- #1

- 10,204

- 1,367

The question is in the title. I believe the answer is yes.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- A
- Thread starter pervect
- Start date

- #1

- 10,204

- 1,367

The question is in the title. I believe the answer is yes.

- #2

- 20,004

- 10,651

The spatial slices are, yes. But not only that, the de Sitter space is maximally symmetric.

- #3

- 4,066

- 1,645

But then I got confused by this paper, which appears to state that Schwarzschild spacetime, which is neither isotropic nor homogeneous, also has a flat foliation. Although they claim that can be observed from the metric, which I can't see, since the transformed metric they present in support of that claim still depends on radius ##r##, which I would have thought implies non-homogeneity.

- #4

- 10,204

- 1,367

- #5

romsofia

- 558

- 253

- #6

PeterDonis

Mentor

- 41,317

- 18,943

I agree.I believe the answer is yes.

- #7

PeterDonis

Mentor

- 41,317

- 18,943

No, a flat slicing is not sufficient for homogeneity and isotropy, as the counterexample of Schwarzschild spacetime with Painleve coordinates shows. Nor is a flat slicing even a necessary condition for homogeneity and isotropy, since a closed FRW universe, for which spatial slices are 3-spheres, is homogeneous and isotropic and has no flat spatial slicing.I would have thought that the existence of a flat slicing for de Sitter space, as described here, implies homogeneity and isotropy.

The paper you cited isn't claiming that Schwarzschild spacetime is homogeneous. They are only claiming that it has a flat spatial slicing. The way to see that from the Painleve metric is to note that the spatial part of the metric (i.e., the part you get when you set ##dt = 0## so all the terms with a ##dt## in them vanish) is the metric for flat Euclidean 3-space.they claim that can be observed from the metric, which I can't see, since the transformed metric they present in support of that claim still depends on radius ##r##, which I would have thought implies non-homogeneity.

- #8

PeterDonis

Mentor

- 41,317

- 18,943

It isn't spelled out explicitly in any treatment I'm aware of, but the existence of global Lorentz transformations (i.e., Lorentz transformations preserving the metric globally, instead of just locally) implies zero geodesic deviation (any two geodesics that start out parallel will always remain parallel), which is equivalent to flatness.I don't recall ever seeing flatness ever spelled out as a requirement for SR

- #9

- 22,472

- 13,406

SR simply assumes a flat spacetime manifold (the pseudo-Euclidean affine space called Minkowski space), which is fixed once and for all. In SR there are global inertial frames and any observer at rest wrt. an inertial frame by assumption describes space as a Euclidean affine space (with all its symmetries of homogeneity and isotropy).

- #10

- 20,004

- 10,651

From what I read in the paper, if you set ##dt=0## you do not get Euclidean space as it does not have a ##dr^2## term. The ##dr## only appears together with a ##dt##.No, a flat slicing is not sufficient for homogeneity and isotropy, as the counterexample of Schwarzschild spacetime with Painleve coordinates shows. Nor is a flat slicing even a necessary condition for homogeneity and isotropy, since a closed FRW universe, for which spatial slices are 3-spheres, is homogeneous and isotropic and has no flat spatial slicing.

The paper you cited isn't claiming that Schwarzschild spacetime is homogeneous. They are only claiming that it has a flat spatial slicing. The way to see that from the Painleve metric is to note that the spatial part of the metric (i.e., the part you get when you set ##dt = 0## so all the terms with a ##dt## in them vanish) is the metric for flat Euclidean 3-space.

- #11

DrGreg

Science Advisor

Gold Member

- 2,438

- 1,701

I'm not familiar with Gullstrand-Painleve coordinates, but a Google search suggests that the formula used in the paper referred to above is wrong: they seem to have omitted a ##dr^2## term.From what I read in the paper, if you set ##dt=0## you do not get Euclidean space as it does not have a ##dr^2## term. The ##dr## only appears together with a ##dt##.

- #12

PAllen

Science Advisor

- 9,046

- 2,281

Just consider setting t constant in the metric from that paper. Then you have Euclidean flat metric left, in polar coordinates. The r of polar coordinates doesn’t make it non flat. In fact, within such a t constant slice, you can trivially transform to Cartesian coordinates, getting the Euclidean flat Cartesian metric.

But then I got confused by this paper, which appears to state that Schwarzschild spacetime, which is neither isotropic nor homogeneous, also has a flat foliation. Although they claim that can be observed from the metric, which I can't see, since the transformed metric they present in support of that claim still depends on radius ##r##, which I would have thought implies non-homogeneity.

- #13

PAllen

Science Advisor

- 9,046

- 2,281

Oops, yes, but the conclusion is still correct. I’ve seen this many times before, so my mind filled in the missing drI'm not familiar with Gullstrand-Painleve coordinates, but a Google search suggests that the formula used in the paper referred to above is wrong: they seem to have omitted a ##dr^2## term.

- #14

George Jones

Staff Emeritus

Science Advisor

Gold Member

- 7,611

- 1,506

From page 60 of Wald's "General Relativity": "Thus, the theory of special relativity asserts thatI don't recall ever seeing flatness ever spelled out as a requirement for SR

(Wald's italics, not mine.)

- #15

PeterDonis

Mentor

- 41,317

- 18,943

I think the paper is using unusual notation; by ##d\Omega^2## I think they mean the metric for Euclidean 3-space in spherical coordinates, not the metric of a 2-sphere in angular coordinates.From what I read in the paper, if you set ##dt=0## you do not get Euclidean space as it does not have a ##dr^2## term. The ##dr## only appears together with a ##dt##.

As others have pointed out, it is easily confirmed by checking other references that the correct metric for Schwarzschild spacetime in Painleve coordinates has flat (Euclidean 3-space) spatial slices of constant coordinate time.

- #16

- 20,004

- 10,651

I think it is a typo. They do write ##r^2 d\Omega^2## so it would be a very strange notation to have the ##r^2## in front when there is no ##r^2## in the ##dr^2## term.I think the paper is using unusual notation; by dΩ2 I think they mean the metric for Euclidean 3-space in spherical coordinates, not the metric of a 2-sphere in angular coordinates.

- #17

- 13,288

- 1,725

To be fair, the textbook on Relativity (Ray D'Inverno'sIt isn't spelled out explicitly in any treatment I'm aware of, but the existence of global Lorentz transformations (i.e., Lorentz transformations preserving the metric globally, instead of just locally) implies zero geodesic deviation (any two geodesics that start out parallel will always remain parallel), which is equivalent to flatness.

”Space and time are represented by a four-dimensional manifold endowed with a symmetric affine connection ##\Gamma^{a}_{~bc}## and a metric tensor ##g_{ab}## satisfying the following:

(i) ##g_{ab}## is non-singular with signature ##-2##;

(ii)## \nabla_c g_{ab} = 0##;

(iii) ##R^{a}_{~b\vert cd} = 0.##"

Quoting just under Axiom II on the same page: "The first axiom defines the

So the flatness in explicitly defining SR is a matter of reading (the good) books.

- #18

PeterDonis

Mentor

- 41,317

- 18,943

Hm, yes, good point.I think it is a typo. They do write ##r^2 d\Omega^2## so it would be a very strange notation to have the ##r^2## in front when there is no ##r^2## in the ##dr^2## term.

Share:

- Replies
- 4

- Views
- 642

- Last Post

- Replies
- 2

- Views
- 353

- Replies
- 4

- Views
- 631

- Last Post

- Replies
- 5

- Views
- 750

- Last Post

- Replies
- 6

- Views
- 2K

- Last Post

- Replies
- 13

- Views
- 135

- Replies
- 35

- Views
- 2K

- Replies
- 61

- Views
- 2K

- Replies
- 63

- Views
- 3K

- Replies
- 13

- Views
- 617