This I didn't know, and this hasn't been evident to me so far from what I've read in the texts. So thanks for clarifying this.
This is also useful and also wasn't evident to me. The book quote I gave in the OP seems to imply an equivalence between atlas and differentiable structure. I...
So then the wrong assumption on my part is that there should be a global differentiable structure for the entirety of the spacetime manifold? Based on what you and @pasmith said, here's what I'm thinking and you can correct me wherever I'm wrong:
Whole spacetime is modeled as a 4D...
Thank you! So then physics-wise, the problem we're trying to solve or the object we're trying to study determines what metric we'll use - so indirectly it will determine the differentiable structure?
I say the above since I've heard different kinds of metric being used to study different kinds...
I'm studying Liang's book on differential geometry for general relativity. There's a para in it talking about a maximal atlas - "Later on, when we talk about a manifold, we always assume that the largest possible atlas has been chosen as the differentiable structure, so that we can perform any...
Thank you. One follow-up thing though - you mentioned that objects in free fall, initially at rest w.r.t. each other, won't remain at rest w.r.t. each other in the presence of "real" gravity. In other words, real gravity causes geodesic deviation that is equivalent to curvature.
But then...
Thanks! If you don't mind, I want to clarify another thing.
I'm trying to understand the motivation of modeling gravitation using spacetime curvature. Here's my limited understanding:
"Fake" gravitational field can be made to vanish using a global coordinate transformation or can be introduced...
So be it - obviously I'm not wrong on purpose. Maybe I'm not too bright? It obviously is useful to me to be corrected so that I figure it out eventually (even if it takes me time). Not everyone is the same and what seems obvious to one may not be to another. At the end of the day, I gain...
It's clear to me at this point. But you have to understand there's a considerable difference between our knowledge levels, and maybe even aptitude since I probably have less of it. Whenever I'm restating something in a different way - it's not because I want to contradict you and others.
It's...
Makes sense. So that means the statement: What Einstein really meant was that small objects, for a small length of time, cannot tell the difference between a gravitational field and an accelerated frame of reference.
is exactly, mathematically correct.
Right, and that's why I got tripped up by the statement that an observer in an infinitesimally small, closed laboratory won't be able to tell the difference whether they're on earth or in an accelerated room in free space.
If they have a way of physically measuring the 2nd-order derivatives...
Thank you! And I see, so practically as Peter said, in an infinitesimally small region of spacetime this experiment will not yield any discernable deviation in the earth gravitation field vs. accelerating in free space cases.
But as a purely thought experiment, if a tiny observer in a tiny...
My bad, I'm being a bit sloppy with my statements.
Is there a way to physically measure these second order effects of difference between the metric and Minkowski, and its first derivative? If so, then in this scenario:
The observer can measure those second order effects, and be able to deduce...
First of all thanks a lot for the above post - it's very helpful.
So even in a tiny region of spacetime, the metric tensor itself and its first derivative won't vanish completely - they'll just be so small that they can be practically ignored.
I'm picturing a "local reference frame at some...
Thanks! So in an infinitesimal neighborhood of an event in curved spacetime, we can diagonalize the metric and set its first derivatives to zero. But we can't make its second derivatives zero. In other words, "locally flat" (only metric is diagonalized) isn't the same as "absence of curvature"...