# MTW Chapter 21 Arcane Language

1. Nov 10, 2017

### TerryW

1. The problem statement, all variables and given/known data
Occasional, MTW's use of language is a bit obscure. I've highlighted a section from p491 (attached) which I just can't make any sense of.

I know about the 20 components of Rαβγδ but what does "these twenty components are arbitrary to the extent of the six parameters of a local Lorentz transformation" actually mean? And are they actually referring to Rαβγδ, because it goes on to say "However, (4)R is the only one of these 14 quantities...." which seems to imply that we aren't talking about the 20 Rαβγδ. So what are the 20 components? And what are the 6 parameters of the local Lorentz transformation?

Also, in the context of a local Lorentz transformation, where gαβ,γ= 0, all the values of Rαβγδ are linear in the second derivatives because Rαβγδ = Γαβδ,γ - Γαβγ,δ because the ΓΓ terms are all zero.

2. Relevant equations
See above

3. The attempt at a solution
See above

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2. Nov 10, 2017

### phyzguy

(1) The six components of an arbitrary Lorentz transformation are three rotations and three boosts. You can rotate around any of the three spatial coordinate axes and boost along any of these same axes. Any Lorentz transformation can be decomposed in this manner.

(2) Remember that you are trying to build a scalar quantity out of the Riemann tensor. I think they are saying that R is the only scalar quantity that you can construct that is linear in the second derivatives of g.

3. Nov 13, 2017

### TerryW

Thanks phyzguy,

This hasn't helped me very much though.

I don't think these (six components of an arbitrary Lorentz transformation) can really be the objects (independent local features of curvature) referred to as they are not of the same type (ie Rαβγδ). When you carry out a Lorentz transformation, you still end up with the Minkowski metric, for which all values of Rαβγδ are zero.

I'm sure that this is what they are saying, but my original post highlighted the fact that all the Rαβγδ objects meet the criterion of being linear in the second derivatives.

So my questions still hold - What are the 20 "curvature invariants"? Why can we ignore 6 of them? How come (4)R is one of the 20 "curvature invariants"?

Regards

TerryW

4. Nov 29, 2017

### TSny

I looked at the paragraph in MTW that you are trying to understand. I have read it many times and I cannot follow it either (probably due to my lack of ability).

For the action principle you need a scalar (invariant) quantity determined by the metric $g_{\mu \nu}$ and its derivatives. Using the Riemann curvature tensor, there are many scalars that you can construct, especially if you include covariant derivatives of the curvature tensor. These scalars are called curvature invariants. You can find some information about these invariants doing a web search.

For example

https://en.wikipedia.org/wiki/Curvature_invariant

At this link http://cds.cern.ch/record/580327/files/0209024.pdf you can see that they mention that there are 14 curvature invariants of zeroth order (i.e., the invariants do not involve derivatives of the curvature tensor).

I have not found a link that proves there are 14 such invariants. In particular, I cannot find an argument that is similar to MTW's argument.

There are some relevant threads on Physics Forums. This one gives a detailed argument for the choice of the Ricci scalar as the Lagrangian

https://www.physicsforums.com/threa...hilbert-action-come-from.449920/#post-3021374

More discussion here

Here are some general comments about the Hilbert action from page 114 of Sean Carroll's notes
https://arxiv.org/pdf/gr-qc/9712019.pdf

Anyway, none of this helps much with deciphering the particular wording of the paragraph in MTW. But even if we could understand the argument for why 14 invariants, it doesn’t seem to matter much. The rest of the paragraph makes no use of the 14. It simply goes on to state that only one of these 14 curvature invariants is linear in the second derivatives of the metric tensor, and this is the Ricci scalar R. So, it seems to me that the argument of MTW concerning the 20 independent curvature components (which somehow implies that there are 14 curvature invariants) is not very important to the final conclusion.

But I'm with you. I don't follow the argument that takes you from 20 independent components of the curvature tensor to 14 curvature invariants.