# Physical meaning of a spacelike geodesic

• I
cianfa72
Yes. That's why I said its intersection with the road's worldsheet may be a spacelike geodesic. It doesn't have to be.
So the definition of "now" is not completely arbitrary though. The requirement as in #33 and #34 must be fulfilled.

Gold Member
It's perhaps worth mentioning, in passing, that there are some circumstances when you can extract "space" from spacetime as a quotient manifold (instead of by foliation). In this special case, "space" is not a submanifold of spacetime, and I suspect that geodesics in the quotient space do not correspond to geodesics in spacetime. Arguably the quotient space represents "physical space" in these cases. (I say "arguably" because I'm not sure everyone agrees.)

(For those not familiar with this, it becomes relevant if you are using a timelike congruence with vorticity, e.g. representing a rotating cylinder in flat spacetime, or more generally the timelike KVF of any stationary-but-not-static spacetime.)

cianfa72
Staff Emeritus
Homework Helper
Gold Member
It's perhaps worth mentioning, in passing, that there are some circumstances when you can extract "space" from spacetime as a quotient manifold (instead of by foliation). In this special case, "space" is not a submanifold of spacetime, and I suspect that geodesics in the quotient space do not correspond to geodesics in spacetime. Arguably the quotient space represents "physical space" in these cases. (I say "arguably" because I'm not sure everyone agrees.)

(For those not familiar with this, it becomes relevant if you are using a timelike congruence with vorticity, e.g. representing a rotating cylinder in flat spacetime, or more generally the timelike KVF of any stationary-but-not-static spacetime.)
Can you give a specific example of this construction?

Gold Member
Can you give a specific example of this construction?
For Born coordinates, which represent a rotating cylinder in flat spacetime, there's an example at Born coordinates § Radar distance in the small on Wikipedia.

The essential idea is that a "point in space" corresponds to a member of the congruence of timelike wordlines (the flow of the KVF in a stationary spacetime). "Lying on the same worldline" defines an equivalence relation on spacetime which allows you construct a quotient space. The metric on the quotient space defines the distance between two "infinitesimally close" points to be the orthogonal distance between two worldlines; in a stationary spacetime that distance is constant.

In practice you take a "stationary" coordinate system ##(t, x^1, x^2, x^3)## in which all components of the metric are independent of ##t##, and "complete the square" on the metric quadratic form to get it in the form$$\text{d}s^2 = A(\text{d}t - \omega_i \, \text{d}x^i)^2 - h_{ij} \, \text{d}x^i \, \text{d}x^j$$and then ##h## is the metric on the quotient space with coordinates ##(x^1, x^2, x^3)##. (In a static spacetime you can put ##\omega_i = 0##.)

Staff Emeritus
Homework Helper
Gold Member
For Born coordinates, which represent a rotating cylinder in flat spacetime, there's an example at Born coordinates § Radar distance in the small on Wikipedia.

The essential idea is that a "point in space" corresponds to a member of the congruence of timelike wordlines (the flow of the KVF in a stationary spacetime). "Lying on the same worldline" defines an equivalence relation on spacetime which allows you construct a quotient space. The metric on the quotient space defines the distance between two "infinitesimally close" points to be the orthogonal distance between two worldlines; in a stationary spacetime that distance is constant.

In practice you take a "stationary" coordinate system ##(t, x^1, x^2, x^3)## in which all components of the metric are independent of ##t##, and "complete the square" on the metric quadratic form to get it in the form$$\text{d}s^2 = A(\text{d}t - \omega_i \, \text{d}x^i)^2 - h_{ij} \, \text{d}x^i \, \text{d}x^j$$and then ##h## is the metric on the quotient space with coordinates ##(x^1, x^2, x^3)##. (In a static spacetime you can put ##\omega_i = 0##.)

I do not see how this is fundamentally different from providing a foliation of spacetime. It is a particular foliation for a stationary spacetime based on the existence of a timelike KVF.

Mentor
It is a particular foliation for a stationary spacetime based on the existence of a timelike KVF.
No, it isn't. The quotient space is not a "surface of constant time" corresponding to the timelike KVF and does not correspond to any foliation of the spacetime. We have had plenty of previous PF threads on this topic.

You can of course construct a foliation of Minkowski spacetime using the Langevin congruence (which is the congruence that is "at rest" in Born coordinates), but that foliation's leaves will be identical to surfaces of constant time in the inertial chart in which the center of rotation is at rest, and of course the geometry of those leaves will be flat. The geometry of the quotient space, however, is not flat, and there is no way to foliate Minkowski spacetime with a "stack" of copies of it.

Gold Member
I do not see how this is fundamentally different from providing a foliation of spacetime. It is a particular foliation for a stationary spacetime based on the existence of a timelike KVF.
But the metric on the quotient space is not the metric that would be induced on a 3-surface within the 4-manifold. In the example I gave, if you foliate into surfaces of constant ##t## you get the surface metric by putting ##\text{d}t = 0## into the 4-manifold metric i.e.$$\text{d}s^2 = -A(\omega_i \, \text{d}x^i)^2 + h_{ij} \, \text{d}x^i \, \text{d}x^j$$But that's not ##h##.

The geodesics of ##h## need not correspond to geodesics in the 4-manifold, or geodesics in a surface of constant ##t##. You can't even map a curve in the quotient space to a curve in 4-space, as one space isn't a subspace of the other.

Staff Emeritus
Homework Helper
Gold Member
But the metric on the quotient space is not the metric that would be induced on a 3-surface within the 4-manifold. In the example I gave, if you foliate into surfaces of constant t you get the surface metric by putting dt=0 into the 4-manifold metric i.e.ds2=−A(ωidxi)2+hijdxidxjBut that's not h.
Sure, but the general question becomes: Why is h of interest in and of itself? What physics can you extract from h and its geodesics?

cianfa72
but that foliation's leaves will be identical to surfaces of constant time in the inertial chart in which the center of rotation is at rest, and of course the geometry of those leaves will be flat.
Just to check my understanding: the inertial chart in which the center of rotation is at rest is actually unique up to Lorentz transformations either of timelike coordinate or any of spacelike coordinate.

Gold Member
2022 Award
Geometry is an important part of physics. I would even say it is central
From a mathematical point of view it looks as if physics is just a subset of geometry (nearly ;-)).

Dale
Mentor
the inertial chart in which the center of rotation is at rest is actually unique up to Lorentz transformations either of timelike coordinate or any of spacelike coordinate.
I'm not sure what you mean by "Lorentz transformations". It's unique up to time translations along the worldline of the center of rotation, and spatial rotations about that axis. But you can't apply a Lorentz boost since that would change which timelike geodesic is "at rest".

cianfa72
But you can't apply a Lorentz boost since that would change which timelike geodesic is "at rest".
Yes, a Lorentz boost is not allowed. Btw either a scaling of time coordinate or a transformation of spatial coordinates but rotations are not allowed as well since they would change the one-way speed of light (the new global chart would not longer be inertial).

Last edited:
Gold Member
Sure, but the general question becomes: Why is h of interest in and of itself? What physics can you extract from h and its geodesics?
The claim was made earlier in the thread that a spacelike geodesic (in spacetime of course) somehow represented the "geometry of physical space" (loosely speaking, whatever that means). I just mentioned this as a counterexample to the converse. I would argue that ##h## represents the "geometry of physical space" for a Langevin observer (e.g.) but the geodesics of ##h## don't have any obvious correspondence (as far as I can tell) with spacelike geodesics of spacetime.

E.g. as a practical experiment, if you were sitting on a rotating disk and took your time to measure its geometry with a ruler or a tape measure (no clocks or simultaneity conventions required because the disk is Born rigid and its geometry is independent of time), the answer you would get would be ##h##.

Mentor
The claim was made earlier in the thread that a spacelike geodesic (in spacetime of course) somehow represented the "geometry of physical space"
The claim I made in post #9 was more limited than that; it was about a spacelike geodesic representing the proper length of an object, not the "geometry of physical space". The former is more of a local concept; the latter is more of a global concept. In the case of the rotating disk, local measurements of the proper lengths of objects rotating with the disk would correspond to short segments of spacelike geodesics orthogonal to the worldlines of those objects, but the way those local measurements fit together globally into a "geometry of space" would correspond to the geometry of the quotient space you described.

cianfa72 and DrGreg
cianfa72
The metric on the quotient space defines the distance between two "infinitesimally close" points to be the orthogonal distance between two worldlines; in a stationary spacetime that distance is constant.
In a stationary-but-not-static spacetime in the "stationary" coordinates by definition the 3D hypersurfaces of constant ##t## all have the same (spatial) geometry. So the distance between neighboring worldlines of the KVF timelike congruence evaluated on each of such hypersurfaces does not change. However it is not the same as the orthogonal distance between them (since the KVF timelike congruence is not hypersurface orthogonal).

Last edited:
Gold Member
2022 Award
I'm not sure what you mean by "Lorentz transformations". It's unique up to time translations along the worldline of the center of rotation, and spatial rotations about that axis. But you can't apply a Lorentz boost since that would change which timelike geodesic is "at rest".
Lorentz transformations in GR are local transformations, i.e., they refer to transformations in the tangent space at one spacetime point. A local inertial frame is defined in the tangent space at one spacetime point and a tetrad as basis. Choosing another tetrad of course leads to another inertial frame, and the corresponding transformation is a Lorentz transformation.

cianfa72
How would you build the optimal road from one city to another? I would build a straight road, that is a spacelike geodesic.
One city and another city (considering them without extend so they are points) do not specify events in spacetime. You have some implicit assumptions. May be you are thinking of the two cities at the same time, which leads to simultaneity conventions. Or you are assuming a stationary spacetime. Or something else.

Gold Member
In a stationary-but-not-static spacetime in the "stationary" coordinates by definition the 3D hypersurfaces of constant ##t## all have the same (spatial) geometry. So the distance between neighboring worldlines of the KVF timelike congruence evaluated on each of such hypersurfaces does not change. However it is not the same as the orthogonal distance between them (since the KVF timelike congruence is not hypersurface orthogonal).
That is all true. In this case there are two candidates for representing "space"; one by foliating 3D submanifolds of constant ##t## and the other is the quotient manifold. The two candidates have different geometries, i.e. different metric tensors. But it is, arguably, the quotient manifold that represents what a family of observers at rest would mutually experience as the "geometry of space".

(In the case of a "static congruence", both methods give the same answer. And in the case of a "non-stationary congruence", it gets more complicated.)

cianfa72
Mentor
Are the only lines with “physical meaning” the lines that particles can follow? I don’t think so, but again “physical meaning” is a slippery concept, so I can see where others disagree.

I think that it is perfectly meaningful physically to say that my table’s leg is perpendicular to the table top. That is a spacelike geometrical relationship even though the individual particles are all timelike. So extending that to spacelike geodesics doesn’t seem unwarranted.

Are the only lines with “physical meaning” the lines that particles can follow? I don’t think so, but again “physical meaning” is a slippery concept, so I can see where others disagree.

I think that it is perfectly meaningful physically to say that my table’s leg is perpendicular to the table top. That is a spacelike geometrical relationship even though the individual particles are all timelike. So extending that to spacelike geodesics doesn’t seem unwarranted.
The issue for me is, that this is a local statement. It makes sense (physical as well) to talk about the part of the tangent space which is orthogonal to your four velocity, your rest space. It is space-like and can be viewed as space near you. Now take a very long space-like geodesic. What is the physical interpretation?

cianfa72
cianfa72
Now take a very long space-like geodesic. What is the physical interpretation?
That's exactly my viewpoint/doubt too.

Gold Member
2022 Award
Are the only lines with “physical meaning” the lines that particles can follow? I don’t think so, but again “physical meaning” is a slippery concept, so I can see where others disagree.

I think that it is perfectly meaningful physically to say that my table’s leg is perpendicular to the table top. That is a spacelike geometrical relationship even though the individual particles are all timelike. So extending that to spacelike geodesics doesn’t seem unwarranted.
The point, however is that this is a local (!) property and thus unproblematic: You measure at some time the angle of two spacelike tangent vectors at one point in spacetime, namely the angle of the leg of the table to any tangent vector to the table-top plane (being at rest relative to the table at the moment of doing this measurement).

Everything must be very carefully specified when it comes to properties at two distant places, i.e., you have to specify a phenomenon and how it is measured by a specific observer, and observations at the end are always local in GR!

martinbn and Dale
Mentor
The issue for me is, that this is a local statement.
The point, however is that this is a local (!) property and thus unproblematic:
Your arguments make sense. I am fine with the local restriction. I would be more likely to object to a claim that non-local things have “physical meaning” than to object about spacelike lines.

Mentor
Lorentz transformations in GR are local transformations, i.e., they refer to transformations in the tangent space at one spacetime point.
If we exclude translations, yes. The way @cianfa72 phrased the post I responded to made it seem like translations were included. If we include translations, we are (implicitly) making use of the fact that an inertial chart on flat Minkowski spacetime can be viewed, not just as a chart on the tangent space at a point, but as a chart on the entire spacetime, since the two are isomorphic.

A local inertial frame is defined in the tangent space at one spacetime point and a tetrad as basis. Choosing another tetrad of course leads to another inertial frame, and the corresponding transformation is a Lorentz transformation.
And, as I said, if this transformation is a spatial rotation (so the timelike vector of the tetrad stays the same and only the spacelike ones change), the "rest frame of the center of rotation", which is what @cianfa72 and I were discussing, is still at rest in the new chart; but if the transformation is a boost, it is not.

cianfa72
cianfa72
The way @cianfa72 phrased the post I responded to made it seem like translations were included. If we include translations, we are (implicitly) making use of the fact that an inertial chart on flat Minkowski spacetime can be viewed, not just as a chart on the tangent space at a point, but as a chart on the entire spacetime, since the two are isomorphic.
Yes, that was exactly my point.

cianfa72
You measure at some time the angle of two spacelike tangent vectors at one point in spacetime, namely the angle of the leg of the table to any tangent vector to the table-top plane (being at rest relative to the table at the moment of doing this measurement).
My mental model for this is to consider an infinite set of firecrackers placed on points on the table-top plane and its legs at rest each other. Based on Einstein’s synchronized clocks co-located with such firecrackers turn on them 'at the same point in time'. Then the firecrackers' explosions will define a set of spacelike separated events. Basically the measurement of the angle between the two spacelike tangent vectors involves events belonging to this set.

Last edited:
Gold Member
2022 Award
If we exclude translations, yes. The way @cianfa72 phrased the post I responded to made it seem like translations were included. If we include translations, we are (implicitly) making use of the fact that an inertial chart on flat Minkowski spacetime can be viewed, not just as a chart on the tangent space at a point, but as a chart on the entire spacetime, since the two are isomorphic.
Sure, SRT has the full symmetry of the Lorentzian affine space, including translations, ##\mathrm{ISO}(1,3)^{\uparrow}##, while in GR in general translations are no symmetries (except you have such symmetries, characterized by spacelike Killing vectors).
And, as I said, if this transformation is a spatial rotation (so the timelike vector of the tetrad stays the same and only the spacelike ones change), the "rest frame of the center of rotation", which is what @cianfa72 and I were discussing, is still at rest in the new chart; but if the transformation is a boost, it is not.
Sure.

Mentor
while in GR in general translations are no symmetries (except you have such symmetries, characterized by spacelike Killing vectors)
Aren’t translations still a local spacetime symmetry, even in the absence of a Killing vector? Isn’t that what it means to have Poincare symmetry locally

vanhees71
Gold Member
2022 Award
That's also true, if you interpret the tangent space at the point under consideration as an affine point space. This affine space then indeed has the full Poincare symmetry, but for general GR spacetimes you usually don't have a translation symmetry.

Dale
cianfa72
That's also true, if you interpret the tangent space at the point under consideration as an affine point space.
As discussed in a recent thread, to endow the underlying set (i.e. the set of spacetime points) of an affine structure the set of tangent spaces at each point must met let me say a 'consistency condition'. Basically vectors belonging to different tangent spaces have to be understood/treated as elements of the same vector space (i.e. the translation vector space) in a such way that axioms of affine space are fullfilled.