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

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The discussion centers on the properties of spacelike and timelike geodesics in both special and general relativity, particularly regarding their extremal characteristics. It is established that timelike geodesics maximize proper time, while spacelike geodesics do not exhibit extremal properties, acting as saddle points instead. The conversation highlights that in curved spacetime, variations of spacelike geodesics can yield both longer and shorter intervals, complicating the notion of minimality. Additionally, it is noted that timelike geodesics may not always be maximal if conjugate points exist along their paths. Overall, the complexities of geodesic behavior in different spacetime geometries are emphasized.
  • #61
JimWhoKnew said:
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".
 
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  • #62
Kostik said:
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.

Kostik said:
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)

Kostik said:
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.

Kostik said:
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 :smile:
 
  • #63
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).
 
  • #64
Kostik said:
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.
 
  • #65
PAllen said:
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.
 
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  • #66
JimWhoKnew said:
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 :smile:
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!
 
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  • #67
Kostik said:
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.
 
  • #68
Kostik said:
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.

Kostik said:
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?

Kostik said:
And thank you for your help on this!
You're welcome
 
  • #69
JimWhoKnew said:
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)!
 
  • #70
JimWhoKnew said:
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##?
 
  • #71
Kostik said:
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.
 
  • #72
JimWhoKnew said:
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.
 
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  • #73
JimWhoKnew said:
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 $$\omega^2 = \frac{m}{r^2(r-2m)} .$$
 
  • #74
Kostik said:
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

$$
v = \frac{\omega r}{\sqrt{1 - 2m / r}}
$$
 
  • #75
Damn, thank you!
 
  • #76
PAllen said:
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.
 
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  • #77
JimWhoKnew said:
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
 

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