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

[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.

 

[cianfa72]

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, why an object initially at rest, starts falling.

Surely mine was, a that level, a pure and simple spatial 2D <-> 3D analogy.

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.

Take an ordinary 2-D surface with no metric defined on it (just a 2D smooth manifold with affine connection defined)

Limiting ourselves to it, how can an 'ant' -- from an operational point of view -- actually 'implement' (let me say step-by-step) the parallel transport of its tangent vector along a 'small' closed path in order to detect the geodesic curvature ? I am not sure there exist actually such a way for the ant to do that without an operative procedure to 'implement' the chosen (mathematical) affine connection structure.

 

[PeterDonis]

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

If you are just talking about "a spacelike surface", then you can just say "a spacelike surface". For what you're asking about, it doesn't matter whether or not that spacelike surface is embedded in a higher-dimensional manifold, or whether that manifold is Riemannian (positive definite metric) or pseudo-Riemannian (like spacetime).

Take an ordinary 2-D surface with no metric defined on it (just a 2D smooth manifold with affine connection defined)

Limiting ourselves to it, how can an 'ant' -- from an operational point of view -- actually 'implement' (let me say step-by-step) the parallel transport of its tangent vector along a 'small' closed path in order to detect the geodesic curvature ?

The term "geodesic curvature" is an oxymoron. As I posted previously, you need to carefully distinguish two kinds of curvature: path curvature (of curves that are not geodesic) and curvature of the manifold itself. Path curvature does not require a metric, only a connection. But manifold curvature does require a metric; without a metric it makes no sense to ask whether a manifold is curved or not. In fact, you can't even tell whether a manifold is spacelike or not without a metric, or whether it is Riemannian or pseudo-Riemannian (like spacetime). All those things require a metric.

In the absence of a metric, with only a connection, you can define parallel transport: that's what the connection is. The connection tells you what vector you get at point B when you take a particular vector at point A and parallel transport it to point B. There is nothing to "implement": if you're given the connection, you're done.

 

[cianfa72]

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 segment. This is possible with or without a metric, but it does require some way to say that two intervals are equal.

How can one check for a midpoint of a line segment (geodesic) without any metric ? The Wiki entry for Schild's ladder seems to assume a Riemannian or pseudo-Riemannian manifold.

The term "geodesic curvature" is an oxymoron.

Definitely, that was a nosense

 

[PeterDonis]

How can one check for a midpoint of a line segment (geodesic) without any metric ?

You can't. AFAIK the Schild's latter construction requires a metric. But note that for a timelike curve, you don't need the Schild's ladder construction to check whether the curve is a geodesic; you just attach an accelerometer to an object that has that curve as its worldline and see what the accelerometer reads.

 

[PAllen]

The term "geodesic curvature" is an oxymoron. As I posted previously, you need to carefully distinguish two kinds of curvature: path curvature (of curves that are not geodesic) and curvature of the manifold itself. Path curvature does not require a metric, only a connection. But manifold curvature does require a metric; without a metric it makes no sense to ask whether a manifold is curved or not. In fact, you can't even tell whether a manifold is spacelike or not without a metric, or whether it is Riemannian or pseudo-Riemannian (like spacetime). All those things require a metric.

I don't think this is correct. You need a metric to define inner products, compute interval, etc. but curvature can be completely defined by a connection. See, for example:
http://mtaylor.web.unc.edu/files/2018/04/appendc.pdf

 

[PeterDonis]

You need a metric to define inner products, compute interval, etc. but curvature can be completely defined by a connection.

Hm, yes, you're right, if you only want to define curvature of a manifold, you don't need a metric on the manifold, just a connection.

However, the Schild's ladder construction would still require a metric since it requires you to find midpoints of curves.

 

[cianfa72]

However, the Schild's ladder construction would still require a metric since it requires you to find midpoints of curves.

That means we are actually restricted to use the Levi-Civita connection from the metric chosen.

 

[PeterDonis]

That means we are actually restricted to use the Levi-Civita connection from the metric chosen.

Yes.

 

[pervect]

How can one check for a midpoint of a line segment (geodesic) without any metric ? The Wiki entry for Schild's ladder seems to assume a Riemannian or pseudo-Riemannian manifold.Definitely, that was a nosense

Actually, it's not.

MTW has a somewhat discussion of Schild's ladder, Wiki has a very short one at https://en.wikipedia.org/w/index.php?title=Schild's_ladder&oldid=910236915.

From the wiki article quoted above, in the "notes" section:

• Schild's ladder requires not only geodesics but also relative distance along geodesics. Relative distance may be provided by affine parametrization of geodesics, from which the required midpoints may be determined.
• The parallel transport which is constructed by Schild's ladder is necessarily torsion-free.
• A Riemannian metric is not required to generate the geodesics. But if the geodesics are generated from a Riemannian metric, the parallel transport which is constructed in the limit by Schild's ladder is the same as the Levi-Civita connection because this connection is defined to be torsion-free.

To go into further depth

It turns out the concept of equal intervals isn't quite the same as having a metric.

The most practical example is a null geodesic. In special relativity, the Lorentz interval , which is given by the metric, between any two points on a null geodesic is zero.

For instance in a Minkowskii metric, where the coordinates are (t,x,y,z), x=ct, y=z=0 is a null geodesic, and $x^2 - c^2 t^2$ = 0 for all points on the curve x=ct.

Things are rather similar locally in GR, but the SR case is simpler and the statements are global.

So while we have a metric, in this case, we can't use just the metric to mark out equal intervals along the null geodesic, because all the intervals are zero. However, such a concept exists, and it turns out to be useful and important. The process of marking out equal intervals along the null geodesic is called "affine parameterization".

The geodesic equation when using the usual formalisms will automatically generate an affinely parameterized geodesic,whether it is null geodesic, or otherwise.

The concept of an affine space might also be helpful here.

The cliff notes version of affine space can be found on wiki at https://en.wikipedia.org/w/index.php?title=Affine_space&oldid=985348011

n mathematics, an affine space is a geometric structure that generalizes some of the properties of Euclidean spaces in such a way that these are independent of the concepts of distance and measure of angles, keeping only the properties related to parallelism and ratio of lengths for parallel line segments.

I'll note though that for the most part, while I know about the existence of affine spaces, I don't actually use them much. I'm much more used to having a metric, and my intuition is attuned to having a metric as well. But if you really want to dig into the math, there are many cases where you don't actually need everything a metric gives you. If you don't want to dig into the math that much and just learn what you need to learn to do GR, sticking with the metric is fine, and is simpler to learn.

 

[PeterDonis]

It turns out the concept of equal intervals isn't quite the same as having a metric.

Yes, this is a valid point; I had forgotten that you can have an affine parameterization along individual geodesics even if you don't have a metric.