GR Metric Meaning: Unpacking Confusion

[Rob Woodside]

I usually just see the decomposition of the Riemann into two parts, the Ricci and the Weyl. I'm not positive what E is here, I assume that G+E = the Ricci and that G is the Einstein, which makes E = R g_ab, IIRC.

Yes and no, The decomposition for the fourth rank curvature takes the decomposition you know:

Curvature = Ricci piece + Weyl

one step further to:

Curvature = Curvature scaler piece + trace free Ricci piece + Weyl

For this post take curvature components as Rabcd. Then Ricci components are Rab = R^sasb and the curvature scaler is R= R^aa = R ^a^bab. The trace free part of Ricci is then Sab = Rab -(1/4) R gab where gab are the metric components. This allows

Eabcd = (1/2)( gac Sbd + gbd Sac -gad Sbc - gbc Sad )

and

Gabcd = (1/12) R (gac gbd - gad gbc )

I've found it easier to calculate with two indices up and two indices down, then one can use antisymmetric kronecker deltas in place of the metric components. For electro vac universes with Fab as the electromagnetic field and *Fab its Hodge dual, E and G become in geometric units

Eabcd = (1/2) ( Fab Fcd + *Fab *Fcd ) , Gabcd = 0

Anyway, if I'm following you correctly, the point you are making is that Riemann tensor (the curvature of space-time), at a distant point away from the sun, is entirely due to the Weyl curvature terms, because R is equal to zero where there is no matter density. And you are calling this the "non-local" component of curvature. This is interesting and useful to know, but it's considerably more advanced then the very simple point I was trying to make about how space-time (and the Earth's surface) can be considered to be locally flat. See my previous post for a fuller explanation.

The R in your last quote is Ricci. Yes, I'm calling the piece of curvature at a point that depends explicitly on the matter at that point "the local curvature" and the remaining piece of curvature at that point which depends on distant matter, "the non local curvature". The curvature outside matter is entirely Weyl. This Weyl depends on distant matter, is divergence free, and so non local. I doubt that this is more advanced than your point about a manifold being nearly flat in the small. Certainly your point is deeper.

 

[pmb_phy]

Sure, Riemann's idea was that in a tiny region around a point would look nearly flat and the smaller the region the better the approximation...so even though things look pretty flat along the geodesic the curvature survives there.

That's quite. Clifford will wrote a paper in which this point is emphasized quite nicely. The calculation he presented showed that a charged particle at rest in a Schwarzschild field, and hence in a curved spacetime, will weigh less than if the spacetime was flat and the gravitational acceleration the same. The electric field couples to the curvature. This illustrates the notion that the electric field of a charged particle is not local and therefore a charged particle cannot serve as a test particle since its pushed off its geodesic.

It is not clear to me how you could slip a curvature detector inside matter without disrupting it.

Excellant point. Conside the Newtonian case (i.e. Newtonian limit of GR). Take the case of a spherical body with a uniform mass distribution. An experimenter wants to measure the tidal forces (spacetime curvature) inside the body. He tunnels into the point of interest and digs out a small spherical cavity, just large enough to contain a very small gradiometer. He fills the tunnel back and and then uses the gradiometer to measure the tidal forces there. He will find that no matter where this point is inside such a body that the instrument measures no tidal forces at all - zero! Hence the experimenters attempt to measure the tidal force destroyed the tidal force altogether because the gravitational field in side such a cavity is perfectly uniform and hence there are no tidal forces. For derivation please see - http://www.geocities.com/physics_world/gr/grav_cavity.htm

Yes I would be interested and I'm trying to get Rindler's new book.

See What is the principle of equivalence? Hans C. Ohanian,
Am. J. Phys. 45, 903 (1977)

Abstract - The strong principle of equivalence is usually formulated as an assertion that in a sufficiently small, freely falling laboratory the gravitational fields surrounding the laboratory cannot be detected. We show that this is false by presenting several simple examples of phenomena which may be used to detect the gravitational field through its tidal effects; we show that these effects are, in fact, local (observable in an arbitrarily small region). Alternative formulations of the strong principle are discussed and a new formulation of strong equivalence (the ''Einstein principle'') as an assertion about the field equations of physics, rather than an assertion about all laws or all experiments, is proposed. We also discuss the weak principle of equivalence and its two complementary aspects: the uniqueness of free fall of test particles in arbitrary gravitational fields (''Galileo principle'') and the uniqueness of free fall of arbitrary systems in weak gravitational fields (''Newton principle'').

I know your e-mail address so if you'd like I can scan the article in an e-mail it to you.

Thanks for the kind words, Pete. Things seem a little less Raunchy here than over at SPR. What do all the letter icons and padlocks icons on the message board mean?

You're welcome. I wish I knew what those things were. I slowly learn how to use the gimmics here. But I think a padlock denotes a locked thread.

Pete

 

[pmb_phy]

The curvature scaler is just the trace of the Ricci tensor.

Okay. Now I see. This part confused me a bit since I'm not used to seeing the term "trace" applied to tensors. I'm used to the term contraction. The term "trace" as applied to matrices means to sum the diagonal components of a matrix. Thus the trace of R_{\alpha\beta} is

R_{00} + R_{11} + R_{22} + R_{33}

which, of course, is not an invariant. However I believe you are referring to the quantity R = R^{\alpha}_{\alpha} where R^{\alpha}_{\beta} = g^{\alpha\mu}R_{\mu\beta}. Is that correct?

I've never liked the term "curvature scalar" referring to R. R can be zero even when the spacetime is curved (e.g. both R = 0 and R_{\alpha\beta} vanish for a Schwarzschild spacetime ... outside the matter/star etc.) and thus can be a tad misleading.

Relevant equations can be found in my paper gr-qc/ 0410043.

Thanks. I'm anxious to read this but am unable to right now. I have back problems which prevents me from sitting in front of my computer for too long and I don't have a printer. However I e-mailed the link to a buddy of mine who will print it out and mail it to me.

I'm sure I'll have more questions when I finish reading your paper but that will be a ways off since I have a lot on my plate right now.

Pete

 

[dextercioby]

Okay. Now I see. This part confused me a bit since I'm not used to seeing the term "trace" applied to tensors. I'm used to the term contraction. The term "trace" as applied to matrices means to sum the diagonal components of a matrix. Thus the trace of R_{\alpha\beta} is

R_{00} + R_{11} + R_{22} + R_{33}

which, of course, is not an invariant. However I believe you are referring to the quantity R = R^{\alpha}_{\alpha} where R^{\alpha}_{\beta} = g^{\alpha\mu}R_{\mu\beta}. Is that correct?
Pete

He's referring to this "baby":
R^{\mu} \ _{\mu} =g^{\mu\nu} R_{\nu\mu}
,which is equal to this "baby"
R_{\mu} \ ^{\mu} =g_{\mu\nu} R^{\nu\mu}
,because the Ricci tensor is symmetric wrt to its components.

Daniel.

 

[pmb_phy]

He's referring to this "baby":
R^{\mu} \ _{\mu} =g^{\mu\nu} R_{\nu\mu}

That's what I just said Dan. Thanks though.

,which is equal to this "baby"
R_{\mu} \ ^{\mu} =g_{\mu\nu} R^{\nu\mu}
,because the Ricci tensor is symmetric wrt to its components.

Daniel.

Yup. I'm quite familiar with that fact too. Thanks anyway.

Pete

 

[Rob Woodside]

Sorry for the confusion, you are right. It is contraction and the indices must be upstairs and downstairs. I've seen things like Ricci = Tr Riemann meaning to contract the first and third indices or what ever convention one is using. I guess one should write Ricci = Con Riemann, but I've never seen it. At least I didn't use "Spur"

Pete, yes I 'd like a copy of Ohanian's paper, if it is not too much trouble. I mentioned this thread to a friend and he said there is nothing local about the curvature ! I exploded with: "The whole point of curvature was that the rotation of a vector parallel transported around a small arbitrary loop was given LOCALLY by the curvature tensor" He responded with: "It depends on second partials of the metric and that's non local" In reply to "Wadayamean?", he pointed out that you can't MEASURE curvature at a point. It requires the separation of neighbouring geodesics and that occurs over a small region and not at a point. The mathematical notion of local is "at a point". The physicist's idea of local is "in a small neighborhood" To sum up: the Curvature tensor is well-defined at each point and changes little over a small neighbourhood and is measured by geodesic separation (or vector rotation around a loop, etc) over this neighbourhood. Does this help or merely add more confusion?

 

[pmb_phy]

Pete, yes I 'd like a copy of Ohanian's paper, if it is not too much trouble.

It's on the way. Check your e-mail. I sent 3 articles. One is Ohanian's article. The other is a criticism of Ohanian's article by Walstad. The last is Ohanian's response to Walstad.

I mentioned this thread to a friend and he said there is nothing local about the curvature !

Yep. Been there, done that. I had the same argument with other people too.

I exploded with: "The whole point of curvature was that the rotation of a vector parallel transported around a small arbitrary loop was given LOCALLY by the curvature tensor" He responded with: "It depends on second partials of the metric and that's non local"

That is entirely an incorrect statement.

In reply to "Wadayamean?", he pointed out that you can't MEASURE curvature at a point.

He's showing that he thinks that local means at a single point where in fact the term local, as the term is used in differential geometry, refers to something in a small neighborhood. In fact recall the definition as given in Differential Geometry, by Kreyszig

(page 2) local – A geometric property is called local, if it does not pertain to the geometric figure as a whole but depends only on the form of the configuration in an (arbitrarily small) neighborhood under consideration

The author actually gives an example stating "For instance, the curvature of a curve is a local property." which readily extends to Riemann curvature as being a local property referring to a small neighborhood. Therefore since derivatives are defined in a limit in small region and not with regard to a single point only it follows that "local" applies to a point and "nearby" surrounding points.

It requires the separation of neighbouring geodesics and that occurs over a small region and not at a point.

Exactly. Bravo.

The mathematical notion of local is "at a point".

Sorry but I have to disagree with that assertion.

Does this help or merely add more confusion?

I wasn't confused in the first place. I was merely pointing out that this point is the subject of debate in GR.

Pete

 

[Rob Woodside]

Thanks Pete, I'll check for the papers on Monday at the office.

I'm surprised at Kreyszig. A derivative senses changes at nearby points and can't be found without knowing what goes on at the nearby points, but it lives at a specific point. His notion of local seems to be in contrast with global. Live and learn.

 

[pmb_phy]

Thanks Pete, I'll check for the papers on Monday at the office.

I'm surprised at Kreyszig. A derivative senses changes at nearby points and can't be found without knowing what goes on at the nearby points, but it lives at a specific point. His notion of local seems to be in contrast with global. Live and learn.

Kreyszig is doing the typical thing - He's speaking about what is normally considered "local" in mathematics, i.e. an \epsilon neighborhood. An \epsilon neighborhood is not a global region. How else could anything be a local property? A propertys is not something a point on a curve has.

This is not unique to Kreyszig. See -- http://mathworld.wolfram.com/Local.html

In the end what really matters is what an author means by "local" when they use it. But its nice to stick to well defined terms rather than to redefine terms to fit our ideas of them.

How would you define local property?

Pete

 

[jcsd]

Now I'm confused I always though that local was defined in the limit.

 

[pmb_phy]

Now I'm confused I always though that local was defined in the limit.

Same here.

If anyone finds a mathmatical sourc/text which states otherwise please let me know.

Pete

 

[Rob Woodside]

How would you define local property?

Pete

The property lives at the point but depends on the neighbourhood. The derivative from first year calculus is the paradigm.