Spacetime curvature observer and/or coordinate dependent?

Passionflower

Spacetime curvature observer and/or coordinate dependent?

In another topic several people suggested that spacetime curvature is not absolute, it apparently depends on the observer and/or coordinate system. Apparently if someone goes fast (whatever that might mean in relativity) curvature is more in the time direction while if someone goes slow (again whatever that might mean) curvature is more in the spatial direction.

[sarcasm on]Seems we got a way to determine absolute motion as we simply have to check the curvature for the two components to see how fast we are going[/sarcasm off]

Any takers?

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Jonathan Scott

Gold Member
Spacetime curvature observer and/or coordinate dependent?

In another topic several people suggested that spacetime curvature is not absolute, it apparently depends on the observer and/or coordinate system. Apparently if someone goes fast (whatever that might mean in relativity) curvature is more in the time direction while if someone goes slow (again whatever that might mean) curvature is more in the spatial direction.

[sarcasm on]Seems we got a way to determine absolute motion as we simply have to check the curvature for the two components to see how fast we are going[/sarcasm off]

Any takers?
That sort of curvature is related to acceleration, and one person's accelerating frame of reference is another's free fall, so it does depend on the frame of reference.

However, "going fast" is never a correct description from one's own point of view, so it's not quite clear what you mean.

If you are describing different test objects moving through the same static fields, the curvature doesn't change but the effect changes with velocity, and it's the other way round from what you say, in that the curvature of space only affects moving things (as should be obvious).

In a weak approximation, using an isotropic coordinate system (with one or more static sources), relativistic coordinate momentum per energy (not quite the same as velocity because the coordinate speed of light changes) changes at a rate g (1 + v2/c2) if the effective Newtonian field is g.

If the test object is moving tangentially to the field, the same rule applies to the velocity, hence the effect that light is deflected by twice the Newtonian acceleration.

Mentz114

Gold Member
Spacetime curvature observer and/or coordinate dependent?

In another topic several people suggested that spacetime curvature is not absolute, it apparently depends on the observer and/or coordinate system.
The components of the Riemann tensor will change value in a local frame according to
$$R_{abcd} = \Lambda^\alpha_a \Lambda^\beta_b \Lambda^\gamma_c \Lambda^\delta_d R_{\alpha\beta\gamma\delta}$$
where a,b,c,d are local frame indexes and α,β,γ,δ are coordinate indexes. The tetrad $\Lambda$ is defined so that
$$\eta_{ab} = \Lambda^\alpha_a \Lambda^\beta_b g_{\alpha\beta}$$
where η is the Minkowski metric.

Apparently if someone goes fast (whatever that might mean in relativity) curvature is more in the time direction while if someone goes slow (again whatever that might mean) curvature is more in the spatial direction.
Nobody said that.
In the case of the Schwarzschild vacuum the velocities are relative to the source.
[sarcasm on]Seems we got a way to determine absolute motion as we simply have to check the curvature for the two components to see how fast we are going[/sarcasm off]
What curvature and what components ? Velocities are relative to the source of the field.

Passionflower

The components of the Riemann tensor will change value in a local frame according to
$$R_{abcd} = \Lambda^\alpha_a \Lambda^\beta_b \Lambda^\gamma_c \Lambda^\delta_d R_{\alpha\beta\gamma\delta}$$
where a,b,c,d are local frame indexes and α,β,γ,δ are coordinate indexes. The tetrad $\Lambda$ is defined so that
$$\eta_{ab} = \Lambda^\alpha_a \Lambda^\beta_b g_{\alpha\beta}$$
where η is the Minkowski metric.
Yes the components change but curvature is absolute.

Do you think we should talk about coordinate effects when we describe curvature or do you think that is wrong or misleading to say the least?

Mentz114

Gold Member
Another example of curvature being dependent on an observer is the Langevin observer in the Born chart, where the 3D submanifold is curved ( non-Euclidean). That curvature depends on the angular velocity.

If the Minkowski metric is written in spherical polar coords it acquires non-zero components of the Riemann tensor ( I can't remember how many - perhaps only one).

Why did you phrase your question in the way that you did if you didn't want to talk about coordinate effects ?

When I refer to curvature I mean the Riemann tensor. Do you think the only important number is the invariant Ricci scalar ?

In answer to your question I would say that observed effects of curvature are observer dependent. For instance the Hagihara observer sees very different tidal effects compared to a Gull-Painleve observer, even though they are both moving in the same Schwarzschild spacetime.

Passionflower

In answer to your question I would say that observed effects of curvature are observer dependent.
So then tell me how does an observer observe the effects of

A) spatial curvature
B) temporal curvature

?

PAllen

If the Minkowski metric is written in spherical polar coords it acquires non-zero components of the Riemann tensor ( I can't remember how many - perhaps only one).
Hmm. How can a tensor (all components, everywhere) vanish in one coordinate system but not in another? Seems to violate the tensor transform law...

Dale

Mentor
Spacetime curvature observer and/or coordinate dependent?

In another topic several people suggested that spacetime curvature is not absolute, it apparently depends on the observer and/or coordinate system. Apparently if someone goes fast (whatever that might mean in relativity) curvature is more in the time direction while if someone goes slow (again whatever that might mean) curvature is more in the spatial direction.

[sarcasm on]Seems we got a way to determine absolute motion as we simply have to check the curvature for the two components to see how fast we are going[/sarcasm off]

Any takers?
There are a bunch of different ways that you might use the word curvature. Several of them are tensor of higher order than 1, so I don't know if you want to call those absolute or not. One of them is a scalar field on spacetime, so that is clearly absolute.

Then one other thing that you could talk about would be the curvature of a worldline. I think that this is the curvature refered to in the other discussion. That curvature is a scalar (the magnitude of the 4-acceleration), but it also depends on the worldline. So I don't know if you want to call that absolute or not.

Mentz114

Gold Member
Hmm. How can a tensor (all components, everywhere) vanish in one coordinate system but not in another? Seems to violate the tensor transform law...
I just checked and I'm referring to an embedded spherical surface, which has one non-zero Riemann component. But probably not relevant to this topic.

So then tell me how does an observer observe the effects of

A) spatial curvature
B) temporal curvature

?
Off topic. It belongs in the other thread. Why don't you answer the points I made in post#5 ?

PAllen

I just checked and I'm referring to an embedded spherical surface, which has one non-zero Riemann component. But probably not relevant to this topic.
Right, that's no problem. A 2-sphere can be embedded in Euclidean 3-space, for example.

Passionflower

Off topic. It belongs in the other thread. Why don't you answer the points I made in post#5 ?
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.

However it seems that quite a few folks on this forum are actively supporting these statements.

So now I wonder why I am even bothering......

Mentz114

Gold Member
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.

However it seems that quite a few folks on this forum are actively supporting these statements.
The geodesic equations show clearly that coordinate acceleration depends on velocity. They contain the Christoffel symbols which, along with their derivatives appear in the curvature tensors.

Your question about measuring temporal curvature brought to mind gravitational time dilation, which depends only on the time parts of the metric. The acceleration needed to remain at constant r in the Schwarzschild vacuum is the Newtonian value divided by the square root of g00. But I'm not sure if that is relevant either.

Anyhow it is late in my timezone so I'm clocking off.

Jonathan Scott

Gold Member
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.

However it seems that quite a few folks on this forum are actively supporting these statements.

So now I wonder why I am even bothering......
If you move in a straight line in local space which is curved (as described by the metric) relative to the coordinate system, then that obviously adds to the acceleration relative to the coordinate system. Just as in ordinary mechanics, if the linear curvature is 1/r, then the resulting acceleration is v2/r. In this case, r = c2/g, so this causes an acceleration of g times v2/c2 (for motion perpendicular to the direction of the field).

Jonathan Scott

Gold Member
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.

However it seems that quite a few folks on this forum are actively supporting these statements.

So now I wonder why I am even bothering......
If you're bothered by the implications for the principle of equivalence of different accelerations for different velocities, that's not actually a problem if you think about it more carefully.

The principle of equivalence holds within a local coordinate system which is small enough to approximate SR. If you consider a local box at rest in the field and view it from an isotropic coordinate system, it appears to be curved, along with local space, towards the source. A light ray crossing the box is twice as curved (at 2g). This means that relative to a local horizontal line across the box, which follows the curvature of space, the light actually appears to be accelerating downwards at only 1g, the same as any material object at rest or moving at any other speed. It's all consistent.

Naty1

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 agree based on prior explanations from a number of people here in the forums:

The curvature or flatness of space (as opposed to spacetime) IS a coordinate dependent idea.

Overall, can be a confusing issue due in large part to language.....

From my [physics forums] notes of posts:

[one perspective]
Any situation involving a rapidly moving massive body's gravitational effect and a 'stationary' observer can be transformed to an equivalent question about the interaction between a rapidly moving observer and a 'stationary' massive body. So all observations relating to a rapidly moving massive body can be answered as if the body is stationary...as if all measures are local. Local measures trump distant measures.

And a further perspective:
For a single electron [ or black hole] as an example, the rest energy density of the electron is the only thing that causes spacetime curvature. The kinetic energy is frame-dependent, just as the velocity is; in the electron's rest frame it is zero, and we can predict all physical observables, like whether the electron forms a black hole, by solving the EFE in the electron's rest frame.
This means that no matter how fast you see a particle whizzing by, it will never become a black hole.
Those simple perspectives clinch the issue for me;there was no dissent in the thread(s).

Other [supporting] perspectives:

[I don't have the original thread(s)]

DrGreg:

"...... let's restrict our attention to 2D spacetime, i.e. 1 space dimension and 1 time dimension, i.e. motion along a straight line. …
In the absence of gravitation, an inertial frame corresponds to a flat sheet of graph paper with a square grid. If we switch to a different inertial frame we "rotate" to a different square grid, but it is the same flat sheet of paper. (The words "rotation" and "square" here are relative to the Minkowski geometry of spacetime, which doesn't look quite like rotation to our Euclidean eyes, but nevertheless it preserves the Minkowski equivalents of "length" (spacetime interval) and "angle" (rapidity).)

If we switch to a non-inertial frame [an accelerated observer] but still in the absence of gravitation), we are now drawing a curved grid, but still on the same flat sheet of paper. Thus, relative to a non-inertial observer, an inertial object seems to follow a curved trajectory through spacetime, but this is due to the curvature of the grid lines, not the curvature of the paper which is still flat.

When we introduce gravitation, the paper itself becomes curved. (I am talking now of the sort of curvature that cannot be "flattened" without distortion. The curvature of a cylinder or cone doesn't count as "curvature" in this sense.) Now we find that it is impossible to draw a square grid to cover the whole of the curved surface. The best we can do is draw a grid that is approximately square over a small region, but which is forced to either curve or stretch or squash at larger distances. This grid defines a local inertial frame, where it is square, but that same frame cannot be inertial across the whole of spacetime.

So, to summarize, "spacetime curvature" refers to the curvature of the graph paper, regardless of observer, whereas visible curvature in space is related to the distorted, non-square grid lines drawn on the graph paper, and depends on the choice of observer...."

DrGreg:
When we talk of curvature in spacetime (either curvature of a worldline, or curvature of spacetime itself) we don't mean the kind of curves that result from using a non-inertial coordinate system, i.e, non-square graph paper in my analogy.

Four acceleration:
[both from pervect I suspect]

In Minkowski coordinates in Special Relativity, 4-acceleration is just the coordinate derivative of 4-velocity with respect to arc-length (proper time), and the 4-velocity is the unit tangent vector of worldline. As the 4-velocity has a constant length its derivative must be orthogonal to it. The 4-acceleration is the curvature vector; orthogonal to the worldline and its length is the reciprocal of the worldline's "radius of curvature".

In non-inertial coordinates in GR, the 4-acceleration is defined as a covariant derivative. This takes into account (and removes) any curvature of spacetime or "apparent curvature" due to using a "non-square grid", and leaves us with curvature that is a property of the worldline itself, not the spacetime or the choice of coordinates. Then everything else I said in the last paragraph is still true in a coordinate-independent sense.

A geodesic has zero 4-acceleration and zero curvature.

Pervect:
If you draw a largish parallelogram on a curved surface, you'll find that the opposite sides might not necessarily be equal in length, even though the sides are parallel. Now, imagine such a parallelogram, but that one of the sides of this parallelogram is the time axis on a space-time diagram. What you'll see is something that looks pretty much like gravitational time dilation. One timelike worldline will be shorter than another, even though they are connected by parallel geodesics.

from another discussion:

The Stress energy Tensor

The "amount of gravity produced" by the object is not a function of its energy alone, it's a function of its stress-energy tensor [ SET] of which energy is only one component. In a frame in which the object is moving, there will be other non-zero components of the SET as well as the energy, and their effects will offset the apparent "effect" of the increase in energy, so the final result will be the same [gravitation] as it is for a frame in which the object is at rest.

[A way to think about this is that in the frame of the object, the 'other' mass is the one that is moving....so the moving object cannot have 'greater gravity.....]

Naty1

Jonathan Scott post #2:

That sort of curvature is related to acceleration, and one person's accelerating frame of reference is another's free fall, so it does depend on the frame of reference.
is entirely correct, but that is not the 'spacetime curvature' as described in my prior post by Dr Greg. These 'acceleration effects' are the curved grids lines as described by Dr Greg, not the 'spacetime' graph paper.

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Naty1

Dalespam:
That curvature is a scalar (the magnitude of the 4-acceleration), but it also depends on the worldline. So I don't know if you want to call that absolute or not.
This too is curved grid lines on gravitational spacetime....so as noted, you have to define to what type of curvature you refer. But for a given worldine seems like it is a scalar invarient.

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:

passionflower:
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.

me:
I agree based on prior explanations from a number of people here in the forums:
I 'wanna' reneg on my agreement, but agree with passionflower's original implied statement:
Spacetime curvature IS NOT observer and/or coordinate dependent?
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

Gold Member
Re: Spacetime curvature observer dependent?

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/research/papers/files/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
$$T_{\mu\nu}=R_{\mu\rho\nu\sigma}U^\rho U^\sigma$$
(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
$$\hat{T}_{ab}=\Lambda^\mu_a \Lambda^\nu_b T_{\mu\nu}$$
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.

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yuiop

Naty1 quoting someone else:
Any situation involving a rapidly moving massive body's gravitational effect and a 'stationary' observer can be transformed to an equivalent question about the interaction between a rapidly moving observer and a 'stationary' massive body. So all observations relating to a rapidly moving massive body can be answered as if the body is stationary...as if all measures are local. Local measures trump distant measures.
This should probably be qualified with the bolded words as:
Any situation involving a rapidly moving massive body's gravitational effect and a 'stationary' inertial observer can be transformed to an equivalent question about the interaction between a rapidly moving inertial observer and a 'stationary' massive body. So all observations relating to a rapidly moving massive body can be answered as if the body is stationary...as if all measures are local. Local measures trump distant measures.
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

Mentor
Re: Spacetime curvature observer dependent?

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.

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:
...What this tells me is that the term "spacetime curvature" is not a good one......
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:

..But whereas potential energy is independent of the frame of reference used, kinetic energy has different values in different frames of reference, and so does total energy.
from THE RIDDLE OF GRAVITATION,1992, Peter Bergmann (a student of Einstein)

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

In GR, the gravitational field g is not really a well-defined thing. By the equivalence principle, you can always pick any point and make g=0 there by adopting a free-falling frame of reference. There isn't even a gravitational field tensor. The closest thing to a gravitational field in GR is the Christoffel symbol, and it doesn't transform as a tensor.
so Ive been concluding now for sometime that the quote {Feynman??}

"shut up and calculate"
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

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.
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:
I find myself sympathetic to both points of view in this thread. They are emphasizing different aspects of the above observations.
seems a better course based on this discussion....bummer...

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

The Riemann tensor is observer independent quantity (being a true tensor), but the tidal forces you measure with rods and accelerometers are only exactly equal to the Riemann tensor for someone who is not accelerating, for someone who is accelerating there is a (usually very tiny) difference between the observer-independent Riemann tensor components and the actual measurements of the gravity gradient, also known as tidal forces.
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:
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.....
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...

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