modest Posted May 24, 2008 Report Share Posted May 24, 2008 CC, I apologize for neglecting our conversation in this thread lately... It has become increasingly apparent that the laws of nature (to which the physical mechanism of gravity must be attached) are most eloquently expressed in terms of a minimum principle (as opposed to a maximum principle) that opens the way to a quasi-complete understanding of a particular phenomenon or aspect of nature (notably that of gravitation). Minimum / maximum principle? Minimum gravitational force is zero newtons. Zero potential is somewhat arbitrarily set at infinity where a test particle is unbound from a mass. These are not a physical mechanism for gravity. Here we get right down to the core of the physical mechanism of the gravitational interaction. What is the relationship between Lagrange points, general relativity, and the physical mechanism of gravity? A Lagrange point is a physical location where a test particle (LP) will orbit a massive body M1 motionless relative to a massive body M2 which is also orbiting M1. Therefore Lagrange points are a three body orbital solution (I don't mean to insult your intelligence, I know you know this). M1, M2, and LP follow the equations of motion in general relativity or Newtonian mechanics. As such, they are an example of a mass / particle interaction via gravity and motion. Lagrange points are not representative or indicative of a physical mechanism of gravity. They are simply the spots where the centripetal acceleration needed to keep something in orbit stationary to m1 is exactly provided by the gravitational force of M1 and M2. While this is something that has been known and accepted since Lagrange first solved these equilibrium points; it is something you disagree with as in this earlier post: I argue that there is a mechanism operational in space responsible for generating equilibrium, independent of centrifugal force. and again recently: 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. While I don’t understand how you are relating L-points to gravity’s cause, I see your argument is reliant on this reinterpretation of the forces involved. I think you’re mistaken on this and I’ll now do my best to demonstrate how. My claim is that a system must be in motion (orbit) for Lagrange points to exist. When L-points do exist, it is because the gravitational ‘force’ exerted by the two masses is exactly balanced by the centrifugal ‘force’ caused by the rotating reference. As my source points out, L1 is the only quasi-exception: Such a system must be in rotation for Lagrange points to exist. If the system is not rotating, point L1 will exist for as long as it takes the two large bodies to fall into each other, but all of the other points require the two large bodies to be in orbit around one another. -source The easiest way to prove this is with things we know about our own solar system (and have tested). The sun/earth L1 is about 1/100th the distance to the sun. The force of gravity is: [math]F = G \frac{M_1M_2}{r^2}[/math] Anything at the earth/sun Lagrange point L1 is pulled by the sun: [math]F_S = \frac{ \left[ 6.674 \times 10^{-11} m^3kg^{-1}s^{-2} \right] \left[ 1.989 \times 10^{30}kg \right]}{ \left[ \frac{99}{100}1.496 \times 10^{11}m \right]^2}[/math] with a force of 0.006052 Newtons And pulled by the earth: [math]F_E = \frac{ \left[ 6.674 \times 10^{-11} m^3kg^{-1}s^{-2} \right] \left[ 5.974 \times 10^{24}kg \right]}{ \left[ \frac{1}{100}1.496 \times 10^{11}m \right]^2}[/math] with a force of 0.000178 Newtons Just by comparing the two we see straight away that the sun is pulling on L1 with more force than the earth. Spacetime is not flat at L1. I can support this with the plot I made in our earlier thread. The plot is of the sun / earth potential. The orange line is L1 and the blue line (potential) is not flat at L1: The difference in gravitational force between the earth and the sun at L1 is 0.006052 - 0.000178 = 0.005874 N. So, why is this spot L1 if there is a net gravitational force of 5.9 thousands of a newton per kilogram toward the sun? Why wouldn’t the spot be closer to the earth where gravitational forces cancel and spacetime is indeed flat? The answer is the centrifugal force. Lagrange found points where the force of gravity is exactly canceled by the centrifugal force of the rotating system. We can calculate the centrifugal force and see if it equals our disparity in gravitation. [math]F_C = \frac {m v^2}{r}[/math] where radius is [imath](99/100) 1.496 \times 10^{11}m[/imath]Velocity is: [math]\frac{2 \pi r}{31556926 s} = 2.95 \times 10^4 m/s[/math] [math]F_C = \frac {v^2}{r}[/math] [math]F_C = \frac {(2.95 \times 10^4)^2}{1.481 \times 10^{11}} = 0.00588 N[/math] Which is what we found for the net gravitational force at L1 (0.005874 N) trying to pull it the other way. In short, there can be no doubt a Lagrange point is a balance between gravity and centrifugal force. There are other ways to prove it. Finding a L-point (as is done here or a different method here) involves finding the balance between gravitational and centrifugal forces (and the Coriolis force at L4 and 5.) I don't see how it is possible to reject this description of the forces involved. All of these examples of Lagrange points that you've shown look like orbital dynamics to me. They are described by Newton's laws of motion. I don't see a mechanism for gravity. -modest Quote Link to comment Share on other sites More sharing options...
coldcreation Posted May 25, 2008 Author Report Share Posted May 25, 2008 I apologize for neglecting our conversation in this thread lately... It is I who apologizes for my three week quasi-absence. Minimum / maximum principle? Minimum gravitational force is zero newtons. Zero potential is somewhat arbitrarily set at infinity where a test particle is unbound from a mass. These are not a physical mechanism for gravity. I was referring to local or relative minima, not the absolute minimum (zero potential) at infinity. Obviously that absolute zero value is an important one, just as the absolute zero of temperatures is important, in that it shows that there exists in nature an fundamental 'ground-state' (in this case for gravity), however unattainable it may be locally. The key here is that local minima mimic the absolute zero potential state, since gravitational force (spacetime curvature) cancels to zero (the local value) at particular locations in the combined fields of massive objects. A Lagrange point is a physical location where a test particle (LP) will orbit a massive body M1 motionless relative to a massive body M2 which is also orbiting M1. Therefore Lagrange points are a three body orbital solution... Yes, but that is not all there is to it. These are locations where the stresses and strains, as well as alleviations or relaxations in the field extending on either side of local minima (and saddle points) are directly and indirectly 'felt' by the massive bodies: a crucial point (pun non-intended). Tides here on earth are a prime example. The shape of the field surrounding massive objects (when two or more are present) is no longer perfectly spherical (as would be the case of a field surrounding one mass when no other mass is present). They are oblong, as in this illustrious :D Coldcreation illustration below (exaggerated greatly to show the direction of deviation). The shape (away from spherical) of the intrinsic fields is significant. It shows that the minima (and saddle points) cannot be neglected when considering gravity. It shows that gravity is not an attractive force operational in a Euclidean space (in the Newtonian sense). At the same time, it sheds light on the framework of GR, which states that "there is no such thing as empty space, i.e. a space without field." (Einstein)' There is space without field. These are points where the gravitational field gradient is locally equal to zero. But too, and more generally, even though spacetime does not claim existence on its own (as in Newton's absolute space), the local (or relative) minima are a structural quality of the field: a quality present in all gravitating systems. And so, in addition to, and in accord with Einstein's statements: "the field is an irreducible element of physical description” and “the totality of physical ‘events’ is thus thought of as being embedded in a four-dimensional continuous manifold" we extend: Field-free space (the local minimum) is the ultimate limit of the gravitational field curvature, the irreducible physical ground state. That is the starting point from where the full description of the mechanism behind gravity must emerge (as we will see shortly). M1, M2, and LP follow the equations of motion in general relativity or Newtonian mechanics. As such, they are an example of a mass / particle interaction via gravity and motion. Agreed, but again, that cannot be the whole picture. Proof is that there exists, hitherto, no quantum theory of gravity that explains all observations, events or phenomena. Nor does GR or NM draw an accurate sketch of phenomena such as galactic rotational curves (which are most often flat, or nearly so) without ad hoc ingredients, additives and a dash of preservatives, e.g., DE, CDM (which most do not even hesitate to swallow). Lagrange points are not representative or indicative of a physical mechanism of gravity. It is well known that in particle physics the search for the so-called Kaluza-Klein graviton is one where, because we cannot shine a light directly on them (to 'see' them), physicists are searching for where they are not, i.e., the missing energy signature in particle interactions. (See here for example: Search for Extra Dimensions using Missing Energy at CDF). This is remotely similar to what I am proposing with respect to Lagrange points (the local minima), though instead of postulating a 5th dimension (or even a graviton) it is the field-free points that need to be illuminated. L-points are thus thought of as gravity mediators. There is no new physics involved. In fact, going beyond the current discussion, but not beyond the scope of this thread, it could be advanced that unification of the "forces of nature" occurs not inside the Plank scale, at energies compatible to the Mother of all explosions (the big bang), but in the low energy regime, as the forces (including the odd-ball gravity) tend to zero (or even at zero). More on that soon. And so (and in short), your conclusion that "Lagrange points are not representative or indicative of a physical mechanism of gravity" is premature, i.e., the relation cannot yet be ruled out. They are simply the spots where the centripetal acceleration needed to keep something in orbit stationary to m1 is exactly provided by the gravitational force of M1 and M2. While this is something that has been known and accepted since Lagrange first solved these equilibrium points; it is something you disagree with as in this earlier post: I argue that there is a mechanism operational in space responsible for generating equilibrium, independent of centrifugal force...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. The reason for my distaste of centrifugal force is captured in the following passages: In classical mechanics' date=' centrifugal force (from Latin centrum "center" and fugere "to flee") is a fictitious force[1'][2] that is associated with the centrifugal effect[3][4], which is an apparent acceleration that appears when describing physics in a rotating reference frame; centrifugal force appears to act on anything with mass considered in such a frame. Centrifugal force is considered to be a fictitious force (or pseudo force) because it is an artifact of acceleration of the rotational frame.[5] As a result, this force does not appear in Newton's laws of motion for an inertial frame of reference (in which the motion of an object is explained by the real impressed forces), because this force does not originate as a physical interaction between real objects. However, when Newton's laws are applied in the rotating frame (instead of an inertial frame), they must be modified by adding the centrifugal force and other pseudo-forces such as the Coriolis force, in order to make the same physical predictions as made by Newton's laws alone in an inertial frame. The centrifugal force depends only on the position and the mass of the object and is always oriented away from the axis of rotation of the rotating frame, whereas the Coriolis force depends on the velocity and mass of the object but is independent of its position.[6] ...Because this centripetal force is combined from only pseudo forces, it is "fictitious" in the sense of having no apparent origin in physical sources (like charges or gravitational bodies); and having no apparent source, it is simply posited as a "fact of life" in the rotating frame, it is just "there". It has to be included as a force in Newton's laws if calculations of trajectories in the rotating frame are to come out right. ...The centrifugal and Coriolis forces are called "fictitious" because they do not appear in an inertial frame of reference.[8] Unlike forces present in an inertial frame, fictitious forces do not express interactions between objects, like nuclear or electromagnetic forces. Instead, they are an expression of accelerations that make a frame non-inertial. In the non-inertial frame, these fictitious forces must be added to the real forces when using Newton's laws (and often are referred to as 'inertial forces' for that reason.[22][23]) Because they do not originate from another object, there is no object to experience an associated reaction force: in this sense fictitious forces do not obey Newton's third law. However, with the addition of fictitious forces, Newton's laws do apply in non-inertial reference frames, such as planets, centrifuges, carousels, turning cars, and spinning buckets. Despite the name, fictitious forces are experienced as very real to those actually in a non-inertial frame. Fictitious forces also provide a convenient way to discuss dynamics within rotating environments, and can simplify explanations and mathematics. And on the score of L1: Unlike the other Lagrangian points, L1 would exist even in a non-rotating (static or inertial) system. Rotation slightly pushes L1 away from the (heavier) Earth towards the (lighter) Moon. The fact that L1 would exist (with its know characteristics and properties) in the absence of orbital motion is compelling. My claim is that a system must be in motion (orbit) for Lagrange points to exist. When L-points do exist, it is because the gravitational ‘force’ exerted by the two masses is exactly balanced by the centrifugal ‘force’ caused by the rotating reference. Again, this statement would appear to contradict your claim: Unlike the other Lagrangian points, L1 would exist even in a non-rotating (static or inertial) system. As my source points out, L1 is the only quasi-exception: I would remove the "quasi" Anything at the earth/sun Lagrange point L1 is pulled by the sun: Anything at the earth/sun Lagrange point L1 is equally pulled by the earth. It is for this reason that a test particle placed at L1 'feels' no gravitational force, at least considering the lateral components of the M1 and M2 fields. Just by comparing the two we see straight away that the sun is pulling on L1 with more force than the earth. Spacetime is not flat at L1. I can support this with the plot I made in our earlier thread. The plot is of the sun / earth potential. The orange line is L1 and the blue line (potential) is not flat at L1: The L1 point is a saddle point (a hypersurface), a stationary point, not an extremum. By definition an objects place there will remain in an unstable equilibrium (like a pencil balancing on its point). If you remove the object (or test particle) the L1 point remains. Because there is either a maximum or a minimum in each direction (perpendicular to and directly on the M1-L1-M2 line), both derivatives are zero, the function levels out, the function decreases to zero, the gravitational field curvature gradient vector is flat. Around the inner Lagrangian L1 point (a critical point of inflection) the gradient increases in one direction and decreases in another. ... All of these examples of Lagrange points that you've shown look like orbital dynamics to me. They are described by Newton's laws of motion. I don't see a mechanism for gravity. The examples of objects placed at Lagrange points that I have shown throughout this thread (and there are many more systems of the kind generally accepted as being associated) do indeed look like orbital dynamics. That's because they are based on orbital mechanics. The principle reason for pointing then out was to show that these gravitating systems possess a strict and ordered geometrical pattern, and that this pattern is ubiquitous throughout the universe, at all scales and energy levels, that this juxtaposition of maxima and minima in N-body gravitating systems is not solely a restricted problem. That this should be so is no surprise. It would have been a surprise had Lagrangian dynamics been operational exclusively in the solar system. What this pattern shows is that Newton's laws of motion is insufficient as a description of nature, since it does not predict for example the existence of L4 and L5 points, and thus it does not predict the observed geometric structure seen in all of the above images. Nor does Newton's laws of motion predict the linear formation present in barred galaxies, or the observed rotational curves of galaxies (Lagrangian dynamics for which is directly responsible). Certainly, the Euler-Lagrange equations concord with Newton's laws of motion regarding L1, L2 and L3, when the fictitious centrifugal force is applied. The pattern shows that systems bounded under the sole influence of gravity are not random collections of objects that happen by chance to be associated orbitally because of some initial condition, because of some natural selection process (though these are bound to exist as well), that these systems are not intrinsically chaotic. They are stable dynamical and evolving. What this means is that GR does not reduce classical mechanics to special case phenomena at low energy and small velocities. Instead, GR must function in unison with classical mechanics at all energies, velocities and scales. Finally, and most importantly, what this Lagrange pattern shows too is that there exists a mechanism inherent and operational in all gravitating systems (i.e., it is universal) that regulates formation, equilibrium, order and longevity; a mechanism that the simple Newtonian solution (gravity as an attractive force) or complex GR solution (with its curved spacetime) do not take into consideration. That is why both NM and GR have not identified the physical mechanism of gravity. There is a mechanism responsible for the interplay between massive bodies, between the fields, the combined fields, the Lagrangian field-free space (or Minkowski spacetime). These, coupled with the fictitious centrifugal force and Laplacian mean motion resonances, combine to form (along with GR) a description of nature that is consistent with observations. (Note, CDM and DE are not mentioned, or needed). This physical mechanism, the source of gravity, is what I hope to identify in this thread. I don't think it is beyond our current understanding of physics to do so. Something has only just begun CC Quote Link to comment Share on other sites More sharing options...
modest Posted May 26, 2008 Report Share Posted May 26, 2008 The reason for my distaste of centrifugal force is captured in the following passages: Yes, but you cannot leave the centrifugal force out of the discussion. You can't find a Lagrange point, keep a Lagrange point, or understand a Lagrange point without the force given to the point from the rotating reference. You are ignoring the main ingredient. And on the score of L1: Unlike the other Lagrangian points, L1 would exist even in a non-rotating (static or inertial) system. Rotation slightly pushes L1 away from the (heavier) Earth towards the (lighter) Moon. The fact that L1 would exist (with its know characteristics and properties) in the absence of orbital motion is compelling. You're not quite getting it. If the earth were NOT rotating around the sun - but was static then L1 would NOT be a million miles away from the earth. It would be much closer to the earth and its position would move as the earth fell toward the sun. Without rotation the whole thing breaks down. The only reason L1 is where it is and acts the way it acts is because the system is rotating and there is a centrifugal force at L1. Again, this statement would appear to contradict your claim:Unlike the other Lagrangian points, L1 would exist even in a non-rotating (static or inertial) system. As my source points out, L1 is the only quasi-exception: I would remove the "quasi" I know you would. This is why I'm trying to explain - spacetime is not flat at L1 unless:the system is not rotating, orM1 = M2The reason I showed how to calculate the forces at L1 was to point out the centrifugal force can't be ignored. It certainly can't be ignored at the other points. It is the whole point of these L-points - it is the reason they act the way they do. Perhaps the reason you are attributing great powers to L points is because you fail to attribute them the normal power of centrifugal force. Anything at the earth/sun Lagrange point L1 is equally pulled by the earth. It is for this reason that a test particle placed at L1 'feels' no gravitational force, at least considering the lateral components of the M1 and M2 fields. This is not exactly right. The combined gravitational force of M1 and M2 cancels the centrifugal force. That is the reason a test particle has no net force. Would you think the earth and sun cancel at L2? The L1 point is a saddle point (a hypersurface), a stationary point, not an extremum. By definition an objects place there will remain in an unstable equilibrium (like a pencil balancing on its point). If you remove the object (or test particle) the L1 point remains. Because there is either a maximum or a minimum in each direction (perpendicular to and directly on the M1-L1-M2 line), both derivatives are zero, the function levels out, the function decreases to zero, the gravitational field curvature gradient vector is flat. The derivative of potential is not zero at Lagrange points. The reason I calculated forces in my last post was to show this. The only exception (which is incidental rather than significant) is L1 when M1 equals M2 or the system is not rotating. I should point out that a not-rotating system is impossible. As such, it is very much a quasi-exception. Around the inner Lagrangian L1 point (a critical point of inflection) the gradient increases in one direction and decreases in another. Yes - this is what I'm saying. Spacetime isn't flat. The gradient increases toward the earth which means (by definition) there is a net gravitational force toward the sun (which is exactly equal to the centripetal force needed to keep something at L1 where it is) The net gravitational force is equal and opposite to the centrifugal force. I calculated it in my last post - they were equal and opposite. A lagrange point is a balance between gravity and forces induced by rotation. -modest Quote Link to comment Share on other sites More sharing options...
coldcreation Posted May 27, 2008 Author Report Share Posted May 27, 2008 Yes, but you cannot leave the centrifugal force out of the discussion. You can't find a Lagrange point, keep a Lagrange point, or understand a Lagrange point without the force given to the point from the rotating reference. You are ignoring the main ingredient. I'm not ignoring the centrifugal force, I just don't like it. Surely, the gravitational interaction is a real one (whether it is a force or not). So in generalizing about Lagrange points, I prefer to speak (perhaps with lack of rigor) of gravity (as in this sentence: The Lagrangian points (...Lagrange point, L-point, or libration point), are the five positions in an orbital configuration where a small object affected only by gravity can theoretically be stationary relative to two larger objects (such as a satellite with respect to the Earth and Moon).) rather than the centrifugal force, or centripetal force, which are only a fictitious forces. But yes it is true that: The Lagrange points mark positions where the combined gravitational pull of the two large masses provides precisely the centripetal force required to rotate with them.. But, the centripetal force is supplied by the gravitational attraction between massive bodies, and is thus implied or inferred from the trajectory of the massive bodies. It is only a force requirement deduced from a trajectory observed. Centripetal force is not included in the forces used to apply Newton's laws F = m a in calculating a trajectories because it is (only) inferred from an established trajectory. It is true that to solve Lagrangian mechanics problems exactly in an Earth-bound reference frame, the coriolis force and the centrifugal force (two fictitious forces) have to be introduced. Note though, the use of coriolis and the centrifugal fictitious force(s) is based on space that have a Euclidean geometry, not curved space time, as in General relativity, within which a free-falling observer, say, in an elevator, would not be able to detect the gravitational force. Therefore, freely-falling reference frames are equivalent to inertial reference frames (see Einstein's equivalence principle). In this discussion, so far, I am attributing the apparent acceleration (centripetal force, along with centrifugal force) to a curved spacetime phenomenon (see Einstein's theory of general relativity). Certainly, the trajectory of, say the Earth (or any other body), can be though of, from the perspective of an observers own reference frame, to be traveling in a straight line (a geodesic) along a path described by curved spacetime (freely falling, accelerating). So fictitious forces are replaced with an effect due to curved spacetime. In general relativity, an observer and test particle in relative motion have differing local rest spaces, and so, a geometric description of centrifugal force in the local rest space of the test particle is required, if you hope to generalize the concept with classical mechanics (i.e., nonrelativistically). Even in classical mechanics, centrifugal force is relative involving a particular motion (a relative one) with respect to an implicit family of Newtonian inertial observers. (In flat spacetime this is interpreted, and generally accepted as a centrifugal force). There is no unambiguous way of separating gravitational force and inertial force in the general relativistic framework. So Lagrange regions, in the context of curved spacetime (i.e., geometrically speaking), may be interpreted as describing locations of gravitational force balance in the local rest-space of the test particle. There is no physical difference, locally, between a real gravitational force and a fictitious inertial force, since freely-falling objects in motion within a gravitational field are understood to be purely inertial (again, locally). In another way, the gravitational field (curved spacetime) and the inertial field are arguably identical: characterized at each point of spacetime by the ten metric tensor coefficients, i.e., the metric field described by (or responsible for) curved spacetime (gravity) is identical with the metric field responsible for the fictitious inertial force. This is why I stress the necessity of combining Lagrangian dynamics with GR. (See CENTRIPETAL ACCELERATION AND CENTRIFUGAL FORCE IN GENERAL RELATIVITY). Concluding remarksThe natural Frenet-Serret machinery from the point of view of an observer family has been presented and linked to the natural geometrization of the usual notion of a centrifugal force in the rest space of a test particle world line (as opposed to a centripetal acceleration in the rest space of the observers). For the most interesting case in which these concepts may be illustrated' date=' namely circular orbits in stationary axisymmetric spacetimes, it has been shown in a related paper that this comoving relative generalized centrifugal force is nothing other than the effect of the orbital rotation relative to gyro-fixed axes of the relative velocity, itself rigidly attached to stationary axisymmetric axes. For equatorial plane orbits in the Kerr spacetime, it agrees with our nonrelativistic intuition about centrifugal force near the circular geodesic orbits. For equatorial plane orbits at r = 3M of a Schwarzschild black hole, this generalized centrifugal force then vanishes due to the phase-locking of gyro-fixed axes with stationary axisymmetric axes. For more general noncircular orbits in stationary spacetimes, or completely general trajectories in any spacetime, this comoving relative generalized centrifugal force is the first such inertial force introduced by decomposing the test particle acceleration in its own rest space, which is exactly how the usual nonrelativistic centrifugal force is experienced.[/quote'] With this in mind, my distaste for centrifugal force (namely the way it artificially balances, or cancels precisely, the gravitational force, leading to a fine-tuning problem inherent in celestial mechanics) can be satisfied. I don't have to swallow it. The fictitious nature of the coriolis force and centrifugal force is easily observed by looking at various scenes from both rotating and stationary reference frames. Check out this vid: Visualization of the Coriolis and centrifugal forces http://www.youtube.com/watch?v=49JwbrXcPjc I wish I could illustrate with such clarity the same phenomenon using curved spacetime. If the earth were NOT rotating around the sun - but was static then L1 would NOT be a million miles away from the earth. It would be much closer to the earth and its position would move as the earth fell toward the sun. Without rotation the whole thing breaks down. The only reason L1 is where it is and acts the way it acts is because the system is rotating and there is a centrifugal force at L1. You bring up a good point. And I agree with most of it. However, the reason L1 acts the way it does is because it is a saddle point region. ...spacetime is not flat at L1 unless:the system is not rotating, orM1 = M2The reason I showed how to calculate the forces at L1 was to point out the centrifugal force can't be ignored. It certainly can't be ignored at the other points. It is the whole point of these L-points - it is the reason they act the way they do. Perhaps the reason you are attributing great powers to L points is because you fail to attribute them the normal power of centrifugal force. Technically you are correct. The true origin of the saddle point where the actual combined gravitational fields cancel is not exactly located at L1. It is however very close. There is no attribution of great powers to L-points. Quite the contrary: these are simply minima (and saddle points) of the field. There is no power, or force, at all. They are important precisely for that reason. The combined gravitational force of M1 and M2 cancels the centrifugal force. That is the reason a test particle has no net force. Would you think the earth and sun cancel at L2? Good question. Geometrically speaking I might be inclined to answer affirmatively, but the shape of the field there might resemble a monkey saddle, as opposed to a horse saddle. I'll study the question further and get back to you on that. Yes - this is what I'm saying. Spacetime isn't flat. The gradient increases toward the earth which means (by definition) there is a net gravitational force toward the sun (which is exactly equal to the centripetal force needed to keep something at L1 where it is) Nevertheless, the true center or origin of the saddle point is not very far away on the same line between M1 and M2. The net gravitational force is equal and opposite to the centrifugal force. I calculated it in my last post - they were equal and opposite. A lagrange point is a balance between gravity and forces induced by rotation. -modest I think my above remarks addressed this issue sufficiently. It is really a question of whether we use classical mechanics in a Euclidean space, or a combination of Lagrange dynamics with GR in a curved spacetime. For sure, one method of deriving the equations of motion for massive bodies in classical mechanics is to set the gravitational force exactly equal to the centrifugal force in a coordinate system that revolves with the object (which is equivalent to setting the total force on the orbiting body equal to zero), otherwise resulting in a straight trajectory in Euclidean space, or a geodesic in a non-Euclidean spacetime. The main difference between the geometrically curved spacetime description of nature and the 'fictitious force' description is that the latter can be reduced to a mere artifact of coordinate motion. With the former, we virtually do away with the inertial frame where physical laws are more easily understood. And, we are left with a coordinate system that corresponds to the local non-Euclidean geometry in accord with GR. It can be argued too that (like centrifugal force) gravitational force is also a fictitious force. Indeed, that is the case when gravity is considered a curved spacetime phenomenon. It is no longer an attractive force. In fact it is no force of nature at all. So the answer to the question: what is the physical mechanism of gravity(?) becomes even more pressing. Let's see what happens when we place fictitious forces (gravity included) to the side... Let's look at the geometry... CC Quote Link to comment Share on other sites More sharing options...
modest Posted May 27, 2008 Report Share Posted May 27, 2008 I'm not ignoring the centrifugal force, I just don't like it. Surely, the gravitational interaction is a real one (whether it is a force or not). So in generalizing about Lagrange points, I prefer to speak (perhaps with lack of rigor) of gravity (as in this sentence: The Lagrangian points (...Lagrange point, L-point, or libration point), are the five positions in an orbital configuration where a small object affected only by gravity can theoretically be stationary relative to two larger objects (such as a satellite with respect to the Earth and Moon).) rather than the centrifugal force, or centripetal force, which are only a fictitious forces. But yes it is true that: The Lagrange points mark positions where the combined gravitational pull of the two large masses provides precisely the centripetal force required to rotate with them.. Gravity can be an effect of spacetime curvature and yet be a 'real' force. Pseudo-forces can be the effect of a rotating reference and yet be a 'real' force. Both are necessary for Lagrange points to exist. There's no point in saying you like one and don't like the other - they are equally necessary and equally real. That said, I do like your goal of trying to eliminate the mysterious nature of non-inertial forces. I personally feel unsatisfied with GR's treatment of this (unsatisfied on a gut level is all). An explanation is needed to deal with non-inertial frames. They won't go away by ignoring them. But, the centripetal force is supplied by the gravitational attraction between massive bodies, and is thus implied or inferred from the trajectory of the massive bodies. exactly true - as far as my understanding It is only a force requirement deduced from a trajectory observed. Centripetal force is not included in the forces used to apply Newton's laws F = m a in calculating a trajectories because it is (only) inferred from an established trajectory. The centripetal force is included as gravity where trajectories are concerned. It is true that to solve Lagrangian mechanics problems exactly in an Earth-bound reference frame, the coriolis force and the centrifugal force (two fictitious forces) have to be introduced. Note though, the use of coriolis and the centrifugal fictitious force(s) is based on space that have a Euclidean geometry, not curved space time, as in General relativity, within which a free-falling observer, say, in an elevator, would not be able to detect the gravitational force. Because gravity is a fictitious force does not mean all fictitious forces are explained by gravity. In fact, they most certainly are not. I will say once again though, I very much like how you're thinking. Let's see where we're going here... Therefore, freely-falling reference frames are equivalent to inertial reference frames (see Einstein's equivalence principle). In this discussion, so far, I am attributing the apparent acceleration (centripetal force, along with centrifugal force) to a curved spacetime phenomenon (see Einstein's theory of general relativity).Certainly, the trajectory of, say the Earth (or any other body), can be though of, from the perspective of an observes own reference frame, to be traveling in a straight line (a geodesic) along a path described by curved spacetime (freely falling, accelerating). So fictitious forces are replaced with an effect due to curved spacetime. Ah, yes, very good. I like this. However, this would seem to support my claim which you have yet to agree with. I'm saying the point L1 (as well as the rest) cannot have flat spacetime. For your description here to be accurate you must concede this point. L1 cannot very well be a null geodesic in a circular orbit and have flat spacetime. There is no unambiguous way of separating gravitational force and inertial force in the general relativistic framework. So Lagrange regions, in the context of curved spacetime (i.e., geometrically speaking), may be interpreted as describing locations of gravitational force balance in the local rest-space of the test particle. There is no physical difference, locally, between a real gravitational force and a fictitious inertial force, since freely-falling objects in motion within a gravitational field are understood to be purely inertial (again, locally). In another way, the gravitational field (curved spacetime) and the inertial field are arguably identical: characterized at each point of spacetime by the ten metric tensor coefficients, i.e., the metric field described by (or responsible for) curved spacetime (gravity) is identical with the metric field responsible for the fictitious inertial force. This is why I stress the necessity of combining Lagrangian dynamics with GR. The equations of motion are already in GR. They describe this situation in the framework you describe just as well as Lagrange described it in Newton's framework. So I'm afraid I still have to say I don't see a physical mechanism of gravity. I'm finding what you're saying much more agreeable - nevertheless I'm still unsure where we're going. With this in mind, my distaste for centrifugal force (namely the way it artificially balances, or cancels precisely, the gravitational force, leading to a fine-tuning problem inherent in celestial mechanics) can be satisfied. I don't have to swallow it. Here you are mistaken. Finding a null geodesic in GR that keeps an equilibrium orbit (two or three bodies) is a fine-tuning problem no different than Newton would show. This has been proven mathematically (which I think I've sourced before) and proven with spacecraft at unstable L-points. Technically you are correct. The true origin of the saddle point where the actual combined gravitational fields cancel is not exactly located at L1. It is however very close. I contend it is not close. For your GR-view of things to be correct it cannot be. It's easy to calculate where it is in any case: [math]F = G \frac{M_1M_2}{R^2}[/math]Force from the sun will equal force from the earth at L1 so:[math]\frac{GM_{(sun)}}{R_{(sun)}^2} = \frac{GM_{(earth)}}{R_{(earth)}^2}[/math] the ratio of earth GM to sun GM is 3 x 10^-6, the square root of which is .00173. Therefore I bet if you plug the numbers into the equation above you'd find that a spot .173% of the distance to the sun has equal force of gravity from the earth and sun. That's about 10% of L1's current distance which is hardly close. Let me think and read some more about rotation in GR before I comment further. I'm sure Einstein's bucket thought experiment of Mach's principle can somehow be related to our discussion. I'll have to think about it. -modest Quote Link to comment Share on other sites More sharing options...
modest Posted June 2, 2008 Report Share Posted June 2, 2008 It sounds like what you’re saying is that the key to the physical mechanism of gravity can be found in the equivalence principle of GR. That is to say: there is a significant reason gravity and inertial acceleration share physical and mathematical descriptions. To demonstrate the link between the two you are looking at Lagrange points where the force given by the gravitational field is exactly equal to the inertial force of rotation. You are perhaps thinking that an investigation into the nature of this equality could be useful in understanding what’s behind GR. If this is indeed a correct characterization - I would suggest simplifying your perspective. Lagrange points do have an equality of gravitational and inertial forces that are reflected in the weak equivalence principle. However, a simple orbit does as well. On this page:Field Equations & Equations of MotionThe equation of motion, a geodesic, is found for a circular orbit by setting the gravitational force equal to the centrifugal force [imath]mg = -(mv^2/r)u[/imath]. It would be easier to investigate this situation rather than a three body solution if I have correctly understand your intent. My problem is: I’m not sure if this kind of inquiry can get a person past the description which I’m sure you’ve heard. When you are accelerating in a rocket - you accelerate through space and feel weight as a result. When you stand on a planet - space moves (because of curved time) through you and you feel weight because of it. So, the equivalence principle seems to describe mass curving spacetime as the mechanism for gravity. The math that is built on that principle (general relativity) works well. In short - If we are looking for a reason mass curves spacetime then I don’t see how we’re getting there by looking at areas in the gravitational field that are equal in gravity and inertial acceleration. But, that certainly doesn't mean it can't be done. I'm just giving you my perspective on what you're saying as best I understand it. -modest Quote Link to comment Share on other sites More sharing options...
Pyrotex Posted June 2, 2008 Report Share Posted June 2, 2008 ...Note that because the field surrounding each individual body (M1 and M2) is elliptical in shape, the actual gravitational force on the surface of each object varies depending on location. ...AHA!! Here's your mistake. The proof requires some facility with simple Calculus--specifically the concept of gradient.No need to thank me. You're welcome.Pyro Quote Link to comment Share on other sites More sharing options...
coldcreation Posted June 5, 2008 Author Report Share Posted June 5, 2008 ...Note that because the field surrounding each individual body (M1 and M2) is elliptical in shape, the actual gravitational force on the surface of each object varies depending on location. ...AHA!! Here's your mistake. The proof requires some facility with simple Calculus--specifically the concept of gradient.No need to thank me. You're welcome. Welcome to the discussion Pyro. The concept of gradient has already surfaced a dozen times in this thread. What specifically would you like to add. Are you saying that the fields surrounding M1 and M2 (of any two-body system) are spherically symmetric, and/or that the actual gravitational force (or field gradient) on the surface of each object (M1 or M2) does not vary depending on location? Where exactly do you see a mistake? The proof of what requires some facility with simple Calculus? For what is there no need to thank you? CC Quote Link to comment Share on other sites More sharing options...
coldcreation Posted June 6, 2008 Author Report Share Posted June 6, 2008 double post removed. Quote Link to comment Share on other sites More sharing options...
coldcreation Posted June 6, 2008 Author Report Share Posted June 6, 2008 Gravity can be an effect of spacetime curvature and yet be a 'real' force. Pseudo-forces can be the effect of a rotating reference and yet be a 'real' force. Both are necessary for Lagrange points to exist. There's no point in saying you like one and don't like the other - they are equally necessary and equally real. That depends on you point of view (an observers reference frame). That said, I do like your goal of trying to eliminate the mysterious nature of non-inertial forces. I personally feel unsatisfied with GR's treatment of this (unsatisfied on a gut level is all). An explanation is needed to deal with non-inertial frames. They won't go away by ignoring them. Excellent point. I have to look into this further. Because gravity is a fictitious force does not mean all fictitious forces are explained by gravity. In fact, they most certainly are not. I will say once again though, I very much like how you're thinking. Let's see where we're going here... Agreed. Ah, yes, very good. I like this. However, this would seem to support my claim which you have yet to agree with. I'm saying the point L1 (as well as the rest) cannot have flat spacetime. For your description here to be accurate you must concede this point. L1 cannot very well be a null geodesic in a circular orbit and have flat spacetime. See "Balance of Gravitational Fields" below. The equations of motion are already in GR. They describe this situation in the framework you describe just as well as Lagrange described it in Newton's framework. So I'm afraid I still have to say I don't see a physical mechanism of gravity. I'm finding what you're saying much more agreeable - nevertheless I'm still unsure where we're going. I will attempt to elaborate further once I'm back at my home base. With this in mind["the natural geometrization of the usual notion of a centrifugal force in the rest space of a test particle world line (as opposed to a centripetal acceleration in the rest space of the observers"], my distaste for centrifugal force (namely the way it artificially balances, or cancels precisely, the gravitational force, leading to a fine-tuning problem inherent in celestial mechanics) can be satisfied. I don't have to swallow it. Here you are mistaken. Finding a null geodesic in GR that keeps an equilibrium orbit (two or three bodies) is a fine-tuning problem no different than Newton would show. This has been proven mathematically (which I think I've sourced before) and proven with spacecraft at unstable L-points. Again, I agree. But I'm not convinced that this fine-tuning problem (present also in GR) is not irremediable. I contend it is not close. For your GR-view of things to be correct it cannot be. It's easy to calculate where it is in any case: [...] you'd find that a spot .173% of the distance to the sun has equal force of gravity from the earth and sun. That's about 10% of L1's current distance which is hardly close. If you look at both illustrations here: Actual Lagrange Point Geometry you will see that L1 and the point H (the true location of gravitational field balance) are extremely close to one another. Balance of Gravitational FieldsIgnore corrections for the Sun's field and the motion of the Earth-Moon barycentre' date=' and treat orbits as circular.There is a point H between L1 and Moon at which there is, from Earth and Moon, no net gravitational force. It is atM / Q2 = m / q2where Q+q is the Earth-Moon separation, R. As M ? 81×m, Q ? 9×q and H is about 10% of the way from Moon to Earth.An unpowered body at H will not be changing its velocity but will be stationary or move in a straight line. But H itself moves in a circle, so the body and H cannot stay together.Conversely, the net pull of the Earth and the Moon will, at L1, hold an unpowered body in a circular orbit with a period of one relevant Month. The body will remain at L1. (Same source as above General Explanation So the point where gravitational fields balance, H, is not 10% of L1's current distance. H is about 10% of the way from the Earth to the Sun, which places it very close to L1. Let me think and read some more about rotation in GR before I comment further. I'm sure Einstein's bucket thought experiment of Mach's principle can somehow be related to our discussion. I'll have to think about it. -modest I'm still getting my ducks in a row too. It's not easy. For example, I always thought (until now) L1 and H were at the same point. It shouldn't change much for the sake of this discussion. The fact remains: there is a point (a saddle point, in this case between massive bodies) where there is no net gravitational force (where the field gradient is zero, and not any other value). Why then is this so important? That will emerge shortly, if it hasn't already. CC Quote Link to comment Share on other sites More sharing options...
modest Posted June 6, 2008 Report Share Posted June 6, 2008 If you look at both illustrations here: Actual Lagrange Point Geometry you will see that L1 and the point H (the true location of gravitational field balance) are extremely close to one another. So the point where gravitational fields balance, H, is not 10% of L1's current distance. H is about 10% of the way from the Earth to the Sun, which places it very close to L1. I agree with the calculations you link to and quote. The difference in position between the actual point L1 and the actual point gravitational force cancels depends on the mass of the two bodies. The closer the two bodies are to each other in mass, the closer the two points are. This relationship continues until M1 = M2 where there is no difference between the actual spot L1 and the place where gravitational force is nill. Notice earlier when I said the only valid exception to L1 not having flat spacetime is when M1 = M2. This means the two points will be closer between the earth / moon system as you showed above than the earth / sun system which has a smaller mass ratio and was what I calculated. The mass ratio (how close the two bodies are in mass) determines this. It is a red herring. // editWe at least agree the two spots are not the same which means L1 cannot and must not have flat spacetime. Which, by the by, you still haven't acknowledged. //edit -modest Quote Link to comment Share on other sites More sharing options...
tierradelfuego Posted June 6, 2008 Report Share Posted June 6, 2008 Hey, Cold -- "I'm still getting my ducks in a row too. It's not easy." Yes, I agree ... but I have been waiting a year and a half! :-) I am glad to read your recent postings. You are finally revealing more details on your broad cosmological model (and by "broad", I include the nuclear/subatomic ramifications), in which I believe you have significant insights. But I am dying to read at least a sketch or an outline of your overall picture! Let's get on with it! Naysayers be damned! :-p I love the idea that the Lagrange points have much overlooked significance, particularly with respect to Arp's work. Dark matter, dark energy -- pfft! "I fart in their general direction!" There are many objects out there that are screaming for a new explanation -- quasi-stellar objects, objects with discordant redshifts, some of the peculiar galaxies, some of the objects that are incorrectly thought to be gravitationally lensed images of other objects, ... anything else? And you are on track for saying something more significant, for which I only have a couple of bits and pieces from your previous postings. To the current dispute ... It seems to me that if one accepts Einstein's equivalency principle, then one should not have a problem with your assertions about a saddle point at L1. That is the precisely the gist of the issue. When someone argues against your point that you are somehow lacking in respect for the centripetal acceleration of M2 towards M1, I would argue the opposite, and in fact, an argument against your point lacks an understanding of the equivalency principle, and hence, its significance to what you are trying to emphasize. Einstein said in GR -- no difference. No difference. No difference. ... My qualm -- with respect to a two-body problem (say, Sun-Earth), are you ignoring the solar system's galactic centripetal acceleration (let's assume an ultra-simple model in which the solar system is orbiting a central galactic mass)? So is not L1 merely an approximate saddle point, hence, L1 is not flat with regard to L1's galactic orbit? And so on ... Milky Way's orbit within the local cluster, etc? However, as one changes to larger and larger frames of reference, does not the Lagrangian points with respect to the objects involved get "flatter" in some "absolute" sense, in the sense of an absolute frame of reference (to the degree that one can employ that concept in a post-Einsteinian cosmology)? And could that be what you are hinting at? For example, in a simple "cluster" of galaxies in which a small galaxy is rotating about a much more massive galaxy, can't the Lagrangian points be described as being "flatter" than the Sun-Earth Lagrangian points? Or the solar system's Lagrangian points with respect to a simple Milky Way model? And the "flatter" a Lagrangian point, the more significant it is within your "cold matter creation model" (my short-hand term)? Just guessing here as to where you are going, but the "flatter" the point, the more prolific its "ability" in matter creation? Hence, the active regions at the Lagrangian points around galactic clusters? Am I on the right track?If so, did I guess too much or reveal too much? I think that I see bits and pieces of the whole picture. I am dying to read an explanation from you for how your model explains the discordant redshifts. I cannot conceive at all how this would be -- but, hey, I should not feel too bad, it's your model, not mine. ;-) --Tierra Quote Link to comment Share on other sites More sharing options...
coldcreation Posted June 7, 2008 Author Report Share Posted June 7, 2008 It sounds like what you’re saying is that the key to the physical mechanism of gravity can be found in the equivalence principle of GR. That is to say: there is a significant reason gravity and inertial acceleration share physical and mathematical descriptions. To demonstrate the link between the two you are looking at Lagrange points where the force given by the gravitational field is exactly equal to the inertial force of rotation. You are perhaps thinking that an investigation into the nature of this equality could be useful in understanding what’s behind GR. Certainly the Einstein equivalence principle is the drum and bass of the gravitational theory that signals a departure from linearity. In other words, gravity is a ‘curved spacetime’ phenomenon; the effects of gravity are equivalent to the effects of living in a curved spacetime. A consequence of EEP is that the laws governing experiments or observations must be independent of velocity in a free-falling frame of reference (Lorentz invariance). All bodies, without exception, fall with the same acceleration in the same gravitational field, and physics in a freely falling frame of reference is independent of location and the velocity of the frames. And so, yes, it is key to elucidating the physical mechanism of gravity. If this is indeed a correct characterization - I would suggest simplifying your perspective. Lagrange points do have an equality of gravitational and inertial forces that are reflected in the weak equivalence principle. However, a simple orbit does as well. It has been known since the advent of general relativity that it is impossible to discover by experiment whether a given system of coordinates is accelerated, or whether its motion is straight and uniform and the observed effects are due to a gravitational field. The implication of the equivalence principle “shatters the concept of inertial systems, as soon as gravitation enters in.” (Einstein 1940, 1954, p.330). So the simple perspective is equivalent to trying to find the answer to the problem by using the wrong approach. On this page:Field Equations & Equations of MotionThe equation of motion, a geodesic, is found for a circular orbit by setting the gravitational force equal to the centrifugal force . It would be easier to investigate this situation rather than a three body solution if I have correctly understand your intent. Easier, for sure, but to pursue this path further would be no different than hunting dogs barking at the base of a tree where they erroneously think their quarry is hiding. My ambition is to do away with the centrifugal force in this context. My problem is: I’m not sure if this kind of inquiry can get a person past the description which I’m sure you’ve heard. When you are accelerating in a rocket - you accelerate through space and feel weight as a result. When you stand on a planet - space moves (because of curved time) through you and you feel weight because of it. So, the equivalence principle seems to describe mass curving spacetime as the mechanism for gravity. The math that is built on that principle (general relativity) works well. It follows from Einstein's equivalence principle, that the degree (or gradient) of gravitational field curvature is directly proportional to the inertial mass of a massive body (or gas cloud), 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. Each gravitational well potential is gauged not solely on mass, but on the degree of displacement from a linear metric that has a unique relative value for all times and all local minima of combined fields.. It is deduced that the acceleration of a freely falling object can be interpreted as having nothing to do with its mass. Understandably, gravity is not an attractive force between objects (it is not a real force at all) but a curved spacetime phenomenon where space between objects (as well as other key locations in the combined fields of massive bodies) is less curved (relative to the standard zero condition there is less deviation from linearity) than in other parts of the field; since the gradient tends to, and attains, the local zero potential along the M1-M2 line. Furthermore, space between objects is curved hyperbolically (viz L1 Lagrange-Euler saddle points) as opposed to quasi-spherically in the surrounding vicinity of massive bodies. These geometrically opposed fields act seemingly against one another. One is the result of tension and stress, the other due to a cancellation of the interacting gravitational fields. Accordingly, let’s be very clear: An apple is not pulled towards the earth, but neither is it pushed from outer space. The latter (where space is seen as a repulsive force) is just as viable observationally (i.e., we are equally entitled to view the situation as such) as the former to describe gravity. But conceptually it fails in the same way as considering gravity an attractive force generated by mass. An apple experiences no force at all during its free-fall. It is thus neither lured from below nor shoved from above. So, why on earth (pun intended) do objects plummet in a gravitational field? Implicitly, general relativity describes this phenomenon correctly (without providing a mechanism): the apple follows the path of least action. It follows the geometrical shape of the curved field, falling ‘down’ the slope or gradient of the spacetime manifold. What follows is that removed from the concept of Newtonian gravitation is (1) the notion of intrinsic ‘force’ that causes objects to accelerate towards an origin O, located at the center of massive bodies. (2) Gravitationally bounded systems in long-term quasi-stable configurations are not maintaining orbits due to the mutual attraction of massive bodies counter-balanced by a fictitious finely tuned centrifugal ‘force.’ It can also be deduced that gravitational curvature is not the 'stretching' of spacetime, but actually a 'compression' of spacetime, since any curvature in the surface results in greater area (not less, as would be the case if spacetime were stretched). This is an important distinction to be made (and to be debated further if need be), since it would appear to imply that the origin of the tension or stress associated with gravity, and more importantly, responsible for curvature (gravity), resides not at the core of massive objects but in the vacuum itself, i.e., the mechanism of gravity bares the hallmark thus of a stress on the spacetime manifold itself, it is directly related to a property of the vacuum itself, as a structural quality of the spacetime manifold. For this reason I see it as essential the relation of local minima (Lagrange points in general) in the grand scope of GR. An analogy, albeit limited, would be the tension associated with the surface of water (surface tension), as mentioned above in this thread. The indentation and stress surrounding the tip of an insects appendage when gliding upon the surface of a pond is meaningless without regard of the original flat surface from which the deviation from linearity and stress are derived. The weight of the insect determines the depth on the indentation, but it is the surface tension, present wether the insect is there or not, that carries the weight. Surely that original surface is more than a simple starting point from which a measurement can be made. Certainly, it is more than a mathematical artifact, for sure it is more than nothing. To remove this fundamental ground-state from the picture, would be to ignore the foundation within which all of physics must be based, the stage or foundation within which all interactions, events, happening and phenomenon transpire. In short - If we are looking for a reason mass curves spacetime then I don’t see how we’re getting there by looking at areas in the gravitational field that are equal in gravity and inertial acceleration. But, that certainly doesn't mean it can't be done. I'm just giving you my perspective on what you're saying as best I understand it. We're not looking for the reason mass curves spacetime, we're looking for the reason spacetime is curved in the presence of mass, rather than remaining flat (Euclidean). This is no play on words. In the former it is implicit that 'something' inherent in mass generates gravity. In the latter it is left open other possibilities. In another way, we need to know the exact properties of spacetime both in the presences of mass-energy and in its absence. We at least agree the two spots are not the same which means L1 cannot and must not have flat spacetime. Which, by the way, you still haven't acknowledged. I would hope we both agree that there is a spot between M1 and M2 where the gravitational fields of each body cancel (the gradient is flat), i.e., that there exists a point, a saddle point (surrounded by a field, the geometric structure of which is hyperbolic) located between M1 and M2 where the gravitational field gradient is equal to zero. There are many references that state this location to be at L1, perhaps for simplicity or lack of rigor. When we examining this point where the effective potential drops to zero, we find a smooth surface, not a sharp peak. This means that from L1 to the surface of M1 or M2 there is a gradual transition in the gradient that tends from a local minimum value of zero to whatever it might be on the surface of a celestial object. The fact emerges: the gravitational field curvature gradient is different along the M1-L1-M2 line than the potential at, say, an angle 90 degrees. It follows that the field surrounding M1 (and M2) is not spherically symmetrical. So regardless of the location where an object would remain at rest du to the fictitious centrifugal force, there is an interaction taking place amid the combined fields that is entirely geometric in origin and structure, that impels us to reassess the age-old notion that e.g., the gravitational force of the Moon pulls on the surface of the Earth causing the tides to rise, and that on the other side of the Earth centrifugal force is responsible. The geometric argument reduces the two fictitious effects to one real effect: spacetime curvature. This will appear nonintuitive for the newcomer, who may argue that gravity has to be greater in the direction of the Moon, not less, otherwise how would the tides rise when the Moon is overhead? The answer is simple as it is elegant: If indeed the structure of the field is responsible for the rise and fall of tides here on earth then this tells us something more about our beloved G-spot (or lack of G-spot). It tells us that gravity is less intense along the M1-L1-M2-L2 line than it is in all other directions. Less, not more intense. (Less is best) The binding mechanism associated with the gravitational interaction geometric in origin, and it is directly associated with a weakness in the field, or at the very least, a weakness in the potential between bodies. Thanks to the cancelation of the field between bodies and the resulting field structure in the immediate (and not so immediate) vicinity. Obviously that's not all. There is something else to be derived from this geometrical conjecture. To be continued... CC Quote Link to comment Share on other sites More sharing options...
Little Bang Posted June 7, 2008 Report Share Posted June 7, 2008 It seems rather strange that no theory has ever explained mass, charge, gravity, and inertia. I don't have a theory that explains everything but I can suggest a relationship between every one of the four. Gravity is the result of time running slower as a particle falls, inertia is the result of time running faster as a particle is accelerated, mass is related to time(frequency) via f = MC^2/h, and I'd bet my bottom dollar that charge is related to frequency as a result of a waveform or waveforms being turned back in on themselves forming a spinning woveform, something like a dog chasing it's tail. That's my story and I'm sticking to it. Albert spent the last forty years of his life search for a connection between EMR and gravity. He had it all along and I don't understand why he didn't see it. TIME Quote Link to comment Share on other sites More sharing options...
modest Posted June 8, 2008 Report Share Posted June 8, 2008 We at least agree the two spots are not the same which means L1 cannot and must not have flat spacetime. Which, by the by, you still haven't acknowledged. We at least agree the two spots are not the same which means L1 cannot and must not have flat spacetime. Which, by the way, you still haven't acknowledged. While I appreciate you correcting my English, what I was actually going for was: "by the bye" which does sound a little ridiculous when you say it three times really fast. But, I would submit that "any road" (also an acceptable substitute for apropos) is even more ridiculous. ;) -modest Quote Link to comment Share on other sites More sharing options...
coldcreation Posted June 9, 2008 Author Report Share Posted June 9, 2008 While I appreciate you correcting my English, what I was actually going for was: "by the bye" which does sound a little ridiculous when you say it three times really fast. But, I would submit that "any road" (also an acceptable substitute for apropos) is even more ridiculous. :) -modest Sorry, my bad.:confused: I'll be back. Quote Link to comment Share on other sites More sharing options...
coldcreation Posted June 9, 2008 Author Report Share Posted June 9, 2008 Hey, Cold -- "I'm still getting my ducks in a row too. It's not easy." Yes, I agree ... but I have been waiting a year and a half! :-) I am glad to read your recent postings. You are finally revealing more details on your broad cosmological model (and by "broad", I include the nuclear/subatomic ramifications), in which I believe you have significant insights. But I am dying to read at least a sketch or an outline of your overall picture! Let's get on with it! Naysayers be damned! :-p Hello Tierra, welcome back. It's been a long time, for sure. Thanks for your post. Certainly some of the concepts discussed here may be applied to a broad cosmological picture. Particularly since gravity is likely responsible for the existence of the beautiful agglomerations of material, with their wispy tendrils and compact entrails, observed throughout the cosmos. Particularly too since all of these simple and complex, hot or cold, compact or spars, active or passive, stable or unstable, large and small gravitating structures are embedded within the vacuum continuum of space. It is possible that once the mechanism responsible for the gravitational interaction has been identified cosmological implications may surface, along with the nuclear/subatomic ramifications, if any. Let's wait and see. I love the idea that the Lagrange points have much overlooked significance... More generally than Lagrange points I think the properties of 'empty space' (or field-free space, local minima) relative to (and in relation with, via the field) that which occupies space, that have been overlooked to some extent, or at the very least, have been looked at ambiguously at times (particularly the concept of curved spacetime). It is this relation that can be perceived in studying the attributes of maxima and minima (along with saddle point regions), and visa versa. To the current dispute ... It seems to me that...understanding...the equivalency principle [is significant] to what you are trying to emphasize. Einstein said in GR -- no difference. No difference. No difference. ... Sorry for quote-mining. I have attempted to respond to this in a reply-post to modest (see above). There is a big difference between the concept of spacetime curvature and the concept of gravity as an attractive force balance by centrifugal 'force.' My qualm -- with respect to a two-body problem (say, Sun-Earth), are you ignoring the solar system's galactic centripetal acceleration (let's assume an ultra-simple model in which the solar system is orbiting a central galactic mass)? So is not L1 merely an approximate saddle point, hence, L1 is not flat with regard to L1's galactic orbit? And so on ... Milky Way's orbit within the local cluster, etc? The fundamental nature of the Sun-Earth saddle point is not modified by the central galactic mass, aside from slight perturbations in its structure, any more than by the presence of the Moon. One thing is for sure, there are saddle point areas between the solar system and other nearby stellar systems, between the spiral arm and the inner galactic region, between the Galaxy and its companions, between the Local Cluster and the core of the Local Supercluster, between the latter and other distant clusters, and so on. However, as one changes to larger and larger frames of reference, does not the Lagrangian points with respect to the objects involved get "flatter" in some "absolute" sense, in the sense of an absolute frame of reference (to the degree that one can employ that concept in a post-Einsteinian cosmology)? And could that be what you are hinting at? For example, in a simple "cluster" of galaxies in which a small galaxy is rotating about a much more massive galaxy, can't the Lagrangian points be described as being "flatter" than the Sun-Earth Lagrangian points? ... Surely, as the scale under consideration increases and the density decreases, the gravitational field gradient becomes smoother, flatter. There however some very massive objects (and groups of) that likely have steep gradient fields. Note too that even though the field around a saddle point is less intense, the trajectory of a test particle will be in accord with the structure of the saddle point area (large or small, steeply inclined or not). As far as "flatter" in some "absolute" sense: Flat is flat. There is no flatter. For this reason the distinction between local minima and global minima, relative minima and absolute minima is blurred. So there is no "absolute frame of reference." The closest we can get to an absolute frame of reference is the local (or relative) minima of any given system. Those minima or surfaces would be equivalent to a flat Minkowki spacetime (at least at one point in the field). Topologically, we have a situation that would appear as a series of troughs, peaks, valleys and planes. Sea level is unattainable, even in an empty universe: There remain residual fluctuations of the lowest possible energy state, the ground energy state) in accord with the laws of nature, i.e., the physical laws. It could be argued that the absolute zero value of gravitational potential (at infinity) represents by definition an ultimate limit inherent in nature (an absolute minimum value of spacetime curvature) precisely because it is unattainable. It could be argued too that local (or relative) zero value of gravitational potential (located at certain Lagrange points). Before moving on to a "post-Einsteinian cosmology" (if at all we will: I'm not sure what that means) there are several points to retain: One-Body (M1): The field of a single object is trivial (but interesting in that it is the simplest case of a gravitational field). If this (say a perfectly spherical star) were the only object in the universe (the Lone Star State :confused:) the field is spherically symmetrical, the gradient of curvature (the gravitational potential) is everywhere the same on its surface. The gradient of curvature falls off inversely and proportionally with the square of the distance away from the objects, in all directions. Two-Body System (M1, M2): The fields (or gravitational wells) of these each of these objects is no longer spherically symmetrical. The gradient of curvature (the gravitational potential) is NOT everywhere the same on the surface of M1 or M2. The inverse square Law would seem not no hold along the M1-L1-M2 line, since the local gradient falls of to zero near L1. The geometric structure of the combined fields between M1 and M2 is hyperbolic, i.e., there is a saddle shaped structure (in reduced dimension) of the field in the vicinity of L1, or H (see H above). In addition, there are other locations in the combined fields of a co-rotating system where material can coalesce and remain there without accreting onto M1 or M2. Thus there is a potential for the formation of a stable N-body system inherent in the combined field of a rotating two-body system. The pattern created when these areas are occupied is typical of Lagrange dynamics. Upon careful examination, this pattern can be seen in a wide variety of locations inside and outside of the solar system. Three-Body Systems and N-Body Systems: Upon careful examination, too, alterations of this pattern can be seen in a wide variety of locations outside of the Galaxy. A two-body system can very easily become a three- or four-body system. These alterations of the original patterns occur for a variety of reasons: depending on material flow, material availability, physical scale of the system and especially because once enough mass collects in particular zones (say at L4 and/or L5) the gravitational well of the new body (or group of objects) interacts with the field of M1 and M2, creation new L1 points and new L4 and L5 zones, where more material can collect, and so on increasing in complexity. Examples of some of these systems are reproduced above. They range in location from local stellar clusters (e.g., Pleiades), galactic nuclei, active galactic nuclei, barred galaxies, barred spiral galaxies, associated objects (e.g., quasars) thought to be gravitationally lensed (e.g., Einstein Cross), to galaxy clusters and galactic superclusters. I think that I see bits and pieces of the whole picture. I am dying to read an explanation from you for how your model explains the discordant redshifts. ... Note Tierra: These patterns (seen reproduced throughout this thread) can be observed in systems that are known to be associated, as well as in systems that display discordant redshifts (I have intentionally avoided these: the Einstein Cross being the exception). The ubiquity of such Lagrangian systems suggests that the physical mechanism involved must be directly related (inextricably attached) to the concept of gravity (spacetime curvature) as described by general relativity (thus cosmology); and so must also be attached to the fundamental laws of nature (the physical laws). CC Quote Link to comment Share on other sites More sharing options...
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