Spacetime curvature observer and/or coordinate dependent?

Mentz114

Gold Member
I can't vouch for its precision, but in the context of an earlier discussion elswhere it made a lot of sense. Can some of you math whizzes comment on whether, in general, an 'invariant' tensor [a true tensor] is so only in inertial conditions? If so, do we still call it 'invariant' because its so close, or not use that terminology.
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 $U^\mu, V^\mu$ and we make the contraction $l=U^\mu V_\mu$, where $V_\mu = g_{\mu\alpha}V^\alpha$. Now apply a coordinate transformation $\Lambda$ to the vectors. The contraction of the transformed vectors is $(\Lambda^a_\mu U^\mu)( \Lambda_a^\mu V_\mu)$. Because $\Lambda$ is not a tensor but a transformation matrix, we can rearrange the last expression to $(\Lambda^a_\mu \Lambda_a^\mu) U^\mu V_\mu = U^\mu V_\mu$.

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, $R=R^\mu_\mu$. Because $R_{\mu\nu}$ is a tensor the contraction is invariant under coordinate tansformation.

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

Last edited:

Naty1

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

Can some of you math whizzes comment on whether, in general, an 'invariant' tensor [a true tensor] is so only in inertial conditions?
I really can't tell....thanks.

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.
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:

[
… 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!!!!

Dale

Mentor
Can some of you math whizzes comment on whether, in general, an 'invarient' tensor [a true tensor] is so only in inertial conditions?
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

Gold Member
Naty1 said:
Can some of you math whizzes comment on whether, in general, an 'invariant' tensor [a true tensor] is so only in inertial conditions?
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 Rμμ.

...but so far the real meaning of 'contraction' has eluded me. Any good online source that comes to mind??
What exactly do you mean by 'real meaning' ? The physical interpretation or the mechanics ?

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

Staff Emeritus
Re: Spacetime curvature observer dependent?

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.
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

Gold Member
Naty1 said:
Are the scalar contractions of the Einstein Stress Energy Tensor invariant....??
The trace is an invariant.

Rab - (R/2) gab = κTab

-> Rab - (R/2) gab = κTab

-> Raa - (R/2) gaa = κTaa

-> R - 2 R = κTaa

-> Taa = -R/κ

using R = Raa and gaa = 4.

Last edited:

PeterDonis

Mentor
Can some of you math whizzes comment on whether, in general, an 'invarient' tensor [a true tensor] is so only in inertial conditions?
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.

Is this why the " stress energy [momentum] tensor" is sometimes referred to as a 'pseudo tnesor'?
I think you are referring to this:

http://en.wikipedia.org/wiki/Stress–energy–momentum_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

Mentor
Re: Spacetime curvature observer dependent?

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".
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....
Is this why the " stress energy [momentum] tensor" is sometimes referred to as a 'pseudo tnesor'? I think you are referring to this:

http://en.wikipedia.org/wiki/Stress%...m_pseudotensor [Broken]
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:
...but so far the real meaning of 'contraction' has eluded me. Any good online source that comes to mind?? What exactly do you mean by 'real meaning' ? The physical interpretation or the mechanics ?
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.....

Last edited by a moderator:

PAllen

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.....
The mechanics is straightforward. A contraction is formed from one or more tensors having (in total) a matched number of contravariant and covariant indexes; you pair these up (multiple ways to do it, each producing, in general, a different contraction), and sum over component products for all combinations matching on the chosen index pairs. Because the transform under diffeomorphism for contravariant indexes is the inverse of of that for covariant indexes, it is trivially guaranteed than any contraction is diffeomorphism invariant. I don't think there is a 'general' physical meaning to contractions. Because of the generality of the above principle, you can readily form contractions that don't have any reasonable physical meaning. The physical meaning of different contractions gets at the interpretation of GR as a physical theory rather than some arbitrary collection formulas from differential geometry.

[Edit: in the above, I am focusing contraction to a scalar not to a lower rank tensor. ]

Last edited:

Mentz114

Gold Member
Naty1, I haven't got much to add to PAllen's #35 except an example.

Raising an index is a contraction. This gab Uc is a rank-3 mixed tensor ( 3 distinct idexes), and if we contract it we get gab Ua or gab Ub. These are the same thing because the metric is symmetric and the result is actually U with a contravariant (raised) index

Ua = gab Ub = ga0 U0+ ga1 U1 +ga2 U2+ga3 U3

which is a linear combination of the covariant components of U.

Naty1

My attempt at a Summary:
[I’ve don’t think I’ve ever seen one so I thought I’d give it a try..]

[#19] Is spacetime curvature observer dependent ?

None of the answers to the OP are precise…none can be unless an agreement is reached on exactly waht component of 'curvature' measurement will be used...and that would provide only a partial answer;the reason is given at the end of this post with some of the great insights insights along the way summarized here first.…

Originally Posted by Mentz114
If we take 'spacetime curvature' to mean gravity then the answer must be 'yes'.
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…. 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.
passionflower asked several questions that I really liked:


So then tell me how does an observer observe the effects of
A) spatial curvature
B) temporal curvature
and 
All I want is to address statements that bodies are affected differently by space and time curvature depending on their velocity. These statements I consider wrong and very misleading.
I’ll try some explanations; these, too will be imprecise, but hopefully again illustrate
The problem; feel free of course pick them apart if you don’t like them:

 Locally, everything stays the same....spacetime stays flat….but as a differential instead of infinitesimal observation is made, curvature becomes apparent. Time dilation enables the observer to make a comparison with a ‘distant’ observer to see their clocks run at different speeds; such an observer may also see a distant inertial observer apparently moving in an arc...with some acceleration, that is some curvature is apparent resulting from the acceleration of either. docAl might say: "yes, that's the effect of different grid lines but drawn on the original curved graph paper.…the original gravitational field. [This is another imprecise explanation.]

 Jonathan Scott provided insights to this one in posts #13,14; I want to try a complementary approach: I think he said different curvatures, accelerations and velocities are all related in curved spacetime.

I am reminded that ‘light follows null geodesics’ is an APPROXIMATION; a good one, but what really follow null geodesics are test particles...ideals with no mass, no energy, no disturbance of the initial gravitational field. So we see that photons of different energy follow slightly different paths. [I got this idea elsewhere from pervect.] So these ‘bodies’ are affected by energy….and you [OP] were worried about just speed affecting curvature’!!

Next, suppose we have a particle with some mass at some velocity relative to an otherwise stationary gravitational field. Send a second particle at the same initial trajectory but, say, double the velocity of the first; or impart it with some spin [angular momentum] but at the original velocity: All the particles will follow different trajectories. In docAl's language, the original gravitational field maintains it's 'curved graph paper' shape, unless it’s evolving, but there is an overlay of different grid lines on that curved graph paper representing ‘spacetime curvature…...due to different energies; hence different paths result. The ‘gravitational curvature’ may remain invariant [depending on how we define that] but the effect of the now differently curved grid lines results in a different overall ‘curvature’.

We seem to have agreed that: An invariant would be something that has exactly the same value in whatever frame it is calculated; And Dalespam and Mentz seem to agree that: Invariant refers to scalars (tensors of rank 0), but Tensors are 'covariant' not invariant.” And Dalespam notes:
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….[
but the Observers WILL agree on the total..] No controversy nor obstacles here.

Is that progress?? It sure helps me, but I don’t know about the OP…where is he??

So even if you all were to agree with all those descriptions, which would be a miracle, and the ones Pallen described in post #23, we are still faced with the problem of ascribing those descriptions of ‘grid lines’ and ’curved graph paper’ [or some other equivalents] to some mathematical component of the EFE.

That seems impossible because if my understanding is correct MTW lays out an absolute impass: [from my post #27, quote courtesy of pervect from elsewhere]

MTW
… 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.”
I take gravitational field to be equivalent to ‘gravity’…or ‘spacetime curvature’ and so am stuck right here. Seems we have come as far as we can….

I do have a final question: Can someone give a sentence or three outline of the difference between ‘invariant’ and ‘covariant’: If I used the explanation above [invariant means an identical value while covariant means components may change among observers but the totals remains the same] would that be a start??? Or is the essential difference something else?

my personal thanks to passionflower, the OP, for starting this...

Mentz114

Gold Member
Naty1 said:
I do have a final question: Can someone give a sentence or three outline of the difference between ‘invariant’ and ‘covariant’: If I used the explanation above [invariant means an identical value while covariant means components may change among observers but the totals remains the same] would that be a start?
I think you've got it. Instead of total use 'contraction'.

We can describe a mathematical expression as covariant, which means it's written in tensors. Doing physics covariantly means that we are not plagued with artifacts induced by coordinate basis changes.

An easy way to remember this is to think of the 4-velocity. The components are physically meaningful but frame dependent, while the contraction with itself is an invariant.

Last edited:

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving