Properties of the Scale Factor

[Arman777]

Does scale factor must be continuous(Con.) and differentiable (Diff.) ? Or can it be one of them or neither ? Physically one expects it to be Con. and Diff. but is there a more rigorous proof.

And as a separate question, if $\dot{a}$ is not continuous/differentiable in some case, does that mean $a$ is also not continuous ?

 

[PeterDonis]

Does scale factor must be continuous(Con.) and differentiable (Diff.) ? Or can it be one of them or neither ? Physically one expects it to be Con. and Diff. but is there a more rigorous proof.

Assuming you mean the scale factor $a(t)$ in the standard FRW metric, yes, it is a continuous and differentiable function of the coordinates (actually of $t$ only in the standard FRW coordinate chart, but that is immaterial for the question you're asking).

As far as "rigorous proof" goes, the function $a(t)$ arises naturally from solving the Einstein Field Equation with appropriate assumptions (homogeneity and isotropy of the metric), and any solution of a differential equation must consist of continuous and differentiable (more precisely, differentiable to at least the degree of the differential equation--the EFE includes second derivatives so any solution must consist of functions that are at least twice differentiable) functions of the coordinates.

 

[Arman777]

I see, thanks.

 

[PAllen]

Assuming you mean the scale factor $a(t)$ in the standard FRW metric, yes, it is a continuous and differentiable function of the coordinates (actually of $t$ only in the standard FRW coordinate chart, but that is immaterial for the question you're asking).

As far as "rigorous proof" goes, the function $a(t)$ arises naturally from solving the Einstein Field Equation with appropriate assumptions (homogeneity and isotropy of the metric), and any solution of a differential equation must consist of continuous and differentiable (more precisely, differentiable to at least the degree of the differential equation--the EFE includes second derivatives so any solution must consist of functions that are at least twice differentiable) functions of the coordinates.

This is all true, but there are common GR usages that don't follow this model. Instead, they take (M,g) - manifold and Lorentzian metric, to be generally a solution, with the field equations holding only where sufficient differentiability exists. This is not perverse, it is done to 'glue' solutions together in computationally tractable way, with understanding there are hopefully 'small' unphysicalities in the result. The Israel junction conditions allow the Einstein tensor to have discontinuities across a boundary, as long as specified conditions on G (Einstein tensor) are satisfied at the boundary, and that the metric first derivatives are continuous.

 

[PeterDonis]

The Israel junction conditions allow the Einstein tensor to have discontinuities across a boundary, as long as specified conditions on G (Einstein tensor) are satisfied at the boundary, and that the metric first derivatives are continuous.

This is true, but note that it does not lead to any discontinuity or lack of differentiability of the scale factor at the boundary, for the case where different FRW regions are "glued" together in this way. (Our best current model of our universe does this, as I understand it, at the end of inflation; it models that event as a discontinuous change in the stress-energy tensor, or more precisely in the equation of state that relates the pressure terms to the energy density--the latter is continuous at the boundary.)

 

[PAllen]

This is true, but note that it does not lead to any discontinuity or lack of differentiability of the scale factor at the boundary, for the case where different FRW regions are "glued" together in this way. (Our best current model of our universe does this, as I understand it, at the end of inflation; it models that event as a discontinuous change in the stress-energy tensor, or more precisely in the equation of state that relates the pressure terms to the energy density--the latter is continuous at the boundary.)

Agreed, just pointing out that your derivation precludes such composite solutions. I think even second derivatives of the scale factor must be continuous because continuity of G is required orthogonal to a junction (roughly speaking), and then isotropy comes into play.

 

[PeterDonis]

your derivation precludes such composite solutions

It doesn't preclude them, I just didn't mention them. In a composite solution, what I described is what would apply to the interior of each region that is being "glued" together, and the junction conditions would apply at the boundaries between the regions.

Note also that it's not really correct to say the field equations "do not hold" at the boundaries. The junction conditions are derived using the field equations. See, for example, the discussion in section 21.13 of MTW.