# Stationary/extremal L for timelike/null/spacelike geodesics

1. Jul 12, 2015

### bcrowell

Staff Emeritus
I would be interested in knowing if others think I have the correct analysis of whether length is stationary and/or extremal in the cases of geodesics that are timelike, null, and spacelike.

Timelike

In Minkowski space, the proper time $\tau=\int \sqrt{g_{ij}dx^i dx^j}$ (+--- metric) is both maximized and stationary. Apparently this need not hold in some cases in GR, if there are multiple geodesics connecting two events; WP references MTW pp 316-319 on this, but I don't have my copy handy. Is there a simple example?

Spacelike

In Minkowski space, with n+1 dimensions for $n\ge 2$, the proper length $\sigma=\int \sqrt{-g_{ij}dx^i dx^j}$ is stationary and is a saddle point. To see that it's a saddle point, pick a frame in which the two events are simultaneous and lie on the x axis. Deforming the geodesic in the xy plane does what we expect according to Euclidean geometry: it increases the length. Deforming the geodesic in the xt plane, however, reduces the length (as becomes obvious when you consider the case of a large deformation that turns the geodesic into a curve of length zero, consisting of two null line segments).

I don't know how much this has to be weakened for GR.

Null

This was discussed here last year: https://www.physicsforums.com/threads/null-geodesic-definition-by-extremisation.768196/ . I agree with some of the posts and disagree with others, and there didn't seem to be a consensus reached at the end.

In this case you have to define what "length" is. You can either take an absolute value, $L=\int \sqrt{|g_{ij}dx^i dx^j|}$, or not, $L=\int \sqrt{g_{ij}dx^i dx^j}$.

If you don't take the absolute value, L need not be real for small variations of the curve, and therefore you don't have a well-defined ordering, and can't say whether L is a max or min or neither.

Regardless of whether you take the absolute value, L doesn't have differentiable behavior for small variations around a null geodesic, so you can't say whether it's stationary; see https://www.physicsforums.com/threads/null-geodesic-definition-by-extremisation.768196/#post-4844366 , which seems to verify this aspect of my analysis.

If you do take the absolute value, then for the geodesic curve, the length is zero, which is the shortest possible. However, one can have nongeodesic curves of zero length, such as a lightlike helical curve about the t axis.

2. Jul 12, 2015

### Staff: Mentor

Sure, just compare a radial geodesic in Schwarzschild spacetime that starts upward from some $r$, rises, then falls back to the same $r$, with a circular orbit geodesic at that same $r$. All you need is to adjust the initial upward velocity of the radial geodesic so that it intersects the circular orbit geodesic again after one orbit. Then the radial geodesic will have a larger elapsed proper time than the circular orbit one between the same two events; so the circular orbit geodesic is not a global maximum of proper time between those events (though it is a local extremum).

3. Jul 12, 2015

### bcrowell

Staff Emeritus
Hmm...thanks, that's a helpful example. Well, it's obvious that in such a case the two geodesics can't both be the global extremum, since their proper times are unequal. Maybe it will clarify things if I give the text from the WP article:

I think what they're saying is that such a geodesic may not even be a local extremum.

It's not obvious to me how to tweak your example to prove this. The two orbits will in general have different periods, so to get them to coincide at a later event in spacetime, we'd probably have to choose a case in which one of them did m orbits while the other did n orbits. But then it's not a family of geodesics that's a function of a continuous parameter.

Last edited: Jul 12, 2015
4. Jul 12, 2015

### bcrowell

Staff Emeritus
Here's an example that is not a local extremum but may be stationary. In a Schwarzschild spacetime, you launch a projectile from a spatial point P and it ends up at P's antipodal spatial point Q. The azimuthal angle $\phi$ about the line PQ was arbitrary, so we have a family of geodesics that are a function of one continuous parameter and that all take the same amount of time to get from P to Q, and by symmetry all of them have the same proper time.

But that still doesn't seem to be an example quite as strong as what the WP article suggests.

Last edited: Jul 12, 2015
5. Jul 12, 2015

### Staff: Mentor

If "extremum" is interpreted so as not to include "saddle point", then any null geodesic would qualify, since a saddle point is a stationary point but not an extremum.

I don't see how one could.

I'm not sure I understand. If the spatial points P and Q are both fixed, and the time it takes the projectile to travel between them is fixed, doesn't that fix a unique geodesic? It doesn't seem like there are any free parameters left.

6. Jul 12, 2015

### bcrowell

Staff Emeritus
They're talking about timelike geodesics. (And as described in my #1, I don't think a null geodesic is stationary.)

Let the Schwarzschild spatial coordinates be $(r,\theta,\phi)$. Spatial point P is at $r=r_0$, $\theta=0$. Q is at $r=r_0$, $\theta=\pi$. Pick some geodesic from P to Q that lies in the $\phi=0$ half-plane. In any other half-plane defined by another choice of $\phi$, there is a geodesic from P to Q that takes the same time.

Last edited: Jul 12, 2015
7. Jul 12, 2015

### Staff: Mentor

I read through the discussion you linked to, which does raise an issue I don't have an immediate answer to; I'll have to check further into this when I get a chance.

Ah, ok got it--because of the rotational symmetry there will be a one-parameter family of geodesics connecting the same two events. The same would be true for the circular orbit geodesics I described in my earlier example.

I'm not sure that these geodesics (either "half-circle" or full circle) would be an example of what's mentioned in the Wikipedia article, because I don't think there are any "nearby" geodesics to any of these that connect the same two events and have a longer proper time. The article references MTW, so I'll have to check it when I get a chance to see what example they had in mind.

8. Jul 12, 2015

### bcrowell

Staff Emeritus
Right, I agree.

Here's a possible way to modify it so it does make a full example of what they're describing. We have some rotating body, and the surrounding spacetime is described by a Kerr metric. The axis of rotation is $\theta=\{0,\pi\}$. Let spatial point P be at $r=r_0$, $\theta=\pi/2$, $\phi=0$, and let Q be similar but with $\phi=\pi$. If the body's rotation was zero, then we would have a family of semicircular trajectories from P to Q that would all have equal proper times. If we turn on the rotation, each of these orbits can be adjusted within its own plane so as to cause all of them to arrive at the same time, but now the proper times will all be different.

Maybe MTW have a simpler example or one that's more obviously correct than mine.

9. Jul 13, 2015

### PAllen

What I've seen claimed (e.g. in Synge's GR textbook) is that for any timelike geodesic, if you pick points on it sufficiently close to each other, the given geodesic between them is unique (and maximal, both locally and globally). Further, in any non-singular region, a sufficiently small ball around a given event will have unique timelike geodesics between the given event any any other events in the ball that can be connected by timelike curves. These features are used to make his 'Two Point World Function' well defined, which he uses heavily to do geometric series expansions without coordinates. However, for arbitrary events in arbitrary spacetimes that can be connected by a timelike world line, there is no guarantee that the geodesic bertween them is maximal, even locally.

This seems consistent with consistent with the analysis above.

10. Jul 13, 2015

### Staff: Mentor

I'm probably being dense, but I don't see how this adjustment can be made. It seems to me that, if the radial coordinate of the orbit is constant, and we pick a particular plane (value of $\theta$), then the orbit is fully determined. In Kerr spacetime, as you say, the proper time as a function of $\theta$ will not be constant; but I don't see that the coordinate time can be made to be constant either.

11. Jul 13, 2015

### PAllen

It seems there is a much simpler example of case a timelike geodesic not being a local maximum. Consider one complete non-circular (not quiet closed) orbit in SC spacetime. Forget about the near radial geodesic that happens to be the global maximum in this case. Consider local deformations of the orbit (non-inertial) such that they (a) go a bit faster at aphelion and slower at peri-helion and (b) the reverse [both (a) and (b) deformations adjusted to meet the geodesic after one 'orbit']. While I haven't calculated it explicitly, it seems clear that (a) deformations may elapse less proper time than the geodesic, while (b) deformations may elapse more. Thus, the geoesic is a saddle point.

12. Jul 13, 2015

### bcrowell

Staff Emeritus
I'm talking about adjusting the orbit so it becomes non-circular. If you aim low and shoot at high speed, you can get to the other side faster. If you aim high and shoot at low speed, you can get there after more time.

13. Jul 13, 2015

### bcrowell

Staff Emeritus
What is SC spacetime? Does SC mean Schwarzschild?

I'm not convinced at all that this example works. We do normally expect timelike geodesics to be local maxima of the proper time. Something special has to be going on for this not to be the case, and from the WP article, it sounds like the something special is for "a pair of widely-separated events to have more than one time-like geodesic that connects them." Since you haven't invoked any such property in your example, it seems unlikely to me that it works out.

14. Jul 13, 2015

### Staff: Mentor

Non-circular, but still passing through two antipodal points at the same radial coordinate? (In that case, the "orbits" would not be geodesic.)

15. Jul 13, 2015

### Orodruin

Staff Emeritus
I do not have a direct example in the case of an indefinite metric, but the standard example for a Riemannian metric would be the longer great circle between two points on a sphere, which is a local saddle point.

This depends on what you mean by "local". If you with "local" mean "proper time between two nearby points", then yes. If you mean "local" in the space of all world-lines, then no. The issue is the same as with the great circle above, the great circle minimises the distance between nearby points, but in the space of curves on the sphere, it is a saddle point.

16. Jul 13, 2015

### bcrowell

Staff Emeritus
What makes you think that? You have two degrees of freedom to play with on the initial conditions, so I don't see why it should be a problem to hit a target.

17. Jul 13, 2015

### bcrowell

Staff Emeritus
Local as in the calculus of variations. Local means that you can change the curve, but you can't move any point on the curve by more than $\epsilon$ (as measured, for example, by coordinate changes). We then expect that the limit of small $\epsilon$, the proper time will change by $O(\epsilon^2)$.

18. Jul 13, 2015

### Staff: Mentor

Because I don't think a geodesic that has a varying radial coordinate can have two antipodal points where the radial coordinate is the same--unless perhaps it is so close to the central body that its perihelion precesses by $\pi$ on every orbit.

19. Jul 13, 2015

### PAllen

What's special is that the given geodesic is not a global maximum, by construction. We have an elliptical orbit, not quite circular (and not elliptical in that it doesn't quite close due to perihelion shift). We know the nearly radial geodesic (as in Peter's example, which I've used many times as well) is the global geodesic. MTW makes the statement that whenever more than one timelike geodesic connects two events, then a 'typical one' is 'usually' a saddle point.

However, I agree the disagreement can't be resolved without a calculation or more formal mathematical argument. Otherwise, we just have my intuition versus yours.

20. Jul 13, 2015

### bcrowell

Staff Emeritus
I don't see why you would think that. Say I position myself at P and raise my cannon to 3 degrees above the horizontal. If my muzzle velocity $v$ is too low, my cannonball will pass under the antipode Q. If too high, it passes over. By the intermediate value theorem, there is guaranteed to be some value of $v$ for which it hits Q.