Error below :grumpy: For a couple of years now, I have been attempting to solve for the values in GR of the time dilation z and the radial and tangent length contractions, L and L_t respectively, which form the metric [tex]c^2 dτ^2 = c^2 z^2 dt^2 - dr^2 / L^2 - d_θ^2 r^2 / L_t^2[/tex] (along a plane) using only two main principles in GR that give rise to the single equation [tex]m L_t^2 / r^2 = z' L[/tex] as demonstrated in this thread. With this equation, we can make a coordinate choice for one of the variables and find a relationship between the other two, or likewise, make a coordinate choice for the relationship between two of the variables and solve precisely for the third, but we are still left to determine the precise values between the two variables we only have a relationship for. I gave up a few times, thinking it impossible to solve precisely for all three variables without applying an additional principle which would give another mathematical relationship between them. I tried various methods, attempting to apply relationships involving circular orbits, angular momentum, the equivalence principle, etc., all in vain and usually just adding even more unknown variables to determine. Einstein applied field equations, but I wanted to find and use something much simpler, some basic principle that would solve for all three. As it turns out, I may have stumbled across just the principle I need to solve for all three precisely, the low Newtonian approximation. The word 'approximation' threw me off for a long time as I wanted precise values and so I was reluctant to use it as a guiding principle, but it appears it may have indeed provided me with the final relationship I have been searching for. Finding the derivatives for the previous equation produces the 00 component of the Ricci tensor as also demonstrated in that thread. If one prefers, we can solve for the three values starting with only the 00 component of the Ricci tensor and reverse engineer it to gain the previous relationship, which can also be demonstrated simply by finding the indefinite integrals of the 00 component which directly gives us the previous equation along with an unknown constant which we find to be m in the low Newtonian approximation. So applying the low Newtonian approximation has already helped us. Let's see what else it can do. The energy relationship in GR as shown in the thread is [tex]z / sqrt(1 - (v/c)^2) = k[/tex] where k is a constant for all r. In the low Newtonian limit as r approaches infinity, for an object falling from rest at r = infinity, we have v^2 = 2 m c^2 / r. Applying this directly to the energy relationship, we would have z = 1 and v = 0 at r = infinity, so K = 1, giving [tex] z = sqrt(1 - 2 m / r) [/tex] Now, we could apply this directly to GR as a coordinate choice for the time dilation at any r, in other words by placing each spherical shell with an invariant value for the time dilation z at the corresponding r defined by the equation, and we can then gain a relationship between L and L_t by doing so, but we cannot solve for each of the contractions separately, whereas we want to solve for all three. So instead, we will define the relationship between z and r as [tex] z(r) = sqrt(1 - 2 m f(r) / r)[/tex] where f is some unknown function of r. This allows z to have any arbitrary relation to r that we choose by using different functions for f, so we have not yet applied a coordinate choice. It is close to the actual relationship in the low Newtonian limit, but all we can say at this point is that f must work toward unity at r = infinity, as must also z, L , and L_t. So we now have four variables to solve for, seemingly making things worse, but let's go with it. By introducing f, we still have a coordinate choice available, so whereas L_t = 1 is generally applied, we will instead use z = L. We assume that this is possible, that whatever the value for z at a particular shell, we can place the shells at each r in such a manner that z = L. It can also be demonstrated weakly how this can be performed, by beginning randomly with a spherical shell at a random r, and then placing consecutive spherical shells both toward r = 0 and r = infinity such that the invariant locally measured distance between two shells is L smaller according to the distant observer, placed such that L is equal to the invariant time dilation z of a particular shell. The entire map can then be slid inwards or outwards together at will without affecting the values of z and L since the distances between the shells remain the same so L remains the same and z is invariant for each shell. Only the value of L_t at each shell will be affected, and we will slide the map to the place where L_t is precisely unity at r = infinity, no more and no less. Okay, so here we go. So far we have made the coordinate choice [tex]z = L = sqrt(1 - 2 m f / r)[/tex] WolframAlpha gives the derivative of z, then, as [tex]z' = m (f - r f') / (r^2 sqrt(1 - 2 m f / r))[/tex] and applying that to our previous relationship, we gain [tex]m L_t^2 / r^2 = m (f - r f') / r^2[/tex] [tex]L_t^2 = f - r f'[/tex] and the derivative of that gives [tex]2 L_t L_t' = - r f"[/tex] Starting back with the original relationship again, we gain [tex]L_t^2 = z z' r^2 / m[/tex] and taking the derivative of that, [tex]2 L_t L_t' = r (r z'^2 + z (r z" + 2 z')) / m[/tex] So we now have the new relationship, [tex]- r f" = r (r z'^2 + z (r z" + 2 z')) / m[/tex] [tex]f" = -(r z'^2 + z (r z" + 2 z')) / m[/tex] Taking the indefinite integral for that using WolframAlpha also, we get ERROR HERE. Wolfram found for the integral of m rather than r for some reason [tex]f' = - log(m) (r z'^2 + z (r z" + 2 z')) + j[/tex], where j is some constant and again to find f, [tex]f = j r - r log(m) z z' - log(m) z^2 / 2 + p[/tex], where p is another constant Starting back with what we originally had for z, [tex]z = sqrt(1 - 2 m f / r)[/tex] [tex]2 m f / r = 1 - z^2[/tex] [tex]f = r (1 - z^2) / (2 m)[/tex] and then substituting, [tex]r (1 - z^2) / (2 m) = j r - r log(m) z z' - log(m) z^2 / 2 + p[/tex] and plugging this equation into Wolframalpha, it finds the solution for z to be [tex]z = sqrt(1 + (c1 / r) e^(r / (m log(m))) + m (1 - 2 j m) log(m) / r - 2 m (j r + p) / r)[/tex], where c1 is yet another constant Now, we want z = 1 precisely at r = infinity. (c1 / r) e^(r / (m log(m))) will become infinite at r = infinity unless c1 = 0, so c1 = 0 and then for any finite value of r, this part drops out and we are left with [tex]z = sqrt(1 + m (1 - 2 j m) log(m) / r - 2 m j - 2 m p / r)[/tex] The value m(1 - 2 j m) log(m) - 2 m p is all constant, so to simplify we will denote it with the variable q, now giving [tex]z = sqrt(1 + q / r- 2 m j)[/tex] At r = infinity, this equation will reduce to [tex]z = sqrt(1 - 2 m j)[/tex] where 2 m j is a constant whereas we want z to be precisely unity, so j = 0. This leaves us with the equation [tex]z = sqrt(1 + q / r)[/tex] and we originally had [tex]z = sqrt(1 - 2 m f / r)[/tex] so [tex]q = 2 m f[/tex] [tex]f = q / (2 m)[/tex] q / (2 m) is a constant, which makes f constant also, so f' = 0. Going back to our original equation for L_t^2, then, we gain [tex]L_t^2 = f - r f'[/tex] [tex]L_t^2 = f[/tex] Since we have determined that f is a constant for all r, then L_t is a constant also for all r when z = 1 at r = infinity. We want L_t = 1 at r = infinity as well, so if L_t = 1 at r = infinity and L_t is a constant for all r, then L_t = 1 for all r. We have now solved for the variable L_t. Now let's solve for the other two. As our coordinate choice, we had [tex]z = L = sqrt(1 - 2 m f / r)[/tex] and we just gained [tex]L_t^2 = f = 1[/tex] giving [tex]z = L = sqrt(1 - 2 m / r)[/tex] We now have exact solutions for all three variables :) Of course, everybody check my math. I just found this solution today and was so excited I had to go ahead and post it. I've gone back through it a couple of times though and it seems pretty straightforward and error free. I never would have thought, however, that all three variables could be determined without applying some other underlying principle, as determined by the full Einstein field equations or otherwise, rather than just using the low Newtonian approximation. Yet apparently, the best that I can tell so far anyway, they can.