Synchronous Reference Frame: Definition and Usage

[cianfa72]

TL;DR Summary

On the conditions to be able to build a synchronous reference frame in any spacetime

Hi,
reading the Landau book 'The Classical theory of Field - vol 2' a doubt arised to me about the definition of synchronous reference system (a.k.a. synchronous coordinate chart).

Consider a generic spacetime endowed with a metric $g_{ab}$ and take the (unique) covariant derivative operator $\nabla_a$ associated to it. From the Frobenius theorem we know that a timelike vector field $\xi^a$ is hypersurface orthogonal (w.rt. the given metric) if and only if the following holds (see for instance Wald appendix C):
$$\xi_{[a} \nabla_b \xi_{c]} = 0$$
So, given the metric and its associated covariant derivative operator $\nabla_a$ , we can solve the above equation for a timelike vector field $\xi^a$. Now if a solution $\xi^a$ exists, then it is hypersurface orthogonal -- namely there is a scalar field, say $t$, such that its $t=const$ level sets are spacelike hypersurfaces foliating the spacetime and orthogonal at each point to the vector field $\xi^a$ solution of the above equation.

If the above is correct, I do not fully understand the point made here wiki - synchronous coordinates -- the same as in Landau book - section 99. It seems that in any spacetime we can always construct a synchronous reference frame starting from a given spacelike hypersurface. Namely take a family of timelike geodesics normal at each point to this spacelike hypersurface. Choose these lines as time coordinate lines and define the time coordinate $t$ as the length $s$ of the geodesic measured starting from the given hypersurface; the result is a synchronous frame.

My concern is that the hypersurfaces we get using this procedure are not for sure orthogonal to the 4-velocity 's timelike geodesics. If we added the constrain that the timelike vector field $\xi^a$ was a timelike Killing vector field of the given spacetime then yes, moving along an isometry that would actually make sense.

What do you think about ? Thank you.

 

[Orodruin]

the result is a synchronous frame.

No, it isn’t. There is no guarantee that your family of geodesics will remain orthogonal to your surfaces because of geodesic deviation.

 

[Orodruin]

Summary:: On the conditions to be able to build a synchronous reference frame in any spacetime

If we added the constrain that the timelike vector field ξa was a timelike Killing vector field of the given spacetime then yes, moving along an isometry that would actually make sense.

There is no guarantee that you can find such a vector field.

 

[PAllen]

No, it isn’t. There is no guarantee that your family of geodesics will remain orthogonal to your surfaces because of geodesic deviation.

Wald, in section 3.3 (p.42-43 of my edition), discusses the construction of synchronous coordinates in arbitrary spacetimes (also called Gaussian Normal). He explains why they are always possible in some finite patch of spacetime, and what types of things cause them to fail be be global. Wald specifically proves that the geodesics do remain orthogonal to the surface family until the validity of the patch becomes impossible due to e.g. intersection of geodesics.

An interesting case study is synchronous coordinates patches for the Kerr metric:

https://arxiv.org/abs/1004.3913

[edit: This paper even gives a link to an earlier paper showing synchronous coordinate patches for as pathological a case as Godel spacetime].

 

[cianfa72]

Wald, in section 3.3 (p.42-43 of my edition), discusses the construction of synchronous coordinates in arbitrary spacetimes (also called Gaussian Normal). He explains why they are always possible in some finite patch of spacetime, and what types of things cause them to fail be be global. Wald specifically proves that the geodesics do remain orthogonal to the surface family until the validity of the patch becomes impossible due to e.g. intersection of geodesics.

Ah...I see. So that actually means that-- starting from a given spacelike hypersurface assigned in a some finite patch of spacetime -- we can always find a geodesic congruence hypersurface orthogonal in that region.

Then, Frobenius and the geodesic condition actually imply the 4-velocity vector field tangent at that geodesic congruence (say $\xi^a$) solves both the equations
$$\xi^a \nabla_a \xi^b = 0, \text{ } \xi_{[a} \nabla_b \xi_{c]} = 0$$ Hence we can claim the above set of equations has always a solution in a some finite region of spacetime. However it could be not extended globally.

Does it make sense ?

 

[Orodruin]

Wald specifically proves that the geodesics do remain orthogonal to the surface family until the validity of the patch becomes impossible due to e.g. intersection of geodesics.

Right, I somehow mixed it up with the standard example I use which is the Schwarzschild metric with its timelike Killing field where following the field for a fixed proper time does not preserve orthogonality but of course the flow of that Killing field is not consisting of geodesics.

 

[PAllen]

Ah...I see. So that actually means that-- starting from a given spacelike hypersurface assigned in a some finite patch of spacetime -- we can always find a geodesic congruence hypersurface orthogonal in that region.

Then, Frobenius and the geodesic condition actually imply the 4-velocity vector field tangent at that geodesic congruence (say $\xi^a$) solves both the equations
$$\xi^a \nabla_a \xi^b = 0, \text{ } \xi_{[a} \nabla_b \xi_{c]} = 0$$ Hence we can claim the above set of equations has always a solution in a some finite region of spacetime. However it could be not extended globally.

Does it make sense ?

Yes it makes sense. Consider, as a trivial case, great circles orthogonal to a given latitude circle. Any part of the sphere can be included in such a construction, but no such vector field can cover the whole sphere - each such vector field reaches two poles to which it cannot be extended.

 

[cianfa72]

Yes it makes sense. Consider, as a trivial case, great circles orthogonal to a given latitude circle. Any part of the sphere can be included in such a construction, but no such vector field can cover the whole sphere - each such vector field reaches two poles to which it cannot be extended.

ok, hence Landau in section 99 was wrong (or rather...he was actually referring to a finite region of spacetime).

 

[PAllen]

ok, hence Landau in section 99 was wrong (or rather...he was actually referring to a finite region of spacetime).

I am sure he was referring to a finite region, and was not wrong. Since it is normal for a coordinate chart not to cover the whole spacetime, it would be worth commenting on the special case where it does.

 

[cianfa72]

I am sure he was referring to a finite region, and was not wrong. Since it is normal for a coordinate chart not to cover the whole spacetime, it would be worth commenting on the special case where it does.

ok, your point is that he had actually built a coordinate chart just for a limited region of spacetime (basically as long as the costruction was still feasible).

 

[PAllen]

ok, your point is that he had actually built a coordinate chart just for a limited region of spacetime (as long as the costruction was still feasible).

Often, the coverage of synchronous coordinates is no smaller than other common coordinates. Besides FLRW solutions, where they are global, consider:

1) Lemaitre coordinates for Schwarzschild geometry. These are synchronous coordinates, and they cover the whole exterior and physically meaningful part of the interior (that is, all that would exist in a BH formed from collapse). Thus, they cover more than the standard Schwarzschild coordinates which cover only exterior In one patch. Of course, there is much of the Kruskal extension that they don’t cover.

2) Kerr solution. Synchronous coordinates can cover one whole exterior.

So, in many common cases, this finite region can be the whole region of interest.

 

[cianfa72]

Sorry, another point related to this. If we start from a given smooth spacelike hypersurface we can always find (by definition) timelike vectors orthogonal to it at each point. Btw that spacelike hypersurface can be always given as the level set of some function (a scalar field) $h$ for a given $h=c$ (constant).

So, Frobenius theorem is actually a condition about vector fields and not about vectors ! Namely take a smooth vector field $\xi^a$ that assigns to the points on the given spacelike hypersurface their orthogonal timelike vectors. Then if and only if the following condition holds:
$$\text{ } \xi_{[a} \nabla_b \xi_{c]} = 0$$ we can find a smooth scalar function $t$ such that its level sets are spacelike hypersurfaces foliating a region of spacetime (this family include the spacelike hypersurface we started with for a specific $t=c$) and the vector field $\xi^a$ is hypersurface orthogonal in that region.

As we said before, we can always find a such vector field $\xi^a$ (that assigns to the points on the given spacelike hypersurface their orthogonal timelike vectors ) solving both
$$\xi^a \nabla_a \xi^b = 0, \text{ } \xi_{[a} \nabla_b \xi_{c]} = 0$$ in a spacetime region surrounding the given spacelike hypersurface we started with (hopefully it extends globally as you said).

 

[PAllen]

Sorry, another point related to this. If we start from a given smooth spacelike hypersurface we can always find (by definition) timelike vectors orthogonal to it at each point. Btw that spacelike hypersurface can be always given as the level set of some function (a scalar field) $h$ for a given $h=c$ (constant).

Yes.

So, Frobenius theorem is actually a condition about vector fields and not about vectors !

Yes (there are multiple formulations of Frobenius, but this is true of the one you are talking about.).

Namely take a smooth vector field $\xi^a$ that assigns to the points on the given spacelike hypersurface their orthogonal timelike vectors. Then if and only if the following condition holds:
$$\text{ } \xi_{[a} \nabla_b \xi_{c]} = 0$$ we can find a smooth scalar function $t$ such that its level sets are spacelike hypersurfaces foliating a region of spacetime (this family include the spacelike hypersurface we started with for a specific $t=c$) and the vector field $\xi^a$ is hypersurface orthogonal in that region.

Here I have some issues. The idea of Frobenius, as I understand it (I am not and expert in this area) is you start with an arbitrary vector field, and you want to know if you can construct a family of hypersurfaces such the the vector field is orthogonal to each. If the given condition is true, then there exists such a foliation.

In contrast, the construction starting from a given spacelike surface, using geodesics orthogonal to it, and then constructing the vector field of tangents of these geodesics, guarantees the Frobenius criterion will be satisfied.

As we said before, we can always find a such vector field $\xi^a$ (that assigns to the points on the given spacelike hypersurface their orthogonal timelike vectors ) solving both
$$\xi^a \nabla_a \xi^b = 0, \text{ } \xi_{[a} \nabla_b \xi_{c]} = 0$$ in a spacetime region surrounding the given spacelike hypersurface we started with (hopefully it extends globally as you said).

Yes.

 

[PeterDonis]

the construction starting from a given spacelike surface, using geodesics orthogonal to it, and then constructing the vector field of tangents of these geodesics, guarantees the Frobenius criterion will be satisfied.

For some neighborhood of the given spacelike surface, yes. But that neighborhood might not be the entire spacetime.

 

[PAllen]

For some neighborhood of the given spacelike surface, yes. But that neighborhood might not be the entire spacetime.

Of course, I mentioned that several times earlier. But many types of coordinate systems are not global. For example, in Schwarzschild spacetime, the coverage of Lemaitre coordinates (which are synchronous coordinate system) is larger than standard Schwarzschild coordinates (but less than Kruskal or even Eddington-Finkelstein).

 

[PAllen]

Here I have some issues. The idea of Frobenius, as I understand it (I am not and expert in this area) is you start with an arbitrary vector field, and you want to know if you can construct a family of hypersurfaces such the the vector field is orthogonal to each. If the given condition is true, then there exists such a foliation.

In contrast, the construction starting from a given spacelike surface, using geodesics orthogonal to it, and then constructing the vector field of tangents of these geodesics, guarantees the Frobenius criterion will be satisfied.

A few further clarififications:

1) A hypersurface orthogonal congruence need not be geodesic (thus a vector field can satisfy Frobenius without having its integral curves be geodesic). Two common examples are the Rindler congruence and the killing vector field of Schwarzschild geometry. However, neither of these is the basis for synchronous coordinates, because as @Orodruin mentioned earlier in this thread, the orthogonal slices bound differing proper times along the congruence. Alternatively, advancing the same proper time along each congruence element from starting orthogonal surface gives a surface not orthogonal to the congruence.

2) However, combining the Frobenius condition with the requirement of geodesic integral curves of the vector field, is sufficient to build a synchronous coordinate patch. Of course, if you start with a spacelike surface, you don't need to apply Frobenius to build a synchronous coordinate patch.

 

[cianfa72]

Thanks all