I The spacetime length of finite spacelike intervals

[cianfa72]

TL;DR Summary

In the context of SR ad GR how interpret and measure the length of finite spacelike intervals

Hello,

I'm aware of the following topic has already been discussed here on PF, nevertheless I would like to go deep into the concept of "finite spacelike interval" in the context of SR and GR.

All us know the physical meaning of timelike paths: basically they are paths followed through spacetime from massive objects. Furthermore we also know how to measure this kind of path length: just use a wristwatch attached to it and then measure the time elapsed from the initial to the final event through the path itself.

What about a "finite spacelike path" ? We know it is not the path followed by any physical object. However, from a local point of view, we can interpret it as the spacelike direction of a spatial axis carried by an observer (basically the spatial axis of a tetrad field)

Then if we consider a finite spacelike interval how should we actually interpret and physically measure it ? Thanks.

 

[mitochan]

Hello. Wikipedia Metre says:
The metre is the length of the path traveled by light in vacuum during a time interval of 1/299 792 458 of a second.

It means length is measured by light speed and second.
You send light and measure time light to comeback by reflection, say t, then space like distance to the reflection point is ct/2.

 

[martinbn]

Then if we consider a finite spacelike interval how should we actually interpret and physically measure it ? Thanks.

This depends on the choice of a spacelike hypersurface so it doesn't have that much physical meaning. It is also impossible to measure. If you are in a special spacetime, for example static, you could do what you would expect from non-relativistic intuition. Or if you have the spacetime metric and two events that belong to a chosen spacelike hypersurface you can compute the spatial distance between them. But as I said this has little significance.

 

[vanhees71]

Fortunately we can obviously measure distances in non-static spacetimes. E.g., in cosmology we can measure the luminosity distance to, e.g., Type 1a supernovae and relate it to the Hubble-Lemaitre red shift and determine the Hubble constant $H_0$, though that's far from being trivial.

 

[cianfa72]

This depends on the choice of a spacelike hypersurface so it doesn't have that much physical meaning. It is also impossible to measure. If you are in a special spacetime, for example static, you could do what you would expect from non-relativistic intuition. Or if you have the spacetime metric and two events that belong to a chosen spacelike hypersurface you can compute the spatial distance between them. But as I said this has little significance.

I hope the following definition of spacelike path makes sense: a path through spacetime having a spacelike vector as tangent vector to each point (event) on it.

As I take it is actually possible measure its length only when choosing a spacelike hypersurface on which the spacelike path lies on, otherwise it actually makes no sense.

 

[martinbn]

I hope the following definition of spacelike path makes sense: a path through spacetime having a spacelike vector as tangent vector to each point (event) on it.

As I take it is actually possible measure its length only when choosing a spacelike hypersurface on which the spacelike path lies on, otherwise it actually makes no sense.

This is the definition of a spacelike curve, and its length is well defined, you don't need hypersurfaces for this. But i don't think that this has any physical implications.

 

[cianfa72]

This is the definition of a spacelike curve, and its length is well defined, you don't need hypersurfaces for this. But i don't think that this has any physical implications.

Therefore for a spacelike curve the length is well defined however it has no physical meaning nor it can be physically measured alike timelike curves

 

[cianfa72]

This is the definition of a spacelike curve, and its length is well defined, you don't need hypersurfaces for this. But i don't think that this has any physical implications.

Thinking again about it: in the context of SR (flat spacetime) we actually 'attach' a physical meaning to spacelike intervals and we are also able to measure their lengths. Take for instance a physical latticework made of rigid rods and clocks in inertial coordinate system and consider the spacelike path made of simultaneous events (w.r.t the inertial coordinate system chosen) belonging to a rod world-tube.

 

[PeterDonis]

in the context of SR (flat spacetime) we actually 'attach' a physical meaning to spacelike intervals

Yes.

we are also able to measure their lengths

Only indirectly. See below.

Take for instance a physical latticework made of rigid rods and clocks in inertial coordinate system and consider the spacelike path made of simultaneous events (w.r.t the inertial coordinate system chosen) belonging to a rod world-tube.

This construction does not directly measure the spacelike path you describe, since no single measuring device can possibly be at both of a pair of spacelike separated events. The measurement relies on the physical latticework already having been set up, clocks already having been synchronized, etc., and all of those processes occur over timelike or null intervals. Only after all that has been done can we infer the length of the particular spacelike path you describe.

 

[cianfa72]

This construction does not directly measure the spacelike path you describe, since no single measuring device can possibly be at both of a pair of spacelike separated events. The measurement relies on the physical latticework already having been set up, clocks already having been synchronized, etc., and all of those processes occur over timelike or null intervals. Only after all that has been done can we infer the length of the particular spacelike path you describe.

Thanks, I believe I got your point. By the way I found the following link about the process can be used to measure the spacelike path length in case of stationary spacetime (i.e. spacetime admitting a timelike Killing symmetry)

https://physics.stackexchange.com/q...-and-lightlike-spacetime-interval-really-mean

 

[PeterDonis]

the process can be used to measure the spacelike path length in case of stationary spacetime

Where is such a process described on the page you linked to?

 

[pervect]

Summary:: In the context of SR ad GR how interpret and measure the length of finite spacelike intervals

Hello,

I'm aware of the following topic has already been discussed here on PF, nevertheless I would like to go deep into the concept of "finite spacelike interval" in the context of SR and GR.

All us know the physical meaning of timelike paths: basically they are paths followed through spacetime from massive objects. Furthermore we also know how to measure this kind of path length: just use a wristwatch attached to it and then measure the time elapsed from the initial to the final event through the path itself.

What about a "finite spacelike path" ? We know it is not the path followed by any physical object. However, from a local point of view, we can interpret it as the spacelike direction of a spatial axis carried by an observer (basically the spatial axis of a tetrad field)

Then if we consider a finite spacelike interval how should we actually interpret and physically measure it ? Thanks.

It's just the proper length of the curve, which is a spatial length. Mathematically, it has the opposite sign as the proper time that you've already discussed, but it's only a sign difference, it's mathematically a similar invariant interval.

The physical interpretation is a bit trickier, you might ask "the length according to what observer(s)"? The theory of timelike congruences comes in handy here, one can regard a timelike congruence of worldlines as an sort of generalized observer. I don't know if you are familiar with time-like congruences, I'm not sure how much to explain - I will assume for now that you are familiar enough with them that I can make my point, rather than digress into an explanation and lose the point I'm trying to make. If you are not familiar the point may get lost, but then I suppose we can try and discuss the background if you find the idea interesting to see how it applies to your specific quesiton.

The point is that if at least one timelike congruence of worldines exists that is everywhere orthogonal to the space-like curve in question, the invariant length of the curve is naturally the spatial length of the curve associated with said time-like congruence.

Mathematically, the time-like congruence of worldlines allows us to separate space and time by a projection operator. A projection operator that maps every event on a worldline in the congruence to the same point "projects" a space-time event into a spatial location. This then allows us to discuss "spatial length", and "spatial geometry", via the induced metric, as we've projected our 4d space-time into a 3d space. It's best for interpreation if the space-time is static.

 

[cianfa72]

Where is such a process described on the page you linked to?

Sorry my fault

https://physics.stackexchange.com/q...eodesics-or-what-is-the-physical-interpretati

 

[cianfa72]

Mathematically, the time-like congruence of worldlines allows us to separate space and time by a projection operator. A projection operator that maps every event on a worldline in the congruence to the same point "projects" a space-time event into a spatial location. This then allows us to discuss "spatial length", and "spatial geometry", via the induced metric, as we've projected our 4d space-time into a 3d space. It's best for interpreation if the space-time is static.

Thanks, I believe that was the point in the link attached so far (sorry I attached the wrong one)

 

[PeterDonis]

Sorry my fault

https://physics.stackexchange.com/q...eodesics-or-what-is-the-physical-interpretati

I don't see any process for directly measuring spacelike path lengths, either in a stationary spacetime or not, described here. Please quote specifically the part that you think does that.

I do see a "theoretical" process described, but the person who posted it immediately admits that it cannot be realized in practice because doing so would require a single measuring device to instantaneously make measurements at spatially separated locations, which is not possible.

The same post describes how having a stationary spacetime does help somewhat, because it at least allows you to measure, not instantaneously, a spacelike path length, by making measurements at different positions at different times and then time translating the results. "Stationary" means, roughly, "things don't change when you time translate", so if, for example, I measure the positions of two ends of a rod at different times, I can assume that the positions don't change with time, and nothing else that would affect the measurements (like the spacetime geometry) does either, so I can "time translate" one of the measurements to the same time as the other and thereby infer the path length along a spacelike curve that describes the rod "at one instant of time". This all works fine, but it still does not count as a direct measurement of a spacelike path length.

 

[PeterDonis]

Mathematically, the time-like congruence of worldlines allows us to separate space and time by a projection operator. A projection operator that maps every event on a worldline in the congruence to the same point "projects" a space-time event into a spatial location. This then allows us to discuss "spatial length", and "spatial geometry", via the induced metric, as we've projected our 4d space-time into a 3d space. It's best for interpreation if the space-time is static.

Unfortunately, even having a static spacetime is not really a help, because the important thing is not whether the spacetime is static but whether the congruence is. There are many timelike congruences in any static spacetime which are not static--because they are not formed from integral curves of the timelike Killing vector field.

What's more, if we want to find geodesic congruences that are static, in general, there will not be any even in a static spacetime. For example, Schwarzschild spacetime is static, but the static congruence is not geodesic; it is the congruence of "hovering" observers, who each stay motionless at the same altitude above the central gravitating mass, and must have nonzero proper acceleration to do so. They can indirectly measure spacelike path lengths between them, using the method I described in my previous post, and the spacelike curves whose lengths they are will be geodesics of the spacelike 3-surfaces in which they lie, but the observers' worldlines themselves will not be geodesics. I'm not aware of any static [Edit: curved] spacetime in which the Killing congruence is a geodesic congruence. (I do know of one example of a stationary, but not static, spacetime whose Killing congruence is geodesic: Godel spacetime. But Godel spacetime is widely viewed as physically unreasonable.)

 

[pervect]

For the purposes of physical insight, my opinion is that the most useful congruences are the ones where the tangents to the worldlines making up the congruence are Killing vectors, i.e. the worldlines of the congruence are integral curves of the Killing vectors. But sometimes one might want other choices. It doesn't matter that much to the point I wanted to make , which was that the congruence is one of the best tools for gaining physical insight, breaking apart space-time into space and time.

Fermi-Normal coordinates are a strong runner up for gaining physical insight, but they're not quite as general and useful (all my opinion) as congruences. But both are very important to get physical insight, unfortunately they're introduced at a rather late date (if at all) in most textbooks.

 

[PeterDonis]

For the purposes of physical insight, my opinion is that the most useful congruences are the ones where the tangents to the worldlines making up the congruence are Killing vectors, i.e. the worldlines of the congruence are integral curves of the Killing vectors.

I agree that such congruences are easy to interpret physically. The problem is that some very important congruences in commonly used spacetimes are not such congruences. For example, the congruence of "comoving" worldlines in FRW spacetime is not a Killing congruence. Also, as I noted before, almost all Killing congruences in known spacetimes are not geodesic congruences, which means that some important intuitions from inertial frames in flat spacetime do not carry over.

(Another issue is that important stationary spacetimes are not static, which means their Killing congruences are not hypersurface orthogonal. That also invalidates some important intuitions and properties.)

 

[pervect]

Congruences with vorticity, such that there is no spacelike hypersurface orthogonal to the congruence, can indeed be non-intuitive. This shows up in the "Erhenfest paradox" problem. One of the strong features of the congruence treatment though is that it is capable of handling Erhenfest's rotating disk within the formalism. It may not be intuitive, but the formalism works in this case. So it's useful for understanding the issue, such that appllying ones knowledge of congruences can give some insight into the rotating disk. In fact, that's one of the reasons I like congruences, there are some other systems which can work under more limited circumstances but are not capable of handling the rotating disk gracefully.

The other point I wanted to make is that the formalism of the latticework of clocks and rods that is often used for special relativity generates a congruence. Every point on the lattice has a worldline. A "point in space", is just one particular worldline of the congruence, the worldline contains all events that occur at any time at that "spatial location", and the projection operator projects this worldine in 4d space to a point in 3d space, giving each event a "location", a mapping from the 4d space-time to a 3d space.

When one pick a specific frame of reference with some particular velocity in SR, one picks a particular lattice structure (of which there are an infinite number of possible lattice structures). Thus the congruence is a natural extension of the idea of a "frame of reference", one that is more general than the lattice structure. And the lattice is just a special case of the more general congruence.

The important case of the FRLW "expanding universe" is another example of where the congruence tool works, and lattices do not, at least not without modifications. The fact that the congruence can handle the FRLW expanding space-time is another example of it's generality.

For a reference that goes into congruences in more detail, I'd suget "A relativists toolkit", by Eric Poisson. I believe it only handles geodesic congruences, which I find unfortunate, many useful congruences are not geodesic congruences , however, as you point out.

 

[PeterDonis]

the formalism of the latticework of clocks and rods that is often used for special relativity generates a congruence

Yes, but it's worth noting that since it's both a geodesic congruence (all of the worldlines are freely falling) and a Killing congruence, it has a set of properties that do not all generalize, and which ones do generalize depends on the particular scenario.

 

[cianfa72]

I don't see any process for directly measuring spacelike path lengths, either in a stationary spacetime or not, described here. Please quote specifically the part that you think does that.

I do see a "theoretical" process described, but the person who posted it immediately admits that it cannot be realized in practice because doing so would require a single measuring device to instantaneously make measurements at spatially separated locations, which is not possible.

Bold in the quote is mine. I believe the point you was referring to is the following: "Very unfortunately all that I said above is theoretical, in the sense that it cannot be realized in practice. This is because "tempus fugit"

The same post describes how having a stationary spacetime does help somewhat, because it at least allows you to measure, not instantaneously, a spacelike path length, by making measurements at different positions at different times and then time translating the results. "Stationary" means, roughly, "things don't change when you time translate", so if, for example, I measure the positions of two ends of a rod at different times, I can assume that the positions don't change with time, and nothing else that would affect the measurements (like the spacetime geometry) does either, so I can "time translate" one of the measurements to the same time as the other and thereby infer the path length along a spacelike curve that describes the rod "at one instant of time". This all works fine, but it still does not count as a direct measurement of a spacelike path length.

From my understanding, the measurement proposed actually measure --indirectly as you pointed out -- the minimum of the lengths in the set of spacelike curves joining the two given (spacelike separated) events constrained to lie down (belonging) to a chosen spacelike hypersurface.

 

[cianfa72]

The point is that if at least one timelike congruence of worldines exists that is everywhere orthogonal to the space-like curve in question, the invariant length of the curve is naturally the spatial length of the curve associated with said time-like congruence.

Not sure to fully understand...Does 'the existence of one timelike congruence of worldlines everywhere orthogonal to the given spacelike curve' actually imply there exists a spacelike hypersurface the given spacelike curve belongs to ?

I think that was actually the assumption in my previously referenced link (basically the $\sum_ {t}$ hypersurface he was talking about)

 

[vanhees71]

The point is we can only do local measurements, which are, as Einstein realized after some years of struggle with the interpretation of diffeomorphism invariance (involving the infamous hole argument), coincidences of events. Spatial distances have to be defined by such measurements. E.g., concerning cosmology most things we measure is electromagnetic radiation (and more recently also gravitational waves), and to define the distance of the source of this radiation you start with distances of not too far objects using the parallax method and then use "standard candles" like Cepheids or Type 1a Supernovae with a known brightness at the source and then you can measure the distance to very far objects by assuming that all the physical laws are valid everywhere and at any time of the universe, i.e., that the brightness at the source of some identified standard candle is the same everywhere and at everytime and then infer the distance from the luminosity at your place. In this way, e.g., the redshift-distance relation and with that the Hubble constant is measured.

It's far from being trivial to take into account all kinds of perturbations like "dust" (where there is a big debate between experts, how to properly take this into account) etc. At the moment there's also some tension between measurements of the Hubble constant using different methods, and the big question is, whether this is a problem with the measurements or whether our "cosmological standard model" ("$\Lambda\text{CDM}$") is really right.

 

[mitochan]

@chianfa
In SR as for time like intervals in rest IFR and other IFR the interval is common, i.e.,
c^2\triangle \tau^2=c^2\triangle t^2- \triangle x^2
so
\triangle x^2= c^2\triangle \tau^2-c^2\triangle t^2
Why don't you apply this formula for length in that IFR?

 

[vanhees71]

If you make your $\Delta$ a $\mathrm{d}$ then that makes sense.

 

[PeterDonis]

I believe the point you was referring to is the following: "Very unfortunately all that I said above is theoretical, in the sense that it cannot be realized in practice. This is because "tempus fugit"

Yes.

From my understanding, the measurement proposed actually measure --indirectly as you pointed out -- the minimum of the lengths in the set of spacelike curves joining the two given (spacelike separated) events constrained to lie down (belonging) to a chosen spacelike hypersurface.

Yes. The constraint that the curves have to lie in the chosen spacelike hypersurface is the key thing that allows the particular spacelike curve in question (the spacelike geodesic connecting the two events) to be the curve of minimum length.

Does 'the existence of one timelike congruence of worldlines everywhere orthogonal to the given spacelike curve' actually imply there exists a spacelike hypersurface the given spacelike curve belongs to ?

There will be an infinite number of possible spacelike hypersurfaces that the given curve belongs to. But only one of them will be orthogonal to the timelike congruence the way the curve itself is. That is the spacelike hypersurface we are interested in.

 

[cianfa72]

Yes. The constraint that the curves have to lie in the chosen spacelike hypersurface is the key thing that allows the particular spacelike curve in question (the spacelike geodesic connecting the two events) to be the curve of minimum length.

Just to be clear: consider two spacelike separated events and a given spacelike curve joining them that lies on the chosen spacelike hypersurface: that curve may not actually be the curve of minimum length in the set of all spacelike curves joining the two given events and belonging to the chosen spacelike hypersurface, right ?

It is our choice to search for the minimum length curve (basically a geodesic of the induced metric on the chosen hypersurface) I believe..

 

[PeterDonis]

consider two spacelike separated events and a given spacelike curve joining them that lies on the chosen spacelike hypersurface: that curve may not actually be the curve of minimum length in the set of all spacelike curves joining the two given events and belonging to the chosen spacelike hypersurface, right ?

Not if you're allowed to choose any spacelike curve you like, no.

It is our choice to search for the minimum length curve (basically a geodesic of the induced metric on the chosen hypersurface) I believe..

The "distance" between two spacelike separated events is defined as the length of the spacelike geodesic connecting them. That's how our intuitive concept of "distance" is made precise in relativity.

 

[cianfa72]

Not if you're allowed to choose any spacelike curve you like, no.

Sorry...I might have some misunderstanding with my poor English

I take it if we are allowed to choose any spacelike curve we like joining the two events --with the constrain that it must belong to the set of spacelike curves lying on the chosen spacelike hypersurface-- it need not necessarily be a geodesic (of the induced metric on the hypersurface itself)

 

[PeterDonis]

I take it if we are allowed to choose any spacelike curve we like joining the two events --with the constrain that it must belong to the set of spacelike curves lying on the chosen spacelike hypersurface-- it need not necessarily be a geodesic (of the induced metric on the hypersurface itself)

If we are allowed to choose any such curve (not just the one of minimal length), then yes, it need not be a geodesic. But what we are "allowed" to choose depends on what you are trying to do. If you are trying to define the "distance" between two events, the geodesic is the only curve it makes sense to choose.

 

[pervect]

Not sure to fully understand...Does 'the existence of one timelike congruence of worldlines everywhere orthogonal to the given spacelike curve' actually imply there exists a spacelike hypersurface the given spacelike curve belongs to ?

I think that was actually the assumption in my previously referenced link (basically the $\sum_ {t}$ hypersurface he was talking about)

I am pretty sure the answer is no, though I am am relying on an example, rather than a reference. The example is the rotating disk, which generates a set of worldlines. There is no hypersurface orthogonal to the worldlines comprising the disk, but one can find a non-closed spacelike curve representing the circumference of the disk that's orthogonal to all the worldlines comprising the disk.

The spacelike curve starts and stops on one particular shared wordline, but the two ends of said spacelike curve are timelike separated, there isn't any spacelike surface holding the whole curve.

So this spacelike curve representing the circumference has a length, and the construct of worldlines represents an array of rulers that fit around the circumference of the disk, each ruler being represented by a pair of worldlines a constant distance apart. So, we can find the length of this spacelike curve, and argue that it represents the circumference of the disk, but we do not have a spatial hypersurface in which the entire curve lies. This is because the ends are timelike separated even though the curve is everywhere spacelike.

 

[PeterDonis]

Does 'the existence of one timelike congruence of worldlines everywhere orthogonal to the given spacelike curve' actually imply there exists a spacelike hypersurface the given spacelike curve belongs to ?

I think that was actually the assumption in my previously referenced link (basically the $\sum_ {t}$ hypersurface he was talking about)

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.

Note, btw, that the spacelike curve @pervect described in his post #31 cannot be constructed the way I just described; it does not lie in any spacelike surface, certainly not a surface of "constant time" in an inertial frame.

 

[Dale]

What about a "finite spacelike path" ? We know it is not the path followed by any physical object. However, from a local point of view, we can interpret it as the spacelike direction of a spatial axis carried by an observer (basically the spatial axis of a tetrad field)

Then if we consider a finite spacelike interval how should we actually interpret and physically measure it ?

So a spacelike spacetime interval is just interpreted as a length. The interpretation is not problematic.

What is a challenge is to decide which spacelike path you want to determine the length of.

If you pick two arbitrary events then there are a few different paths that you can calculate. One is the spacelike path from one event to another in some reference frame where they are simultaneous. This has the advantage of not just being a general length, but specifically the distance between the two events in the chosen coordinate system. It has the disadvantage of being defined with respect to a reference frame. Another option is to pick the geodesic from one event to the other. This has the advantage of being completely frame independent. However, it may not be unique.

 

[Dale]

I'm not aware of any static spacetime in which the Killing congruence is a geodesic congruence.

Surely Minkowski spacetime fits that description, no?

 

[PeterDonis]

Surely Minkowski spacetime fits that description, no?

Yes, I meant to say static curved spacetime. I'll edit the post.

 

[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.