Robertson-Walker metric and time expansion

[grelf]

The Robertson-Walker metric applies a time-dependent scale factor to model the expansion of the universe. The scale factor is only applied to the spatial coordinates (in the frame of the "comoving observer"). That is not covariant and it is hard to see how c could remain constant if the same factor is not applied to the time coordinate. So my question is: is there any evidence that time does not expand at the same rate as space?

 

[bcrowell]

Hi, grelf,

Welcome to Physics Forums!

The scale factor is only applied to the spatial coordinates (in the frame of the "comoving observer"). That is not covariant

Covariance is a local property of the laws of physics, not a global property of the solutions to those laws.

and it is hard to see how c could remain constant if the same factor is not applied to the time coordinate.

Cosmological expansion is not just a change of coordinates. If it were, then it would be undetectable. One of the basic principles of GR is that you can make any arbitrary, smooth change of coordinates, and it doesn't change the physical interpretation.

So my question is: is there any evidence that time does not expand at the same rate as space?

I don't think expansion of time is a notion that's well defined. If it means a nonlinear (but smooth) distortion of the time coordinate, then it's not observable. If you have in mind something observable, what observations do you have in mind?

-Ben

 

[Chalnoth]

The Robertson-Walker metric applies a time-dependent scale factor to model the expansion of the universe. The scale factor is only applied to the spatial coordinates (in the frame of the "comoving observer"). That is not covariant and it is hard to see how c could remain constant if the same factor is not applied to the time coordinate. So my question is: is there any evidence that time does not expand at the same rate as space?

To expand a little bit on bcrowell's point, if you also multiplied the time coordinate by the scale factor, then what you'd have is just a coordinate transformation of Minkowski space-time, so that there would be no expansion at all, just static, flat space-time.

 

[TrickyDicky]

The Robertson-Walker metric applies a time-dependent scale factor to model the expansion of the universe. The scale factor is only applied to the spatial coordinates (in the frame of the "comoving observer"). That is not covariant and it is hard to see how c could remain constant if the same factor is not applied to the time coordinate. So my question is: is there any evidence that time does not expand at the same rate as space?

To expand a little bit on bcrowell's point, if you also multiplied the time coordinate by the scale factor, then what you'd have is just a coordinate transformation of Minkowski space-time, so that there would be no expansion at all, just static, flat space-time.

Try this, transform the FRW metric with k=0 to its conformal form, in this particular case you can use conformal time as the time coordinate:you get a line element in terms of conformal time with a scale factor that also multiplies the time coordinate and it is actually a coordinate transformation of Minkowski spacetime with conformal time. (No expansion at all if conformal time is used instead of cosmological time)ds^2 = a(t)(dt^2 - dx^2 - dy^2 - dz^2)

Also, in the conformal form of the FRW line element coordinate c is isotropic and has the constant value c everywhere.

http://arxiv.org/PS_cache/arxiv/pdf/1103/1103.4743v1.pdf

 

[TrickyDicky]

See also:
http://www.tapir.caltech.edu/~chirata/ph217/lec02.pdf
eq. (23) and adjunct text.

 

[TrickyDicky]

Cosmological expansion is not just a change of coordinates. If it were, then it would be undetectable.

I show above that for the case of a spatially flat FRW universe actually it is just a change of coordinates. Just by switching back and forth between cosmic and conformal time coordinate you have spatial expansion or a static spacetime multiplied by a scala conformal factor (expanding space-time).
Expansion in itself is indeed directly undetectable. Certainly we detect redshift, but attributing it to expansion is purely model-dependent ( the prevailing model by the way).

One of the basic principles of GR is that you can make any arbitrary, smooth change of coordinates, and it doesn't change the physical interpretation.

bapowel, chalnoth? anyone disagrees with any of this?

 

[Chalnoth]

bapowel, chalnoth? anyone disagrees with any of this?

That's exactly correct. But the change of coordinates to a flat spacetime is only possible with the FRW metric if you have an empty universe (because it was a flat spacetime to begin with).

I believe that the problem with your solution is that the time (t) that is the argument of the scale factor function is not the coordinate being used any longer, so that when you run through Einstein's equations, they come out non-zero.

In any event, it is a general result within General Relativity that no coordinate transformation can change the curvature scalar R, which is non-zero for any curved space-time. Since R is non-zero in the FRW universe (unless said universe is empty), this transformation cannot give a flat space-time.

 

[bapowell]

bapowel, chalnoth? anyone disagrees with any of this?

Sounds good to me as well, with the caveat that this refers to a global change of coordinates (diffeomorphisms applied to the whole manifold). Locally, one can always transform away the gravitational field (the equivalence principle) leading to different (gravitational) physics.

 

[TrickyDicky]

That's exactly correct. But the change of coordinates to a flat spacetime is only possible with the FRW metric if you have an empty universe (because it was a flat spacetime to begin with).
I believe that the problem with your solution is that the time (t) that is the argument of the scale factor function is not the coordinate being used any longer, so that when you run through Einstein's equations, they come out non zero.

Sure, any flat spacetime corresponds to empty spacetime. But what I was highlighting in the conformal form of FRW metric with k=0 is not that, this is not a flat spacetime, but a Minkowski spacetme multiplied by a conformal scale factor -function of conformal time that may include a density parameter so it has non zero curvature scalar.
I was rather pointing out that this conformal form is compatible with a static spacetime or a expanding spacetime with space and time components exapnding by the same factor, which amounts to the same thing.

In any event, it is a general result within General Relativity that no coordinate transformation can change the curvature scalar R, which is non-zero for any curved space-time. Since R is non-zero in the FRW universe (unless said universe is empty), this transformation cannot give a flat space-time.

As I said the above mentioned transformation doesn't give a flat spacetime, but a static spacetime with curvature dependent on the density parameter encoded in the scale factor.

 

[TrickyDicky]

Sounds good to me as well, with the caveat that this refers to a global change of coordinates (diffeomorphisms applied to the whole manifold).

I'm not sure in what sense this is a caveat. It is a perfectly valid change of coordinates (conformal time is widely used in cosmology). The manifold is the same.
General covariance (diffeomorphism invariance) of GR ensures the form of all physical laws under arbitrary differentiable coordinate transformations.

 

[bapowell]

I'm not sure in what sense this is a caveat. It is a perfectly valid change of coordinates (conformal time is widely used in cosmology). The manifold is the same.
General covariance (diffeomorphism invariance) of GR ensures the form of all physical laws under arbitrary differentiable coordinate transformations.

It's a caveat in that the statement as made is only true for global coordinate changes. A local coordinate transformation very much changes the physics.

 

[TrickyDicky]

It's a caveat in that the statement as made is only true for global coordinate changes. A local coordinate transformation very much changes the physics.

I don't see how a change to conformally flat spacetime coordinates is a local change of coordinates. It is not. The physics here doesn't change at all.
See http://arxiv.org/pdf/1103.4743 section 3

 

[bapowell]

I don't see how a change to conformally flat spacetime coordinates is a local change of coordinates. It is not. The physics here doesn't change at all.
See http://arxiv.org/pdf/1103.4743 section 3

I'm not referring to any specific coordinate transformation, conformal or otherwise. There must be some confusion -- I was remarking only on the quote made by bcrowell --which was a general statement about coordinate transformations -- and nothing else.

 

[PAllen]

It's a caveat in that the statement as made is only true for global coordinate changes. A local coordinate transformation very much changes the physics.

That's interesting. Can you clarify what you mean by a local versus global coordinate transform? Suppose I have a smooth mapping that, outside of some 4-ball of spacetime is the identity. Are you saying this can change physics? How? (Every invariant quantity derivable from the metric will be unchanged). I must be mis-understanding what you mean.

 

[bapowell]

Perhaps I'm unclear on what constitute a "change in physics." By local transformation, I have in mind the coordinate transformation at a single point on the manifold that brings you to the tangent space. I was referring to this as a transformation that "changes physics" because it removes the gravitational field (sends a general metric g_{\mu \nu} to Minkowski). I'm merely referring to the equivalence principle here. This is of course different from a global diffeomorphism, under which spacetimes describing the same gravitational physics form equivalence classes. This is what everyone has in mind when they refer to the coordinate invariance of GR, and what bcrowell was referring to. I'm sure everyone's clear on this and so it was probably unnecessary for me to be so nit picky about this point.

 

[Chalnoth]

Of course, if you really want to be pedantic, it is coordinate covariance, not coordinate invariance. Invariance would mean that if you change the coordinates, all the output numbers you get remain the same. Covariance means that you change the numbers, and many quantities you measure will be different (e.g. you'll have different energy, momentum, etc.), but the relationships between them remain the same.

One way of saying the same thing is that I can start in some coordinate system, perform a transformation into another, do a bunch of calculations, and then transform back, and the answer I get is guaranteed to be identical to the answer I get if I just leave everything in the original coordinate system (provided I'm careful not to divide by zero anywhere).

 

[PAllen]

Of course, if you really want to be pedantic, it is coordinate covariance, not coordinate invariance. Invariance would mean that if you change the coordinates, all the output numbers you get remain the same. Covariance means that you change the numbers, and many quantities you measure will be different (e.g. you'll have different energy, momentum, etc.), but the relationships between them remain the same.

One way of saying the same thing is that I can start in some coordinate system, perform a transformation into another, do a bunch of calculations, and then transform back, and the answer I get is guaranteed to be identical to the answer I get if I just leave everything in the original coordinate system (provided I'm careful not to divide by zero anywhere).

Actually, I was intentionally saying invariants derived from the metric because any actual measurement I can think of involves contraction or integration resulting in a quantity that is invariant. Even for seemingly coordinate dependent quantities, like KE, the way I look at is that if you specify a measuring device following a particular world line measuring KE of some object, the result is invariant. Thus, if the tensors are covariant, the observables are invariant.

 

[Chalnoth]

Actually, I was intentionally saying invariants derived from the metric because any actual measurement I can think of involves contraction or integration resulting in a quantity that is invariant. Even for seemingly coordinate dependent quantities, like KE, the way I look at is that if you specify a measuring device following a particular world line measuring KE of some object, the result is invariant. Thus, if the tensors are covariant, the observables are invariant.

I don't think that's right. Just to take a really simple example: consider the round-trip light travel time measured at the source by, oh, bouncing a laser off of the Moon from an observatory on the Earth. I can change this quantity trivially by just changing my time coordinate around. But, of course, the number I get will always differ by a simple coordinate transformation.

 

[PAllen]

I don't think that's right. Just to take a really simple example: consider the round-trip light travel time measured at the source by, oh, bouncing a laser off of the Moon from an observatory on the Earth. I can change this quantity trivially by just changing my time coordinate around. But, of course, the number I get will always differ by a simple coordinate transformation.

I don't agree. Consider an actual clock. If you build in some notion of time measurement that differs from proper time, then modeling the physics of the clock as specified will allow computation of its behavior as an invariant. I think I'm making a fairly trivial statement - the result of an actual measurement is coordinate independent. Changing coordinates used for analysis cannot change what reading a device will produce or what the result of a particle interaction is. Thus I claim all actual observables in GR are invariant; any quantity that is not invariant is not actually an observable.

 

[TrickyDicky]

I don't agree. Consider an actual clock. If you build in some notion of time measurement that differs from proper time, then modeling the physics of the clock as specified will allow computation of its behavior as an invariant. I think I'm making a fairly trivial statement - the result of an actual measurement is coordinate independent. Changing coordinates used for analysis cannot change what reading a device will produce or what the result of a particle interaction is. Thus I claim all actual observables in GR are invariant; any quantity that is not invariant is not actually an observable.

I don't think this is right, the distinction between covariance and invariance Chalnoth made is well established.

 

[Chalnoth]

I don't agree. Consider an actual clock. If you build in some notion of time measurement that differs from proper time, then modeling the physics of the clock as specified will allow computation of its behavior as an invariant. I think I'm making a fairly trivial statement - the result of an actual measurement is coordinate independent. Changing coordinates used for analysis cannot change what reading a device will produce or what the result of a particle interaction is. Thus I claim all actual observables in GR are invariant; any quantity that is not invariant is not actually an observable.

Except that by defining a clock (including its motion), you've picked out a particular coordinate system. I could easily pick a different clock that it accelerated differently, and it will record a different time. But the times recorded on the two clocks will be simply-related by the coordinate transform between one clock and the other.

 

[PAllen]

I don't think this is right, the distinction between covariance and invariance Chalnoth made is well established.

I think what I am saying is also well established. Tensor quantities don't represent observables in GR. Only contractions or integrations of them do, resulting in invariant observables.

 

[PAllen]

Except that by defining a clock (including its motion), you've picked out a particular coordinate system. I could easily pick a different clock that it accelerated differently, and it will record a different time. But the times recorded on the two clocks will be simply-related by the coordinate transform between one clock and the other.

You can choose to look at it that way. I think it makes more sense to say the observable - the reading of a particular clock - can be described in a coordinate independent invariant way; and that all real observables should be defined this way. For example:

1) device measures KE of a particle; this is dot product of device 4-velocity at interaction event with particle 4 momentum, minus norm of 4-momentum. Coordinate independent and invariant. No matter what coordinates I use, I conclude the same thing about what the device measures.

2) clock measures round trip time of radar signal to the moon. Proper time along clock world line from emission to reception event. Different clock will be different. Both measurements coordinate independent, invariant. Of course they depend on the clock (particularly, its state of motion), but they don't depend on labels I attach to points on a manifold.

I stand by the point of view that any real observable must be an invariant in GR.

 

[TrickyDicky]

I think what I am saying is also well established. Tensor quantities don't represent observables in GR. Only contractions or integrations of them do, resulting in invariant observables.

Define "observable in GR" and give some examples of them. Is momentum a "GR observable"?

 

[Chalnoth]

You can choose to look at it that way. I think it makes more sense to say the observable - the reading of a particular clock - can be described in a coordinate independent invariant way; and that all real observables should be defined this way. For example:

1) device measures KE of a particle; this is dot product of device 4-velocity at interaction event with particle 4 momentum, minus norm of 4-momentum. Coordinate independent and invariant. No matter what coordinates I use, I conclude the same thing about what the device measures.

2) clock measures round trip time of radar signal to the moon. Proper time along clock world line from emission to reception event. Different clock will be different. Both measurements coordinate independent, invariant. Of course they depend on the clock (particularly, its state of motion), but they don't depend on labels I attach to points on a manifold.

I stand by the point of view that any real observable must be an invariant in GR.

Hmmm, I suppose. However, this particular statement of invariance is just a statement of consistency: if this weren't true, then the theory in question would predict different experimental observations just depending upon how you did the calculation. Any theory that did that would clearly be wrong.

I suppose the more accurate statement, then, would be that tensors in General Relativity are covariant, while contractions of tensors to scalar quantities are invariant (this follows, if I remember correctly, from the covariance of tensors).

 

[TrickyDicky]

I stand by the point of view that any real observable must be an invariant in GR.

This seems a very trivial statement that has nothing at all to do with GR.

 

[Chalnoth]

Define "observable in GR" and give some examples of them. Is momentum a "GR observable"?

No, it isn't. Imagine that I have a sensor attached to a clock, that ouputs on a digital display the time that a photon hits this sensor based upon this clock. The observable is the output on the digital display.

 

[PAllen]

Define "observable in GR" and give some examples of them. Is momentum a "GR observable"?

No, momentum per se is not an observable. A measured momentum is an observable. Defining the measuring process (most importantly, the world line of the measuring device) gives an invariant definition of the measured momentum. This all gets back the fundamental principle that results of measurements and interactions are invariant. What shows up on a dial, a picture, a particle detector are all invariants.

 

[PAllen]

I suppose the more accurate statement, then, would be that tensors in General Relativity are covariant, while contractions of tensors to scalar quantities are invariant (this follows, if I remember correctly, from the covariance of tensors).

Correct. And this is almost exactly the wording I have used a few times: observables are contractions or integrations of contractions. And yes, all such are invariants in the mathematical sense.

 

[TrickyDicky]

I don't actually see the connection between saying that a scalar measure obviously is an invariant (otherwise what 's the use of measuring anything?) with the previous discussion or with the OP.

 

[PAllen]

I don't actually see the connection between saying that a scalar measure obviously is an invariant (otherwise what 's the use of measuring anything?) with the previous discussion or with the OP.

I agree. Apologize for this. Chalnoth challenged a particular statement of mine, so we discussed until we have some mutual understanding - of a point with no connection to the thread.

 

[TrickyDicky]

To come back to the thread discussion, I was suggesting that spatial expansion (which is not an observable to use the terms recently discussed, whilst for instance spectral frequency is an observable), in the FRW metric with k=0 appears to be a coordinate property because it can be tranformed away thru a global change of coordinates and that is the kind of thing GR general covariance seems to be stating: that for something to be physical it must not depend on a particular coordinate system.
Might this be a hint that only the FRW metric with k different than 0 correspond to our universe?
Is the LCDM model only valid for k=0 or does it admit some small spatial curvature?

 

[Chalnoth]

To come back to the thread discussion, I was suggesting that spatial expansion (which is not an observable to use the terms recently discussed, whilst for instance spectral frequency is an observable), in the FRW metric with k=0 appears to be a coordinate property because it can be tranformed away thru a global change of coordinates and that is the kind of thing GR general covariance seems to be stating: that for something to be physical it must not depend on a particular coordinate system.
Might this be a hint that only the FRW metric with k different than 0 correspond to our universe?
Is the LCDM model only valid for k=0 or does it admit some small spatial curvature?

Well, sort of. The space-time curvature scalar R, which manifests itself as expansion in that particular coordinate system, remains the same no matter what coordinates you choose. So if you transform to some other coordinates, you'll still have the expansion after a fashion, it will just look like something else in the new coordinates.

 

[TrickyDicky]

Well, sort of. The space-time curvature scalar R, which manifests itself as expansion in that particular coordinate system, remains the same no matter what coordinates you choose. So if you transform to some other coordinates, you'll still have the expansion after a fashion, it will just look like something else in the new coordinates.

But the Ricci scalar doesn't have to manifest itself as expansion in ay coordinates, any non empty universe is going to have a curvature scalar whether they are static, contracting or expanding , I don't know why you relate expansion to the Ricci scalar. You can have expansion in a flat universe without curvature scalar (like Milne model).

 

[Chalnoth]

But the Ricci scalar doesn't have to manifest itself as expansion in ay coordinates, any non empty universe is going to have a curvature scalar whether they are static, contracting or expanding , I don't know why you relate expansion to the Ricci scalar. You can have expansion in a flat universe without curvature scalar (like Milne model).

Well, this goes back to what I was saying early in the thread, where I explicitly stated that except for an empty universe, you can't just transform the expansion away. The expansion is directly related to the space-time curvature (in fact, in a spatially-flat FRW universe, the expansion is the space-time curvature). You may be able to write down a metric that doesn't necessarily look like it has expansion in it, but because it is related to a real, invariant quantity, the expansion will still have to manifest itself somehow no matter what coordinates you choose.

 

[TrickyDicky]

The expansion is directly related to the space-time curvature (in fact, in a spatially-flat FRW universe, the expansion is the space-time curvature). You may be able to write down a metric that doesn't necessarily look like it has expansion in it, but because it is related to a real, invariant quantity, it will still have to manifest itself somehow no matter what coordinates you choose.

I still don't see the relation, forget empty universes, how is the scalar curvature or spacetime curvature related to expansion in the static Einstein universe?

 

[Chalnoth]

I still don't see the relation, forget empty universes, how is the scalar curvature or spacetime curvature related to expansion in the static Einstein universe?

Well, that universe isn't flat, so the relationship isn't direct. I *believe* that the curvature scalar is related to:

R \propto H_0^2 (1 - \Omega_k)

...though I suppose I'd have to do the calculations again to be sure (I'm a bit reluctant, because it is a bit of a lengthy calculation, and there are most likely some factors of the scale factor that I'm leaving off).

 

[TrickyDicky]

Well, that universe isn't flat, so the relationship isn't direct. I *believe* that the curvature scalar is related to:

R \propto H_0^2 (1 - \Omega_k)

...though I suppose I'd have to do the calculations again to be sure (I'm a bit reluctant, because it is a bit of a lengthy calculation, and there are most likely some factors of the scale factor that I'm leaving off).

You say that scalar curvature is intrinsically associated to expanding space, I'm just giving you examples where that is not the case, so it can't be a generalizable fact.
I don't know why that would be a feature only of spatially flat non empty universes. Scalar curvature is an invariant of different manifold modelling non-empty universes regardless of their particular configuration.

 

[Chalnoth]

You say that scalar curvature is intrinsically associated to expanding space, I'm just giving you examples where that is not the case, so it can't be a generalizable fact.
I don't know why that would be a feature only of spatially flat non empty universes. Scalar curvature is an invariant of different manifold modelling non-empty universes regardless of their particular configuration.

Well, assuming the equation I put together above is reasonably-correct (It may be a bit off, but I'm pretty sure it has to take a very similar form), then you can see how the expansion must be directly-related to the curvature in a flat universe, and how there is a relationship, though not as direct, in a non-flat one.

 

[TrickyDicky]

Well, assuming the equation I put together above is reasonably-correct (It may be a bit off, but I'm pretty sure it has to take a very similar form), then you can see how the expansion must be directly-related to the curvature in a flat universe, and how there is a relationship, though not as direct, in a non-flat one.

Aha, and when H_0=0? like in static manifolds with non zero R?

 

[Chalnoth]

Aha, and when H_0=0? like in static manifolds with non zero R?

Then we're not talking about an expanding universe, are we?

But in any event, if you're talking about a FRW universe, that situation is dynamically unstable and thus not really important.

 

[TrickyDicky]

Then we're not talking about an expanding universe, are we?

Exactly, my point is that you can't link the mere existence of the invariant scalar curvature to expansion as you did in previous posts to justify there is spatial expansion in the flat FRW metric even when it can be shown that in conformal coordinates there is no expansio wrt time. You can't assume it is a expanding universe beforehand regardless of the features of the metric or the whole discussion is meaningless.

 

[Chalnoth]

Exactly, my point is that you can't link the mere existence of the invariant scalar curvature to expansion

Um, I didn't. My point was that it goes the other way. If you have an expanding, non-empty universe, then you necessarily have non-zero curvature that is directly related to said expansion. That means that you can't just coordinate-transform the expansion away: you might hide it, but it will still appear in some form in your equations.

 

[Imax]

Can a Robertson-Walker Metric behave like a Kerr Metric near a black hole?

 

[Chalnoth]

Can a Robertson-Walker Metric behave like a Kerr Metric near a black hole?

No. The FRW metric assumes a homogeneous, isotropic universe, so it can't include black holes. To add matter that isn't perfectly smooth to the FRW metric, we make use of perturbation theory, where we start with FRW (where all matter is perfectly evenly-distributed), and add small deviations from that. This tends to work reasonably-well at large scales in the universe, but once you start to get down to the sizes of gravitationally-bound objects, such as galaxy clusters, this approximation breaks down. Black holes are way, way more dense than things like galaxy clusters, and there just isn't an easy way to place one in an FRW universe, except in very special cases (such as a universe with only a black hole and a cosmological constant).

 

[Imax]

No. The FRW metric assumes a homogeneous, isotropic universe, so it can't include black holes. To add matter that isn't perfectly smooth to the FRW metric, we make use of perturbation theory, where we start with FRW (where all matter is perfectly evenly-distributed), and add small deviations from that. This tends to work reasonably-well at large scales in the universe, but once you start to get down to the sizes of gravitationally-bound objects, such as galaxy clusters, this approximation breaks down. Black holes are way, way more dense than things like galaxy clusters, and there just isn't an easy way to place one in an FRW universe, except in very special cases (such as a universe with only a black hole and a cosmological constant).

That’s what I'm wondering about. A cosmological model based on an FRW metric can't see trees for its forest.

 

[Chalnoth]

That’s what I'm wondering about. A cosmological model based on an FRW metric can't see trees for its forest.

Well, one of the nice things is that the local details of the distribution of matter (e.g. black holes) don't have much of any impact at all on the larger-scale structure. In the regime where you can make use of perturbation theory to predict observations, observations match theory incredibly well.

At somewhat smaller scales, where perturbation theory starts to break down (this is still at a scale larger than galaxy clusters), we can make use of numerical simulations. These work quite well until you start getting into the regime where the physics of normal matter becomes important, which is around the size of galaxies. Once you get to the size of galaxies, our ability to predict what cosmological observations should look like from theory starts to break down (e.g. it's hard to predict the relative abundance of small and large galaxies). This isn't so much because we can't put black holes into the FRW metric, but instead because the physics of normal matter, and supermassive black holes in particular, are incredibly, obscenely difficult to get right.

 

[Imax]

So, a cosmological model based on an FRW metric + perturbation theory can fit with observations of our universe, but it has problems when going down to galactic scales, or around super massive black holes?

An FRW metric assumes a homogeneous and isotropic universe. Can time be different at different points in the metric, something like a twin paradox?

 

[Chalnoth]

So, a cosmological model based on an FRW metric + perturbation theory can fit with observations of our universe, but it has problems when going down to galactic scales, or around super massive black holes?

Well, the main problem here is that gravity is non-linear on smaller scales. Linearity means that if you change the system by some amount, the end result you get is always proportional to the amount you change the system. This is really important for doing calculations, because if a system is linear, you can vastly simplify those calculations. And it works really well for our universe on large scales. Linear theory predicts the CMB to tremendous accuracy, and even predicts much of the large-scale distribution of matter.

But when you start to get to smaller scales, the non-linearity of gravity becomes important. No longer is the output proportional to the input, but once you get enough matter concentrated into a small enough area, it will just keep collapsing in on itself. That kind of behavior simply cannot be modeled with a linear approximation, so what we do is make use of N-body simulations, where you imagine that matter is made up of a number of particles, and directly compute the relative gravitational attraction between all of them. This kind of calculation is relatively straightforward, but for large numbers of particles, it is extremely slow. But it works rather well at intermediate scales.

The problem with the N-body simulations, however, is that it considers matter to be made of particles, and thus this sort of simulation can't deal with gas physics. For that, we need to add another layer of complexity: hydrodynamic simulations. Here you're not only computing the gravitational interactions, but are also modeling the matter as an interacting gas, and that is where things get horribly complex, because things like supermassive black holes and supernova explosions have tremendous impacts upon the nearby gas, but we still don't know about the full behavior of these objects in the first place.

Don't get me wrong, we have made a lot of progress in understanding these amazing objects and events. It's just that figuring out what the physics we know today tells us about them is incredibly difficult to work out.

The real take-away here is just that even if we assume that the physics at work at smaller scales in the universe is completely known, we just haven't gotten to the point yet where we can say what the physics we know implies. However, at large scales, the calculations are much, much easier, so we can say with a great deal of certainty what the physics we know implies. So if you want to ask questions about whether or not experiment matches theory in terms of cosmology, your best bet is to look at the largest-scale observations you can.

An FRW metric assumes a homogeneous and isotropic universe. Can time be different at different points in the metric, something like a twin paradox?

Well, the time coordinate is arbitrary, so you can define it however you like. But there isn't much use in doing that, so typically we just do the easy thing and consider equal-time slices of the universe. Such equal-time slices, by definition, age at the same rate relative to one another. It makes the math easy, and it makes understanding the results easy.

 

[WannabeNewton]

So, a cosmological model based on an FRW metric + perturbation theory can fit with observations of our universe, but it has problems when going down to galactic scales, or around super massive black holes?

An FRW metric assumes a homogeneous and isotropic universe. Can time be different at different points in the metric, something like a twin paradox?

I'm not sure I understand you. The FRW metric is a time - dependent metric so it does evolve with time but if you take any hypersurface of time then all points on the hypersurface will be the same distance from t = 0 as defined by the metric.