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

[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.

 

[Kostik]

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.

True. Unlike in the case of the sphere in 3-space, in this case we have the flexibility to vary the radius as well. I might play around with that. I would be surprised if only "major arcs" were saddle points of proper time, but nor "minor arcs".

 

[JimWhoKnew]

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 $$K = (1-2m/R)(dt/d\phi)^2.$$

I got a different proportionality factor. Note that your $K$ and "1" don't have the same dimensions.

Since $dt/d\phi$ is the inverse of the angular velocity $\omega$, we have $$K = (1-2m/R)(R^3/m) = R^3/m - 2R^2.$$

The relativistic expression for $\omega$ in Schwarzschild coordinates should be used. Are you sure it is $\sqrt{m/R^3}$ ? (my memory is not to be trusted)

this condition may restrict the minimum orbit radius

It is well known that the geodesics of circular orbits at $r=3m$ are null, so we should expect a restriction.

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.

Come on, give me some credit

 

[PAllen]

I pose a question I don’t have an answer to at this time. In calculus of variations, the normal reason a geodesic may fail to be a local extremum is the occurrence of conjugate points. If you consider a geodesic starting from a point, it will be an extremum to an end point before its first conjugate point. In GR, a common case of a non-extremal geodesic is when two geodesics intersect at two points. Between the points of intersection, only one of the geodesics will be extremal. So my question is whether there is any relation between intersecting geodesics and conjugate points? At this moment, I don’t have time to think about it, so want to ask it before I forget the idea (I am of that age where that is too common).

 

[JimWhoKnew]

True. Unlike in the case of the sphere in 3-space, in this case we have the flexibility to vary the radius as well.

By passing from the 2 dimensions of the sphere's surface to 4 dimensional spacetime we added 2 dimensions. so in principle there should be 3 "independent directions" in which variations are allowed.

 

[martinbn]

I pose a question I don’t have an answer to at this time. In calculus of variations, the normal reason a geodesic may fail to be a local extremum is the occurrence of conjugate points. If you consider a geodesic starting from a point, it will be an extremum to an end point before its first conjugate point. In GR, a common case of a non-extremal geodesic is when two geodesics intersect at two points. Between the points of intersection, only one of the geodesics will be extremal. So my question is whether there is any relation between intersecting geodesics and conjugate points? At this moment, I don’t have time to think about it, so want to ask it before I forget the idea (I am of that age where that is too common).

In a intuitive non rigorous way conjugate points are the points where nearby geodesics intersect. To precise statements is the existence of a Jacobi field that vanishes at the endpoint.

 

[Kostik]

I got a different proportionality factor. Note that your $K$ and "1" don't have the same dimensions.

The relativistic expression for $\omega$ in Schwarzschild coordinates should be used. Are you sure it is $\sqrt{m/R^3}$ ? (my memory is not to be trusted)

It is well known that the geodesics of circular orbits at $r=3m$ are null, so we should expect a restriction.

Come on, give me some credit

Thank you, I lost the $r^2$ factors in the metric for the angular coordinates. Re-doing it, I find the same formula holds but with $$K=\frac{R}{m}-2 \, .$$ Thus, the condition $K>1$ is equivalent to $R>3m$. (This is much nicer than the cubic equation for $R$ I had before.)

I think it's OK to use $$\omega^2 = \left( \frac{d\phi}{dt} \right)^2 = \frac{m}{r^3} \, .$$ After all, the circular orbit is the same in GR as in Newtonian mechanics.

And thank you for your help on this!

 

[JimWhoKnew]

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.

When I wrote "give me some credit" I intended to hint that I obviously needed no help in noting the reversal on my own. But "likes" are nice too.

 

[JimWhoKnew]

Re-doing it, I find the same formula holds but with $$K=\frac{R}{m}-2 \, .$$

There is still a missing $R$ in the numerator. $\frac{d^2 \tau}{d\eta^2}$ should have dimensions of time.

I think it's OK to use $$\omega^2 = \left( \frac{d\phi}{dt} \right)^2 = \frac{m}{r^3} \, .$$ After all, the circular orbit is the same in GR as in Newtonian mechanics.

I can't look it up right now (yes, I'm aware I have access to the internet), but as I remember, circular orbits can be either stable, unstable or semi-stable. Do they all obey the Kepler form? Can you provide a reference?

And thank you for your help on this!

You're welcome

 

[Kostik]

There is still a missing $R$ in the numerator. $\frac{d^2 \tau}{d\eta^2}$ should have dimensions of time.

Yes, I forgot there is an $R$ in front of the integral sign (which affects nothing)!

 

[Kostik]

I can't look it up right now (yes, I'm aware I have access to the internet), but as I remember, circular orbits can be either stable, unstable or semi-stable. Do they all obey the Kepler form? Can you provide a reference?


You're welcome

I also recall that circular orbits are unstable if $r < 6m$ (?). But does this affect the constant $K$?

 

[PAllen]

I also recall that circular orbits are unstable if $r < 6m$ (?). But does this affect the constant $K$?

The instability may suggest that even a short orbital segment is a saddle point rather than extremal for such an orbit.

 

[JimWhoKnew]

I can't look it up right now (yes, I'm aware I have access to the internet), but as I remember, circular orbits can be either stable, unstable or semi-stable. Do they all obey the Kepler form? Can you provide a reference?

I carried out the calculation, aided by chapter 25 in MTW, and got that Kepler's Law $$\omega^2 = \left( \frac{d\phi}{dt} \right)^2 = \frac{m}{r^3}$$ is indeed the general result for circular orbits also in Schwarzschild coordinates.

 

[Kostik]

I carried out the calculation, aided by chapter 25 in MTW, and got that Kepler's Law $$\omega^2 = \left( \frac{d\phi}{dt} \right)^2 = \frac{m}{r^3}$$ is indeed the general result for circular orbits also in Schwarzschild coordinates.

Actually, using $v^2 = m/(r-2m)$ I am getting $$\omega^2 = \frac{m}{r^2(r-2m)} .$$

 

[PeterDonis]

Actually, using $v^2 = m/(r-2m)$ I am getting $$\omega^2 = \frac{m}{r^2(r-2m)} .$$

No, that's not correct, because in Schwarzschild spacetime, the implicit substitution $v = \omega r$ that you are using is not correct. The correct relationship is

$$
v = \frac{\omega r}{\sqrt{1 - 2m / r}}
$$

 

[Kostik]

Damn, thank you!

 

[Orodruin]

The instability may suggest that even a short orbital segment is a saddle point rather than extremal for such an orbit.

It does not. In a sufficiently small neighbourhood, spacetime will be approximately Minkowski - sufficiently so to ensure geodesics between timelike separated events in such neighbourhoods are local maxima of proper time.

 

[romsofia]

I carried out the calculation, aided by chapter 25 in MTW, and got that Kepler's Law $$\omega^2 = \left( \frac{d\phi}{dt} \right)^2 = \frac{m}{r^3}$$ is indeed the general result for circular orbits also in Schwarzschild coordinates.

In case you'd like some online sources for this in the future, here is a link: (for the circular orbit case) https://sites.science.oregonstate.edu/physics/coursewikis/GGR/book/ggr/ocircle.html

For general quick discussion on orbits: https://sites.science.oregonstate.edu/physics/coursewikis/GGR/book/ggr/orbits.html