Can space be curved in relation to an absolute straight space?

[Ibix]

Maybe it is no use to talk about physical existence of such absolute straight space of our interest until we have a physical maneuver to check it or it becomes a marvelous hypothesis to explain a lot of phenomena.

Yes. This is the point about general relativity. You can define manifolds without reference to an embedding, in terms of set theory. And no physics requires any of the concepts that embedding a manifold in a higher dimensional space provides. So we have a situation where adding a space in which to embed spacetime does nothing except let us embed spacetime.

Occam's razor strongly suggests that you drop the idea, lest it lead you in completely the wrong direction if and when evidence inconsistent with GR turns up.

 

[m4r35n357]

Is this the sense in which space is curved?

Can anyone comment on this please

Briefly, if a circle has circumference $2 \pi r$ then the space is flat. Otherwise the space is not flat.
Alternatively, if the angles of a triangle add up to $\pi$ then the space is flat, otherwise it is not.
[EDIT - circumference not diameter!]

 

[PeterDonis]

I am interested in physics which deals existing things.

How do you test, experimentally, whether something "exists" or not?

I have no idea about relation between logically required and physical existence.

"Physical existence" is not a precise concept so this is not surprising. But this thread is about what is logically required to define the concept of curvature of a manifold.

 

[DaveC426913]

I'm trying to answer the OP's question without using extrinsic curvature at all. We keep coming back to that, which is likely only enhancing his belief that it is required.

You and I are in a spaceship and we both start out 1 million kilometers apart and on parallel paths.
We fly past the sun.
Each of us maintains a course that is straight to several decimal places.

After passing the sun, we discover that we are actually converging, even though neither of us has changed our heading.
When we finally meet, we measure our angle of intersection and draw a diagram of the triangle formed by our two starting points and our intersection.

The angles at our starting points are both 90 degrees (we were alongside each other and moving parallel) but our intersection is much less than 90. In curved space, parallel lines can converge (or diverge). When the angles of a triangle do not add up to 180, we know we are embedded in curved space.

We know, without external reference that we have been moving through curved space. Notice that there was no 3D surface upon which we moved, or made reference to. We simply flew straight on our parallel courses.

Alternately, if we used each other to establish parallelicity, we would find that we would have to make course corrections to our heading in order to remain at the same distance apart. We would have to veer off our straight course away from each other just to keep our 1 million km distance.

 

[PeterDonis]

Each of us maintains a course that is straight to several decimal places.

A key part of this is to define, physically, how this is done. The way it is done is to stay in free fall, as accurately as you can measure. That is important because you can measure whether you are in free fall without any input from outside the spaceship.

(The only sticky bit is this: how did we know we were moving in straight lines?)

It's not "sticky", just important to make clear. See above.

 

[DaveC426913]

The way it is done is to stay in free fall, as accurately as you can measure. That is important because you can measure whether you are in free fall without any input from outside the spaceship.

Nicely done.

 

[tophatphysicist]

Paulo Constantino, the purpose of my response is to just encourage you in your pursuit of understanding. Your intuition seems to be driving you to pursue some underlying fundamental aspect of space. I’ve tried thinking through the problem from the viewpoint implied by your questioning. So far I haven’t developed a string of thoughts that I feel would provide you with a satisfactory response. Yet I can’t help but feel that your questions are worthy of discussion here.

It was of course very necessary to first make sure you understand the intrinsic and extrinsic curvature concepts. I think that by now you’ve had ample instruction about that. We who have been through a doctoral program in Relativity had that drummed into our heads early in our grad school courses. Yet, I am still sympathetic with the direction your intuition drives you. Further discussions genuinely trying to respond to your fundamental question, expanding beyond the mathematical definitions of curvature and the “mass tells space how to curve, and space tells mass where to go” would probably soon get out of the intended bounds for discussion on this Forum. If it makes you feel any better, I’m sure there are other members of the forum who share your wonder about the fundamental nature of space.

 

[bahamagreen]

A key part of this is to define, physically, how this is done. The way it is done is to stay in free fall, as accurately as you can measure. That is important because you can measure whether you are in free fall without any input from outside the spaceship.
It's not "sticky", just important to make clear. See above.

Even in free fall aren't the craft still subject to their mutual gravitational attraction causing their paths to converge?
Measurements in free fall without any input from outside the ship... include Weyl and Ricci?

 

[PeterDonis]

Even in free fall aren't the craft still subject to their mutual gravitational attraction causing their paths to converge?

In GR, gravity is not a force. The craft are in free fall, moving on geodesics of the spacetime they are in. The geometry of that spacetime is affected by the stress-energy of the craft, which makes it curved, so the geodesics the craft are moving on converge.

Measurements in free fall without any input from outside the ship... include Weyl and Ricci?

No; measuring that you are in free fall, by itself, doesn't tell you the curvature (if any) of the spacetime you are in. It just tells you that you are moving on a geodesic of whatever the spacetime geometry is.

If the two ships want to know whether they are converging, diverging, or neither, they can exchange light signals. In the case @DaveC426913 described, the light signal exchange will tell them that they start out neither converging nor diverging (i.e., at rest relative to each other), but then start converging. That tells them that the spacetime they are in must be curved, because in flat spacetime, if they started out at rest relative to each other and were both in free fall, they would stay at rest relative to each other forever.

None of these measurements require any "embedding" of the spacetime in a higher dimensional space. But the second measurement (exchanging light signals) differs from the first in that neither spaceship can make it on its own; the two need to cooperate, unlike the measurement of being in free fall, which each ship can do on its own, with no outside input.

 

[DParlevliet]

What I want to say is that it seems to me impossible for something to be curved unless it is curved in relation to absolute straightness.

I think PauloConstantino is looking for a more fundamental answer, about against which reference space is curved.
My answer would be: against an mathematical Euclidean space, which is absolute straight, connected to the masses causing the curvature.
Measuring is easy: sit on the mass (earth) under a tree and see an apple fall.

 

[Dale]

My answer would be: against an mathematical Euclidean space, which is absolute straight, connected to the masses causing the curvature.

Your answer is not consistent with the mathematics of general relativity.

 

[PeterDonis]

I think PauloConstantino is looking for a more fundamental answer, about against which reference space is curved.

And if that is the answer he is looking for, then such an answer does not exist, because the concept of curvature that is used in GR is intrinsic curvature, which, as has already been pointed out a number of times in this thread, is not defined against any reference space. It is an intrinsic property of the manifold and its metric and does not require comparison with anything else.

My answer would be: against an mathematical Euclidean space

And this answer is wrong. See above.

Measuring is easy: sit on the mass (earth) under a tree and see an apple fall

That doesn't tell you that spacetime is curved; if you in a rocket accelerating at 1 g in flat spacetime, and you release an apple, it falls the same way it would if the rocket were sitting at rest on the Earth's surface.

Spacetime curvature is tidal gravity, so you need to measure tidal gravity in order to measure spacetime curvature.

 

[Ibix]

I think PauloConstantino is looking for a more fundamental answer, about against which reference space is curved.

We all agree with you that that is what he wants. Unfortunately he's badly confused about how curvature is defined in relativity - there is no such reference. Or at least, we've no evidence of one and we don't need one to describe any known physics. So he's doomed to disappointment.

 

[DParlevliet]

Then I am doomed to disappointment too. I will think about it.

 

[A.T.]

Then I am doomed to disappointment too.

It's all about expectations management. You should not expect all naming conventions in math and science to be intuitive. Here in particular, using the same term "curvature" for the extrinsic and the intrinsic one might not have been ideal.

 

[DParlevliet]

Allright, I understand that because spacetime is an intrinsic curvature it does not need a reference. However basically both extrinsic and intrinsic curvature are curved against what an observer sees, its mathematical Euclidean space. This probably is not useful for GR theory and calculation, but can be used for understanding spacetime curvature by layman (as I am). If GT would be extrinsic curved, like a cylinder, it would also cause gravity (of a very different shape). I suppose (until he answers) that PauloConstantino understands space as what we see, so if space is curved then our reference, our observations are curved too.

Anyway, to rephrase I think that spacetime is also curved against the mathematical Euclidean space of an observer. This is not the reference, but a self chosen reference. This approximation shows easily the effects of curvature and is useful for simple calculations, like the radius of the curvature on the Earth's surface. But indeed this is not useful for GT theory and will fail in details and extreme situations.

 

[A.T.]

This probably is not useful for GR theory and calculation, but can be used for understanding spacetime curvature...

Or rather misunderstanding it.

If GT would be extrinsic curved, like a cylinder, it would also cause gravity (of a very different shape).

Nope. Any extrinsic curvature of space-time would undetectable within space-time, per definition.

 

[PeterDonis]

basically both extrinsic and intrinsic curvature are curved against what an observer sees, its mathematical Euclidean space.

This is not correct. Measuring intrinsic curvature does not require comparison with anything else. This has already been pointed out several times in this thread.

to rephrase I think that spacetime is also curved against the mathematical Euclidean space of an
observer

This is personal speculation and is off topic here. Please review the PF rules.

 

[DParlevliet]

Any extrinsic curvature of space-time would undetectable within space-time, per definition.

Indeed then my understanding of spacetime curvature was wrong. So I suppose that if a space is curved extrinsic, an observers Euclidean space is curved in the same way, so he does see this curvature. Or perhaps better: the observers Euclidean space does not have (does not see) the dimension in which it is extrinsic curved. An intrinsic curvature is visible inside the observers Euclidean space.

Still we see the effect of spacetime curvature as a curvature in our "straight" Euclidean space, isn't it? When on Earth I am looking to a photon passing Earth I see a curvature in his path, which I can measure. I suppose that is curved space as it shows to me. Or is my Euclidean space curved?

Please don't be irritated about my statements. It is just a way to figure out how it works and I appreciate your answers. Everybody is using models and I am searching for an easy one which perhaps does not follow exact GT calculation, bus is also not really wrong.

 

[PeterDonis]

So I suppose that if a space is curved extrinsic, an observers Euclidean space is curved in the same way

No. Forget about extrinsic curvature of spacetime; the concept is not used at all in GR and has no physical meaning.

Or perhaps better: the observers Euclidean space does not have (does not see) the dimension in which it is extrinsic curved

No. There is no "observer's Euclidean space".

An intrinsic curvature is visible inside the observers Euclidean space.

No. There is no "observer's Euclidean space".

Still we see the effect of spacetime curvature as a curvature in our "straight" Euclidean space, isn't it?

No. There is no "straight Euclidean space" in the sense you are using the term. Spacetime curvature is tidal gravity. We observe tidal gravity without having or needing any "straight Euclidean space" to compare with.

When on Earth I am looking to a photon passing Earth I see a curvature in his path, which I can measure.

No. The photon is traveling on a geodesic of spacetime, i.e., a straight line in a curved geometry. The apparent "curvature" of its path in space is an artifact of using a particular coordinate chart, i.e., a particular splitting of spacetime into "space" and "time". Different splittings give different apparent "curvatures", but none of them are what we are talking about when we talk about spacetime curvature.

The way to see spacetime curvature with photons is to compare the paths of different nearby photons--for example, take two photons which both make close approaches to the Sun, but at slightly different altitudes. The photons will converge as they approach the Sun and then diverge as they fly back out away from the Sun. That is due to tidal gravity, i.e., spacetime curvature. But you need the comparison of both photons' paths to see it.

I am searching for an easy one

The simplest model that will give you reliable answers is the one I stated above: spacetime curvature is tidal gravity.

 

[Nugatory]

...the observers Euclidean space ... our "straight" Euclidean space...

What is this "observer's Euclidean space" of which you speak? There is no Euclidean space outside of geometry textbooks describing a mathematical abstraction that does not exist in the real world - the universe we live in is not Euclidean.

 

[DParlevliet]

...a mathematical abstraction..

Indeed I suppose that this is what I mean. I notice a curvature of (existing) space against what "I see"', which is probably a mathematical (non existing) straight XYZ "space". Perhaps this is not the curvature of GT calculations, but still results in a curvature in what I see.

 

[Ibix]

Indeed I suppose that this is what I mean. I notice a curvature of (existing) space against what "I see"', which is probably a mathematical (non existing) straight XYZ "space".

This seems to me to be a statement that spacetime is curved because it's not flat. Which is true, but not very helpful.

What we do is determine the components of the Riemann tensor. If they're all zero then we call the space flat. Otherwise we call it curved.

 

[PeterDonis]

I notice a curvature of (existing) space against what "I see"'

This makes no sense. What you "see" is the curved spacetime. You don't "see" anything else.

 

[Dale]

Indeed I suppose that this is what I mean. I notice a curvature of (existing) space against what "I see"', which is probably a mathematical (non existing) straight XYZ "space". Perhaps this is not the curvature of GT calculations, but still results in a curvature in what I see.

One thing that you need to understand about extrinsic curvature is that it always occurs in a higher dimensional flat space. If I recall correctly, even a simple spacetime like the Schwarzschild metric would require a 6D flat space for embedding.

So it wouldn't be a "straight XYZ space", but a straight UVWXYZ space. Do you "see" that?

 

[Nugatory]

I notice a curvature of (existing) space against what "I see"', which is probably a mathematical (non existing) straight XYZ "space".

What you "see" (which doesn't mean what you literally "see" - that would be a two-dimensional image on the retina of your eyes - it means what your brain figures out about the world around you based on that two-dimensional image on the retina of your eyes) is not a flat Euclidean space. It's a curved space in which the curvature is small enough that it generally goes unnoticed.

 

[Orodruin]

always occurs in a higher dimensional flat space

Is this really true? I would think you could define extrinsic curvature for any embedding regardless of the flatness of the space you are embedding in.

 

[Dale]

Is this really true? I would think you could define extrinsic curvature for any embedding regardless of the flatness of the space you are embedding in.

Well, I have to admit that I have not done an exhaustive study of the literature, but every embedding theorem that I have read about is embedding in a flat space. That may be selection bias since I usually search for "isometric" too.

 

[Ibix]

You must be able to embed manifolds in curved manifolds - e.g. the Earth's surface in spacetime. And it certainly seems to be extrinsically curved. Or am I being naive?

 

[Orodruin]

You must be able to embed manifolds in curved manifolds - e.g. the Earth's surface in spacetime. And it certainly seems to be extrinsically curved. Or am I being naive?

There certainly is nothing to stop you from doing such an embedding and your example for the Earth's surface (or rather its world tube) is a good one.

I would think that the reason you see embedding existence theorems with flat space is that the curved space is rather non-interesting (just embed $M$ in $M\times N$ where $N$ is any manifold).

 

[DParlevliet]

Most people look to the world with a straight mathematical XYZ-geometry, as Nugatory mentioned. That is the reference for what we "see". The problem is that they think this is equal to space, so get confused when space is curved.

Returning to my first statement: I notice that space is curved against a straight mathematical XYZ-geometry. Perhaps this is not the same value or structure as GT and not useful for GT calculations, but it is caused by space curvature. So it still is a right statement that space is curved with reference to a straight mathematical XYZ-geometry.

 

[Dale]

So it still is a right statement that space is curved with reference to a straight mathematical XYZ-geometry.

No, it is not. Please see above for details. The minimal straight space is more than 3 dimensional.

 

[trainman2001]

Interesting discussion as always. What I can't wrap my head around is that space curvature around a mass is in all directions 360 degrees in every conceivable orientation, but the rubber sheet analogy is still in 3D space and is only in one plane. It's as if the rubber sheet was everywhere around the mass and that I find very hard to visualize. I completely accept the outcome that mass curves space which is what we perceive as gravity, but I just can't get the imagery right in my mind. Does anyone have something to put me out of my angst?

 

[Nugatory]

Interesting discussion as always. What I can't wrap my head around is that space curvature around a mass is in all directions 360 degrees in every conceivable orientation, but the rubber sheet analogy is still in 3D space and is only in one plane. It's as if the rubber sheet was everywhere around the mass and that I find very hard to visualize. I completely accept the outcome that mass curves space which is what we perceive as gravity, but I just can't get the imagery right in my mind. Does anyone have something to put me out of my angst?

Try to forget that you ever saw the rubber sheet analogy. If it's not helping you, you don't need it.

Yes, the picture is drawn in three-dimensional space, but you can't draw a picture showing three-dimensional curvature in a three-dimensional space; this is why you're finding it impossible to visualize the curvature being the same in all directions and orientations. But even if you could make it past that hurdle you'd still be stuck because we aren't working with curved three-dimensional space. We're working with curved four-dimensional spacetime and there isn't any time axis in the rubber sheet analogy.

 

[DaveC426913]

but you can't draw a picture showing three-dimensional curvature in a three-dimensional space;

Why not?