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

[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.

 

[pervect]

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

The ant starts out at point, and picks out a direction in which to walk. This direction can be represented as a vector. He then proceedes to walk in that direction. While he walks, he also "parallel transports" the vector, representing the direction in which to walk, along with himself. He continues to walk in the direction the vector points. He doesn't need information other than the connection at the points he visits in order to do this - the value of the connection along his path gives him all he needs.

 

[pervect]

A fine point that concerns me a bit, so I'll give a quote from Wald, "General Relativity", pg 34.

The quick summary is to provide a reference and check on my statement that the parallel transport process depends only on the value of the connection coefficients along the curve, and to flesh out the process a bit.

We start by writing the parallel transport equations for a vector $v^a(\tau)$ along a curve with a tangent $t^a$. We will further assume that the manifold is labelled with some coordinates, and an associated coordinate basis for vectors.

In terms of components in the coordinate basis, and the parameter $\tau$ along the curve

$$\frac{dv^\nu}{d\tau} + \sum_{\mu,\lambda}t^\mu \, \Gamma^\nu{}_{\mu\lambda} v^\gamma = 0$$

This shows that the parallel transport of $v^a$ depends only on the values of $v^a$ on the curve, so we may consider the prallel transport properties of a vector defined only along the curve as opposed to a vector fields.

The connection as represented in the coordinate basis are the Christoffel symbols denoted by the symbols $\Gamma$. $\Gamma^a{}_{bc}$ maps two vectors to a number, but it doesn't transform as a tensor, so it's best considered as a map from two vectors to a number, said map defines the connection in the particular coordinates and coordinate basis chosen.

This snippet shows mathematically how to parallel transport a vector along a curve, though it's a bit terse. And it fleshes out how an "ant" can perform the process knowing the connection.

To turn this parallel transport equation for a vector $v^a$ along a curve $t^a$ into the equations for a geodeisc, we only needed to to equate $t^a$ and $v^a$, as a geodesic parallel transports the tangent vector of a curve along the curve. This, for a geodesic, we have:

$$\frac{dv^\nu}{d\tau} + \sum_{\mu,\lambda}v^\mu \, \Gamma^\nu{}_{\mu\lambda} v^\gamma = 0$$

I have assumed that $v^a$ is a unit vector, though I suppose we're considering the case where $v^a$ doesn't necessarily have a length, so this assumption is probably not needed.

For a space or space-time of dimension n, this is just n linear differential equations in the components $v^0, v^1, ...v^{n-1}$

For a flat space-time and cartesian coordinates, the Christoffel symbols are all zero, and you find that the components of the vector $v^a$ are constant.

 

[cianfa72]

The ant starts out at point, and picks out a direction in which to walk. This direction can be represented as a vector. He then proceeds to walk in that direction. While he walks, he also "parallel transports" the vector, representing the direction in which to walk, along with himself. He continues to walk in the direction the vector points. He doesn't need information other than the connection at the points he visits in order to do this - the value of the connection along his path gives him all he needs.

Surely, so to carry out the parallel transport it is enough for the ant to know the connection just at the points he visits. That means that if the manifold has a metric then the ant can derive the connection from it, otherwise 'someone' has to assign it at least at points he visits.

 

[A.T.]

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).

My takeaway from page 1 was that accelerometers cannot be pointlike, just "local" in the sense of: arbitrarily small, but not zero sized. So an accelerometer "looks off its path" in space time:

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

 

[vanhees71]

Sure, "local" in this case means a space (or space-time) region across which you can assume the gravitational field being homogeneous, i.e., the tidal forces are negligible.

 

[pervect]

Surely, so to carry out the parallel transport it is enough for the ant to know the connection just at the points he visit. That means that if the manifold has a metric then the ant can derive the connection from it, otherwise 'someone' has to assign it at least at points he visit.

Knowing the metric on the path is not in general sufficient for the ant to navigate a geodesic. Knowing the metric, and the first partial derivatives of the metric with respect to the coordinates is sufficient, given that one is assuming the Levi-Civita connection. Knowing the connection is sufficient, and given a metric, the Levi-Civita connection can be calculated from the (inverse) metric and it's partial derivatives. And the inverse metric can be calculated from the metric.

The issue here the value of a function f(x) at a point is not sufficient to know it's derivative at the point, i.e f'(x). But if you know the value of f(x) in some neighborhood, and not just at a point, you can calculate it's derivative from f(x).

The issue with the connections is a multi-dimensional case of the same issue. Knowing the value of the metric at a point doesn't give the partial derivatives or the values of the connection coefficients/Christoffel symbols. But knowing it in a neighborhood does.

Note that it's possible and sometimes useful to use some other connection than the Levi-Civita connection, though it's unnecessary to use any other connection in GR. It's a bit off topic, so I won't go into it more unless asked.

 

[pervect]

Something else that might be useful to point out. In an orthonormal basis representing the "proper reference frame" in GR (not a coordinate basis), the reading of an accelerometer at a point with constant coordinates is given by the values of certain of the connection coefficients / Christoffel symbols at that point.

To be specific, $\Gamma^\hat{0}{}_{\hat{j}\hat{0}} = a^j$. See for instance MTW, pg 330. Here it is assumed that the coordinate $x^{\hat{0}} = \tau$, $\tau$ being the proper time, and that $a^1, a^2, a^3$ are the readings of the accelerometers measuring the components of the proper acceleration in the three spatial directions.

Thus, if we know the value of the connection coefficients in an orthonormal basis, we know the accelerometer readings of an accelerometer with constant coordinates. Further, we can see , as I argued previously, that in general we need to know the metric and it's partial derivatives to calculate these readings.

Unsurprisingly, in an inertial frame, or for a particle following a geodesic, the accelerometer readings above will all be zero.

Some of the other connection coefficients can be processed by the Levi-Civita symbol to give the rotation of the basis, $\omega^i$ as well, though I won't give the details.

 

[PeterDonis]

My takeaway from page 1 was that accelerometers cannot be pointlike, just "local" in the sense of: arbitrarily small, but not zero sized. So an accelerometer "looks off its path" in space time

When we describe an object (more precisely, a test object, an object whose presence does not affect the spacetime geometry) using a worldline in a spacetime model, we are treating it as being "pointlike" even though no actual object is pointlike. An accelerometer is treated as "pointlike" in the same sense in the model, even though no actual accelerometer is pointlike. So in the model, an accelerometer does not need to "look off the path" in spacetime.

In actual spacetime, any accelerometer, like any object at all, has finite size; but in actual spacetime, no real object is a "test object" in the strict sense--every object has some finite amount of stress-energy and has some finite effect on the spacetime geometry, even if it's too small to matter. So the model we use is idealized in these respects.

 

[Ibix]

So in the model, an accelerometer does not need to "look off the path" in spacetime.

Yes - but isn't examining that idealisation relevant to this question? As pervect points out, velocity and acceleration are perfectly well defined at a point in a classical theory, but actually measuring them requires you to look at a small region. At least, I can't think of an accelerometer that can work at one event. For a concrete example, imagine an accelerometer consisting of a mass attached by six springs to the walls of a box. I can't take a snapshot of the state of the accelerometer and deduce anything about my acceleration. For example, if I've just turned off my rockets the mass will be out of equilibrium for a small time, so the mere fact that the springs are stretched doesn't tell me that I'm accelerating.

 

[PeterDonis]

isn't examining that idealisation relevant to this question?

It depends on what kind of answer the OP wants.

If the OP wants to know how the models we use work, which will be a pretty good for practical purposes, though simplified, description of how the actual world works, then we want to look at the models as they are used, idealizations and all.

If the OP wants to know how the actual world works in all details, then we need to examine how the models are idealized and what aspects of the actual world those idealizations leave out.