When did spacetime form?

[KJW]

10 hours ago, joigus said:

Symmetry of partial derivatives is a consequence of continuity

I have a question for you:

Suppose I have a metric that is pathological in some way. But there exists a coordinate transformation that is everywhere continuous and invertible from this metric to the Minkowskian metric. However, due to the pathological nature of the metric, the Riemann tensor is not everywhere zero. Is the spacetime described by the metric flat?

 

[MigL]

Are we still talking Physics, or pure Mathematics ?

My second year Math prof, Dr Srivastava, an expert in partial differential equations, used to say Mathematicians dislike Physicists; they impose too many 'roadblocks' on our beautiful Maths.

[Markus Hanke]

15 hours ago, joigus said:

Mm... Yes, you're right. Genady actually said it (if I understood correctly). That, as Gμν=8πGTμν , any Gμν obtained from the metric connection by means of the gμν you give yourself can be used as the Tμν just after dividing by 8πG .

Exactly. Which of course is obvious, if you think about it…

4 hours ago, KJW said:

But there exists a coordinate transformation that is everywhere continuous and invertible from this metric to the Minkowskian metric.

For this transformation to be a valid diffeomorphism, it needs to be differentiable (in addition to being invertible). If it’s only continuous but not differentiable, then I think this spacetime isn’t flat. 

[KJW]

3 hours ago, Markus Hanke said:

For this transformation to be a valid diffeomorphism, it needs to be differentiable (in addition to being invertible). If it’s only continuous but not differentiable, then I think this spacetime isn’t flat. 

Ok. This would appear to resolve the inconsistency I attempted to create by having a metric that can be coordinate-transformed to the Minkowskian metric but with a Riemann tensor that is not everywhere zero. I was thinking that a choice had to be made between the coordinate transformation and the Riemann tensor.

However, because the metric:

[math]\text{diag}(−1, 1, 1, 1 + H(z) z^2 + H(−z) z^4)[/math]

is everywhere continuous and invertible, the coordinate transformation to the Minkowskian metric is a diffeomorphism.

 

[Markus Hanke]

9 hours ago, MigL said:

Are we still talking Physics, or pure Mathematics ?

I think the main message is that spacetime can, under certain circumstances, appear pathological in some way, eg for certain observers, yet still remain physically reasonable. That’s an important point actually.

The reverse is true also - a metric may appear reasonable, yet still not necessarily be physically possible.

[joigus]

 

22 hours ago, KJW said:

But, we are not dealing with mixed partial derivatives. So, even if Rijkl=−Rjikl doesn't hold in general due to discontinuity, Rzzzz=−Rzzzz=0 must still hold.

 

13 hours ago, KJW said:

I have a question for you:

Suppose I have a metric that is pathological in some way. But there exists a coordinate transformation that is everywhere continuous and invertible from this metric to the Minkowskian metric. However, due to the pathological nature of the metric, the Riemann tensor is not everywhere zero. Is the spacetime described by the metric flat?

 

Ok. For the record, I think you're right that Genady's metric just describes flat spacetime, and that this argument that I'm giving is really clutching at straws. I was exposed to Schwarz's/Clairaut/Young theorem in the past and I remembered that if a function is not C² (never mind the only non-continuous derivatives being the diagonal ones) the conditions of the theorem are no longer satisfied and it could happen that some devilish argument gives you a contradiction. I just don't know. Thereby my question "are you sure?" which I should have formulated more clearly. My intuition tells me that f_xy and f_yx could give you problems, but f_xx and f_xx is the same thing no matter what the continuity status is, because x is the same thing as x, as you rightly imply.

Another part that makes me think that you're right even from the POV of a rigorous proof is that the distributions involved are Heaviside's, the delta, and its derivative. And those are extraordinarily well-behaved as long as you never integrate them against functions that don't fall to zero fast enough at infinity. All functions we use in physics fall to zero pretty fast, so no problem there.

As to your question for me, if by "pathological" you mean "not continuous", it's not possible to have a composition of two functions, one continuous, and the other not, that gives you a continuous function. So my answer would be no.

What does happen sometimes is that, trying to solve EFE, we get bad coordinate systems, and we must use singular coordinate transformations that mend the "singularities" that were never there in the first place. I do know that that's what happens with the Kruskal-Szekeres coordinates. After all, the composition of two singular transformations can restore continuity, exactly as 1/u (singular at u=0) with u=1/x (singular at x=0) becomes just x (continuous everywhere).

2 hours ago, Markus Hanke said:

I think the main message is that spacetime can, under certain circumstances, appear pathological in some way, eg for certain observers, yet still remain physically reasonable. That’s an important point actually.

It indeed is.