# Missing step in the derivation of the Robertson-Walker metric

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## Summary:

How can one drop "dtdr" terms in RW metric

## Main Question or Discussion Point

To arrive at the Robertson-Walker metric for a spatially homogeneous and isotropic cosmology, one first writes down the the metric for spatial sections i.e. constant t surfaces,

2 = d2 +f2(r) (dθ2 + sin2θ dφ2),

where f(r) can take only 3 special forms, and then one promptly writes the metric for general case when cosmic time, t, is non-constant to be the RW form,
ds2 = -dt2 + a2(t) dσ2.

My question is how can one rule out "dtdr" terms in the second step?

PeterDonis
Mentor
To arrive at the Robertson-Walker metric for a spatially homogeneous and isotropic cosmology, one first writes down the the metric for spatial sections i.e. constant t surfaces
No, one doesn't. One first shows that the metric can be written as $ds^2 = - dt^2 + a^2 (t) d \sigma^2$, and then one investigates what forms $d\sigma^2$ can take. See, for example, the discussion in Chapter 8 of Carroll's lecture notes on GR:

https://arxiv.org/abs/gr-qc/9712019

PAllen
No, one doesn't. One first shows that the metric can be written as $ds^2 = - dt^2 + a^2 (t) d \sigma^2$, and then one investigates what forms $d\sigma^2$ can take. See, for example, the discussion in Chapter 8 of Carroll's lecture notes on GR:

https://arxiv.org/abs/gr-qc/9712019
Thanks, I have looked at Carroll's notes in the past but I am not sure he explains why spatial homogeneity and isotropy implies the metric can be foliated R × Σ (to quote Carroll "this translates into the statement that the universe can be foliated into spacelike slices such that each slice is homogeneous and isotropic."). This is why I called it a missing step. Is there a way to see that ∂t is orthogonal to spatial sections?

PAllen
Thanks, I have looked at Carroll's notes in the past but I am not sure he explains why spatial homogeneity and isotropy implies the metric can be foliated R × Σ (to quote Carroll "this translates into the statement that the universe can be foliated into spacelike slices such that each slice is homogeneous and isotropic."). This is why I called it a missing step. Is there a way to see that ∂t is orthogonal to spatial sections?
It can always be done. See my links above, and you can search on synchronous coordinates for more detailed derivations.

To add a little, you can first get rid of those metric terms in any manifold by choice of coordinates:
https://en.wikipedia.org/wiki/Coordinate_conditions#Synchronous_coordinateshttps://en.wikipedia.org/wiki/Synchronous_frame#Synchronous_coordinates
Then, arguments from homogeneity and isotropy allow the simplification of the space space metric terms.
Thanks. However it seems problematic to me that such a gauge (frame) can be chosen in all situations, as it would imply all stationary metrics ought to be static. E.g. if it can be chosen for the Kerr metric, then the Kerr metric would contain no dt dφ terms which would make it a static metric instead of a stationary metric. Is there some restriction under which these g0t terms can be set to zero globally?

PeterDonis
Mentor
I am not sure he explains why spatial homogeneity and isotropy implies the metric can be foliated R × Σ (to quote Carroll "this translates into the statement that the universe can be foliated into spacelike slices such that each slice is homogeneous and isotropic.")
He refers to equation 7.2; that equation and the discussion surrounding it should make that step clearer.

PeterDonis
Mentor
it seems problematic to me that such a gauge (frame) can be chosen in all situations, as it would imply all stationary metrics ought to be static
No, it doesn't, because "static" has a frame-independent meaning, that the timelike Killing vector field is hypersurface orthogonal. Finding synchronous coordinates is not the same thing. See below.

if it can be chosen for the Kerr metric, then the Kerr metric would contain no dt dφ terms which would make it a static metric instead of a stationary metric
No, because the vector field $\partial_t$ in these coordinates, the one that is orthogonal to the spacelike hypersurfaces, is not the same as the timelike Killing vector field of the spacetime. There is no family of spacelike hypersurfaces in Kerr spacetime that is orthogonal to the timelike KVF; that's why Kerr spacetime is only stationary and not static.

PAllen
Thanks. However it seems problematic to me that such a gauge (frame) can be chosen in all situations, as it would imply all stationary metrics ought to be static. E.g. if it can be chosen for the Kerr metric, then the Kerr metric would contain no dt dφ terms which would make it a static metric instead of a stationary metric. Is there some restriction under which these g0t terms can be set to zero globally?
In these coordinates, the metric is never stationary. That is, the timelike kvf in a stationary spacetime will take a very complex form in synchronous coordinates. Look up two special cases: lemaitre coordinates, which are synchronous coordinates for Schwarzschild spacetime, and the Milne cosmology. The former completely hides the stationary character, though the kvf still exists. In Milne cosmology you have metric expansion in a flat spacetime manifold.

He refers to equation 7.2; that equation and the discussion surrounding it should make that step clearer.
Yes, I am aware of this as well, but he does not really make it any clearer. Instead, Just says it is a "powerful theorem" that such a choice can always be made and that "Proving the theorem is a mess", and then in turn refers to Weinberg's Chapter 13 which is on Maximally symmetric spaces! (Although RW is not a maximally symmetric metric). So my question is what is a "non-messy" i.e. straightforward and physical way to prove that RW does not contain a dt dr piece.

PAllen
In these coordinates, the metric is never stationary. ....
This is an over statement. It is true if any matter is present. But with no matter, there are exceptions, the most obvious being Minkowski coordinates in flat spacetime.

PeterDonis
Mentor
In these coordinates, the metric is never stationary.
I'm not sure what you mean by this. "Stationary" is a coordinate-independent geometric property: the existence of a timelike Killing vector field.

This is an over statement. It is true if any matter is present. But with no matter, there are exceptions, the most obvious being Minkowski coordinates in flat spacetime.
Thanks but not what I am looking for. I get what you are trying to get at, that perhaps I can choose such a gauge regardless of the symmetry situation. But my question has to specifically do with symmetries, in particular global foliations of the metric in the presence of a certain set of symmetries. In the case of RW the (maximal) symmetries of spatial sections dictate that the foliations can be made orthogonal to ∂t vector. But thanks, anyways.

PeterDonis
Mentor
Just says it is a "powerful theorem" that such a choice can always be made and that "Proving the theorem is a mess", and then in turn refers to Weinberg's Chapter 13 which is on Maximally symmetric spaces!
Well, "homogeneous and isotropic" means that the spacetime can be foliated by a set of spacelike hypersurfaces which are maximally symmetric spaces.

my question is what is a "non-messy" i.e. straightforward and physical way to prove that RW does not contain a dt dr piece
There might not be one if you don't regard the proof of the theorem Carroll references in 7.2 as "straightforward and physical".

Carroll's discussion following 7.2 gives what might be a more intuitive description of the general case, which can be specialized to the FRW case as follows: from the fact that the spacetime is homogeneous and isotropic, we know that it can be foliated by spacelike 3-surfaces which are maximally symmetric. That only leaves one more coordinate, the timelike coordinate, so all we have to do to put the metric in the standard FRW form is:

(1) Choose the timelike tangent vector field $\partial_t$ to be everywhere orthogonal to the spacelike hypersurfaces (the "powerful theorem" ensures that we can always do this); and

(2) Scale the $t$ coordinate so that $g_{tt} = - 1$ everywhere.

Physically, this amounts to defining the integral curves of $\partial_t$ as the worldlines of "comoving" observers who always see the universe as homogeneous and isotropic (that's what it means for their 4-velocities, the $\partial_t$ tangent vectors, to be orthogonal to the maximally symmetric spacelike hypersurfaces), and then scaling $t$ so it is the same as the proper time of such comoving observers (since the elapsed proper time along any two comoving worldlines must be the same between the same pair of spacelike hypersurfaces).

PAllen
I'm not sure what you mean by this. "Stationary" is a coordinate-independent geometric property: the existence of a timelike Killing vector field.
I am distinguishing a coordinate feature (metric coefficients not involving a timelike coordinate in chosen coordinates) from the invariant feature (stationary spacetime). Not necessarily a good choice of term (metric being stationary), but wikipedia uses it in discussing the feature of synchronous coordinates that they generally make a metric look time dependent even if it is a stationary spacetime.

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PAllen
Thanks but not what I am looking for. I get what you are trying to get at, that perhaps I can choose such a gauge regardless of the symmetry situation. But my question has to specifically do with symmetries, in particular global foliations of the metric in the presence of a certain set of symmetries. In the case of RW the (maximal) symmetries of spatial sections dictate that the foliations can be made orthogonal to ∂t vector. But thanks, anyways.
My point is you can first choose synchronous coordinates, by fiat. In a general manifold, they may cover only some large region; and to cover the maximally continued manifold you may need several such patches. Note, in any such large region, the foliations are orthogonal to the ∂t vector. Then bring in symmetry arguments that further simplify the metric form, and allow global statements about the foliation. The key point is that there is nothing special about about being able to remove the dtdr term, per se. Global coverage of the simplified metric basically follows from the global symmetry statements - a failure of coverage would represent of failure of global homogeneity or isotropy. (Perhaps something like the theorem in Weinberg is needed to make this argument bulletproof; but it seems pretty intuitive - what is it about a failure location that is different from other locations - how can that square with global homogeneity?).

As an aside, here is a reference that derives synchronous coordinates for Kerr spacetime (Gaussian is just another word for this type of coordinate choice):

https://www.researchgate.net/publication/45913433_Gaussian_Coordinate_Systems_for_the_Kerr_Metric
They discuss the issue of incomplete coverage.

This is an over statement. It is true if any matter is present. But with no matter, there are exceptions, the most obvious being Minkowski coordinates in flat spacetime.
Thanks but not what I am looking for. I get what you are trying to get at, that perhaps I can choose such a gauge regardless of the symmetry situation. But my question has to specifically do with symmetries, in particular global foliations
Well, "homogeneous and isotropic" means that the spacetime can be foliated by a set of spacelike hypersurfaces which are maximally symmetric spaces.

There might not be one if you don't regard the proof of the theorem Carroll references in 7.2 as "straightforward and physical".

Carroll's discussion following 7.2 gives what might be a more intuitive description of the general case, which can be specialized to the FRW case as follows: from the fact that the spacetime is homogeneous and isotropic, we know that it can be foliated by spacelike 3-surfaces which are maximally symmetric. That only leaves one more coordinate, the timelike coordinate, so all we have to do to put the metric in the standard FRW form is:

(1) Choose the timelike tangent vector field $\partial_t$ to be everywhere orthogonal to the spacelike hypersurfaces (the "powerful theorem" ensures that we can always do this); and

(2) Scale the $t$ coordinate so that $g_{tt} = - 1$ everywhere.

Physically, this amounts to defining the integral curves of $\partial_t$ as the worldlines of "comoving" observers who always see the universe as homogeneous and isotropic (that's what it means for their 4-velocities, the $\partial_t$ tangent vectors, to be orthogonal to the maximally symmetric spacelike hypersurfaces), and then scaling $t$ so it is the same as the proper time of such comoving observers (since the elapsed proper time along any two comoving worldlines must be the same between the same pair of spacelike hypersurfaces).
So what I claimed in the question is correct, first one has to write down the maximally symmetric spatial sections and then add time to the mix.

Thanks but Carroll's take is okay as you mention for intuitive purposes but not helpful or satisfactory to me as I have noted it relegates the ability to drop "dt dr" terms to some powerful theorem without making it obvious how/why. For example, let's say I put in a dt dr term. Is there a coordinate transformation I can do to redefine the time variable, t, in a way which does not affect/touch the spatial at all other than just getting rid of the dt dr term. If I write down such a redefinition of t, say t →t'=t'(t,r), then I can indeed set gtr=0, but the coefficient of dr2 term gets changed :( and the spatial sections do not look maximally symmetric form anymore! It will be helpful if I start from a gauge where gtr is non-vanishing and then by a coordinate transformation I can make it vanish without touch the spatial sections at all!

PS: I understand the physical consequence of that gauge choice as you say comoving observers trace out integral curves of ∂t orthogonal to spatial sections, but that is AFTER I have made the choice. Couple of lines of math which makes such a gauge choice obvious for maximal symmetric spatial sections would go a long way to persuade me.

PeterDonis
Mentor
So what I claimed in the question is correct, first one has to write down the maximally symmetric spatial sections and then add time to the mix.
If you mean, first one has to assume that the spacelike hypersurfaces are maximally symmetric, yes, that is correct, because the theorem requires the submanifolds (which in this case are the spacelike hypersurfaces) to be maximally symmetric. But you don't have to make any other assumption about the particular form of the metric of the spacelike hypersurfaces, beyond maximal symmetry. In particular, you don't have to choose coordinates on the spacelike hypersurfaces that let you write their metric in the specific form you gave in the OP.

it relegates the ability to drop "dt dr" terms to some powerful theorem without making it obvious how/why
Well, then, there is no proof that meets your requirements for being "obvious". Sorry.

let's say I put in a dt dr term. Is there a coordinate transformation I can do to redefine the time variable, t, in a way which does not affect/touch the spatial at all other than just getting rid of the dt dr term
There must be a transformation that gets rid of the dt dr term, by the theorem. But if you mean, does the theorem explicitly tell you how to construct such a coordinate transformation, I don't think it does. But I am not familiar with the detailed proof of the theorem, so I don't know for sure.

Also, there is no guarantee that such a transformation will leave the spatial part of the metric unchanged, nor would it need to. See below.

If I write down such a redefinition of t, say t →t'=t'(t,r), then I can indeed set gtr=0, but the coefficient of dr2 term gets changed :( and the spatial sections do not look maximally symmetric form anymore!
You are missing two crucial points.

First, by "look maximally symmetric", you appear to mean "look like the spatial metric in the OP". But the way you wrote the spatial metric in the OP is not the only way to write the metric for a maximally symmetric 3-surface.

Second, if there is a dt dr term in the metric, then the spacelike hypersurfaces of constant $T$ in such a coordinate chart (I'm capitalizing $T$ here to make it clear that it's a different coordinate) will not be the same as the hypersurfaces of constant $t$ in standard FRW coordinates. Which means the hypersurfaces of constant $T$ in such a chart will not be maximally symmetric, since only the hypersurfaces of constant $t$ in standard FRW coordinates are.

In other words, you can't just wave your hands and say "put a dt dr term in the metric" and change nothing else. You have to actually look at how such a choice of coordinates would describe the spacetime geometry.

...But you don't have to make any other assumption about the particular form of the metric of the spacelike hypersurfaces, beyond maximal symmetry. In particular, you don't have to choose coordinates on the spacelike hypersurfaces that let you write their metric in the specific form you gave in the OP.
Choosing these coordinates for spatial slices was not my issue - I wrote them down explicitly as an example/illustration. I am well aware I can use other coordinates, say cartesian type coordinates x,y,z to label the spatial sections instead. I was more concerned about the dtdx pieces.

There must be a transformation that gets rid of the dt dr term, by the theorem. But if you mean, does the theorem explicitly tell you how to construct such a coordinate transformation, I don't think it does.
That is what is exactly I am looking for, *the* transformation which leads me to the desired gauge from a general form.

Second, if there is a dt dr term in the metric, then the spacelike hypersurfaces of constant $T$ in such a coordinate chart (I'm capitalizing $T$ here to make it clear that it's a different coordinate) will not be the same as the hypersurfaces of constant $t$ in standard FRW coordinates. Which means the hypersurfaces of constant $T$ in such a chart will not be maximally symmetric, since only the hypersurfaces of constant $t$ in standard FRW coordinates are.
That is not what I am getting at. I am aware that I can contrive coordinates where the symmetry won't be manifest at all. But here I am asking a subclass of coordinate choices or gauges where dt dr term is there in addition to the usual terms, namely,
ds2 = - dt2 + b(t) dt dr + a2(t) dσ2.
In such case the spatial sections, i.e. t= constant will leave you with just the spatial part, the dt dr term vanishes when you take t= constant and such sections better be homogeneous or isotropic - otherwise they will not manifest the symmetries. I am talking about such restricted case,s so to say "subgauges" where the spatial symmetry is manifest. My issue now is to find a gauge transformation which will kill the dt dr term without affecting the spatial part.

PeterDonis
Mentor
I am asking a subclass of coordinate choices or gauges where dt dr term is there in addition to the usual terms
But if the dt dr term is there, you are not using "the usual" coordinates, so you should not expect "the usual" terms to be there. As I said, you can't just wave your hands and say "put in a dt dr term" and not change anything else.

In such case the spatial sections, i.e. t= constant will leave you with just the spatial part, the dt dr term vanishes when you take t= constant and such sections better be homogeneous or isotropic
No, they will not be. That's my point. Surfaces of constant $T$ in your hypothetical chart will not be the same as surfaces of constant $t$ in the standard FRW chart, because you changed the chart.

For an example of such a chart on FRW spacetime, see this thread:

I strongly suggest that you take the the time to work out the relationship between the chart described there, and the standard FRW coordinate chart, and how they both describe the same spacetime geometry, just with different choices of the time coordinate. Also I strongly suggest that you look at the spatial part of the metric in this chart and note that it is flat (despite the fact that this metric can describe a closed FRW spacetime, where the spacelike hypersurfaces in standard FRW coordinates are 3-spheres), and think about what that means.

otherwise they will not manifest the symmetries
There is no requirement that every coordinate chart on a spacetime with certain symmetries must manifest those symmetries.

I am talking about such restricted case, so to say "subgauges" where the spatial symmetry is manifest. My issue now is to find a gauge transformation which will kill the dt dr term without affecting the spatial part.
Before you even try this, you need to first demonstrate that a chart which manifests the spatial symmetry, but also has a dt dr term in the metric, even exists. I strongly suspect that no such chart exists.

To put this another way: try working the problem you see from the other end. Try starting with the standard FRW chart, and seeing if you can find a coordinate transformation to another chart that keeps the spatial symmetry manifest but also puts a dt dr term in the metric. I strongly suspect you will not be able to.

PAllen
If you want full rigor on these questions, I suggest the following, which gives a definition of homogeneity and isotropy in terms of kvfs, and shows that some imprecise definitions in common use are not quite adequate. Physics books typically gloss over such details, but this fills in all the gaps:

https://arxiv.org/abs/1610.05628
(In one section, as an aside, it demonstrates existence of synchronous coordinates in a neighborhood by avoiding use any assumptions about the curvature of the spatial slices; it notes this in passing).

(Another aside is that the author's literature review suggests that Weinberg is perhaps the only source adequately addressed these questions previously; and that there is most likely no simpler approach).

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kimbyd
Gold Member
2018 Award
Thanks. However it seems problematic to me that such a gauge (frame) can be chosen in all situations, as it would imply all stationary metrics ought to be static. E.g. if it can be chosen for the Kerr metric, then the Kerr metric would contain no dt dφ terms which would make it a static metric instead of a stationary metric. Is there some restriction under which these g0t terms can be set to zero globally?
I'm pretty sure any off-diagonal terms not related to the FLRW metric by a coordinate transformation would break isotropy.

You note, for instance, that the $dt d\phi$ term in the Kerr metric relates to the rotation of the black hole. If we had rotation for the FLRW metric then it would cease to be isotropic, because rotation picks out a particular direction as special.

Another way to examine it is to look at the other side of the equation, the stress-energy tensor. It's not too difficult to prove that the stress-energy tensor of a perfect fluid can be represented as a simple diagonal tensor, with diagonal components related to the pressure and energy density, with the pressure represented as a single number due to the isotropy of the system. Off-diagonal components of this tensor would indicate twisting forces or movement in a particular direction which would break the isotropy. You can, of course, represent this tensor in a non-diagonal form, but the point is that there are coordinates where it is diagonal.

So, we have a diagonal tensor on the right represented by two functions, which indicates we only have two independent equations available. This indicates that the general solution to the Einstein equations would be a two-parameter solution possibly combined with a vacuum solution*. A two-parameter solution has been discovered, so we know that we can't have a more complicated solution without an additional vacuum solution added.

In principle, the vacuum solution could contain many parameters. But how many such vacuum solutions would be isotropic? I think it would be pretty easy to show that the inclusion of such non-diagonal elements is always going to break the isotropy, as every one of them represents anisotropic behavior, and you can't combine them to create isotropic behavior without a change in coordinates that may make the stress-energy tensor anisotropic.

This isn't a rigorous proof, of course. But hopefully it shows that the idea is at least reasonable.

* Note that the solutions can't trivially add, as the Einstein equations are not linear. But in the search for a solution we could start with the linear approximation and expand out from there.

I'm pretty sure any off-diagonal terms not related to the FLRW metric by a coordinate transformation would break isotropy.

You note, for instance, that the $dt d\phi$ term in the Kerr metric relates to the rotation of the black hole. If we had rotation for the FLRW metric then it would cease to be isotropic, because rotation picks out a particular direction as special.
Thats actually not true, you can put a dtdr term, it does not violate spatial isotropy or homogeneity. dtdφ of course does, which is why I did not put in the metric I mentioned in my original post.

But if the dt dr term is there, you are not using "the usual" coordinates, so you should not expect "the usual" terms to be there. As I said, you can't just wave your hands and say "put in a dt dr term" and not change anything else.
I am not putting a dtdr term by hand, the metric with the dt dr term still respects spatial isotropy and homogeneity. It is easy to check it, just set dt=0, and the spatial section you are left with will be homogeneous and isotropic. Not sure what you mean by "usual coordinates", what I meant are the "usual terms" where dtdr term does not feature at all.

No, they will not be. That's my point. Surfaces of constant $T$ in your hypothetical chart will not be the same as surfaces of constant $t$ in the standard FRW chart, because you changed the chart.
It is actually trivial to see it that they will - the moment you put dt=0, it kills both dt^2 and dtdr terms, leaving the same spatial section metrics.

Before you even try this, you need to first demonstrate that a chart which manifests the spatial symmetry, but also has a dt dr term in the metric, even exists. I strongly suspect that no such chart exists.

To put this another way: try working the problem you see from the other end. Try starting with the standard FRW chart, and seeing if you can find a coordinate transformation to another chart that keeps the spatial symmetry manifest but also puts a dt dr term in the metric. I strongly suspect you will not be able to.
Yes, I too have the same suspicions, but doing the other way around kinda assumes the solution which I feel is like back-calculating or reverse-engineering. I would rather keep the more general form/gauge and then eliminate terms to arrive at the special form by symmetry constraints.

PAllen