Friedmann, "On the curvature of space" (1922) and "On the possibility of a world with constant negative curvature" (1924), chose a hyperbolic line element in which the spatial part was the hemisphere model equipped with a Riemann metric. There were two unknown parameters, R2 and M 2, that multiplied the space and time components of the metric, respectively, and were required to satisfy the Einstein field equations. The spatial part was supposed to represent a homogeneous and isotropic universe, with a (incomplete) reference given to Bianchi's "Lessons in Differential Geometry".
In a "non-stationary world", the parameters, R and M, were considered by Friedmann to be functions of time only. According to him, "[t]he problem becomes finding two functions R and M which satisfy the Einstein equations." He found that M could be set equal to one, "without loss of generality", leaving the radius of curvature, R, as the sole function of time. Does this have a meaning?
The "limit space" of the Riemann metric (embedded in an (n+1)-dimensional Euclidean space), ds2 = R2(dx12+...+dxn2+dy2)/y2, consists of boundary points where the denominator of the metric vanishes, y=0, implying that they are infinitely far from the interior. These points are known as "points at infinity". In the heated plane analogy, the edge, y=0, would be considered as infinitely "cold".
Although Robertson was well aware of the analogy with the 2D and 3D heat metrics [pp. 182ff, 192ff in his "Relativity and Cosmology"], yet, when it came to writing down his metric, he "split" the sole parameter, R, into two parameters: a constant "curvature index", k, which could take only 3 values (0,+1,-1), and the "radius of curvature", R, which now became the "radius of the universe", and a function of time. According to both Friedmann and Robertson (and about everyone else since then), R(t) had to satisfy Einstein's field equations. The "curvature" is now k/R2 [p.221 of Robertson's book]: a ratio of the curvature index and the expansion function of the universe. What kind of curvature is this?
Furthermore, what is the purpose of determining the time evolution of R(t) when we know already that the model is one of constant negative curvature in which the boundary is infinitely far away? If an observer is located in the plane or hemisphere, he (or she) will never reach the boundary (because he (or she) shrinks along with his (or her) measuring rods). So what is the meaning of R(t) tending to infinity, while the curvature index is still constant, k=-1? Does it say that the "curvature" tends to zero while still being hyperbolic with "constant" negative curvature?
The definitions of "curvature" do not finish here. Einstein's equations introduce two additional curvatures---the Ricci tensor and the scalar curvature---which are completely extraneous to the hyperbolic model of a uniform and homogeneous universe. From Einstein's equations comes the result that the (constant) density of matter of the universe is proportional to the "curvature", -1/R2, and that there exists a "critical density" for k=0!
All this brings to mind Zel'dovich's remark:
"If we start from the Friedmann model of the world, the state of the world can be characterized by the mean radius of curvature of space. The curvature of space is a local concept....Since in the Friedmann model of the world the radius changes in the course of time, the conclusion is drawn that the physical constants also change in the course of time. This pseudological view, however, cannot withstand criticism: the Friedmann solution has a constant curvature of space only when one makes the approximation of a strictly uniform distribution of the matter density!...a dependence of the constants on the local value of curvature would lead to great differences in the constants at the Earth's surface and near the Sun, and so on, and hence is in complete contradiction with experience."
Robertson -Walker generally allows an ideal homogenous universe with uniformly distributed mass as gas and dust in the positive curvature, or uniformly distributed negative energy in the case of negative curvature. Much of science does not embrace negative curvature by anything other than dark energy. My work is suggesting that electromagnetic radiation and local inertial fields contribute to negative curvature, and make dark energy unnecessary. There are substantial reasons for these opinions in the other metric solutions and in the way Lagrangian is used in predicting actions. RW only needs the two parameters. But care is needed in how the parameters are constructed.
Einstein equations are more general and recognize possibility of mass concentrations which require additional parameters.
Your reference to original work on negative curvature is helpful to my project, and helps to resolve arguments I had a few days ago. Much value is found in old references to original work of famous pioneers, because some of the best parts are not taught in college.
You are right about the complete lack of references to original papers. You've got to go back and read what the authors wrote and the logic (or lack thereof) they used.
Regarding the two parameters in the Friedmann solution, there is only k=-1 which can be obtained by introducing a scale factor [c.f. Barankin, "Heat flow and non-Euclidean Geometry" Am Math Monthly (1942) 4-14, or "A New Perspective on Relativity" Sec. 2.1.1]. The introduction of a matter density is not part of the model of the half-plane or the hyperbolic disc. It was Friedmann's desire to connect these models with the Einstein equations (an impossible mismatch). It is the same as adding the energy constraint in the permutation of balls among urns to obtain the Boltzmann factor. The energy constraint is completely extraneous to the permutation problem! [c.f. "Statistical Physics: A Probabilistic Approach"]
Dear Bernard Lavenda,,
I want to know your view on the lines 'Einstein's equations introduce two additional curvatures-the Ricci tensor and the scalar tensor'. Is it not the incompleteness of the Einstein's equations in the sense that, the full curvature tensor is missing in the Einstein's equations ? As a result, we are trying to add a quantity like cosmological constant, the variable constant of gravitation etc. in the Einstein's equations according to our suitability.
Regards
Dear Ramesh,
Rather, than being incomplete, the Ricci tensor is an average of the collection of Gaussian curvatures in the Riemann metric. (Remember, it was Riemann's idea to generalize Gauss's definition of curvature to more than two dimensions.)
How do we know that the collection of Gaussian curvatures is complete? We don't. For instance, if we want to determine only the radial metric in the Schwarzschild metric setting all changes in the angle variables equal to zero, we don't find any restriction to the radial coordinate (i.e. no black hole).
The measurement of Gaussian curvature begins with quadratic deviations from flat space so that the Riemann tensor depends on second derivatives of the metric. Said differently, curvature will depend on quadratic, and higher, variations. This is why Einstein's pseudo-tensor of the gravitational field has nothing to do with curvature since it only depends on first derivatives of the metric, and if curvature is related to mass-energy, then it has nothing to do with it.
In the Schwarzschild solution, the Gaussian curvatures in the different directions are non-vanishing. What is vanishing is their sum, the Ricci tensor. If you contract the Ricci tensor the scalar curvature vanishes. But, if you contract the Ricci tensor with a unit timelike vector, by Einstein's equations you get the spatial scalar curvature which is non-vanishing. And in terms of Einstein's equations, this is equal to the energy density. This is often stated as energy-mass warps space and is responsible for curvature.
But, the space scalar curvature of the Schwarzschild metric is non-zero whereas the total scalar curvature vanishes. The energy-stress tensor in the Schwarzschild solution is identically zero. So what then is the relation between curvature and mass-energy?
Kind regards,
Bernard
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