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#1
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The Math is still Not Ready
On Jan 5, 8:54 am, Tom Roberts wrote:
Here is General Relativity: On a 4-d Lorentzian manifold M, G = T where G is the Einstein curvature tensor and T is the energy-momentum tensor. Please allow Koobee Wublee reminds Tom where that overly simplified equation[s] above come from. Let’s follow Hilbert’s footsteps and pull out the following so-called Lagrangian out of Hilbert’s ass. ** L = (R / K + rho) sqrt(-det[g]) Where ** L = Lagrangian ** R = Ricci scalar ** K = Constant ** rho = Mass density ** [g] = The metric (a 4x4 matrix) ** det[] = Determinant of a matrix For the language of convention in this case, [A] means a matrix with elements [A]_ijk... or [A]^ijk... The field equations can be derived in just one step by taking the partial derivative of the Lagrangian above with respect to [g^-1]^ij where [g^-1], a matrix, is the inverse of [g], another matrix, and after setting each of the partial derivative to null, the result is the following relationships of matrices. ** [R] – R [g] / 2 = K rho [g] / 2 Where ** [R] = Ricci tensor (another 4x4 matrix) You would call the following. ** [G] = [R] – R [g] / 2 ** [T] = K rho [g] / 2 Thus, ** [G] = [T] Koobee Wublee would also like to remind Tom that the above equation has never been tested with any experimentations, and the best Tom can hope for is the following where the energy momentum tensor is null. ** [G] = 0 Where ** [T] = 0 Using only diagonal [g], the equation above simplifies into the following where the effect of the ever so celebrated trace term is nullified. The null Ricci tensor was basically Nordstrom’s work where Schwarzschild had been working on the solution for years. That is why within a couple months after Hilbert presented the field equations, Schwarzschild published a solution. ** [R] = 0, first proposed by Nordstrom as the field equations Where ** R [g] / 2 = The trace term shrug To get SR from GR: Riemann = 0 Top(M) ~ R^4 where Riemann is the Riemann curvature tensor on M, and Top(M) is the topology of M. Nonsense, Tom. If the Riemann tensor is null, the Ricci tensor must be null as well in which you end up with the null Ricci tensor above where you can solve for the Schwarzschild metric and other equally valid solutions that are able to degenerate into Newtonian law of gravity at weak curvature in spacetime. shrug The best way to get SR from GR is to set the gravitating mass, M, to 0, duh! shrug ** ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2 Where ** U = G M / c^2 / r [Note that approximations are important in applying the theory, as this is PHYSICS, not math. Based on your writings around here, and your aversion to any intellectual effort, I estimate you will never understand this.] Tom, in GR, physics = math, and math = physics. So, start understanding the mathematics involved instead of wishing for what you believe in. shrug Faith should not come into any equations of science, no? shrug |
#2
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The Math is still Not Ready
On Jan 6, 10:52 am, Jimmy Kesler wrote:
Koobee Wublee wrote: For the language of convention in this case, [A] means a matrix with elements [A]_ijk... or [A]^ijk... three indices matrix? Yes, Jimmy. Three indices indicate rank-three matrices. Riemann curvature tensors are rank-four matrices while both the Ricci and the metric tensors are rank-two matrices. shrug where are the ten field equation then R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R + g_{\mu \nu} \Lambda = {8 \pi G \over c^4} T_{\mu \nu} The field equations actually involves with 4x4 matrices. Yes, they are all the familiar rank-two matrices. These 4x4 matrices should give you 16 elements, and each element is a partial differential equations. When space and time are not intertwined like the scientific communities have decided time and space should intertwine, all correlations of elements between time and space become null. That is what you are thinking of only 10 equations. However, for almost all practical applications, only diagonal metrics are considered. In doing so, there are only 4 differential equations you have to worry about. shrug and how come the expansion is metric tensor dependent, ie the universe expand accordingly more at the regions with intense curvature (presence of matter); why not the universe expands flat? what tells it must expand like this? Not sure what you are talking about, but the expansion of the universe can be found with the de Sitter and the Friedman-LeMaitre-Robertson- Walker metrics where both do not satisfy Newtonian law of gravity at short distances. You can apply Koobee Wublee’s theorem of parallel variations to find a modified de Sitter metric that satisfies the Newtonian law of gravity. You can also modify the Schwarzschild metric to allow an expanding universe at cosmic scales. shrug this was the first question and also, i dont understand a half of a metric tensor, lol, what on earth is a half of a metric tensor!! Half a metric tensor is simply ([g] / 2) where [g] is a 4x4 matrix describing spacetime. Not sure what is not to understand? shrug |
#3
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The Math is still Not Ready
On Sat, 5 Jan 2013 21:36:26 -0800 (PST), Koobee Wublee
wrote: On Jan 5, 8:54 am, Tom Roberts wrote: Here is General Relativity: On a 4-d Lorentzian manifold M, G = T where G is the Einstein curvature tensor and T is the energy-momentum tensor. Please allow Koobee Wublee reminds Tom where that overly simplified equation[s] above come from. Let’s follow Hilbert’s footsteps and pull out the following so-called Lagrangian out of Hilbert’s ass. ** L = (R / K + rho)sqrt(-det[g]) sqrt(-det[g])? Why should it be necessary to first make the determinant negative? (we can all see the algebraic requirement of course). Don't you have any suspicions about such a fictitious looking term? I have pointed this out befo the metric tensor g is invalid. The term g00 = -1 is purely fraudulent, an arrangement calculated to avoid the product ict x ict and make it look like other real dimensions: e.g. ct x ct. This is gloatingly described in Gravitation by MTWheeler, "Farewell to ict". I think we can agree that it is invalid to make major changes in the coefficients of a matrix like g, just to make up for the defects in the vector field. g is Diagonal and is meant strictly for stretching, but at the same time With a negative determinant it is thereby inadvertently converting positive volumes into negative ones, which is clearly impermissible. It is regrettable that this duplicity has not been challenged anywhere, but it should be up for discussion. The time coordinate has to be retained as ict and it can never legally be promoted as an additional dimension that can be matched up with the real XYZ. Snip Faith should not come into any equations of science, no? shrug No. John Polasek |
#4
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The Math is still Not Ready
On Mar 17, 2:15 pm, wrote:
On Sat, 5 Jan 2013, Koobee Wublee wrote: Please allow Koobee Wublee reminds Tom where that overly simplified equation[s] above come from. Let’s follow Hilbert’s footsteps and pull out the following so-called Lagrangian out of Hilbert’s ass. ** L = (R / K + rho)sqrt(-det[g]) sqrt(-det[g])? The determinant of the metric is negative. So, sqrt(-det[g]) is a real number. shrug Why should it be necessary to first make the determinant negative? (we can all see the algebraic requirement of course). That is because nothing can travel beyond the speed of light. shrug Don't you have any suspicions about such a fictitious looking term? Yes, of course. shrug I have pointed this out befo the metric tensor g is invalid. The term g00 = -1 is purely fraudulent, an arrangement calculated to avoid the product ict x ict and make it look like other real dimensions: e.g. ct x ct. [g]_00 (your g00) is not -1. It is +1 --- (1 – 2 U) thing. shrug This is gloatingly described in Gravitation by MTWheeler, "Farewell to ict". If nothing can travel beyond the speed of light, the signature of the metric ought to be (+1, -1, -1, -1). shrug I think we can agree that it is invalid to make major changes in the coefficients of a matrix like g, just to make up for the defects in the vector field. The fault of GR starts way before the construction of spacetime. shrug g is Diagonal and is meant strictly for stretching, but at the same time With a negative determinant it is thereby inadvertently converting positive volumes into negative ones, which is clearly impermissible. [g] (your g) does not have to be diagonal. It is made diagonal to simplify the already complex math. If [g] is not diagonal, it would be relatively impossible to solve for the field equations. shrug It is regrettable that this duplicity has not been challenged anywhere, but it should be up for discussion. The time coordinate has to be retained as ict and it can never legally be promoted as an additional dimension that can be matched up with the real XYZ. There is no ict thing if the signature is (+1, -1, -1, -1). shrug Faith should not come into any equations of science, no? shrug No. amen |
#5
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The Math is still Not Ready
On Sun, 17 Mar 2013 21:10:16 -0700 (PDT), Koobee Wublee
wrote: On Mar 17, 2:15 pm, wrote: On Sat, 5 Jan 2013, Koobee Wublee wrote: If nothing can travel beyond the speed of light, the signature of the metric ought to be (+1, -1, -1, -1). shrug Look at what you have gone and done, proved negative volume. This format is just as flaky. John Polasek |
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