Sign of Expansion Scalar in Expanding FLRW Universe

[Onyx]

TL;DR

Why is the trace of the extrinsic curvature tensor negative?

Considering the FLWR metric in cartesian coordinates:

$ds^2=-dt^2+a^2(t)(dx^2+dy^2+dz^2)$

With $a(t)=t$, the trace of the extrinsic curvature tensor is $-3t$. But why is it negative if it's describing an expanding universe, not a contracting one?

 

[PeterDonis]

the trace of the extrinsic curvature tensor is $-3t$.

Are you interested in the trace of the extrinsic curvature tensor or the expansion scalar?

Note that these are properties of two different things. The expansion scalar is a property of a congruence of worldlines, in this case the congruence of comoving worldlines. This is positive in an expanding FRW universe.

The trace of the extrinsic curvature tensor is a property of spacelike surfaces of constant comoving coordinate time. It is negative because $a(t) = t$ means you are looking at the Milne universe, whose surfaces of constant comoving coordinate time are negatively curved hyperboloids. (Strictly speaking, the negative curvature of such hyperboloids is intrinsic curvature, not extrinsic curvature, but since the hyperboloids in this particular case are embedded in flat Minkowski spacetime--the Milne universe is simply flat Minkowski spacetime in unusual coordinates--the extrinsic and intrinsic curvatures are the same.) Edit: corrected in post #11.

 

[Onyx]

Are you interested in the trace of the extrinsic curvature tensor or the expansion scalar?

Note that these are properties of two different things. The expansion scalar is a property of a congruence of worldlines, in this case the congruence of comoving worldlines. This is positive in an expanding FRW universe.

The trace of the extrinsic curvature tensor is a property of spacelike surfaces of constant comoving coordinate time. It is negative because $a(t) = t$ means you are looking at the Milne universe, whose surfaces of constant comoving coordinate time are negatively curved hyperboloids. (Strictly speaking, the negative curvature of such hyperboloids is intrinsic curvature, not extrinsic curvature, but since the hyperboloids in this particular case are embedded in flat Minkowski spacetime--the Milne universe is simply flat Minkowski spacetime in unusual coordinates--the extrinsic and intrinsic curvatures are the same.)

This is helpful information because I actually thought the trace of the extrinsic curvature tensor and expansion scalar were the same. I read on Wikipedia that the expansion scalar measures the rate of change of a matter ball's volume wrt the time of a comoving observer. But that's not what the trace of the extrinsic curvature is?

 

[Onyx]

Are you interested in the trace of the extrinsic curvature tensor or the expansion scalar?

Note that these are properties of two different things. The expansion scalar is a property of a congruence of worldlines, in this case the congruence of comoving worldlines. This is positive in an expanding FRW universe.

The trace of the extrinsic curvature tensor is a property of spacelike surfaces of constant comoving coordinate time. It is negative because $a(t) = t$ means you are looking at the Milne universe, whose surfaces of constant comoving coordinate time are negatively curved hyperboloids. (Strictly speaking, the negative curvature of such hyperboloids is intrinsic curvature, not extrinsic curvature, but since the hyperboloids in this particular case are embedded in flat Minkowski spacetime--the Milne universe is simply flat Minkowski spacetime in unusual coordinates--the extrinsic and intrinsic curvatures are the same.)

This is helpful information because I actually thought the trace of the extrinsic curvature tensor and expansion scalar were the same. I read on Wikipedia that the expansion scalar measures the rate of change of a matter ball's volume wrt the time of a comoving observer. But that's not what the trace of the extrinsic curvature is?

Also is "comoving coordinate time" different from coordinate time $t$?

 

[PeterDonis]

I read on Wikipedia

You should not be trying to learn science from Wikipedia. Try a textbook on GR or a textbook on cosmology. Sean Carroll's online lecture notes on GR give a good, if brief, introduction to cosmology and the standard FRW spacetime models used by cosmologists.

 

[PeterDonis]

the expansion scalar measures the rate of change of a matter ball's volume wrt the time of a comoving observer.

More precisely, the expansion scalar associated with the congruence of worldlines of comoving observers measures the rate of change of the volume of a ball of matter that contains a fixed number of comoving observers, with respect to those observers' proper time.

But that's not what the trace of the extrinsic curvature is?

No. The trace of the extrinsic curvature is the trace of the extrinsic curvature tensor of a spacelike surface of constant comoving coordinate time. The specific extrinsic curvature tensor in question will depend on the specific FRW model being used. The case $a(t) = t$ corresponds to the Milne universe, which is an edge case FRW model in which the matter density is zero everywhere, i.e., it is flat Minkowski spacetime in unusual coordinates.

is "comoving coordinate time" different from coordinate time ?

"Comoving coordinate time" is coordinate time in comoving coordinates, i.e., the standard FRW coordinates used in cosmology.

 

[Onyx]

More precisely, the expansion scalar associated with the congruence of worldlines of comoving observers measures the rate of change of the volume of a ball of matter that contains a fixed number of comoving observers, with respect to those observers' proper time.No. The trace of the extrinsic curvature is the trace of the extrinsic curvature tensor of a spacelike surface of constant comoving coordinate time. The specific extrinsic curvature tensor in question will depend on the specific FRW model being used. The case $a(t) = t$ corresponds to the Milne universe, which is an edge case FRW model in which the matter density is zero everywhere, i.e., it is flat Minkowski spacetime in unusual coordinates."Comoving coordinate time" is coordinate time in comoving coordinates, i.e., the standard FRW coordinates used in cosmology.

So in this particular metric, coordinate time and comoving coordinate time are the same?

 

[PeterDonis]

So in this particular metric, coordinate time and comoving coordinate time are the same?

In this particular coordinate chart on this particular metric, coordinate time is defined to be the same as proper time for comoving observers, and comoving observers have constant spatial coordinates; so these coordinates are often called "comoving coordinates", hence the term "comoving coordinate time".

 

[Onyx]

So in this particular metric, coordinate time and comoving coordinate time are the same?

Because I thought there would have to be a transformation like R=a(t)r, or something like that, to get a comoving time.

 

[PeterDonis]

I thought there would have to be a transformation like R=a(t)r, or something like that, to get a comoving time.

Why would you think that?

 

[PeterDonis]

I actually thought the trace of the extrinsic curvature tensor and expansion scalar were the same.

They are different geometric objects, as I have said. However, it can be shown that, for the particular case of comoving observers in an FRW spacetime and the spacelike surfaces of constant comoving coordinate time, they are numerically equal. See, for example, Section 9.3 of Wald.

This means, of course, that your original statement that, for the case $a(t) = t$, the trace of the extrinsic curvature is $- 3t$ is incorrect. The problem is that the metric you have written down in the OP is not consistent with $a(t) = t$. The metric you wrote down has flat spacelike surfaces of constant time; but the Milne universe, with $a(t) = t$, does not; they are, as I said before, hyperboloids with negative intrinsic curvature.

That also means that I need to correct myself, since my earlier statement that the extrinsic and intrinsic curvatures of the Milne hyperboloids were the same was incorrect. The extrinsic curvature of the hyperboloids is positive.

 

[Onyx]

Why would you think that?

Because the comoving observer would be physically moving (presumably in the $r$ coordinate) to match the expansion. However I just remembered that I'm used to thinking about $c=1$, so I guess I forgot that that would just be $cd\tau$ instead of $d\tau$. So then I guess it would be the same.

 

[Onyx]

Because the comoving observer would be physically moving (presumably in the $r$ coordinate) to match the expansion. However I just remembered that I'm used to thinking about $c=1$, so I guess I forgot that that would just be $cd\tau$ instead of $d\tau$. So then I guess it would be the same.

Okay, this answers my questions, thank you. Just as a quick aside, if we were instead considering something like the Kerr metric, where $d\tau$ does not equal $dt$ for an observer comoving to match the rotation, would the extrinsic curvature tensor then depend on something different from $t$?

 

[Ibix]

Because the comoving observer would be physically moving (presumably in the $r$ coordinate) to match the expansion.

Not in the usual co-moving coordinates. That's what co-moving means - observers at constant spatial coordinates move along with the expansion.

Okay, this answers my questions, thank you. Just as a quick aside, if we were instead considering something like the Kerr metric, where $d\tau$ does not equal $dt$ for an observer comoving to match the rotation, would the extrinsic curvature tensor then depend on something different from $t$?

Depends which spacelike slices you pick. In the case of Kerr spacetime, which is stationary, you would reasonably pick spacelike slices that do not change over time; thus their curvature cannot depend on the time coordinate.

 

[Onyx]

Not in the usual co-moving coordinates. That's what co-moving means - observers at constant spatial coordinates move along with the expansion.

Depends which spacelike slices you pick. In the case of Kerr spacetime, which is stationary, you would reasonably pick spacelike slices that do not change over time; thus their curvature cannot depend on the time coordinate.

I can't tell whether or not that is already true about the spacelike slices foliated by coordinate time. I just assumed that since the Kerr metric has a cross term, like the alcubierre metric, then the surfaces would change, but I'm not sure.

 

[Ibix]

I can't tell whether or not that is already true about the spacelike slices foliated by coordinate time.

Do they depend on $t$? If not, they can't be changing with time.

The Alcubierre metric depends on $t$ through the quantity $x_s(t)$.

 

[PeterDonis]

Because the comoving observer would be physically moving (presumably in the $r$ coordinate) to match the expansion.

No, they wouldn't. The whole point of comoving coordinates, as I have already said (see post #8), is that comoving observers have constant spatial coordinates.

I just remembered that I'm used to thinking about $c=1$, so I guess I forgot that that would just be $cd\tau$ instead of $d\tau$. So then I guess it would be the same.

The choice of units in which $c = 1$ is irrelevant to the above point.

 

[PeterDonis]

if we were instead considering something like the Kerr metric, where $d\tau$ does not equal $dt$ for an observer comoving to match the rotation, would the extrinsic curvature tensor then depend on something different from $t$?

The extrinsic curvature tensor of what set of spacelike slices? If you mean spacelike slices of constant coordinate time in the usual Boyer-Lindquist coordinates for Kerr spacetime, the metric is not a function of coordinate time in these coordinates so the extrinsic curvature of spacelike slices of constant coordinate time will not change with coordinate time.

It seems to me that you are trying to understand a relatively advanced topic without the necessary underlying basic knowledge. You did mark this thread as "B" level, which I think correctly indicates your current level of knowledge; but the questions you are asking are not "B" level questions; they are "I", or in some cases possibly even "A" level questions. You can't expect to gain a good understanding of these things if you do not have an "I" or "A" level background knowledge in the subject.

 

[Onyx]

The extrinsic curvature tensor of what set of spacelike slices? If you mean spacelike slices of constant coordinate time in the usual Boyer-Lindquist coordinates for Kerr spacetime, the metric is not a function of coordinate time in these coordinates so the extrinsic curvature of spacelike slices of constant coordinate time will not change with coordinate time.

It seems to me that you are trying to understand a relatively advanced topic without the necessary underlying basic knowledge. You did mark this thread as "B" level, which I think correctly indicates your current level of knowledge; but the questions you are asking are not "B" level questions; they are "I", or in some cases possibly even "A" level questions. You can't expect to gain a good understanding of these things if you do not have an "I" or "A" level background knowledge in the subject.

Well, I figured that if coordinate time is transformed to something which eliminates the off-diagonal term in the metric, then that would be the proper time coordinate of someone moving to match the rotation. That is the foliation I was thinking of.

 

[PeterDonis]

if coordinate time is transformed to something which eliminates the off-diagonal term in the metric

You can't eliminate the off-diagonal term in the Kerr metric.

That is the foliation I was thinking of.

There is no such foliation in Kerr spacetime.

 

[PeterDonis]

if coordinate time is transformed to something which eliminates the off-diagonal term in the metric, then that would be the proper time coordinate of someone moving to match the rotation

Even if we consider this in general, not just for Kerr spacetime (where, as I have noted, there is no way to do what you are suggesting), your implied assumption here, that a diagonal metric means coordinate time is the same as proper time for observers with constant spatial coordinates, is false. For that to be true, you must also have $g_{tt} = - 1$ everywhere, as it is in FRW spacetime in standard FRW coordinates.

Again, you appear to me to be trying to understand advanced topics without having the required basic background knowledge. That is not a good idea. You need a good foundation before you can build a building.

 

[PeterDonis]

the extrinsic curvature tensor

Is there a particular reason why you are interested in the trace of the extrinsic curvature tensor? This is not normally a good way to understand a spacetime geometry.

 

[Onyx]

Well, I figured that if coordinate time is transformed to something which eliminates the off-diagonal term in the metric, then that would be the proper time coordinate of someone moving to match the rotation.

your implied assumption here, that a diagonal metric means coordinate time is the same as proper time for observers with constant spatial coordinates, is false.

I was actually assuming that they would be different.

 

[Onyx]

Is there a particular reason why you are interested in the trace of the extrinsic curvature tensor? This is not normally a good way to understand a spacetime geometry.

The original reason for my interest has to do with the topic of one of my previous threads: the Alcubierre metric. I think in the original paper it calls the trace the "expansion of the volume elements associated with the eulerian observers." The last forum cleared up for me that there is no "proper volume" here, since the metric is non-stationary, which makes sense. I think the eulerian observers it refers to are moving with a speed of $v_s$ to match the ship, but I'm not sure if they are supposed to be inside or outside the bubble. Basically, I'm confused as to "who's" volume elements it is talking about.

 

[PeterDonis]

I was actually assuming that they would be different.

Then I don't understand what you were trying to describe. But since it is impossible to have a diagonal metric in Kerr spacetime anyway, whatever you were trying to describe would appear to be impossible. (Note, btw, that in Kerr spacetime, "observers moving with the rotation" have different angular velocities, relative to an observer at rest at infinity, at different spatial points. These observers are also called zero angular momentum observers, or ZAMOs, in the literature.)

 

[PeterDonis]

The original reason for my interest has to do with the topic of one of my previous threads: the Alcubierre metric.

This one?

https://www.physicsforums.com/threa...-internal-volume-to-distant-observer.1045283/

 

[Onyx]

This one?

https://www.physicsforums.com/threa...-internal-volume-to-distant-observer.1045283/

Yes, I think that's the one.

 

[PeterDonis]

I think that's the one.

The only reference to extrinsic curvature that I find in that thread is this from your final post:

someone else told me that the trace of the extrinsic curvature tensor is a coordinate-invariant description of the curvature

This is, first, a vague reference ("someone told me...", as opposed to "here is a paper which says..."), and second, is incorrect, or at least unacceptably vague (what "curvature" is it supposed to be a description of? Not the curvature of spacetime).

The usual way the extrinsic curvature tensor arises in GR, as you will see from the chapter in Wald that I referenced earlier, is in the initial value formulation of GR. This is, as I said before, a fairly advanced topic (which is why it is Chapter 9 in Wald, after 8 chapters covering more fundamental items that give a foundation for understanding what is in Chapter 9). But in any case, it does not seem to be a good thing to be looking at to try and ground any of the things you have been asking about in the actual math of GR.

 

[Ibix]

The usual way the extrinsic curvature tensor arises in GR, as you will see from the chapter in Wald that I referenced earlier, is in the initial value formulation of GR.

As I recall, Alcubierre's original paper is couched in terms of the ADM formalism, with a foliation, lapse, and shift. So in that context they are probably talking about the extrinsic curvature of the leaves of Alcubierre's foliation. That foliation is, of course, a calculational convenience that may have little (directly) to do with any actual observations.

 

[PeterDonis]

in that context they are probably talking about the extrinsic curvature of the leaves of Alcubierre's foliation

Yes.

That foliation is, of course, a calculational convenience that may have little (directly) to do with any actual observations.

Indeed.