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

• I

## 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 ?

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PeterDonis
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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.

• vanhees71
Dale
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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.

• romsofia and vanhees71
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)?

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PeterDonis
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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.

• vanhees71
Dale
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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.

• vanhees71, A.T. and cianfa72
pervect
Staff Emeritus
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 travelling 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
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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.

• vanhees71 and cianfa72
Dale
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An accelerometer is local.

• vanhees71
pervect
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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
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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

• vanhees71 and PeterDonis
pervect
Staff Emeritus
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
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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
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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.

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
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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.

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.

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vanhees71
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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
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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.

• vanhees71 and cianfa72
PeterDonis
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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.

• vanhees71 and cianfa72
PeterDonis
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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.

• vanhees71
PeterDonis
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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
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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.

• vanhees71
PeterDonis
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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.

• vanhees71
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 ?