From: darshanbeniwal11@gmail.com   
      
   On Wednesday, 12 August 2020 14:20:49 UTC+5:30, rockbr...@gmail.com  wrote:   
   > I see, for the coupling coefficient for Einstein's field equations,   
   > different things in the literature even though one choice appears   
   > natural and obvious.   
   >   
   [[Mod. note -- Quoted text trimmed. -J.T.]   
   >   
   > Nobody, that I am aware of, ever takes the line element with   
   > dimensions A = LT. So k = +/- 8 pi G/c^4 makes no sense at all. The   
   > natural choices are +/- 8 pi G/c^3 with A = L^2 and the line element   
   > being regarded as a measure for spatial distance; or +/- 8 pi G/c^5   
   > with A = T^2 and the line element being regarded as a measure of   
   > proper time.   
   >   
   > Of these two, the only one consistent with the condition   
   > [Lambda] = 1/L^2 is A = L^2, |eta_{00}| = c^2 and k = +/- 8 pi G/c^3.   
      
   First of all, as per my knowledge This dimensional analysis   
   does not depend on the system of coordinates neither on the metric tensor as   
   g_mn has no dimension.   
      
   So the Einstein field equation is G_mn+lambda*g_mn=kT_mn. And we   
   have to calculte the dimension of k. As G_mn is already defined as   
   R_mn-1/2R. The scalar curvature (R) is a contraction of the Ricci   
   tensor (R_mn). The units do not alter through a contraction. The   
   Ricci tensor is therefore a contraction of the Riemann tensor and   
   all terms (R, R_mn) are the dimension of lenght (IInd derivative   
   of field).   
      
   [[Mod. note -- Line-wrapping adjusted so equations aren't broken   
   across lines. Alas, there were garbled 8-bit characters whose meaning   
   wasn't clear; I've replaced them with "???". -J.T.]   
      
   And hence, [G_mn]=1/L^2; [Lambda]=1/L^2, and [kT_mn]=1/L^2.   
   T_mn is energy density of dimension ML^2T^(???)/L^3, which   
   implies   
   [k]=(L^3/(MT^(-2)L^2))*(1/L^2)-->[k]=T^2/(ML) ...............(1)   
      
   On the other hand, [G]=L^3/(MT^2) and   
   [c]=L/T---->[G/c^4]=(L^3/(MT^2))/(T^4/L^4)   
          ---->[G/c^4]=T^2/(ML) .............(2)   
      
   From Equ (1) and (2), we can conclude that only k=8*pi*G/c^4 consistant   
   with the  Einstein Field equation even with the Einstein-Hilbert Action.   
      
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