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New form for GR's gravitational red shift & a novel interpretationof the underlying physics 
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Dec1306, 05:00 AM

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[Moderator's note: the message is posted because it had to be
an enormous work to write it. LM] I am posting an unusual version of the standard relation for the gravitational red shift embodied in general relativity. This new version uses the KleinGordon equation, sans psi notation, and gives results which are exactly identical to those yielded by the fully relativistic GRS equation in standard texts. The proof of equivalence can be found in Part II below. I emphasize here that my KG approach is NOT an <alternative> to Einstein's well confirmed general relativity or his gravitational red shift result in particular. Indeed, his standard result can be transformed algebraically (though tediously!) to obtain this new KG approach and vice versa without loss of information. My equivalent rendering simply opens a window on an unexpected and selfconsistent reinterpretation of some of the basic physics of objects in a gravitational field. This new viewpoint both supports and flows from considerations of a larger toy model where particle rest mass and hbar increase directly with gravitational potential, with important consequences for the uncertainty principle. However, the energy content of a particle _at rest_ DECREASES with increase in gpotential  that energy content going increasingly over into the gravitational field itself, until at a black hole's event horizon the particle's own energy is effectively zero. This is one more way of saying black holes "have no hair". Also, for reasons given below, the gravitational fine structure 'constant' should approach unity near the event horizon of a black hole.]                          << Part I >> Below I will show a self consistent way of recasting the stan dard gravitational red shift (GRS) of general relativity which gives identically the same results as GR: [A] Assume that a photon, characteristic of a specific atomic / nuclear transition Q, is moving upward in the gravitational well of a mass M. The photon will experience a GRS in its wavelength, as cor rectly predicted by general relativity. The fully relativistic relation giving the fractional change in the energy of that photon, having been emitted at distance r1 from the center of the mass and traveling to a receiver at a greater distance r2, is (1) (1  Rs/r1)^.5  (1  Rs/r2)^.5 =  = (1  Rs/r1)^.5 (1  Rs/r2)^.5 1   (1  Rs/r1)^.5 Below I will show that E2 1   = delta E'/E' = E1 (1  Rs/r2)^.5 1   (1  Rs/r1)^.5 where E1 is the photon's energy measured at r2, and E2 is the expected energy of an identical transition Q photon if it were both emitted and then measured at r2. Rs is the mass's Schwarzschild radius, = 2GM / Co^2 . For E' , see [B] below. Following is what I will show to be an exactly equivalent relation, which gives predictions completely identical to those of (1). How ever, the nature of the second relation's variables seems to allow an unusual interpretation of the physics involved, and suggests that certain fundamental parameters may vary with gravitational potential with no apparent contradiction implied for local physics. This note will only touch on some of this. The second relation (3 below) incorporates the KleinGordon eq. for a particle's relativistic energy E where generally (2) E^2 = (mCo^2)^2 + (pCo)^2 , and p = relativistic momentum of a particle with rest mass m. Co is velocity of light in fieldfree space. [B] Using (2) : I consider the total KG energy E' of each of two identical par ticles, eg: two hydrogen atoms, <at rest> at two different eleva tions in a gravitational field potential phi, E'1 associated with particle 1 at lower elevation r1, and E'2 with particle 2 at higher elevation r2. Each particle is in an inertial frame with respect to the gravitational field. (3) E'1  E'2 E'2  = 1   = delta E'/E' = E'1 E'1 [ (m2*C2^2)^2 + (M2v2C2)^2 ]^.5 1   [ (m1*C1^2)^2 + (M1v1C1)^2 ]^.5 C1 and C2 are the local velocities of light, < Co, due to the action of M's gravitational field at r1 and r2, where (4) C1,2 = Co[ 1  2GM/(Co^2*r1,2) , from general relativity, and m1 and m2 are the 'rest' or 'invariant' masses of the two particles, which will be seen to differ as a function of the magnitude of the field at r1 and r2. Also (5) m1C1 = m2C2 , (6) M1,2 = m1,2 / [1  2GM / (Co^2*r1,2) ]^.5 , and (7) v1,2 = [2GM / (Co^2*r1,2) ]^.5 * C1,2 and p = (M1,2) * (v1,2) , effectively a form of relativistic potential momentum, purely as a function of the local gfield. [C] Rel. (3) turns out to give identically the same prediction as (1), and indeed both are shown below to be algebraically identical (see Part II below), and seem to reflect different ways of looking at the same phenomenon. The physical interpre tation of (3) is of course open, but I propose the following: When mass particle 1 is at rest at r1 in the gravitational field of M, its rest mass is greater than that of identical particle 2 at rest at r2 (> r1), by a factor C2 / C1, or equivalently a factor of (8) 1  (2GM / Co^2*r2)  . 1  (2GM / Co^2*r1) On the other hand, the total rest mass energy E2 (=m2(C2)^2) of particle 2 is greater than that of particle 1 by the same factor. This means that when particle 1 emits photon 1 for a character istic emission line Q, photon 1 has proportionately LESS energy than an equivalent transition Q photon 2, emitted and measured entirely at r2 higher up in the gravitational well. This also means that the photon itself can be seen as having constant energy throughout its trajectory. This is of course quite different from the usual interpretation whereby the photon loses an amount of energy equal to E'2  E'1 to the gravitational field along the way from r1 to r2. As an example, this would mean that a Lymanalpha photon emitted by a hydrogen atom at r1 could be seen as having in trinsically less energy than a Lymanalpha photon emitted by an identical hydrogen atom higher up at r2. This leads immed iately to a further result...Since we know already from experi ments (eg: Pound, Rebka, Snider) that photon 1's wavelength lambda1 is larger than that of photon 2 by the same factor given by (8), we can also say that since generally lambda = h / mC^2 = h C / E , for the photon, then Planck's constant h at r1 is actually (9) h1 = (E1 lambda1) / C1 = 1  (2GM / Co^2*r2)  * h2 . 1  (2GM / Co^2*r1) That is, the value of Planck's constant would then vary directly with the local gfield. It needs to be emphasized that, with one exception, none of these dimensional parameter variations with gravity can be observed locally in a lab, even in principle, since all of the measuring apparatus at any level in a gfield is also changed commensurately (local measuring rods, clocks, etc.), along with the quantity being measured. Thus the simultan eously changed apparatus will be blind to the changes in val ues of fundamental parameters. The key exception is the pho ton since, from this new viewpoint, both its energy E and wave length lambda are truly constant and are preserved over its path as long as it interacts only with the gravitational field. A point needing further exploration: near the event horizon of a black hole, the hugely increased value of Planck's 'constant' would greatly enhance the local importance of the Heisenberg un certainty principle, where (delta X) (delta momentum_x) = or > ihbar . Also, with the plausible argument that G is truly constant, the local value of the gravitational fine structure 'constant' GFS = hbar*C / (GMp^2) for a particle Mp should _decline_ to somewhere near unity at the event horizon. What interesting effects might we expect from these changes on all local classical and quantum physics?                          << Part II >> Demonstration of the formal equivalence of (1) and (3) : From E1E2 (1  Rs/r2)^.5  = 1   E1 (1  Rs/r1)^.5 and Rs = 2GM/Co^2 , we have (10) Rs/r1 = 2GM/r1Co^2 and let this equal A. Similarly, Let Rs/r2 = 2GM/r2Co^2 and let this equal B. Thus (1) is (11) (1  Rs/r2)^.5 (1  B)^.5 1   = 1   . (1  Rs/r1)^.5 (1  A)^.5 Adding 1, then multiplying by 1 and squaring gives (1  B) (1  A)^2 (1  B)^3  =  x  = (1  A) (1  B)^2 (1  A)^3 (1  A)^2 (1 3B + 3B^2  B^3)  . (1  B)^2 (1 3A + 3A^2  A^3) And since "3B" =  4B + B, and "3B^2" = +6B^2  3B^2 , and "B^3" = 4B^3 + 3B^3 (& similarly for A), by substituting we get (1  A)^2 (14B+6B^24B^3+B^4+B3B^2+3B^3B^4)  


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