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How it is possible to infer too much physical meaning from mathematical tools By J.P. Baugher October 2012 Ver 1.0 Let me know if you catch any mistakes! From Poisson Equation f 2 For a function f given on 2 or 3 On boundary of g “First, since u = constant satisﬁes the homogeneous problem with f = 0, g = 0, it is clear that a solution to a Neumann problem can only be unique up to an additive constant.” This means take the derivative of g ( ) ' g ' 2 2 0 g ' 2 2 1 g' f 2 Minus sign switched since convention had normal vectors pointing out from a volume f 2 Typical field drawing… Presence of au seems irrelevant since disappears when derivative taken, arbitrary value would give same solution ( ) ' g ' 2 2 0 g ' 2 2 What does solution “only unique up to an additive constant” mean? Example given as Euclidean 2D slice of scalar values in R3 space for center point. g 10 10 10 10 10 10 10 10 10 1 1 2 1 2 3 2 3 3 11 11 12 11 12 13 12 13 13 Second derivative… ( ) ' g ' 2 2 10 10 10 10 10 10 10 10 10 Regardless of magnitudes, no gradient exists…. 1 1 2 1 2 3 2 3 3 11 11 12 11 12 13 12 13 13 These become identical second derivatives But our understanding is metaphysical. We are implying more certainty than our tool allows… g g Second derivative… g 10 10 10 10 10 10 10 10 10 Regardless of magnitudes, no gradient exists…. 1 1 2 1 2 3 2 3 3 11 11 12 11 12 13 12 13 13 These are identical second derivatives g 10 10 10 10 10 10 10 10 10 Regardless of magnitudes, no 0 gradient exists here, but we could model this as…. 9 9 8 1 1 2 9 8 7 1 2 3 8 7 7 2 3 3 …however the directional derivatives are still identical whether au is taken into consideration or not. What about Gauss’ Theorem? 2 f Theorem also implies too much certainty… 1 2 ( C ) Integration of a constant over a differential volume has no effect… What about the whole volume? 2 f Not a problem since second derivatives identical…but… There is another relationship that must hold to demonstrate that there are no unique solutions…so not only... 1 2 ( C ) …but also (need to preserve total u in volume, scalar values of u important)... 1 2 ( C ) This means that although the theorems are formulaically correct, the most we can agree on is that, by convention, the “field” is directed towards or away from the source since it could be considered as either attracting… …or reducing a repulsion. If au exists, what does it physically represent? 0 To state definitively that some physical parameter represented by function f causes, through action at a distance, an “attraction” represented by the Poisson equation cannot be supported by the equations themselves. This is a metaphysical description brought about by ignoring au. f 2 Is f over “there”causal to the appearance (from nothing) of this vector “here”? or… 0 is f over “there” causal to a change to what is already “here”? Why is this important? General Relativity… For regular General Relativity, the Einstein tensor Guv is equated to the stress energy tensor of matter kTuv. The T00 term is modeled to be energy density from a perfect fluid tensor. In the weak field approximation to Newtonian gravity, we now must use the Poisson equation to derive our understanding of how gravity is “attractive”. So… 1 R Rg G T 2 Why is this important? T00 term only for velocities much slower than c we can infer an approximation… T T00 …so that we get Poisson’s equation for gravity… T00 4 G 2 Why is this important? Great approximation to “attractive” Newtonian gravity for most of 20th century as long as one does not worry about “dark matter”… Cosmological Constant problem General Relativity is derived so that in the presence of matter (energy, mass) there is curvature, but if this mass or energy vanishes, so too must the curvature… R 0 Why is this important? Although there is much debate about the cosmological constant, in Unimodular relativity it is known simply as a constant of integration… 1 R Rg G g 2 For the derivation of GR, this term must be set to zero since Einstein had already equated Guv to kTuv. Why is this important? It became apparent through observations though, as accurate as GR is, for some reason there appears to be a tiny value for . And so, in spite of the mathematical rigor used to derive the field equation using Ruv=0, our understanding demanded that we must consider it possible for to have a value. In the late 1960s, Zel’dovich postulated that perhaps it represented an energy density of the vacuum. Although it is possible to measure the value there are no accepted methods to understand either its magnitude or even physical presence. Why is this important? Using our previous argument of the Poisson equation and Unimodular Relativity, we ask: Using Unimodular theory, let us define a multiple of the metric as g G L where Guv is the Einstein tensor and Luv is a tensor we name the Lorentz tensor. Why is this important? We can see that if =0, then the negative of the Lorentz tensor is equivalent to the Einstein tensor (reduces to GR).. L G Thus let us enforce Ruv=0, so that the field equation can be stated as 1 R Rg G g L 2 Why is this important? Note that since we are enforcing Ruv=0, no extraneous multiple of the metric, such as is allowed. Therefore no “vacuum solutions” or cosmological models such as deSitter space exists within this field equation… This model presents the same challenge that was required of Guv, explaining the Newtonian approximation. Although I have made much progress, there will probably be great push back on explanation of the perfect fluid tensor. The end goal is to compare rigorously… G G00 g L g00 L00 …approximates to… …approximates to… g g …for example… 10 10 10 10 10 10 10 10 10 0 …for example… 0 1 1 2 1 1 2 1 2 3 1 2 3 2 3 3 2 3 3 9 9 8 1 1 2 9 8 7 1 2 3 8 7 7 2 3 3 Between these two approximations, we can develop an understanding of why the following is so paradoxical… Equation from General Relativity: An Introduction for Physicists , Hobson, Efstathiou, Lasenby. For a spherical mass M (note that the r vectors are unit vectors) GM c r g 2 r r r 3 2 We have from modified EFE. Why is the first vector so many orders of magnitude larger than the second? Why isn’t the vector from exactly zero? What does it represent? What is dark energy? g A 0 is not derived as being completely independent of radius. If this also applies to then vector A may decrease in magnitude at a greater rate than Newtonian. When a mass achieves the radius of where vector A is zero, what occurs at radius greater than this? Does gravity become repulsive after: 6G V 1/3 r( ) 2 c res