5 ms·

I don't think this works in GR the way you think it works (regardless of what 'physical circle' means), aren't you always able to change coordinates for the spa

I don't think this works in GR the way you think it works (regardless of what 'physical circle' means), aren't you always able to change coordinates for the space to be locally flat? (This always confuses me, can someone that knows GR shed some light on this? I would greatly appreciate it. I've been trying to teach myself GR a while back from the lectures of Frederic Schuller and Alex Flournoy that are available on youtube but it's hard without being able to ask questions)

krastanov 5y ago
"Locally flat" means flat only very close to the point of consideration. That is the mathematical definition, the "practical" definition is "flat only as long as you do not stray further away than some characteristic measure of distance defining your problem and accuracy requirements". So as long as your circle is very small then the " experimentally measured" pi and the math pi are the same. The moment your circle becomes big enough to pass through interesting (measurable) gravitational effects, the experimental value will be different. If anything, experimentally measuring that difference is one way to experimentally measure spacetime curvature.
- ithinkso 5y ago
  That's why I asked about GR specifically, to have a common framework to work with. Without it you can say words like experimentally, practical and talking about differences between different pis whatever that means but this will only lead to misunderstandings at best and nonsense at worst
shadowgovt 5y ago
Not in accelerating frames of reference, which includes both rotating frames (such as spinning objects) and frames with high gravitational distortion. Spinning black holes have a measured circumference that is not 2 x pi x r. (Consequences of the accelerating frame of reference include the spaghettification in a black hole... We can't use a mathematical conversion to go to a frame of reference where the subject isn't being spaghettified).
- ithinkso 5y ago
  Why not? (to the first sentence, locally is locally) But I think I have found the source of my confusion in my original post, I was thinking about manifolds only in the topological sense and for some reason forgot that when there is a metric then we have to preserve distances. No 'mug and a donut' type of shenanigans.
  - shadowgovt 5y ago
    As distance to center of black hole approaches zero, curvature approaches infinity. There is no "locally" one can constrain small enough to observe no curvature upon approach to a singularity... Given two particles arbitrarily close together at arbitrary velocity (< c) relative to the center of a black hole, there is a distance from the center at which they will encounter unequal acceleration.
joppy 5y ago
In Riemannian geometry (i.e. GR) each point has some local invariants associated to it, a famous one being curvature. "Local" here means that it only depends on the metric in an arbitrarily small disc/ball around the point. So the answer is no, in GR you can't always change coordinates so that space is locally flat. (This is in contrast to symplectic geometry for example, where each point does look locally like the model space).
- ithinkso 5y ago
  Ok, yeah. We obviously can't get rid of a curvature. I think SE answer [1] explains what I assumed incorrectly to mean 'locally flat'. I meant it so that you can make a metric look like Minkowski metric at a point p, basically existence of normal coordinates. Of course second derivatives (and thus curvature) won't vanish in the neighborhood. Case closed, thank you. [1] https://physics.stackexchange.com/a/157019 https://physics.stackexchange.com/a/157019

krastanov 5y ago

"Locally flat" means flat only very close to the point of consideration. That is the mathematical definition, the "practical" definition is "flat only as long as you do not stray further away than some characteristic measure of distance defining your problem and accuracy requirements". So as long as your circle is very small then the " experimentally measured" pi and the math pi are the same. The moment your circle becomes big enough to pass through interesting (measurable) gravitational effects, the experimental value will be different. If anything, experimentally measuring that difference is one way to experimentally measure spacetime curvature.

ithinkso 5y ago
That's why I asked about GR specifically, to have a common framework to work with. Without it you can say words like experimentally, practical and talking about differences between different pis whatever that means but this will only lead to misunderstandings at best and nonsense at worst

shadowgovt 5y ago

Not in accelerating frames of reference, which includes both rotating frames (such as spinning objects) and frames with high gravitational distortion. Spinning black holes have a measured circumference that is not 2 x pi x r. (Consequences of the accelerating frame of reference include the spaghettification in a black hole... We can't use a mathematical conversion to go to a frame of reference where the subject isn't being spaghettified).

ithinkso 5y ago
Why not? (to the first sentence, locally is locally) But I think I have found the source of my confusion in my original post, I was thinking about manifolds only in the topological sense and for some reason forgot that when there is a metric then we have to preserve distances. No 'mug and a donut' type of shenanigans.
- shadowgovt 5y ago
  As distance to center of black hole approaches zero, curvature approaches infinity. There is no "locally" one can constrain small enough to observe no curvature upon approach to a singularity... Given two particles arbitrarily close together at arbitrary velocity (< c) relative to the center of a black hole, there is a distance from the center at which they will encounter unequal acceleration.

joppy 5y ago

In Riemannian geometry (i.e. GR) each point has some local invariants associated to it, a famous one being curvature. "Local" here means that it only depends on the metric in an arbitrarily small disc/ball around the point. So the answer is no, in GR you can't always change coordinates so that space is locally flat. (This is in contrast to symplectic geometry for example, where each point does look locally like the model space).

ithinkso 5y ago
Ok, yeah. We obviously can't get rid of a curvature. I think SE answer [1] explains what I assumed incorrectly to mean 'locally flat'. I meant it so that you can make a metric look like Minkowski metric at a point p, basically existence of normal coordinates. Of course second derivatives (and thus curvature) won't vanish in the neighborhood. Case closed, thank you. [1] https://physics.stackexchange.com/a/157019 https://physics.stackexchange.com/a/157019