# Example of curved spacetime with flat space layers

curved spacetime is a famous geometrical and physical object. on the other hand there are debates on if space itself is curved or not. but is it really possible to a curved spacetime that all of its spatial layers be flat?
more precisely: every freely falling observer at every instance sees a 3-dimensional space. is it possible that all of this 3-dimensional spaces are flat while spacetime itself is curved?
is there a any explicit metric by this property?

• I don't know about the plural of "layers". I believe that there could be just a single layer of space. I've seen it plausibly claimed, on this site, that gravity itself is only dependent on the curvature of time, and, given the fact that even a plane or cube can rotate, it seems like it would be technically possible, although a rectilinear universe would seem so unnaturally artificial that objections raised against the acceptance of infinity within physics (as by Hossenfelder at youtube.com/watch?v=Bq9xR5PUs6s ) might seem implausible until hearing her video. – Edouard Jan 18 at 21:59
• Physicists like Hilbert, and even Godel, have objected to the inclusion of infinity in physics, but I don't think anyone's objected to eternality, at least in one temporal "direction" (the past-to-future one). – Edouard Jan 19 at 2:07
• There does seem to be one tiny flaw in Hossenfelder's video, 8 min. into it: A laser beam being swung back & forth while directed at a section of a distant wall might, when the wall would be rotated so that the beam would be striking an edge of its surface, appear to be travelling at infinite speed, "except that the wall would not have a perfectly flat surface". The circumvention would require the wall to be formed of an exotic non-particulate substance (or of a substance made with rectilinear particles), although only as a smooth wall might've been pictured before microscopes were invented. – Edouard Jan 21 at 20:31
• Re my previous comment, some acceptability of physical singularities would have to be implied by Roger Penrose's receipt of a Nobel Price for his 1970 singularity theorem, late in 2020. – Edouard Mar 3 at 20:23

## 5 Answers

The Friedmann metric with $$k=0$$ is spatially flat. It is consistent with current cosmological observations.

• At gresham.ac.uk/lecture/transcript/print/to-infinity-and-beyond , in a paragraph headed "How dark matter affects the CMB", it was claimed that satellite observations show the observable region to be flat only to within 1% or 2% of perfection, which sounds more like what Charles Francis said, and is more consistent with the mishmash of spatial and temporal terminology which has characterized the verbiage of physics for a century. (If I sound bitter, it's becuz I'm lame at math.) The data, however, is from 1992. Is there something more recent, or has matter thrown things off? – Edouard Jan 19 at 4:14
• @Edouard Wikipedia :“detailed observations of the cosmic microwave background made by the Planck spacecraft [between 2009 and 2013] ... [show] that the Universe is flat to within 0.5 percent” – G. Smith Jan 19 at 4:28
• @Edouard Note that the parameters for the current cosmological model have $\Omega_\Lambda$ and $\Omega_m$ adding to 1.0000, the critical value for flatness. But the two values do have errors of 0.0062. – G. Smith Jan 19 at 4:37

is it possible that all of this 3-dimensional spaces are flat while spacetime itself is curved? is there a any explicit metric by this property?

Sure, it's called the Friedman-Robertson-Walker metric and it describes the large-scale evolution of our Universe: $$ds^2 = -dt^2 + a^2(t) (dx^2 + dy^2 + dz^2)$$ On any surface of constant $$t$$, the metric is that of flat Euclidean space. The Einstein equation gives an equation of motion for the so-called scale factor $$a(t)$$.

(Technically, this is only one of three possible forms of the FRW metric; there are also versions where the surfaces of constant $$t$$ are curved. But the version above is the simplest one, and also appears to be the one that describes our Universe.)

• as it could be seen an observer who sees the metric similar to what you have written sees spatial layers flat. but what about other observers? an observer who is falling freely by this metric slices the spacetime to different layers, are you sure these layers are also flat? @Michael – moshtaba Jan 18 at 22:22

As @G. Smith said,

The Friedmann metric with $$k=0$$ is spatially flat,

but it is worth noting that this is an idealisation which depends on a perfectly uniform matter distribution. As such, we know it is an approximation to reality. It is only flat on a slice of constant cosmic time, which means it is only flat for observers with constant position in comoving coordinates (i.e. stationary in the CMB frame). For moving observers it would not be flat.

Also, when we consider observers in free fall, we are likely to be talking of observers in the vicinity of a gravitating body. Space in the vicinity of a gravitating body cannot be perfectly flat, although typically space curvature is so small that it is not directly measurable.

So when we debate whether space may be flat, we mean flat on cosmological scales. We know it is not perfectly flat.

• please for the moment forget source of curvature and consider spacetime as a curved geometric manifold by (-1,1,1,1) metric. is it true that every freely falling observer slices this spacetime to flat spatial slices (at every instance of time) or it is false almost for every observer? (I mean rather than very rare observers who see metric of spacetime as standard form of Freidmann metric. am I correct that they are very rare?) @Charles – moshtaba Jan 18 at 22:40
• @G.Smith, I beg your pardon – Charles Francis Jan 19 at 7:16
• @CharlesFrancis --Well, couldn't space be Friedmanniacally flat if every object had a double that, whenever the object would go left, would go right, etc.? The gravitating bodies might also have remote twins (perhaps necessarily in separate observable regions, with their separations evolved to prevent interference by malignant species) to balance their effects. In fact, that's sounding too much like the effects of quantum entanglement to be true, unless maybe such entanglement is that source of time which Ekaterina Moreva & other quantum physicists have considered that it might be. – Edouard Jan 22 at 19:14

Schwarzschild metric describing an isolated non-rotating black hole (and also spacetime outside isolated spherically symmetric body) admits “slicing” of spacetime in flat spatial layers with the help of Gullstrand–Painlevé coordinates: $$g = \left(1-\frac{2M}{r} \right)\, dt^2- 2\sqrt{\frac{2M}{r}} dt dr - dr^2-r^2 \, d\theta^2-r^2\sin^2\theta \, d\phi^2.$$ The geometry of a slice $$t=\text{const}$$ is a flat 3D space in spherical coordinates: $$dl^2=dr^2+r^2 \, d\theta^2+r^2\sin^2\theta \, d\phi^2.$$

Metric of rotating black hole (Kerr metric), however, does not admit spatially flat slicing.

Note, that there seems to be some misconceptions that this question displays:

more precisely: every freely falling observer at every instance sees a 3-dimensional space. is it possible that all of this 3-dimensional spaces are flat while spacetime itself is curved?

1. Slicing of spacetime into spatial layers by itself does not have anything to do with observers. If we consider observers staying at the fixed values of spatial coordinates as metric evolves with time (so-called fiducial observers), then such observers generally would not be free-falling.

2. What an individual observer sees is not the geometry of spatial slice. Instead one should remember that light propagates with a finite velocity, so the light rays arriving to observer at a specific moment $$t=t_0$$ carry information from the past slices. Moreover, these light rays would generally be bent by gravitational field (even when spatial geometry is perfectly flat).

A simple example of "spatial" slices not sharing the curvature of the background space is Euclidean space sliced into concentric spheres. This is similar to what happens in FLRW geometries. Like the spheres, the FLRW spatial slices are generally extrinsically curved (lines in them aren't geodesics).

I can't think of an example of a curved space with flat slices that is so easy to visualize, because it would need to be three-dimensional and to be easily visualizable it would need to be embedded in a flat background of four or more dimensions. But you could think of a conformally flat space with a metric of $$a(z)^2(dx^2+dy^2+dz^2)$$ or $$a(t)^2(dx^2+dy^2-dt^2)$$. It's obvious that slices of constant $$z$$ or $$t$$ are flat, and pretty easy to see that they are extrinsically curved iff $$a'\ne0$$ (because parallel paths on opposite sides of the surface have different lengths). Spatially flat FLRW cosmologies can be written in this form. The time coordinate is called conformal (cosmological) time, and is usually written with $$η$$ rather than $$t$$.

• as it could be seen an observer who sees the metric similar to what you have written sees spatial layers flat. but what about other observers? an observer who is falling freely by this metric slices the spacetime to different spatial layers (for every instance), are you sure these layers are also flat? – moshtaba Jan 18 at 22:27
• @moshtaba I think you're trying to port the special-relativistic concept of "observers" to general relativity where it doesn't work very well. Human cosmologists use FLRW coordinates even though Earth's velocity in those coordinates is nonzero, because any other coordinates would be incredibly inconvenient. Even in SR you don't need to use coordinates in which you're at rest, it's just less inconvenient than in GR. But if you did slice the FLRW spacetime differently, indeed those slices would not be flat. – benrg Jan 18 at 22:39
• you mean even following question doesn't work very well in GR: a freely falling observer records all events of spacetime which are Simultaneous by time 0 of his clock. this is a 3dimensional manifold. it is flat of not? (for a given metric) – moshtaba Jan 18 at 22:50
• @moshtaba A single clock doesn't define a surface of simultaneity. You need a family of clocks. In SR you can imagine an implicit family of Einstein-synchronized clocks but in GR there's no way to do that, you need to specify where the clocks are and how they're synchronized. Here's another answer I wrote about this. – benrg Jan 18 at 23:01