I The spacetime length of finite spacelike intervals

[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