Galaxy recession and Universe expansion

[cianfa72]

I see that I did incorrectly use "coordinate speed" later on in that post. I have edited it to correct those statements.

Ok thanks, now it makes sense.

 

[cianfa72]

More precisely, as an example, every such observer would measure the CMB to have the same temperature in all directions. That is an indication that the universe is seen to be isotropic. And if we pick out one particular spacelike hypersurface, at the event on each such worldline that is in that hypersurface, each observer would measure the CMB temperature to have the same numerical value. That is an indication that the universe is seen to be homogeneous.

I think it is mathematically described from the FLRW metric that on spacelike hypersurfaces of constant cosmological time $t$ is homogeneous and isotropic.

 

[PeterDonis]

I think it is mathematically described from the FLRW metric that on spacelike hypersurfaces of constant cosmological time $t$ is homogeneous and isotropic.

No, it isn't; that doesn't even make sense.

 

[cianfa72]

No, it isn't; that doesn't even make sense.

It doesn't make any sense to say that a metric (restricted on spacelike hypersurfaces) is homogeneous and isotropic ?

 

[PeterDonis]

It doesn't make sense to say that a metric (restricted on spacelike hypersurfaces) is homogeneous and isotropic ?

I thought your post #37 was saying that $t$ (the coordinate) is homogeneous and isotropic. That doesn't make sense.

It makes sense to say that the geometry of a spacelike hypersurface of constant $t$ is homogeneous and isotropic, so if by "metric" you mean "geometry" then it would also make sense to say the metric is homogeneous and isotropic.

It also makes sense to say that the distribution of stress-energy on such a spacelike hypersurface is homogeneous and isotropic; that's the original physical statement that is used to derive the FRW spacetime geometry.

 

[cianfa72]

It makes sense to say that the geometry of a spacelike hypersurface of constant $t$ is homogeneous and isotropic, so if by "metric" you mean "geometry" then it would also make sense to say the metric is homogeneous and isotropic.

From a mathematical point of view, the homogeneous property of a Riemannian manifold (that is any spacelike hypersurface of constant cosmological time) boils down to the existence of a transitive isometry map on it.

What about the mathematical definition of isotropic for that manifold ?

 

[PeterDonis]

From a mathematical point of view, the homogeneous property of a Riemannian manifold (that is any spacelike hypersurface of constant cosmological time) boils down to the existence of a transitive isometry map on it.

Yes.

What about the mathematical definition of isotropic for that manifold ?

It means that, if we pick a point in the manifold, there is an SO(3) isometry centered on that point.

Note that it is possible to have a manifold that is isotropic about one point, but not homogeneous. However, as should be evident, if a manifold is homogeneous, and we can show that it is isotropic about one point, then it must be isotropic about every point.

 

[cianfa72]

It means that, if we pick a point in the manifold, there is an SO(3) isometry centered on that point.

SO(3) is the special continuous group of linear transformations that preserves orthogonality. In this context the orthogonality is w.r.t. the (positive definite) metric tensor defined on each point on spacelike hypersurfaces.

An element of SO(3) group acts on vectors of tangent vector space defined at each point. Therefore, I think, the definition of isotropy at a point does require SO(3) group elements act on tangent space's vectors as isometries.

 

[PeterDonis]

SO(3) is the special continuous group of linear transformations that preserves orthogonality.

SO(3) is the group of rotations in 3 dimensional space. These are linear transformations and they do preserve the orthogonality, but they are hardly the only such group of transformations. So I don't see where you are getting "the" (bolded in the quote above) from.

An element of SO(3) group acts on vectors of tangent vector space defined at each point.

Yes.

Therefore, I think, the definition of isotropy at a point does require SO(3) group elements act on tangent space's vectors as isometries.

That's what I said.

 

[PeterDonis]

An element of SO(3) group acts on vectors of tangent vector space defined at each point.

Note that, while this is correct, it is not the only possible way to interpret the SO(3) group. It can also be interpreted as a three-parameter group of Killing vector fields on the spacetime. A spacetime with such a group of KVFs is called "spherically symmetric", and will be isotropic about at least one point, but possibly only one. For example, a spacetime containing a single non-rotating body surrounded by vacuum will be spherically symmetric, and will be isotropic about the center of mass of the body, but not about any other point.

In the case of FRW spacetime, however, the spacetime is isotropic about every point. That means it is spherically symmetric no matter which point we choose as our center of symmetry. It also means that there is a different SO(3) group of KVFs centered on each such point. In other words, a much stronger set of properties than just an SO(3) isometry on tangent spaces.

 

[cianfa72]

SO(3) is the group of rotations in 3 dimensional space. These are linear transformations and they do preserve the orthogonality, but they are hardly the only such group of transformations. So I don't see where you are getting "the" (bolded in the quote above) from.

Ok, so you are saying that there could be other groups of possibly non-linear transformations that preserve orthogonality and length as well.

Note that, while this is correct, it is not the only possible way to interpret the SO(3) group. It can also be interpreted as a three-parameter group of Killing vector fields on the spacetime. A spacetime with such a group of KVFs is called "spherically symmetric", and will be isotropic about at least one point, but possibly only one.

Ok, so integral curves of such KVFs lie on spacelike hypersurfaces, I believe.

Sorry, when you say "point" you actually mean an event in spacetime.

 

[DrGreg]

there could be other group of possibly non-linear transformations that preserve orthogonality and length as well

... or groups of linear transformations that do preserve orthogonality but don't preserve length

 

[PeterDonis]

integral curves of such KVFs lie on spacelike hypersurfaces, I believe.

They lie within each spacelike hypersurface of constant FRW coordinate time.

when you say "point" you are actually mean an event in spacetime.

Yes. But we can also consider points in a spacelike hypersurface of constant FRW coordinate time, since the integral curves of the KVFs in question like entirely within such surfaces.

 

[cianfa72]

In the case of FRW spacetime, however, the spacetime is isotropic about every point. That means it is spherically symmetric no matter which point we choose as our center of symmetry. It also means that there is a different SO(3) group of KVFs centered on each such point. In other words, a much stronger set of properties than just an SO(3) isometry on tangent spaces.

That means the three-parameters family of KVFs (that form a Lie algebra) are the infinitesimal generators of a group of isometries centered on any point on each spacelike hypersurface of constant cosmological time. Such groups are different instances each isomorphic to SO(3).

 

[PeterDonis]

the three-parameters family of KVFs (that form a Lie algebra) are the infinitesimal generators of a group of isometries centered on any point on each spacelike hypersurface of constant cosmological time.

Of SO(3) KVFs, yes. Note that the isometries associated with homogeneity, the spatial translations in a spacelike hypersurface of constant time, are also generated by a three-parameter group of KVFs, but the group is not SO(3).

Such groups are different instances each isomorphic to SO(3).

I guess you could think of them that way, although I don't think the term "instances" appears in this connection in the literature.

 

[cianfa72]

Of SO(3) KVFs, yes. Note that the isometries associated with homogeneity, the spatial translations in a spacelike hypersurface of constant time, are also generated by a three-parameter group of KVFs, but the group is not SO(3).

Sorry, such KVFs form a Lie algebra (w.r.t. the Lie bracket). Are themselves elements of SO(3) group, or are the isometries they generate actually elements of groups each isomorphic to SO(3) ?

 

[PeterDonis]

such KVFs form a Lie algebra

Sure, any group of KVFs forms a Lie algebra.

Are themselves elements of SO(3) group

A Lie algebra is not the same as a Lie group. Elements of a Lie algebra can be generators of a Lie group.

or are the isometries they generate actually elements of groups each isomorphic to SO(3) ?

It would be really helpful if you would be specific about which isometries you are talking about.

In FRW spacetime, any spacelike surface of constant FRW coordinate time $t$ has two three-parameter groups of isometries, generated by two three-parameter groups of KVFs.

One three-parameter group, the one associated with isotropy at every point, is associated with the group SO(3).

The other three-parameter group, the one associated with homogeneity, is associated with a group that depends on the specific FRW model. In a spatially flat model, the group is E(3) (I think that's right--it's the three-parameter group of translations in Euclidean 3-space). In a spatially open model, the group is H(3) (again, I think that's right--it's the three-parameter group of translations in hyperbolic 3-space). In a spatially closed model, I believe the group is SO(4).

 

[cianfa72]

A Lie algebra is not the same as a Lie group. Elements of a Lie algebra can be generators of a Lie group.

Yes, so the "set" of three-parameters KVFs associated with the isotropy at every point (on any spacelike hypersurface of constant FRW coordinate time) form a Lie algebra and generate isometries that form a Lie group isomorphic to SO(3).

One three-parameter group, the one associated with isotropy at every point, is associated with the group SO(3).

Yes, I was referring to this three-parameter group.

 

[cianfa72]

Sorry to bother you with these details. We said FRW spacetime is spherically symmetric regardless the point we choose as center of symmetry. The geometry of any spacelike hypersurface of constant cosmological time can be only one of the three type: flat (Euclidean), hyperbolic, spherical.

In any case there will be 6 linearly independent KVFs (one three-parameters "set" of KVFs for translations and one three--parameter "set" of KVFs for rotations about one chosen point).

The notion of linearly independent KVFs actually means that all the coefficients of their linear combination that give the null vector field are null (zero).

Now each other "set" of KVFs that are generators of the isometry group of rotations about any other point on that manifold can be given as linear combination of the first 6 KVFs. In other words, from the three-parameters translational KVFs and from the "set" of three-parameters rotational KVFs about one point, we get the set of three-parameters rotational KVFs about any other point.

 

[PeterDonis]

We said FRW spacetime is spherically symmetric regardless the point we choose as center of symmetry.

Yes.

The geometry of any spacelike hypersurface of constant cosmological time can be only one of the three type: flat (Euclidean), hyperbolic, spherical.

Yes.

In any case there will be 6 linearly independent KVFs (one three-parameters "set" of KVFs for translations and one three--parameter "set" of KVFs for rotations about one chosen point).

Yes.

The notion of linearly independent KVFs actually means that all the coefficients of their linear combination that give the null vector field are null (zero).

I'm not sure what you mean by this. "Linearly independent" means that none of the KVFs can be expressed as a linear combination of the others.

Now each other "set" of KVFs that are generators of the isometry group of rotations about any other point on that manifold can be given as linear combination of the first 6 KVFs. In other words, from the three-parameters translational KVFs and from the "set" of three-parameters rotational KVFs about one point, we get the set of three-parameters rotational KVFs about any other point.

No, this is not correct. A linear combination of a translation and a rotation is not a rotation about a different point.

What is correct is that you can Fermi-Walker transport the "rotation" KVFs at one point along the integral curves of a translation to another point, and you will get the "rotation" KVFs at the other point. But that's not the same as forming a new KVF from a linear combination of the translation and rotation.

 

[cianfa72]

No, this is not correct. A linear combination of a translation and a rotation is not a rotation about a different point.

I think I misinterpreted it. The set of KVFs on a manifold is not a vector space, therefore we can't get for instance a KVF for rotation about a different point as linear combination of overall 6 translation KVFs and rotation KVFs at another point.

What is correct is that you can Fermi-Walker transport the "rotation" KVFs at one point along the integral curves of a translation to another point, and you will get the "rotation" KVFs at the other point.

Maybe the result is the same as Lie dragging the rotation KVFs at one point along the integral curves of translation to another point.

 

[PeterDonis]

The set of KVFs on a manifold is not a vector space

Correct.

 

[PeterDonis]

Maybe the result is the same as Lie dragging the rotation KVFs at one point along the integral curves of translation to another point.

Actually, Lie dragging might be the correct transport law here. I believe we had a previous thread where we showed that Lie transport and Fermi-Walker transport are not necessarily the same.

 

[cianfa72]

The set of KVFs on a manifold is not a vector space, therefore we can't get for instance a KVF for rotation about a different point as linear combination of overall 6 translation KVFs and rotation KVFs at another point.

I've some doubts about this. From the above 7 KVFs are not linearly independent (see also Sean Carroll Spacetime and geometry - section 3.9).
What does it mean ? We said that the set of KVFs on a manifold is not a vector space (w.r.t. the pointwise sum of vector fields).

Perhaps the point is that if we take 7 KVFs then we can always write down any of them by "some kind of operations" acting on the other 6 KVFs. For example, as in post #58, we get the rotation KFV about a point by Lie dragging the rotation KVFs about a different point.

On the other hand, we can always find 6 KFVs such that none of them can be obtained as linear combination of the other 5 and does not exist a "some kind of operation" to get it from the others.

 

[cianfa72]

Another point related to the definition of isotropy. As in Sean Carroll Spacetime and geometry section 8.2, an universe spatially homogeneous and isotropic can be foliated into maximally symmetric spacelike hypersurfaces .

In section 8.1 the isotropy condition at a point on any such 3D spacelike manifold $M$ goes as follows:

More formally, a manifold $M$ is isotropic around a point $p$ if, for any two vectors $V$ and $W$ in $T_pM$ there is an isometry of $M$ such that the pushforward of $W$ under the isometry is parallel with $V$ (not pushed forward).

This definition should be equivalent to the existence of spacelike KVFs associated to those isometries (the pushforward is actually the differential of the isometry map evaluated at the point $p$).

 

[PeterDonis]

7 KVFs are not linearly independent

More precisely, you can't find 7 linearly independent KVFs at a particular point in one spacelike hypersurface of constant FRW coordinate time. You can only find 6.

What does it mean ? We said that the set of KVFs on a manifold is not a vector space

A vector space is defined as a space satisfying a particular set of axioms. Look up those axioms and then check whether the set of KVFs on a manifold satisfies them.

Perhaps the point is that if we take 7 KVFs then we can always write down any of them by "some kind of operations" acting on the other 6 KVFs

If you are talking about linearly independent, look up the definition of that. That definition is not in terms of "some kind of operations". There is one particular operation involved. Look it up and see.

we get the rotation KFV about a point by Lie dragging the rotation KVFs about a different point.

This has nothing to do with either linear independence or forming a vector space. (Which are two different things, btw.)

This definition should be equivalent to the existence of spacelike KVFs associated to those isometries

Yes.

 

[cianfa72]

More precisely, you can't find 7 linearly independent KVFs at a particular point in one spacelike hypersurface of constant FRW coordinate time. You can only find 6.

Ah ok, so the the notion of linearly independence applies to KVFs "evaluated" at specific points.

A vector space is defined as a space satisfying a particular set of axioms. Look up those axioms and then check whether the set of KVFs on a manifold satisfies them.

Re-thinking about it, I believe the set of KVFs on a manifold is a real vector space. Namely the pointwise sum of two KVFs is a KFV (i.e. the sum is closed) as the scalar moltiplication for a real number (together with the Lie bracket such vector space is promoted to an algebra).

 

[PeterDonis]

the notion of linearly independence applies to KVFs "evaluated" at specific points.

No, it applies to vector fields in general. The reason I said "at a specific point" is that the KVFs associated with isotropy at different points are different KVFs.

Re-thinking about it, I believe the set of KVFs on a manifold is a real vector space. Namely the pointwise sum of two KVFs is a KFV (i.e. the sum is closed) as the scalar moltiplication for a real number (together with the Lie bracket such vector space is promoted to an algebra).

Those aren't the only vector space axioms.

 

[cianfa72]

Those aren't the only vector space axioms.

I could be wrong but an algebra requires an underlying structure of vector space on a field -- Lie algebra.

 

[PeterDonis]

I could be wrong but an algebra requires an underlying structure of vector space on a field -- Lie algebra.

That is true, but irrelevant to what I said.

 

[cianfa72]

That is true, but irrelevant to what I said.

Sorry, I don't understand. If the set of KVFs on a manifold is a Lie algebra then it is a vector space on $\mathbb R$ w.r.t. the pointwise sum of KVFs.

 

[PeterDonis]

If the set of KVFs on a manifold is a Lie algebra then it is a vector space

Ah, I see. Yes, that's correct.

However, I still think it would be a good idea for you to check that the set of KVFs satisfies all of the vector space axioms, not just the ones you mentioned.

 

[cianfa72]

However, I still think it would be a good idea for you to check that the set of KVFs satisfies all of the vector space axioms, not just the ones you mentioned.

My point is the following: start by noting that vector fields on a manifold form themselves a vector space on $\mathbb R$, therefore it suffices to show that KVFs are closed both w.r.t. the vector field sum and the product for a scalar in $\mathbb R$.

We can use the following definition: $X$ is a KVF iff:
$$g(\nabla_YX,Z)+g(Y,\nabla_ZX)=0,\ \forall Y,Z\in\Gamma(TM).$$ Using the linear property of covariant derivatives and bi-linearity of metric tensor $g$ we can show that both the aforementioned conditions are satisfied.

 

[PeterDonis]

vector fields on a manifold form themselves a vector space on $\mathbb R$, therefore it suffices to show that KVFs are closed both w.r.t. the vector field sum and the product for a scalar in $\mathbb R$.

No, it doesn't. You also have to show that the identity element of the vector space is in the set of KVFs. That might seem trivial, but it at the very least should give some food for thought.

 

[cianfa72]

No, it doesn't. You also have to show that the identity element of the vector space is in the set of KVFs. That might seem trivial, but it at the very least should give some food for thought.

Yes, the identity element of the vector field vector space is the null vector field $0$. If we "insert" it into post#68 definition we get $$\nabla_Y0 =0, \nabla_Z0=0,\ \forall Y,Z\in\Gamma(TM)$$ hence $$g(0,Z)=0, g(Y,0), \ \forall Y,Z \in\Gamma(TM)$$