Dynamic Equilibrium of the Universe and Subsystems

[coldcreation]

Here we are finally getting to the crux of the problem. The view expressed above (in the post by modest) is largely the mainstream view. It cannot be correct. Because of the error intrinsic in the above interpretation of both GR and Lagrange dynamics (regarding libration points, geometrical field structure, equilibrium, etc) an entire host of astronomical (and cosmological, yet to be discussed) problems have not been resolved: rotational curves, "anomalies" associated with spacecraft - SOHO etc., barred galaxy structure and longevity, the equilibrium of bounded gravitating systems, the mechanism of tides here on earth, unification between GR and QM, unification with gravity and the other so-called forces of nature, the cold-dark matter problem, the DE problem, the lambda problem, and so on.

 

All the above have remained elusive for the very reason that the structure of the gravitation field between massive objects has been erroneously visualized. That is why I have written many times that in order to reconcile these difficulties, the physical mechanism of the gravitational interaction must be elucidated, and the Lagrange mechanics must be generalized. In order to elucidated the gravitational mechanism we must first understand what is 'empty' field-free space (what would be the properties at say the Lagrange inner libration point L1), and how this space with a zero value of gravity affects gravitating bodies.

 

For this to be true you'd have to say all bodies of equal mass at any Lagrange point have equal potential gravitational energy to some reference.

 

All bodies (or test particles) have the same potential at L1. It is a zero gravitational potential. This would be true for a test particle placed at any L1 point in the universe (and there are many, to say the least). The reference frame is spacetime itself, the vacuum substratum.

 

 

That can't be true. The point L1 between two orbiting stars may have very high potential if it is in a gravity well. The stars may be near a super-massive black hole. That L1 from our reference wouldn't have the same potential as our earth-sun L1. We can confirm this with gravitational lensing. All the galaxies of a cluster make one gravitational lens. The Lagrange points don't make holes in the lens.

 

It is easily shown that the gravitational potential at the earth-moon L1 point and the gravitational potential at the earth-sun L1 point are identical: equal to zero. So in reduced dimension, the 'peaks' in the field (as opposed to the potential wells of both M1 and M2, which may vary in 'depth') at all L1 points throughout the universe are at the same 'height.'

 

Indeed we have two superposing realities in nature: one is the curved spacetime phenomenon (characterized by gravity), the other has a Euclidean or Minkowski signature (characterized by Lagrange points). One cannot be independent of the other. These peaks and troughs in the field are not polar opposites, but act as if they were. The difference between them is that all wells have their own value depending on mass, and all 'peaks' (L1) have the same value (zero). Note: L4 and L5 are not strictly speaking peaks in the field (since they are tadpole-like shape). L2 and L3 are very likely as is L1 (and are just as important in the maintainance of equilibrium).

 

 

The opposite of gravity is NO gravity.

 

 

Said differently, the opposite of curvature is NO curvature. Unfortunately gravitational lensing, as a method of testing, is too week an effect to detect this boundary condition. There are, however, other means by which we can test these contentions empirically, aside from the straight forward maths.

 

 

I don't see any reason why general relativity would be wrong on this. The Lagrange point is where the field is flat - not zero. Besides, null geodesics continue to diverge after L1 all the way to infinity.

 

Again, GR is not wrong on this. All this does is set a limit, a lower bound, on the gravitational potential (when and where the field strength, or curvature, is equal to zero). Indeed, the Lagrange point (L1) is a point at which the field is flat - zero. Without the ultimate limit of curvature— an absolute zero of gravity, a fundamental surface, pure unadulterated space, an elemental state with a value fundamentally constant and equal to zero—there would be no stress on the field, no tension. And so gravity would still be spacetime curvature but without the mechanism to explain where is comes from, or how it works.

 

We will see later how the relation lambda-Lagrange point manifests itself, and why the value for both the former and the latter are zero (before moving on the cosmological considerations).

 

 

 

 

CC

[snoopy]

 

Again, GR is not wrong on this. All this does is set a limit, a lower bound, on the gravitational potential (when and where the field strength, or curvature, is equal to zero). Indeed, the Lagrange point (L1) is a point at which the field is flat - zero. Without the ultimate limit of curvature— an absolute zero of gravity, a fundamental surface, pure unadulterated space, an elemental state with a value fundamentally constant and equal to zero—there would be no stress on the field, no tension. And so gravity would still be spacetime curvature but without the mechanism to explain where is comes from, or how it works.

 

 

CC

 

Interesting are you going to discuss the Interplanetary Transport Network as well ?

 

I find that bit fascinating...

 

It would be good if you could give it a mention.

 

Still not sure where exactly you are going with this but I am enjoying the ride.

 

Peace

:eek_big:

[modest]

Here we are finally getting to the crux of the problem. The view expressed above (in the post by modest) is largely the mainstream view. It cannot be correct. Because of the error intrinsic in the above interpretation of both GR and Lagrange dynamics (regarding libration points, geometrical field structure, equilibrium, etc) an entire host of astronomical (and cosmological, yet to be discussed) problems have not been resolved: rotational curves, "anomalies" associated with spacecraft - SOHO etc., barred galaxy structure and longevity, the equilibrium of bounded gravitating systems, the mechanism of tides here on earth, unification between GR and QM, unification with gravity and the other so-called forces of nature, the cold-dark matter problem, the DE problem, the lambda problem, and so on.

 

A good way to get an idea ignored is saying it solves these problems. Unfortunate, but true.

 

That is why I have written many times that in order to reconcile these difficulties, the physical mechanism of the gravitational interaction must be elucidated, and the Lagrange mechanics must be generalized. In order to elucidated the gravitational mechanism we must first understand what is 'empty' field-free space (what would be the properties at say the Lagrange inner libration point L1), and how this space with a zero value of gravity affects gravitating bodies.

 

All bodies (or test particles) have the same potential at L1. It is a zero gravitational potential. This would be true for a test particle placed at any L1 point in the universe (and there are many, to say the least). The reference frame is spacetime itself, the vacuum substratum...

 

 

So in reduced dimension, the 'peaks' in the field (as opposed to the potential wells of both M1 and M2, which may vary in 'depth') at all L1 points throughout the universe are at the same 'height.'

 

Not only isn't this true, it can't be true. My representation of a 2-D slice of spacetime follows. 'A' is the usual GR result. 'B' is my interpretation of the point L1 having zero gravitational potential. The height at any spot of the black like is equal to the numerical value of gravitational potential energy - it is zero toward the top of the pic and some negative value toward the bottom. The derivative or slope of the line equals the force of gravity at that point. Yellow is the sun, Blue is the earth.

 

 

The slope at L1 in both these is zero. No gravitational forces would be felt at L1. However, the slope next to L1 is large in your example. There would be a large force of gravity. This is what tidal forces look like and your tidal forces would be too strong.

 

It is easily shown that the gravitational potential at the earth-moon L1 point and the gravitational potential at the earth-sun L1 point are identical: equal to zero.

 

The value of potential is GM/r. Please show how the values are either zero or equal.

[coldcreation]

The Euler-Lagrange equation will give you or will reduce to Newton’s second law. In various circumstances it is more convenient or easier to solve Lagrangian mechanics, however, this is an issue of math - the results are the same. The pioneer and other anomalies (and let’s even add galaxy rotation or dark matter to the list) these anomalies do not agree with Newtonian predictions and therefore necessarily don’t agree with Lagrangian mechanics.

 

Certainly there is a relation between Lagrange equations and Newtons. After all, Lagrange used Newton's equations in his own work. However the results and conclusion derived by Lagrange are not the same, as you seem to be implying. See this from Wiki:

 

History and concepts

In 1772' date=' the Italian-French mathematician Joseph-Louis Lagrange was working on the famous three-body problem when he discovered an interesting quirk in the results. Originally, he had set out to discover a way to easily calculate the gravitational interaction between arbitrary numbers of bodies in a system, because Newtonian mechanics conclude that such a system results in the bodies orbiting chaotically until there is a collision, or a body is thrown out of the system so that equilibrium can be achieved. The logic behind this conclusion is that a system with one body is trivial, as it is merely static relative to itself; a system with two bodies is very simple to solve for, as the bodies orbit around their common center of gravity. However, once more than two bodies are introduced, the mathematical calculations become very complicated. A situation arises where you would have to calculate every gravitational interaction between every pair of objects at every point along its trajectory.

 

Lagrange, however, wanted to make this simpler. He did so with a simple hypothesis: The trajectory of an object is determined by finding a path that minimizes the action over time. This is found by subtracting the potential energy from the kinetic energy. With this way of thinking, Lagrange re-formulated the classical Newtonian mechanics to give rise to Lagrangian mechanics. With his new system of calculations, Lagrange’s work led him to hypothesize how a third body of negligible mass would orbit around two larger bodies which were already in a near-circular orbit. In a frame of reference that rotates with the larger bodies, he found five specific fixed points where the third body experiences zero net force as it follows the circular orbit of its host bodies (planets).[/quote']

 

Indeed Lagrange was the first to recognize and stress the importance of neutral points amid the space surrounding gravitating bodies. The method that Lagrange employed was different from all prior attempts to clarify the 3-body problem. Lagrange's discovery not only confines the number of possible structures that can exist, it demonstrates a regularity and pattern among those systems that do exist.

 

As a consequence, there emerge non-Newtonian contributions that are consistently included in the dynamics at the echelon of geometric description. Recall that the general relativistic approach applies to all coordinate systems, and that similarly to Newtonian mechanics gravity is everywhere present (i.e., spacetime is never flat). By ignoring parts of the gravitational field, e.g., those where no acceleration is present (where gravitational curvature is zero), we restrict our vision to a coordinate system that is merely a special limiting case of a more general devise.

 

 

 

If the two intrinsic fields (the Earth’s and the Sun’s) cancel at L1, the value of the field is zero at that point.

[True]. Well, depends what you mean 'value of the field'. It would be flat. But, any object there would have potential gravitational energy compared to the sun and the earth.

 

By 'value of the field' it is meant the value of curvature or quantity of curvature, the 'force' of gravity, in this case at a particular point, say, L1. The value of the field is zero. There is no gravitational force at L1, and so it is gravity-free. The field shape around L1, in reduced dimensions, is like a Pringles potato chip, a saddle. No visualize that in three or four dimensions and you will have a realistic picture of the hyperbolic space at and around L1..

 

A test particle placed there would zero potential gravitational energy compared to the sun and the earth.

 

 

The gradient in the Earth’s field, consequently, begins close to the Newtonian value, but the closer the ship moves toward the Sun, the greater the deviation from the inverse square law. In other words, the further removed an object from Earth, along the Sun-Earth (or Moon-Earth) lines, the weaker the field becomes, until its value approaches zero at L1.

 

Not exactly. You wouldn’t say the gradient in the earth’s field - but in the total field. The earth plus the sun. It is true that the ‘gradient’ in the field (or the change in potential as GR would prefer) is zero at L1. There is no gravitational force because the vectors cancel. The distinction that needs made, however: Earth’s field is not weakening at a greater than expected rate. Earth is affecting space time at L1 just as much as if the sun were not there. The reason space time is flat at L1 is because the sun effectively ‘bends’ it the other direction.

 

Not at all. The Earth is inside its own gravitational well (a curved spacetime well, the change in potential), which in turn is located inside the gravitational well of the Sun. In the direction of the Sun toward L1 (along the line connecting the center of M1 and M2) it is the gradient of the Earths fields which affects objects. On the other side of L1 the gradient of the Sun affects objects.

 

The result is that a test particle at rest on the L1 point will experience no net acceleration—consequently, the net gravitational force is zero at L1. Note that the cancellation of the gravitational force field at L1 is not due to the balance between an attractive force and the centrifugal acceleration away from the orbital axis. It is due to the equivalence of the value of curvature at this intersection. Space curvature at the saddle point L1 is neutral, equal to zero. The resulting state at L1 is very close to that of a true vacuum, empty of gravitational field. In the absence of gravity, L1 has a lower energy (or force) than any other location within the combined field of M1 and M2.

 

 

 

As I sourced above, Lagrangian dynamics predicts nothing different from Newtonian mechanics.

 

As I sourced above, Lagrangian dynamics predicts something different from Newtonian mechanics.

 

 

 

It is thus GR combined with the Lagrange geometric scheme that will (and does) provide the solution. How? There is a generalization of the Lagrange formula, and a restriction on GR (the boundary condition I wrote about earlier) to be made. The combination of both determines the dynamics of self-gravitating systems.

 

What boundary condition?

 

The boundary condition is flat, gravity-free space. If we consider an absolute scale for gravity (spacetime curvature), similar to the Kelvin temperature scale, the boundary condition is absolute zero.

 

Indeed, this condition is missing from both GR and Newtonian gravitation. It is particularly important that this boundary be included into the framework of GR since it is believed, erroneously, that this boundary can be surpassed, i.e., that there could exist a different type of curvature (with a negative sign, a false vacuum-like spacetime).

 

This boundary does not set a limit on the maxima of curvature (say, in the vicinity of a neutron star, or black hole for those who believe in them), but it does set a limit on the minima of curvature. This is as fundamental as it is primordial if every we are to understand gravity and the dynamics that accompanies it.

 

The boundary condition is an extension of GR that needed to be incorporated if resolution is to be made in the understanding of gravity and it’s relation not just to things but also to empty space as a generator of stability, equilibrium and symmetry, rather than as a fictitious medium or intermediate state (or even ‘nothing’) halfway between two extremes.

 

 

PS. modest, your illustration B above it not at all what I have been describing. Please modify your post: either remove illustration "B" or remove you phrase "B" is your idea." B is horrendous, it is not my idea, and "A" is the erroneous mainstream view. 'A' should be modified by lifting the L1 saddle point slightly higher (not pointy, but rounded as you cirrectly drew). The height of that 'hill' should attain the Euclidean plane (which you would have drawn if M1 and M2 were absent). I haven't yet figured out how to include illustrations here otherwise I could have avoided this misunderstanding ("B") by providing illustrations myself (I have plenty of those).

 

CC

[Erasmus00]

As I sourced above, Lagrangian dynamics predicts something different from Newtonian mechanics.

 

Umm... no it doesn't. The lagrangian formulation of mechanics gives the exact same differential equations that Newton's formulation gives(and Hamilton's, and Helmholz's, etc). It is nothing new. See Goldstein's Classical Mechanics book, for instance.

 

However, sometimes looking at an equivalent formulation reveals a way of looking at a problem that you wouldn't have seen from the original formulation. The lagrange points exist exactly the same in Newton's equations as Lagrange's, but Lagrange's formulation makes it easier to see that they must exist.

-Will

[modest]

PS. modest, your illustration B above it not at all what I have been describing. Please modify your post: either remove illustration "B" or remove you phrase "B" is your idea." B is horrendous, it is not my idea, and "A" is the erroneous mainstream view. 'A' should be modified by lifting the L1 saddle point slightly higher (not pointy, but rounded as you cirrectly drew). The height of that 'hill' should attain the Euclidean plane (which you would have drawn if M1 and M2 were absent). I haven't yet figured out how to include illustrations here otherwise I could have avoided this misunderstanding ("B") by providing illustrations myself (I have plenty of those).

 

I have edited my post to reflect that 'b' is my depiction of a point L1 that has zero potential relative to the earth and sun. It is my point, after all, that L1 would have to be raised out of the sun's gravity well. Earth is in the sun's gravity well - so L1 would stick up past everything around it.

 

I guess to stick with the rest of your description, L1 would also have to stick up out of the Milky Way's gravity well - we'd have to raise it even higher.

 

If you scan your drawing or take a pic of it with a digital camera, you can upload it as an attachment. That would be helpful.

 

-modest

[modest]

As a consequence, there emerge non-Newtonian contributions that are consistently included in the dynamics at the echelon of geometric description.

The contribution is that it's easier to solve. The answer is the same. Is there any way I could prove this to you? I have a felling solving both ways won't do it - that is after all what they do on the link I gave. I wish there were some way of proving something to you. Sources don't work, showing a solution by solving doesn't work. :)

By 'value of the field' it is meant the value of curvature or quantity of curvature, the 'force' of gravity, in this case at a particular point, say, L1. The value of the field is zero. There is no gravitational force at L1,

I agree. The force of gravity is zero. A test particle wouldn't be accelerated toward anything. Even without centrifugal force when everything is static - there is no gravitational force.

A test particle placed there would zero potential gravitational energy compared to the sun and the earth.

Here is the problem. The value of potential is different from the force of gravity. If you plot the potential of spacetime then the force is the rate of change of that potential. Force is the derivative of the potential. Saying something has zero force of gravity is saying the spacetime is flat or that point on the function is flat. Saying it has zero potential is describing numerically the position of the point. They are different.

 

The point L1 is not zero in potential. The point of zero potential is usually set arbitrarily at infinity. One thing is certain - L1 cannot have the same potential as infinity or the surface or center of the mass.

'A' should be modified by lifting the L1 saddle point slightly higher (not pointy, but rounded as you cirrectly drew). The height of that 'hill' should attain the Euclidean plane (which you would have drawn if M1 and M2 were absent).

 

The potential gravitational energy is equal to the work done lifting an object higher in a gravity well. I've done another pic. This one shows two stars near a black hole. The stars are labeled M1 and M2. L1 is between them. We will consider zero potential at X which is infinitely far from the black hole. You would say L1 has zero potential, I'd say not and demonstrate why:

The potential of a mass in the black hole's gravity well is found by how much work it takes someone standing at X with a rope tied to that mass to pull it out. The potential at L1 is found by the person at X measuring that work. (1) If someone does that in this example it is easy to see that it will take effort to move a mass from L1 to X. Therefore a mass at L1 must, by all convention and all theory, must have potential gravitational energy.

 

I would also suggest that there really isn't any other way to draw this diagram. The consequences of all L1 points having the same potential would be very odd. Spacetime would be so curved I doubt anything (including photons) could reach the spot.

 

-modest

[snoopy]

Umm... no it doesn't. The lagrangian formulation of mechanics gives the exact same differential equations that Newton's formulation gives(and Hamilton's, and Helmholz's, etc). It is nothing new. See Goldstein's Classical Mechanics book, for instance.

 

However, sometimes looking at an equivalent formulation reveals a way of looking at a problem that you wouldn't have seen from the original formulation. The lagrange points exist exactly the same in Newton's equations as Lagrange's, but Lagrange's formulation makes it easier to see that they must exist.

-Will

 

 

I have to agree with Will here they are all the same equations in a different format.

 

But watching CC and modest going at it hammer and tongs is very entertaining.

 

Peace

:)

[modest]

But watching CC and modest going at it hammer and tongs is very entertaining.

 

Peace

:)

 

Thank you snoop. Wait 'til I pull out the bazooka.

 

If one thing is certainly in dynamic equilibrium - it's CC and I in disagreement :)

 

PS, do you think there is such a thing as an orbit that will never fail?

 

-modest

[snoopy]

PS, do you think there is such a thing as an orbit that will never fail?

 

-modest

 

It depends on long 'never' is given the dynamism of the Universe every orbit must fail because the system of bodies has changed but I do believe in long lived 'fixed' orbits.

 

I am not sure if that helps any though.

 

Peace

:)

[coldcreation]

I have edited my post to reflect that 'b' is my depiction of a point L1 that has zero potential relative to the earth and sun. It is my point, after all, that L1 would have to be raised out of the sun's gravity well. Earth is in the sun's gravity well - so L1 would stick up past everything around it.

 

I guess to stick with the rest of your description, L1 would also have to stick up out of the Milky Way's gravity well - we'd have to raise it even higher.

 

As I wrote before, L1 (in 2-dimensions) would attain or touch at one point, on the top of the hill, the plane that would be your line if there were no mass present (without M1 and M2: the same value that would be found at infinity). I will show you with a drawing.

 

If you scan your drawing or take a pic of it with a digital camera, you can upload it as an attachment. That would be helpful.

OK, how do I upload an attachment?

[modest]

OK, how do I upload an attachment?

 

When you are writing your post - click on the paper clip. It is to the right of the fonts and font sizes. It will open up the 'Manage Attachments' window where you can follow the instructions. It is very much like putting an attachment in an email.

 

By the way, let me give props to hypography for all the extra file and image and latex and other options. :)

 

-modest

[coldcreation]

This is an illustration, in reduced dimensions, of the relation between M1 - L1 - M2.

 

 

The dotted line, which represents a flat Euclidean plane (i.e., a space devoid of matter).

 

L1 attains and is located on this plane (i.e., the value of spacetime curvature at L1 is zero).

 

L2 and L3 have been left out of this illustration. They would also attain the same value, zero curvature (they would touch the dotted line).

 

M1 in this illustration is the less massive of the two bodies. Had M1 and M2 been of equal mass L1 would be located in the center and also touching the line at one point.

 

On either side of M1 and M2 the field curvature diminishes according to the inverse square law tending to infinity (thus never attaining the zero value).

 

The gravitational wells represented here (the curved lines espousing M1 and M2) are greatly exaggerated.

 

Again, a test particle introduced into the field at L1 will either move towards M1 or M2 with the slightest perturbation. Directly at this junction, L1, a test particle at rest will experience no net acceleration—consequently, the net gravitational force is zero at L1. Note that the cancellation of the gravitational force field at L1 is not due to the balance between an attractive force and the centrifugal acceleration away from the orbital axis. It is due to the equivalence of the value of curvature (which is nul) at this intersection. Space curvature at the saddle point L1 is thus equal to zero. The resulting state at L1 is very close to that of a true vacuum, empty of gravitational field. In the absence of gravity, L1 has a lower energy (or gravitational force) than any other location within the combined field of M1 and M2 (excluding for now other L-points).

 

 

This would be true for all L1 points in the universe...

 

 

 

CC

[modest]

The dotted line, which represents a flat Euclidean plane (i.e., a space devoid of matter).

 

Now put a planet in one of the star's gravity wells. Put a point L1 between the planet and the star. Raise L1 to the 'euclidean plane'. You should see the problem clearly.

 

Note that the cancellation of the gravitational force field at L1 is not due to the balance between an attractive force and the centrifugal acceleration away from the orbital axis. It is due to the equivalence of the value of curvature (which is nul) at this intersection.

 

I have agreed with this. The point L1 has no acceleration of gravity because spacetime is flat - not because of centrifugal forces.

 

-modest

[coldcreation]

Now put a planet in one of the star's gravity wells. Put a point L1 between the planet and the star. Raise L1 to the 'euclidean plane'. You should see the problem clearly.

 

There is no problem. M1 in my illustration could very well be the Earth.

 

M1 and M2 could very well be any 2-body system (e.g., Earth-Moon, Earth-Sun, etc.). This is a general illustration (again excluding L2 and L3).

 

Note: the fact that L1 points (all L1 points in the universe) are located on that plane is a stricking conclusion, the consequences of which are extremely important. I can provide you with a list.

 

It is very easy to prove empirically that fact. There is a simple observational test that can be carried out with a minimal cost.

 

The degree (or amount) of gravitational field curvature is directly proportional to the inertial mass of a massive bodies (see the equivalence principle), not because the mass itself produces the gravity field, but because the spacetime manifold contours itself (there is a displacement from linearity) accordingly and proportionally to the mass of an object in relation to a hitherto trivial or inconsequential lower-threshold: spacetime with a Minkowski signature (the dotted line in the illustration).

 

Each gravitational well potential is gauged not solely on mass, but on the degree of displacement from a linear metric that has a unique value for all times and all L1 locations.

 

This is not a test of the validity of Newtonian mechanics, of Lagrange dynamics, or even of Einstein's equivalence principle, it is a test of the boundary condition of GR.

 

 

 

 

CC

[modest]

There is no problem. M1 in my illustration could very well be the Earth.

 

M1 and M2 could very well be any 2-body system (e.g., Earth-Moon, Earth-Sun, etc.). This is a general illustration (again excluding L2 and L3).

 

You will not address my diagrams or put a third mass in your own diagram. I think you know my objection and you know what I'm asking. You also must know you can't draw the three mass diagram. It will look like my diagram B. You also can't solve U=GM/r to get zero potential at L1 - let alone all L1s. You cannot find a source to support that L1 has zero potential. You frankly cannot explain how or why this would work.

 

Draw a diagram of Andromeda, the Milky Way, Earth, and the Sun with all relevant L1s. If it looks appropriate, I'll support your idea.

 

-modest

[coldcreation]

You will not address my diagrams or put a third mass in your own diagram. I think you know my objection and you know what I'm asking. You also must know you can't draw the three mass diagram. It will look like my diagram B. You also can't solve U=GM/r to get zero potential at L1 - let alone all L1s. You cannot find a source to support that L1 has zero potential. You frankly cannot explain how or why this would work.

 

Draw a diagram of Andromeda, the Milky Way, Earth, and the Sun with all relevant L1s. If it looks appropriate, I'll support your idea.

 

-modest

 

I will answer your queries tomorrow. I can assure you that the 3-body problem will not resemble your diagrams A or B.

 

I will provide links to show the potential is zero at L1. I will explain why it works that way and not otherwise.

 

As far as illustrating an entire galaxy, that would be quite an endeavor. I will do my best though (I actually have simplified versions already). I would like to see you do that with the mainstream diagram A.

 

Perhaps we should solve the 3-body problem first. And this time, we should include L2 and L3.

 

 

CC