Undergrad Question about Parallel Transport

[cianfa72]

To reiterate, the tank can measure its rotation rate with a gyroscope internal to its hull and program its turret to counter-rotate at the same rate. No external references are needed.

As far as I understand this, using a gyroscope internal to the tank's hull and programming the turret to counter-rotate at the same tank's rotation rate w.r.t. it, one basically "implements" the parallel transport of the turret's attached gun along the path being taken from the tank.

Only when the tank takes the geodesic path along the Earth, it doesn't rotate at all w.r.t. the gyroscope, hence the programmed turret's counter-rotate rate vanishes.

As you pointed out, there is no external reference involved. In a sense, let me say, the gyroscope takes in account internally the metric connection (Levi-civita). Suppose to pick a different connection on the Earth as sphere (e.g. compass connection). In that case tank gyroscope's relative rotation no longer gives the relevant information about the curvature of the path being taken, hence it isn't good to use to program the turret's counter-rotation in order to parallel transport the gun according the chosen non-metric connection.

 

[A.T.]

As far as I understand this, using a gyroscope internal to the tank's hull ...

As I explained in post #25, a gyroscope doesn't work well in this 2D-space analogy.

 

[cianfa72]

As I explained in post #25, a gyroscope doesn't work well in this 2D-space analogy.

Yes, as @Orodruin said in #27, we can assume a sort of 2D gyroscope (i.e. free to rotate only around one axis).

 

[A.T.]

Yes, as @Orodruin said in #27, we can assume a sort of 2D gyroscope (i.e. free to rotate only around one axis).

As I said in post #28, constraining a standard gyroscope axis to stay tangential to the surface won't work here.

An actual 2D gyroscope would (in the idealized sense) be what the OP proposed: Make the turret rotation relative to the hull free and frictionless with turret center of mass on the axis of that rotation. This is actually the simplest idealised mechanical analogy I can think of.

 

[cianfa72]

An actual 2D gyroscope would (in the idealized sense) be what the OP proposed: Make the turret rotation relative to the hull free and frictionless with turret center of mass on the axis of that rotation. This is actually the simplest idealised mechanical analogy I can think of.

Ah ok. So suppose to start with the turret's gun pointing in one given direction. Then, along the path being taken from the tunk hull on Earth's surface, the turret's gun is parallel transported (according to the Levi-Civita connection). Suppose now the gun initially points along the direction of the path being taken from the tank hull. Then only if that path is geodesic will the gun stay aligned with it along the travel.

 

[cianfa72]

As far as I can tell, the parallel transport on the round sphere, when considering it embedded in 3D ambient euclidean space, works as follows.

Take a curve C on the sphere starting from point p and a vector v in the tangent space at p. Move v along C pointing in the same direction within the 3D ambient space (i.e. apply to it the "plain" parallel transport in euclidean space). Then take its orthogonal projection on the tangent plane at any point along the curve. Done, what you get is the parallel transport of v along C on the round sphere according to the Levi-Civita connection (i.e. metric compatible torsion-free affine connection).

 

[pervect]

A limited notion of parallel transport that works on a sphere is, IMO, a good place to start.

One start with the notion that great circles are geodesics of the sphere. (If I'm being pendatic, which is sort of required to avoid having someone point out the oversimplification, I'll add that a great circle is a geodesics of the Levi-Civita connection on the sphere. In the context of GR, use of the Levi-Civita connection is almost always implied, so one often omits specifying it.. And it's useful to note that these great circles are the paths of shortest distance between two points on a sphere (assuming the points aren't antipoidaal. On a sphere, this doesn't affect them being the paths of shortest distance, but things can get messy in general. I'm not going to discuss that further in this post.

This notion of "shortest distance" mostly generalizes to "geodesics of the Levi-Civita connection". But I'm not going to attempt to go into detail.

I'll add one interesting bit of complexity, though. A great circle is different from a course of constant compass heading. The curves of constant compass headings are called loxodromes, and they're not particularly relevant, but - I find them interesting enough to mention, and they serve as an example of something that is NOT a geodesic and not a great circle, but still useful. You can see the wiki entries on https://en.wikipedia.org/wiki/Loxodromic_navigation, and the related entry on "rhumb lines". I'll even add that one could find a connection that is not the Levi-Civita connection where rhumb lines are the geodesics, though it may need to exclude points on the north and south pole. But that's not really relevant to General Relativity, where we always use the Levi-Civita connection, which also has the fortuitous property of being a path of shortest distance.

The notion of parallel transport along a segment of great circle (geodesic) on a sphere is then simple. You hold a spear, that represents a vector, at a constant angle to your course. You're transporting the spear with you, that's the "transport" part of parallel transport, you're "transporting" a vector along a path.

When you reach the junction of two different segments, you leave the spear in place, as you change course from one segment to the next in your course.

This procedure will let you parallel transport your spear, which represents a vector, along a closed curve on a sphere, and it will be sufficient to show you that the spear winds up pointing in a different direction after you traverse a closed loop.

If you use some spherical trig, you can find that the amount of rotation of the spear is proportional to the enclosed area of a spherical triangle. This generalizes to more complex polygons, though I don't have an accessible proof or reference for this statement. This is related to the notion of the sum-of-angles of a spherical triangle.

This concept of parallel transport generalizes to higher dimensions, and even to curves in space-time (worldlines) rather than curves in space, but probably the generalization isn't so straightforwards, and this short post won't be enough to perform it. At that point, you may need to study a bit of differential geometry. Physics textooks will talk and motivate parallel transport, but they won't go so far as to prove existence or uniqueness. This feeling of lack of p roof of existence and uniquness may be bothersome, all I can say is that there are answers in math if it bothers you enough, and if it doesn't bother you enough you just have to take it on faith that parallel transport exists.

 

[pervect]

For a general way to get an intuitive feelings for geodesics that is not too advanced, I'll put in a plug for the "sector model" approach of Kraus and Zahn. There are a number of papers on sector models, the one that google finds specific to geodesics is https://arxiv.org/pdf/1804.09828, "Sector models—A toolkit for teaching general relativity: II. Geodesics", C Zahn and U Kraus. It might be better of course to start with part I, depending.

This link provides a resource that is peer reviewed and a bit more formal than my rather lax suggestion in my previous post on this thread. And it's a bit longer and better illustrated than anythhing I'm willing to write here on PF, which is good for explanatory detail, though bad for getting people to read the whole thing. Still, it's out there if readers have enough interest.

Of course, the original topic was not drawing geodesics, but parallel transport. Fortunately, given the ability to draw and extend geodesics, there's a rather simple and purely geometric construction known as "Schild's ladder" that can peform this task.

There's a good write up in MTW's graduate level textbook, "Gravitation". Which is great, and the specific section may not be graduate level , but it is of course a bit specialized, even though I like it a lot.

The only other reference I have at the moment for this geometric construction is wikipedia's https://en.wikipedia.org/wiki/Schild's_ladder and the related wiki's article on the Levi-Civita parallelogram, https://en.wikipedia.org/wiki/Levi-Civita_parallelogramoid.

The TL/DR version is that using the proper geometric construction , one can draw parallelograms using geodesic curves (that conceptually replace straight lines) which perform the operation of parallel transport.

As usual in GR, we are talking specifically about using the Levi-Civita connection, outside of GR people may occasionally use geodesics that use a different connection.

 

[A.T.]

And it's useful to note that these great circles are the paths of shortest distance...

... or longest distance, if you go in the opposite direction. The general property is that geodesics are paths of stationary distance.

 

[PAllen]

... or longest distance, if you go in the opposite direction. The general property is that geodesics are paths of stationary distance.

No, not longest distance - you can deviate from a non minimal great circle path and make it longer - by any amount with squiggles. In straight Riemannian geometry, for two points and a neighborhood containing them, when sufficiently small, the geodesic connecting them within the neighborhood is minimal. For GR the same statement holds except that it only applies to timelike geodesics, and substitute maximal. Spacelike geodesics are ALWAYS saddle points, with no minimum or maximum property.

 

[A.T.]

No, not longest distance

Right, the longer part of the great circle is a saddle. It's easier to remember that each part is a path of stationary length.

 

[cianfa72]

In straight Riemannian geometry, for two points and a neighborhood containing them, when sufficiently small, the geodesic connecting them within the neighborhood is minimal. For GR the same statement holds except that it only applies to timelike geodesics, and substitute maximal. Spacelike geodesics are ALWAYS saddle points, with no minimum or maximum property.

As far as I can understand, in GR, the spacelike geodesics are (always) saddle points when considering them in the overall 4D spacetime. On the other hand, if you take a spacelike hypersurface and restrict yourself to consider only the (spacelike) geodesics w.r.t. the induced Riemannian metric on it, then, in small neighborhood containing any two points, they are always minimal.

 

[PAllen]

As far as I can understand, in GR, the spacelike geodesics are (always) saddle points when considering them in the overall 4D spacetime. On the other hand, if you take a spacelike hypersurface and restrict yourself to consider only the (spacelike) geodesics w.r.t. the induced Riemannian metric on it, then, in small neighborhood containing any two points, they are always minimal.

True, but that is just a recipe for constructing a pure Riemannian manifold, that happens to a submanifold. So it is actually covered in what I said.

 

[PAllen]

Right, the longer part of the great circle is a saddle. It's easier to remember that each part is a path of stationary length.

Actually, if you take any small part of the geodesic, and a small neighborhood around that part, then that geodesic segment is minimal - even though a bigger piece is not minimal.

 

[cianfa72]

True, but that is just a recipe for constructing a pure Riemannian manifold, that happens to a submanifold. So it is actually covered in what I said.

You mean, taking a spacelike hypersurface (endowed with the induced metric) is basically a recipe to build a 3D Riemannian manifold.

 

[PAllen]

You mean, taking a spacelike hypersurface (endowed with the induced metric) is basically a recipe to build a 3D Riemannian manifold.

Yes.

 

[A.T.]

Actually, if you take any small part of the geodesic, and a small neighborhood around that part, then that geodesic segment is minimal - even though a bigger piece is not minimal.

Yes, that is another way to state the key property of a geodesic.

Great circles are often used as an simple example, but without other examples they can be is misleading:
- Since the sphere is closed you have two different geodesics connecting two points
- Since the sphere is so simple, the "small" part of the geodesic, that is minimal, is unusually "big" (up to half the circumference).

 

[cianfa72]

True, but that is just a recipe for constructing a pure Riemannian manifold, that happens to a submanifold. So it is actually covered in what I said.

This raises the following point. Let's take a spacelike geodesic of the 4D Lorentzian spacetime. There exists a spacelike hypersurface containing it, hence it is a (spacelike) geodesic on that hypersurface with the (Riemmanian) induced metric.

What about the other way around. Under which conditions a geodesic on a spacelike hypersurface is a (spacelike) geodesic of the 4D manifold ?

 

[PAllen]

This raises the following point. Let's take a spacelike geodesic of the 4D Lorentzian spacetime. There exists a spacelike hypersurface containing it, hence it is a (spacelike) geodesic on that hypersurface with the (Riemmanian) induced metric.

I am not sure that is always true. Consider, by analogy, a 3-sphere embedded in Euclidean 4 space. It can contain a geodesic of the 4-space that is not a geodesic of the 3-sphere. I don’t immediately see any reason that couldn’t happen in a Lorentzian 4 space.

[edit - see below for more discussion. This example is not correct.]

What about the other way around. Under which conditions a geodesic on a spacelike hypersurface is a (spacelike) geodesic of the 4D manifold ?

One way is if the the surface is constructed of e.g. all spacelike geodesics through a point, orthogonal to some timelike vector. I have no idea what the most general criterion might be. Generally, I would say this is rarely true.

 

[cianfa72]

I am not sure that is always true. Consider, by analogy, a 3-sphere embedded in Euclidean 4 space. It can contain a geodesic of the 4-space that is not a geodesic of the 3-sphere. I don’t immediately see any reason that couldn’t happen in a Lorentzian 4 space.

No, maybe I wasn't clear. For any geodesic of the 4D Euclidean space you can always find an hypersurface containing it (on this hypersurface that curve is a geodesic of the induced metric).

 

[PAllen]

No, maybe I wasn't clear. For any geodesic of the 4D Euclidean space you can always find an hypersurface containing it (on this hypersurface that curve is a geodesic of the induced metric).

I am saying this is not necessarily true. True is that you can find a hypersurface containing it such it is a geodesic of the hypersurface, but you also may find a hypersurface containing it such that it is NOT a geodesic of the hypersurface. I gave an example of this.

[edit: turns out that example doesn't work, see later post]

 

[pervect]

It's always safer (though less understandable) to say that geodesics (of the Levi-Civita connection) are extremal rather than maximums or minimums. But for the Riemannian case, where one is not also dealing with time (and distances are always non-negative), it's a common simplification to say "shortest distance", though as I recall it does require that one limit oneself to what's called the "local convex neighborhood". Basically, antipoidal points make things complicated for the minimum distance idea, this is a problem that occurs where you start to have multiple geodesic paths connecting the same two points. For instance, if two towns are separated by a mountain, there's probably one geodesic path that goes over the mountain, and one that goes around it (just as there are multiple geodesics from the north pole to the south pole), and they may not be the same length in general (they're the same length on a sphere, but not on an ellipsoid, for instance).

I find "shortest distance" more intuitive, but if one can accept the idea of "the same direction" and or the idea of parallel transport, geodesics can be and usually are defined that way. However, one has to know what parallel transport is in order to use it to define geodesics. And I find Schild's ladder very, very helpful in describing parallel transport and motivating it's existence and the conditions under which it is also unique - but it does require one to be able to draw and extend geodesics where that's possible.

We haven't even gotten into geodesic incompleteness, which is another obstacle to worry about in the frameork I outlined, but I think it's unhelpful to make things too complicated for what 's supposed to be a popular treatment. It's a question of how much it's good to oversimplify, and how bad it is if people don't realize there are some extra complexities lurking hidden in the math.

 

[PAllen]

I am not sure that is always true. Consider, by analogy, a 3-sphere embedded in Euclidean 4 space. It can contain a geodesic of the 4-space that is not a geodesic of the 3-sphere. I don’t immediately see any reason that couldn’t happen in a Lorentzian 4 space.

What I was thinking of here was based on the 3-sphere embedded in $E^4$ being able to contain a surface that is metrically $E^2$. However, such a contained surface, while intrinsically flat, has extrinsic curvature in $E^4$ (as well as within the 3-sphere). As such, its contained geodesics are not geodesics of the $E^4$. Realizing this, I tried to come up with a correct example and failed. I now present an argument @cianfa72 's original claim is true: That if a hypersurface of a Lorentzian spacetime contains a geodesic of that spacetime, then that geodesic is also a geodesic of the hypersurface per the induced metric.

The argument is based on parallel transport. The spacetime geodesic parallel transports its tangent vector. Supposing that within a surface containing it, it is not a geodesic. Note that, since the hypersurface contains the geodesic, its tangent vectors must also be within the the tangent space at those points for the hypersurface as well. But somehow, using the induced metric, the transport of one of these tangents ceases to be tangent to the geodesic. While this would be possible (of course) applying an arbitrary metric to the hypersurface, I believe it is impossible using the induced metric.

 

[cianfa72]

What I was thinking of here was based on the 3-sphere embedded in $E^4$ being able to contain a surface that is metrically $E^2$. However, such a contained surface, while intrinsically flat, has extrinsic curvature in $E^4$ (as well as within the 3-sphere). As such, its contained geodesics are not geodesics of the $E^4$.

Ok, this is actually an example of the other way around. Namely geodesics of that 3-sphere's contained surface fail to be a geodesics of the $E^4$.

But somehow, using the induced metric, the transport of one of these tangents ceases to be tangent to the geodesic. While this would be possible (of course) applying an arbitrary metric to the hypersurface, I believe it is impossible using the induced metric.

Ah ok, yes. Actually in my post #48 I was thinking of spacelike hypersurfaces, though.

 

[A.T.]

It's always safer (though less understandable) to say that geodesics (of the Levi-Civita connection) are extremal rather than maximums or minimums.

'Extremal' doesn't include saddles, like the longer part of a great circle.

The safe statement for an arbitrary long geodesic is that it's a path of stationary length.

But the key that defines geodesics is that for small enough sections they become paths of extremal length.

 

[PAllen]

Ok, this is actually an example of the other way around. Namely geodesics of that 3-sphere's contained surface fail to be a geodesics of the $E^4$.

No, that's not the issue I realized. You have a piece of $E^2$ embedded in $S^3$, all embedded in $E^4$. I was thinking that a geodesic of the $E^2$ is a case of a geodesic of $E^4$ contained in $S^3$ that is not a geodesic of $S^3$. What I ignored was that the $E^2$ in this case, while intrinsically flat, has extrinsic curvature in the $E^4$ (as well as in the $S^3$). As a result, its geodesics are not geodesics of the $E^4$.

Ah ok, yes. Actually in my post #48 I was thinking of spacelike hypersurfaces, though.

My parallel transport argument about why a geodesic of the whole spacetime must also be a geodesic of a submanifold (with induced metric) that contains it, applies perfectly well to spacelike hypersurfaces.

 

[cianfa72]

No, that's not the issue I realized. You have a piece of $E^2$ embedded in $S^3$, all embedded in $E^4$. I was thinking that a geodesic of the $E^2$ is a case of a geodesic of $E^4$ contained in $S^3$ that is not a geodesic of $S^3$. What I ignored was that the $E^2$ in this case, while intrinsically flat, has extrinsic curvature in the $E^4$ (as well as in the $S^3$). As a result, its geodesics are not geodesics of the $E^4$.

Ah ok, so you (mistakenly) thought you had found a (counter) example consisting of a geodesic of $E^4$ which is not a geodesic of the 3-sphere (embedded within $E^4$) that contains it.

My parallel transport argument about why a geodesic of the whole spacetime must also be a geodesic of a submanifold (with induced metric) that contains it, applies perfectly well to spacelike hypersurfaces.

Ok, good.

 

[pervect]

'Extremal' doesn't include saddles, like the longer part of a great circle.

The safe statement for an arbitrary long geodesic is that it's a path of stationary length.

But the key that defines geodesics is that for small enough sections they become paths of extremal length.

I'm not doubting your word on this, but a reference would be handy.I thought "extremal" and "stationary" were synonyms. For an I level example, f(x)=x^2 and f'(x)=x^3 are both extremal at x=0, because the derivatives df/dx and df'/dx both vanish. But the first is a minimum, the second is a saddle point. For the multidimensonal case, I thought that extremal and stationary both just involved replacing setting the ordinary derivative of a function of one variable equal to zero, with the expanded notion of setting all the partial derivatives of a multivariable function to zero. However, I'm not finding much online on the issue to show where I might be going wrong, or how to remove the coordinate dependence baked into my crude formulation.

 

[A.T.]

I thought "extremal" and "stationary" were synonyms.

"Extrema" is a generalization of "minima" and "maxima" only, and doesn't include "saddles" which are "stationary" but not "extremal".

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

,,,stationary points that are not local extrema—are known as saddle points.