# Logic of GR as mathematically derived

by grav-universe
Tags: derived, logic, mathematically
 P: 406 Well sure, you could do it like that too I suppose, but only if you like doing things the easy way.
 P: 406 Ah, okay. For the past week I have been trying to incorporate assumption C for the conservation of angular momentum m_p v'_t r', where m_p here is the mass of the particle, which is considered invariant so divided out for convenience, leaving the constant of motion P = v'_t r', which is measured the same locally at every shell depending upon how a particle is originally set in motion. Okay, so by reverse engineering the solution to the SC metric in another thread, I had gained P = v'_t r' = v'_t r / sqrt(1 - 2 m / r), which I took to mean r' = r / L, using the inferred radius by extending a local ruler at r all the way along r radially that is contracted radially by a factor of L = sqrt(1 - 2 m / r). One problem I have had with that, though, upon seeing that the quantities z, L / dr, and L_t / r are invariant for a particular shell (or coordinate independent), is that P is not invariant as it is currently expressed, although it should be. I realize now that with the tangent motion of a particle, the local shell is not using some locally inferred radius r', but rather C' / (2 pi), where C' is the locally measured circumference of the shell at r. C' / (2 pi) ≠ r / L here of course, so with the distant observer inferring a circumference of the shell of C = 2 pi r, the local static observer with a tangent contraction of rulers of L_t, will physically measure C' = (2 pi r) / L_t by placing infinitesimal rulers end to end around the circumference, which is invariant. So to conserve the locally measured angular momentum of a particle, we would have p'_angular = p'_t r' = [m_p v'_t / sqrt(1 - (v'/c)^2)] (C' / 2 pi) = constant where from assumption B we gained sqrt(1 - (v'/c)^2) = z / K, so = m_p v'_t K ((2 pi r / L_t) / 2 pi) / z = m_p v'_t K (r / L_t) / z and upon dividing out the invariant m_p and constant of motion K, we gain another constant of motion P = v'_t r / (L_t z) which still works out to P = v'_t r / sqrt(1 - 2 m / r) as found by reverse engineering SC, but is now invariant. This is the new corrected equation for assumption C. I will have to go back to a couple of other threads where I got this far and then got stuck at this point also.
P: 406
 Quote by grav-universe Let me ask this. This Wiki link has the derivation for the Schwarzschild solution. For the 3 lines contained in the link under "Using the field equations to find A(r) and B(r)" and shown below, what is each saying physically? $$\rm{4 \dot{A} B^2 - 2 r \ddot{B} AB + r \dot{A} \dot{B}B + r \dot{B} ^2 A=0}$$ $$\rm{r \dot{A}B + 2 A^2 B - 2AB - r \dot{B} A=0}$$ $$\rm{- 2 r \ddot{B} AB + r \dot{A} \dot{B}B + r \dot{B} ^2 A - 4\dot{B} AB=0}$$
Well, since L_t / r is an invariant, and those equations should represent invariants, I tried simply replacing r with r / L_t in the equations. They all work out to zero with Schwarzschild, as they should anyway since L_t is just 1, but they also all work out to zero with GUC, where L_t is non-unity. GUC is the same as SC but all of the shells are moved uniformly closer to the center while keeping the same coordinate distance between the shells and cutting out the volume inside the event horizon, so it has the same value for L at any r as SC, so the same z and L for any shell, and so the same A and B, only represented with the conversion for r1 instead of r. Apparently then, L_t / r is the correction factor for A and B when changing coordinates systems of the form r1 = r - n m from SC, simply sliding the shells uniformly inward or outward, where n has any numerical value. They didn't work out for EIC for some reason though, not sure why. I need those same equations but including C for tangent length.
 P: 406 Ahah! Yay. :) I can now see where one of the solutions to the Ricci tensors comes from. From the relationship we found before m L_t^2 = (dz / dr) L r^2 which for convenience I will write dz / dr as just z' with second derivatives double primed. We can re-arrange to gain L_t^4 = z'^2 L^2 r^4 / m^2 The variables in the tensors are A = 1 / L^2 B = - z^2 and B' = d(-z^2) = - 2 z z' z' = - B' / (2 z) z'^2 = B'^2 / (4 z^2) = - B'^2 / (4 B) so we can rewrite the relationship once more to L_t^4 = - B'^2 r^4 / (4 m^2 A B) Finding the derivative for that using Wolfram, we get d(L_t^4) = d[- B'(r)^2 r^4 / (4 m^2 A(r) B(r))] = r^3 B' (r B A' B' + A(r B'^2 - 2 B (r B" + 2 B')) / (4 m^2 A^2 B^2) Now, if as a coordinate choice, we make L_t = 1, then its derivative is zero. So now we have (0) (4 m^2 A^2 B^2) / (r^3 B') = 0 = r B A' B' + A(r B'^2 - 2 B (r B" + 2 B')) r B A' B' + r A B'^2 - 2 r A B B" - 4 A B B' = 0 This is the same as the third Ricci tensor solution in the last post. :) Now I just need something similar to find the other two.
 P: 406 In terms of solving for C(r), for cases where C(r) is non-unity, with the metric of the form ds^2 = A(r) dr^2 + B(r) c^2 dt^2 + C(r) dO^2 r^2 {as Wiki has it but with c^2 drawn out of B} we would have C(r) = 1 / L_t^2, giving d[L_t^4] = d[1 / C^2] = - 2 C' / C^3 and d[- B'^2 r^4 / (4 m^2 A B)] = r^3 B' (r B A' B' + A(r B'^2 - 2 B (r B" + 2 B'))) / (4 m^2 A^2 B^2) from before, so - 2 C' / C^3 = r^3 B' (r B A' B' + A(r B'^2 - 2 B (r B" + 2 B'))) / (4 m^2 A^2 B^2) - 8 m^2 A^2 B^2 C' / (r^3 C^3 B') = r B A' B' + A(r B'^2 - 2 B (r B" + 2 B'))
 P: 406 Actually, here's a simpler one involving C. From the last post, we had 1 / C^2 = -B'^2 r^4 / (4 m^2 A B) 1 = -B'^2 C^2 r^4 / (4 m^2 A B) d(1) = 0 = d[-B'^2 C^2 r^4 / (4 m^2 A B)] = r^3 C B'(r B C A' B' + A(r C B'^2 - 2 B(r C B" + r B' C' + 2 C B'))) / (4 m^2 A^2 B^2) r B C A' B' + r A C B'^2 - 2 r A B C B" - 2 r A B B' C' - 4 A B C B' = 0

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