I see, for the coupling coefficient for Einstein's field equations, different   
   things in the literature even though one choice appears natural and obvious.   
      
   The one seen in numerous Wikipedia articles and on a wall in Leiden uses c^4.   
   I have a problem with that, and I'll explain what's wrong with it.   
      
   The Einstein-Hilbert action is, up to a proportionality, the integral of R   
   root(|g|) d^4 x, where g is the determinant of the metric, and R the curvature   
   scalar associated with the metric; the volume element being d^4 x = dt ^ dx ^   
   dy ^ dz when using    
   local coordinates with t for time, (x,y,z) for space. The proportionality is   
   inverse to the gravitational coupling, k.   
      
   Let M, L, T denote the dimensions respectively of mass, length and time; so   
   [x] = [y] = [z] = L, [t] = T, [d^4 x] = L^3 T and [action] = ML^2/T. Assume   
   for the line element [g_{mn} dx^m dx^n] = A, where x^0 = t, x^1 = x, x^2 = y,   
   x^3 = z, where A is left    
   to be determined.   
      
   Denoting the dimensions [x^m] = [m] for brevity (with [0] = T, [1] = [2] = [3]   
   = L), then for the metric, [g_{mn}] = A/[mn] and inverse metric [g^{mn}] =   
   [mn]/A. Thus, [g] = A^4/[0123]^2 and [root(|g|) d^4 x] = A^2.   
      
   For the connection coefficient [Gamma_{mnr}] = A/[mnr], and [Gamma^m_{nr}] =   
   [m]/[nr], since Gamma^m_{nr} = Gamma_{snr} g^{ms} is raised with the inverse   
   metric. (We're using the summation convention here and below).   
      
   Thus [R^s_{rmn}] = [s]/[rmn] and [R_{mn}] = 1/[mn], where R_{mn} =   
   R^s_{msn}.   Finally, [R] = 1/A, since R = g^{mn} R_{mn} is contracted with   
   the inverse metric.   
      
   Combining, we get [integral R root(|g|) d^4 x] = A. So, in order for this to   
   produce an action, the proportionality has to be ML^2/AT and the coupling   
   coefficient (which is inverse to this) is [k] = AT/ML^2.   
      
   As an aside, Einstein's equation with the cosmological coefficient Lambda is:   
   G_{mn} + Lambda g_{mn} = k T_{mn}, where T_{mn} is the stress tensor. Since   
   G_{mn} = R_{mn} - 1/2 g_{mn} R, then [G_{mn}] = 1/[mn]. Consequently, for the   
   cosmological    
   coefficient, we must have [Lambda] = 1/A. Normally, it is taken to have   
   dimensions [Lambda] = 1/L^2, which requires A = L^2.   
      
   For the Minkowski metric [eta_{mn}] = A/[mn], we have [eta_{00}] = A/T^2.   
   Combining this with the Newton gravitational coefficient [G] = L^3/MT^2, and   
   light speed [c] = L/T, this yields [eta_{00} 8 pi G/c^5] = AT/ML^2 = [k]. And   
   we assume that |eta_{00}|    
   is a power of c. Therefore,   
      
      k = +/- eta_{00} 8 pi G/c^5.   
      
   For k = +/- 8 pi G/c^n, if n = 3, 4 or 5, then we have:   
      
   (A, [Lambda], |eta_{00}|) = L^2, 1/L^2, c^2, for n = 3;   
   (A, [Lambda], |eta_{00}|) = LT, 1/LT, c, for n = 4; or   
   (A, [Lambda], |eta_{00}|) = T^2, 1/T^2, 1, for n = 5.   
      
   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.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)