# What Do We Mean By Flat Space

1. Oct 31, 2014

### shounakbhatta

Hello All,

Pertaining to General Relativity, what do we mean when we tell a space is flat or space is curved?

Is it that sum of the angles of a triangle = 180?
Or is it something else.

I was listening to a lecture of Prof.Susskind, when he takes a piece of paper, makes certain points on the paper and then folds it, up/down. He tells that this is not curved. He reasons that as because the distance between two neighboring points is still the same (when the paper is folded).

Is there any way to say that my table top is flat and the other one is curved?

Thanks and regards

2. Oct 31, 2014

### Simon Bridge

Flat space is where the metric is diagonal - where there is no intrinsic curvature, and, thus, no gravitational effects.
One of the consequences is that the sum of angles of a triangle is pi ... at this level it is best to work in radians.
Another is that parallel lines do not intersect.

A sphere has this kind of curvature, but a cylinder does not. A simple test is to see if there is any way to cut it so it flattens out into a plane without distortion.
http://www.geom.uiuc.edu/docs/doyle/mpls/handouts/node21.html

3. Oct 31, 2014

### shounakbhatta

Diagonal metric in the sense? Can you please give an example.

4. Oct 31, 2014

### ShayanJ

Does it work in Minkowski space-time too?
It seems to me that flat space-time has a slightly different geometry than Euclid's!

I think the right way of saying this is: In flat space, there exist pairs of lines, called parallel lines, that don't intersect.

5. Oct 31, 2014

### Staff: Mentor

That is indeed something that happens in curved space, although it is not the defining feature. The defining feature of a flat space is that the Riemann curvature tensor is 0.

This tensor describes how strongly dependent on the path is the parallel transport of vectors. If the curvature tensor is 0 then parallel transport is not path dependent and we call the space flat.

6. Oct 31, 2014

### stevendaryl

Staff Emeritus
That doesn't seem like the best way to put it, because it's often (usually? always?) possible to choose coordinates so that the metric is diagonal, even in curved spacetime. For example, the Schwarzschild metric is diagonal.

If you said the metric is diagonal with constant components, that would do it.

7. Oct 31, 2014

### Staff: Mentor

That is a sufficient condition, but not a necessary one. For example, the Euclidean plane is flat, but after the coordinate transformation $x'=x$ and $y'=x^2-y$ the metric will have off-diagonal and non-constant components.

Of course this further supports your point, that looking at whether the metric is diagonal isn't the right place to start.

8. Oct 31, 2014

### m4r35n357

That is a good way of looking at to start with, it but with the following very important proviso: you must be able to make the triangle arbitrarily big, otherwise you could just be in a region of very slight curvature. As others have pointed out, to be precise, for space the metric should be k X the identity matrix, and for spacetime k X the Minkowski metric.

9. Nov 1, 2014

### Simon Bridge

That should be in the sense that the Reimann curvature tensor is 0 ;) ... Did you have a go reading the references I gave you?

An example of a flat spacetime metric can be found in: http://en.wikipedia.org/wiki/Metric_tensor_(general_relativity)#Flat_spacetime
... and many other places for the googling.

Try it and see.

... I was being too glib yes. OTOH: I also gave references to elaborate on what I was talking about so I didn't need nor intend to provide an exact definition. There are just so many available for a quick google.

10. Nov 1, 2014

### Ben Niehoff

Spacetime being flat has nothing whatsoever to do with whether the metric tensor is diagonal. It is neither a necessary nor a sufficient condition (there are both diagonal metrics that a not flat, and flat metrics that are not diagonal).

A spacetime is flat if and only if the Riemann tensor vanishes*. Operationally, this means that a vector parallel-transported around a closed curve will always return to itself.

* Note: this is a local condition, so this only tells you the spacetime is locally flat (i.e., within a given coordinate patch). There may be global issues to consider, such as conical singularities: if you parallel-transport around the tip of a cone, your vector will return tilted by the angular deficit of that cone.

11. Nov 1, 2014

### pervect

Staff Emeritus
"Curved" and "Flat" are not very precise when used in popularizations, but in the context of General Relativity, "curved" usually means that the Riemann curvature tensor vanishes.

The sum-of-angles of a triangle technique works to determine the flatness of Euclidean spatial slices, but the necessary concepts of angle do not translate well into the Minkowski geometry of space-time. For instance, if we replace "angle" with "rapidity" because angles add and rapdities also add, one must face up to the fact that the angle / rapidity between a beam of light and either a spacelike or timelike geodesic is infinite, while Euclidean angles are always between 0 and 2pi radians (or if you prefer 0 and 360 degrees).

Setting the condition that parallel transport around a closed curve does not change the orientation of a vector is equivalent to the condition that the Riemann tensor vanishes and thus will work to determine the curvature of either Euclidean space or Minkowskian space-time, but it requires one to understand the concept of parallel transport.

There is a geometrical construction known as "Schild's ladder" that can provide a semi-intuitive geometrical way of defining parallel transport (at least as long as torsion is absent, which is presumed to be true in General Relativity). See http://en.wikipedia.org/w/index.php?title=Schild's_ladder&oldid=586460884

To use Schild's ladder one must have some formal or informal understanding of geodesics, specifically how to draw and extend a geodesic between two points (the points can be presumed to be close enough together so that said geodesic is unique), and also how to take the midpoint of a geodesic segment.

The underlying motivation of Schild's ladder is that given the absence of torsion, the sides of a quadrilateral whose opposing sides have equal lengths / lorentz intervals are "parallel", and that by drawaing an infinite number of such quadrilaterals, one can parallel transport a vector.

Perhaps, but in the context of General relativity it is important to note that the intrinsic geometry is independent of whether or not the paper is folded. Thus, the concepts of flatness that is useful in general relativity will not change if you fold the paper.

See for instance the general wiki discussion of curvature http://en.wikipedia.org/w/index.php?title=Curvature&oldid=631099092
and note that the Gaussian curvature described in said article is the intrinsic sort which is useful in General Relativity. You will note that there are various different definition of "extrinsic curvature", but those are not applicable to general relativity, as they would require one to be "outside the universe" to observe, while the methods of intrinsic curvature give a number based on measurements we can actually perform.

12. Nov 1, 2014

### Matterwave

Ben is right of course. I just want to add that the Riemann tensor is a measure of intrinsic (and local, as Ben mentioned) curvature. This means that it measures curvature independent of any particular embedding that may be performed on the space. The most basic way to look at intrinsic curvature, in my opinion, is by looking at geodesic deviation. If two geodesics which begin parallel remains parallel then the space is flat. If two initially parallel geodesics converge or diverge then the space is curved. In other words, we can see if Euclid's 5th postulate (parallel postulate) holds or not (but we have to then replace "straight line" with "geodesic" since a "straight line" doesn't have a rigorous definition in a non-Euclidean space).

If the space is embedded in a higher dimensional space, for example a cylinder embedded in 3-D Euclidean space, then there is the additional notion of extrinsic curvature present. The basic way to look at extrinsic curvature is to look at how normal vectors change as you move along the surface. If the normal vector doesn't change, the the space is extrinsically flat. An object can be intrinsically flat, but you can embed it in such a way as to make it curved. A cylinder, for example, is intrinsically flat, but extrinsically curved. An object which is intrinsically curved, though, won't be able to be embedded in such a way as to turn out extrinsically flat (AFAIK, someone correct me if I'm wrong here).

13. Nov 1, 2014

### atyy

How about if we add diagonal and constant in at least one set of coordinates?

14. Nov 2, 2014

### Matterwave

By "in at least one set of coordinates" do you mean specifically that we are restricting ourselves to using coordinate bases? Using a tetrad basis, the metric will be $\eta_{\mu\nu}$ regardless of curvature.

15. Nov 2, 2014

### atyy

I didn't but I should. Does it work then or do we really have to use the curvature tensor to define flatness?

16. Nov 2, 2014

### Matterwave

I think it will work then. In a coordinate basis, the Riemann curvature tensor is composed of first and second order derivatives of the metric. A constant metric will have derivatives of all orders being 0, and will therefore mean 0 intrinsic curvature. In fact, I am not sure about the requirement that the metric be diagonal. It seems to me that the important requirement is that it is constant so that its derivatives vanish. Probably, any constant metric (in coordinate bases) can be diagonalized by a simple coordinate transformation (and not just at a point, since that can be done for any metric). And then a simple re-scaling of the basis vectors will give you the Minkowski metric. I'm not 100% sure of this statement.

17. Nov 2, 2014

### Staff: Mentor

Yes.

This seems right to me.

18. Nov 2, 2014

### Ben Niehoff

Guys, "There exist coordinates in which the metric (in coordinate basis) is constant" is equivalent to "The Riemann tensor vanishes".

However, the latter condition is simpler to apply in generality, because the change of coordinates needed in the first case might not be at all obvious.

In general, to find a change of coordinates that brings the metric into a given form, you must integrate a system of nonlinear PDEs. And a priori, you don't even know if a solution exists, because presumably you don't know whether your metric is flat! However, using the Riemann tensor, you just need to differentiate twice and do some algebra.