Rest length in general relativity

[pmb_phy]

JesseM - Quick question; Is it your contention that the quantity ds², whose square root is the proper distance between two points, is coordinate dependant? I.e. that a coordinate transformation will change its value?

Pete

 

[Fredrik]

You do realize that you must also transform the components of the metric when you do a coordinate transformation, right? This may be the reason that you got the wrong result. However I'm not sure of this because I have a sneaking suspicion that this transformation leaves the metric unchanged. Not sure. Check this out if you have the time.

You're right. The coordinate change t'=t+vx, x'=x changes the components of the metric to

$$\begin{pmatrix}-1 && -2v\\0 && 1\end{pmatrix}$$

so

$$ds^2=-dt^2+dx^2=-dt'^2-2v dt' dx'+dx'^2$$.

Since dt'=dt=0 along the relevant path, the proper length remains the same. A coordinate change that doesn't change which slices of space-time we consider space can't possibly change the proper length. To change the proper length, you have to change what path to use in the integral.

Proper length is only ambiguous when it's not clear which slices of space-time we should think of as space at different times.

 

[JesseM]

JesseM - Quick question; Is it your contention that the quantity ds², whose square root is the proper distance between two points, is coordinate dependant? I.e. that a coordinate transformation will change its value?

Pete

That's a very good question, and thinking about it helps me clarify the argument I'm making. In curved spacetime ds would normally refer to an infinitesimal line element, but if you take a specific spacelike curve between two events, you can integrate ds and get a coordinate-independent length, yes? So certainly if you pick a particular spacelike curve which is bounded on one end by an event on the worldline of the left edge of an object, and bounded on the other end by an event on the worldline of the right edge of an object, then you will get a unique answer for the length along that particular curve. The problem is that in curved spacetime you can pick an infinite variety of spacelike curves whose ends correspond to events on opposite sides of the object, and for each spacelike curve, you can find a coordinate system in which the object is at rest and where every event on the spacelike curve has the same time-coordinate. So if "length" is understood as the spatial distance from one end to another at a particular instant in time, in GR this doesn't pick out a unique curve precisely because there's an infinite variety of coordinate systems which define "instant in time" (i.e. simultaneity) differently.

 

[Fredrik]

What I said in #72 can't be right. The metric must be symmetric. I don't have time to figure out what I did wrong right now, but I will in a couple of hours.

 

[Fredrik]

I forgot to transpose a matrix in the calculations. Now I get

$$g=\begin{pmatrix}-1+v^2 && -v\\-v && 1\end{pmatrix}$$

so that

$$ds^2=-dt^2+dx^2=-(1-v^2)dt'^2-2v dt'dx'+dx'^2$$.

 

[Mentz114]

Jessem,

The problem is that in curved spacetime you can pick an infinite variety of spacelike curves whose ends correspond to events on opposite sides of the object, ...

If the length is defined as the time taken for light to travel the path, then you integrate along the null geodesic, and there is no ambiguity in the choice of integral.

M

 

[JesseM]

Jessem,


If the length is defined as the time taken for light to travel the path, then you integrate along the null geodesic, and there is no ambiguity in the choice of integral.

M

If you integrate ds along a null geodesic isn't it always zero?

 

[Mentz114]

If you integrate ds along a null geodesic isn't it always zero?

We won't be integrating the proper interval, only the spatial or temporal part, they being equal in the case of light. Either will give us the time and distance. Since we measure times, we will integrate the temporal part between our readings, t_0 and t_1.

Am I making sense ?

M

 

[JesseM]

We won't be integrating the proper interval, only the spatial or temporal part, they being equal in the case of light. Either will give us the time and distance. Since we measure times, we will integrate the temporal part between our readings, t_0 and t_1.

But unlike ds, integrating only the spatial part would be coordinate-dependent, no?

 

[Mentz114]

But unlike ds, integrating only the spatial part would be coordinate-dependent, no?

I don't know, but yes I suspect it. I'll work on it when my stock of old envelopes is restored.

M

 

[gel]

You do realize that you must also transform the components of the metric when you do a coordinate transformation, right? This may be the reason that you got the wrong result.

The result was right. The metric components are different expressed in terms of x',t' than in x,t, as I wrote. The length is different because the path with t=0 as different from the path with t'=0.
...I think this is what Frederick was saying in post #72

 

[snoopies622]

Suppose we have flat space and (t,x,y,z) are standard inertial coordinates...Change to coords (t',x',y',z') with x'=x, y'=y, z'=z and t' = t + 0.6 x/c.

Is this legitimate? It looks like if $$\frac{dx}{dt}=c$$ then $$\frac{dx'}{dt'}=\frac{c}{1.6}$$

Isn't the speed of light invariant in a flat space?

 

[pmb_phy]

The result was right. The metric components are different expressed in terms of x',t' than in x,t, as I wrote. The length is different because the path with t=0 as different from the path with t'=0.
...I think this is what Frederick was saying in post #72

How did the path change? I had the chance to sit down with pen and paper and do the math yet but I will get around to it. Perhaps tonight!

Pete

 

[pmb_phy]

ok, let me have a go instead.
Suppose we have flat space and (t,x,y,z) are standard inertial coordinates.
Suppose that there is a long and thin rod at rest with respect to these coords, at y=z=0 and 0 <= x <= 1, so it has length 1.
Change to coords (t',x',y',z') with x'=x, y'=y, z'=z and t' = t + 0.6 x/c.
Note, the rod is still at rest in the new coord frame, and occupies the same region of space. However, proper length (s) along a curve with y=z=0 is,
$$ds^2 = dx^2-c^2dt^2 = (dx')^2 - (c dt'-0.6 dx')^2$$
Along a slice of constant t', use dt'=0 to get
$$ds^2 = (dx')^2 - 0.36 (dx')^2=(0.8 dx')^2$$

So, the length of the rod measured using the (t',x',y',z') coord system is 0.8, not 1.

Okay. I'm working on this problem tonight.

The proper length of the rod, $d\sigma$, is related to the spacetime interval ds² by

$d\sigma$ = $\sqrt{-ds^2}$

Recall that the proper length of the rod is 10, not 1.

I figured out what the problem is. And I'm sooo embarrassed! The problem here is in the false assumption that the value of spacetime interval will remain invariant under all coordinate transformations. That is incorrect. The spacetime interval is a Lorentz invariant. This means that its value remains unchanged upon a Lorentz transformation. "diffeomorphism invariance" refers to the fact that tensor equations retain their form under an arbitrary coordinate transformation. It does not mean that scalars remain unchanged under arbitrary coordinate transformations. The spacetime interval remains invariant under Lorentz transformations. That doesn't mean that it remains invariant under an arbitrary coordinate transformation. E.g. if you employed a Galilean transformation then there's no reason to expect that the spacetime interval will remain unchanged. (sigh) I should have know that! I guess I'm getting old! :)

Pete

 

[Fredrik]

Is this legitimate? It looks like if $$\frac{dx}{dt}=c$$ then $$\frac{dx'}{dt'}=\frac{c}{1.6}$$

Isn't the speed of light invariant in a flat space?

It's equal to c in all inertial frames, but this isn't a Lorentz transformation, so the new coordinate system isn't an inertial frame. (Yes, it's legitimate).

 

[Fredrik]

The result was right. The metric components are different expressed in terms of x',t' than in x,t, as I wrote. The length is different because the path with t=0 as different from the path with t'=0.
...I think this is what Frederick was saying in post #72

Actually when I looked at your transformation and saw that it only changes the time coordinate, I assumed (incorrectly) that it would tilt the t axis and leave the spatial axes unchanged. That's why I expressed myself the way I did. I thought it was obvious that the path doesn't change here (again, incorrectly), so I thought that by calculating the line element to be "something*dt'+something*dt'dx'+1*dx'", I was proving that the length doesn't change. Of course the calculation was wrong too. Not just the one in #72, but also my "correction" in #75.

I found the mistake today, and corrected it. Now I'm pretty sure I have the correct answer. I haven't had anyone else look at it, but at least the result comes out the same regardless of whether I'm using the primed or unprimed coordinates.

The coordinate change you suggested is

$$\begin{pmatrix}t'\\x'\end{pmatrix}=\Lambda\begin{pmatrix}t\\ x\end{pmatrix}$$

with

$$\Lambda=\begin{pmatrix}1 && v\\0 && 1\end{pmatrix}$$

and v=0.6. (I'm setting c=1). Assume that the world lines of the endpoints of the rod in the unprimed system are just the lines x=0 and x=1. When we say that the length of the rod is 1, we really mean that the proper length along a line of constant t from x=0 to x=1 is 1. The new coordinate system treats a different set of "slices" of space-time as "space". A line of constant t' is a line of constant t+vx. So put t+vx=k and find out the k of the line of constant t' that goes through t=0,x=1 (an arbitrary point on the world line of the right end of the rod). It's obviously v. So the line is t=-vx+v and it intersects the world line of the left end of the rod (x=0, remember) at t=v,x=0.

So the new path is a straight line from t=v,x=0 to t=0,x=1. Since it's a straight line, we don't have to do an integral. We can just write dt=0-v=-v, dx=1-0=1 and [itex]ds^2=-dt^2+dx^2=-v^2+1[/tex]. So the proper length along a path of constant t' is 0.64 (if we take v to be 0.6).

We can do this in the primed coordinates too. The components of the metric in the new frame are

$$g'=(\Lambda^{-1})^T\eta\Lambda^{-1}=\begin{pmatrix}-1 && v\\v && 1-v^2\end{pmatrix}$$

so the line element can be expressed as

$$ds^2=-dt^2+dx^2=-dt'^2+2vdt'dx'+(1-v^2)dx'^2$$.

The coordinates of the endpoints of the path are just t'=v,x'=0 and t'=v,x'=1, so dt'=0 and dx'=1, and that gives us the same result as before.

 

[Fredrik]

I think the coordinate change that gel came up with shows very effectively that proper length is a problematic concept even in Minkowski space. We don't have to consider shape-shifting objects in non-stationary curved space-times to see that there's a problem. The problem is present even in special relativity, even when every part of the object is stationary at all times in some inertial frame.

The question is, is there a way around the problem? In this particular case there is. It makes sense to define the proper length to be the integral of $\sqrt{-ds^2}$ along a path of constant time in the inertial frame where every part of the object is stationary.

If the different parts of the objects are moving so that there's no such inertial frame, then choose one where the center of mass is stationary.

Things get a lot more complicated in GR, so I'm not at all convinced that there's a way to get around the problem there. I would have to see a proof to believe that there is.

 

[JesseM]

I have a question which might be relevant to the issue of whether there's any meaningful way to define "rest length" in general relativity. If we have two points with a spacelike separation, is it meaningful to talk about a spacelike "geodesic" between them? It's not obvious to me that the concept of spacelike geodesics would make sense. Thinking about flat spacetime, if you draw a straight line between two events with a spacelike separation, which would presumably be the geodesic if one exists at all, I don't think this path would minimize the value of ds integrated along it--consider that if you have two points A and B with a spacelike separation, you can find a third point C such that A and B both lie on the surface of C's past light cone, and a path which went along the light cone from A to C and then back from C to B should have 0 length when you integrate ds along it, and it's possible to find spacelike paths which are arbitrarily close to this path. On the other hand, the straight-line path in flat spacetime also doesn't seem to maximize the value of ds integrated along it, since the value of ds integrated along a spacelike path is just the length of the path in the surface of simultaneity that contains it, and obviously in a given surface of simultaneity, a squiggly path between two points has a greater length than a straight-line path.

But I don't know, I may be thinking about this wrong. If it is possible to find a unique spacelike geodesic path between two events with a spacelike separation, then maybe we could define the "rest length" at a given moment by picking an event A on the worldline of one end of the object, and then finding the event B on the worldline of the other end such that the length of the geodesic path between A and B is maximized (in analogy with the flat spacetime case where the length of an object is maximized in its own rest frame).

 

[Fredrik]

(I haven't tried to find a rigorous argument, but...) I don't think "spacelike geodesic" makes sense. You probably have to slice up a region of spacetime into spacelike hypersurfaces first, and then you can talk about geodesics in each slice.

That "slicing" is of course exactly what a coordinate system does, and there are always many possible choices. I would be surprised if there's a "natural" choice even for a steel rod, and very surprised if there's a natural choice for a spinning soft rubber bag that's shaped like the statue of liberty and is half filled with water.

 

[pmb_phy]

I have a question which might be relevant to the issue of whether there's any meaningful way to define "rest length" in general relativity. If we have two points with a spacelike separation, is it meaningful to talk about a spacelike "geodesic" between them?

Absolutely. A geodesic in flat spacetime, in Lorentz coordinates, is a straight line. Likewise, any straight line is a geodesic. This is readily seen if one considers the requirement of being a geodesic; that the "length" of the worldline be stationary. When this is the case the geodesic equation holds along that worldine. Consider the geodesic equation; in flat spacetime in Lorentz coordinates all of the Christoffel symbols will be zero. This will yield the equation of a line.

It's not obvious to me that the concept of spacelike geodesics would make sense. Thinking about flat spacetime, if you draw a straight line between two events with a spacelike separation, which would presumably be the geodesic if one exists at all, I don't think this path would minimize the value of ds integrated along it--...

It actually does minimize the length of the path.

consider that if you have two points A and B with a spacelike separation, you can find a third point C such that A and B both lie on the surface of C's past light cone, and a path which went along the light cone from A to C and then back from C to B should have 0 length when you integrate ds along it, and it's possible to find spacelike paths which are arbitrarily close to this path. On the other hand, the straight-line path in flat spacetime also doesn't seem to maximize the value of ds integrated along it, since the value of ds integrated along a spacelike path is just the length of the path in the surface of simultaneity that contains it, and obviously in a given surface of simultaneity, a squiggly path between two points has a greater length than a straight-line path.

You're using Euclidean intuition in a Minkowski geometry. You can't determine the length of a worldine merely by looking at how long it is. E.g. for a time like worldline a squiggly path connecting two events has a smaller value of proper time than a straight worldline between the two.

Regarding your example of null geodesics. Its possible to connect any two events on a timelike geodesic by two straight null geodesics.

Pete

 

[Hurkyl]

I don't think "spacelike geodesic" makes sense.

What's wrong with the obvious definition?

 

[Fredrik]

What's wrong with the obvious definition?

I don't know. Maybe nothing. What's the obvious definition?

This is what I'm thinking right now:

In 1+1 dimensional Minkowski space, a geodesic between space-like separated events A and B can be defined as the curve from A to B that maximizes the integral of $\sqrt{dx^2-dt^2}$ along the path. This is a straight line, because every other path gets a bigger contribution from -dt², and that makes the result smaller.

This definition seems to make sense in 1+1 dimensions, but if we add more spatial dimensions, the integral can be made as large as we want by having the path take a long detour into the other dimensions. So it doesn't seem to make sense in 3+1 dimensions. Maybe there's a way around it, but I don't see one right now.

 

[Hurkyl]

The definition you quoted actually states that a geodesic is a local optimum: it isn't required to be a global optimum of the arclength function. For a nice example embedded in Euclidean space, consider the sphere: there are exactly two geodesics between any two (non-antipodal) points on the sphere. Their union is the great circle passing through those two points. One of them is the shortest path between the two points, the other is not. (But it is a local minimum: any local perturbation gives a longer path)

Incidentally, I was thinking about parallel transport: a geodesic is a curve whose tangent vector has length 1, and remains a tangent vector under parallel transport.

Either way, my point is that nothing breaks down in the definition of geodesic here.

Edit: ignore this post, read this one instead.

 

[JesseM]

You're using Euclidean intuition in a Minkowski geometry. You can't determine the length of a worldine merely by looking at how long it is. E.g. for a time like worldline a squiggly path connecting two events has a smaller value of proper time than a straight worldline between the two.

But I wasn't considering the length of just any path by looking at how long it was, I was specifically considering a spacelike path which lies in a single SR surface of simultaneity (the same surface that the straight-line path lies in). Is ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2 for this path not just equal to the spatial length of the path in the frame that uses this definition of simultaneity? After all, in the coordinate system where the path lies on a single surface of simultaneity, dt will always be zero. If that's right, that would be an argument to suggest why a geodesic doesn't maximize ds for spacelike paths, but that's OK because you said it minimized it.

Regarding your example of null geodesics. Its possible to connect any two events on a timelike geodesic by two straight null geodesics.

Yes, but for timelike geodesics, a geodesic is not supposed to minimize $$\sqrt{-ds^2}$$ integrated along it, it's supposed to maximize it. My argument about connecting two events by a path consisting of two null geodesics was meant to show why I didn't think ds integrated along a spacelike path would be minimized. But maybe the answer has something to do with Hurkyl's distinction between being a "local optimum" and a "global optimum of the arclength function".

In any case, it occurs to me that even if you can find a unique geodesic between any two spacelike separated events, there is a problem with defining the length of the object at the time of event A on the worldline of one end by finding the event B on the other end such that the length of the geodesic between them is maximized. Although this definition would correspond to the rest length for a rigid object in SR, I don't think it'd make much sense for an object in SR whose length is changing; so in curved spacetime this definition would at best only be useful in the case where the object was rigid (the internal stresses at any point inside the object don't change over time), and where you could pick any event on the worldline for A and the "length" would always be the same. This might work in a static spacetime but I'd guess it wouldn't work otherwise.

 

[jostpuur]

In general relativity, we cannot even define what it means for things at two different locations in space-time to be at rest with each other, so "observer at rest with the mesaured object" is, strictly speaking, nonsense.

What justification do you have for this assertion?

I suppose it's the usual: You need to do parallel transporting to compare velocities at different space time points. The relative velocity depends on the chosen path of parallel transport.

This is altogether different. Parallel transport has nothing to do with proper distance.

We need the rest frame in order to choose the spatial path along which the proper distance is calculated. We need to know relative velocities to solve the rest frame, and then we need parallel transport to solve relative velocities.

 

[jostpuur]

general relativity question about inertial frames

This isn't like SR where there is a preferred way to construct an observer's "rest frame" based on the fact that there's a special set of coordinate systems where the laws of physics take the same form, in GR all coordinate systems are on equal footing as far as physics is concerned, no?

I've understood that the theory of gravitation is formulated so that all, also non-inertial, coordinate sets are equal, but I wouldn't be so sure about all physics. If we want to do stuff with electromagnetism, shouldn't we use something like inertial frames? I'm guessing: Wouldn't frames where geodesic paths are straight lines be the inertial frames?

 

[MeJennifer]

In a Minkowski spacetime a geodesic is defined as the longest path between two events not the shortest path. In a Lorentzian spacetime the principle is similar but in some cases of curvature the longest path is only locally the longest path. Lightlike geodesics, also called null geodesics, have a zero length in spacetime.

Note that spacetime is more a chronometric than a geometric description of reality. Distances are derived in GR, they are not primary.

 

[JesseM]

In a Minkowski spacetime a geodesic is defined as the longest path between two events not the shortest path.

This is true for timelike paths, but pmb_phy seems to say it's the opposite for spacelike paths. And as I said, it seems to me that if you look at two events with a spacelike separation and draw a squiggly path between them which lies entirely in the surface of simultaneity which contains both, this will have a greater spatial length (in the inertial coordinate system which defines simultaneity this way) than a straight-line path (which also lies within this surface of simultaneity). And in the inertial coordinate system where this surface of simultaneity has constant t, then dt is going to be 0 for every line element on the path, so ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2 reduces to ds^2 = dx^2 + dy^2 + dz^2, so the spatial length of the path in this coordinate system is the same as ds integrated along the path.

 

[pmb_phy]

But I wasn't considering the length of just any path by looking at how long it was, ...

I was referring to the statement you made, i.e.

a squiggly path between two points has a greater length than a straight-line path.

What did you mean by "longer"?

I was specifically considering a spacelike path which lies in a single SR surface of simultaneity (the same surface that the straight-line path lies in). Is ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2 for this path not just equal to the spatial length of the path in the frame that uses this definition of simultaneity?

The proper distance is defined for such a path, yes.

After all, in the coordinate system where the path lies on a single surface of simultaneity, dt will always be zero. If that's right, that would be an argument to suggest why a geodesic doesn't maximize ds for spacelike paths, but that's OK because you said it minimized it.

To be precise, a geodesic is a worldline for which "s" has a stationary value.

Yes, but for timelike geodesics, a geodesic is not supposed to minimize $$\sqrt{-ds^2}$$ integrated along it, it's supposed to maximize it.

I was merely giving you an example of a timelike geodesic for which events can be connected by two null worldlines.

Pete

 

[gel]

So the proper length along a path of constant t' is 0.64 (if we take v to be 0.6).

Very minor point here, the proper length along constant t' is 0.8. I think you missed a square root. I only mention it because I was careful to pick numbers for which the square root worked out nicely :/

Things get a lot more complicated in GR, so I'm not at all convinced that there's a way to get around the problem there. I would have to see a proof to believe that there is.

For a one dimensional object like a long thin rod or a piece of string, then you can come up with a definition of its length.
Each point on the object has a 4-velocity (tangent to its world line). Given any space-time point in the world-sheet of the object, you can pass a space-like curve through it such that its tangent vector remains orthogonal to the local 4-velocity of the object.
As long as the object is smoothly embedded in space, the existence and uniqueness of the curve will follow from uniqueness of solutions to first order diff eqns.

Just use the proper length to such a curve to define the length of your rod/string. Of course, it can still vary in time but that's unavoidable.

 

[gel]

This is true for timelike paths, but pmb_phy seems to say it's the opposite for spacelike paths. And as I said, it seems to me that if you look at two events with a spacelike separation and draw a squiggly path between them which lies entirely in the surface of simultaneity which contains both, this will have a greater spatial length...

A spacelike geodesic neither minimizes nor maximizes the length, even locally. Which you can see by perturbing it in either a timelike or spacelike direction. However its length will be stationary (to first order under small perturbations). In any case, a geodesic is defined more generally as a curve which parallelly transports its own tangent vector.

 

[pmb_phy]

A spacelike geodesic neither minimizes nor maximizes the length, even locally. Which you can see by perturbing it in either a timelike or spacelike direction. However its length will be stationary (to first order under small perturbations). In any case, a geodesic is defined more generally as a curve which parallelly transports its own tangent vector.

There are two equivalent definitions of a geodesic. One is, as you've said, a curve which parallel transports its tangent, the other is a curve which has a stationary value for its "length". Each is, equivalently, a more general definition.

Pete

 

[gel]

There are two equivalent definitions of a geodesic. One is, as you've said, a curve which parallel transports its tangent, the other is a curve which has a stationary value for its "length". Each is, equivalently, a more general definition.

when I said "more generally" I was referring to the fact that geodesics only require the concept of parallel transport to be defined. This can be defined by a metric, but only requires the existence of a connection, which makes it the more general definition.

 

[pmb_phy]

when I said "more generally" I was referring to the fact that geodesics only require the concept of parallel transport to be defined. This can be defined by a metric, but only requires the existence of a connection, which makes it the more general definition.

However I could likewise say that geodesics require only the concept of a metric to be defined. This can be defined by an affine connection but only requires the existence of a metric, which makes it the more general definition. :)

Pete

 

[JesseM]

I was referring to the statement you made, i.e.

a squiggly path between two points has a greater length than a straight-line path.

What did you mean by "longer"?

I think if you looked at the context of that quote, you can see I was talking about a squiggly path through the surface of simultaneity which contained both events:

On the other hand, the straight-line path in flat spacetime also doesn't seem to maximize the value of ds integrated along it, since the value of ds integrated along a spacelike path is just the length of the path in the surface of simultaneity that contains it, and obviously in a given surface of simultaneity, a squiggly path between two points has a greater length than a straight-line path.

In that context, I just meant having a longer spatial length in the coordinate system which used that definition of simultaneity.

To be precise, a geodesic is a worldline for which "s" has a stationary value.

Does this mean that all small perturbations to the path change s in the same way, i.e. for a given path, either all small perturbations increase s, or else all small perturbations decrease s? (as suggested by Chris Hillman's post here) If so, is it possible to come up with examples of timelike paths where all small perturbations increase s (increase the proper time), or examples of spacelike paths where all small perturbations decrease s? Or do all timelike geodesics maximize the proper time with respect to small perturbations, and all spacelike geodesics minimize the length with respect to small perturbations? This review paper does seem to say that spacelike geodesics minimize s in some sense, in section 2.2, if I'm interpreting the language correctly:

2.2. Special properties of geodesics in spacetimes depending
on their causal character. We will mean by co–spacelike any geodesic such
that the orthogonal of its velocity is a spacelike subspace at each point, that is: all
the geodesics in the Riemannian case and timelike geodesics in the Lorentzian one.

...

Timelike and co–spacelike geodesics. It is well known that conjugate points
along a timelike (resp. Riemannian) geodesic in a Lorentzian (resp. Riemannian)
manifold cannot have points of accumulation. Even more:

(1) Any timelike geodesic maximizes locally d in a similar way as any Riemannian
geodesic minimizes locally its corresponding d. Nevertheless, there are two
important differences:

– Riemannian geodesics minimize locally length among all the smooth
curves connecting two fixed points p, q. Nevertheless, the timelike ones
maximize only among the causal curves connecting p, q;

Yes, but for timelike geodesics, a geodesic is not supposed to minimize $$\sqrt{-ds^2}$$ integrated along it, it's supposed to maximize it.

I was merely giving you an example of a timelike geodesic for which events can be connected by two null worldlines.

You didn't give an example, just stated that it would be possible to do so--but anyway, I agree (you just need an event C between A and B such that A is on the past light cone of C and B is on the future light cone of C). Still, as I said, my argument was trying to show that there could always be a path with smaller s, whereas for timelike geodesics my understanding was that the geodesic maximizes s, at least compared to small perturbations. But perhaps the answer here is that this 0-s path is not itself a spacelike path, and it's possible to find a separate spacelike path which is minimal with respect to small perturbations?