The spacetime length of finite spacelike intervals

[cianfa72]

The $\Sigma_t$ spacelike hypersurface in that link is not "assumed", it is constructed. But it is constructed only from a single observer's 4-velocity at a given event, not from a congruence of worldlines. The construction being described in that link basically amounts to: treat the 4-velocity at the given event as the timelike basis vector of an inertial frame; then the spacelike hypersurface is just the "surface of constant time" in the same inertial frame that contains the given event.

I believe you were referring to the following sentence there:

So, let me define space-like geodesic for my purpose as follows, which will be a local notion: Take an observer, which simply shall be a time-like vector $U$ at an event P of space-time M. Let $V$ be the orthogonal complement of $U$ (a three-dimensional space of space-like directions at P). Let $\Sigma$ be the image of a small neighborhood of 0 in $V$ under the exponential mapping (that is the set of events that are connected to P by small geodesics that are orthogonal to u at P).

$U$ should be the observer 4-velocity at the given event P you were talking about and $V$ the orthogonal complement of $U$ in P (basically the set of spacelike vectors in the tangent space at P). $\Sigma$ is defined as the image of a neighborhood of the zero vector under the exponential map however it has actually just a local extent, I believe.

Is this the inertial frame the $\Sigma_t$ spacelike hypersurface represents the "surface of constant time" of ? Thanks.

 

[PeterDonis]

I believe you were referring to the following sentence there:

So, let me define space-like geodesic for my purpose as follows, which will be a local notion: Take an observer, which simply shall be a time-like vector $U$ at an event P of space-time M. Let $V$ be the orthogonal complement of $U$ (a three-dimensional space of space-like directions at P). Let $\Sigma$ be the image of a small neighborhood of 0 in $V$ under the exponential mapping (that is the set of events that are connected to P by small geodesics that are orthogonal to u at P).

Yes.

$U$ should be the observer 4-velocity at the given event P you were talking about and $V$ the orthogonal complement of $U$ in P (basically the set of spacelike vectors in the tangent space at P).

Yes.

$\Sigma$ is defined as the image of a neighborhood of the zero vector under the exponential map however it has actually just a local extent, I believe.

That depends on the spacetime. The vectors in the tangent space are local; but exponentiating them basically means "extend geodesic curves in the directions of the vectors as far as you can". That is not limited to a local region of spacetime.

What will happen in a general curved spacetime when you do the exponential map thing is that the local coordinate chart you are using to describe the curves you produce will no longer work. But in flat Minkowkski spacetime, this problem does not arise; you can do the exponential map thing all the way out to infinity and there will be no problem at all.

Is this the inertial frame the $\Sigma_t$ spacelike hypersurface represents the "surface of constant time" of ?

In flat spacetime, yes, what the construction described amounts to is considering the 4-velocity $U$ to be the timelike basis vector of an inertial frame at the chosen event, and then constructing the spacelike hypersurface of constant time in that frame that contains the chosen event. The "exponential map" thing in this case just means "draw all possible geodesics radiating out in the three spacelike dimensions from the chosen event that are orthogonal to $U$".

 

[cianfa72]

That depends on the spacetime. The vectors in the tangent space are local; but exponentiating them basically means "extend geodesic curves in the directions of the vectors as far as you can". That is not limited to a local region of spacetime.

What will happen in a general curved spacetime when you do the exponential map thing is that the local coordinate chart you are using to describe the curves you produce will no longer work. But in flat Minkowkski spacetime, this problem does not arise; you can do the exponential map thing all the way out to infinity and there will be no problem at all.

ok but in the scenario described there on that link, the spacetime in not flat thus even though that exponential map thing cannot be extended all the way out nevertheless it can be extended at least over a "finite" region starting from the given event.

Therefore I think the entire procedure described there actually makes sense only if the two chosen events are such that the exponentiating of the spacelike vectors belonging to the tangent space at the given (first) event is actually able to produce (spacelike) geodesic curves that extend up to the second one.

 

[PeterDonis]

in the scenario described there on that link, the spacetime in not flat thus even though that exponential map thing cannot be extended all the way out nevertheless it can be extended at least over a "finite" region starting from the given event.

No, that's not correct. The exponential map thing, as I said, goes as far as it can go while still staying in the spacetime. That is true even if the spacetime is curved. For example, in Schwarzschild spacetime, you can do the exponential map thing for spacelike vectors orthogonal to the 4-velocity of a "hovering" observer all the way out to infinity.

What will happen in a curved spacetime, as I said, is that you will not be able to use a local inertial coordinate chart to describe the spacelike curves you get by doing the exponential map thing, beyond a small local region of spacetime in which the Minkowski metric in that local inertial coordinate chart is a good enough approximation to the actual metric.

I think the entire procedure described there actually makes sense only if the two chosen events are such that the exponentiating of the spacelike vectors belonging to the tangent space at the given (first) event is actually able to produce (spacelike) geodesic curves that extend up to the second one.

The limiting factor here is not any limitation of how far the exponential map extends. It is whether any of the spacelike geodesics you get by doing the exponential map starting from the first event actually contain the second event. In other words, it's not enough for the two events to be spacelike separated. The second event has to actually lie in the particular spacelike hypersurface you get by doing the exponential map thing starting from the first event. You will get some spacelike hypersurface by doing that, and it will extend all the way across the spacetime, but it might not contain the second event.

 

[cianfa72]

What will happen in a curved spacetime, as I said, is that you will not be able to use a local inertial coordinate chart to describe the spacelike curves you get by doing the exponential map thing, beyond a small local region of spacetime in which the Minkowski metric in that local inertial coordinate chart is a good enough approximation to the actual metric.

Maybe I misunderstood the use of term "inertial frame" in your post #32. It should be in the sense of "local inertial frame" at that given event I believe.

 

[PeterDonis]

Maybe I misunderstood the use of term "inertial frame" in your post #32. It should be in the sense of "local inertial frame" at that given event I believe.

If the spacetime is curved, then of course "inertial frame" has to mean "local inertial frame" since that's the only kind there is in a curved spacetime.

However, that does not mean the exponential map construction is limited to the local inertial frame. Within the local inertial frame, the construction will give the natural "surface of constant time" in local inertial coordinates on the local inertial frame. But the construction itself will extend all the way across the spacetime, as I said. It just won't be describable in the local inertial coordinates outside the local inertial frame.

There is also no guarantee, either in flat or curved spacetime, that the spacelike hypersurface generated by the construction, even though it is orthogonal to the 4-velocity $U$ at the chosen event, will be orthogonal to any other 4-velocity vectors of other worldlines in the same congruence.

 

[cianfa72]

In flat spacetime, yes, what the construction described amounts to is considering the 4-velocity $U$ to be the timelike basis vector of an inertial frame at the chosen event, and then constructing the spacelike hypersurface of constant time in that frame that contains the chosen event. The "exponential map" thing in this case just means "draw all possible geodesics radiating out in the three spacelike dimensions from the chosen event that are orthogonal to $U$".

Sorry for still bother you...I'm in trouble with the last sentence "draw all possible geodesics radiating out in the three spacelike dimensions from the chosen event that are orthogonal to $U$". To me it lacks of physical interpretation.

Which could be (if any !) the physical "process/procedure" to follow in order to implement it ?

Thanks in advance

 

[PeterDonis]

I'm in trouble with the last sentence "draw all possible geodesics radiating out in the three spacelike dimensions from the chosen event that are orthogonal to $U$". To me it lacks of physical interpretation.

Which could be (if any !) the physical "process/procedure" to follow in order to implement it ?

There isn't a direct physical process to draw a spacelike curve. It's something you do in a mathematical model, not in the actual world. The point of the operation described is to pick out a particular spacelike hypersurface in the model that corresponds with some set of events of interest in the real world.

 

[cianfa72]

There isn't a direct physical process to draw a spacelike curve. It's something you do in a mathematical model, not in the actual world. The point of the operation described is to pick out a particular spacelike hypersurface in the model that corresponds with some set of events of interest in the real world.

I try again from a basic point of view...

Take the event P in physical spacetime (we know it does exist regardless the mathematical model !). Now consider the set of all paths followed by massive objects radiating out of P in all possible directions having all possible velocities (up to the speed of light). They should represent all timelike paths starting from event P in the spacetime mathematical model. Add now to it the set of all possible light rays paths radiating out of P in all possible direction. This one basically represents the (future) light-cone at event P.

From my understanding all other spacetime's events not reachable from P that way (via timelike or light-like paths from P as defined above) are defined as spacelike separated from P and furthermore there exist spacelike paths joining them from P. Events belonging to these spacelike paths are themselves spacelike separated from P.

Does it make sense ? Thanks.

 

[PeterDonis]

Does it make sense ?

Yes, all of this is correct.

 

[DrGreg]

Does it make sense ? Thanks.

Well, actually to be precise you'd also have to exclude the past light cone as well, i.e. all paths, from the past, followed by massive objects or light rays arriving at P.

 

[cianfa72]

Well, actually to be precise you'd also have to exclude the past light cone as well, i.e. all paths, from the past, followed by massive objects or light rays arriving at P.

Sure, definitely.

The spacelike hypersurface in that link is not "assumed", it is constructed. But it is constructed only from a single observer's 4-velocity at a given event, not from a congruence of worldlines. The construction being described in that link basically amounts to: treat the 4-velocity at the given event as the timelike basis vector of an inertial frame; then the spacelike hypersurface is just the "surface of constant time" in the same inertial frame that contains the given event.

Thus, based on the above posts, exponentiating the spacelike vectors belonging to the orthogonal complement at P of the single observer 4-velocity we construct the spacelike hypersurface $\Sigma_0$ we are interested in.

Take now another event Q that lies on that hypersurface $\Sigma_0$. To perform the procedure we were talking of we need to find out (if any !) a congruence of timelike paths intersecting $\Sigma_0$ in P and Q such that they are orbits of spacetime timelike Killing vector field (we assume a stationary spacetime).

By very definition, moving along one of them, the spacetime metric does not change. This way we can actually construct a family of hypersurfaces $\Sigma_t$ with the same fixed spatial metric. Now starting from the event "corresponding" of P in a $\Sigma_t$ through the congruence, we are actually able to build a chain of rods from it to the event Q in $\Sigma_0$. The shortest rods'chain between them should give indirectly the length of the spacelike geodesic joining the event P with the event Q that lies on $\Sigma_0$.

Does it sound right ?

 

[PeterDonis]

exponentiating the spacelike vectors belonging to the orthogonal complement at P of the single observer 4-velocity we construct the spacelike hypersurface $\Sigma_0$ we are interested in.

That we might be interested in, depending on, well, what we are interested in.

See further comments below.

Take now another event Q that lies on that hypersurface $\Sigma_0$. To perform the procedure we were talking of we need to find out (if any !) a congruence of timelike paths intersecting $\Sigma_0$ in P and Q such that they are orbits of spacetime timelike Killing vector field (we assume a stationary spacetime).

I've lost track of "the procedure we were talking of". But generally, we don't look for a congruence of timelike worldlines; we will generally already have one that is either specified in or can be deduced from the problem statement. The question is whether the spacelike hypersurface $\Sigma_0$, which by construction is orthogonal to the congruence at P (because by construction it is orthogonal to the 4-velocity at P of the worldline of the congruence that passes through P), is orthogonal to the congruence at Q. It might be, or it might not; this will depend on the congruence and the spacetime geometry.

It is also worth noting that just specifying a 4-velocity at P is not sufficient to specify a single worldline at P. It is sufficient to specify a single geodesic worldline at P, but there will be an infinite number of different non-geodesic worldlines passing through P that have the same 4-velocity at P. That is why we generally need to have a specific congruence of worldlines specified, or deducible from, the problem statement.

In fact, it is not even always true that a given 4-velocity at P is sufficient to specify (assuming one exists) a single worldline at P that is an orbit of a Killing vector field. For example, in Minkowski spacetime, a given 4-velocity at P does specify a single geodesic worldline passing through P, which also happens to be the orbit of a Killing vector field; but there are also an infinite number of non-geodesic worldlines with the same 4-velocity at P, namely, the worldlines of uniform proper acceleration $a$, where $0 < a < \infty$; and all of those are also orbits of Killing vector fields (different ones in each case). I believe the same holds in de Sitter and anti-de Sitter spacetimes. In Schwarzschild spacetime, however, there is only one timelike KVF, so at any given event P there will be at most one Killing worldline for a given 4-velocity, and that only if we pick the right 4-velocity (the stationary one).

By very definition, moving along one of them, the spacetime metric does not change. This way we can actually construct a family of hypersurfaces $\Sigma_t$ with the same fixed spatial metric.

The construction of $\Sigma_0$ we have been using will only give you this family of hypersurfaces if the timelike KVF is hypersurface orthogonal, i.e., if the spacetime is static, not just stationary. In a spacetime that is stationary but not static, such as Kerr spacetime, we can do the construction of $\Sigma_0$ at one event, and get a spacelike hypersurface orthogonal to the particular Killing worldline that passes through that event, but that hypersurface $\Sigma_0$ will not be orthogonal to any other Killing worldlines, and it will not be possible to find a family of hypersurfaces containing $\Sigma_0$ that foliates the spacetime. There is a family of spacelike hypersurfaces that foliates the spacetime, but it is not orthogonal to the timelike KVF.

Now starting from the event "corresponding" of P in a $\Sigma_t$ through the congruence, we are actually able to build a chain of rods from it to the event Q in $\Sigma_0$. The shortest rods'chain between them should give indirectly the length of the spacelike geodesic joining the event P with the event Q that lies on $\Sigma_0$.

In the case of a static spacetime, yes, you can construct such a chain of rods, and I believe the shortest such chain will define a spacelike geodesic between P and Q that lies in $\Sigma_0$. My only reservation is that I am not positive that the spacelike curve so defined will always be a geodesic of $\Sigma_0$, considered as a 3-surface in its own right. That will be the case in Schwarzschild spacetime and for all of the static KVFs in Minkowski spacetime, but I am not positive it is true in general.

 

[cianfa72]

The construction of $\Sigma_0$ we have been using will only give you this family of hypersurfaces if the timelike KVF is hypersurface orthogonal, i.e., if the spacetime is static, not just stationary. In a spacetime that is stationary but not static, such as Kerr spacetime, we can do the construction of $\Sigma_0$ at one event, and get a spacelike hypersurface orthogonal to the particular Killing worldline that passes through that event, but that hypersurface $\Sigma_0$ will not be orthogonal to any other Killing worldlines, and it will not be possible to find a family of hypersurfaces containing $\Sigma_0$ that foliates the spacetime. There is a family of spacelike hypersurfaces that foliates the spacetime, but it is not orthogonal to the timelike KVF.

ok, let's assume accordingly spacetime is static (as you said timelike KVF is hypersurface orthogonal). Therefore starting from an event P we can actually pick a single observer passing through it having 4-velocity $U$ exactly the same as the timelike KVF at P. Then we construct the $\Sigma_0$ hypersurface at P using the 'exponentiating thing' and now there exist a family of spacelike hypersurfaces $\Sigma_t$ orthogonal to the timelike KVF that include $\Sigma_0$.
Pick then another event Q that lies on $\Sigma_0$.

In the case of a static spacetime, yes, you can construct such a chain of rods, and I believe the shortest such chain will define a spacelike geodesic between P and Q that lies in $\Sigma_0$. My only reservation is that I am not positive that the spacelike curve so defined will always be a geodesic of $\Sigma_0$, considered as a 3-surface in its own right. That will be the case in Schwarzschild spacetime and for all of the static KVFs in Minkowski spacetime, but I am not positive it is true in general.

I believe such shortest chain of rods will actually define a spacelike geodesic between the "corresponding" of P in some $\Sigma_{t=T}$ through the timelike KVF-based congruence and the event Q that lies on $\Sigma_0$. The length of such spacelike geodesic should be the same as the length of the spacelike geodesic between P and Q that lies on $\Sigma_0$ -- in this sense it indirectly measure the length of the spacelike geodesic between P and Q that lies on $\Sigma_0$, I think.

About your "reservation"...Do you mean the shortest chain of rods between the corresponding of P in $\Sigma_{t=T}$ and Q on $\Sigma_0$ might be a spacelike geodesic only w.r.t. the 4D spacetime metric ?

Thanks for your time.

 

[pervect]

ok, let's assume accordingly spacetime is static (as you said timelike KVF is hypersurface orthogonal). Therefore starting from an event P we can actually pick a single observer passing through it having 4-velocity U exactly the same as the timelike KVF at P. Then we construct the Σ0 hypersurface at P using the 'exponentiating thing' and now there exist a family of spacelike hypersurfaces Σt orthogonal to the timelike KVF that include Σ0.
Pick then another event Q that lies on Σ0.I believe such shortest chain of rods will actually define a spacelike geodesic

The shortest chain of rods will be a 3-space geodesic, but it's unclear if it will be a 4-space geodesic.

In general 3-space and 4-space geodesics are different. It would be interesting to consider a specific static space-time, say the Schwarzschild space-time, and ask if the 3-space geodesics are also 4-space geodesics for that space-time. I'm afraid I don't know the answer though.

So you might have to be a bit careful about assuming the shortest chain of rods is a geodesic if by geodesic you mean a 4-space geodesic.

An advanced note. GR used only the Levi-Civita connection, and the use of this connection is the reason that curves of extremal length are geodesics in the associated space and/'or space-time. We can just add the use of the Levi-Civita connection to the other assumptions we've already made.

[add]. The fundamental reason why things get a bit tricky here is the relativity of simultaneity. The simultaneity conventions have to match, along with everything else.

 

[pervect]

I did some calculations, and it looks like in the case of the Schwarzschild space-time, the 4- geodesics on the space [t,r,theta,phi] on a hypersurface of constant Schwarzschild time t will also be 3-geodesics in the 3d space [r,theta,phi] obtained by projecting the 4-d spacetime to a 3d space by omitting the time parameter t.

So the 4-geodesics on a hypersurface of constant t should be the same curves as the 3-geodesics which extrremize (minimize, in this case) the spatial distance as measured by rods or rulers. These rods are rulers are those of the static observer in this static space-time. Lorentz contraction (among other issues) means that we have to be careful to talk about the state of motion of our rods and rulers, a moving ruler is not the same as a stationary one.

I believe this will be true for any static space-time, as coordinate systems exist for such a space-time where none of the metric coefficients are functions of time, implying that those Christoffel symbols in these coordinates without any time components should be the same between the 4-metric and the induced 3-metric. Since the Christoffel symbols determine the geodesic equation, in these cases the 4-geodesic curve with dt/dtau = 0 (which implies t=constant) should be a 3-geodesic.

I do believe that in the FRLW cosmology, which is not static, one needs to distinguish between the 4-geodesics and the 3-geodesics. This is based on a memory of old calculations, not a textbook reference or even a fresh calculation.

Also, this is oriented towards static observers in the static space-times. If we consider a moving observer in the static space-time, the timelike worldline of the moving observer will still generate a hypersurface via the exponential map of the vectors orthogonal to the timelike worldine of the moving observer. Let's consider the Schwarzschild space-time again for definiteness and ease of discussion. The hypersurfaces generated via the exponential map method for the moving observer will be a different hypersurface than a hypersurface of constant Schwarzschild time t.

Basically, the hypersurface generated by the exponential map method for the moving observer will be an "instant of time" in Fermi-Normal coordinates associated with the moving observer. Fermi-Normal coordinates are the coordinates that use the process of the exponential map of a set of tangent vectors orthogonal to some base worldline, so studying them will give some insight into this process. However, the Fermi-Normal coordinate time t-fermi won't be the same as the Schwarzschild time coordinate t-Schwarzschild. Hypersurfaces of constant t-fermi will be different hypersurfaces that hypersurfaces of constant t-Schwarzschild. This is an example of the relativity of simultaneity, something I mentioned in my previous post.

 

[PeterDonis]

The fundamental reason why things get a bit tricky here is the relativity of simultaneity. The simultaneity conventions have to match, along with everything else.

Yes, and in the particular case of a static spacetime, there is a particular simultaneity convention picked out, namely the one defined by the spacelike 3-surfaces orthogonal to the timelike KVF. See further comments below.

in the case of the Schwarzschild space-time, the 4- geodesics on the space [t,r,theta,phi] on a hypersurface of constant Schwarzschild time t will also be 3-geodesics in the 3d space [r,theta,phi] obtained by projecting the 4-d spacetime to a 3d space by omitting the time parameter t.

Yes.

I believe this will be true for any static space-time, as coordinate systems exist for such a space-time where none of the metric coefficients are functions of time

That property (the existence of a chart in which none of the metric coefficients are functions of $t$) is true for any stationary spacetime, not just static spacetimes. But the key additional feature of a static spacetime is that you can find such a chart in which there are no $dt dx^i$ cross terms, where $x^i$ are the spatial coordinates. The hypersurface orthogonality of the timelike KVF is what enables this to be the case. And in such a chart, the 3-metric on the spacelike hypersurfaces orthogonal to the timelike KVF will be the same as the 4-metric on the spacetime, with the $dt$ term omitted. And that is what enables the same argument that you make for Schwarzschild spacetime to be made for any static spacetime.

So it looks like I was too cautious about this in my post #48 earlier.

 

[cianfa72]

The shortest chain of rods will be a 3-space geodesic, but it's unclear if it will be a 4-space geodesic.

In general 3-space and 4-space geodesics are different. It would be interesting to consider a specific static space-time, say the Schwarzschild space-time, and ask if the 3-space geodesics are also 4-space geodesics for that space-time. I'm afraid I don't know the answer though.

So you might have to be a bit careful about assuming the shortest chain of rods is a geodesic if by geodesic you mean a 4-space geodesic.

I believe a similar topic has been already covered here https://www.physicsforums.com/threa...-spacelike-related-events.992261/post-6377987

 

[cianfa72]

So the 4-geodesics on a hypersurface of constant t should be the same curves as the 3-geodesics which extremize (minimize, in this case) the spatial distance as measured by rods or rulers. These rods are rulers are those of the static observer in this static space-time. Lorentz contraction (among other issues) means that we have to be careful to talk about the state of motion of our rods and rulers, a moving ruler is not the same as a stationary one.

Regarding spatial distance as measured by rods & rulers I think it applies what @Peter said in post #9 since no single measuring device can possibly be at both of a pair of spacelike separated events. Assuming static spacetime, that was actually the point of build physically a chain of rods starting from an event P on $\Sigma_{t=T}$ and ending on the event Q that lies on $\Sigma_{t=0}$ and searching for the minimum length of the rods chain joining them (in the limit of smaller and smaller rod lengths)

 

[cianfa72]

Regarding spatial distance as measured by rods & rulers I think it applies what @Peter said in post #9 since no single measuring device can possibly be at both of a pair of spacelike separated events. Assuming static spacetime, that was actually the point of build physically a chain of rods starting from an event P on $\Sigma_{t=T}$ and ending on the event Q that lies on $\Sigma_{t=0}$ and searching for the minimum length of the rods chain joining them (in the limit of smaller and smaller rod lengths)

I hope I got it correctly...
Thank you

 

[pervect]

Regarding spatial distance as measured by rods & rulers I think it applies what @Peter said in post #9 since no single measuring device can possibly be at both of a pair of spacelike separated events. Assuming static spacetime, that was actually the point of build physically a chain of rods starting from an event P on $\Sigma_{t=T}$ and ending on the event Q that lies on $\Sigma_{t=0}$ and searching for the minimum length of the rods chain joining them (in the limit of smaller and smaller rod lengths)

Physically, a rod is a pair of worldlines, one worldline for each end of the rod. The distance between close-together worldlines, which is the length of the rod, is defined by the round trip travel time of light signals, as a consequence of the SI defintion of the meter. If the distance is not changing with time, that's pretty much all you need. Though I suppose I have to note that what you essentially get from this is a 3d metric, one still has the futher step of interpreting the 3d metric as something physical. The metric gives the distance between close-togehter points. The more general problem is to find the distant between points that are not necessairly close togheter. However, going into that more would be a digression, I think - this post is already too long.

That's for the static and stationary cases that we have been discussing. If the distance is changing, then you need to address the issue of simultaneity. The distance at some instant is given by the round-trip travel time, but which instant? The round trip travel time takes some time, but conceptually the changing distance is a function of time, and we need to assign a value to the distance "now" at some particular event. And, as you point out, no measuring device can be at both ends of the rod at the same time. Thus, the measurement of distances that change with time inherently requires a simultaneity convention. Simultaneity is and must be a matter of convention - the relativity of simultaneity via Einstien's train thought experiment demonstrates this. While it is a matter of convention, many well-known and commonly used formulas will not work correctly if the convention is not followed.

Textbooks tend not to discuss distances. They give the underlying math, and leave it to the student to figure out the physical interpretation. Wald is an example of this approach - he gives the simple formula for determining the length of a space-like curve, and leaves it at that.

My own generic approach to distances is to note that the concept natively applies to 3d spaces, not to 4d space-time. And I espouse the projection technique for projecting a 4d space-time into a 3-d space. Then the process of defining distance in the resulting 3d space is mostly straightforwards, exept possibly for the point I mentioned about finding the "distance" between points that are not close together.

Complexities can arise with the projection process, the process is inherently observer dependent, and the notion of an observer in GR gets rather involved. Misner, for instance, suggests abandoning it entirely. Unfortunately this may not be helpful to someone who is trying to find a physical interpretation of the theory that has some intuitive meaning to them. Thus, as an alternative, as I mentioned before, I suggest that a natural generalization of the lattice of clocks and rods used in classical mechanics and in special relativity is a time-like congruence of worldlines. This time-like congruence can then perform the projection process, and serve as the role of "the observer".

 

[PeterDonis]

I espouse the projection technique for projecting a 4d space-time into a 3-d space.

There are two issues with this technique.

The first is that it only works for stationary spacetimes. There is at least one important class of spacetimes, the FRW spacetimes, which are not stationary.

The second is that there are actually two different kinds of "projection" that you could mean. The first is the one we have been discussing thus far: in a static spacetime, i.e., one in which the timelike KVF is hypersurface orthogonal, you can view the family of orthogonal hypersurfaces as "space", since they all have the same geometry. However, in a spacetime that is stationary but not static, the orthogonality property no longer holds globally and any interpretation of a hypersurface that is orthogonal to one particular worldline at one particular event as "space" becomes problematic.

The other kind of "projection" is to form a quotient space, i.e., to consider the manifold of "points" that each represent a stationary worldline, i.e., an integral curve of the timelike KVF. This is an approach advocated by a number of sources, for example, for dealing with things like the Ehrenfest paradox and making sense of the concept of the "spatial geometry" of an object like a rotating disk. However, such a quotient space will in general not correspond to any actual spacelike hypersurface in the actual spacetime, so its physical interpretation requires considerable care and can easily lead to misconceptions.

 

[pervect]

The projection technique I am suggesting works just fine for FRLW space-times if one gives it the proper congruence, the congruence of worldlines of observers who see the CMB as isotropic.

The split of space-time into space+time is inherently observer dependent. Specifying a congruence of worldlines (not necessarily a Killing vector field, though that is certainly a good choice if one exists) cleanly specifies a particular projection that maps the 4-d spacetime manifold to a 3d spatial manifold without time. Basically , all points on the integral curve of the congruence are regarded as "being at the same location in space at different times".

The technique does not demand hypersurface orthogonality. For an instance of its application to the rotating disk, see "The Relative Space: Space Measurements on a Rotating Platform" by Ruggiero, https://arxiv.org/abs/gr-qc/0309020. While the technique defines a 3d space, it does not and cannot define an orthogonal hypersurface as none exists. So it does not result in a coordinate system by itself - it splits space-time into a space part and a time part, but it doesn't go so far as to assign time coordinates to events.

Tensors can be decomposed via the method of a congruence of worldlines. This is trivial for a vector, for other tensors it's more complicated. For the rank 4 Riemann curvature tensor, the Bel decomposition, https://en.wikipedia.org/wiki/Bel_decomposition, is the end result. I find that this decomposition greatly aids in the physical in interpretation of the Riemann tensor.

 

[PeterDonis]

The projection technique I am suggesting works just fine for FRLW space-times if one gives it the proper congruence

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".

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.

The technique does not demand hypersurface orthogonality.

Not if you use the quotient space method, no. That's what the reference you gave by Ruggiero is doing. But that means that, as I said, there are two techniques being referred to by the term "projection method", not one.

it does not result in a coordinate system by itself - it splits space-time into a space part and a time part, but it doesn't go so far as to assign time coordinates to events.

This is not correct as a description of the quotient space method.

First, that method does result in coordinates on the quotient space. The Ruggiero paper does exactly that for the quotient space obtained for the rotating disk. (True, this is an easy case because the coordinates can simply be carried over from the spacetime--but the metric on the quotient space is not the same as the metric on the spacetime with the time coordinate taken away.) These coordinates are coordinates on a 3-manifold, so yes, they don't include a "time" coordinate--any such coordinate would be meaningless anyway because of the way the quotient space is obtained (see below).

Second, a quotient space is not obtained by "splitting spacetime into a space part and a time part". It is obtained, as I said before, by treating each worldline in the congruence as a point in the quotient space, and investigating the geometry of the resulting manifold. See the definition at the top of p. 9 of the Ruggiero paper, and the discussion immediately following. (Note that on the previous page, confusingly, Ruggiero does talk about "local space-time splitting", but, as he notes at the bottom of that page, all this is just to "suggest" the definition at the top of p. 9--it is not part of that definition or part of the construction of the quotient space.)

 

[vanhees71]

Of course you can only locally synchronize clocks (and in general not along a closed path), but I don't see, what's wrong with the standard treatment in Landau and Lifshitz Sects. 84 and 97? In Sect. 97 a "synchronous reference system" is elegantly constructed using the Hamilton Jacobi equation for time-like geodesics.

 

[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