About global inertial frame in GR - revisited

[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.

 

[cianfa72]

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.

You mean in almost such cases (i.e. spacetime types) in which there are more than one timelike KVF, only one of them is hypersurface orthogonal.

 

[PeterDonis]

You mean in almost such cases (i.e. spacetime types) in which there are more than one timelike KVF, only one of them is hypersurface orthogonal.

No. I mean that in almost all cases where there is a timelike KVF at all in the spacetime, there is only one such KVF. And in most cases where there is a timelike KVF, it is not hypersurface orthogonal--i.e., most stationary spacetimes are not static.

 

[PeterDonis]

Sorry for my poor english: what do you mean with "where there is a timelike KVF at all in the spacetime" ?

If a timelike KVF exists.

 

[cianfa72]

No. I mean that in almost all cases where there is a timelike KVF at all in the spacetime, there is only one such KVF.

So, in almost all cases when a timelike KVF exists, there is actually just/only one. Then it may or may not be hypersurface orthogonal.

 

[PeterDonis]

So, in almost all cases when a timelike KVF exists, there is actually just/only one. Then it may or may not be hypersurface orthogonal.

Yes.

 

[cianfa72]

Another point related to the Frobenius condition in the covector field/differential one-form formulation.

Wald in appendix B.3 claims:

Let $T^*$ a smooth specification of 1-dimensional subspace of one-forms. Then the associated $n-1$ dimensional subspace $W$ of the tangent space at each point admits integral submanifolds if and only if for all $\omega \in T^*$ we have $d\omega = \sum_{\alpha} \mu^{\alpha} \wedge v^{\alpha}$, where each $\mu^{\alpha} \in T^*$

Now each $\mu^{\alpha}$ and $v^{\alpha}$ should be covector field/one-forms. So why they are written like vectors -- i.e. with an upper index instead of a lower index ?

 

[Orodruin]

Another point related to the Frobenius condition in the covector field/differential one-form formulation.

Wald in appendix B.3 claims:Now each $\mu^{\alpha}$ and $v^{\alpha}$ should be covector field/one-forms. So why they are written like vectors -- i.e. with an upper index instead of a lower index ?

$\alpha$ enumerates the one-form. It is not a one-form index.

 

[cianfa72]

$\alpha$ enumerates the one-form. It is not a one-form index.

So in that specific case we've just one one-form, therefore the Frobenius's condition becomes $d\omega=\omega \wedge v$ for a one-form $v$. This condition is equivalent to $\omega \wedge d\omega = 0$.