On Thursday, August 13, 2020 at 7:51:55 PM UTC-5, Savin Beniwal wrote:   
   > T_mn is energy density   
      
   No it's not. And that's where your mistake ultimately comes from. T_mn is   
   *not* even a tensor density at all, but a tensor. The correct statement is   
   that it is #T_mn = root(|g|) T_mn is the density; and it is this which has the   
   various densities (mass/   
   energy density, pressure) as its components; not T_mn.   
      
   Second, the densities you're thinking of (those that are associated with the   
   transport laws for mass, momentum and energy) come out of the (1,1) tensor   
   density, not the (2,0) tensor density. It is #T^0_0 which is mass-energy   
   density, not #T_00. The    
   native form of that tensor density (before any metric is applied to raise or   
   lower indexes) is (1,1), not (2,0); this corresponding to the tensor that   
   gives the (N-1)-current associated with the local diffeomorphism given by the   
   vector field X = X^m @_m:   
      
   T_X = boundary-integral (#T^m_n X^n) (@_m _| d^N x)   
      
   where @_m = @/@x^m and _| is the contraction operator and d^N x the N-form   
      
   d^N x = dx^0 ^ dx^1 ^ ... ^ dx^{N-1}   
      
   So, let's go back over this in further detail ... and generalize it to N   
   dimensions ... because the factor 8 pi is also wrong, except for N = 4, and   
   that also needs to be brought up. Oftentimes the literature glibly keeps 8 pi   
   in when generalizing to    
   higher dimensions, forgetting about where it came from - people like Mansouri   
   have pointed out the error and provided the correct analysis, which I'll   
   replicate below   
      
   On Thursday, August 13, 2020 at 7:51:55 PM UTC-5, Savin Beniwal wrote:   
   > 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.   
      
   Everything has dimensions - including even things that "don't". Actually, it   
   is *my* analysis, not yours, which is independent of what choice of dimensions   
   one ascribes to the coordinates, and covers all possibilities in a single   
   universal framework -    
   which also happens to subsume yours. Yours is heavily-dependent on the   
   particular choices (already accounted for my mine) and so sheds no additional   
   light, but only obscures the light already shed.   
      
   Your choice corresponds to [g_mn] = 1, "no dimension" meaning "dimensionless"   
   and [x^m] = 1 = [x^n]. That's a special case of [g_mn dx^m dx^n] = A, [x^m] =   
   [m] and [x^n] = [n] corresponding to the selection A = 1, [m] = 1 = [n].   
      
   But your assumption is wrong: coordinates *do* have dimensions: space-like   
   coordinates are generally taken with dimension L, time-like with T. But it   
   doesn't matter, since my analysis doesn't depend on how they're set, though   
   yours did. (Indeed, that's    
   part of what's proven by the analysis - so it doesn't need to be assumed up   
   front, as you erroneously thought it had to be).   
      
   (It also contradicts what you later said below, which amounts to the choice   
   [0] = [1] = ... = [N-1] = L and A = L^2).   
      
   In general, as per the analysis the dimensions of [g_mn] are A/[mn], while   
   those of the inverse metric are [g^mn] = [mn]/A (which - once again - includes   
   your analysis as a special case and subsumes it).   
      
   The coefficient kappa is given first and foremost by the action principle via   
      
   S = integral 1/(2 kappa) R root(|g|) d^N x   
      
   analyses involving the stress tensor are not actually relevant at all and have   
   no say on the final outcome. Quite the opposite, it's the other way around:   
   the stress tensor too comes out of the action principle as follows.   
      
   The right hand side of Einstein's equations are determined from the action   
   principle by variation with respect to g^{mn}.   
      
   #T_{mn} = -2 @(#L)/@g^{mn};   
   @ = partial derivative operator   
      
   where #L and #T are tensor DENSITIES, #L being the Lagrangian density,   
   normally written L root(|g|).   
      
   The units for #T are given directly in these terms:   
      
   ML^2/T = [S] = [g^{mn}] [#T_{mn}] [d^N x] = [mn]/A [#T_{mn}] [01...(N-1)]   
      
   hence   
      
   [#T_{mn}] = A ML^2/(T [01...(N-1)] [mn])   
      
   With the normal choices A = L^2, [0] = T, [1] = [2] = [3] = L, this works out   
   to [#T_{mn}] = ML/T^2 1/[mn] - which yield the expected result for the entire   
   tensor density [#T_{mn} dx^m (x) dx^n] = ML/T^2.   
      
   (x) denote the tensor product operator (circle with x inside).   
      
   When the indices are put back in native form, this yields   
      
   [#T^m_n] = [g^{mr}] [#T_{nr}] = [mr]/A A ML^2/(T[01...(N-1)][nr])   
   or   
   [#T^m_n] = ML^2/(T[01...(N-1)]) [m]/[n]   
      
   For the energy density (m=0=n) and pressure (m=n=1,...,(N-1)), this produces   
   the result ML^{3-N}/T^2 or for N = 3, M/(LT^2), as expected. This is the same   
   dimensions as for the entire tensor density:   
      
   [#T^m_n @_m (x) dx^n] = ML^2/(T[01...(N-1)]).   
      
   > The scalar curvature (R) is a contraction of the Ricci tensor   
      
   with [R] = 1/A, as I already showed; with [R_mn] = 1/[mn].   
      
   The contract changes the dimension, except in the case where [0] = [1] = ... =   
   [N-1] = root(A). Your "doesn't change the dimension" is the special case   
   corresponding to A = 1 (or later, below, to A = L^2), so is already subsumed.   
      
   > all terms (R, R_mn) are the dimension of lenght   
      
   which contradicts what you earlier stated that all the quantities have "no   
   dimensions" (i.e. are dimensionless). What you're now saying here - *this*   
   statement - corresponds to the choice:   
      
   A = L^2, [0] = [1] = ... = [N-1] = L.   
      
   With *that* choice   
      
   [#T_{mn}] = ML^{2-N}/T = [#T^m_n]   
      
   while with your previous choice (A = 1, [0] = [1] = ... = [N-1] = 1):   
      
   [#T_{mn}] = ML^2/T = [#T^m_n].   
      
   > And hence, [G_mn]=1/L^2; [Lambda]=1/L^2, and [kT_mn]=1/L^2.   
      
   which is the special case of the analysis I provided:   
      
   [G_mn] = 1/[mn] = [kappa T_mn].   
      
   corresponding to your choice [m] = L = [n].   
      
   > T_mn is energy density of dimension ML^2T^(???)/L^3   
      
   which is wrong: it's not a density at all, #T_mn is.   
      
   In fact, I did the same analysis as you did long ago and ended up making that   
   same mistake resulting in the same wrong conclusion with c^4 instead of c^3.   
      
   #T_mn is the density, not T_mn.   
      
   The dimensions of T_mn are actually given as follows   
      
   [#T_{mn}] = A ML^2/(T [01...(N-1)] [mn])   
   [root(|g|)] = A^{N/2}/([01...(N-1)])   
      
   hence   
      
   [T_mn] = A^{1-N/2} ML^2/(T [mn])   
      
   For N = 4 with the choice A = L^2, this yields   
      
   [T_mn] = M/(T[mn])   
      
   Thus   
      
   1/[mn] = [G_mn] = [kappa] [T_mn] = M/(T[mn])   
      
   [kappa] = T/M   
      
   And since [G] = L^3/(MT^2) and [c] = L/T, this shows that   
      
   [kappa] = [G/c^3], not [G/c^4].   
      
   In general   
      
   1/[mn] = [G_mn] = [kappa] A^{1-N/2} ML^2/(T [mn])   
      
   or   
      
   [kappa] = T A^{N/2-1}/(ML^2).   
      
   The N-dimensional gravitational coefficient, which we'll call G', is that   
   which yields an inverse r^{N-2} field for a point-like source of mass M of the   
   form   
      
   g = G' M/r^{N-2}   
      
   and so has dimensions   
      
   [G'] M/L^{N-2} = L/T^2   
   or   
   [G'] = L^{N-1}/MT^2   
   or   
   [G'/c^3] = L^{N-4}T/M   
      
      
   [continued in next message]   
      
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