# B How do we determine flatness?

1. Nov 29, 2016

### newjerseyrunner

I'm not sure if this is meant for a math or cosmology section but GR is what had me on this train of thought so I'll post it here.

What is flat space? Does that simply mean that the universe as a whole has the same geometry as we have here on Earth (minus the gravity well?). Or is it something more fundamental that I'm missing.

Basically, if the universe was curved in such a way that there were three right angles in a triangle, would we have defined that as flat? Is flatness itself a relative term to the observer? If so, why is the observation that the universe appears perfectly flat not expected, wouldn't we only to be able to detect gradients?

2. Nov 29, 2016

### Staff: Mentor

One thing we need to be clear about: Are you asking about flatness in space, or flatness in spacetime?

Spatial flatness is determined by looking for violations of Euclidean geometry: Do the interior angles of a triangle add up to anything but 180 degrees? Do parallel lines intersect if you extend them far enough? Does the Pythagorean theorem not work quite right? Do all of these effects become greater as you try larger distance scales? If so, you're in a curved space.

It's possible for space to be flat although spacetime is curved, so we have to check for that separately. Spacetime curvature is determined by looking for tidal effects and geodesic deviation. What trajectories do objects in freefall follow?
No.

3. Nov 29, 2016

### Staff: Mentor

A manifold is called flat if it obeys Euclidean geometry. There are other defining properties, but that is the easiest one to understand.

4. Nov 29, 2016

### Bandersnatch

...using this as an example, if you had a triangle with three 90 degree angles, that would most definitely not be what you have in an Euclidean (flat) space, where, as you may recall, all triangles have angles that add up t 180 degrees. You can get 3x90° angles e.g. on a curved surface of a sphere - put one apex at the pole, and two at the equator, and with some fiddling you can easily draw a triangle that'll have 270° of internal angles.

So, if we can measure some tirangles in the universe, and if these triangles prove to have >180 degrees of internal angles, then it shows that the space in which these angles were drawn is curved inwards (like a sphere; negative curvature). If these angles are >180° in total, the curvature is positive (saddle-like). All you need to do to do it, is to find an observable that is at a known distance and should have a calculable proper diameter. If the actual observable it's smaller than that, then the triangles in the universe must be curving inwards (closed universe). If it's larger, then the universe is open. If they're as expected, it's flat.

The observable used for this purpose are the baryonic acoustic oscillations in the CMBR. The distance to CMBR and the expected size of the oscillations are known, so it's just a matter of comparing the predicted size to the observed one.

So far, the size of BAOs seems to fit the predicted size for flat geometry, within the error bars.

5. Nov 29, 2016

### Battlemage!

Okay this topic is out of my depth, but wouldn't the kronecker delta as the metric tensor imply flat space? Is there some relationship there? I thought I had heard something like that before, and it makes a kind of sense to me.

6. Nov 29, 2016

### Staff: Mentor

Yes, but the converse does not follow. The components of the metric tensor depend on the coordinates you're using, and they only come out to be $\delta_{ij}$ if you're using Cartesian coordinates. (For an easy but really important example, compare the components of the metric tensor for a two-dimensional Euclidean plane when you use polar $r$,$\theta$ coordinates and Cartesian $x$,$y$ coordinates).

Note that this is flat space. Flat spacetime is Minkowskian not Euclidean, and even using Cartesian $x$, $y$, $z$, $t$ coordinates the metric components are not quite equal to $\delta_{ij}$ because the sign of $g_{tt}$ will be opposite to the sign of $g_{xx}$, $g_{yy}$, and $g_{zz}$.

7. Nov 30, 2016

### vanhees71

Flatness can be defined for manifolds that do not even have a metric. An affine connection is sufficient. It's locally flat if the curvature tensor vanishes everywhere.

Another example of a non-Euclidean space that is flat is the Minkowski space of special relativity. It's pseudo-Euclidean since the fundamental bilinear form is not positive definite but has signature (1,3) or equivalently (3,1).

8. Nov 30, 2016

### A.T.

The deviation from 180° would depend on the size of the triangle, so I don't think we would have given 270° a special name.

9. Nov 30, 2016

### epovo

What would the metric look like in such spacetime?

10. Nov 30, 2016

### Ibix

The flat space FRW metric is one such. $$ds^2=dt^2-a (t)\left (dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2\right)$$Note that the thing in brackets is a regular Euclidean metric expressed in polar coordinates, although the scale changes with time.

11. Nov 30, 2016

### vanhees71

Please, refer to FLRW metric. Don't omit Lemaitre, who is only regularly ignored, because he published his results in French in a not so well-known journal :-(.

12. Nov 30, 2016

### Mister T

Is a flat space mathematically possible where the sum of the interior angles of a triangle is always, say 270° rather than 180°, regardless of the size of the triangle?

13. Nov 30, 2016

### Grinkle

I opine no. Humans thought the earth was flat (I guess some still might) using the same definition of flatness that we are talking about in this thread. The knowledge we now have that the earth is curved and observing that if one draws a triangle on the surface of the earth its angles will not add up to 180 degrees does not cause us to re-define what we mean by flat. People who live on a visually observably curved part of the earth's surface do not grow up intuitively confused about what other people living far away call 'flat'.

I suppose our brain is wired to give special significance to the direction perpendicular to earth's gravity and from that we understand the concept of flat.

14. Nov 30, 2016

### Staff: Mentor

The first question would be, is such a space mathematically possible at all, without even asking whether it is "flat"?

15. Dec 1, 2016

### pervect

Staff Emeritus
I think you can get a triangle with three right angles on a continuously differentiable Riemannian manifold if the triangle encloses a singularity, but not otherwise. For an example, consider a cone made by cutting out one quadrant of a plane, and gluing the cut edges together. We'll draw the triangle on the flattened cone, before we glue it together.

Straight lines on the paper will be geodesics on the cone. The tip of the cone is not part of the manifold, it's a singularity, because it's geodesically incomplete. The red lines are straight lines on the cut and flattened cone, and hence geodesics. I believe that drawn as shown, the red lines form a triangle on the cone. But this isn't from a textbook example or anything, so reader beware.

It'd be interesting to see a more mathematical analysis of this simple case. Triangles not including the central singularity at the origin will have the usual 180 degree sum of angles. I believe we'll always get 270 degrees for the sum of angles if the triangle includes the singularity, but I don't have a formal proof. My informal intuition (I'll stop short of calling it a proof) involves parallel transporting a vector around the triangle, and observing that said vector doesn't rotate if the triangle doesn't include the singularity, but does rotate 90 degrees when the vector is transported around the "edge".

Note that the singularity is geodesically incomplete, which is pretty much the definition of a singularity as I recall. If you have a geodesic (a straight line on the cut paper) that reaches the central singularity, it's undefined how to continue said straight line afterwards.