Spacetime curvature observer and/or coordinate dependent?

Naty1

passionflower...I think your original question deserves a lot more discussion....It has taken me a long time to develop a perspective with help from several experts here who agree with you....this discussion has so far not caused me to change my mind.

post #15:

I 'wanna' reneg on my agreement, but agree with passionflower's original implied statement:
that GRAVITATIONAL spacetime curvature [curved graph paper as described by docAl] is invarient, but acceleration overlays [curved grid lines drawn on the curved graph paper] are
not.

How can you experts agree on such an issue when no one entity in the Einstein Field Equations can claim preeminence in describing 'gravity'.....

Also, if no matter how fast you see a particle whizzing by, it will never become a black hole, how can it change 'gravitational curvature'? Such speed does not add to gravitational curvature yet you generally PERCEIVE the spacetime as being curved .....the gravitational spacetime curvature component [curved graph paper] must be considered frame independent.

Mentz114

Is spacetime curvature observer dependent ?

If we take 'spacetime curvature' to mean gravity then the answer must be 'yes'.

1. A freely falling observer feels no gravity, but a stationary one does. This is because the elements of curvature that depend on the first derivatives of the metric can be transformed away for a point on a geodesic worldline. See for instance Fermi coordinates here http://en.wikipedia.org/wiki/Fermi_coordinates or Riemann coordinates here http://users.monash.edu.au/~leo/rese...s/lcb96-01.pdf. The latter is a lot better than the Fermi page.

2. Tidal effects ( also gravity ) can be different for different observers in the same spacetime. The tidal tensor in the coordinate basis is defined as
[tex]
T_{\mu\nu}=R_{\mu\rho\nu\sigma}U^\rho U^\sigma
[/tex]
(which already contains an observer dependent vector U) and being a tensor can be transformed to a local frame ( which may be geodesic or not) by the usual tetrad
[tex]
\hat{T}_{ab}=\Lambda^\mu_a \Lambda^\nu_b T_{\mu\nu}
[/tex]
and since the tidal effects are components of T, they can change without violating the tensorial properties of T. I gave an example in an earlier post of two observers who see different tidal effects in the Schwarzschild vacuum.

There could be something wrong with my logic, in which case, please politely point it out.

yuiop

Naty1 quoting someone else:
This should probably be qualified with the bolded words as:
and there probably should be a few more qualifying statement in there. Anyway, without any qualification it is not always true. For example a stationary observer in the field of a rotating gravitational body is not equivalent to a moving observer in the field of a non rotating gravitational body.

PeterDonis

If by "gravity" you mean "particular effects of gravity", then yes. As you point out, particular effects of gravity on particular observers will always be dependent on the observer's 4-velocity.

However, one could also mean by "spacetime curvature" or "gravity" something that is *not* observer-dependent, such as: the metric, the Riemann curvature tensor, or other tensors derived from it such as the Weyl, Ricci, or Einstein tensors. Most of the time I see the term "spacetime curvature" in the literature, it seems to refer to one of these objects, usually the Riemann curvature tensor.

So it seems to me that this is a question of terminology, not physics. Whether or not "spacetime curvature" is observer-dependent depends on what you define "spacetime curvature" to mean. What this tells me is that the term "spacetime curvature" is not a good one, and should be avoided whenever one is trying to be precise; instead, one should directly talk about something with a precise meaning, so that it is obvious whether or not it is observer-dependent.

Naty1

Mentz..post #19:...I don't have the mathematical skills to agree or disagree...hopefully someone will do one or the other....

yuiop..post#20: good clarification. [I updated my notes !]

PeterDonis:#21: That is the conclusion I reached in quoting previous threads above.

But even beyond that, it seems to me [and I don't claim a complete mathematical insight into GR] that GR has infirmities similar to those of quantum mechanics. In QM people have different physical interpretations of some of the mathematics...the collapse of the wave function, the meaning and relationship different models within string theory, and ADS/CFT correspondnce, and on and on; in GR is it seems a lot of people have similar difficulty because time and distance don't have precise meanings, because energy of the gravitational field isn't amenable to a precise interpretation, and because 'g' doesn't even have a specific description with the EFE....

As an example of GR difficulties, I posted "Relativity of potential energy" over two years ago when I knew even less than now:

http://www.physicsforums.com/showthr...13#post2549413


from THE RIDDLE OF GRAVITATION,1992, Peter Bergmann (a student of Einstein)

and was, frankly, rocked by the answers.....including this from BenCrowell:

so Ive been concluding now for sometime that the quote {Feynman??}

applies to not only QM but GR.

PAllen

Following Peter Donis's suggestion to define terms in relation to mathematical objects, let's make sure at least that there is no disagreement on the following:

1) Curvature scalars ( e.g. Ricci, Kretschmann) are invariant and thus coordinate and observer independent. (The value of Kretschmann curvature invariant is that it is not identically zero for vaccuum regions in GR. Thus it can give a scalar measure of vaccuum curvature).

2) Curvature tensors (e.g. Riemann, Ricci, Weyl) are most properly viewed as intrinsic geometric features of the manifold, and can be studied with coordinate free (and thus, obviously, observer independent) techniques. However, their expression as components is obviously coordinate dependent. Different observers (following different world lines), performing measurements influenced by e.g. the tt component of such an object (expressed in the local frame of that observer) will make different measurements (because the tt components will be different).

3) In a geometry that has natural symmetries (e.g. spherical, axial), you can choose (but don't have to) make statements about motion in coordinate systems that manifest the symmetry. In this sense, you can give meaning to statements like: two observers in different states of motion in such a coordinate system are more or less affected (for some measurement) by different curvature or metric components.

Thus, I find myself sympathetic to both points of view in this thread. They are emphasizing different aspects of the above observations.

PAllen

Actually, the context of the discussion Naty got that quote from was purely boosted non-rotating black hole versus stationary black hole. Thus, SC geometry in two different coordinate systems.

Naty1

PAllen:
seems a better course based on this discussion....bummer...

I also found this description in my notes from an earlier discussion:

I can't vouch for its precision, but in the context of an earlier discussion elsewhere it made a lot of sense. Can some of you math whizzes comment on whether, in general, an 'invarient' tensor [a true tensor] is so only in inertial conditions? If so, do we still call it 'invarient' because its so close, or not use that terminology.

I've read in these forums of a theory of gravity which gives the same predictions as GR but uses the torsion defect instead of curvature. Does such a 'torsion' viewpoint offer any insights into these differing views of 'cuvature'?

Pallen: Is this why the " stress energy [momentum] tensor" is sometimes referred to as a 'pseudo tnesor'?.....I've understood just the component origins of that statement...how it arises in the Einstein Field Equations...

Mentz114

The tensorial property is the invariance of scalar contractions. For instance a 4-vector is a rank-1 tensor. Suppose we have two 4-vectors [itex]U^\mu, V^\mu[/itex] and we make the contraction [itex]l=U^\mu V_\mu[/itex], where [itex]V_\mu = g_{\mu\alpha}V^\alpha[/itex]. Now apply a coordinate transformation [itex]\Lambda[/itex] to the vectors. The contraction of the transformed vectors is [itex](\Lambda^a_\mu U^\mu)( \Lambda_a^\mu V_\mu)[/itex]. Because [itex]\Lambda[/itex] is not a tensor but a transformation matrix, we can rearrange the last expression to [itex](\Lambda^a_\mu \Lambda_a^\mu) U^\mu V_\mu = U^\mu V_\mu[/itex].

What makes U,V tensors is that despite having their components changed by a transformation, their contraction remains the same. The same logic can be applied to any contraction of indexed objects. The objects in question are tensors iff their scalar contractions are invariant.

You must have come across the Ricci scalar which is the self-contraction of the Ricci tensor, [itex]R=R^\mu_\mu[/itex]. Because [itex]R_{\mu\nu}[/itex] is a tensor the contraction is invariant under coordinate tansformation.

(I am not a "math whizz", this is standard textbook stuff.)

Naty1

Mentz: is that a yes or no to this question:

I really can't tell....thanks.


I've seen that enough to know it is 'standard'...but so far the real meaning of 'contraction'
has eluded me. Any good online source that comes to mind??

Are the scalar contractions of the Einstein Stress Energy Tensor invarient....?? I seem to recall several sources indicating it is a 'pseudo tensor'....

Pervect provided this quote elsewhere from MTW:

[QUOTE][
… nowhere has a precise definition of the term “gravitational field” been
given --- nor will one be given. Many different mathematical entities are
associated with gravitation; the metric, the Riemann curvature tensor, the curvature scalar … Each of these plays an important role in gravitation theory, and none is so much more central than the others that it deserves the name “gravitational field.”
/QUOTE]

I rely on that so I don't get too discouraged!!!!

DaleSpam

I am not sure this is a direct answer to your question, but as much as possible I try to use the word invariant only to refer to scalars (tensors of rank 0). It is OK to use the word to refer to tensors of higher rank as they can be considered to be geometric objects which do not change under a coordinate basis change, but it can cause confusion since the components of a higher rank tensor do change under a coordinate basis change. It is probably more correct to call them covariant, but then that can be confusing in the case of one or more contravariant indices. I don't think that the usage is standardized yet.

Mentz114

A tensor is always a tensor, whatever frame it is transformed to. Tensors are not invariant, but 'covariant'. An invariant would be something that has exactly the same value whatever frame it is calculated in, like R^μ_μ.

What exactly do you mean by 'real meaning' ? The physical interpretation or the mechanics ?

It is a tensor. So the scalars formed by contracting with itself and contracting with another rank-2 tensor will be invariant.

This is OK
http://en.wikipedia.org/wiki/Tensor_contraction and a search for 'tensor contraction' throws up a lot of stuff.

pervect

It also depends on what means by "observer dependent". Certainly the numerical values of the components of a tensor can vary when you change frames - in fact, they must change, to satisfy the tensor transformation laws.

I agree that in talking with a reader with some expertise, it's better to be more specific. "Space-time curvature" could mean any of a number of things.

On the other hand, talking to a reader without any expertise, it's not only not especially helpful to be technically precise, it's probably actually conuter-productive. Unfamiliar words tend to scare the new readers - sometimes, it seems to scare them away from grasping simple basic concepts (like geodesic deviation) that they probably could grasp with an approach that was less formal and more "chatty".

As far as the original question about whether or not spatial curvature doubles the deflection of light, there are a number of different approaches that all seem to converge on the answer "yes, it does".

The best documented approach doesn't use tensors at all, but the PPN parameter beta, which can be and is described as the amount of spatial curvature induced by a unit rest mass, for instance in MTW's "Gravitation". (Note that here we see yet another possible meaning for the term spatial curvature via this usage.)

One can then state that deflection of light would be half of what GR predicts if the PPN parameter beta (the spatial curvature) was zero, instead of unity, the value that GR predicts.

THere are other approaches as well, for instance on can decompose the Riemann tensor using the Bel decomposition, and arrive at similar results regarding light deflection from the geodesic deviation equation - in this case that half of the geodesic deviation is due to the electrogravitc part of the tensor, and the other half is due to the topogravitic parts.

The Bel decomposition requires a timelike vector field to define the decomposition - i.e. to define the split of space-time into space and time. The obvious choice for this vector field is the timelike killing vectors associated with a static metric.

Mentz114

The trace is an invariant.

R^ab - (R/2) g^ab = κT^ab

-> R^a_b - (R/2) g^a_b = κT^a_b

-> R^a_a - (R/2) g^a_a = κT^a_a

-> R - 2 R = κT^a_a

-> T^a_a = -R/κ

using R = R^a_a and g^a_a = 4.

PeterDonis

I don't claim to be a math whiz, but I think the answer to this question is "no". Tensors are tensors in curved spacetime as well as in flat spacetime, in the presence of gravity as well as in its absence. They work the same way (with the one key proviso that in curved spacetime, you have to make sure you use covariant derivatives instead of just ordinary derivatives when writing tensor equations) and have the same invariance properties in both cases.

I think you are referring to this:

http://en.wikipedia.org/wiki/Stress%...m_pseudotensor

which is indeed a pseudo-tensor, but is *not* what appears on the RHS of the Einstein Field Equation; that is the "stress-energy tensor" (yes, the terminology can be confusing), which is a true tensor.

What makes the other thing a "pseudo-tensor" is that it tries to incorporate "energy stored in the gravitational field" as well as the ordinary stress-energy that appears on the RHS of the EFE; but there is no way to define "energy stored in the gravitational field" as a true tensor.

PeterDonis

Yes, it's a judgment call which approach to use. What tends to flip me over into "need to use more specific, precise terminology" mode is when I find different people (one of whom may well be me) using the same word to refer to different things, and then arguing past each other.

Also, sometimes people will ask questions that really shouldn't be answered in the terms they were posed in, because those terms make implicit assumptions that can hinder understanding. But again, I agree it's a judgment call.

Naty1

hey, thanks a lot guys...especially:

PeterDonis....
yes....your answer a big help, I could have read that for another ten years and never deciphered it on my own. I think that must have been where I did mislead myself.

Mentz:
I'd be happy to start with either....physical would be an easier start......but the wikipedia explanation is 'above my paygrade"...I'll check out 'mathpages' online.....