# The FRW universe is not asymptotically flat? Its mass?

+ 3 like - 0 dislike
606 views

The Friedman-Robertson-Walker (FRW) metric in the comoving coordinates $(t,r,\theta,\varphi)$ which describes a homogeneous and isotropic universe is $$ds^2\,= -dt^2+\frac{a(t)^2}{1-kr^2}\,dr^2 + a(t)^2 r^2\,\Big( d\theta^2+\sin^2 \!\theta \,d\varphi^2 \Big)$$ where $k$ is the curvature normalized into $\{-1\,,0\,,+1\}$ which refers to a closed, flat and open universe, respectively; and $a(t)$ is the scale factor.

My question is, this FRW metric is NOT asymptotically flat at spatial infinity $r\to+\infty$, isn't it? Thus, we can not calculate the so-called ADM mass (Arnowitt-Deser-Misner), right? If so, how to get the mass of the matter content from the metric?

Note: I do not mean the trivial $m=\rho V$, I mean the mass obtained from the FRW metric.

The matter/material content determines geometry/metric, and reversely the metric reflects the matter content. So I'm trying to recover the material mass (not including the gravitational energy) from the FRW metric.

This post imported from StackExchange Physics at 2015-03-23 09:10 (UTC), posted by SE-user David

retagged Mar 23, 2015
Why do you think there is a well defined mass?

This post imported from StackExchange Physics at 2015-03-23 09:10 (UTC), posted by SE-user MBN
@MBN Thank you, MBN. Yes. I mean the mass of the matter content, not that of the gravitational field.

This post imported from StackExchange Physics at 2015-03-23 09:10 (UTC), posted by SE-user David
Oh, I read your question again and had to delete my answer since it gave only the $\rho V$ answer. I think there is no good definition of what you would like. It would probably be possible only for one type of source, i.e. for radiation a larger universe would be more energetic, but for dark energy it would be less energetic. Etc.

This post imported from StackExchange Physics at 2015-03-23 09:10 (UTC), posted by SE-user Void
@Void Thank you for the comment, Void.

This post imported from StackExchange Physics at 2015-03-23 09:10 (UTC), posted by SE-user David
@David I tried to think about the whole thing a bit more and decided to rewrite and undelete the answer. It is now as close to a complete answer as I could get.

This post imported from StackExchange Physics at 2015-03-23 09:10 (UTC), posted by SE-user Void

+ 3 like - 0 dislike

The question is what do we need the matter content of the universe for. As I understand it, in the usual case we want to find the conserved quantity associated with a certain conserved current gained by the projection of the energy-momentum tensor into a Killing vector, as for example in the paper by Abott and Deser.

The requirement of asymptotical behaviour seems to be connected only with the elimination of boundary terms, i.e. of the flux of the spatial part of the 4-vector current "through infinity". The integration of the divergence of the four-vector over the whole space-slice then gives us a time conservation of the space-slice sum of the zeroth component of the 4-vector.

In the FLRW models, there is however no time-like Killing vector and thus no kind of conserved Killing energy or matter content. The only possibility of a preferred time-like vector is through the orthogonal vector to all the three space-like Killing vectors. This is the four-vector used for the definition of matter density $\rho$ and the trivial (using Einstein equations) $$m=\rho V = \int \frac{3}{8 \pi} \frac{\dot{a}^2 + k}{a^2} dV_{sp} \,.$$ Where $dV_{sp}$ is just the spatial part of the volume element.

But by definition, there cannot be a purely matter-based conserved energy and thus and ADM-sense matter content. I also believe that the main problem with this kind of generalization is also the fact that the ADM mass is designed to portray the energy content of an "isolated" space-time or at least a space-time with a certain dominating background. A cosmological situation defines such a dominant background and thus the notion of ADM mass should not be expected to hold up.

This post imported from StackExchange Physics at 2015-03-23 09:10 (UTC), posted by SE-user Void
answered Aug 19, 2014 by (1,645 points)
Great! Thank you very very very much for the detailed explaination! It is very helpful for me.

This post imported from StackExchange Physics at 2015-03-23 09:10 (UTC), posted by SE-user David

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar\varnothing$sicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.