Deriving Essential Quantities from Metric Tensor for GR Calculations

In summary: The vertical line is a partition between the "electric" part of the Riemann, and the "magnetic" part.So the electric part of the Riemann is made of the 3 2x2 matrices (01), (02), and (03) while the magnetic part of the Riemann is made of the 3 2x2 matrices (23), (13), and (12).In MTW, they write the electric part of the Riemann as E = *R* | *R*, where *R* is the Levi-Civita symbol.The (01) part of the "Electric" part of the Riemann is$$E_{01} = R
  • #1
Arman777
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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?
 
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  • #2
I would say Christoffel symbols, and also you can try to implement something related to the geodesic equations.
 
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  • #3
Kretschmann scalar, ##R^{abcd}R_{abcd}##
 
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  • #4
Thanks, I ll try those as well.
 
  • #5
Ibix said:
Kretschmann scalar, ##R^{abcd}R_{abcd}##
What is the physical relevance of this for the simplest GR models, such as Schwarzschild and Kerr spacetimes?
 
  • #6
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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  • #7
dextercioby said:
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.
 
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  • #8
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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  • #9
Arman777 said:
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.
 
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  • #10
pervect said:
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}##?
 
  • #11
Ibix said:
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.
 
  • #12
romsofia said:
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.
 
  • #13
Ibix said:
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!
 
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  • #14
Ibix said:
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.
 
  • #15
pervect said:
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.
pervect said:
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...
 
  • #16
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.
 
  • #17
pervect said:
It took me quite a while to find that - the index in MTW is useless :(.
Glad it's not just me!
pervect said:
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.

Will have a re-read of your post.
 
  • #18
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, [itex]{-\hspace{-.75ex}G}[/itex]=*Riemann* (analog of Maxwell=*Faraday), which has components
[tex] {-\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)} [/tex]
The Einstein curvature tensor, which is symmetric (exercise 13.11)
[tex]
G^{\beta}{}_{\delta}=
{-\hspace{-.75ex}G}^{\mu\beta}{}_{\mu\delta};
\qquad G_{\beta\delta}=G_{\delta\beta}
\qquad\mbox{(13.47)
}[/tex]

[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.

See also, p. 326 (Exercise 13.11)
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 [itex] S_{[ab][cd][ef]}[/itex] 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.
 
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  • #19
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.
 
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  • #20
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.
 
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1. What is the metric tensor in general relativity?

The metric tensor is a mathematical object used in general relativity to describe the curvature of spacetime. It contains information about the distances and angles between points in spacetime, and is essential for making calculations in general relativity.

2. How is the metric tensor used to derive essential quantities in general relativity?

The metric tensor is used to calculate the essential quantities in general relativity, such as the curvature of spacetime and the geodesic equations. By plugging the metric tensor into the Einstein field equations, we can solve for these quantities and gain a better understanding of the behavior of matter and energy in the universe.

3. What are some common essential quantities derived from the metric tensor in general relativity?

Some common essential quantities derived from the metric tensor include the Ricci tensor, the Ricci scalar, and the Einstein tensor. These quantities are used to describe the curvature of spacetime and the effects of matter and energy on this curvature.

4. How is the metric tensor related to the concept of spacetime curvature?

The metric tensor is directly related to the concept of spacetime curvature in general relativity. The components of the metric tensor describe the geometry of spacetime, and changes in these components indicate changes in the curvature of spacetime. This allows us to understand how matter and energy affect the curvature of spacetime.

5. What are some challenges in using the metric tensor for GR calculations?

One of the main challenges in using the metric tensor for general relativity calculations is its complexity. It involves multiple components and equations, and can be difficult to work with. Additionally, the metric tensor may not always be known or easily obtainable, making it challenging to use in certain situations. Understanding the physical meaning of the components and equations of the metric tensor can also be a challenge for scientists.

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