Rosetta orbits and phase space

In summary: When it is on the same spatial location (and has same mass, momentum and energy) it has to follow the same trajectories. This is because the velocity vector at this location points in the same direction. So if a trajectorie in...
  • #1
Oliver321
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5
I was recently working on the two body problem and what I can say about solutions without solving the differential equation. There I came across a problem:
Lets consider the Kepler problem (the two body problem with potential ~1/r^2). If I use lagrangian mechanics, I get two differential equations. Now if I plot the phase portrait it looks like a vectorfield in the x-y-plane. Trajectories on this phase plane describe (under certain conditions) ellipses and circles. But they never cross and are closed!
Now if I consider the two body problem with an other potential, say ~1/r^5. Its told that if the potential is not like ~1/r^2 or 1/r there are not necessarily closed orbits but possibly rosetta orbits (https://en.wikipedia.org/wiki/Rosetta_orbit). But here the trajectories cross if I plot it in 2D phase space like in the kepler problem. But trajectory can never cross in phase space! How is this possible?
1598167000407.png


My solution: If I have a n-body problem the phase space is not a 2D vector field anymore. So the trajectories don’t cross at all, it’s only because of the projection on the plane. But I can’t find a solution to the problem if I use a potential like ~1/r^5. The motion should be planar and only two body’s are considered. So I should get no crossing of trajectories in 2D phase space.

Thanks for every awnser!
 
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  • #2
Oliver321 said:
But here the trajectories cross if I plot it in 2D phase space like in the kepler problem.
How did you define the phase space? What are the phase variables?
 
  • #3
A.T. said:
How did you define the phase space? What are the phase variables?

You can take x and y or r and phi, but it should be planar coordinates. On every point in this field there is a vector giving the velocity.
 
  • #4
Oliver321 said:
You can take x and y or r and phi, but it should be planar coordinates. On every point in this field there is a vector giving the velocity.
So how many dimensions does the phase space have?
 
  • #5
A.T. said:
So how many dimensions does the phase space have?

I think that’s a definition ( at least I read it differently from different sources). Some would define a 2D phase space with a vector on every point, some would define a 4D phase space with 4 axes and no vectors. The result should be the same?
 
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  • #6
Oliver321 said:
... some would define a 4D phase space with 4 axes and no vectors.
And that's where you cannot have intersections.
 
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  • #7
A.T. said:
And that's where you cannot have intersections.

Thanks for your answer!
But nevertheless: I could plot it like a vector field in a plane. And there should be no intersection, because that would violate uniqueness.
Or to say it another way: If the trajectories don’t cross in 4D, then I cannot ‘reduce‘ it to a 2D vector fiel with constant vectors. But why should the latter be the case (mathematically I cannot find a reason)?
 
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  • #8
Oliver321 said:
I think that’s a definition ( at least I read it differently from different sources). Some would define a 2D phase space with a vector on every point, some would define a 4D phase space with 4 axes and no vectors. The result should be the same?

I'm not sure what you mean by this. As far as I am aware, the dimensionality of the phase space equals the number of phase variables. Since your original problem takes place in a 2D Euclidean space, that means that you have two position coordinates and two momentum coordinates, so the corresponding phase space is 4-dimensional.

At any point in the phase space, you can associate a vector whose components are the rates of change in the phase variables; the result is the associated vector field to the phase space.

I think you must be clear on the distinction between the Euclidean space in which the particles are moving, and the phase space. The trajectories in the Euclidean space can cross themselves, but the trajectories in the phase space cannot.

Oliver321 said:
You can take x and y or r and phi, but it should be planar coordinates. On every point in this field there is a vector giving the velocity.

I don't think the velocity in the Euclidean space is described by a vector field. The acceleration, however, could be described by a vector field.
 
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  • #9
Oliver321 said:
But nevertheless: I could plot it like a vector field in a plane. And there should be no intersection, because that would violate uniquenes.
Why should the body not be allowed to visit the same spatial location twice? Your expectation makes no sense to me.
 
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  • #10
A.T. said:
Why should the body not be allowed to visit the same spatial location twice? Your expectation makes no sense to me.

When it is on the same spatial location (and has same mass, momentum and energy) it has to follow the same trajectories. This is because the velocity vector at this location points in the same direction. So if a trajectorie in a two body problem crosses itself at a later time and goes in a different direction, there has to be a change in energy, momentum, etc. that happend.
It could visit the same location twice, but has to follow the same path as before and therefore it has to be a closed orbit, not a rosetta orbit.
 
  • #11
Oliver321 said:
When it is on the same spatial location (and has same mass, momentum and energy) it has to follow the same trajectories. This is because the velocity vector at this location points in the same direction. So if a trajectorie in a two body problem crosses itself at a later time and goes in a different direction, there has to be a change in energy, momentum, etc. that happend.

That's not true. The particle is not moving in a velocity field, there is no reason why the velocity should be the same at consecutive times that the particle is at the same position.
 
  • #12
Oliver321 said:
This is because the velocity vector at this location points in the same direction.
Why?
 
  • #13
etotheipi said:
That's not true. The particle is not moving in a velocity field, there is no reason why the velocity should be the same at consecutive times that the particle is at the same position.

Thank you for the answer!
I have calculated following system of differential equations for the kepler problem (I am not sure if all prefactors are right, I don’t have my notes with me):

1598211709681.jpeg

If I put in the same coordinates r and theta I always get the same velocity. What am I misunderstanding?
 
  • #14
Oliver321 said:
I have calculated following system of differential equations for the kepler problem...
If I put in the same coordinates r and theta I always get the same velocity.
Just because this is true for potentials with closed orbits, doesn't mean it has to be true for other potentials.
 
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  • #15
A.T. said:
Just because this is true for potentials with closed orbits, doesn't mean it has to be true for other potentials.

And exactly this is my problem. If I try my derivation with another potential (which is not time dependent) I get a solution of the same kind which is not time dependent either. I don’t understand how this is fitting together.
 
  • #16
etotheipi said:
I think you must be clear on the distinction between the Euclidian space in which the particles are moving, and the phase space. The trajectories in the Euclidian space can cross themselves, but the trajectories in the phase space cannot.

This.

In configuration space (what etotheipi calls "Euclidian space") paths can cross. In phase space, they cannot.

Consider a simpler case: a 1-d pendulum. In configuration space, every point but the endpoints is visited twice in a full oscillation: once coming from the left and once coming from the right. In phase space, which has two variables, on an x-p plot the path is a single, non-intersecting loop.

Why can't you have an intersecting path in phase space? Because it means that a particle with a given x and p will have two possible directions it can take from then on. How does it know?
 
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  • #17
Thanks, yes "configuration space" was the term I was looking for. I also managed to spell Euclidean wrong...

I think @Oliver321 misunderstands what is meant by a phase space; I strongly suspect that every time the term has been used he actually means the configuration space. It would be worth making sure OP is on the same page before going any further with the Kepler problem.
 
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  • #18
Thank you for the help!
In the summer I read a book about nonlinear dynamics and there the term phase space was used in this context. So sorry if i was on the wrong page!

But nevertheless I think, I have found the problem in my calculation: There is a square root in my r equation, so no point in configuration space (i hope I used the term right; I mean caretsian space ) is related to a unique velocity. There is an ambiguity because of the square root. So I am forced to go into higher „dimensions“ and the interpretation of a vectorfiel is in this context wrong.
But am I right with this assumption (if not, the whole book would be wrong and I don’t think that’s the case): If every point in space is correlated to exactly one velocity (so it can be portrait as a vectorfield), there could be no intersections between trajectories. This would violate uniqueness (the particle would not know where to go next).
In the simple pendulum example x is not correlated to a unique velocity so I have to go to higher dimensions (using x and v). And in my case it is the same. But if x would be correlated with only one velocity, I could picture it like a vectorfield on the line.
 
  • #19
Vanadium 50 said:
In phase space, which has two variables, on an x-p plot the path is a single, non-intersecting loop.

Following @Vanadium 50's comment, here's a plot of a simple 1-D pendulum in phase space.

24537282.png


Notice how at small values of ##\{\theta,p_{\theta}\}##, which describe back and forth motion of the bob, an ellipse pops up (credit to David Tong for the plot). Paths never cross.

There's a Wolfram Project (with the code available) about the simple pendulum here.
 
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  • #20
Oliver321 said:
If every point in space is correlated to exactly one velocity (so it can be portrait as a vectorfield), there could be no intersections between trajectories.
Yes. No intersections at non-zero angle, but eventually re-running the same path again.
 
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  • #21
I really thank you for your helping hand!

I had a bit of trouble but I think I worked it out:
My posted equation is a first order differential equation. But if I use for example lagrange, I get a second order equation for r. I reduced it by considering energy conservation to the first order equation on the top. But since it is not possible to simply reduce a second order to a first order, the ambiguity appeared in form of the square root. So if I should plot the first order equation, considering both signs and if I plot the second order equation, I should get the same result, and this was the case:
1598262206429.jpeg

So in fact I was really mislead by the square root to think I can plot it simply in a 2D vector field!

But another question: I used pplane to plot vector fields but in my opinion it does not look great. Is there a program which can plot vector fields (of differential equations) with more settings to choose from or do I have to program it myself?
 

1. What is a Rosetta orbit?

A Rosetta orbit is a type of orbit used by spacecraft to rendezvous with a comet. It involves a series of complex maneuvers that allow the spacecraft to match the speed and trajectory of the comet, ultimately placing it in a stable orbit around the comet.

2. How is a Rosetta orbit different from other spacecraft orbits?

A Rosetta orbit is unique because it involves multiple gravity assists and flybys of other celestial bodies in order to reach the comet. This allows the spacecraft to conserve fuel and adjust its trajectory as needed to successfully orbit the comet.

3. What is phase space in relation to Rosetta orbits?

Phase space is a mathematical concept used to describe the position and momentum of a system, such as a spacecraft in orbit. In the case of Rosetta orbits, phase space is used to calculate the trajectory and velocity of the spacecraft as it approaches and orbits the comet.

4. How does the shape of a comet affect its phase space and orbit?

The shape of a comet, as well as its rotation and composition, can greatly affect its phase space and orbit. These factors can influence the gravitational pull and trajectory of the comet, which in turn can impact the spacecraft's orbit and require adjustments to maintain a stable orbit.

5. What challenges do scientists face when planning and executing a Rosetta orbit?

Planning and executing a Rosetta orbit can be challenging due to the complex nature of the maneuvers and the unpredictable behavior of comets. Scientists must carefully calculate and adjust the spacecraft's trajectory to ensure a successful orbit, while also accounting for potential hazards such as dust and gas emissions from the comet.

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