Alternative definitions of geodesic

1. Oct 15, 2015

bcrowell

Staff Emeritus
I teach both physics and math at a community college, and I've volunteered to give a short talk for students at our weekly math colloquium that has to do with curvature and non-curvature singularities in relativity. This is a tall order, given that I can't even assume that all the students will have had calculus. I'm hoping to make it the kind of talk where everyone at least understands some of it. Just to make things harder, the culture of these colloquia is that there should be at least a taste of rigor, e.g., speakers are usually expected to give a full and rigorous statement of something like one central definition and/or one main theorem or conjecture.

The basic concept I'm trying to get across is geodesic incompleteness, and for that purpose I'd like to be able to give a fairly elementary, but rigorous, definition of a geodesic. The standard definition is that it's a curve that parallel-transports its own tangent vector, but this definition requires quite a bit of preliminary machinery. One could leave geodesic as a fundamental notion, as in Euclid, but I think this only really works in contexts like hyperbolic geometry.

A third option might be to talk about geodesics as curves of extremal or stationary metric length. This may not be the best approach if you're trying for full generality (won't work for spacelike or null geodesics), but it might be easy to state and understand for timelike geodesics. It has the advantage that it's intuitively appealing and easy to explain at an intuitive level, and then if the formalized definition is too confusing, people will still have gotten the general idea.

The main technical issue seems to be that we can have conjugate points. Hawking and Ellis have a good statement of the main result on pp. 110-111. Here are some nonrigorous paraphrases of some of the preliminary notions. They define a timelike geodesic $\gamma(t)$ from q to p as maximal if it can't be lengthened by making small variations. Point p and q on a certain geodesic are said to be conjugate if the geodesic can be perturbed infinitesimally to yield another geodesic that also passes through p and q (p. 97). They then have:

Now we only expect to have a finite number of conjugate points conjugate to a given point in any finite segment of a geodesic. (H&E state on p. 111 that "conjugate points are isolated." I assume this is what they mean, although they don't say why it has to be so. Presumably this relates to smoothness.) So can I define a timelike geodesic as follows?

A timelike geodesic $\gamma$ is defined as a timelike curve of maximal length, in the following sense. Given points $a$ and $b$ on $\gamma$, one can always find a set of points $h_0$,... $h_n$, also on $\gamma$, with the following properties:

(1) $h_0=a$ and $h_n=b$. The points are laid out in order from 0 to n according to the parameter used to define $\gamma$.

(2) The length of each segment, from $h_i$ to $h_{i+1}$, is maximal in the sense defined above.

(3) The curve is twice differentiable.

Does this work? I'm thinking that if there are points conjugate to a or b, we can just put h's at them, and also fill in further h's between these, so there will never be two successive h's that are conjugate to each other. Condition 3 is intended to ensure that we can't have a situation where segment $h_ih_{i+1}$ is a geodesic, and so is $h_{i+1}h_{i+2}$, but $h_ih_{i+2}$ is not, because there's a kink at $h_{i+1}$.

Last edited: Oct 15, 2015
2. Oct 15, 2015

PAllen

I think that works (or at least, Synge claims a result, without proof, in his GR book, that justifies it). What he states is that if p and q are made close enough, the geodesic between them, within some neighborhood containing p and q, is unique (the neighborhood is to rule out 'long way around' geodesics in closed spacetimes). Then, if it is a timelike geodesic, it is maximal in the neighborhood (again, for p and q made sufficiently close, the only exception is a long ctc).

3. Oct 15, 2015

Staff: Mentor

I assume you mean "maximal"?

It looks OK to me. The only question I have is how much detail you expect to have to go into regarding conjugate points, if someone asks why you're doing all this in small segments instead of just looking at the whole curve at once.

4. Oct 15, 2015

andrewkirk

The lay definition of geodesic that I find very intuitive is that it's the path you follow if you start off in a certain direction and then just 'follow your nose'. That's a folksy formulation of the 'parallel transport the tangent vector' definition, but I don't think it actually introduces any misconceptions.

A nice historical version of the more general notion of parallel transport is the (mythical?) ancient Chinese 'south-pointing chariot'. I don't think anybody knows whether one was ever built, but the idea was to put mechanisms on the chariot, connected to the wheels, so that it parallel transports the direction that a figure on top of the chariot is pointing, as it rolls along, including turning corners, going up and down hills etc. The chariot notion doesn't deal with the Earth's curvature. I suppose that's because the Ancient Chinese hadn't done enough differential geometry to work out that parallel transporting a vector around a loop on a sphere will make it point in a different direction when you get back to the start.

I think a modern version of that is the gyroscope, of the type used in aeroplanes. If we spin up a gyroscope that points in the direction you want to set off, and then we always follow the direction pointed by the gyroscope, we will be tracing out a geodesic, because the gyroscope's vector is being parallel-transported.

Another easily-graspable definition of a timelike geodesic is as a path with the property that an observer travelling along that path will feel perfectly weightless for the entire trip. ie it is a 'free-fall'. On reflection, this definition is superior to the above because it deals with spacetime geodesics, whereas the above only works for spatial geodesics. That is, the 'follow your nose' definition doesn't specify the spatial speed at which you traverse the path.

A nice example of the impact of curvature, that requires no calculus, is that if you start on the equator at longitude 90, spin up a gyroscope to point due North, travel to the North Pole, then down to the equator at longitude 0 (passing through Greenwich), then back to the starting point, the gyro will be pointing West. This is easily grasped using a pointing finger and moving one's hands around in the air. It's even easier if one has a large sphere, like a beach ball, handy.

The proposed definition in the OP looks like it will work. But I have no proof handy that there will only be a finite number of conjugate points between any two conjugate points. The nice thing about the parallel transport definition is that one can then just present the minimal property as a theorem stating that any curve of minimal length between two points will be a geodesic. There does not have to be a unique minimum, and not every geodesic between a given two points need be minimal.

Last edited: Oct 15, 2015
5. Oct 15, 2015

SlowThinker

I can't really say if your definition is correct, but I can tell that most people will have no idea what it means.
I can kind-of understand a pair of conjugate points as two focuses (foci?) of a gravitational lens. Even if I don't quite understand why these would have to be points rather than line segments.
But having a shortest path that has no conjugate points... ugh.
If you can spend 30+ minutes on the idea, it might make sense to some.

6. Oct 15, 2015

Staff: Mentor

I would stick with the standard definition. The preliminary machinery is not too onerous. I think it will be less difficult and more intuitive than the suggested alternative.

7. Oct 16, 2015

pervect

Staff Emeritus
The simplest thing I think one can say that doesn't have counterexamples (that I know of, at least) is that if we start out with the assumption there is no torsion - a standard assumption in GR (but not for instance a standard assumption in ECSK gravity), then when two points are "sufficiently close" together, a spatial geodesic is a curve of minimal proper length, while a time-like geodesic is a curve of maximal proper time.

This is already getting fairly complicated, and a mathemetician might possibly further object to the use of "minimum" rather than "greatest lower bound". It also doesn't really address what "sufficiently close" means in a rigorous manner, I believe it means "there exists a local convex neighborhood" around any point in which the statement is true.

That said, I find the basic idea of the curve of shortest length (or longest time) is much more intuitive and much more likely to be understood by a lay audience than a curve that parallel transports itself.

A useful examples of the need for the caveats with the "shortest/longes curve" definition is to consider two towns separated by a very tall hill. For a sufficiently tall hill, it will be shorter to go around the hill than straight over it. However, the longer path "over the hill" is still a geodesic path, even though it's not the shortest.

Good luck with your talk, whatever approach you decide is best.

8. Oct 16, 2015

Staff: Mentor

I would do parallel transport and tangent vectors graphically. Just show pictures of parallel transport and tangent vectors. People will understand the concept geometrically quite easily even if the symbols are a little intimidating.

9. Oct 16, 2015

bcrowell

Staff Emeritus
Actually this isn't true for the spacelike case. Example: In Minkowski space, pick a frame in which two events are simultaneous and lie at distance 1 from one another. Suppose we take the geodesic connecting them and deform it into an isosceles triangle with sides of slope b. If the triangle lies in the tx plane, then the length is $\sqrt{1-b^2}$. If the triangle lies in the ty plane, then the length is $\sqrt{1+b^2}$.

Right. An example I find helpful is the following one on the sphere. Let A, B, and C be three points that form an equilateral triangle on a great circle. Let AB be the shortest path between A and B, and BC the shortest path between B and C. Then ABC is a geodesic, but is not the shortest path between A and C.

Last edited: Oct 16, 2015
10. Oct 16, 2015

bcrowell

Staff Emeritus
Yeah, that works, but it's sort of like sketching what rigor would look like without actually doing the rigor.

11. Oct 16, 2015

Staff: Mentor

Oh. I thought that is what you were going for.

12. Oct 16, 2015

PAllen

The first part is not true in pseudo-Riemannian manifold. Even for arbitrarily close points connected by a spacelike geodesic, there is a lightlike (non-geodesic) path between that is obviously of 0 interval (this path will have the feature that, in any 1x3 foliation, it will go backwards in time, but it is still a path). Then, a nearby spaceilike path of arbitrarily small proper length (again, it must go back in time in any local foliation). Nutshell: spacelike paths have no extremal properties whatsoever of the 4-manifild if it is Minkowskian. Only timelike paths have extremal properties. I believer Synge's result, that I quoted, is the tightest that can be made. For spacelike geodesic between sufficiently close points, there is a neighborhood uniqueness property, but no extremal property.

13. Oct 16, 2015

andrewkirk

Despite the discussion having moved away from the 'minimising' definition, I wanted to follow through on this to see what constraints were needed for it to be valid.

John Lee provides formal proofs of the necessary results in his 'Riemannian Manifolds' in Chapter 6, subsection 'Geodsesics and minimizing curves'. Proposition 6.10 states:
[A geodesic ball around $p$ is a neighbourhood of $p$ that is diffeomorphic (say by diffeomorphism $\phi$) to an open ball in $\mathbb{R}^n$.
The radial geodesic from $p$ to $q$ is the pre-image under $\phi$ of the geodesic in $\mathbb{R}^n$ from $\phi(p)$ to $\phi(q)$]

Although Lee's proofs are for Riemannian Manifolds, they do not appear to use any properties that are not also possessed by pseudo-Riemannian manifolds, ie spacetime. Essentially the proofs rely on (earlier proofs of) the existence of a neighbourhood of $p$ with normal coordinates, and the uniqueness of the exponential map. For this application we need to interpret 'minimising' as referring to minimising the curve length over the set of all qualifying timelike curves, and we replace each instance of $\mathbb{R}^n$ by 'Minkowski Space'.

The 'close enough' property to which Synge refers would be the necessity for $q$ to be in a geodesic ball around $p$. The bijectiveness of the ball's diffeomorphism $\phi$ forbids it from including a point antipodal/conjugate to $p$. So in particular, $q$ cannot be conjugate to $p$.

With proposition 6.10 under our belt we can proceed as follows:

Let $\gamma:I\to M$ be a timelike geodesic such that $\gamma(0)=a,\ \gamma(1)=b$. For each point $\gamma(t)$ on the geodesic there is a collection of diffeomorphisms of neighbourhoods of $\gamma(t)$ to open balls in $\mathbb{R}^n$. Let $S(t)$ be the subset of those diffeomorphisms for which the image contains the unit open ball in $\mathbb{R}^n$, and let $D(t)$ be the set of domains of diffeomorphisms in $S(t)$

Then $\bigcup_{t\in I}D(t)$ is an open covering of the geodesic. Since the geodesic is compact, there exists a finite subcover of the geodesic by geodesic balls. If we choose our $h_i$s to be the centres of those balls, then the geodesic is the unique differentiable curve that minimises the distance through $a,b$ and all the finite set of $h_i$s.

EDIT: I realised we'd need to replace the open balls in Minkowski space by rectangles of the form $\{\vec{x}\ : \ \forall \alpha\in\{0,1,2,3\}\ |x^\alpha-(\phi(p))^\alpha|<\epsilon\}$, because open balls in Minkowski space will contain unlimited expanses of spacelike and lightlike separated points. The geodesic balls in $M$ would become rectangles that are the pre-images under $\phi$ of those rectangles in Minkowski space.

Last edited: Oct 16, 2015
14. Oct 17, 2015

bcrowell

Staff Emeritus
I think this can be strengthened quite a bit. For points spacelike in relation to one another and sufficiently close together, I think the geodesic between them has vanishing variation. That is, if you move points on the geodesic by epsilon, the length changes by something on the order of epsilon squared. (Of course this is not a rigorous formulation, e.g., you need some machinery to define how you measure epsilon.)

15. Oct 17, 2015

bcrowell

Staff Emeritus
@andrewkirk: Thanks very much for posting the info from Lee and discussing how it could be generalized to a semi-Riemannian space. That's very helpful. Your post refers to a theorem, but I don't see where you ever state the actual content of the theorem...?

It seems to me that this isn't going to work. As a counterexample, consider a spacetime describing a black hole forming by gravitational collapse. If you look at the Penrose diagram for this type of spacetime, you can see that the whole thing is diffeomorphic to an open ball in $\mathbb{R}^4$ (or in Minkowski space, since diffeomorphism doesn't depend on the metric). So by this definition, the geodesic ball around any point would be the whole spacetime. But it is certainly true that this spacetime contains conjugate points. (In fact, conjugate points occur generically in almost any physically realistic spacetime.) So I think maximal length within a geodesic ball may be a necessary condition for a timelike curve to be a geodesic, but it cannot be a sufficient one.

The definition I proposed in my OP was my attempt to make the "close enough" property rigorous. It is much more restrictive than the one you propose, and therefore I think it's more likely to allow us to state necessary and sufficient conditions for a curve to be a geodesic -- which is what we need for a definition.

I don't think this matters, because the definition you give of a geodesic ball only depends on diffeomorphism, and a diffeomorphism on Euclidean space is also a diffeomorphism on Minkowski space. But are you sure you're not paraphrasing the definition of a geodesic ball incorrectly? If the definition only depends on diffeomorphism, then I don't see why it would be called a *geodesic* ball, since geodesics are a metrical notion.

Last edited: Oct 17, 2015
16. Oct 17, 2015

bcrowell

Staff Emeritus
Googling for definitions of a geodesic ball in Riemannian spaces, what I seem to find is that it's the set of all points that you can reach from a given point by traveling along a geodesic for less than some fixed distance.

17. Oct 17, 2015

PAllen

Vanishing variation, yes, but still no extremal property. Pick any neighborhood of p, q, and the spacelike geodesic between them, however small. Then, imagine a zigzag (in time, in e.g. Minkowski space in standard coordinates) light path within the neighborhood, connecting them. Then a spacelike zigzag path can approach arbitrarily close to this. Note, the zig zags can be made arbitrarily small to stay within any tubular neighborhood of the geodesic. Thus I claim:

For any p,q, (however 'close'), and any neighborhood containing p,q and the spacelike geodesic between them (however small and tight to the geodesic), there exists a spacelike (non-geodesic) path with length arbitrarily close to zero between them, that remains in the neighborhood.

Last edited: Oct 17, 2015
18. Oct 17, 2015

bcrowell

Staff Emeritus
Actually, doesn't your zigzag example in #17 prove that I was completely wrong, and there's not even a vanishing variation?

19. Oct 17, 2015

PAllen

As I understand it, the variation being zero means that:

Given any function h that is zero at p and q, and continuous, and the one parameter family curves formed by adding ah to the given curve, the rate of change of the integral with 'a' goes to zero as 'a' goes to zero. The generality is that it doesn't matter what h you choose. In general, this variation being zero is only a necessary condition for any kind extremal property, even a local one, for arbitrary integrals. Actions and positive definite metric distances have special properties that allow this condition to be sufficient.

Note that my example does not violate this. You can take h to be a function such that for 'a' = 1, it produces any particular small amplitude zig zag path. Still, as 'a' goes to zero, the variation with 'a' of the proper length will be quadratic in 'a', for a spacelike geodesic (and not for some arbitrary spacelike curve).

20. Oct 17, 2015

bcrowell

Staff Emeritus
Right, that makes sense. Equivalently, this definition requires both the displacement and the derivative of the displacement to be of order a.

But it would be perhaps equally natural to consider variations such that only the displacement has to be of order a, not its derivative. Then spacelike curves are not stationary.