I Check for geodesically-followed path in a coordinate-free way

[cianfa72]

TL;DR Summary

Can a free body check for its geodesically followed path in spacetime ?

Hi,

My question can result a bit odd.

Consider flat spacetime. We know that inertial motions are defined by 'zero proper acceleration'. Suppose there exist just one free body in the context of SR flat spacetime (an accelerometer attached to it reads zero). We know that 'zero proper acceleration' means geodesic path followed in spacetime.

The point is: in a coordinate-free way (no reference frame involved at all) is the 'body' able to check that the path it is following in spacetime is actually a geodesic one ?

 

[PeterDonis]

in a coordinate-free way (no reference frame involved at all) is the 'body' able to check that the path it is following in spacetime is actually a geodesic one ?

Sure, just have an accelerometer attached to the body and verify that it reads zero. Or anything equivalent to that--such as, if the "body" is a human, verifying that he or she feels no weight.

 

[Dale]

Suppose there exist just one free body in the context of SR flat spacetime (an accelerometer attached to it reads zero). We know that 'zero proper acceleration' means geodesic path followed in spacetime.

The point is: in a coordinate-free way (no reference frame involved at all) is the 'body' able to check that the path it is following in spacetime is actually a geodesic one ?

I guess I don’t understand your question. It seems like you already answered it. You check to see if it is a geodesic by using an accelerometer.

 

[cianfa72]

Trying to better explain my point consider the 2D euclidean plane geometry and take a straight line. Locally -- from the point of view of a 'human' body following the straight path -- it is possible to check at each point on the path if its tangent vector is equal or not to the 'parallel transported' one.

Does exist a similar check to be carried out for the taken spacetime path (out of the reading of the accelerometer)?

 

[PeterDonis]

Locally -- from the point of view of a 'human' body following the straight path -- it is possible to check at each point on the path if its tangent vector is equal or not to the 'parallel transported' one.

How would you do this?

Does exist a similar check to be carried out for the taken spacetime path (out of the reading of the accelerometer)?

This has already been answered: if the reading of the accelerometer is zero, the path is a geodesic.

 

[Dale]

Does exist a similar check to be carried out for the taken spacetime path (out of the reading of the accelerometer)?

What you appear to be asking is if there is anything other than an accelerometer that can measure proper acceleration. But anything that measures proper acceleration is, by definition, an accelerometer.

 

[pervect]

Trying to better explain my point consider the 2D euclidean plane geometry and take a straight line. Locally -- from the point of view of a 'human' body following the straight path -- it is possible to check at each point on the path if its tangent vector is equal or not to the 'parallel transported' one.

Does exist a similar check to be carried out for the taken spacetime path (out of the reading of the accelerometer)?

A non-rigorous and incomplete but still mathematical answer is yes, assuming I understand the intent of the question. A human being can't really live on a plane, of course. It is sufficient, though, to have a notion of distance on the plane, in particular, it is sufficient that your human being can measure the distance between two points he can touch at the same time.

He does have to do some surveying work, though. If your problem definition prevents him from doing such surveying of the geometry, things get murky. I am assuming, though, that your human being can survey all the points he can touch as he moves down the path of which he wishes to determine if it's straight.. It may not be necessary that he carries some little flags that he can drive into the path, but it'd be helpful, and that's how I imagine him doing the procedure. Then, by doing the survey of the points he can reach as he moves along the path, he can determine if he took the shortest path. For instance if we have three points A, B, and C, only if the distance from A to C is the sum of the distance from A to B and B to C are the points ABC collinear.

I suppose, actually, that he only needs to be able to measure distances between points on his path to pull this off, though this is my own personal conclusion and not from a textbook. It's probably better if he can measure the distance between all the points he can reach, even if these points he can reach are off the path, though it may not be necessary.

The oversimplified answer here is that the straight line on the plane is the shortest distance between two points, and that by the process of surveying the route, the human being can determine what this path is. On the flat (no curvature) planet, the straight line path is unique no matter how long the path is.

If you want a better more general answer, you need the concepts of a connection. And it's a specific connection, the Levi-Civita connection, that makes a staight line the shortest distance between two points. The notion of a straight line as the shortest distance between two points is most familiar, but it's possible to have consistent mathematical defintions that don't have this property. They're not so useful on a plane, but they can be useful under the right circumstances. I won't get into where, unless asked, as it'd be a digression.

With the full advanced treatment, it is the connection determines parallel transport, and parallel transport determines "straighness" of a line by defintion, the fact that a vector traveling along a straight line must parallel tranpsort itself. Additionally a metric defines a unique connection, the metric corresponding the "human being's" ability to measure the distance between two points he can touch. However, while the metric defines the Levi-Civita connection, one still has to decide to adopt it and not use some other sort of connection as it's mathematicallyi consistent not to. WHen you use GR, you do make this assumption, however.

 

[PeterDonis]

He does have to do some surveying work, though.

This surveying work is not local, and the OP was asking for a local criterion.

if we have three points A, B, and C, only if the distance from A to C is the sum of the distance from A to B and B to C are the points ABC collinear.

The distances along the path that the observer travels will always meet this criterion; they have to, by definition.

The question the OP is trying to answer is whether the path the observer travels is a geodesic, and whether there is a local way of knowing that. Obviously you can know it if you know the global geometry and how the observer's path fits into that global geometry. But that's not a local criterion.

 

[Dale]

An accelerometer is local.

 

[pervect]

This surveying work is not local, and the OP was asking for a local criterion.

It depends on what you mean by local. For instance, if you have a scale, to weigh an object, it can't deflect to measure acceleleration unless it has some spatial extent.

Can you really put an accelerometer on a point, with no spatial extent? I suspect not, but I could be mistaken. I haven't seen anything definitive.

However, to me, local does not necissarily mean pointlike, but it means a limited spatial extent. Hence the analogy - "as long as your arms can reach".

The question the OP is trying to answer is whether the path the observer travels is a geodesic, and whether there is a local way of knowing that. Obviously you can know it if you know the global geometry and how the observer's path fits into that global geometry. But that's not a local criterion.

There is a local, by my defintion at least, , way of telling if an observer's path is following a straight line on a plane, or a geodesic. For instance, one might use the Euler-Lagrange equations. Or for the plane, one can measure the distance to the point behind one's position, and the distance to the point ahead of one, to a high (second order) level of precision, and ensure that the distance's add.

However, I am assuming that local includes a large enough region where one can at least evaluate derivatives, i.e. that local includes the "local neighborhood".

Neighborhoods are included in topological spaces, and hence in manifolds, so it's not a huge requirement.

 

[Dale]

However, to me, local does not necissarily mean pointlike, but it means a limited spatial extent. Hence the analogy - "as long as your arms can reach".

My understanding is also that local is not point like but merely at a scale too small for curvature to be noticed. You can certainly make accelerometers that small

 

[pervect]

It depends on what you mean by local. For instance, if you have a scale, to weigh an object, it can't deflect to measure acceleleration unless it has some spatial extent.

Can you really put an accelerometer on a point, with no spatial extent? I suspect not, but I could be mistaken. I haven't seen anything definitive.

However, to me, local does not necissarily mean pointlike, but it means a limited spatial extent. Hence the analogy - "as long as your arms can reach".

There is a local, by my defintion at least, , way of telling if an observer's path is following a straight line on a plane, or a geodesic. For instance, one might use the Euler-Lagrange equations. Or for the plane, one can measure the distance to the point behind one's position, and the distance to the point ahead of one, to a high (second order) level of precision, and ensure that the distance's add.

However, I am assuming that local includes a large enough region where one can at least evaluate derivatives, i.e. that local includes the "local neighborhood".

Neighborhoods are included in topological spaces, and hence in manifolds, so it's not a huge requirement.

I would go futher, and say that without some concept of "neighborhood", one would not be able to tell anything about a space, not even it's dimenson. It'd be a collection of points, without structure, unless and until one adds in the notion of "neighborhood".

 

[PeterDonis]

It depends on what you mean by local.

I was using that term in the way @Dale described: over small enough scales that curvature of the manifold can be neglected.

 

[PAllen]

I was using that term in the way @Dale described: over small enough scales that curvature of the manifold can be neglected.

I agree with your point, but to play Devil's advocate, Synge never agreed with this notion. He insisted curvature is defined at a single point, and is, in fact, the limit as scaled measurements are reduced to zero extent; thus, the smaller the region of evaluation, the more constant (rather than decreasing) curvature measurements become. My disagreement with this, from a physical perspective, is that the direct measurements are scaled by inverse area in taking these limits. The direct measurements go to zero much faster than linear. It is only the scaling by inverse area that allows them to have a nonzero limiting value.

 

[cianfa72]

He does have to do some surveying work, though. If your problem definition prevents him from doing such surveying of the geometry, things get murky. I am assuming, though, that your human being can survey all the points he can touch as he moves down the path of which he wishes to determine if it's straight..

That's the point: I do not know which definition assume/prefer to define -- locally -- (as long as it make sense !) the notion of geodesic for the path the observer travels in spacetime

I was using that term in the way @Dale described: over small enough scales that curvature of the manifold can be neglected.

ok, assuming 'local' as 'small enough scale' or simply a neighborhood of a point, I believe it is possible as @pervect described:

Or for the plane, one can measure the distance to the point behind one's position, and the distance to the point ahead of one, to a high (second order) level of precision, and ensure that the distance's add.

 

[vanhees71]

To put it more formal, you can say that an observer is described locally by a tetrad such that the time-like basis vector is tangential to his world line and the three spacelike basis vectors along the trajectory are determined by Fermi-Walker transport such that the observer's so defined reference frame is rotation free. Then s/he can use an accelerometer (nowadays nearly all of us carry one in our pockets in form of a smartphone ;-)) at rest relative to this reference frame to establish whether his/her world line is a geodesic. Of course the accelerometer must be small enough such as that tidal forces are negligible.

 

[cianfa72]

To put it more formal, you can say that an observer is described locally by a tetrad such that the time-like basis vector is tangential to his world line and the three spacelike basis vectors along the trajectory are determined by Fermi-Walker transport such that the observer's so defined reference frame is rotation free.

I believe the physical content of 'rotation-free' should be the same: an acclerometer attached to the so defined observer's reference frame actually read zero.

 

[vanhees71]

Yes, if the observer is freely falling, i.e., moving on a geodesic. I think the question is more puzzling than one might think in the first place. I think the point is how to operationally define a non-rotating frame of reference with respect to which you have to put the accelerometer at rest, because only then it reading 0 is equivalent for the observer being on a geodesic. In other words, the observer has to establish a (momentary) local inertial frame first. The question is, how to do this, if he is not knowing whether he is in free fall or not!

 

[PeterDonis]

Synge never agreed with this notion. He insisted curvature is defined at a single point

In the sense that curvature is a tensor (the Riemann tensor), yes, this is true: that tensor is an object in the tangent space at a single point. But this tensor involves derivatives, and derivatives require that you know how things change in an open neighborhood of the single point.

 

[PeterDonis]

I believe the physical content of 'rotation-free' should be the same: an acclerometer attached to the so defined observer's reference frame actually read zero.

No, it isn't. You can have an inertial worldline with a rotating tetrad. And, conversely, you can have an accelerated worldline with a non-rotating tetrad.

Proper acceleration has to do with the timelike vector in the tetrad--the tangent vector to the worldline. Rotation has to do with the spacelike vectors in the tetrad; "non-rotating" means that the directions of the spacelike vectors are fixed with respect to the directions indicated by three mutually perpendicular gyroscopes that are carried along the worldline.

 

[PeterDonis]

I think the point is how to operationally define a non-rotating frame of reference with respect to which you have to put the accelerometer at rest, because only then it reading 0 is equivalent for the observer being on a geodesic.

You do that with gyroscopes. Mathematically, you do it by testing for Fermi-Walker transport. You can do that regardless of whether the worldline itself is a geodesic or not.

 

[PeterDonis]

for the plane, one can measure the distance to the point behind one's position, and the distance to the point ahead of one, to a high (second order) level of precision, and ensure that the distance's add.

See my response to your earlier statement along these lines in the latter part of post #8.

 

[Dale]

I think the point is how to operationally define a non-rotating frame of reference with respect to which you have to put the accelerometer at rest, because only then it reading 0 is equivalent for the observer being on a geodesic.

When I say “accelerometer” I am referring to a 6 degree of freedom accelerometer which measures rotation as well as linear acceleration. So an accelerometer reading of 0 already includes 0 rotation as well as 0 linear acceleration. As far as how such devices measure rotation, you can use e.g. ring interferometers or even gyroscopes.

 

[PeterDonis]

When I say “accelerometer” I am referring to a 6 degree of freedom accelerometer which measures rotation as well as linear acceleration.

I don't think this is a good usage of the term "accelerometer", because, as I said in post #20, proper acceleration and rotation are different things. If an accelerometer is supposed to measure just proper acceleration, then your 6 degree of freedom device is not just an accelerometer; it has additional capabilities (measuring rotation, i.e., difference between the actual motion of the spacelike vectors of a tetrad vs. Fermi-Walker transport) that do not involve measuring proper acceleration.

 

[cianfa72]

ok thanks all.

I've now another doubt related in some way to the first. Consider two timelike worldlines (paths taken in spacetime from two physical bodies) in the context of SR flat spacetime.
In flat spacetime the geodesic deviation is zero thus two free bodies left initially at rest remain at rest.

As before consider two such 'human' bodies: from a physical point of view which is the meaning to be at rest each other ? How can each 'human' body check if it stay at rest on not respect to the other body ?

 

[PeterDonis]

from a physical point of view which is the meaning to be at rest each other ?

The obvious way to test this is to exchange round-trip light signals. If the round-trip travel time, as measured by either observer, stays the same, the two observers are at rest relative to each other.

 

[cianfa72]

The obvious way to test this is to exchange round-trip light signals. If the round-trip travel time, as measured by either observer, stays the same, the two observers are at rest relative to each other.

Does the same test also apply for GR (not zero geodesic deviation between the two geodesic paths through spacetime) ?

 

[PeterDonis]

Does the same test also apply for GR (not zero geodesic deviation between the two geodesic paths through spacetime) ?

Yes. The difference is that, if spacetime is curved, two objects that are at rest relative to each other cannot both be traveling on geodesics; at most one can be (and in most cases neither of them will be). For example, two objects at different constant heights above the Earth (say on different floors of a tall building) will be at rest relative to each other, but neither one will be traveling on a geodesic; both will have nonzero proper acceleration.

 

[cianfa72]

The obvious way to test this is to exchange round-trip light signals. If the round-trip travel time, as measured by either observer, stays the same, the two observers are at rest relative to each other.

Is theoretically possible that just one round-trip light travel time (thus for just one of the two observers) stays the same ?

 

[PeterDonis]

Is theoretically possible that just one round-trip light travel time (thus for just one of the two observers) stays the same ?

No. If the round-trip light travel time is constant for one observer, it must be constant for the other. The actual time as measured by each observer's clock might be different (in fact it will be in general), but the fact that the time does not change from one round trip to the next must be the same for both.

 

[pervect]

That's the point: I do not know which definition assume/prefer to define -- locally -- (as long as it make sense !) the notion of geodesic for the path the observer travels in spacetimeok, assuming 'local' as 'small enough scale' or simply a neighborhood of a point, I believe it is possible as @pervect described:

Or for the plane, one can measure the distance to the point behind one's position, and the distance to the point ahead of one, to a high (second order) level of precision, and ensure that the distance's add.

I'd say that a concept of local is a fundamental assumption. The assumption lies in the definition of topological spaces, further assumption and conditions are added to the notion of a toplogical space to define a manifold. GR is built on the manifold structure. There are a couple of ways to define topological spaces, one uses the idea of neighborhoods, the other uses the idea of "open sets" or "open balls". See for instance wiki , https://en.wikipedia.org/w/index.php?title=Topological_space&oldid=974881289. This may give you some idea, but is probably not enough to learn the topic. This topic, known as point set topology, is not usually covered in physics textbooks, but in math texts. However, Wald, "General Relativiy", has some brief discussion of the math in his appendices.

The usual definition to define the dimensionality of a space, which allows one to determine that a set of points is a plane and not a line, for instance, uses these topological concepts. There is a 1:1 correspondence between the number of points on a plane, and the number of points on a line, one can create an invertible mapping between all the points on a plane, and all the points on a line. But they are organized differently, and it is the concept of the neighborhoods or open balls that organizes an infinite set of points in a matter that one can determine the dimensionality of the structure, whether it be a line, plane, or of higher dimension. While one can map the plane to a line, one cannot make an invertible map from a plane to a line that preserves the neighborhood structure.

 

[cianfa72]

Yes. The difference is that, if spacetime is curved, two objects that are at rest relative to each other cannot both be traveling on geodesics; at most one can be (and in most cases neither of them will be).

Which is an example of one object traveling along a geodesic and the other not -- assuming they are at rest to each other (as checked by round-trip light travel time) ?

 

[PeterDonis]

Which is an example of one object traveling along a geodesic and the other not

If one is in fact traveling on a geodesic; but as I said, in most cases in a curved spacetime neither object will be.

 

[Dale]

I don't think this is a good usage of the term "accelerometer", because, as I said in post #20, proper acceleration and rotation are different things. If an accelerometer is supposed to measure just proper acceleration, then your 6 degree of freedom device is not just an accelerometer; it has additional capabilities (measuring rotation, i.e., difference between the actual motion of the spacelike vectors of a tetrad vs. Fermi-Walker transport) that do not involve measuring proper acceleration.

Be that as it may, if there is a need to identify a non-rotating frame in a discussion where “accelerometers” have already been specified then it is nearly trivial to use the 6-degree-of-freedom-non-accelerometers instead.

 

[PeterDonis]

if there is a need to identify a non-rotating frame in a discussion where “accelerometers” have already been specified then it is nearly trivial to use the 6-degree-of-freedom-non-accelerometers instead.

In practice, yes, but conceptually there is still a distinction between "proper acceleration" and "non-rotating", which I've already described. The fact that there happens to be a single experimental device that measures both does not change that.

 

[vanhees71]

I guess by this you mean something like the Gavity Probe B setup?

https://en.wikipedia.org/wiki/Gravity_Probe_B

 

[cianfa72]

The question the OP is trying to answer is whether the path the observer travels is a geodesic, and whether there is a local way of knowing that. Obviously you can know it if you know the global geometry and how the observer's path fits into that global geometry. But that's not a local criterion.

Sorry to resume this post. I was wondering if the "analogy" between spacetime and 2D surface really makes sense. From what we said it is possible for an observer traveling along a path through spacetime to know if its path is actually a geodesic without (locally) looking off its path (basically it suffices to have an accelerometer attached to it).

What about an observer (let me say an 'ant') traveling along a path on a 2D surface ? From the discussion above it seems, otherwise, it must perform (locally) measurements off its path (e.g. tracing locally a 'geodesic' triangle and measuring the sum of its angles). By the way that means a metric has to defined for the 'manifold surface' (otherwise it makes no sense to talk about angle measurements). Thus locally -- from an operational point of view -- how does the ant detect the 'curvature' (if any) in the case a metric is not defined for the surface ? I'm not sure if 'geodesic deviation' could help in that case...

 

[pervect]

Sorry to resume this post. I was wondering if the "analogy" between spacetime and 2D surface really makes sense. From what we said it is possible for an observer traveling along a path through spacetime to know if its path is actually a geodesic without (locally) looking off its path (basically it suffices to have an accelerometer attached to it).

What about an observer (let me say an 'ant') traveling along a path on a 2D surface ? From the discussion above it seems, otherwise, it must perform (locally) measurements off its path (e.g. tracing locally a 'geodesic' triangle and measuring the sum of its angles). By the way that means a metric has to defined for the 'manifold surface' (otherwise it makes no sense to talk about angle measurements). Thus locally -- from an operational point of view -- how does the ant detect the 'curvature' (if any) in the case a metric is not defined for the surface ? I'm not sure if 'geodesic deviation' could help in that case...

It seems to me that we can tell if three points are colinear on a plane without going off the path, by insisting that if we have three points A, B, C, the distance from A to C is the sum of the distance from A to B and B to C.

This isn't the general case, but seems to be somewhat of a counterexample, depending on exactly what one means by "off the path".

 

[PeterDonis]

It seems to me that we can tell if three points are colinear on a plane without going off the path, by insisting that if we have three points A, B, C, the distance from A to C is the sum of the distance from A to B and B to C.

This will be true along any path, geodesic or not, since you are restricting yourself to only measuring distances along the path.

 

[PeterDonis]

I was wondering if the "analogy" between spacetime and 2D surface really makes sense.

Within its limits, it does. But you have found one of the limits. See below.

From what we said it is possible for an observer traveling along a path through spacetime to know if its path is actually a geodesic without (locally) looking off its path (basically it suffices to have an accelerometer attached to it).

Yes. But note that there is no such thing as an "accelerometer" for spacelike paths. And in the analogy with ordinary 2-D space, all paths in the latter space are spacelike. So the analogy between spacetime and ordinary 2-D space breaks down here; spacetime has timelike paths, and you need timelike paths for the concept of "accelerometer" as a way to measure path curvature without "going off the path" to make sense.

 

[cianfa72]

And in the analogy with ordinary 2-D space, all paths in the latter space are spacelike. So the analogy between spacetime and ordinary 2-D space breaks down here; spacetime has timelike paths, and you need timelike paths for the concept of "accelerometer" as a way to measure path curvature without "going off the path" to make sense.

Thus, following your reasoning, the 2-D space is actually the "analogue" of the 3-D space of an observer at a given coordinate time $t$ (assuming a chart has been chosen for a spacetime region around it). The set of the spacelike separated points (events) from it defines its 3-D rest space at that given (coordinate) time. To detect the curvature (if any) of spacelike paths starting from it (basically the curvature of the 3-D space locally around it as defined before) he must actually perform measurements involving a neighborhood around it.

 

[A.T.]

Sorry to resume this post. I was wondering if the "analogy" between spacetime and 2D surface really makes sense. From what we said it is possible for an observer traveling along a path through spacetime to know if its path is actually a geodesic without (locally) looking off its path (basically it suffices to have an accelerometer attached to it).

What about an observer (let me say an 'ant') traveling along a path on a 2D surface ? From the discussion above it seems, otherwise, it must perform (locally) measurements off its path (e.g. tracing locally a 'geodesic' triangle and measuring the sum of its angles).

An analogy is not something where everything is the same, just some aspects. The analogous part here is that both paths are geodesics. However, experimentally testing for following geodesics in space-time is different than experimentally testing following geodesics on a surface.

 

[PeterDonis]

Thus, following your reasoning, the 2-D space is actually the "analogue" of the 3-D space of an observer at a given coordinate time $t$

No, it isn't. The analogy simply breaks down.

 

[cianfa72]

No, it isn't. The analogy simply breaks down.

Why not ? Just to make it simple, take for instance a 'static' spacetime and a congruence of observer's timelike paths through it. We can choose a coordinate system such that events having a given coordinate time $t$ belong to a spacelike hypersurface (orthogonal to each observer's timelike tangent vector). It should represent the "space" at that given coordinate time $t$.

That submanifold is a 3D space (actually a Riemann manifold with a positive definite inner product) thus observers can detect locally its geometry performing local measurements involving neighborhoods around them (basically off their own path).

 

[PeterDonis]

Why not ?

Because it doesn't work. The analogy breaks down.

take for instance a 'static' spacetime and a congruence of observer's timelike paths through it. We can choose a coordinate system such that events having a given coordinate time belong to a spacelike hypersurface (orthogonal to each observer's timelike tangent vector). It should represent the "space" at that given coordinate time .

We all know how taking a 3-D spacelike surface of constant coordinate time works. That is not the issue.

The issue is that doing that is irrelevant to the analogy we were discussing between an ordinary 2-D curved surface, which has a positive definite metric, and 1 x 1 spacetime, which does not. That analogy breaks down when you try to find something in the 2-D curved surface case that corresponds to an accelerometer in the spacetime case. No amount of discussion of how to take a 3-D spacelike surface of constant time from a 4-D spacetime will fix that problem.

 

[pervect]

There are some matters of terminology here to my mind as far as what the question means.

Suppose you have function y(x). Suppose you want to find dy/dx. You might say that dy/dx is defined at a point, but you might also say that to evaluate dy/dx, you need to have the value of y(x) in some neighborhood of x, so that you can calculate the limit of y(x+dx)-y(x)/dx and take the derivative.

In order to be able to define the notion of a differentiable function, you need to have a concept of neighborhood. And you need to have a specail sort of function, one that is continuous.

I think settling this simple case first would shed some light about the more complex cases being considered, which I'll talk about next. Does a derivative of a simple function of one variable require a knowledge of the neighborhood, or is it defined at a point?

Things get slightly more complicated with geodesics assuming the presence of a metric, but the principles are similar.

One way of writing the geodesic equation, using a metric, is that $\nabla_u u = 0$. Here u is the 4-velocity, the tangent to a curve in space-time. $\nabla_u$ is a sort of a derivative operator, the directional derivative.

It's a bit of a tricky question as to whether the directional derivative operator really needs to know about points not on the curve. I've heard arguments that it does not, however, so I think the matter might need some serious consideration, probably by someone more up on the foundational mathematics than I am.

You don't need a metric to define geodesics though. A connection is sufficient, I think.

Parallel transport can be defined with a connection without a metric, for instance using the geometric construction of Schild's ladder. And parallel transport can be used to tell if a spatial slice is curved or not - in a curved space, parallel transport around a closed loop for all possible closed loops is only possible in the absence of curvature.

You do need a closed loop to use this notion of flatness vs curvature, though, so this approach requires the neighborhood around the point to determine the presence or absence of curvature. However, one doesn't need a metric. I believe there may be a requirement that the space is affine - Schild's ladder, for instance, requires that one be able to find the midpoint of a line segement. This is possible with or without a metric, but it does require some way to say that two intervals are equal.

 

[PeterDonis]

One way of writing the geodesic equation, using a metric, is that $\nabla_u u = 0$. Here u is the 4-velocity, the tangent to a curve in space-time. $\nabla_u$ is a sort of a derivative operator, the directional derivative.

But, as you note further on, you don't need a metric to define $\nabla_u$ (or the $\nabla$ operator in general). A connection is sufficient.

It's a bit of a tricky question as to whether the directional derivative operator really needs to know about points not on the curve.

I have been assuming that, mathematically speaking, it doesn't.

However, there is an additional issue of what the mathematical operator $\nabla_u$ represents, physically--or, to put it the other way around, how we construct a physical measuring device that realizes the operator $\nabla_u$. For a timelike curve, it's simple: an accelerometer. But AFAIK there is no corresponding direct physical realization of $\nabla_u$ for a spacelike curve (or an null curve). Which means that, while we can evaluate $\nabla_u$ directly for timelike curves (just attach an accelerometer to an object that has that curve as its worldline), we have no corresponding way of doing that for spacelike (or null) curves. So we have to evaluate path curvature (which is what $\nabla_u$ represents mathematically) indirectly for those curves, and those indirect evaluations involve points not on the curve, which can give the (IMO misleading) impression that evaluating path curvature always requires points not on the curve (which IMO it doesn't for timelike curves).

Parallel transport can be defined with a connection without a metric, for instance using the geometric construction of Schild's ladder. And parallel transport can be used to tell if a spatial slice is curved or not - in a curved space, parallel transport around a closed loop for all possible closed loops is only possible in the absence of curvature.

Here a different notion of "curvature" is being used: curvature of spacetime, not curvature of a particular curve in spacetime. The two are distinct. To evaluate the curvature of spacetime does always require knowledge of physical quantities in an open neighborhood--you can't do it along a single curve. But that is a separate question from the question of what it takes to evaluate path curvature of a single curve.

 

[pervect]

Here a different notion of "curvature" is being used: curvature of spacetime, not curvature of a particular curve in spacetime. The two are distinct. To evaluate the curvature of spacetime does always require knowledge of physical quantities in an open neighborhood--you can't do it along a single curve. But that is a separate question from the question of what it takes to evaluate path curvature of a single curve.

Yes, I agree. It's only necessary to be able to compute the directional derivative to calculate the path curvature. Calculating the curvature of a manifold is a bit more involved. The methods are very similar for a spatial manifold, which is Riemannian, or a space-time manifold, which is pseudo-Riemannian.

The OP was interested in both, I think - he talked about measuring the sum of the angles of geodesic triangles. I believe this is equivalent to the usual formulation in terms of parallel transport - the sum of the angles of a geodesic triangle is 180 degrees if parallel transporting a vector along a closed loop does not change a vector.

However, it'd be more productive to look at the simpler case first, I think. The directional derivative is the easier concept to get a handle on, I think, and I'd encourage the OP to do some background reading on the topic as I think it could help sharpen up and define his questions.

 

[cianfa72]

The issue is that doing that is irrelevant to the analogy we were discussing between an ordinary 2-D curved surface, which has a positive definite metric, and 1 x 1 spacetime, which does not. That analogy breaks down when you try to find something in the 2-D curved surface case that corresponds to an accelerometer in the spacetime case. No amount of discussion of how to take a 3-D spacelike surface of constant time from a 4-D spacetime will fix that problem.

Maybe I was unclear about it. The analogy I have been talking about since post #37 is not between an 1 + 1 spacetime and an ordinary 2-D curved surface (it definitely does not apply for the reasons you highlighted). My point is about the 'spatial' part of a 2 + 1 spacetime and the 'spatial' part of 'our' 3 + 1 spacetime (taking in account obviously their different dimensions: 2 for the first and 3 for the latter).

 

[A.T.]

My point is about the 'spatial' part of a 2 + 1 spacetime and the 'spatinal' part of 'our' 3 + 1 spacetime (taking in account obviously their different dimensions: 2 for the first and 3 for the latter).

Of course you can make the purely spatial analogy 2D <-> 3D. But to explain how gravity works in GR, you have to include the time dimension. For example to explain why an object initially at rest starts falling.