Conditions for spacetime to have flat spatial slices

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  • #1
PeterDonis
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In another thread,

https://www.physicsforums.com/showpost.php?p=2973770&postcount=45,

some questions came up about what the conditions are for a spacetime to admit flat spatial slices, and for a spacetime to have a time-independent "scale factor" (see definition below). These questions seemed interesting enough to warrant a new thread.

One key definition:

(A) The term "scale factor" is here extended from what I believe is its normal usage, which refers to the coefficient a(t) in front of the spatial part of the FRW metric in at least one of its "standard" forms. Physically, a(t) tracks the "comoving distance" between nearby geodesics that are "at rest", meaning that they are the worldlines of observers who see the universe as homogeneous and isotropic. By analogy, we can extend the term to refer to the "comoving distance" between nearby worldlines "at rest" in other spacetimes (with some caveats to the definition of "comoving", since the "nearby worldlines at rest" won't always be geodesics), and specifically to the fact that, in the other spacetimes we will be considering, there is no time-dependence in the "scale factor", meaning, roughly speaking, that the "size" of a given region of space does not change with time.

The reason this came up in the other thread was my use of the term "spatially flat" to describe the FRW spacetime with k = 0. Passionflower pointed out that, since the scale factor in this spacetime is time-dependent, even though each hypersurface of constant FRW coordinate time is flat, the "size" of a given region of space changes from one spatial slice to another, so, for example, a geometric figure "drawn" using geodesics in one spatial slice would have "expanded" in a later slice. This raised the question of what would be required of a spacetime to ensure that this didn't happen.

Two proposed conditions, based on the discussion in the other thread, are:

(1) For a spacetime to admit a metric with a time-independent scale factor, it must be stationary. The "canonical" example here is the Kerr spacetime, which is stationary but not static, and which does not admit any metric with flat spatial slices. It does, however, have the property that none of the metric coefficients depend on the time (in any of the standard coordinate charts--this is, of course, a restatement of the fact that the spacetime is stationary), which means that the "scale factor" is constant in time; contrast this with, for example, any of the FRW spacetimes, which have a scale factor that varies with time (see above).

(2) For a spacetime to admit a metric with flat spatial slices *and* a time-independent scale factor, it must be static. The "canonical" example here is, of course, Schwarzschild spacetime, which admits a coordinate chart (the Painleve chart) with flat spatial slices, and which also shares the time-independent scale factor property with the Kerr metric (of which it is a special case). (Of course there are other charts for this spacetime as well, which do not have flat spatial slices, but that doesn't matter for our purposes here as long as there is *some* chart that does.)

Condition (1) seems straightforward: a time-independent scale factor requires a time-independent metric.

I'm not sure about condition (2), though, because I'm not sure the interior portion of Schwarzschild spacetime--with r < 2M--qualifies as "static". (This portion is covered by the Painleve chart, so we can't finesse the issue by only considering the exterior portion.) The definitions in Wald are, briefly, that "stationary" means there is a timelike Killing vector field, and "static" means that field is hypersurface orthogonal. In the exterior of Schwarzschild spacetime, [itex]\partial_{t}[/itex] meets both these conditions, but it doesn't in the interior since it's no longer timelike. As far as I know, there is no other Killing vector field that *is* timelike in the interior (let alone hypersurface orthogonal).

There is one additional condition that we didn't propose an answer for in the other thread:

(3) What is required for a spacetime to admit a metric with flat spatial slices (but not necessarily a time-independent scale factor)? The "canonical" example here would be the FRW spacetime with k = 0, which has flat spatial slices but a scale factor that varies with time.

The key property that enables the construction of the metric for FRW spacetime in simple form is that it is isotropic; however, that condition can't be the right one for admitting flat spatial slices because it is too strong--Schwarzschild spacetime is not isotropic (it can be represented in so-called "isotropic coordinates", but that's not the same thing--the radial direction is still fundamentally different from the other two spatial directions). So maybe "spherically symmetric" is the right condition? I believe FRW spacetimes meet the textbook definition for that, even though they're not usually thought of in that way.

Can any of the experts here shed any more light on conditions (1), (2), and (3)?
 

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  • #2
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I'm not sure about condition (2), though, because I'm not sure the interior portion of Schwarzschild spacetime--with r < 2M--qualifies as "static". (This portion is covered by the Painleve chart, so we can't finesse the issue by only considering the exterior portion.)
People often confuse (I am not saying you do) things here:

There are two Schwarzschild solutions one is a vacuum solution and then there is an interior solution which is obviously not a vacuum solution. But the confusion comes because often the region 0 < r < 2M of the vacuum solution is also called interior.

The interior solution is static, however the vacuum solution is only static for 2M < r < infinity.
 
  • #3
PAllen
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For the purposes of this discussion, I have another question about what you want 'spatially flat' to mean. From discussion in another thread I initiated, it is clear that you can have valid coordinate systems with no timelike coordinate at all (and this is possible even in Minkowski spacetime of SR). So, if we're going to talk about something like time, and 'slices of constant time', we need to agree on a coordinate independent definition. Perhaps we need to say that a spatial slice is locally Euclidean, as bcrowell suggested in my other thread. I'm sure others can suggest better definitions, but just a coordinate t being constant means nothing.
 
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PeterDonis
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People often confuse (I am not saying you do) things here:

There are two Schwarzschild solutions one is a vacuum solution and then there is an interior solution which is obviously not a vacuum solution. But the confusion comes because often the region 0 < r < 2M of the vacuum solution is also called interior.

The interior solution is static, however the vacuum solution is only static for 2M < r < infinity.
Good point; I should clarify that here by "Schwarzschild interior solution" I mean specifically the interior *vacuum* solution, i.e., the vacuum solution for r < 2M.
 
  • #5
PeterDonis
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So, if we're going to talk about something like time, and 'slices of constant time', we need to agree on a coordinate independent definition. Perhaps we need to say that a spatial slice is locally Euclidean, as bcrowell suggested in my other thread. I'm sure others can suggest better definitions, but just a coordinate t being constant means nothing.
Roughly speaking, by "time" I mean "a coordinate that's timelike". :wink: I guess a somewhat more precise way of saying that would be that the vector field generated by the partial derivative with respect to the coordinate must be a timelike vector field. I think (but am not positive) that's enough to guarantee that any slice of the spacetime with a constant value of such a coordinate must be a spacelike hypersurface.
 
  • #6
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I'm not sure if this will help any, but here goes:

"Flat" is a little ambiguous, but usually people mean that the Riemann tensor vanishes everywhere.

Occasionally one will see the term used loosely in some other sense.

I'm not sure if saying that "flat" means that the opposite sides of a parallelogram are always equal in length is equivalent to the above definition or not, though I suspect it's very close even if it turns out to be slightly "off". Of course, you'll need some notion of "parallel" to make this work.

If you take a spatial slice (setting a timelike coordinate constant is one way of slicing space-time, it doesn't really matter how you do it as long as you manage it somehow), a metric is induced on the spatial slice by your space-time metric. Basically you can use the space-time metric to compute the lengths of curves that lie entirely in the spatial slice, and this is sufficient to determine the distances between points on that space-like slice and hence the metric on the space-like surface.

It's not clear to me how to tell if a given metric allows a flat space-like slice to be taken, however.
 
  • #7
PAllen
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I'm not sure if this will help any, but here goes:

"Flat" is a little ambiguous, but usually people mean that the Riemann tensor vanishes everywhere.

Occasionally one will see the term used loosely in some other sense.

I'm not sure if saying that "flat" means that the opposite sides of a parallelogram are always equal in length is equivalent to the above definition or not, though I suspect it's very close even if it turns out to be slightly "off". Of course, you'll need some notion of "parallel" to make this work.

If you take a spatial slice (setting a timelike coordinate constant is one way of slicing space-time, it doesn't really matter how you do it as long as you manage it somehow), a metric is induced on the spatial slice by your space-time metric. Basically you can use the space-time metric to compute the lengths of curves that lie entirely in the spatial slice, and this is sufficient to determine the distances between points on that space-like slice and hence the metric on the space-like surface.

It's not clear to me how to tell if a given metric allows a flat space-like slice to be taken, however.
Agreeing with the ambiguity referenced here, it is easy to have a purely spacelike hypersurface on which the normal notions of spatial flatness fail (e.g. the triangle inequality is not true, even locally, everywhere; for example, I believe this is true for 'natural' spacelike slices of an accelerated frame in SR; this is because a little bit of mixed signature metric components are mixed into the induced 3-metric). So there must be some agreed on definition of a flat spatial slice. Pervect's parallellogram definition is a good one. For my purposes, I was more interested in triangle inequality (which amounts, I think, to positive definite spatial metric). (The triangle inequality guarantees that a spatial geodesic is a true local minimum).
 
  • #8
PeterDonis
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If you take a spatial slice (setting a timelike coordinate constant is one way of slicing space-time, it doesn't really matter how you do it as long as you manage it somehow), a metric is induced on the spatial slice by your space-time metric. Basically you can use the space-time metric to compute the lengths of curves that lie entirely in the spatial slice, and this is sufficient to determine the distances between points on that space-like slice and hence the metric on the space-like surface.
This is the procedure for finding a "flat spatial slice" that I was thinking of. For example, if we take the Painleve metric and set dt = 0 (where t is the Painleve t-coordinate), the induced 3-metric on the resulting t = constant hypersurface is the metric for Euclidean 3-space in spherical coordinates. I think we're safe in saying that Euclidean 3-space is flat. :wink:

"Flat" is a little ambiguous, but usually people mean that the Riemann tensor vanishes everywhere.
Using the procedure above, this would mean, I take it, "the Riemann tensor of the induced 3-metric on the spacelike slice vanishes everywhere." We also have to add the proviso that PAllen mentioned, that the induced 3-metric can't be pseudo-Riemannian, it has to be Riemannian (i.e., signature +++, which I think captures the idea that it needs to be a "purely spatial" metric--I use the term "Riemannian" instead of "positive definite" because of course we can still have zero distances if we calculate the metric from a point to itself). I think that will work as the definition of a "flat spatial slice" for this discussion.

It's not clear to me how to tell if a given metric allows a flat space-like slice to be taken, however.
Darn it!
 
  • #9
bcrowell
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The reason this came up in the other thread was my use of the term "spatially flat" to describe the FRW spacetime with k = 0. Passionflower pointed out that, since the scale factor in this spacetime is time-dependent, even though each hypersurface of constant FRW coordinate time is flat, the "size" of a given region of space changes from one spatial slice to another, so, for example, a geometric figure "drawn" using geodesics in one spatial slice would have "expanded" in a later slice. This raised the question of what would be required of a spacetime to ensure that this didn't happen.
This sounds like what the volume expansion [itex]\theta[/itex] and expansion tensor [itex]\theta_{ab}[/itex] are designed for. I'd suggest using those tools rather than reinventing the wheel. The relevant definitions are given in Wald and in Hawking and Ellis.

I'm not really seeing how this relates to the question of flat spatial slices.
 
  • #10
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Pervect said:
It's not clear to me how to tell if a given metric allows a flat space-like slice to be taken, however.
I think, one has to find a coordinate transformation so that the metric written in the new coordinates has a spatial part which is E3.

It's possible to do that in the Schwarzschild and FLRW spacetimes.

This coordinate transformation of the FRW ( in Cartesian coords)
[tex]
\begin{align*}
dt'&=dt\\
dx'&=dx-\frac{2\,x}{3\,t}dt\\
dy'&=dy-\frac{2\,y}{3\,t}dt\\
dz'&=dx-\frac{2\,z}{3\,t}dt\\
\end{align*}
[/itex]

transforms the FLRW metric into -

[tex]
\left[ \begin{array}{cccc}
\frac{4\,{z}^{2}+4\,{y}^{2}+4\,{x}^{2}-9\,{t}^{2}}{9\,{t}^{2}} & -\frac{2\,x}{3\,t} & -\frac{2\,y}{3\,t} & -\frac{2\,z}{3\,t} \\\
-\frac{2\,x}{3\,t} & 1 & 0 & 0 \\\
-\frac{2\,y}{3\,t} & 0 & 1 & 0 \\\
-\frac{2\,z}{3\,t} & 0 & 0 & 1}\end{array} \right]
[/itex]

which has the required E3 spatial slices and clearly the frame-field defined by the differentials represents a comoving observer of some sort.
 
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  • #11
PeterDonis
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This sounds like what the volume expansion [itex]\theta[/itex] and expansion tensor [itex]\theta_{ab}[/itex] are designed for.
<sound of me hitting myself over the head and saying "D'oh!">

Yes, you're right, I need to refresh my memory on the definitions of those and the required conditions for them to vanish.

I'm not really seeing how this relates to the question of flat spatial slices.
Not flat spatial slices per se, but the question of whether the term "spatially flat" is appropriate when the scale factor is not time-independent, or whether that term has connotations of a constant (in time) scale factor (which would mean a vanishing expansion) as well as no spatial curvature on each individual spatial slice.
 
  • #12
PeterDonis
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I think, one has to find a coordinate transformation so that the metric written in the new coordinates has a spatial part which is E3.

It's possible to do that in the Schwarzschild and FLRW spacetimes.
I was looking for some more general set of conditions that determine when this is possible for a spacetime, something like the ones I proposed in the OP (for example, is spherical symmetry necessary? is it sufficient?).
 
  • #13
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I was looking for some more general set of conditions that determine when this is possible for a spacetime, something like the ones I proposed in the OP (for example, is spherical symmetry necessary? is it sufficient?).
I think the actual symmetry is irrelevant, but obviuosly the more there is the easier to analyse. I was editing my post while you were posting, sorry if this causes confusion.

[edit]

It seems clear that a transformation of the type

[tex]
\begin{align*}
dt'&=dt\\
dx'&=dx-f_1(x,y,z,t)dt\\
dy'&=dy-f_2(x,y,z,t)dt\\
dz'&=dx-f_3(x,y,z,t)dt\\
\end{align*}
[/tex]

where [itex]f_i(x,y,z,t)[/itex] are chosen to reflect the symmetry of the metric. The tensor product of that coframe basis will always give a spatial part diag(1,1,1).

I doubt if this is general enough to cover all cases, though.
 
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  • #14
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PeterDonis said:
(3) What is required for a spacetime to admit a metric with flat spatial slices
At the risk of becoming a bore. To find a transformation that gives a flat spatial part of the metric in a more general case, where the metric is diag( -g00, g11, g22, g33), a transformation of this form is required, ( I'm choosing Cartesian coords and abusing notation)

[tex]

\begin{align*}
dt'&=f_0(x,y,z,t)dt\\
dx'&=dx-f_1(x,y,z,t)dt\\
dy'&=dy-f_2(x,y,z,t)dt\\
dz'&=dx-f_3(x,y,z,t)dt\\
\end{align*}

[/tex]

with the additional constraint that [itex]g_{00}=-f_0^2+f_1^2+f_2^2+f_3^2[/itex] which recovers the original g00.

Given a (diagonal) metric this becomes an algebraic problem.
 
  • #15
PAllen
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At the risk of becoming a bore. To find a transformation that gives a flat spatial part of the metric in a more general case, where the metric is diag( -g00, g11, g22, g33), a transformation of this form is required, ( I'm choosing Cartesian coords and abusing notation)

[tex]

\begin{align*}
dt'&=f_0(x,y,z,t)dt\\
dx'&=dx-f_1(x,y,z,t)dt\\
dy'&=dy-f_2(x,y,z,t)dt\\
dz'&=dx-f_3(x,y,z,t)dt\\
\end{align*}

[/tex]

with the additional constraint that [itex]g_{00}=-f_0^2+f_1^2+f_2^2+f_3^2[/itex] which recovers the original g00.

Given a (diagonal) metric this becomes an algebraic problem.
I was aware that you could diagonalize any metric at a point, but not globally. Then your argument shows you can construct a local euclidean slice given the local diagonalization, which is obvious. It seems to provide a recipe for a global flat spacial slice only if global diagonalization is possible. Isnt that a fairly specialized condition for a metric to have?
 
  • #16
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I was aware that you could diagonalize any metric at a point, but not globally. Then your argument shows you can construct a local euclidean slice given the local diagonalization, which is obvious. It seems to provide a recipe for a global flat spacial slice only if global diagonalization is possible. Isn't that a fairly specialized condition for a metric to have?
You could be right.

We can always find a Minkowski frame locally by using the coordinate coframe but that isn't the same thing as finding a transformation that transforms the spatial part of the metric to diag(1,1,1) or the equivalent. I don't really know what the latter means. I think the best way to look for such a transformation is to use the frame-field method, and that's what I've been on about ( rather feebly).

The transformation dxu -> sqrt(guu)dxu means we can rewrite the metric as diag(-1,1,1,1) immediately, but this is interpreted as a local frame.

This is probably not helping to establish which spacetimes can admit E3 spatial slices, so I'm going to bed.
 
  • #17
JDoolin
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Roughly speaking, by "time" I mean "a coordinate that's timelike". :wink: I guess a somewhat more precise way of saying that would be that the vector field generated by the partial derivative with respect to the coordinate must be a timelike vector field. I think (but am not positive) that's enough to guarantee that any slice of the spacetime with a constant value of such a coordinate must be a spacelike hypersurface.
I was reviewing a book I read some time ago "Relativity Visualized" and came to wonder what this means: "a coordinate that's timelike"

In this book, Epstein makes all of his space-time diagrams as "proper-time vs. space" Thus, the equation we're all familiar with

[tex]d\tau^2=dt^2-dx^2[/tex]​

is turned on its head:

[tex]dt^2=d\tau^2+dx^2[/tex]​

Within this context, the (horizontal) x-component represents the position, the (vertical) \tau component represents the age of the particle, and the lengths of arcs represent the actual amount of time passed for a particle from the frame of an inertial observer.

More importantly, the effects of a "stretching" of space-proper-time becomes somewhat visualizable, using one's knowledge of euclidian geometry.

I just thought I would bring this up here, since you said "a coordinate which is timelike," these two "time-like" coordinates, \tau and t, perform differently.

Another question I would ask, though is whether both of them are actually coordinates. The value of the proper-time, \tau actually represents a property of an object passing through the space, while the value of t represents an actual coordinate value in an inertial reference frame. (In tensor lingo, would this be saying that t is contravariant, while \tau is covariant?)

When Epstein works with geometries in [itex](x,\tau)[/itex] coordinates, he seems to be mixing coordinates of different types; a location in space, and a property of a particle at that location. Under this context, he stretches the space-"proper-time" and is able to make quite a few interesting predictions. However, at no point does he actually stretch the coordinate time, t. There's a rule in the book that the length of the paths is equal to the actual time passed, and it seems like he sticks with it throughout.

Is it possible that all of this stretching of space-time in general relativity is actually stretching of (coordinate space, proper time), while it leaves unstretched (coordinate space, coordinate time?)

If one is weightless inside a windowless room, he cannot tell whether he is in flat space, where his clock is reading coordinate time, or if he is in a gravitational field, where his clock is reading proper-time. But when one measures the motion of the planets with respect to the positions of distant galaxies, he should be able to determine the speed of coordinate time; even if his local clocks are measuring proper-time; and he can measure the speed of his own clocks relative to this universal coordinate time.
 
  • #18
PAllen
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Another question I would ask, though is whether both of them are actually coordinates. The value of the proper-time, \tau actually represents a property of an object passing through the space, while the value of t represents an actual coordinate value in an inertial reference frame. (In tensor lingo, would this be saying that t is contravariant, while \tau is covariant?)

----

If one is weightless inside a windowless room, he cannot tell whether he is in flat space, where his clock is reading coordinate time, or if he is in a gravitational field, where his clock is reading proper-time. But when one measures the motion of the planets with respect to the positions of distant galaxies, he should be able to determine the speed of coordinate time; even if his local clocks are measuring proper-time; and he can measure the speed of his own clocks relative to this universal coordinate time.
I would have thought one of the few things universally agreed is that a clock measures proper time in its own world line, always. Note, proper time is not covariant, it is invariant: all observers and coordinates produce the same value for it beyond purely unit conventions (that is the same value given some local defintion of second).

Also, if on the scale of the room, there area deviations from Euclidean spactial geometry, these are directly measurable using, e.g. pieces of string pulled tight between different points and measuring angles and lengths.
 
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  • #19
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Also, if on the scale of the room, there area deviations from Euclidean spactial geometry, these are directly measurable using, e.g. pieces of string pulled tight between different points and measuring angles and lengths.
Could you demonstrate this by showing us the calculations for the deviations from Euclidean spatial geometry?

If you have not done this I urge you to do this as you then will realize that the effect is beyond anything we can directly measure with state of the art measuring apparatus let alone with pieces of strings.
 
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  • #20
PAllen
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Could you demonstrate this by showing us the calculations for the deviations from Euclidean spatial geometry?

If you have not done this I urge you to do this as you then will realize that the effect is beyond anything we can directly measure with state of the art measuring apparatus let alone with pieces of strings.
Read my hypothesis:

"Also, if on the scale of the room, there area deviations from Euclidean spacial geometry,"

Either the deviations are or are not signficant at lab scales. I am well aware that we are never likely to have a lab at a location where this is true.
 
  • #21
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Read my hypothesis:

"Also, if on the scale of the room, there area deviations from Euclidean spactial geometry,"

Either the deviations are or are not signficant at lab scales. I am well aware that we are never likely to have a lab at a location where this is true.
So can you show me your calculations?
 
  • #22
PAllen
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So can you show me your calculations?
What's to calculate? Do the internal angle of a triangle add up to 180 detrees? Are there deviations from the pythagorean theorem at 1000 meter scales compared to 10 meter scales?
 
  • #23
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What's to calculate?
Let's say the Euclidean deviation for an elevator 3 meters high on the surface of a non rotating planet with surface area X and Mass Y? Are you going to use Schwarzschild coordinates or do you think these will be meaningless, if so, how will you approach the calculation otherwise?
 
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  • #24
PAllen
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Let's say the Euclidean deviation for a 3 meter heigh elevator on the surface of a non rotating planet with surface area X and Mass Y? Are you going to use Schwarzschild coordinates or do you think these will be meaningless, if so, how will you approach the calculation otherwise?
Two separate questions here: real world measurments and modeled measurements. In the real world of a lab scale, I wouldn't worry about coordinates or metric at all. I would, for example, set up a 1 km triangle of laser pointers and measure angles with protractor.

For modeled measurements, attempt to compute only invariants that modeled some physical situation. Then, I wouldn't care about the coordinates. For the example above, the trickiest question, which I don't completely know how to solve is computing an observed spacial angle between a pair of null geodesics.
 
  • #25
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For modeled measurements, attempt to compute only invariants that modeled some physical situation. Then, I wouldn't care about the coordinates. For the example above, the trickiest question, which I don't completely know how to solve is computing an observed spacial angle between a pair of null geodesics.
You would not care about coordinates? Ok. So how would you approach it without using coordinates? Say we got three points arranged in a triangle, how are you going to model that?

You disagree we need coordinates to solve just about any problem in GR?

I am not totally sure but it seems you have the wrong ideas about using coordinates. Coordinates without a metric are not very useful but together with the metric they are very useful. While it is true that a particular coordinate value does not necessarily translate into a directly measurable quantity, we can use these values, or their differences, together with the metric, to calculate physically meaningful results.
 
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