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| Tags: curvature, orbits, spacetime, tensor |
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#1
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We cannot calculate the tensor for every point on an entire orbit (or
even less than that.) General Relativity does not take us forward experimentally rather theoretically. Calculating an orbit with it cannot be done to completion. We have approximation. Mitch Raemsch Twice Nobel Laureate 2008 |
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#3
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On Mar 29, 6:01*pm, The TimeLord wrote:
Am Fri, 28 Mar 2008 15:10:10 -0700 schrieb in in sci.physics.relativity: We cannot calculate the tensor for every point on an entire orbit (or even less than that.) *General Relativity does not take us forward experimentally rather theoretically. Calculating an orbit with it cannot be done to completion. We have approximation. Why would you say something that silly? Of course you can calculate the tensor for the entire orbit. There is a different tensor for every point along an eliptical orbit. For a circular orbit you would be right. There is infinite calculations to be done. Since the orbit is inertial, it's just a geodesic defined by the metric; and that is one of the simplest things to calculate. The fact is orbit motion speeds up toward perehlion and slows down toward apehelion. Mitch Raemsch Twice Nobel Laureate 2008 I doubt it. I won for unmoving acceleration and wave function collapse. Mitch Raemsch Twice Nobel Laureate 2008 |
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#4
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Am Sat, 29 Mar 2008 19:12:35 -0700 schrieb in
: On Mar 29, 6:01Â*pm, The TimeLord wrote: Am Fri, 28 Mar 2008 15:10:10 -0700 schrieb in in sci.physics.relativity: We cannot calculate the tensor for every point on an entire orbit (or even less than that.) Â*General Relativity does not take us forward experimentally rather theoretically. Calculating an orbit with it cannot be done to completion. We have approximation. Why would you say something that silly? Of course you can calculate the tensor for the entire orbit. There is a different tensor for every point along an eliptical orbit. For a circular orbit you would be right. There is infinite calculations to be done. If you mean that the (one) tensor changes value throughout the orbit, I'll agree, but I don't see that the orbit *requires* more than one tensor. As far as recalculating the tensor, that's really what computers are for. At any rate, to get a closed form solution just start with the metric and calculate the geodesic of the orbit: simple. Since the orbit is inertial, it's just a geodesic defined by the metric; and that is one of the simplest things to calculate. The fact is orbit motion speeds up toward perehlion and slows down toward apehelion. So what, it's still included in the geodesic. Also the precession of the orbit is included. So I don't see the problem. Mitch Raemsch Twice Nobel Laureate 2008 I doubt it. I won for unmoving acceleration and wave function collapse. Mitch Raemsch Twice Nobel Laureate 2008 You are such a b---sh--er! The Nobel committee hasn't made the 2008 announcements yet and your name is not on the previous list: http://topics.nytimes.com/top/news/s...es/index.html? inline=nyt-classifier Given that you are such a liar, I doubt you will ever be a Nobel Laureate. |
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