# B Physical meaning of Curvature

1. Aug 27, 2016

### Megaton

I'm new at this, trying to understand the overall picture of GR

From what I understand space does not actually curve but rather test particles tracing out their world lines in Euclidean space as compared to ones tracing out their world lines near a source of gravity is the physical meaning of curvature? And this is described mathematically using the Riemannian-curvature tensor?

Is this correct at all, I'm just trying to understand the overall idea of GR before trying to attempt the complex math

Thanks

2. Aug 27, 2016

### haushofer

Space(time!) does curve. It's what Einstein calls 'gravity'. The curvature of spacetime is described by the Riemann tensor. The Einstein equations put on a constraint on this Riemann tensor, describing the gravitational field (= spacetime geometry).

3. Aug 27, 2016

### pervect

Staff Emeritus
I wouldn't say that "space does not actually curve". But when I think about the statement, I'm not sure how to disprove it. Which is an indication that it doesn't have a well defined empirical meaning, if it did, it'd be testable.

Test particles do trace out world lines in both Euclidean spaces and non-Euclidean spaces, but this doesn't distinguish one from the other.

What distinguishes the Euclidean space from the non-Euclidean one is that particles following geodesics in the flat Euclidean space-time don't change their relative separation, but particles following geodesics in a curved space-time do. This is called geodesic deviation.

Of course, you need to have some idea of what a geodesic is to make sense of this notion. For the purposes of understanding GR, it's sufficient to regard a geodesic as the shortest distance between two nearby points, though this is somewhat oversimplified, especially the "nearby" part. But it should give you some idea of what a geodesic is. Alternatively, you can define a geodesic more generally by the notion of parallel transport, but that may be more abstract and less familiar. And for GR it won't make much difference, but in some other theories, it matters.

It's also helpful to consider the meaning of curvature as applied to space and not space-time as a warm-up exercise. A plane is flat, the surface of a globe is curved. Parallel lines drawn on a plane never intersect and keep a constant distance. On a sphere, the equivalent of straight lines (great circles) must intersect, and in general parallel lines don't maintain a constant distance on a curved surface.

So yet another way of thinking about non-Euclidean geometry is to look at Euclid's parallel postulate.

4. Aug 27, 2016

### Jonathan Scott

There are two types of "curvature" involved.

When GR people talk about "curvature", they usually mean intrinsic curvature which is caused by the presence of mass. Intrinsic curvature is like the fact that if you go round a closed path within the surface of a ball, the total angle you have turned through in the plane of the surface is not exactly 360 degrees. According to the Einstein Field Equations, this curvature is zero except at the location of some mass or equivalent energy. The resulting distortions mean that it is impossible to map local space-time in the vicinity of a mass directly to a conventional Euclidean coordinate system, in the same way that the surface of the Earth cannot be represented on a flat map without distorting scale factors, angles or both. In this sense, space-time really is curved.

They also explain free fall motion (geodesic motion) by a different sort of curvature, like the curvature of an orbit. This is the curvature of a geodesic path relative to a chosen practical coordinate system. An object which starts at rest has a path which starts off parallel to the time axis and then curves in the direction of the gravitational field. An object which moves sideways relative to a gravitational field also has its path curved by the curvature of space, so the rate of change of direction is proportional to the speed and the resulting acceleration is proportional to the square of the speed. The curvature with respect to time and space in conventional coordinates is approximately equal for weak fields, and is equivalent to the Newtonian expression $g/c^2$ where $g$ is the Newtonian gravitational field. The resulting acceleration for a particle moving at speed $v$ perpendicularly to the field is $g (1 + v^2/c^2)$, which is why light is deflected twice as much by the sun as would be predicted from the Newtonian acceleration. This second form of curvature can be transformed away by looking at the situation from the point of view of the falling object, but if one is calculating orbits or similar that point of view is not very useful.

5. Aug 27, 2016

### Ibix

As haushofer has said, it is wrong (or, at least, critically incomplete) to say that space curves - spacetime curves. That's not pedantry, because the curvature in time-like directions is important to understanding why things at rest with respect to a gravitating body start moving.

I think you might be referring to the classic (and unhelpful) picture of a planet sitting in a dent on a plane. Partly this is unhelpful because it's showing spatial curvature only. But it is also mixing up intrinsic curvature (which is what GR cares about) and extrinsic curvature (which has no application in GR). What am I talking about?

That rubber sheet is a 2d surface embedded in a 3d Euclidean space. It can have intrinsic curvature, which means that the geometry of the surface is non-Euclidean. For example, straight parallel lines on the surface might cross if they pass near the dent (pervect's example of great circles on the surface of a 2-sphere is better). But the surface can also have extrinsic curvature, which is to say that lines in the surface are not straight lines in terms of the 3d space. For example, a straight line in 3d Euclidean space cannot lie in the surface of a 2-sphere.

Things with intrinsic curvature must have extrinsic curvature if they are embedded in a Euclidean space (I think...). But the reverse is not true. I can roll a piece of paper (the archetypal Euclidean plane) into a cylinder. The cylinder has extrinsic curvature in any direction except its axial direction, but the intrinsic curvature is still zero - triangle angles add to 180, $c=2\pi r$, etc. The only oddity is that features are periodic in one direction, which is a topological oddity not a geometric one.

In GR we only care about intrinsic curvature because there's no evidence of anything outside the spacetime for spacetime to be embedded in, so nothing to measure extrinsic curvature with respect to. That's possibly what you mean by "not really curved". Spacetime is really curved in the intrinsic sense (and this is what the Riemann tensor describes, yes). It can be dangerous to visualise a slice through spacetime embedded in 3-space because one can visualise (e.g.) the cylinder and think that it's obviously curved when it isn't in the relativistic sense.

To check whether a spacetime is curved you do need to examine geodesics, or (better) compute curvature invariants, or otherwise do experiments that rely wholly on measurements made inside the spacetime you are considering.

It's probably worth noting something that Peter mentioned the other day - at least one candidate for a quantum gravity theory says that spacetime is flat, and that all the curvature effects are simply our interpretation of a spin-2 quantum field. So maybe spacetime doesn't "really" curve. But for the purposes of GR you can say that it does really curve, as long as you are careful about what you mean by "curve".

6. Aug 27, 2016

### Ibix

Is this correct? I thought that the vacuum solution permitted non-zero Weyl curvature. For example Wikipedia cites a non-zero Kretschmann scalar everywhere in a Schwarzschild spacetime.

https://en.m.wikipedia.org/wiki/Kretschmann_scalar

7. Aug 27, 2016

### Jonathan Scott

By the "curvature" in the context of the Einstein Field Equations I was of course referring to the (trace-reversed) Ricci tensor, not the full Riemann curvature. Close enough?

8. Aug 27, 2016

### tionis

Where is that thread? Do you know which theory he was referring to?

9. Aug 27, 2016

### Ibix

10. Aug 27, 2016

### Ibix

Sorry for the pedantry, but I think if the OP is asking about curvature being "actual" or not, we need to be a bit careful.

OP: The Riemann tensor describes curvature. It can be split up and/or summarised in various meaningful ways. The Ricci tensor is one such summary, and it is, as Jonathan says, zero everywhere the stress-energy tensor is zero. But parts of the Riemann that are lost in that particular summary can be non-zero, and that can be reflected in other summaries of the Riemann, such as the Weyl tensor and Kretschmann scalar that I mentioned.

I'm well outside the bounds of a "B" thread here, so I'll just say that "curvature" unqualified can mean a lot of different things, some of which are relevant to GR, and some of which aren't, and some of which can be zero even when others aren't. Fun, isn't it?

11. Aug 27, 2016

### Megaton

Wow thanks for all your answers, boy GR is weird...weird but true

12. Aug 28, 2016

### Staff: Mentor

No, because the "Euclidean space" you are referring to does not exist, so there's no way to make the comparison you describe.

Physically, spacetime curvature is tidal gravity. So if you think about how you would observe tidal gravity, that is how you would observe spacetime curvature. (Pervect mentioned geodesic deviation, which is another way of saying tidal gravity.)

Not if we're considering timelike geodesics, which are the ones we need to consider if we're talking about geodesic deviation for test objects, i.e., tidal gravity. A timelike geodesic is the longest "distance" (proper time) between nearby points (events).

13. Aug 28, 2016

### Staff: Mentor

This doesn't change anything with regard to the topic of this thread, because the flat background spacetime in this theory is unobservable; the spacetime geometry we actually observe is the curved geometry that results from the effects of the spin-2 field.

14. Aug 28, 2016

### Ibix

Understood. What I was getting at was what you wrote at the bottom of this post (now my search seems to be working better). I took that to mean that curvature wasn't "real" in the quantum theory of gravity that you refer to in that post, but on a re-read just now I think you may have been making a slightly more general point (related to the first paragraph of pervect's #3 above), that curvature is arguably only an interpretation of the maths, albeit definitely the most popular and probably the most straightforward.

15. Aug 28, 2016

### haushofer

It's called massless Fierz-Pauli theory. See e.g. Hinterbilcher's notes on massive gravity, or Ortin's "gravity and strings".

16. Aug 28, 2016

### haushofer

A stringy analog I like is the idea of a laserbeam being a coherent state of photons. Similarly, in string theory (or the iteration of Fierz-Pauli) you get a coherent state of gravitons resulting in spacetime geometry at the classical scale. Geometry is as such an emergent effect, similar to LQG.

17. Aug 28, 2016

### Staff: Mentor

No.

Yes. "Spacetime curvature" is a way of describing tidal gravity, which is something we observe directly. But there's nothing that requires us to use that particular description. We use it because it works and because it has proven very fruitful.

18. Aug 28, 2016

### tionis

This is great. Do you mean Riemannian geometry can be interpreted as other than curved, or are you referring to a quantum theory? Doesn't Riemann geometry forces us to accept that curvature is the only interpretation?

19. Aug 28, 2016

### Staff: Mentor

No. Riemannian geometry is a mathematical model. A mathematical model can never force you to adopt it as an interpretation of physical observations.

20. Aug 29, 2016

### tionis

Yes, but isn't Riemannian geometry exclusively about curved spaces? Do you mean within its framework there are other possible ways to interpret the geometry in a way that doesn't involve curves?

21. Aug 29, 2016

### Staff: Mentor

No. But nothing forces you to use Riemannian geometry to interpret the physical observation of tidal gravity.

22. Aug 29, 2016

### tionis

But if I were to use Riemannian geometry to describe tidal gravity, does it force me to use curves and nothing else?

23. Aug 29, 2016

### Ibix

If I understand correctly, if you attempt to draw a triangle near a massive body, you will find that the angles do not add to 180. The GR interpretation of that is that spacetime is curved and there's no such thing as a straight line in this context. The flat space plus quantum gravity field interpretation is that the gravity field always curves any attempt at a straight line. There is nothing gravitationally neutral that you can use to follow a straight line so you're stuck with curves - although straight lines do, in principle, exist.

Think of drawing a large "triangle" on a globe. Then represent it on a Mercator projection. The angles sum to the same (>180). The first drawing, a "triangle" made of great circles, is GR. The second, a "triangle" made of curved lines (curved because of interaction between the lines and the gravity field) is this quantum gravity interpretation. There's no experimental way to differentiate them within the domain of applicability of GR and the maths is the same.

Please bear in mind that this is no more than (my understanding of) pervect's post that I linked in #9, and I don't know any more than that. If this is a helpful analogy I'd wait for the nod from someone who actually knows what they're talking about before assuming it's right...

24. Aug 29, 2016

### tionis

But not in the context of general relativity, correct?

25. Aug 30, 2016

### Ibix

Correct - there is no flat background there. My reading of what Peter and pervect said about the quantum theory on a flat background is that you can at least conceptualise straight lines on the flat background, but you could never verify whether or not you were following one.

As before, I'm simply paraphrasing what I've read on this thread and the post of pervect's that I linked, because I think your question was more "how can a field cause effects we can take to be geometry" than the interpretation issue I think Peter was adressing. Don't assume I'm right about any of it (except the GR has no straight lines - I'm sure about that bit).

26. Sep 1, 2016

### Staff: Mentor

More or less, yes.