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#51
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The Motion of the Perihelion of Mercury
On Jan 3, 12:21*pm, Raghar wrote:
On Dec 28 2008, 11:49*pm, wrote: THE MOTION OF THE PERIHELION OF MERCURY In his general relativity calculation of the motion of the perihelion of Mercury Albert Einstein had only taken into account the gravitational actions between the Sun and the Mercury, which he also assumed as two points. What will be, according to the theory of general relativity, the value of the motion of the perihelion of Mercury if the gravitational actions of all the planets in the solar system are taken into account and also it is taken into account that the Sun is a little oblate? Have any done these calculations? Well van Fandern did this analysis:http://www.metaresearch.org/cosmolog.../PerihelionAdv... TvF writes crank nonsense, and it doesn't address the original question. Also there are flyby anomalies, which are possibly more accuratehttp://www2.phys.canterbury.ac.nz/editorial/Anderson2008.pdf Completely irrelevant to the original question. Its latitude dependence suggests that the Earth’s rotation may be generating an effect much larger than the frame dragging effect of General Relativity, the Lense-Thirring effect |
#52
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The Motion of the Perihelion of Mercury
On Jan 3, 3:02 am, Eric Gisse wrote:
On Jan 2, 10:42 pm, Koobee Wublee wrote: On Jan 2, 12:03 am, Eric Gisse wrote: ds^2 = dx^2 + dy^2 + dz^2 ds^2 = dr^2 + r^2(d\theta^2 + sin^2(\theta)d\phi^2) ds^2 = dr^2 + r^2 d\theta^2 + dz^2 ds^2 = da^2 + db^2 + dz^2 They all describe the same thing, yet "look different". But I can find a coordinate transformation relating all of them to eachother, so they are the same thing. Your notation is screwed up. The r of the 2nd is not the same as r of the 3rd. The phi of the 2nd is the same as theta of the 3rd. Otherwise, they can all represent the same geometry. shrug See? You are a **** hair's distance away from understanding. What is that again? THE LABELS DO NOT MATTER. Wrong! The labels do matter. shrug The COORDINATES do not matter. They all represent the _same_ geometry. Wrong again. To describe a geometry, you need both the labels (coordinate) and how the labels are connected (metric). shrug This is not what I have been talking about. It looks like very few have understood what I am saying. That is why I feel very lonely to be the few who after Riemann and perhaps Christoffel that have understood the curvature business very well. shrug So where did you properly learn this "curvature business"? All I'm asking is for the book you learned it from. Is that asking so much? Just my intelligence which you lack. shrug What I am talking about is that the following geometries are different given the same dr, dLongitude, dLatitude. ** ds1^2 = g1_11 dr^2 + g1_22 r^2 dO^2 ** ds2^2 = g2_11 dr^2 + g2_22 r^2 dO^2 Where ** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2 Except the coordinate labels are not the same in both coordinate systems. What do you mean they are not the same? I have chosen them to be the same. shrug Calculate the surface area of a sphere - don't just write it down, CALCULATE it. I did. Just look up on my older post. What coordinate transformation? u is any function of r. This is very basic mathematics. shrug Yes, and it is called a "coordinate transformation". You start out in the (t,u,\theta,\phi) coordinate system and end up in the (t,r,\theta, \phi) coordinate system. I did not start out in anything. I merely presented these two independent geometries of spacetime without showing any favor of one to the other. You are still dreaming and jacking off in your mathemagical Realm. In doing so, probably wishing Einstein can join you. shrug Let me write down the two very different spacetime in a more concise way. ** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2 ** ds2^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2 THE GEOMETRY ds1 IS NOT THE SAME AS ds2. THAT IS BECAUSE THE METRIC OF ds1 IS NOT THE SAME AS ds2, BUT THE COORDINATE SYSTEM DESCRIBING BOTH ds1 and ds2 IS EXACTLY IDENTICAL. If they are "exactly identical" they should give the same values for the surface area of the sphere, as #1 is in Schwarzschild isotropic coordinates which has spheres of constant (r,t) of surface area 4 pi r^2. They are not identical. shrug Let us see you calculate the surface area of #2. I would also like you calculate the Ricci scalars of both. You'll notice something interesting. Calculating the area does not make any difference. To any observer observing the geometry of any curved spacetime, the answer is always going to be 4 pi r^2. shrug Can you dig this, punk? I don’t think you possess any minute intelligence to get after all these years, and I really don’t care. shrug With any matheMagical tricks, show me how you can coordinate-transform from one equation to the other and what justify one solution is more unique than the others. Remember that the coordinate system is the same with both equations. Except the coordinate systems are different Hey, I am the one presenting the problem. Both ds1 and ds2 share the same coordinate system. They are all solved from the set of field equations with the same coordinate system. shrug - the second one is displaced from the origin by an amount K, which you can see immediately by looking at the coefficients of the angular components. Or by calculating the surface area. Or by calculating the Ricci scalar. Or by looking at the determinant of the metric. Nonsense! There is no displacement of anything. There is no coordinate transformation. shrug I cannot claim one solution is more unique than the other, which is the whole point of my explanation. I just prefer Schwarzschild in isotropic coordinates. I can easily convert from Schwarzschild and every /OTHER/ "different" solution to yours - just chain the transformations. Hold your bullsh*t! Both ds1 and ds2 are spherically symmetric in isotropic coordinate system. Again, for the n’th time, these solutions are static, spherically symmetric, and asymptotically flat. I HAVE SHOWN YOU SOLUTIONS THAT ARE STATIC, SPHERICALLY SYMMETRIC, ABUT NOT ASYMPTOTICALLY FLAT. That proves Birkhoff’s theorem false by example. shrug Your ranting is because you refuse the God you worship is a nitwit, a plagiarist, and a liar. Yes, Einstein was nobody. He was a nitwit, a plagiarist, and a liar. Despite the nitwit, the plagiarist, and the liar did not come up with all these nonsense, whatever is credited to this nitwit, this plagiarist, and this liar is utter nonsense in the first place. shrug |
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The Motion of the Perihelion of Mercury
On Jan 3, 9:09 am, George Hammond wrote:
Correct, he admits to "making a typo" which is why there are two different versions of the Kooby Metric floating around. Oh, there are more than that. In the following post, you have been entertained with a solution that is static, spherically symmetric, but not asymptotically flat. http://groups.google.com/group/sci.p...1526793d?hl=en Of course, allow me to correct my typo. The spacetime I was talking about is: ds^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2 In the original one, you discovered and I have independently confirmed that it is simply a radial coordinate transformation of Schwarzchild given by r=R-K with K=2m. So the original Kooby Metric is nothing but the Schwarzchild Metric in disguise. Oh, it looks like you have been eating the same sh*t with Gisse. shrug In the second case his "Typo Version" of the Kooby Metric is not a coordinate transformation of Schwarzchild and naturally therefore R_uv does not vanish as you discovered and hence it is not a valid solution of the EFE's. shrug [Rest of whining crap snipped] So, the great reverend Hammond is also a nitwit. He had no problem to go to bed with Einstein Dingleberries to save his grace. shrug |
#54
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The Motion of the Perihelion of Mercury
On Sat, 3 Jan 2009 19:29:47 -0800 (PST), Koobee Wublee
wrote: ......... That proves Birkhoff’s theorem false by example. shrug [Hammond] "proves Birkhoff's theorem false by example".....aaahahaha ha hah hah aaaahaha ha ha ha ha...... Your ranting is because you refuse the God you worship is a nitwit, a plagiarist, and a liar. Yes, Einstein was nobody. He was a nitwit, a plagiarist, and a liar. Despite the nitwit, the plagiarist, and the liar did not come up with all these nonsense, whatever is credited to this nitwit, this plagiarist, and this liar is utter nonsense in the first place. shrug [Hammond] Oh here we go ... turns out you're not just a moron, you're a jive ass Natzi too....aaahahaha ha hah hah aaaahaha ha ha ha...... ===================================== HAMMOND'S PROOF OF GOD WEBSITE http://geocities.com/scientific_proof_of_god mirror site: http://proof-of-god.freewebsitehosting.com GOD=G_uv (a folk song on mp3) http://interrobang.jwgh.org/songs/hammond.mp3 ===================================== |
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The Motion of the Perihelion of Mercury
On Sat, 3 Jan 2009 19:37:34 -0800 (PST), Koobee Wublee
wrote: snip [Koobee Wublee] ......... That proves Birkhoff’s theorem false by example. shrug [Hammond] "proves Birkhoff's theorem false by example".....aaahahaha ha hah hah aaaahaha ha ha ha ha...... ===================================== HAMMOND'S PROOF OF GOD WEBSITE http://geocities.com/scientific_proof_of_god mirror site: http://proof-of-god.freewebsitehosting.com GOD=G_uv (a folk song on mp3) http://interrobang.jwgh.org/songs/hammond.mp3 ===================================== |
#56
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The Motion of the Perihelion of Mercury
On Jan 3, 6:29*pm, Koobee Wublee wrote:
On Jan 3, 3:02 am, Eric Gisse wrote: On Jan 2, 10:42 pm, Koobee Wublee wrote: On Jan 2, 12:03 am, Eric Gisse wrote: ds^2 = dx^2 + dy^2 + dz^2 ds^2 = dr^2 + r^2(d\theta^2 + sin^2(\theta)d\phi^2) ds^2 = dr^2 + r^2 d\theta^2 + dz^2 ds^2 = da^2 + db^2 + dz^2 They all describe the same thing, yet "look different". But I can find a coordinate transformation relating all of them to eachother, so they are the same thing. Your notation is screwed up. *The r of the 2nd is not the same as r of the 3rd. *The phi of the 2nd is the same as theta of the 3rd. Otherwise, they can all represent the same geometry. *shrug See? You are a **** hair's distance away from understanding. What is that again? THE LABELS DO NOT MATTER. Wrong! *The labels do matter. *shrug Prove it. Write something with two different labels, and PROVE that you can measure something to be different. I have an idea - let's see your proof that "other solutions" that satisfy the conditions of Birkhoff's theorem produce something other than 43 arcsec/centry for Mercury's perihelion precession. You did, after all, assert that to be the case. The COORDINATES do not matter. They all represent the _same_ geometry. Wrong again. *To describe a geometry, you need both the labels (coordinate) and how the labels are connected (metric). *shrug True - to write the metric down, you need coordinates. Except the coordinates do not matter as I can easily transform from one system to the next and write the metric down in the new coordinate system. As has been done repeatedly. As has the explanation to you. Consider that you are just too stupid to understand, like Androcles and rbwinn before you. This is not what I have been talking about. *It looks like very few have understood what I am saying. *That is why I feel very lonely to be the few who after Riemann and perhaps Christoffel that have understood the curvature business very well. *shrug So where did you properly learn this "curvature business"? All I'm asking is for the book you learned it from. Is that asking so much? Just my intelligence which you lack. *shrug So you just expect me to believe you woke up with this knowledge one morning? I find it more likely that you arrived at this particular level of understanding after trying to study modern general relativity textbooks. Except you failed, and invented all these arguments to protect your ego. Why else would you get so defensive and insulting when I ask a simple question like "where did you learn what you know?". What I am talking about is that the following geometries are different given the same dr, dLongitude, dLatitude. ** *ds1^2 = g1_11 dr^2 + g1_22 r^2 dO^2 ** *ds2^2 = g2_11 dr^2 + g2_22 r^2 dO^2 Where ** *dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2 Except the coordinate labels are not the same in both coordinate systems. What do you mean they are not the same? *I have chosen them to be the same. *shrug Then calculate the surface area of a sphere using the metric. Calculate the surface area of a sphere - don't just write it down, CALCULATE it. I did. *Just look up on my older post. Nope - you _asserted_ it was 4 pi r^2. You never _derived_ it. In fact, you seem quite incapable of proving something as simple as this. Why is the only person who has "truly understood" differential geometry incapable of deriving the surface area of a sphere? [...same idiocy as in 2006...] http://groups.google.com/group/sci.p...c?dmode=source JanPB explained this to you before with much greater patience than I can muster. You used the same arguments then as you use now. I cannot claim one solution is more unique than the other, which is the whole point of my explanation. I just prefer Schwarzschild in isotropic coordinates. I can easily convert from Schwarzschild and every /OTHER/ "different" solution to yours - just chain the transformations. Hold your bullsh*t! *Both ds1 and ds2 are spherically symmetric in isotropic coordinate system. *Again, for the n’th time, these solutions are static, spherically symmetric, and asymptotically flat. ....and the same. How quickly you forget that this has all been explained to you in detail before. With the same metrics no less. http://groups.google.com/group/sci.p...7?dmode=source I HAVE SHOWN YOU SOLUTIONS THAT ARE STATIC, SPHERICALLY SYMMETRIC, ABUT NOT ASYMPTOTICALLY FLAT. *That proves Birkhoff’s theorem false by example. *shrug Except you didn't prove the solution wasn't asymptotically flat because you never calculated something like CURVATURE which would actually prove your point. Plus I showed you the coordinate transformation to navigate between your "different solution" and Schwarzschild. Your ranting is because you refuse the God you worship [...] Funny how you are always the one to bring up God in these little conversations. |
#57
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The Motion of the Perihelion of Mercury
On Jan 3, 8:09*am, George Hammond wrote:
[snip] * *Bear in mind, as a mere 25 year old you still think he's a sincere Physics buff.... NO, I DO NOT. He is an arrogant idiot who has been repeatedly shown he can't do something as simple as calculate the surface area of a sphere, all the way to being unable to understand a derivation of the field equations. [snip] |
#58
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The Motion of the Perihelion of Mercury
On Jan 3, 6:37*pm, Koobee Wublee wrote:
On Jan 3, 9:09 am, George Hammond wrote: * *Correct, he admits to "making a typo" which is why there are two different versions of the Kooby Metric floating around. Oh, there are more than that. *In the following post, you have been entertained with a solution that is static, spherically symmetric, but not asymptotically flat. How do you know it is not asymptotically flat when you did not even consider calculating curvature components? [snip] |
#59
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The Motion of the Perihelion of Mercury
On Sun, 4 Jan 2009 00:29:58 -0800 (PST), Eric Gisse
wrote: What I am talking about is that the following geometries are different given the same dr, dLongitude, dLatitude. ** *ds1^2 = g1_11 dr^2 + g1_22 r^2 dO^2 ** *ds2^2 = g2_11 dr^2 + g2_22 r^2 dO^2 Where ** *dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2 Except the coordinate labels are not the same in both coordinate systems. What do you mean they are not the same? *I have chosen them to be the same. *shrug Then calculate the surface area of a sphere using the metric. Calculate the surface area of a sphere - don't just write it down, CALCULATE it. I did. *Just look up on my older post. Nope - you _asserted_ it was 4 pi r^2. You never _derived_ it. [Hammond] Spherical surface area in flat space is obtained by integrating the surface element cos(theta)d(theta)d(phi) in sperical coordinates over phi and theta for constant r which results in the ancient formula Area=4 pi r^2. In curved space of course the metric must be used if the phi and theta basis vectors are not orthogonal. However in the Schwarzchild and Kooby metrics there is only radial space warping and no angularspace warping (radially symmetric metric), therefore the surface element is still a simple rectangle and not a parallogram as in the general case. Since integrating phi from 0-2pi and theta from 0-pi over the angular part of the flat space metric or even the Schwarzchild Metric: r^2 cos(theta)d(theta)d(phi) results in 4 pi r^2i, we see that the area of an ordinary sphere in flat space, or even a sphere in Schwarzchild space is 4 pi r^2 for a fixed value of r. In the case of the Kooby Metric however, the angular part is not multiplied by r^2, it is multiplied by (R+K)^2 therefore the same surface integration results in an area equal to: 4 pi (R+K)^2 for a fixed value of R; the new Kooby radial coordinate. Notice of course that substituting the coordinate transformation R=r-K transforms thisresult back to the well known Schwarzchild result area=4 pi r^2 again. ===================================== HAMMOND'S PROOF OF GOD WEBSITE http://geocities.com/scientific_proof_of_god mirror site: http://proof-of-god.freewebsitehosting.com GOD=G_uv (a folk song on mp3) http://interrobang.jwgh.org/songs/hammond.mp3 ===================================== |
#60
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The Motion of the Perihelion of Mercury
On Jan 4, 12:29 am, Eric Gisse wrote:
On Jan 3, 6:29 pm, Koobee Wublee wrote: On Jan 3, 3:02 am, Eric Gisse wrote: THE LABELS DO NOT MATTER. Wrong! The labels do matter. shrug Prove it. Write something with two different labels, and PROVE that you can measure something to be different. I have said that many times over. To describe the geometry, one must choose a coordinate system first and then define how the set of coordinates is related to the geometry. shrug I have an idea - let's see your proof that "other solutions" that satisfy the conditions of Birkhoff's theorem produce something other than 43 arcsec/centry for Mercury's perihelion precession. You did, after all, assert that to be the case. Well, it is time consuming. I have lacked any motivation to do so after discovering GR is a total nonsense based on voodoo mathematics. shrug The COORDINATES do not matter. They all represent the _same_ geometry. Wrong again. To describe a geometry, you need both the labels (coordinate) and how the labels are connected (metric). shrug True - to write the metric down, you need coordinates. Oh, wow! That is an improvement after all these years. shrug Except the coordinates do not matter as I can easily transform from one system to the next and write the metric down in the new coordinate system. Again, you are stuck in your matheMagical wonderland of coordinate transformation. There is no such coordinate transformation. You are schizophrenic. shrug As has been done repeatedly. As has the explanation to you. There is no credible coordinate transformation done besides total nonsense. shrug Consider that you are just too stupid to understand, like Androcles and rbwinn before you. The Einstein Dingleberries have no problems bouncing these crackpots up and down the tennis courts, but Koobee Wublee remains unchallenged. Your whining crap does not count. shrug So where did you properly learn this "curvature business"? All I'm asking is for the book you learned it from. Is that asking so much? Just my intelligence which you lack. shrug So you just expect me to believe you woke up with this knowledge one morning? Well, it is up to you. I have also studies a lot about the nonsense which you eagerly swallowed. shrug I find it more likely that you arrived at this particular level of understanding after trying to study modern general relativity textbooks. Except you failed, and invented all these arguments to protect your ego. There are flaws in GR. I have discovered them, and I have advanced my studies. In the meantime, you are stuck in what they have told you. That is why you remain a multi-year super-senior. shrug Why else would you get so defensive and insulting when I ask a simple question like "where did you learn what you know?". I am not getting defensive. I just don’t think my educational background is any of your business. shrug Except the coordinate labels are not the same in both coordinate systems. What do you mean they are not the same? I have chosen them to be the same. shrug Then calculate the surface area of a sphere using the metric. I have told you the answer more than once. To an observer, the surface area of a sphere is always (4 pi r^2) regardless how curved up space is. shrug Calculate the surface area of a sphere - don't just write it down, CALCULATE it. I did. Just look up on my older post. Nope - you _asserted_ it was 4 pi r^2. You never _derived_ it. [Remaining whining crap snipped] You are just trolling me. shrug JanPB explained this to you before with much greater patience than I can muster. You used the same arguments then as you use now. Speaking of the one who calls me mentally unstable, and yet he would have no problem of swimming naked in a public lake. Mr. Bielawski is an Einstein Dingleberry who swallows everything the collective group of Einstein Dingleberries decides what physics should be without any scientific methodology. You are also one of them. You have also become a henchman trying to force that crap down on anyone’s throat. shrug I cannot claim one solution is more unique than the other, which is the whole point of my explanation. I just prefer Schwarzschild in isotropic coordinates. I can easily convert from Schwarzschild and every /OTHER/ "different" solution to yours - just chain the transformations. Hold your bullsh*t! Both ds1 and ds2 are spherically symmetric in isotropic coordinate system. Again, for the n’th time, these solutions are static, spherically symmetric, and asymptotically flat. [snipped whining crap] So, you have no answers as usual besides whining nonsense. I HAVE SHOWN YOU SOLUTIONS THAT ARE STATIC, SPHERICALLY SYMMETRIC, ABUT NOT ASYMPTOTICALLY FLAT. That proves Birkhoff’s theorem false by example. shrug Except you didn't prove the solution wasn't asymptotically flat because you never calculated something like CURVATURE which would actually prove your point. Plus I showed you the coordinate transformation to navigate between your "different solution" and Schwarzschild. There is no need to talk about this further. It is obvious that you do not understand what asymptotically flat means. shrug Your ranting is because you refuse the God you worship [...] Funny how you are always the one to bring up God in these little conversations. That is because you have decided Einstein as your God. shrug Your education is totally based on faith. shrug |
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