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Logic of GR as mathematically derived 
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#37
Dec412, 03:39 PM

P: 429

Well sure, you could do it like that too I suppose, but only if you like doing things the easy way.



#38
Dec812, 10:59 AM

P: 429

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. 


#39
Dec2912, 09:11 AM

P: 429




#40
Jan113, 06:16 PM

P: 429

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 rearrange 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. 


#41
Jan113, 07:21 PM

P: 429

In terms of solving for C(r), for cases where C(r) is nonunity, 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')) 


#42
Jan113, 09:32 PM

P: 429

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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