I About global inertial frame in GR - revisited

[cianfa72]

TL;DR Summary

Fermi Normal hypersurfaces orthogonal to the FLRW congruence of comoving observer's worldlines

Hi, reading this old thread I'd like a clarification about the following:

No, for FLRW solutions, the hypersurfaces orthogonal to the congruence of comoving observers have the property that geodesics of the surface are not geodesics of the spacetime. This follows immediately from the fact that the Fermi Normal surface at an event on a comoving world line is defined by the collection of spacetime orthogonal geodesics, and that this is not the same as a slice of constant cosmological time. But the cosmological slice is tangent to the Fermi slice at each event on the origin comoving observer for the comoving slice. Since there can be exactly one geodesic through a given point with given tangent (trivially follows from parallel transport definition of geodesic), the cosmological slice cannot be made up of geodesics of the spacetime, since these define the Fermi Normal slice.

Fermi Normal hypersurface at an event on a comoving FLRW worldline is defined by the collection of spacetime orthogonal geodesics. Such geodesics should be spacelike since they are orthogonal to the timelike comoving worldlines at each event along them. On the other hand since FLRW comoving congruence's worldlines are hypersurface orthogonal then such spacelike hypersufaces should be the same as the Fermi Normal ones, I believe.

Am I wrong ? Thank you.

 

[Orodruin]

You are wrong. The Fermi Normal coordinates are guaranteed to be orthogonal to the cosmological time direction only in the point used to define them whereas the hypersurfaces of constant cosmological time are orthogonal to the cosmological time direction everywhere by construction.

Edit: Compare with the following example in Riemannian geometry. Take polar coordinates on $\mathbb R^2$. The curves of constant $r$ are everywhere orthogonal to the radial direction. The normal coordinates based on a particular point has the circle’s tangent as the coordinate line. This is orthogonal to the radial direction only at the point used to define the normal coordinates and nowhere else.

Edit 2: See also https://www.physicsforums.com/insights/coordinate-dependent-statements-expanding-universe/ In particular section 3 where the coordinate transformation is constructed explicitly to leading non-trivial order.

 

[cianfa72]

The Fermi Normal coordinates are guaranteed to be orthogonal to the cosmological time direction only in the point used to define them whereas the hypersurfaces of constant cosmological time are orthogonal to the cosmological time direction everywhere by construction

Ah ok, so spacelike hypersurfaces of constant cosmological time cannot be made of spacelike geodesics of spacetime.

 

[Orodruin]

Ah ok, so spacelike hypersurfaces of constant cosmological time cannot be made of spacelike geodesics of spacetime.

Well, it depends on the spacetime. Minkowski space is perfectly conforming to RW coordinates with $a = 1$.

 

[cianfa72]

Well, it depends on the spacetime. Minkowski space is perfectly conforming to RW coordinates with $a = 1$.

Yes, in that thread we were talking about FRW spacetime. Which are the RW coordinates in Minkowski spacetime ?

 

[Orodruin]

Yes, in that thread we were talking about FRW spacetime. Which are the RW coordinates in Minkowski spacetime ?

Minkowski space is a FLRW spacetime (with no energy content).

 

[cianfa72]

Minkowski space is a FLRW spacetime (with no energy content).

Sorry, I have no idea what the RW coordinates are...

 

[Orodruin]

A Robertson-Walker spacetime is a spacetime that is foliated by homogeneous and isotropic spatial hypersurfaces such that you can construct coordinates where the metric takes the form
$$
ds^2 = dt^2 -a(t)^2 d\Sigma^2
$$
with $d\Sigma^2$ being the metric of the homogeneous and isotropic spatial surfaces. These are the Robertson-Walker coordinates.

A RW spacetime that additionally satisfies the Friedmann equations for some energy content would be a FLRW spacetime.

 

[cianfa72]

Ah ok, so your point is that Minkowski spacetime is a RW spacetime with $a(t)=1$. Therefore for Minkowski spacetime the hypersurfaces of RW constant cosmological time are actually hypersurfaces made up of spacelike geodesics of spacetime (i.e. spacelike hypersurfaces of constant cosmological time in RW coordinates are exactly the Fermi Normal hypersurfaces built at events along RW coordinate time congruence).

Right ?

 

[PeterDonis]

Minkowski space is perfectly conforming to RW coordinates with $a = 1$.

By this I assume you are referring to the trivial case where $a = 1$ for all $t$, meaning that both Friedmann equations are simply $0 = 0$ and there is no actual expansion, correct? For this case "RW coordinates" are the same as ordinary inertial coordinates, whose spacelike surfaces of constant $t$ are of course just Euclidean 3-spaces spanned by geodesics of the spacetime.

I bring this up because there is another FRW solution on Minkowski spacetime, or more precisely on the future light cone of the origin in Minkowski spacetime, the Milne model:

https://en.wikipedia.org/wiki/Milne_model

This solution also has zero energy content, but it is expanding: the comoving worldlines are all possible timelike geodesics coming from the origin. The surfaces of constant $t$ in this model are hyperbolas of constant timelike interval from the origin, which are not geodesics of the spacetime.

 

[Orodruin]

By this I assume you are referring to the trivial case where a=1 for all t, meaning that both Friedmann equations are simply 0=0 and there is no actual expansion, correct? For this case "RW coordinates" are the same as ordinary inertial coordinates, whose spacelike surfaces of constant t are of course just Euclidean 3-spaces spanned by geodesics of the spacetime.

Yes.

I bring this up because there is another FRW solution on Minkowski spacetime, or more precisely on the future light cone of the origin in Minkowski spacetime, the Milne model:

https://en.wikipedia.org/wiki/Milne_model

This solution also has zero energy content, but it is expanding: the comoving worldlines are all possible timelike geodesics coming from the origin. The surfaces of constant t in this model are hyperbolas of constant timelike interval from the origin, which are not geodesics of the spacetime.

Indeed. I also brought up the Milne model in the Insight I linked earlier. It is a great example of what appears as cosmological redshift in one coordinate system is just your regular Doppler shift in another.

 

[cianfa72]

BTW, the take-home of that thread was that even in some curved spacetime we can define a global inertial coordinate chart. E.g. in Godel spacetime there is a timelike geodesic congruence that has zero expansion and shear. Therefore we can assign fixed spatial coordinates to each worldline in the congruence defining a global chart in which the congruence's timelike geodesics are "at rest". Morever the condition of zero expansion and shear for the congruence does mean "zero geodesic deviation for those geodesics".

The above conditions seem to met the requirements of a global inertial coordinate chart (see the definition in post#9 there).

 

[PeterDonis]

BTW, the take-home of that thread was that even in some curved spacetime we can define a global inertial coordinate chart.

No, it wasn't. See below.

E.g. in Godel spacetime there is a timelike geodesic congruence that has zero expansion and shear.

But not zero vorticity, so the chart based on this congruence is not inertial. This was brought up in the previous thread, but you apparently have forgotten it. I suggest re-reading the caution I gave in post #79 of the previous thread.

 

[Ibix]

But not zero vorticity, so the chart based on this congruence is not inertial. This was brought up in the previous thread, but you apparently have forgotten it. I suggest re-reading the caution I gave in post #79 of the previous thread.

The consequence of that being that planes constructed from infinitesimal planes locally orthogonal to the congruence don't close, similar to the issues with naively constructing a simultaneity convention from local MCIFs on a rotating disc in flat spacetime?

 

[PeterDonis]

The consequence of that being that planes constructed from infinitesimal planes locally orthogonal to the congruence don't close, similar to the issues with naively constructing a simultaneity convention from local MCIFs on a rotating disc in flat spacetime?

I believe so, yes.

 

[cianfa72]

Sorry, maybe I missed the point. We said that a global inertial chart requires that accelerometers at rest in it measure zero proper acceleration (i.e. their paths are timelike geodesics). Then we added in post#9 the constraint that the distance between those inertial accelerometers doesn't change (i.e. zero expansion and shear for those accelerometer's geodesic timelike paths).

What is the purpose of the further constraint of zero vorticity for the above congruence (i.e. hypersurface orthogonal condition for that congruence) ? Thanks in advance.

 

[PeterDonis]

We said that a global inertial chart requires that accelerometers at rest in it measure zero proper acceleration (i.e. their paths are timelike geodesics). Then we added in post#9 the constraint that the distance between those inertial accelerometers doesn't change (i.e. zero expansion and shear for those accelerometer's geodesic timelike paths).

Yes, and in post #40 I added the zero vorticity requirement.

What is the purpose of the further constraint of zero vorticity

Nonzero vorticity means that there will be "fictitious forces" present--the worldlines of at least some freely falling objects will not be straight lines in the chart. In the vast majority of cases, this also means that the chart itself will not be geodesic--worldlines at rest in the chart will have nonzero proper acceleration, so they won't meet the geodesic requirement anyway. But in the case of Godel spacetime, there is a geodesic congruence that still has nonzero vorticity and for which the spatial paths of geodesics not in the congruence will not be straight lines in the corresponding chart. So to be sure we are specifying an inertial chart, we need to also specify zero vorticity for the congruence the chart is based on.

 

[PeterDonis]

i.e. hypersurface orthogonal condition for that congruence

@Ibix in post #14 brought up another relevant point, which is that if the congruence is not hypersurface orthogonal, it will be impossible to construct a global chart using it, because the surfaces of "constant time" will not close. The example he was implicitly referring to was the Born chart on Minkowski spacetime.

 

[cianfa72]

But in the case of Godel spacetime, there is a geodesic congruence that still has nonzero vorticity and for which the spatial paths of geodesics not in the congruence will not be straight lines in the corresponding chart.

You mean the timelike geodesic congruence of worldlines at rest in standard coordinate chart in Godel spacetime. In this coordinate chart the spatial paths of timelike geodesics not in the congruence are not straight lines, however.

 

[PeterDonis]

You mean the timelike geodesic congruence of worldlines at rest in standard coordinate chart in Godel spacetime.

Yes.

In this coordinate chart the spatial paths of timelike geodesics not in the congruence are not straight lines, however.

Yes.

 

[cianfa72]

So to be sure we are specifying an inertial chart, we need to also specify zero vorticity for the congruence the chart is based on.

So this further constraint make sure that every timelike geodesic either in the timelike geodesic congruence the coordinate chart is based on or not in it, will always have straight spatial path in that coordinate chart.

 

[PeterDonis]

So this further constraint make sure that every timelike geodesic either in the geodesic congruence the coordinate chart is based on or not in it, will always have straight spatial path in that coordinate chart.

A straight worldline, not just a straight spatial path.

 

[cianfa72]

A straight worldline, not just a straight spatial path.

Sorry, in post#17 you talked about spatial paths of timelike geodesics not in the congruence in that coordinate chart.

 

[PeterDonis]

Sorry, in post#17 you talked about spatial paths of timelike geodesics in that coordinate chart.

I said both "worldlines" and "spatial paths" in that post. But actually only "worldlines" is necessary, because if they are straight lines, the spatial paths (the projection of the worldline into a spacelike surface of constant time) will also be straight lines.

 

[cianfa72]

But actually only "worldlines" is necessary, because if they are straight lines, the spatial paths

So this basically amounts to check if the spatial coordinates of the timelike geodesics are linearly dependent with constant coefficients on the coordinate time along the path in spacetime.

 

[PeterDonis]

So this basically amounts to check if the spatial coordinates of the timelike geodesics are linearly dependent with constant coefficients on the coordinate time along the path in spacetime.

Yes.

 

[cianfa72]

I've some difficulty to grasp the concept of coordinate time. From a physical point of view, since it corresponds to timelike paths, it should be related to the proper time along those timelike paths "at rest" in the coordinate chart.

In other words we can calculate the chart's coordinate time of an event that occurs along the timelike path at rest in the chart from the reading of the clock "carried" by its implied observer.

 

[Orodruin]

I've some difficulty to grasp the concept of coordinate time. From a physical point of view, since it corresponds to timelike paths, it should be related to the proper time along those timelike paths "at rest" in the coordinate chart.

In other words we can calculate the chart's coordinate time of an event that occurs along the timelike path at rest in the chart from the reading of the clock "carried" by its implied observer.

Generally, coordinate time $t$ is only related to the proper time $\tau$ of an observer with fixed spatial coordinates as $\tau = f(t)$ where $f$ is some monotonically increasing function with $f’(t)$ is the square root of the factor in front of $dt^2$ in the line element. However, in RW coordinates that factor is one and so $\tau = t$, ie, the coordinate time is the proper time of the comoving observers. This is by construction.

At the same time, for example, this is not the case for stationary observers in a Schwarzschild spacetime.

 

[cianfa72]

Generally, coordinate time $t$ is only related to the proper time $\tau$ of an observer with fixed spatial coordinates as $\tau = f(t)$ where $f$ is some monotonically increasing function with $f’(t)$ is the square root of the factor in front of $dt^2$ in the line element.

So, how do other observers at rest in the coordinate chart with fixed spatial coordinates "assign" coordinate time to events occurring along their worldlines ?

 

[Ibix]

In other words we can calculate the chart's coordinate time of an event that occurs along the timelike path at rest in the chart from the reading of the clock "carried" by its implied observer.

You can do, but the calculation may depend on your position and time.

A simple analogy on the surface of the Earth: if you start on the equator and move north, your latitude coordinate is roughly your distance travelled divided by $111\ \mathrm{km/°}$. On the other hand, if you start on the Greenwich meridian and move only East then your longitude coordinate is roughly $111\sec(\mathrm{latitude})\ \mathrm{km/°}$ (using the geographer's definition of latitude that is zero at the equator).

Just as these coordinates are related to $\int\sqrt{g_{ab}u^au^b}dt$, but not necessarily trivially, so is coordinate time.

 

[Orodruin]

So, how do other observers at rest in the coordinate chart with fixed spatial coordinates "assign" coordinate time to events occurring along their worldlines ?

That would depend on the coordinates being used. There is nothing actually physical about the concept of a coordinate time. It only holds physical significance in relation to whatever process is used to assign it.

 

[cianfa72]

Just as these coordinates are related to $\int\sqrt{g_{ab}u^au^b}dt$, but not necessarily trivially, so is coordinate time.

ok, you're saying that from the spatial position (spatial coordinates) of an event in the chart and the elapsed proper time along a timelike path joining the (0,0,0,0) event with the given event we can calculate back the coordinate time $t$ for it.

 

[cianfa72]

It only holds physical significance in relation to whatever process is used to assign it.

That was my point. A coordinate chart for spacetime (or for the Earth surface) is defined from the process employed to assign those coordinates to events (points on the Earth surface as in the example from @Ibix ).

 

[Ibix]

ok, you're saying that from the spatial position (spatial coordinates) of an event in the chart and the elapsed proper time along the timelike path with fixed spatial coordinates joining the t=0 hypersurface with the given event we can calculate back the coordinate time $t$ for it.

With the bolded changes above, yes.

 

[cianfa72]

ok, you're saying that from the spatial position (spatial coordinates) of an event in the chart and the elapsed proper time along the timelike path with fixed spatial coordinates joining the t=0 hypersurface with the given event we can calculate back the coordinate time for it.

ok, that's basically like moving on the Earth only East along a parallel with a fixed latitude of an amount of longitude starting from Greenwich meridian (all points on Greenwich meridian have longitude 0 like $t=0$ hypersuface events).

 

[cianfa72]

However, in RW coordinates that factor is one and so $\tau = t$, ie, the coordinate time is the proper time of the comoving observers. This is by construction.

Therefore in this case the proper time $\tau$ along each of the comoving observer paths is equal to the coordinate time $t$. That means clocks carried by comoving observers are synchronized on a specific spacelike hypersurace (the $t=\tau=0$ hypersurface).

From a physical point of view what does it mean ?

 

[Orodruin]

From a physical point of view what does it mean ?

Nothing. It just means you picked a particular simultaneity convention in which the space foliation consists of homogeneous and isotropic spaces based on the symmetry of your spacetime.

 

[cianfa72]

Nothing. It just means you picked a particular simultaneity convention in which the space foliation consists of homogeneous and isotropic spaces based on the symmetry of your spacetime.

You mean the spatial metric on the foliation's spacelike hypersurfaces (associated to that particular simultaneity convention) is actually homogeneous and isotropic.

Edit: I would say there exists a simultaneity convention for the timelike congruence of comoving observers such that the spatial metric on each spacelike hypersurface of constant coordinate time $t$ is homogeneous and isotropic.

 

[cianfa72]

Another point related to this. Consider a stationary spacetime -- i.e. there is at least one timelike KVF with a spacelike foliation associated to it. Then take the one-form $\omega$ you get contracting the tensor metric with the KVF evaluated at each point.

Now if $\omega \wedge d\omega=0$ everywhere, then there exists a function $f$ such that its level hypersurfaces are orthogonal to the KVF at each point.

In the adapted coordinate chart in which the timelike direction $\partial t$ is the same as the direction of KVF at each point (and spacelike directions belong to the $f=const$ spacelike hypersurfaces), are the metric coefficients still indipendent from $t$ ?

 

[PeterDonis]

In the adapted coordinate chart in which the timelike direction $\partial t$ is the same as the direction of KVF at each point (and spacelike directions belong to the $f=const$ spacelike hypersurfaces), are the metric coefficients still indipendent from $t$ ?

I don't know what you mean by "still". The definition of "adapted coordinate chart" is the chart in which $\partial_t$ points in the direction of the KVF at each point, and it is trivial to show that in this chart, the metric coefficients are independent of $t$. This is true regardless of how you choose the other coordinates.

 

[cianfa72]

it is trivial to show that in this chart, the metric coefficients are independent of $t$. This is true regardless of how you choose the other coordinates.

Have you a proof of this? Thanks.

 

[PeterDonis]

Have you a proof of this?

As I said, it is trivial. I suggest trying to work it out for yourself. Start by writing down Killing's equation in the adapted chart. (Or, for a quicker method, look at the Lie derivative definition of a Killing vector field and consider what it reduces to in the adapted chart.)

 

[cianfa72]

look at the Lie derivative definition of a Killing vector field and consider what it reduces to in the adapted chart.

By definition $\mathcal L_{\partial_t} \, {g}=0$ in any adapted chart (i.e. in any chart where $\partial_t$ points along the KVF at each point). Pick any other coordinates for spacetime. Then the Lie derivative of metric tensor $g$ along $\partial_t$ is just $$(\partial_t \, g_{ab})dx^a \otimes dx^b$$ To get the null tensor all partial derivatives w.r.t. coordinate $t$ of metric components must vanish. Therefore all metric components in any adapted coordinate chart must be indipendent from $t$.

 

[PeterDonis]

By definition $\mathcal L_{\partial_t} \, {g}=0$ in any adapted chart

Yes.

Pick any other coordinates for spacetime.

Why would you do that? You're trying to prove something about adapted coordinates.

Then the Lie derivative of metric tensor $g$ along $\partial_t$

In other coordinates the KVF is not $\partial_t$. There might not even be a $t$ coordinate in other coordinates.

You are making this way too complicated. In an adapted coordinate chart what does $\mathcal{L}_{\partial_t} \ {g}$ reduce to? (Hint: it's the $\partial_t$ of something. What?)

 

[cianfa72]

Why would you do that? You're trying to prove something about adapted coordinates.

Sorry, maybe I was unclear: I meant the 3 coordinates other than the $t$ coordinate pointing along the KVF at each point.

In an adapted coordinate chart what does $\mathcal{L}_{\partial_t} \ {g}$ reduce to? (Hint: it's the $\partial_t$ of something. What?)

As said before, in any adapted chart it reduces to derivatives $\partial_t$ of the metric tensor components $g_{ab}$. Hence, to get the null tensor, the metric tensor components in any adapted chart must be indipendent from $t$.

 

[PeterDonis]

I meant the 3 coordinates other than the coordinate pointing along the KVF at each point.

Ok. But you don't actually have to assume anything about the other coordinates at all.

As said before, in any adapted chart it reduces to derivatives $\partial_t$ of the metric tensor components $g_{ab}$.

With the clarification above, I now see that this is what you were saying, yes. And since $\mathcal{L}_{\partial_t} \ {g} = 0$, we must have $\partial_t \ g_{ab} = 0$.

 

[cianfa72]

But you don't actually have to assume anything about the other coordinates at all.

Yes, definitely.

Therefore, said that a static spacetime is also stationary, there exists at least a timelike KVF. Then if we take the covector field/one-form $\omega$ resulting by contracting the tensor metric $g$ with the timelike KVF at each point, we get $\omega \wedge d\omega =0$ -- i.e. $\omega$ results to be integrable (there are functions $f$ and $t$ such that $\omega = fdt$).

Now the values of function $t$ on the spacelike hypersurfaces $t=const$ can be taken as timelike coordinate time $t$ of the adapted coordinate chart -- i.e. the coordinate chart adapted to the timelike KVF with spacelike coordinates along the KFV's orthogonal spacelike hypersurfaces foliating the spacetime.

 

[PeterDonis]

Therefore, said that a static spacetime is also stationary, there exists at least a timelike KVF. Then if we take the covector field/one-form $\omega$ resulting by contracting the tensor metric $g$ with the timelike KVF at each point, we get $\omega \wedge d\omega =0$ -- i.e. $\omega$ results to be integrable (there are functions $f$ and $t$ such that $\omega = fdt$).

Now the values of function $t$ on the spacelike hypersurfaces $t=const$ can be taken as timelike coordinate time $t$ of the adapted coordinate chart -- i.e. the coordinate chart adapted to the timelike KVF with spacelike coordinates along the KFV's orthogonal spacelike hypersurfaces foliating the spacetime.

Yes, all of this follows from the fact that the timelike KVF in a static spacetime is hypersurface orthogonal.

 

[cianfa72]

Yes, all of this follows from the fact that the timelike KVF in a static spacetime is hypersurface orthogonal.

But in a static spacetime, every timelike KVF is hypersurface orthogonal or there is at least one of such timelike KVF ?

 

[PeterDonis]

But in a static spacetime, every timelike KVF is hypersurface orthogonal or there is at least one of such timelike KVF ?

In almost all such cases, there is only one such KVF. The only example that comes to mind where there is more than one is flat Minkowski spacetime.