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Re-Examination of Globally Flat Space-Time

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Re-Examination of Globally Flat Space-Time 1 Recent further research regarding the theory of inertial centers (unpublished) 1, Michael R. Feldman ∗ 1 Michael R. Feldman Private researcher, New York, NY, United States of America E-mail: [email protected] ∗ Geodesic solutions Our geodesic equation is given by (see (72) of [1]) 0= U a U b ∇a To find solutions, we start by expressing this equation in terms of Minkowski coordinates (T,X,Y,Z), where the spatial coordinate origin of this system is located at the position of our inertial center (i.e. r = 0 (X,Y,Z) = (0, 0, 0)). Since our metric in terms of this Minkowski coordinate parametrization takes the→ form dχ2 = c2dT 2 + dX2 − − i i X 1 2 3 where i = 1, 2, 3 and X = X,X = Y,X = Z, our metric components in Minkowski coordinates gµν c are all constants. Thus, our affine connection tensor for Minkowski coordinates is zero: Γab Minkowski = 0. Our equations of motion in Minkowski coordinates are then found to be | d2T d2Xi = 0 and = 0 (1) dσ2 dσ2 where σ χ = √Λ r(τ)dτ for massive geodesics in the theory of inertial centers (see the introduction or equation→ (57) of [1]).· Examining only our spatial components, we see that R d2Xi d dT dXi 0= = dσ2 dσ dσ dT d2T dXi dT 2 d2Xi = + dσ2 dT dσ dT 2 which leads us to d2Xi =0 (2) dT 2 where we have used the equation of motion for our Minkowski time coordinate T . The solutions to (2) are of the form i i i arXiv:1312.1182v2 [physics.gen-ph] 3 Jan 2014 X (T )= v T + X0 (3) i where X0 is our spatial Minkowski coordinate location when T = 0. Nevertheless, we know from the form of our affine parameter χ that we wish to express this coordinate parametrization of our geodesics in terms of the more physically applicable radial Rindler coordinates, where the coordinate transformations from the Minkowski coordinate parameterization employed above are r T = sinh(√Λt) c X = r cosh(√Λt) sin θ cos φ Y = r cosh(√Λt) sin θ sin φ Z = r cosh(√Λt)cos θ 2 Using these transformations in (3), we find vX r cosh(√Λt) sin θ cos φ = r sinh(√Λt)+ X c 0 vY r cosh(√Λt) sin θ sin φ = r sinh(√Λt)+ Y c 0 vZ r cosh(√Λt)cos θ = r sinh(√Λt)+ Z c 0 Solving for r by squaring each one of these equations and adding the results together vi 2 r2 cosh2(√Λt)= r sinh(√Λt)+ Xi c 0 i X where we choose our time coordinates T and t to coincide for our initial conditions (i.e. t =0 T =0 ⇐⇒ and [dt = dT ] t=T =0). This requirement forces our Minkowski constant c to be equal to √Λr0 where r0 here is the initial| radial position of our object along the geodesic and √Λ is fixed by the value of the Hubble constant. Remember that our inertial time coordinate t progresses at the same rate as the clock of a stationary observer located the same distance away from our inertial center as our object. This will hold true when the object is located at any particular position along its path, since at each radial location, 2 2 2 2 our line element for a stationary observer reduces to Λr dτS = Λr dt (see (57) and (1) of [1]). Thus, coordinate time t is physically equivalent to the elapsed− time measured− by an observer carried with the object in question but in the limit where the speed of this ‘carried’ observer relative to the inertial center in question is zero at all points along the traversed path (i.e. the ‘carried’ observer remains stationary at each point that the object reaches). Expanding the expression above, vi 2 vi r2 cosh2(√Λt)= r2 sinh2(√Λt)+2 Xir sinh(√Λt) + (Xi)2 c c 0 0 i X But (Xi )2 = (Xi(T ))2 = (Xi(t))2 = r2 0 |T =0 |t=0 0 i i i X X X Using this relation, vi vi 2 r2 +2r sinh(√Λt) Xi = r2 cosh2(√Λt) sinh2(√Λt) 0 c 0 − c i i X X vi 2 = r2 cosh2(√Λt) sinh2(√Λt) 1 1+ − − c i X vi 2 = r2 cosh2(√Λt) sinh2(√Λt) + sinh2(√Λt) 1 − − c i X vi 2 = r2 1 + sinh2(√Λt) 1 − c i X 2 i 2 R However, for R = i(X ) , we see that v (T )= dR/dT is given by P dXi RvR = Xi = Xivi dT i i X X 3 Yet, since dXi/dT is a constant for all T , we have Xivi = RvR = RvR = r vr (4) 0 |T =0 |t=0 0 0 i X r where v0 = dr/dt t=0 is the initial radial velocity of the object in the inertial system. Plugging in above, we find | vi 2 vr 0= r2 1 + sinh2(√Λt) 1 2r sinh(√Λt) r 0 r2 − c − · 0 c − 0 i X Solving for r(t), r r i v0 √ v0 2 2 √ 2 √ v 2 c sinh( Λt) ( c ) sinh ( Λt) + [1 + sinh ( Λt)(1 i( c ) )] r(t)= r0 ± − q 1 + sinh2(√Λt)(1 ( vi )2) − i c P When t = 0, we want our expression to reduce to r0 and not Pr0. Therefore, we take only the positive root for our solution. − i 2 Additionally, we see that i(v ) can be expressed in terms of the induced Riemannian metric com- ponents hij for a space-like hypersurface in our original Minkowski coordinates in the following manner: P i 2 i j 2 (v ) = hij v v = v0 i i,j X X However, this is simply the squared magnitude of the vi’s in Minkowski coordinates along this hypersur- face. Furthermore, since our time coordinates coincide at t = T = 0, v0 can be interpreted physically as the initial speed of our inertially traveling object measured in terms of the clock of a stationary observer located at r = r . In other words, since vi = dXi/dT = dXi/dT , 0 |T =0 dXi dXj dXi dXj v2 = h = h 0 ij dT dT ij dT dT i,j i,j T =0 X X ∂Xi dx˜k ∂Xj dx˜l = h ij ∂x˜k dT ∂x˜l dT i,j k,l T =0 X X ∂Xi ∂Xj dx˜k dx˜l = h ij ∂x˜k ∂x˜l dt dt k,l i,j t=0 X X dx˜k dx˜l = h˜ (5) kl dt dt k,l t=0 X k where h˜kl are induced Riemannian spatial metric components along the t = T = 0 hypersurface andx ˜ are spatial coordinates. Both tilded expressions are in radial Rindler coordinates. Plugging back into our expression for r(t) where we remember to take only the positive root, we find for the radial position of our object r r v0 √ 2 √ v0 2 v0 2 c sinh( Λt)+ 1 + sinh ( Λt)(1 + ( c ) ( c ) ) r(t)= r0 − (6) 1 +q sinh2(√Λt)(1 ( v0 )2) − c 4 where c = √Λr0 and dx˜k dx˜l (vr)2 v2 = (vr)2 h˜ 0 − 0 0 − kl dt dt k,l t=0 X = (vr)2 (vr)2 + r2 (vθ)2 + sin2 θ (vφ)2 0 − 0 0 0 0 0 = r2 (vθ)2 + sin2 θ (vφ)2 (7) − 0 0 0 0 Furthermore, our normalization condition for the ‘four-velocity’ along these geodesics is k = g U aU b − ab where U µ = dxµ/dσ and 0 null geodesics k = 1 time-like geodesics We can express this normalization condition in our original Minkowski coordinates in the following man- ner: dT 2 dXi 2 dT 2 dT 2 k = c2 + = c2 (vi)2 = c2 v2 − − dσ dσ − dσ − − dσ − 0 i i X X which provides us with the constraints that one would expect v0 = c massless objects (8) v0 <c massive objects (9) Returning to our coordinate transformations, the angular positions of our object as a function of t are found to be y v0 √ 1 c r(t) sinh( Λt)+ Y0 φ(t) = tan− x v0 √ c r(t) sinh( Λt)+ X0 φ √ v0 1 r sinh( Λt) c r0 sin θ0 = φ0 + tan− r θ (10) √ v0 v0 r0 sin θ0 + r sinh( Λt)[ c sin θ0 + c r0 cos θ0] and z 1 v0 Z0 θ(t) = cos− tanh(√Λt)+ c √ r(t) cosh( Λt) r θ 1 v0 v0 r0 cos θ0 = cos− tanh(√Λt) cos θ r sin θ + (11) c 0 − c 0 0 √ r cosh( Λt) where X(t) = X = r sin θ cos φ (12) |t=0 0 0 0 0 Y (t) = Y = r sin θ sin φ (13) |t=0 0 0 0 0 Z(t) = Z = r cos θ (14) |t=0 0 0 0 and vX = vx = vr sin θ cos φ + vθr cos θ cos φ vφr sin θ sin φ (15) 0 0 0 0 0 0 0 0 − 0 0 0 0 Y y r θ φ v = v0 = v0 sin θ0 sin φ0 + v0 r0 cos θ0 sin φ0 + v0 r0 sin θ0 cos φ0 (16) vZ = vz = vr cos θ vθr sin θ (17) 0 0 0 − 0 0 0 5 One can obtain initial velocity expressions (15)-(17) by plugging our spatial coordinate transformations i into each of our dX /dT T =0 terms. We have yet to address| the relationship between our time coordinate t and the proper time of our object τ in this particular inertial reference frame.
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