What is the physical relevance of this for the simplest GR models, such as Schwarzschild and Kerr spacetimes?Kretschmann scalar, ##R^{abcd}R_{abcd}##
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.What is the physical relevance of this for the simplest GR models, such as Schwarzschild and Kerr spacetimes?
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.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?
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}##?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.
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.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}##?
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.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.
Personally, I've never even heard of this decomposition. So, if you figure out anything cool about it, please post about it!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.
The notation is weird. As I recall from memory, it's something likeOff 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}##?
That looks exactly right from my reading of the code and the definition of the Hodges dual.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.
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...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,
Glad it's not just me!It took me quite a while to find that - the index in MTW is useless :(.
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.the preview function wasn't available
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]