I Physical meaning of a spacelike geodesic

[DrGreg]

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.)

 

[Dale]

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.

 

[martinbn]

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?

 

[vanhees71]

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!

 

[Dale]

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.

 

[PeterDonis]

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]

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.

 

[vanhees71]

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.

 

[Dale]

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]

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.

 

[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.

 

[vanhees71]

We were talking about one tangent space at one fixed point in spacetime. An affine space is flat, but general-relativistic spacetime at presence of gravitational fields is not.