On Wednesday, May 27, 2020 at 9:18:40 AM UTC-6, smitra wrote: > > On 27-05-2020 11:07, Alan Grayson wrote: > > On Tuesday, May 26, 2020 at 6:24:32 PM UTC-6, Brent wrote: > > > >> On 5/26/2020 6:49 AM, Alan Grayson wrote: > >> > >> On Tuesday, May 26, 2020 at 5:51:50 AM UTC-6, Alan Grayson wrote: > >> > >> On Sunday, May 24, 2020 at 4:49:48 PM UTC-6, Brent wrote: > >> > >> On 5/24/2020 11:21 AM, Alan Grayson wrote: > >> > >> On Sunday, May 24, 2020 at 8:51:35 AM UTC-6, Alan Grayson wrote: > >> > >> On Saturday, May 23, 2020 at 12:06:33 PM UTC-6, Brent wrote: > >> > >> On 5/22/2020 11:25 PM, Alan Grayson wrote: > >> > >> On Friday, May 22, 2020 at 11:03:40 PM UTC-6, Brent wrote: > >> > >> On 5/22/2020 9:48 PM, Alan Grayson wrote: > >> > >> On Friday, May 22, 2020 at 9:05:23 PM UTC-6, Brent wrote: > >> > >> On 5/22/2020 6:26 PM, Alan Grayson wrote: > >> > >> On Monday, May 18, 2020 at 3:28:40 PM UTC-6, Alan Grayson wrote: > >> Suppose the universe is a hyper-sphere, not expanding, and an > >> observer travels on a closed loop and returns to his spatial > >> starting point. His elapsed or proper time will be finite, but what > >> is his coordinate time at the end of the journey? TIA, AG > >> > >> It's not a dumb question IMO. If you circumnavigate a spherical > >> non-expanding universe, what happens to coordinate time at the end > >> of the journey? Does something update the time coordinate? Or does > >> it somehow miraculously(?) remain fixed? TIA, AG > > > > Are you supposing the universe is a 3-sphere? In that case It's just > > like going around a circle. The degree marks on the circle are > > coordinates, they have no physical meaning except to label points. So > > if you walk around the circle you measure a certain distance (proper > > time) but come back to the same point. > > > > Or are you supposing it's a 4-sphere so that all geodesics are closed > > time-like curves? I don't know how that would work. I don't think > > there's any solution of that form to Einstein's equations. > > > > Brent > > > > I'm supposing a 4-sphere and (I think) closed time-like curves. The > > traveler returns presumably to his starting position, but is the time > > coordinate unchanged? AG > > > > I don't think there's any very sensible answer in that case. Goedel > > showed there can be solutions with closed time-like curves if the > > universe is rotating. But solutions of GR don't have any dynamic > > connection to matter and the entropy of matter. In the same spirit > > there could be a solution to quantum field theory that was close > > around the time like curve...in which case you'd experience "Groundhog > > Day"...including your thoughts. > > > > Brent > > > > What does entropy have to do with this problem? AG > > > > Increasing entropy points the direction of time. > > > > Brent > > > > Let me pose the question another way: Is coordinate time ever updated? > > AG > > > > Or say, in the Twin Paradox, the elapsed or proper time for the > > traveling twin is less than for the Earth-bound twin, but when they > > meet, do they share the same coordinate time? AG > > > > Yes. Coordinates are labels for points, so if you're together with > > your twin, you both are at the same point in spacetime and that point > > only has one label in any given coordinate system. > > > > Brent > > > > Since time is just ONE of the 4 labels for spacetime points, can they > > be assigned at random? What specific function do they satisfy? AG > > > > How is the time coordinate chosen such that the Lorentz distance > > between spacetime points is meaningful? AG > > > > The proper distance/duration is an invariant, it doesn't depend on the > > coordinate system. > > > > Brent > > > > I think the invariance of proper distance/duration a direct result of > > the Lorentz transformation, and is one of the results of SR. If that's > > the case, is it used in GR to derive EFE's? TIA, AG > > > > The Lorentz transform results from demanding that ds^2 for a flat > space-time is an invariant. It's easy to derive this, as you know > rotations and translations leave the ordinary Euclidic metric invariant, > the relative minus sign between time and space means that instead of > cos(theta) and sin(theta), you get cosh(theta) and sinh(theta) in > transforms that mix time and space. > > Saibal >
But why would you want ds^2 to be invariant? The answer IMO, is that the LT leaves the SoL invariant as we change coordinate systems, and using this requirement is sufficient for deriving the LT. AG -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/6819598d-dea4-47d3-b37a-bcaa3c36e45f%40googlegroups.com.
