Spacetime curvature observer and/or coordinate dependent?

[Mentz114]

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

Raising an index is a contraction. This g^ab U_c is a rank-3 mixed tensor ( 3 distinct idexes), and if we contract it we get g^ab U_a or g^ab U_b. These are the same thing because the metric is symmetric and the result is actually U with a contravariant (raised) index

U^a = g^ab U_b = g^a0 U₀+ g^a1 U₁ +g^a2 U₂+g^a3 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'.

PeterDonis adds:
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:

[1]

So then tell me how does an observer observe the effects of
A) spatial curvature
B) temporal curvature

and [2]

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:

[1] 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.]

[2] 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??

Here comes the bad news:

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]

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.