Erasmus00 Posted April 1, 2008 Report Share Posted April 1, 2008 This is an illustration, in reduced dimensions, of the relation between M1 - L1 - M2. http://hypography.com/forums/attachment.php?attachmentid=2168&d=1206995855 The dotted line, which represents a flat Euclidean plane (i.e., a space devoid of matter). Making this work requires modifying general relativity pretty severely. -Will Quote Link to comment Share on other sites More sharing options...
coldcreation Posted April 1, 2008 Report Share Posted April 1, 2008 This is an illustration, in reduced dimensions, of the relation between M1 - L1 - M2. http://hypography.com/forums/attachment.php?attachmentid=2168&d=1206995855 The dotted line, represents a flat Euclidean plane (i.e., a space devoid of matter). Making this work requires modifying general relativity pretty severely. -Will In what way do you think GR would be modified "severely." I was under the impression that this modification would be very slight, i.e., it would not affect the results of observational tests that have shown GR to be valid. This modification would be no greater than having a value of lambda set at zero, or say, fixing a limit as to the maximum density (and thus curvature) of a black hole and its surroundings - based on observational data of course (as opposd to arbitrarily). It would, however, change substantially the interpretation of a large quantity of astronomical phenomena (including celestial mechanics), beginning right here in the solar system and extending to scales of galaxies and galaxy clusters, since these L1 points would be operational between all gravitating objects and systems not matter how large. Indeed, earlier renderings of the field had peaks scattered about with differing values just as the gravity wells themselves, each with its own value, where a perfect balance between centrifugal force and the attraction of gravity were finely tuned by some initial condition or highly selective random process. At the very least, it would have the effect of reducing the major discrepancies between the current picture of gravity (the so-called odd-ball force of nature) together with the cosmological constant (another conundrum of modern physics), in that it leads to a description that describes the underlying dynamics of the 4-dimensional spacetime manifold both in the absence of gravity (curvature) and in the presence of massive bodies where curvature is generated. It also opens the door to a full fledged description of the physical mechanism associated with the gravitational interaction. Perhaps a modification of this genre is what Einstein had been searching for in his attempts to unify GR with QM, gravity with the other forces of nature. Perhaps not. In light of the simplicity of this concept it is easy to see how it could be overlooked. The concept is worth exploring, since it does not violate either GR, Lagrange dynamics, or Newtonian mechanics, either locally or globally (not to mention cosmological consideration). It is especially worth exploring since the means with which to test the concept are inexpensive and readily available. Perhaps, too, with this kind of modification, we will finally be able to dump the bunk from physics (e.g., DE, CDM). CC Quote Link to comment Share on other sites More sharing options...
HydrogenBond Posted April 1, 2008 Report Share Posted April 1, 2008 One thing that is not often mentioned in the dynamic equilibrium is the affect of the energy output from the other three forces of nature. For example, if there was no energy output stemming from the sun, gravity would make it smaller. It may collapse to a piece of neutron density. The heat fluffs it up, so the space-time well is more spread out. Wouldn't a cold tiny piece of neutron density change the orbits of the planets? And isn't the heat and the gravity fluff helping to define the total affect needed for the current orbits? Could an orbital anomaly, be caused by simply tweaking the amount of gravity fluff, such as a little sputter in the fusion engine. Quote Link to comment Share on other sites More sharing options...
coldcreation Posted April 1, 2008 Report Share Posted April 1, 2008 You will not address my diagrams or put a third mass in your own diagram. Your diagram A is intrinsically unstable. You cannot explain the observed equilibrium of gravitating systems without resorting to a finely-tuned centrifugal force to counter gravity. On scales compatible with galaxies you will require some form of CDM in addition to centrifugal force. And on larger scales you will require DE. I've already commented on diagram B. 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. I will attach a 3-body solution shortly. Your diagram B is a 2-body illustration. I don't see how a 3-body system would look like your 2-body system. 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. The potential gradient expressed in this diagram is the gravitational force (spacetime curvature). The Lagrangian point L1 is a "critical point" of this potential. The force is zero. And so, it follows, that the gravitational potential is null at the L1 zero-gradient point. A test particle place at L1 has zero potential. Acceleration and relative velocity both equal zero at L1. The value (degree or quantity) zero of gravitational field curvature at L1 is a generalized potential extrema. Equations 13, 14, 15 may or may not have the answer you are looking for. This link too may help[] though they use parameter mu (a curve fit) in order to show how to find L1, L2 and L3, which is not our intension. A more promising approach than to solve U=GM/r is to consider the problem physically, by using the symmetries of a simple two body system (where M1 and M2 are of equal mass) as a guide to the solution. The following sequence represent the maxima (M1 and M2) and minima (L1 = 0) of the gravitational field potential. At L1 the force (or curvature) vanishes. M1-5--4---3----2-----1------0------1-----2----3---4--5-M2 Note: the L1 point would exist even when M1 and M2 are at rest, i.e., it is not a velocity dependent potential. Further more, the curvature of the effective potential in reduced dimensions is a saddle shape (not present in the linear representation above). Now, look at the line of force: --->-->->M1<-<--<----<------0------>---->-->->M2<-<--<--- Here, interestingly, L1 could be interpreted as repulsive, i.e., as if it counters gravity. If a particle place at L1 had gravitational potential (as you claim modest: see your illustration A), the slightest motion away from a perfectly circular relative orbits (i.e., when M1 and M2 move closer together or further apart as in elliptical orbits) the potential of the test particle would vary and thus tend to accelerate due to the force of gravity: spacetime at L1 would not be flat (there would be gravity there, the force of which would oscillate). The gravitational potential itself would be dynamically unstable. Small departures from equilibrium would grow exponentially. You frankly cannot explain how or why this would work. You're starting to sound like Red Butler. :rolleyes: I will be back to explain how and why this concept works and how it explains the stability associated with gravitating systems and sub-systems without fine-tuning or selective initial conditions. CC Quote Link to comment Share on other sites More sharing options...
coldcreation Posted April 2, 2008 Report Share Posted April 2, 2008 I would like to point out, just so it is clear, that the potential gradient expressed in this diagram is the gravitational force (spacetime curvature) cross-section at the line connecting the centers of each massive body M1 and M2 (and of course L1). Off of this cross-section plane the field would tend to resemble diagram A by modest. I assume that was clear but I thought I would reiterate just in case. Those who have been predisposed to the conventional presentation and its interpretation (see diagram A by modest) may be dismayed when they read that the cosmological constant is operational in the local environment of the solar system (indeed at L1, L2 and L3), that the mechanism involved in the equilibrium process observed on local scales is also responsible for stability on scales compatible with superclusters, and that empty space itself is not an acquiescent, malleable, flexible nothingness, but an unyielding physical state in which the fundamental properties can be expressed in terms of minimum or maximum values and magnitudes (where the minimum curvature is at L-point and the maxima at the surface of massive objects). Apprehension will continue unless it is recognized that the new viewpoint entails an academic reorientation and a slight but challenging reassessment of the grounds, concepts and interpretations of gravity and ?, founded on the fundamental laws of physics and based on empirical observation. I argue that there is a mechanism operational in space responsible for generating equilibrium, independent of centrifugal force. Here, we pull together responses to several images in terms of quasi-equilibrium, gravitational stability, zero-velocity surfaces in configuration space, collisionless self-gravitating systems, zero-point potentials and hyperstable solutions at specific points of an orbit from a relaxed physical point of view. In this discussion other possibilities should also be considered: the corroboration and definition of the exact mechanism responsible for the gravitation interaction as related to the cosmological constant (which by definition is related to empty space somehow in opposition to gravity), and the morphogenesis of system evolution corresponding to the minimum of potential wells (located primarily at L1, L2, L3). These are supported by geometrical arguments along with basic concepts and principles in novel logical sequence that reduces the traditional problems of partial definitions and vicious circles. The principles and results targeted in this thread are primarily developed for macroscopic systems and equilibrium states. Stability in microscopic systems and nonequilibrium states could be discussed more extensively as well, but for simplification purposes, we should stick to macroscopic systems. What will emerge is that we live in a highly stable, non-expanding dynamic universe, one in which evolution is vastly different than generally assumed. What I hope to illuminate are some physical aspects of the vacuum that are fundamentally important for understanding what is to follow: how massive bodies and their intrinsic gravitational fields interact in space and how the geometric structure of space itself (defined by the cosmological constant and its role at Lagrange points) plays a leading role in the equilibrium process. The scene will be finally finally set, once the main areas of confrontation are established, for a complete discussion of the stances behind Coldcreation and his interpretation of the mechanism involved in the stability process of quasi-equilibrium self-gravitating systems and the way they interrelate. Our modus operandi should be to correlate the concepts and empirical evidence that form the basis of the natural laws supported by observational and experimental evidence. Something has only just begun CC Quote Link to comment Share on other sites More sharing options...
modest Posted April 2, 2008 Author Report Share Posted April 2, 2008 Your diagram A is intrinsically unstable.Yes, it is unstable. The Lagrange point L1 is unstable which is why it takes a lot of fuel for spacecraft to remain there. From: The Lagrange Points – Montana PhysicsOf the five Lagrange points, three are unstable and two are stable. The unstable Lagrange points - labelled L1, L2 and L3 - lie along the line connecting the two large masses….The L1 point of the Earth-Sun system affords an uninterrupted view of the sun and is currently home to the Solar and Heliospheric Observatory Satellite SOHO. The L2 point of the Earth-Sun system will soon be home to the MAP Satellite and (perhaps) the Next Generation Space Telescope. The L1 and L2 points are unstable on a time scale of approximately 23 days, which requiress satellites parked at these positions to undergo regular course and attitude corrections.I've already commented on diagram B.Your comments were that my diagram A was wrong and B was not your description. To alleviate your perception of my incapability of drawing a diagram of potential gravitational energy, I’ve solved and plotted them for you review: The equation [imath]-GM/r[/imath] and [imath]-3GM/2R[/imath] (for center of mass) is solved with answers following. The complete solution is in the attached excel document. The units of distance are based on au while potential is based on km. distance from sun(au) Potential U --------------------- ----------- 0.9866 -899.345 0.9867 -899.285 0.9868 -899.225 0.9868 -899.165 0.9869 -899.106 0.9870 -899.046 0.9870 -898.986 0.9871 -898.926 0.9872 -898.866 0.9872 -898.807 0.9873 -898.747 0.9874 -898.687 0.9874 -898.627 0.9875 -898.568 0.9876 -898.508 0.9876 -898.448 0.9877 -898.389 0.9878 -898.329 0.9878 -898.270 0.9879 -898.210 0.9880 -898.150 0.9880 -898.091 0.9881 -898.031 0.9882 -897.972 0.9882 -897.913 0.9883 -897.853 0.9884 -897.794 0.9884 -897.734 0.9885 -897.675 0.9886 -897.616 0.9886 -897.556 0.9887 -897.497 0.9888 -897.438 0.9888 -897.379 0.9889 -897.319 0.9890 -897.260 0.9890 -897.201 0.9891 -897.142 0.9892 -897.083 0.9892 -897.024 0.9893 -896.965 0.9894 -896.906 0.9894 -896.847 0.9895 -896.788 0.9896 -896.729 0.9896 -896.670 0.9897 -896.611 0.9898 -896.552 0.9898 -896.493 0.9899 -896.435 0.9900 -896.376 0.9900 -896.317 0.9901 -896.258 0.9902 -896.200 0.9902 -896.141 0.9903 -896.083 0.9904 -896.024 0.9904 -895.966 0.9905 -895.907 0.9906 -895.849 0.9906 -895.790 0.9907 -895.732 0.9908 -895.673 0.9908 -895.615 0.9909 -895.557 0.9910 -895.499 0.9910 -895.440 0.9911 -895.382 0.9912 -895.324 0.9912 -895.266 0.9913 -895.208 0.9914 -895.150 0.9914 -895.092 0.9915 -895.034 0.9916 -894.977 0.9916 -894.919 0.9917 -894.861 0.9918 -894.803 0.9918 -894.746 0.9919 -894.688 0.9920 -894.631 0.9920 -894.573 0.9921 -894.516 0.9922 -894.458 0.9922 -894.401 0.9923 -894.344 0.9924 -894.287 0.9924 -894.230 0.9925 -894.173 0.9926 -894.116 0.9926 -894.059 0.9927 -894.002 0.9928 -893.945 0.9928 -893.888 0.9929 -893.832 0.9930 -893.775 0.9930 -893.719 0.9931 -893.662 0.9932 -893.606 0.9932 -893.550 0.9933 -893.494 0.9934 -893.437 0.9934 -893.381 0.9935 -893.326 0.9936 -893.270 0.9936 -893.214 0.9937 -893.158 0.9938 -893.103 0.9939 -893.048 0.9939 -892.992 0.9940 -892.937 0.9941 -892.882 0.9941 -892.827 0.9942 -892.772 0.9943 -892.718 0.9943 -892.663 0.9944 -892.609 0.9945 -892.555 0.9945 -892.500 0.9946 -892.447 0.9947 -892.393 0.9947 -892.339 0.9948 -892.286 0.9949 -892.232 0.9949 -892.179 0.9950 -892.126 0.9951 -892.074 0.9951 -892.021 0.9952 -891.969 0.9953 -891.917 0.9953 -891.865 0.9954 -891.813 0.9955 -891.762 0.9955 -891.711 0.9956 -891.660 0.9957 -891.610 0.9957 -891.559 0.9958 -891.509 0.9959 -891.460 0.9959 -891.411 0.9960 -891.362 0.9961 -891.313 0.9961 -891.265 0.9962 -891.217 0.9963 -891.170 0.9963 -891.123 0.9964 -891.077 0.9965 -891.031 0.9965 -890.986 0.9966 -890.941 0.9967 -890.897 0.9967 -890.854 0.9968 -890.811 0.9969 -890.769 0.9969 -890.728 0.9970 -890.687 0.9971 -890.648 0.9971 -890.609 0.9972 -890.572 0.9973 -890.535 0.9973 -890.500 0.9974 -890.466 0.9975 -890.433 0.9975 -890.402 0.9976 -890.372 0.9977 -890.344 0.9977 -890.318 0.9978 -890.294 0.9979 -890.272 0.9979 -890.253 0.9980 -890.236 0.9981 -890.222 0.9981 -890.212 0.9982 -890.205 0.9983 -890.203 0.9983 -890.204 0.9984 -890.211 0.9985 -890.224 0.9985 -890.243 0.9986 -890.270 0.9987 -890.306 0.9987 -890.351 0.9988 -890.408 0.9989 -890.479 0.9989 -890.566 0.9990 -890.673 0.9991 -890.803 0.9991 -890.963 0.9992 -891.159 0.9993 -891.401 0.9993 -891.704 0.9994 -892.088 0.9995 -892.582 0.9995 -893.235 0.9996 -894.124 0.9997 -895.394 0.9997 -897.327 0.9998 -900.590 0.9999 -907.174 0.9999 -927.044 1.0000 -980.869 1.0001 -926.926 1.0001 -906.936 1.0002 -900.234 1.0003 -896.853 1.0003 -894.801 1.0004 -893.413 1.0005 -892.404 1.0005 -891.633 1.0006 -891.020 1.0007 -890.518 1.0007 -890.097 1.0008 -889.736 1.0009 -889.421 1.0009 -889.143 1.0010 -888.894 1.0011 -888.668 1.0011 -888.463 1.0012 -888.273 1.0013 -888.098 1.0013 -887.934 1.0014 -887.779 1.0015 -887.634 1.0015 -887.496 1.0016 -887.365 1.0017 -887.239 1.0017 -887.119 1.0018 -887.003 1.0019 -886.891 1.0019 -886.783 1.0020 -886.678 1.0021 -886.576 1.0021 -886.477 1.0022 -886.380 1.0023 -886.286 1.0023 -886.193 1.0024 -886.102 1.0025 -886.014 1.0025 -885.926 1.0026 -885.840 1.0027 -885.756 1.0027 -885.672 1.0028 -885.590 1.0029 -885.509 1.0029 -885.429 1.0030 -885.350 1.0031 -885.272 1.0031 -885.195 1.0032 -885.118 1.0033 -885.042 1.0033 -884.967 1.0034 -884.892 1.0035 -884.819 1.0035 -884.745 1.0036 -884.672 1.0037 -884.600 1.0037 -884.528 1.0038 -884.457 1.0039 -884.386 1.0039 -884.316 1.0040 -884.245 1.0041 -884.176 1.0041 -884.106 1.0042 -884.037 1.0043 -883.969 1.0043 -883.900 1.0044 -883.832 1.0045 -883.764 1.0045 -883.697 1.0046 -883.630 1.0047 -883.563 1.0047 -883.496 1.0048 -883.429 1.0049 -883.363 1.0049 -883.297 1.0050 -883.231 1.0051 -883.165 1.0051 -883.100 1.0052 -883.034 1.0053 -882.969 1.0053 -882.904 1.0054 -882.840 1.0055 -882.775 1.0055 -882.710 1.0056 -882.646 1.0057 -882.582 1.0057 -882.518 1.0058 -882.454 1.0059 -882.390 1.0059 -882.326 1.0060 -882.263 1.0061 -882.199 1.0061 -882.136 1.0062 -882.073 1.0063 -882.010 1.0064 -881.946 1.0064 -881.884 1.0065 -881.821 1.0066 -881.758 1.0066 -881.695 1.0067 -881.633 1.0068 -881.570 1.0068 -881.508 1.0069 -881.446 1.0070 -881.384 1.0070 -881.321 1.0071 -881.259 1.0072 -881.197 1.0072 -881.135 1.0073 -881.074 1.0074 -881.012 1.0074 -880.950 1.0075 -880.889 1.0076 -880.827 1.0076 -880.765 1.0077 -880.704 1.0078 -880.643 1.0078 -880.581 1.0079 -880.520 1.0080 -880.459 1.0080 -880.398 1.0081 -880.337 1.0082 -880.275 1.0082 -880.214 1.0083 -880.154 1.0084 -880.093 1.0084 -880.032 1.0085 -879.971 1.0086 -879.910 1.0086 -879.850 1.0087 -879.789 1.0088 -879.728 1.0088 -879.668 1.0089 -879.607 1.0090 -879.547 1.0090 -879.486 1.0091 -879.426 1.0092 -879.365 1.0092 -879.305 1.0093 -879.245 1.0094 -879.185 1.0094 -879.124 1.0095 -879.064 1.0096 -879.004 1.0096 -878.944 1.0097 -878.884 1.0098 -878.824 1.0098 -878.764 1.0099 -878.704 1.0100 -878.644 1.0100 -878.584 1.0101 -878.524 1.0102 -878.464 1.0102 -878.404 1.0103 -878.345 1.0104 -878.285 1.0104 -878.225 1.0105 -878.165 1.0106 -878.106 1.0106 -878.046 1.0107 -877.986 1.0108 -877.927 1.0108 -877.867 1.0109 -877.808 1.0110 -877.748 1.0110 -877.689 1.0111 -877.629 1.0112 -877.570 1.0112 -877.510 1.0113 -877.451 1.0114 -877.392 1.0114 -877.332 1.0115 -877.273 1.0116 -877.214 1.0116 -877.154 1.0117 -877.095 1.0118 -877.036 1.0118 -876.977 1.0119 -876.917 1.0120 -876.858 1.0120 -876.799 1.0121 -876.740 1.0122 -876.681 1.0122 -876.622 1.0123 -876.563 1.0124 -876.504 1.0124 -876.445 1.0125 -876.386 1.0126 -876.327 1.0126 -876.268 1.0127 -876.209 1.0128 -876.150 1.0128 -876.091 1.0129 -876.032 1.0130 -875.973 1.0130 -875.915 1.0131 -875.856 1.0132 -875.797 1.0132 -875.738 1.0133 -875.679 1.0134 -875.621 This covers 4 million kilometers with the earth at the middle. A plot of this range was my intention when drawing A and the actual plot is as follows:In this graph, the sun would be far off to the left. The earth is about the size of a pixel in the center of the dip. Distance on the bottom is au from the sun. Notice the actual value of potential U on the left side of the graph. If the potential is zero at infinity and relative to the sun / earth only - then the value around earth is in the range of -800 to -1000. If we were to consider the Milky Way, the value would be lower (or more negative) This eliminates the possibility of zero potential at L1. We can, in fact put zero potential at L1 and see what it looks like: If we zoom out the problem is even more clear. Here is the appropriate plot of potential from .1 au to 1.7 au (from the sun): and this is what happens when we set the value of L1 to zero potential: The L1 point moves up to what you would call the ‘Euclidian plane’. You must see why this is wrong. It would require too much energy to reach L1. We have been to L1 so we know it doesn’t have this much difference in value to earth’s surface. It not only looks wrong (your statement when plotted) it is wrong by measurement. Now that this has been properly plotted, it is no longer possible for you to say: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 The problem is even more apparent when three masses are considered. I illustrated the case of 2 stars close to a black hole in a post above. You ignored the example. You would not be able to draw the example I drew in that post. It would end up with a very pronounced bump like ‘B’ has. Also, you seem to have a misunderstanding here:The potential gradient expressed in this diagram is the gravitational force (spacetime curvature). The Lagrangian point L1 is a "critical point" of this potential. The force is zero. And so, it follows, that the gravitational potential is null at the L1 zero-gradient point. A test particle place at L1 has zero potential. Acceleration and relative velocity both equal zero at L1. The value (degree or quantity) zero of gravitational field curvature at L1 is a generalized potential extrema. Equations 13, 14, 15 may or may not have the answer you are looking for. This link too may help[] though they use parameter mu (a curve fit) in order to show how to find L1, L2 and L3, which is not our intension.I have explained why and how potential is different than force. They are different in value and concept. I’ve explained how they are related - one being the derivative of the other. I've explained the importance and significance of that: It is possible for flat space time to have non-zero potential. The value of the slope is not equal to the numerical value at a point in the function. A horizontal line does not have to be at zero. //edit I should note to avoid confusion that a LaGrange point L1 doesn't have to have flat spacetime unless the two masses are equal. In the case of the Earth and Sun shown above the point L1 (at about 1 million miles) is not flat because the system is rotating. If Earth and the Sun were static - not rotating - held apart by a rod - then the point L1 would be closer to the earth and would have flat spacetime. I haven't looked this up but it seems correct. So, Potential U and force are not the same. You cannot equate the two as being the same. If U was zero anywhere the field were flat the consequences would be as I plotted above. How else can I explain this? I’ve show it mathematically. I’ve explained the problem verbally in many posts now. I’ve sourced it. I’ve drawn pics. I've plotted it. I think, perhaps, you just don’t believe me because you see me as an adversary. It is also noteworthy that what you've explained is not consistent with GR. If you determine that I'm incorrect here and hold to your idea then please try what I've asked: Put two masses in a gravity well (like a galaxy) and show the point L1 of the two masses. Using your pic as an example: -modest Quote Link to comment Share on other sites More sharing options...
modest Posted April 2, 2008 Author Report Share Posted April 2, 2008 I forgot to attach the excel document in my last post which I promised, so here it is. -modest Quote Link to comment Share on other sites More sharing options...
coldcreation Posted April 2, 2008 Report Share Posted April 2, 2008 Here is a reduced dimension 3-body system showing L1 points in relation to the straight dotted line (representing flat spacetime). This is a slice through the line conecting the center of three bodies which are aligned. (Other Lagrange points are omitted for simplicity). Note: the curved dotted line represents the curvature of M1 in the absence of, or perpendicular to, M2 and M3. Note too, that this diagram of the gravitation field potential is greatly exagerated, i.e., all lines, curved and straight (dotted or not) would be much closer together, consistent with a weak gravitational force. [Edited to add] Your illustrations above modest are far more exagerated, that is part of the reason why you get strange peaks at L1 when it reaches the Minkowski plane value. Without that gross distortion you would see that the difference between the 'top' of the 'hills' and the trough, where reside the massive bodies, and too, between the potential on and off the M1, L1, M2 line) is very small. In other words, the less exagerated the potential difference, the easier it becomes to appreciate de deviation from linearity, and the approach to linearity, with respect to the real world. As far as your black hole diagram modest: what you have illustrated is model that cannot be tested empirically (by observation of by experiment), i.e., it is thus speculative in origen. I prefer to discuss situations where we can test predictions locally rather than guess as to what may be (or may not be) the case under extreme conditions (that may or may not exist in nature). But to answer your question anyway, yes, L1 point near a BH and its companions would reach zero potential as well. However, the result would not look like your diagram B. The curve would be smooth and continuous, but the situation would surely be extreme. I will be back to remark on your previous post... CC Quote Link to comment Share on other sites More sharing options...
modest Posted April 2, 2008 Author Report Share Posted April 2, 2008 Here is a reduced dimension 3-body system showing L1 points in relation to the straight dotted line (representing flat spacetime). This is a slice through the line conecting the center of three bodies which are aligned. (Other Lagrange points are omitted for simplicity). Note: the curved dotted line represents the curvature of M1 in the absence of, or perpendicular to, M2 and M3. Thank you for doing that CC. I can now say: M2 is not in M1's gravity well. The force of gravity is the slope in your line. Do you see how M2 can't feel M1's gravity - it can't be in orbit around it. How does that work? Are you now saying an object doesn't have to be in another object's gravity well to be in orbit around it? As far as your black hole diagram modest: what you have illustrated is model that cannot be tested empirically It doesn't have to be a black hole. It can be any gravity well. It can be a galaxy. And it can be tested. It was tested with SOHO. Also, gravitational lenses don't have holes at all the Lagrange points. One galaxy cluster makes one lens. That one observation completely invalidates your theory. //editAlso, the slope of the line on the right side of M1 is now steeper than the left side (as your idea necessitates). This means the force of gravity felt would be stronger on the right side of M1 than the left. That's impossible.//edit Your idea has fallen apart CC. You are drawing three masses supposedly all interacting with each other and yet none are in any of the other's gravity well. I believe you are capable of seething the problem with that. -modest Quote Link to comment Share on other sites More sharing options...
coldcreation Posted April 2, 2008 Report Share Posted April 2, 2008 Thank you for doing that CC. I can now say: M2 is not in M1's gravity well. The force of gravity is the slope in your line. Do you see how M2 can't feel M1's gravity - it can't be in orbit around it. How does that work? Are you now saying an object doesn't have to be in another object's gravity well to be in orbit around it? You are beginning to understand something, though I am not sure to what extent. Each object has its own gravity well. At the same time, say, the earth and moon, etc., are locked into the general field of the sun (nothing new there). Look at the curved dotted line of my illustration. That is the potential you write about. It is still there, though it is modified along the line connecting L1 to the centers of M1 and M2. So orbit? Yes they do. It doesn't have to be a black hole. It can be any gravity well. It can be a galaxy. And it can be tested. It was tested with SOHO. Also, gravitational lenses don't have holes at all the Lagrange points. One galaxy cluster makes one lens. That one observation completely invalidates your theory. SOHO hasn't carried out the test needed to be done to resolve the question. It was not desiged to test the boundary condition of GR at L1, so it has not proved or disproved what we are discussing. //editAlso, the slope of the line on the right side of M1 is now steeper than the left side (as your idea necessitates). This means the force of gravity felt would be stronger on the right side of M1 than the left. That's impossible.//edit My next illustration will include the L2 and L3 points. You will see that your objection is unfounded. Your idea has fallen apart CC. You are drawing three masses supposedly all interacting with each other and yet none are in any of the other's gravity well. I believe you are capable of seething the problem with that. -modest Again, your objection is unfounded. All massive bodies are in their own well. And in addition, say, planets are in the global well of the Sun. Perhaps you should look at an illustration viewed from above showing the five L-points to see more or less the mainstream interpretation of the field surronding two massive bodies. My view is hardly any different than this. Again, the curved line you see (that makes the objects look seperated completely) is only a cross section of one line in a spherical 4D field. So far nothing has fallen apart. I will attempt to clarify the situation further, since I see it has not yet been understood entirely. CC Quote Link to comment Share on other sites More sharing options...
Erasmus00 Posted April 2, 2008 Report Share Posted April 2, 2008 In what way do you think GR would be modified "severely." There is no way that both your conjecture AND Einstein's field equations can be correct. Einstein's field equations set the local values of "gravitational potential"(at least where it is sensible to talk about such things-weak field limit). You can't arbitrarily adjust the potential without modifying the field equations. The modification required for every L1 type point to look like your graphs would be severe. This modification would be no greater than having a value of lambda set at zero, or say, fixing a limit as to the maximum density (and thus curvature) of a black hole and its surroundings - based on observational data of course (as opposd to arbitrarily). Adding a lambda is a trivial modification to Einstein's equations, the simplest extension. Fixing the maximum density of a black hole, however, would also require a large modification to the Einstein field equations. The prediction of total gravitational collapse is very robust- modifying the equations to remove this feature would be difficult without dramatically altering the theory. it leads to a description that describes the underlying dynamics of the 4-dimensional spacetime manifold both in the absence of gravity (curvature) and in the presence of massive bodies where curvature is generated. General relativity already does this. The concept is worth exploring, since it does not violate either GR, Lagrange dynamics (the least action principle), or Newtonian mechanics, either locally or globally (not to mention cosmological consideration). It certainly violates GR, as discussed (which of course violates Newtonian gravity). Now, as for the least action principle- without specifying an action, the principle is devoid of content. You can either specify the Newtonian gravitational action, for instance, or the General Relativistic action (or the electrodynamic action, etc). Perhaps your idea could have an action developed for it, but I don't see how because regardless of how you calculate potential you are arbitrarily setting certain points to 0. -Will Quote Link to comment Share on other sites More sharing options...
coldcreation Posted April 2, 2008 Report Share Posted April 2, 2008 There is no way that both your conjecture AND Einstein's field equations can be correct. Einstein's field equations set the local values of "gravitational potential"(at least where it is sensible to talk about such things-weak field limit). You can't arbitrarily adjust the potential without modifying the field equations. The modification required for every L1 type point to look like your graphs would be severe. One of the characteristic of potential energy is that the zero point can be set arbitrarily; just as the origin O of a coordinate system. From that origin all other values of potential energy will be measured relative to that zero value. This however, is not what I am doing. I use the zero of gravitational potential energy at infinity as a reference, where the gravitational force approaches zero. This zero value is absolute in that gravitational potential energy reaches (in principle) an ultimate limit, where gravity is null, equal to zero. I am not setting the zero value arbitrarily, nor am I setting it at an arbitrary location, but at the inner Lagrangian L1 libration point where it is well known that the gravitational fields of M1 and M2 cancel out to zero. What is not well known - and this is the concept that needs to be tested empirically - is that the gravitational potential energy of the entire system (2-body, 3-body, N-body, etc.) tends to zero and attains it at L1 points. In other words, there is a physical and geometric logic to the choice of the zero point of gravitational potential energy at L1. Unfortunately, the problem cannot be unambiguously understood or resolved by visualizing a cross section of a system in reduced-dimension, since all four spatiotemporal dimensions are involved. But again, there are ways of testing it empirically which will either falsify it concord. Adding a lambda is a trivial modification to Einstein's equations, the simplest extension. Fixing the maximum density of a black hole, however, would also require a large modification to the Einstein field equations. The prediction of total gravitational collapse is very robust- modifying the equations to remove this feature would be difficult without dramatically altering the theory. True, and however much Einstein disliked the concept of singularities he could not conveniently do away with them. However, what interests me is the other end of the spectrum: that of the least gravitational curvature (the zero of gravitational potential energy). Basically, what I am doing is making a trivial modification, a simple yet fundamental extension, to celestial mechanics (thus of the theory that describes this dynamics; GR). It consists of setting the total value of gravitational potential energy (curvature) to zero wherever two or more interacting fields (e.g., of M1 and M2) cancel out (e.g., at L1, L2 and L3). What this concept (or conjecture) states, very simply, is that there are locations in the combined gravitational fields of massive bodies (particularly at L1 points) where the zero of gravitational potential energy is attained. And it is attained naturally because of its position in the combined gravitational field where the gravitational fields of two (or more) interacting fields cancel to zero. Indeed, one of the results of this conjecture is that the intensity of the gravity field drops off more quickly along the M1 - L1 - M2 line than would be expected according to the inverse square law, or than is the case in any other direction away from M1 or M2. In another way, and non-intuitively, the force of gravity is weaker between massive objects than it is in other direction. There is thus a relaxation of the field along the M1 - L1 - M2 line. So, for example, a large body of water, say, the Indian Ocean, would tend to feel less gravitational force exerted upon it when the Moon or the Sun passes directly overhead, than otherwise. What changes is the geometric structure of the Earth's gravitational well, in this example. So the Moon's gravitational force is not reaching inside the Earths causing a pull on the ocean surface. The Earth's gravitational wells remains unviolated and intact. It is the shape of the field that is modified (by the presence of the Moon) and thus the gravitational potential energy in the direction of the Moon (and the Sun), causing the tides to rise. Note too that because the same phenomenon is operational on the opposite side of the Earth, with respect to L2, there is a high tide there as well, due to a modification in the geometric structure of the field, a relaxation, which modifies the inverse square law. So yes, this modification to the law of gravity is trivial since the observations are the same whether one considers gravity to be an attractive force or a curved spacetime phenomenon. Where the nontrivial difference manifests itself, between the two opposing views (the mainstream interpretation and this conjecture) is on scales compatible with galaxies and up from there, where rotational curves become (virtually flat) incompatible with the standard model. A new theory of gravity is not required. However, a modification (or extension) is. It sets a fixed lower bound (or upper bound, depending on whether you consider gravity positive or negative) on the intensity of the gravitational field at specific collinear points (which affect the geometric structure of each potential well in the system under consideration), equal to zero gravitational potential energy, and operational in all gravitating systems. Again, zero gravitational potential energy is defined here as having a value equal to the value of gravity at infinity. ...it leads to a description that describes the underlying dynamics of the 4-dimensional spacetime manifold both in the absence of gravity (curvature) and in the presence of massive bodies where curvature is generated.General relativity already does this. It is meant that the concept leads to a description of the underlying physical mechanism associated with the gravitational interaction: something general relativity does not do. The concept is worth exploring, since it does not violate either GR, Lagrange dynamics (the least action principle), or Newtonian mechanics, either locally or globally (not to mention cosmological consideration). It certainly violates GR, as discussed (which of course violates Newtonian gravity). As noted, it does not violate GR. It takes the concept of the cosmological constant (a property of empty space that acts in opposition to gravity) and applies it locally. Note here that lambda is a geometric property of spacetime (curvature-free, or gravity-free), not a real force since it is operational at a points that possess zero gravitational potential energy. So both lambda and gravity, in this light, are consistent with Einstein's general principle of relativity. However, a modification involving the geometrical structure of the field, the intensity, or degree of curvature, at specific locations in the combined fields of massive objects is required. If you would like, I can elaborate on the connection between lambda and Lagrange points, if it is not already obvious. ...regardless of how you calculate potential you are arbitrarily setting certain points to 0. As discussed, the zero gravitational potential energy set at L1 is far from arbitrary. CC Quote Link to comment Share on other sites More sharing options...
Erasmus00 Posted April 3, 2008 Report Share Posted April 3, 2008 In other words, there is a physical and geometric logic to the choice of the zero point of gravitational potential energy at L1. It still violates Einstein's equations. Using Einstein's equations I can calculate (if I pick a reference of 0 at infinity) the potential at L1. To set it to 0 is not a small modification, its a large and violent one. Basically, what I am doing is making a trivial modification, a simple yet fundamental extension, to celestial mechanics (thus of the theory that describes this dynamics; GR). Its far from trivial! It destroys the structure of the theory! You can't say "Einstein's equations are valid everywhere EXCEPT lagrange points." Either Einstein's equations are valid, or they are not. Since they don't get a potential of 0 at lagrange points (which you require) in your view we have to find new equations. Your theory gives no way to actually find the gravitational field anywhere! -Will Quote Link to comment Share on other sites More sharing options...
modest Posted April 3, 2008 Author Report Share Posted April 3, 2008 One thing that is not often mentioned in the dynamic equilibrium is the affect of the energy output from the other three forces of nature. For example, if there was no energy output stemming from the sun, gravity would make it smaller. I agree HB. That certainly is a kind of dynamic equilibrium. Without the pressure from nuclear fusion the sun would collapse - as it will one day.It may collapse to a piece of neutron density. The heat fluffs it up, so the space-time well is more spread out.When the sun collapses into a white dwarf, it will have electron degeneracy pressure. If your wondering what would happen without that pressure - would the density be comparable to a neutron star, I don't know. What do you think?Wouldn't a cold tiny piece of neutron density change the orbits of the planets? And isn't the heat and the gravity fluff helping to define the total affect needed for the current orbits? Could an orbital anomaly, be caused by simply tweaking the amount of gravity fluff, such as a little sputter in the fusion engine.I've heard on this subject that if the sun were replaced by a black hole of the same mass the planets wouldn't be the wiser. If the orbits were affected, I don't think it would be by much. Are you thinking differently? -modest Quote Link to comment Share on other sites More sharing options...
modest Posted April 3, 2008 Author Report Share Posted April 3, 2008 CC, Do you think non-rotating masses hold to your theory? Would two masses that are static and perhaps separated by a rod (so they don't collide), would their L1 point have zero potential or gravitational potential equal to a mass at infinity? If M1 and M2 in your original pic were not rotating about one another it would look similar or the same, yes? -modest Quote Link to comment Share on other sites More sharing options...
coldcreation Posted April 3, 2008 Report Share Posted April 3, 2008 What needs to be determined empirically is whether the value of gravitational potential energy at Lagrange points (L1 saddle-points) represent the local minima (of combined gravitational fields) or whether the value at those libration points represents the equivalent to the global minima: equal to the zero of gravitational potential energy at infinity. The two views, so far here presented, are that L1 has a local minima (the mainstream view) and (the Coldcreation view) that L1 has represents a global minima. So how do we test this empirically, in light of the fact that Einstein's equations tell us there is a local minima yet observations are ambiguous (in that there are two possible ways of explaining the same observations)? It turns out there is a very simple test that can be carried out which will determine empirically which interpretation is correct. In fact, there are several tests that can be carried out easily (and at relatively low cost) capable of distinguishing between two very different interpretations of general relativity. Again, it must be determined the value of gravitational potential energy at L1 (or L2): is it zero or nonzero (with respect to the value at infinity)? Do interacting fields cancel out to a zero local value at L1 or is there a residual nonzero potential energy? 1. Test of the field value at L1 with a Space-Borne Hydrogen Maser. Spacetime Curvature: redshift, blueshifted light signals emitted from the Lagrange points L1 or L2, and from here on earth: Gravitational Red-Shift in Nuclear Resonance. Mission: A spaceship launched from earth should travel in the direction of L1 (L2 will do just as nicely), along the line connecting say the center of the earth and the center of the sun (the earth-moon will do likewise just as nicely). The spacecraft will arrive at L1 (where it is to be docked). The probe must be equipped with a receiver, as well as a device comparable to the one used in the Pound–Rebka experiments (see Mössbauer effect) that can emit signals to be received here on earth, along the same connecting line, near the earth’s equator (this point varies according to season). Important note: the mission should be carried out directly at L1, not in a halo orbit. The 14 keV gamma rays emitted by iron-57 during its transition to its base state should prove to be sufficient for this observational test. The fractional change or difference in energy associated with gravitational redshift over a distance comparable to the L1 – Earth will be large and measurable. Using a space-borne hydrogen maser will increase the accuracy of the measurements. Coldcreation Predictions: (a) Earth-bound signals departing from L1 arriving here on earth will be less blueshifted than a signal sent from the same distance at any other point off of the plane of the solar system. But that will not be necessary. Predictions for the two competing models give different results from that very L1 (or L2) point. (The exact figures still need to be determined). (:rolleyes: Outgoing L1-bound signals from earth to the ship docked at L1 will be less redshifted than the standard model would permit. The exact figures for both models still need to be determined mathematically before launch. According to the standard model, the emitted signal from L1 should display a greater value for blueshift (the signal should be more blueshifted) since the field curvature is greater along that line. i.e., if I understand correctly, gravity fields are thought to cancel or become neutral without actually attaining a gravity-free space at L1 but maintaining a value of curvature equal to whatever that value was upon cancellation (something that make no sense since cancellation or neutralization imply no value at all, i.e., zero curvature, zero gravity). Conclusions: An earth-bound light signal emitted from the craft at L1 will arrive slightly sooner than expected. Inversely, L1-bound light signals emitted from earth will arrive at L1 also slightly sooner than expected according to the standard model than would be the case if spacetime curvature at L1 were not equal to zero with a smooth gradation leading up to it. (The exact figures still need to be determined). 2. Time Dilation: Identical clocks placed at L1 (or L2) as well as other points of equal distance, and on earth to determine the field potentials. Aboard the same spacecraft let us place a stable clock (hydrogen maser, see Shapiro delay) identical to one placed on earth along the line connecting the center of the sun, L1 and the center of the earth. Predictions: Gravitational potential will be equal zero at L1. Thus, clocks should run at different rates at L1 (or L2) than from any other point at the same distance off of the plane of the solar system. Predictions for the two competing models give different results from an L1 point where the gravitational field potential energy value is zero and one where the value is nonzero. This is because coordinate time and proper time diverge as the gravitational field value (or strength) increases. Clocks placed at weak gravitational potentials run faster than clocks placed at deep (or strong) gravitational potentials, which run slower. The standard model and the model proposed here require highly distinct boundary conditions, so a differentiation between the two should be observable and straightforward. When the emission is displaced significantly from L1, but at the same distance from earth, the signal experiences the standard general relativistic gravitational time dilation according a gravitational field that diminishes according to the inverse square law. Coordinate time and proper time will not diverge as much along the sun–L1–earth (or earth–L1–moon) line as predicted by the standard model. There should be a route-dependent difference in underlying clock rates. The fractional difference in the rate of identical clocks between models where L1 is gravity-free and the standard model where the sun’s field passes through L1 (i.e., when L1 is not gravity-free or curvature-free) should be close to one part in 10^8. The difference (the exact figure still needs to be determined) should be large enough to measure with the technology today available. A stabilized cesium atomic clock with high nominal fractional frequency stability should be sufficient. The results will show the combined gravity fields to be spherically asymmetric, that the combined field between massive bodies is continuous (there is a smooth hyperbolic gradation or curve) and becomes asymptotically Minkowskian (i.e., flat) as the field tends toward L1 (as distance tends toward infinity from massive bodies, albeit, at L1 (and L2) the zero value is attained). If the probe is docked beyond L1 (or L2,) as long as it is on the same line, the effect should still be observable, well within the margin of error (to be determined). Further testing: it could be of significant use to check the value of curvature at different location around the Lagrange L1 point to determine if the shape of the field surrounding L1 is truly a smooth saddle-shape or if there is a sharp peak at L1. It would seem, a priori, that the standard model needs the peak at some value of gravity, and that Coldcreation needs the smooth gradation of the field to zero (for lambda to be operational). Note: This is not a test of the field equations, but rather, a test of the boundary conditions of GR. Until this test is carried out (or something similar), I don't see how we can advance in this discussion, since a tenable solution to the dynamic equilibrium of the universe and its subsystems depends on the outcome of this investigation. CC Quote Link to comment Share on other sites More sharing options...
modest Posted April 3, 2008 Author Report Share Posted April 3, 2008 It takes more energy to escape the solar system than it does to get to a Lagrange point in the solar system therefore your theory is proven false. No need for any atomic clocks. Or, do you not realize that zero potential is MORE potential than is currently predicted (and proven by missions like SOHO) at L1. Do you think NASA would not have noticed it took much more energy than predicted to get a spacecraft to L1? Or, do you not understand that there is no difference between the amount of energy needed to get there and the value of gravitational potential? The potential at L1 has been measured! It is not open for debate and it is not zero Instead of proposing a billion dollar mission to find something out - let me suggest wikipedia first. -modest Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.