A Minimal property of Spacelike geodesics in GR/curved spacetime?

[Kostik]

You have to look at the general form of the variation to conclude this. As has already been indicated, it is only true in some cases. Other cases will be saddle points.

In SR, it's clear that a timelike geodesic maximizes $\int ds = \int \sqrt{dt^2 - dx^2 - dy^2 - dz^2}$. Can you give a concrete example in GR of a gravitational field and a timelike geodesic where the geodesic is a saddle point of the integral $\int ds$?

 

[Orodruin]

In SR, it's clear that a timelike geodesic maximizes $\int ds = \int \sqrt{dt^2 - dx^2 - dy^2 - dz^2}$. Can you give a concrete example in GR of a gravitational field and a timelike geodesic where the geodesic is a saddle point of the integral $\int ds$?

@PAllen already did in post #23

 

[Orodruin]

He said both geodesics were local maxima. Neither one is a local saddle point.

Read again.

 

[Kostik]

@PAllen already did in post #23

He said both geodesics were local maxima. Neither one is a local saddle point.

 

[Kostik]

Read again.

Sorry, I see now. So, in GR, neither timelike nor spacelike geodesics are local extrema. That's interesting.

 

[Kostik]

By an argument similar to @Orodruin’s sphere case, the orbital geodesic is actually a saddle point, not even a local maximum.

Can you explain descriptively how to vary these two timelike geodesics (the orbital and the ballistic ones) to make $\int ds = \int \sqrt{g_{\mu\nu}dx^\mu dx^\nu}$ larger and smaller? It's hard to visualize.

More specifically, we can assume the metric is $$ds^2 = (1-2m/r)dt^2 - (1-2m/r)^{-1}dr^2 - r^2d\theta^2 - r^2\sin^2\theta d\phi^2 \,.$$ Can you describe how to vary the geodesics (orbits) to make $\int ds$ larger and smaller?

 

[A.T.]

By an argument similar to @Orodruin’s sphere case, the orbital geodesic is actually a saddle point, not even a local maximum.

What about the geodesic worldline for the center of a spherical mass? Is it a local minimum or a saddle point?

 

[Orodruin]

What about the geodesic worldline for the center of a spherical mass? Is it a local minimum or a saddle point?

Geodesics are always a local maximum or saddle point (remember that the geodesic in SR maximizes proper time).

In this case it would depend on the time interval. If events are close enough it is always a max.

However, for longer times I believe it will be a saddle point. The center of the spherical mass has the lowest gravitational potential and therefore the maximal time dilation for a stationary observer. Perturb the stationary geodesic by going slightly off-center where time dilation is lower, stay there until you accumulate enough proper time and then go back and this will result in a larger proper time.

 

[Kostik]

@Orodruin How do we know that the geodesic in SR maximizes proper time? The zig-zag argument shows that there are always variations with arbitrarily small proper time. But how do we know in SR that a timelike geodesic is NOT a saddle point?

EDIT: I think I have it. The geodesic equation implies $v^\mu$ is constant. Any variation from a straight line timelike path increases the proper time.

 

[PeterDonis]

Can you explain descriptively how to vary these two timelike geodesics (the orbital and the ballistic ones) to make $\int ds = \int \sqrt{g_{\mu\nu}dx^\mu dx^\nu}$ larger and smaller?

The ballistic geodesic is a local (and global) maximum. Only the orbital one is a saddle point. Even in curved spacetime there will be some timelike geodesics which are maxima. Just not all of them.

 

[PeterDonis]

for longer times I believe it will be a saddle point

Not always. For example, the radial ballistic geodesic in the example @PAllen gave in post #23 is a local maximum.

 

[PAllen]

Not always. For example, the radial ballistic geodesic in the example @PAllen gave in post #23 is a local maximum.

Actually, it is a global maximum, but that is quite a chore to prove. You have to bring in global topology, but in this case, assuming standard Kruskal topology, it is true.

 

[Orodruin]

Not always. For example, the radial ballistic geodesic in the example @PAllen gave in post #23 is a local maximum.

The discussion was about the geodesic at the center of the spherical body, not about the ballistic geodesic. I was replying to and quoted @A.T. ‘s post that specified this.

 

[PAllen]

Not always. For example, the radial ballistic geodesic in the example @PAllen gave in post #23 is a local maximum.

I think @Orodruin assumed that, and was only referring to the orbital geodesic, which is analogous to the larger great circle path on a sphere. @Orodruin described something similar to what I had in mind to argue for a saddle point. My version was to transfer at near c to a slightly larger radius, then non-inertially move in a circle there at the speed (in coordinate terms; so a slightly lower ancular speed than the orbit) of the orbit for most of the way, than, just in time to meet that orbital closure event, move at near c back. This procedure will fail for small segments of the orbit, but for some large enough segment, it will work; in particular, for the whole orbit.

 

[Orodruin]

@Orodruin How do we know that the geodesic in SR maximizes proper time? The zig-zag argument shows that there are always variations with arbitrarily small proper time. But how do we know in SR that a timelike geodesic is NOT a saddle point?

Without loss of generality, go to the frame where the events occur at the same spatial point. The proper time for any curve between the points is then
$$
\tau = \int_{t_0}^{t_1} \sqrt{1 - \dot x(t)^2} dt \leq t_1 - t_0$$ with equality if and only if $\dot x(t) = 0$ for all $t$.

Edit: The further restriction is that $x(t_1) = x(t_0)$.

Edit 2: Fixed typo, ”ane” -> ”and”

 

[Orodruin]

I think @Orodruin assumed that, and was only referring to the orbital geodesic, which is analogous to the larger great circle path on a sphere. @Orodruin described something similar to what I had in mind to argue for a saddle point.

I was referring to the geodesic at the center of the body, which is what @A.T. asked about in the post I quoted.

 

[PAllen]

Sorry, I see now. So, in GR, neither timelike nor spacelike geodesics are local extrema. That's interesting.

No, in GR, some timelike geodesics are saddle points and some are local maxima. The usual case is that for sufficiently close points, where one is in the future of the other, there will be a unique timelike geodesic that is a local maximum. For points not so close, there will often be more than one geodesic, and one of them will be local maximum (in realistic cases, it will also be a global maximum).

 

[cianfa72]

@Orodruin already gave an example on a sphere where there are two geodesic paths between two points, only one of which is minimal.

Actually I believe it was @martinbn who gave the example of the sphere in post #10.

 

[PAllen]

Without loss of generality, go to the frame where the events occur at the same spatial point. The proper time for any curve between the points is then
$$
\tau = \int_{t_0}^{t_1} \sqrt{1 - \dot x(t)^2} dt \leq t_1 - t_0$$ with equality if ane only if $\dot x(t) = 0$ for all $t$.

Edit: The further restriction is that $x(t_1) = x(t_0)$.

Simpler than what I was starting to write up - what I know of as the Legendre condition distinguishing whether you have a possible local minimum versus a possible local maximum for a variation.

 

[JimWhoKnew]

Can you explain descriptively how to vary these two timelike geodesics (the orbital and the ballistic ones) to make $\int ds = \int \sqrt{g_{\mu\nu}dx^\mu dx^\nu}$ larger and smaller? It's hard to visualize.

More specifically, we can assume the metric is $$ds^2 = (1-2m/r)dt^2 - (1-2m/r)^{-1}dr^2 - r^2d\theta^2 - r^2\sin^2\theta d\phi^2 \,.$$ Can you describe how to vary the geodesics (orbits) to make $\int ds$ larger and smaller?

In accord with @PAllen's observation in #23 , I think we can slightly generalize @Orodruin's example here to show that the orbital geodesic is a saddle for angles larger than $\pi$.

To do so, note that the orbital geodesic at the equator $\theta=\frac \pi 2$ and at constant $r=R$ can be parametrized by $\phi$. So if we take varied paths with the same form $$\theta(\phi) = \frac \pi 2 + \eta f(\phi)$$we get $$\tau[\theta(\phi)] = \int_0^{a\pi} \sqrt{\left(1-\frac{2m}{R}\right)\left(\frac{dt}{d\phi}\right)^2 -R^2 \eta^2 f'(\phi)^2-R^2 \cos^2(\eta f(\phi))} ~d\phi$$Since $\frac{dt}{d\phi}$ is constant, the first and second derivatives of $\tau[\theta(\phi)]$ wrt $\eta$ when evaluated at $\eta=0$, are proportional to @Orodruin's, and the same further reasoning applies.

Edit: corrected typo in equation

 

[Kostik]

Without loss of generality, go to the frame where the events occur at the same spatial point. The proper time for any curve between the points is then
$$
\tau = \int_{t_0}^{t_1} \sqrt{1 - \dot x(t)^2} dt \leq t_1 - t_0$$ with equality if and only if $\dot x(t) = 0$ for all $t$.

Edit: The further restriction is that $x(t_1) = x(t_0)$.

Why $x(t_1) = x(t_0)$? That can't be: a geodesic in SR is motion in a constant direction with constant speed. This condition will not be satisfied. But otherwise, the logic seems sound.

 

[PeterDonis]

Why $x(t_1) = x(t_0)$?

Because of the first sentence in the post you quoted:

Without loss of generality, go to the frame where the events occur at the same spatial point.

 

[Kostik]

Oh, of course, but in the reference frame of the particle traveling along the geodesic, $dx = 0$, so this condition is superfluous.

 

[PeterDonis]

Oh, of course, but in the reference frame of the particle traveling along the geodesic, $d\bf{x} = 0$, so this condition is superfluous.

Remember that @Orodruin was answering your question about how you know that the geodesic maximizes proper time. That means you can't just consider curves for which $dx = 0$ as it does in that frame for the geodesic. So you have to impose the condition $x (t_1) = x (t_0)$ on the endpoints.

 

[Orodruin]

Oh, of course, but in the reference frame of the particle traveling along the geodesic, $dx = 0$, so this condition is superfluous.

It is not, you are considering variations around the geodesic. Those will not have dx = 0.

 

[Kostik]

In accord with @PAllen's observation in #23 , I think we can slightly generalize @Orodruin's example here to show that the orbital geodesic is a saddle for angles larger than $\pi$.

To do so, note that the orbital geodesic at the equator $\theta=\frac \pi 2$ and at constant $r=R$ can be parametrized by $\phi$. So if we take varied paths with the same form $$\theta(\phi) = \frac \pi 2 + \eta f(\phi)$$we get $$\tau[\theta(\phi)] = \int_0^{a\pi} \sqrt{\left(1-\frac{2m}{R}\right)\left(\frac{dt}{d\phi}\right)^2 -R^2 \eta^2 f'(\phi)^2-R^2 \cos^2(\eta f(\phi))} ~d\phi$$Since $\frac{dt}{d\phi}$ is constant, the first and second derivatives of $\tau[\theta(\phi)]$ wrt $\eta$ when evaluated at $\eta=0$, are proportional to @Orodruin's, and the same further reasoning applies.

Edit: corrected typo in equation

This is interesting - so a circular orbit in spacetime is a saddle point of the proper time if the angular extent of the orbit is $> \pi$, but not $< \pi$?

While this is intuitively clear for the length of a major arc vs. a minor arc on the sphere, it's not intuitive for the proper time of an orbit.

Also, I assume the same results holds for an elliptical orbit, except we will have to choose a different function $f(\phi)$ to show that the proper time is not a maximum?

 

[Kostik]

Digging out a post from my first year at PF, now 11 years old:

Post in thread 'Understand the major arc connecting two points on a sphere'
https://www.physicsforums.com/threa...ng-two-points-on-a-sphere.767025/post-4829841

In that post I showed explicitly that the major arc is indeed a saddle point.

Really nice demonstration. Where do you actually need the boundary conditions $f(0) = f(a\pi) = 0$?

 

[Kostik]

It is not, you are considering variations around the geodesic. Those will not have dx = 0.

I think of your argument as follows: For any timelike path from $P$ to $Q$, in any inertial coordinate system $x^\mu$ , we have $$\Delta\tau = \int ds = \int \sqrt{dt^2 - d\textbf{x}^2 } = \int \sqrt{1-v^2} \, dt \le \int dt = t_Q - t_P \,\, ,$$ with equality if $v=0$. However, $v=0$ if and only if $x^\mu$ are the coordinates of the frame of reference of a particle moving with velocity $\textbf{v}$, which is a geodesic.

Is this essentially correct?

 

[JimWhoKnew]

This is interesting - so a circular orbit in spacetime is a saddle point of the proper time if the angular extent of the orbit is $> \pi$, but not $< \pi$?

The specific example says that there is a saddle for $\Delta\phi >\pi$. It doesn't prove the impossibility of saddles for all $\Delta\phi\leq\pi$, since only variations in the $\theta$ direction were considered.

While this is intuitively clear for the length of a major arc vs. a minor arc on the sphere, it's not intuitive for the proper time of an orbit.

You may try to visualize it as taking the major arc of a ring and uniformly stretching it in a fourth orthogonal dimension, as to make it a segment of a helix. Then stretch the variations accordingly.

The "mathematical intuition" comes from @Orodruin's example. By restricting ourselves to constant $r$ and uniform $dt$ (i.e. constant $\frac{dt}{d\phi}$), we are left with the above relation $$\tau[\theta(\phi)] = \int_0^{a\pi} \sqrt{\left(1-\frac{2m}{R}\right)\left(\frac{dt}{d\phi}\right)^2 -R^2 \eta^2 f'(\phi)^2-R^2 \cos^2(\eta f(\phi))} ~d\phi$$The $g_{tt}$ dependent term is the same for all considered paths, while we expect from @Orodruin's example that the contribution from the angular terms $$-R^2 \eta^2 f'(\phi)^2-R^2 \cos^2(\eta f(\phi))$$can be made either larger or smaller relative to the geodesic.

Also, I assume the same results holds for an elliptical orbit, except we will have to choose a different function $f(\phi)$ to show that the proper time is not a maximum?

Orbits in Schwarzschild geometry are not exactly elliptical in general (remember the precession of Mercury). In non-circular orbits, both $g_{tt}$ and $g_{rr}$ vary along the geodesic (and varied paths), which makes calculations considerably more complex. From the circular case, I'd expect saddles for sufficiently small eccentricities. I suppose that treatment of the general case may be found in literature.

I apologize for frequently re-editing my posts. I have the LaTeX rendering issue.

 

[Kostik]

In accord with @PAllen's observation in #23 , I think we can slightly generalize @Orodruin's example here to show that the orbital geodesic is a saddle for angles larger than $\pi$.

To do so, note that the orbital geodesic at the equator $\theta=\frac \pi 2$ and at constant $r=R$ can be parametrized by $\phi$. So if we take varied paths with the same form $$\theta(\phi) = \frac \pi 2 + \eta f(\phi)$$we get $$\tau[\theta(\phi)] = \int_0^{a\pi} \sqrt{\left(1-\frac{2m}{R}\right)\left(\frac{dt}{d\phi}\right)^2 -R^2 \eta^2 f'(\phi)^2-R^2 \cos^2(\eta f(\phi))} ~d\phi$$Since $\frac{dt}{d\phi}$ is constant, the first and second derivatives of $\tau[\theta(\phi)]$ wrt $\eta$ when evaluated at $\eta=0$, are proportional to @Orodruin's, and the same further reasoning applies.

Doing the work, I get $$\frac{d^2 \tau}{d\eta^2}(0) = \frac{1}{\sqrt{K-1}}\int_0^{a\pi} \left[ f(\phi)^2 - f'(\phi)^2 \right] d\phi $$ where $$K = (1-2m/R)(dt/d\phi)^2.$$ In order that $\tau(\eta)$ has a possible inflection point at $\eta = 0$, we must have $K>1$. Since $dt/d\phi$ is the inverse of the angular velocity $\omega$, we have $$K = (1-2m/R)(R^3/m) = R^3/m - 2R^2.$$ Thus, $K>1$ requires that $$R^3/m - 2R^2 - 1 > 0 .$$ The first term dominates the second two terms even for very small $R$. (But if $m$ is very large, this condition may restrict the minimum orbit radius.)

Note that the two $f(\phi)$ used by @Orodruin in his major arc example to demonstrate that the geodesic is a saddle point of the proper length now reverse roles to show that the circular orbit is a saddle point of the proper time.