Are two-body orbits with a cosmological constant unstable?

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  • #1
kimbyd
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Summary:
In the two-body problem, circular orbits can be stable with a cosmological constant. But can elliptical orbits?
For background, consider this paper, which describes circular orbits for the two-body problem in the presence of a cosmological constant:
https://arxiv.org/abs/1906.05861

What they describe is a system with three regimes of behavior: stable circular orbits below a certain radius, unstable circular orbits above that radius but below another, and no orbits beyond that second radius.

What I'm wondering is: do elliptical orbits screw this up? My searches so far haven't come up with an answer. My reasoning for why elliptical orbits might always exhibit instability in the presence of a cosmological constant starts with the circular case. In particular, it's possible to have a stable circular orbit for any spherically-symmetric potential with the right derivatives.

But an elliptical orbit spends some of its time closer to the center of mass, some of its time further away. The elliptical orbit always spends more of its time further away from the center of mass than closer. Might this not lead to the objects in elliptical orbit gaining some small impulse away from one another?

This wouldn't immediately cause the orbit to get larger in average radius, but I believe it would cause the orbit to become more eccentric over time. Eventually the eccentricity would grow large enough that that it enters the unstable regime and the objects actually do go apart (unless the objects crash due to their finite radii).

Anyway, if elliptical orbits are unstable in the presence of a cosmological constant, it would have to be a very, very slight instability for something like our solar system, so much so that I'd bet it's unmeasurable at present.
 

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  • #2
Orodruin
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Looking at the paper, they end up with a normal effective potential problem. Just a slightly different one from the standard two-body problem due to the presence of an additional ##r^2## term (as there is a linear ##r## term in the derivative of the potential).

We can therefore conclude:
- There are no elliptical orbits. Orbits will not close due to the additional term. Of course, orbits that are so close that the cosmological constant is almost negligible will be close to elliptical but not quite close.
- It is perfectly possible to have stable non-circular orbits.
- Orbits that are not stable will have the system separate within one radial oscillation.
 
  • #3
kimbyd
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Well, yes, the orbits would not be precisely elliptical. Much like observed orbits such as Mercury's, the perihelion of their orbits would precess.

Do you have a source for what is required for orbital stability in this case?
 
  • #4
Orodruin
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All non-circular orbits are stable because only the circular orbits occur at the extreme points of the effective potential. The reason circular orbits can be unstable is that they can correspond to maxima of the effective potential. In that case, ##r## constant and consequently ##\dot r = \ddot r = 0## is a formal solution to the EoM, but it is unstable.

For non-circular orbits however, the energy is not equal to the effective potential at one of its maxima. Therefore, small deviations from the orbit will lead to another (typically also non-circular) orbit.
 
  • #5
kimbyd
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All non-circular orbits are stable because only the circular orbits occur at the extreme points of the effective potential. The reason circular orbits can be unstable is that they can correspond to maxima of the effective potential. In that case, ##r## constant and consequently ##\dot r = \ddot r = 0## is a formal solution to the EoM, but it is unstable.

For non-circular orbits however, the energy is not equal to the effective potential at one of its maxima. Therefore, small deviations from the orbit will lead to another (typically also non-circular) orbit.
That description doesn't conform to what I see in the paper, which is that circular orbits are unstable for a range of radii.
 
  • #6
Orodruin
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That description doesn't conform to what I see in the paper, which is that circular orbits are unstable for a range of radii.
The radius of the circular orbit depends on the angular momentum, which will be different for different circular orbits. For each value of angular momentum you may have two or zero circular orbits (depending on whether the cosmological constant is large enough to remove the stable orbit for that value of angular momentum).

When I talk about the two circular orbits above that refers to the stable and unstable orbits for a fixed angular momentum.

The effective potential is on the form
$$
V = -\frac kr + \frac J{r^2} - \lambda r^2
$$
where the first term is the regular Kepler potential, the second the angular momentum barrier, and the third (inverse harmonic oscillator) arises from the cosmological constant term. The angular momentum potential and CC potential are both repulsive and dominate at short/long distances respectively. If the CC is small enough, there will be a region where the Kepler potential dominates, leading to an overall attractive potential in a middle region. If this is the case, two boundaries between attractive/repulsive potential develop where the effective force from the potential is zero - corresponding to the circular orbits. For the lower circular orbit the potential is repulsive below and attractive above and the orbit is therefore stable. For the upper circular orbit it is the other way around.
 
  • #7
Orodruin
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To elaborate on that:

For zero angular momentum (and a non-zero cc), there is a single extremum for the effective potential (which is just equal to the Kepler + cc potentials). As you add even the tiniest of angular momenta, the angular momentum barrier will prohibit r=0 and the effective potential also develops a minimum close to r=0. As you gradually increase angular momentum, the minimum moves to larger r while the maximum moves to smaller r. Eventually as angular momentum grows large enough they will meet at which point there will no longer be any extrema of the effective potential (and therefore no circular obits). The range of r over which the minimum moved is the range of radii with stable circular orbits. The range of r over which the maximum moved is the range of radii with unstable circular orbits. The radii above the radius for the unstable orbit for zero angular momentum (essentially where the Kepler potential just balances out the cc potential) do not have any circular orbits at all.
 
  • #8
PAllen
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I thought there was no such thing as elliptical orbits in GR at all. Noncircular orbits don’t close - all noncircular orbits are in principle like mercury’s.

Maybe outside the scope of this discussion is that no orbits whatsoever are really long term stable, in GR, due to gravitational radiation.
 
  • #9
Orodruin
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I thought there was no such thing as elliptical orbits in GR at all. Noncircular orbits don’t close - all noncircular orbits are in principle like mercury’s.
Well, the linked paper uses a quasi-Newtonian approximation so if the cc is set to zero, orbits would indeed be elliptical.

As I already mentioned, using a non-zero cc will ruin the ellipticity of these orbits.

Maybe outside the scope of this discussion is that no orbits whatsoever are really long term stable, in GR, due to gravitational radiation.
True, but again the paper is using the quasi-Newtonian approximation.
 
  • #10
PAllen
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Well, the linked paper uses a quasi-Newtonian approximation so if the cc is set to zero, orbits would indeed be elliptical.

As I already mentioned, using a non-zero cc will ruin the ellipticity of these orbits.


True, but again the paper is using the quasi-Newtonian approximation.
I am not familiar with the quasi-Newtonian approximation. I do know some about the post-Newtonian approximation (PPN) which can be carried out to different orders of correction, e.g. 2.5 is commonly used. PPN includes gravitational radiation and general perihelion advance at higher orders (it was used to get initial waveforms for BH inspiral gravitational radiation before numerical relativity was practical). I wonder why they didn't use this? Maybe the stability questions are harder to analyze in the PPN framework?
 
  • #11
kimbyd
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Reading the paper a bit more and working through some of the math, I buy that there are stable non-circular orbits within a certain radius (the paper only mentions that they'll be confined to the region where the gravitational force is attractive). This makes intuitive sense from an energy conservation point of view.

Where I'm still getting stuck on are the unstable orbits, in particular two things you mentioned, Orodruin:
1) That non-circular orbits are necessarily stable.
2) If a system is not stable, it will separate within one radial oscillation.

The unstable circular orbits occupy a fairly substantial region, from about ##r=0.63 r_0## outward to ##r_0##, if ##r_0## is the radius at which the gravitational force is zero due to the opposing attractive and repulsive components balancing. Within this region, the gravitational force is still attractive, but the circular orbits occupy a maximum of the effective potential.

I'm still unclear as to what these unstable orbits would look like in practice (i.e., how perturbations in the orbit would grow over time). And because they occupy such a large region, one would expect there to be non-circular orbits that occupy the same region. Shouldn't those also be unstable? What would that instability mean about the orbits?
 
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  • #12
Orodruin
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I'm still unclear as to what these unstable orbits would look like in practice (i.e., how perturbations in the orbit would grow over time).
Taking a small perturbation in the radial position, the bodies will either separate or enter an orbit that varies widely in radius. From very close to coming back to near the max of the effective potential. Since the effective potential will be close to flat near the max, the system will spend a relatively long amount of time there. For relatively short times, the perturbation will however grow exponentially in both cases (toward larger radii for the unbound case and smaller radii in the bound one).

And because they occupy such a large region, one would expect there to be non-circular orbits that occupy the same region. Shouldn't those also be unstable?
The thing is that those orbits are not on the edge of being bound energetically. A small perturbation will simply send the bodies into a different non-circular orbit relatively similar to the previous one. For the unstable circular orbits you are really sitting on the edge of being bound or not. There is however one possible non-circular orbit that is in some sense unstable and that is the orbit which has just enough energy to climb the effective potential to the max (and this will take infinite time). Perturbing this solution will either lead to a bound state or an unbound one. These solutions in essence correspond to the solutions with exactly the escape velocity in the regular Kepler problem. The difference being that the potential max is at a finite radius rather than at infinity.
 
  • #13
kimbyd
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Taking a small perturbation in the radial position, the bodies will either separate or enter an orbit that varies widely in radius. From very close to coming back to near the max of the effective potential. Since the effective potential will be close to flat near the max, the system will spend a relatively long amount of time there. For relatively short times, the perturbation will however grow exponentially in both cases (toward larger radii for the unbound case and smaller radii in the bound one).
I went ahead and wrote a numerical simulation to see what these orbits look like, and they basically confirm this. When the orbit is in the unstable circular orbit region, it will either drop into a non-circular orbit that varies pretty significantly in radius (by more than a factor of two), or reach separation. The non-circular orbits were entirely contained within the unstable circular orbit radius.

Either case will occur within typically less than one orbit (if the numbers were really close they could take a little bit longer, but not much: maybe 1.2 orbits).

For orbits well within the radius where there are stable orbits, even pretty significant (say, 10%) changes in the initial velocity vector didn't really change the orbital shape all that much, as you'd expect. The orbit remained very close to circular if the initial parameters were close to circular.


The thing is that those orbits are not on the edge of being bound energetically. A small perturbation will simply send the bodies into a different non-circular orbit relatively similar to the previous one. For the unstable circular orbits you are really sitting on the edge of being bound or not. There is however one possible non-circular orbit that is in some sense unstable and that is the orbit which has just enough energy to climb the effective potential to the max (and this will take infinite time). Perturbing this solution will either lead to a bound state or an unbound one. These solutions in essence correspond to the solutions with exactly the escape velocity in the regular Kepler problem. The difference being that the potential max is at a finite radius rather than at infinity.
This can't be quite right. It's easy to set up an initial position and velocity for an orbit that cannot be circular (just by virtue of the position and velocity vectors not being perpendicular), so I tested it out. What happened is that as I got the orbital parameters closer and closer to the critical point between the orbit being bound vs. unbound with non-circular starting parameters, the orbit would first drop into an approximately-circular orbit before then dropping into a highly-eccentric orbit or separating. And the closer to that critical point I got, the closer to circular that initial orbit got.

So I think that's close to what you're saying here, but not exactly. The exactly critical orbits really appear to be circular, and whenever a non-circular orbit approaches the point of criticality it, too, approaches a circular orbit. But because of the instability, there can be a very large difference in the orbital shape with only small perturbations in the initial parameters. So an orbit that doesn't look at all circular can be in some sense "close" to a critical circular orbit.
 
  • #14
Orodruin
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So I think that's close to what you're saying here, but not exactly. The exactly critical orbits really appear to be circular, and whenever a non-circular orbit approaches the point of criticality it, too, approaches a circular orbit. But because of the instability, there can be a very large difference in the orbital shape with only small perturbations in the initial parameters. So an orbit that doesn't look at all circular can be in some sense "close" to a critical circular orbit.
Well, yes. A near critical non-circular orbit will naturally spend quite a lot of time near the unstable circular orbit as there will not be a lot of effective kinetic energy around there. The exactly critical non-circular orbit will do exactly one radial oscillation before exponentially approaching the circular one. I don’t think this violates anything of what I said.
 
  • #15
kimbyd
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If by "spend quite a lot of time" you mean typically less than one orbit, yes. But it has to be really, really close to the point of criticality to make an appreciable fraction of a circular orbit (I had to get something like 8 decimal places of accuracy in the initial conditions for the orbit to even make it half way around a circle).

It also doesn't need to take one radial oscillation before approaching the circular one. Sometimes it can appear to take a little longer, sometimes less. But "typically within one orbit" is probably correct.
 
  • #16
Orodruin
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If by "spend quite a lot of time" you mean typically less than one orbit, yes. But it has to be really, really close to the point of criticality to make an appreciable fraction of a circular orbit (I had to get something like 8 decimal places of accuracy in the initial conditions for the orbit to even make it half way around a circle).
I have not talked about the orbits relative to to the angular coordinate, just in terms of a radial oscillation (which does not need to be the same as a full 2pi orbit since the orbit is not elliptical). The effective potential barrier should make the radial velocity the lowest when passing the circular orbit. If you are not exactly on the right energy, the radial approach should follow either a sinh (if it overshoots) or a cosh behavior close to the circular orbit radius.

It also doesn't need to take one radial oscillation before approaching the circular one. Sometimes it can appear to take a little longer, sometimes less. But "typically within one orbit" is probably correct.
I don’t think you mean the same as I do when I say radial oscillation. By radial oscillation I mean the path from one periapsis to the next, not completing a 2pi angle. If you have two periapsides, then there must also be an apospsis in the orbit and the orbit will turn at that radius every orbit.
 
  • #17
kimbyd
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I have not talked about the orbits relative to to the angular coordinate, just in terms of a radial oscillation (which does not need to be the same as a full 2pi orbit since the orbit is not elliptical). The effective potential barrier should make the radial velocity the lowest when passing the circular orbit. If you are not exactly on the right energy, the radial approach should follow either a sinh (if it overshoots) or a cosh behavior close to the circular orbit radius.
That makes sense. The lobes for the bound orbits close to criticality look like precessing ellipticals with the outer part appearing squashed.


I don’t think you mean the same as I do when I say radial oscillation. By radial oscillation I mean the path from one periapsis to the next, not completing a 2pi angle. If you have two periapsides, then there must also be an apospsis in the orbit and the orbit will turn at that radius every orbit.
I see, but I don't think that changes what I said by much. It can easily take less than a full radial oscillation. It just depends upon the initial conditions. But yes, according to my tests it only appears to approach the closest orbital point at most once, and will rapidly approach the circular orbit if it's close to criticality on its next journey towards its maximal radius.
 
  • #18
Orodruin
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It can easily take less than a full radial oscillation. It just depends upon the initial conditions.
Well, yes, that is why I said ”at most one”. Technically it is not even an oscillation. Any unbound orbit (unless exactly critical) extended both backwards and forwards in time will approach from infinity, monotonically decrease its radius until reaching periapsis, and then monotonically increase the radius which will tend to infinity. Exactly where in this orbit you ”enter” at time zero will depend on your initial conditions. If your initial conditions specify ##\dot r(0) < 0## your future orbit will include the periapsis. If not it will not do so.
 
  • #19
kimbyd
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So, I had some slightly wrong conclusions from my simulation because there were some lingering inaccuracies. I improved the approximation algorithm so that now in the Newtonian case it will perform over a hundred orbits of an elliptical orbit with no visible change in its shape (the previous algorithm was causing orbits to become noticeably more circular over time). So I'm pretty confident it's doing the right thing now. The old algorithm was preventing some cases where the test object would make multiple passes before escaping. With the new algorithm, it can definitely take much longer to escape. So the conclusion I wrote earlier about it separating within about one orbit is just wrong.

Anyway, here's an example of a non-circular orbit that is close to criticality. It makes 3 passes of the closest point in its orbit before escaping (I can tune the parameters so it'll take a dozen passes or more, but it's much harder to see what's going on). As the orbit gets closer and closer to criticality, the lobes get "fatter", so it spends more and more of its time close to what is presumably an unstable circular orbit.
 

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  • #20
Orodruin
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Anyway, here's an example of a non-circular orbit that is close to criticality. It makes 3 passes of the closest point in its orbit before escaping
That is physically impossible with the model used in the paper. To me it sounds like you are still hitting some numerical instabilities (which are very likely to occur close to criticality).

Edit: Of course, this just goes to show just how unstable the unstable solution is.
 
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  • #21
kimbyd
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That is physically impossible with the model used in the paper. To me it sounds like you are still hitting some numerical instabilities (which are very likely to occur close to criticality).

Edit: Of course, this just goes to show just how unstable the unstable solution is.
There are definitely instabilities remaining. Each time iteration introduces a tiny error, so it's as if the orbit is being continuously perturbed by a small amount. But that should make it less likely to orbit multiple times before escape, not more. As was the case when the model was biased towards circular orbits in my first attempt. In that earlier attempt, it would very much either escape quickly or drop into a much more stable orbit. It had to, because after the first orbit it would be pulled away from the critical point by the bias in the model.

How are you reaching the conclusion that the non-circular orbits must separate so quickly? I don't think you've explained that.
 
  • #22
kimbyd
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By the way, in the current simulation, energy is conserved by construction and never changes throughout the run. Angular momentum is very slightly not conserved, shifting by something like a few parts per hundred million each orbit. The bias seems to be towards lower angular momentum, which would again tend to make orbits less likely to escape after multiple iterations.
 
  • #23
Orodruin
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How are you reaching the conclusion that the non-circular orbits must separate so quickly? I don't think you've explained that.
It is a basic energy argument from the effective equation of motion for the radial coordinate. Energy conservation leads to
$$
E = m\frac{\dot r^2}2 + V_{\rm eff}(r)
$$
with ##E## being a constant of motion. The radial velocity ##\dot r## can only change sign where ##V_{\rm eff}(r) = E##. If ##E## is larger than the potential barrier, then ##\dot r## cannot change sign when reaching the barrier from below and it will continue to grow and therefore escape. In other words, the first time you approach the barrier with sufficient ##E## to escape you will escape. If you do not have enough energy to climb the barrier, then you will have a turning point and the orbit will reflect back towards lower ##r##.

In essence, in one-dimensional motion in classical physics, you will either have enough energy to pass the barrier the first time you approach it or never.
 
  • #24
kimbyd
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Update: I made another small tweak by dropping the energy conservation requirement. The simulation now seems to conserve the combination of angular momentum and energy better than it did before. And up to the precision of 64-bit floating point, orbits always separate within one radial oscillation if they're going to separate at all. So you're probably right, that it was the enforcement of energy conservation which messed up the simulation.
 
  • #25
kimbyd
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It is a basic energy argument from the effective equation of motion for the radial coordinate. Energy conservation leads to
$$
E = m\frac{\dot r^2}2 + V_{\rm eff}(r)
$$
with ##E## being a constant of motion. The radial velocity ##\dot r## can only change sign where ##V_{\rm eff}(r) = E##. If ##E## is larger than the potential barrier, then ##\dot r## cannot change sign when reaching the barrier from below and it will continue to grow and therefore escape. In other words, the first time you approach the barrier with sufficient ##E## to escape you will escape. If you do not have enough energy to climb the barrier, then you will have a turning point and the orbit will reflect back towards lower ##r##.

In essence, in one-dimensional motion in classical physics, you will either have enough energy to pass the barrier the first time you approach it or never.
That makes sense, thanks.
 
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