied by MTW's usage of the "hat" symbol in the tensor indices.

If one consider R_{ab**}, when a=b the tensor component is zero, by the Bianchi identies. Therefore, the only possible pairs of values of (ab) are (01, 02, 03, 21, 23, 13).

There is a duality relationship between (01) and (23), between (02) and (13), and between (03) and (12).

In an orthonormal basis, MTW writes

$$R = \begin{bmatrix} E & | & B \\ - \ - \ - \\B^{T} & | & F \end{bmatrix}$$, so E is a 3x3 matrix, B is a 3x3 matrix, and F is a 3x3 matrix.

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The notation is weird. As I recall from memory, it's something like

$$*R_{klab} = \epsilon_{ijkl} R^{ij}{}_{ab}$$

so one does not actually taking the dual of R to get the magnetic part of the Riemann, one divedes the tensor into two parts, and one takes the hodges dual of only the front part.
That looks exactly right from my reading of the code and the definition of the Hodges dual.
There is a discussion of the "Electric part of the Riemann" and the "Magnetic Part of the Riemann" in Misner, Thorne, Wheeler's "gravitation", on pg 360, exercise 14.14,
Ah. I sometimes feel like I don't know how to use the index in MTW because every time I can't find something in it someone comes along with a page reference...

Staff Emeritus
It took me quite a while to find that - the index in MTW is useless :(.

I edit my post a few times - the preview function wasn't available, so I had to fix up a few things, and I expanded the text as I fixed them up, so it might be worth re-reading.

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It took me quite a while to find that - the index in MTW is useless :(.
the preview function wasn't available
It changed a few weeks ago - the button in the top right of the editor that looks like a magnifying glass over a piece of paper toggles the editor between edit and preview mode. Enter some LaTeX, then click/tap the button and you'll see it render along with any other formatting. Click the button again to go back to editing.

Homework Helper
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The "double-dual" of Riemann (that is, the "left and right dual" of Riemann)
is defined in MTW, p. 325 [13.5. METRIC-INDUCED PROPERTIES OF RIEMANN], written as a "dashed-G".
The "double-dual" of Riemann has zero divergence, and the Einstein tensor is a trace of that double-dual of Riemann.

The double dual of Riemann, ${-\hspace{-.75ex}G}$=*Riemann* (analog of Maxwell=*Faraday), which has components
$${-\hspace{-.75ex}G}^{\alpha\beta}{}_{\gamma\delta}\equiv \displaystyle\frac{1}{2}\epsilon^{\alpha\beta\mu\nu}R_{\mu\nu}{}^{\rho\sigma}\frac{1}{2}\epsilon_{\rho\sigma\gamma\delta} =-\frac{1}{4} \delta^{\alpha\beta\mu\nu}_{\rho\sigma\gamma\delta}R_{\mu\nu}{}^{\rho\sigma} \qquad\mbox{(13.46)}$$
The Einstein curvature tensor, which is symmetric (exercise 13.11)
$$G^{\beta}{}_{\delta}= {-\hspace{-.75ex}G}^{\mu\beta}{}_{\mu\delta}; \qquad G_{\beta\delta}=G_{\delta\beta} \qquad\mbox{(13.47) }$$

[Hodge-]Duals are introduced on p.88 (Exercise 3.14),
the "permutation tensor" [https://en.wikipedia.org/wiki/Kronecker_delta#Generalizations (essentially determinants with Kronecker-deltas as elements), see also Wald B.2.12 ; be aware of signature conventions ] on p.87.

and p. 343 [14.2 FORMING THE EINSTEIN TENSOR].

This suggests a pattern (that I can't make more precise right now):
• the "Faraday" tensor (which gives Faraday and Gauss-for-B) is like a "curvature" [not spacetime, but https://en.wikipedia.org/wiki/Connection_(vector_bundle)#Curvature ] for massless-spin-1; the "Maxwell" tensor (its Hodge dual [MTW p.114]) is divergence-free and is associated with the sources of the field.
• the Riemann curvature (for massless spin-2),
the double-dual-of-Riemann [which could be called "the [4-index] Einstein Curvature"] is divergence-free and is associated with the sources
• presumably,
a massless spin-3 field would have 3 symmetric antisymmetric-pairs $S_{[ab][cd][ef]}$ and its triple-dual (Hodge on each antisymmetric pair) have the properties of... ?
... probably best studied group-theoretically.

... but this would get away from the original question.

Kolmo, Ibix, vanhees71 and 1 other person
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For what it's worth, the electrogravitic tensor E[##\vec{X}##] is the one I use the most. Basically, by picking the vector field ##\vec{X}##, it's integral curves define the motion of some reference observer, which also defines "the direction of time", and an associated (local) simultaneity convention. We don't need to know how to take the left and right duals to compute E.

If one has a static observer in a static metric, the magnetogravitic tensor ##B[\vec{x}]## is zero. This is similar to the electrostatic case in E&M. And it can be used as an approximation as well, similar to the electrostatic approximation in E&M. The decomposition process basically somewhat similar to the way we break E&M's Faraday tensor into electric and magnetic fields. It's not quite the same, there is no direct equivalent to the topogravitic part of the decomposition in E&M.

Note that we need to specify some observer to decompose the Faraday tensor into the electric part and the magnetic part - this is the role of the vector field ##\vec{X}##, to specify the observer.

Next, in exercise 14.14 of MTW, the second part of the exercise shows that in a vacuum, the topogravitic part of the tensor, called F in the exercise, is equal to the electrogravitic part, E.

So under the proper circumstances, the nonzero components of the 256 component Riemann are described by the 9 component E. I'll talk more about how that can be reduced even further.

E has a simple physical interpretation - the proper choice of coordinate axes, parallel to the eigenvectgors of the matrix, diagonalizes it. The eigenvalues in the diagonalized form corrrespond to the eignevalues.

So the eigenvectors of E define the principal stress axes of some test observer that is experiencing what we might call "tidal forces", and the eignevalues are the magnitude of these tidal forces.

This is a bit of a digression from the original point, which was basically that GRTensor could be an inspiration for things to calculate, though calclulating everything it does would probably be more ambitious than the OP wants.

Also, a change of basis (at least being able to switch between a coordinate basis and an orthonormal basis) are very useful things to be able to calculate.

For more specifics, see for instance http://grtensor.phy.queensu.ca/Griihelp/grt_objects.help and http://grtensor.phy.queensu.ca/Griihelp/grt_operators.help. The "objects" are self contained, the operators require additonal arguments, typically arguments that specify an observer via a vector field.

Kolmo, vanhees71, Ibix and 1 other person
Gold Member
I am currently working on a friendly GUI, and it is almost done. Without typing a single code, one will obtain the tensors/scalars by typing the metric tensor components and choosing the coordinate system. I will implement some well-known metrics as well. I'll share the page when it's done.

robphy