PAllen said:
If its origin is only partly geometric, as the paper argues using the cosmon field, then tidal gravity is no longer defined by geodesic deviation.
Ok, having looked at the paper some more, let me try to describe how this would have to work, based on the math in the paper.
In the standard cosmology model, the "spacetime geometry" term is ##R \sqrt{g}##, where ##R## is the Ricci scalar and ##g## is the determinant of the metric tensor. (I'm using the paper's notation, which does not appear to be concerned with the sign of ##g##; strictly speaking, the determinant ##g## is negative so the factor in the Lagrangian should be ##\sqrt{-g}## since that factor must be real, and that's how it will appear in most GR textbooks.)
In the paper's model, the corresponding term in the Lagrangian is ##\chi^2 R \sqrt{g}##, where ##\chi## is the "cosmon" field. So if we view ##R## and ##g## as describing, not the actual physical metric (the thing that describes distances and times measured with physical rods and clocks), but a "background" metric that is not directly observable, then the paper's description of the model as describing a static spacetime in which masses continually increase (because the "Planck mass" in the model continually increases with ##\chi##) makes sense, if we bear in mind that the "static spacetime" geometry is
not the spacetime geometry that is directly observed; for example, the timelike geodesics of this background spacetime are
not curves that describe the worldlines of objects with zero proper acceleration, and the curvature tensor derived from this background spacetime metric does
not describe actual deviation of the worldlines of actual physical objects that have zero proper acceleration.
However, that then leaves the obvious question: what
are the direct physical observables in the paper's model? As far as I can tell, this question is never directly addressed in the paper. We are never told, for example, how to describe the motion of a freely falling object--an object with actual, physical zero proper acceleration--in this model. What is its worldline? It isn't a geodesic of the "background" spacetime geometry, we know that. But we are never told what it
is, except for the general statement that the model "represents the same physics" as standard cosmology. But if the model represents the same physics, then all physical observables should be the same, and that includes the physical observables that increase without bound as the Big Bang singularity is approached.
The paper's explanation of how the singularity gets removed is that the curvature scalar ##R## of the "background" spacetime geometry remains finite and bounded as ##t \rightarrow - \infty## (which corresponds to ##t \rightarrow 0## in the standard cosmology). But as we have just seen, that curvature scalar does not represent any physical observable. We are never told what, in the paper's model,
does represent the relevant physical observable as ##t \rightarrow \infty## in the model, much less how such an observable remains finite and bounded--which, if the paper's claim that its model "represents the same physics" as standard cosmology is true, cannot be the case.
So I am still not convinced that the paper actually presents a solution that is free of
physical singularities. As far as I can tell, the "field redefinition" trick just obfuscates the situation by making the unphysical "background" spacetime Ricci scalar stay finite and bounded, and then trying to ignore the fact that this scalar does not represent the actual relevant physical observable.