Timeline for Coordinates for FLRW metric
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 23, 2021 at 16:43 | comment | added | Shashaank | I have a small doubt. dt will be equal to the proper time measured by which observer. Like in the Schwarzschild coordinates dt will be equal to the proper time of the observer at infinity or in the gullstrand coordinates dt will be equal to the proper time of a radially falling observer. Similarly here dt would be equal to the proper time measured by which observer | |
| Aug 13, 2020 at 0:14 | comment | added | benrg | Your metric doesn't have a singularity at $R=1/H$. The $t$ coordinate becomes lightlike there, and spacelike beyond, but it's never singular. The cosmological horizon generally isn't at $R=1/H$ anyway, except when $Ω=Ω_Λ$. Also, the $dR\,dt$ term isn't evidence of physical frame dragging, it's just a coordinate artifact. You can tell by the fact that it's linear in $R$ at the origin. Any first-order deviation from the Minkowski metric at the origin can always be removed by a change of coordinates. For best behavior at the origin you'd need a non-FLRW time coordinate, which starts to get messy. | |
| Jan 26, 2015 at 20:12 | comment | added | Zo the Relativist | Explicitly, if you're measuring the distance along some path $\gamma$, then the distance will be: $I = \int \sqrt{-g}\epsilon_{abcd}v^{a}w^{b}z^{c}dx^{d}$ where $v,w,z$ are three unit vectors normal to $\gamma$. | |
| Jan 26, 2015 at 20:10 | comment | added | Zo the Relativist | If you have two galaxies, and you're measuring the distance between them in some reference frame, the result will be coordinate-independent. | |
| Jan 26, 2015 at 20:07 | comment | added | Shadumu | @JerrySchirmer say I measure the distance between two galaxy using the FRW metric in some coordinate. Certainly I will not get the same value and the value could be changing differently in different coordinates, not necessarily expanding. | |
| Jan 26, 2015 at 20:00 | comment | added | Zo the Relativist | Also, @user3229471, note that the scale factor $a(t)$ is not an observable of the theory -- it is set by a choice of the time coordinate and by an initial scale for the universe. The Hubble parameter, though, is an observable. | |
| Jan 26, 2015 at 19:59 | comment | added | Zo the Relativist | @Christoph: fixed. | |
| Jan 26, 2015 at 19:58 | comment | added | Zo the Relativist | @user3229471: you have to talk about what measurement you're making, at the end of the day. What I did here was take the dynamics out of the $r$ coordinate, and put them in the $t$ coordinate. In the ADM language, all of the physics in this version of the FRLW metric lives in the lapse and shift. | |
| Jan 26, 2015 at 19:57 | history | edited | Zo the Relativist | CC BY-SA 3.0 |
edited body
|
| Jan 26, 2015 at 19:45 | comment | added | Shadumu | It could be that in some coordinates the universe appears not to be expanding as a(t). Though all the physics being unchanged, how we interpret them will depend on coordinates. Is there a notion of expanding universe which does not depend on coordinates? | |
| Jan 26, 2015 at 18:28 | comment | added | Christoph | you've missed some substitutions $r\to R$... | |
| Jan 26, 2015 at 18:18 | comment | added | Zo the Relativist | But note that all coordinate systems are equal, at base. This is just a different way to look at the physics of this coordinate system. | |
| Jan 26, 2015 at 18:17 | history | answered | Zo the Relativist | CC BY-SA 3.0 |