The spacetime length of finite spacelike intervals

[cianfa72]

Perhaps this is part of what you intended to refer to by the phrase "projecting a 4d space-time into a 3-d space". But to me, that phrase implies that there is just one "3-d space" obtained, not an infinite continuum of them.

That was basically the point to assume a static spacetime not just a stationary one.

 

[PeterDonis]

That was basically the point to assume a static spacetime not just a stationary one.

No, you don't need a static spacetime for the "3-d spaces" to all be identical, geometrically. You just need a stationary spacetime. But @pervect was saying his method applies to FRW spacetime, which is not even stationary, so the "3-d space" at each instant of time is different, geometrically, from all the others. That's what I was referring to.

What a static spacetime gives you that a stationary spacetime does not is that the "3-d spaces" are all orthogonal to the worldlines in the congruence. That is what allows the "orthogonal" projection method to end up giving you those "3-d spaces" as its result. But even if the spacetime is just stationary, there will still be a family of "3-d spaces" that foliate the spacetime and that are all identical geometrically. They just won't be orthogonal to the worldlines in the congruence, so they won't match up with the local 3-surfaces orthogonal to each worldline.

Also see my remarks in response to @pervect about quotient spaces. (Note that the quotient space method of "projection" does not require hypersurface orthogonality, so it works for stationary spacetimes, not just static ones.)

 

[cianfa72]

What a static spacetime gives you that a stationary spacetime does not is that the "3-d spaces" are all orthogonal to the worldlines in the congruence. That is what allows the "orthogonal" projection method to end up giving you those "3-d spaces" as its result. But even if the spacetime is just stationary, there will still be a family of "3-d spaces" that foliate the spacetime and that are all identical geometrically. They just won't be orthogonal to the worldlines in the congruence, so they won't match up with the local 3-surfaces orthogonal to each worldline.

With congruence are you referring to the orbits of (one of) the timelike KVFs of the spacetime supposed to be stationary ?

 

[PeterDonis]

With congruence are you referring to the orbits of (one of) the timelike KVFs of the spacetime supposed to be stationary ?

I'm referring to whatever family of timelike worldlines is being used in the "projection" construction. If the spacetime is stationary, then the obvious congruence to use is the orbits of the timelike KVF, yes. But in FRW spacetime, for example, there is no timelike KVF, but there is still an "obvious" congruence to use, namely the family of worldlines of "comoving" observers, who always see the universe as homogeneous and isotropic. Even though the spacetime is not static, this congruence is still hypersurface orthogonal, however; the family of spacelike 3-surfaces of constant FRW coordinate time is orthogonal to the congruence of "comoving" worldlines. So if we pick any event on any particular comoving worldline, and pick out the spacelike vectors orthogonal to the 4-velocity of the worldline at that event, and do the "exponential map" construction starting with those spacelike vectors, we will end up with the entire spacelike 3-surface of constant FRW coordinate time that contains our chosen event.

 

[cianfa72]

Sorry to resume this thread...I have been reading it again and again...

But in FRW spacetime, for example, there is no timelike KVF, but there is still an "obvious" congruence to use, namely the family of worldlines of "comoving" observers, who always see the universe as homogeneous and isotropic. Even though the spacetime is not static, this congruence is still hypersurface orthogonal, however; the family of spacelike 3-surfaces of constant FRW coordinate time is orthogonal to the congruence of "comoving" worldlines.

Therefore in FRW spacetime even though the congruence of worldlines of "comoving" observers is hypersurface orthogonal however the 3-metric of these "3-d" family of spacelike hypersurfaces is not the same (since that congruence of "comoving" observers is not a KVF congruence).

Coming back to what you said...Consider a stationary spacetime: that does mean there exist (at least) one timelike KVF. Take the congruence associated to a chosen timelike KVF. Using the 'exponentiating thing' build the spacelike hypersurface othogonal to a chosen worldline in that congruence starting from an event P on it (by the way: the 3d spacelike hypersurface orthogonal to a given timelike vector at event P is unique). Consider the intersection of that hypersurface with another worldline member of the chosen congruence. Call this event Q.

Now only if spacetime is static then the following will be true at the same time:

1. the 'exponentiating thing' built at Q will give the same spacelike hypersurface we started with at P
2. there will be a family of spacelike hypersurfaces orthogonal to the timelike KVF congruence that foliate the spacetime and includes the spacelike hypersurface we stared with
3. the "3-d" metric on each spacelike hypersurface of the family will be the same (i.e. all of them will have the same 3-d spatial geometry)

 

[pervect]

But the "space" projected changes with "time". So you can't obtain a single "3-d space" this way; you can only obtain an infinite continuum of "3-d spaces", each labeled with a "time".

Yes, I agree. But it's unavoidable, due to the relativity of simultaneity. This may be unintuition, but even in SR, the relativity of simultaneity implies that observers with different states of motion have different notions of space and time.

Also, I should add, only rigid congruences are really equivalent to rigid rulers. And the congruence of comoving observers above isn't rigid, so it doesn't correspond naturally to any system of rigid rulers. But it corresponds to the commonly defined cosmological distance that's the standard in papers which describe the cosmological distance to far away events.

The isotropic CMB congruence not the only congruence one might use - the Fermi Walker congruences assocaited with the worldline of some particular observer are useful in the right circumstances, basically "small" areas around the worldline of some observer as well. So they're useful to explain why Woody Allen's mother was right when she said "Brooklyn is not expanding".
[/quote]

Perhaps this is part of what you intended to refer to by the phrase "projecting a 4d space-time into a 3-d space". But to me, that phrase implies that there is just one "3-d space" obtained, not an infinite continuum of them.

I didn't mean to imply such a thing - it can['t be done :(.

I'll have to study the rest of your response more closely, it will require some thought.

 

[PeterDonis]

Therefore in FRW spacetime even though the congruence of worldlines of "comoving" observers is hypersurface orthogonal however the 3-metric of these "3-d" family of spacelike hypersurfaces is not the same (since that congruence of "comoving" observers is not a KVF congruence).

Correct. The scale factor is different on each spacelike hypersurface.

only if spacetime is static then the following will be true at the same time:

1. the 'exponentiating thing' built at Q will give the same spacelike hypersurface we started with at P
2. there will be a family of spacelike hypersurfaces orthogonal to the timelike KVF congruence that foliate the spacetime and includes the spacelike hypersurface we stared with
3. the "3-d" metric on each spacelike hypersurface of the family will be the same (i.e. all of them will have the same 3-d spatial geometry)

Correct. In fact we can make the stronger statement that only in a static spacetime will 1. and 2. be true at all. (In a stationary spacetime that is not static, you can still find a family of spacelike hypersurfaces for which 3. is true, but no such family of hypersurfaces will satisfy 1. or 2.)

 

[cianfa72]

Thanks all