0
$\begingroup$
  1. The FLRW metric is known to be conformally flat, i.e., it is conformally related to the Minkowski metric. How I read this, is that it makes the FLRW metric expressible in static form. Is this correct?

  2. According to Florides [1], the only FLRW spacetimes that are expressible in static form are the 6 FLRW spacetimes of constant curvature, that is, Minkowski, Milne, deSitter (flat, open, closed) and Anti-deSitter.

The first and second statement combined suggest the FLRW spacetime can only be of one of these 6 forms. Is this correct?

The evolving concordance model is not among them. So what to make of this?

[1] Florides, P.S. The Robertson-Walker metrics expressible in static form. Gen Relat Gravit 12, 563–574 (1980).

$\endgroup$
17
  • $\begingroup$ If $H$ is independend of $t$ like it is in the metrics you mentioned above you can use static coordinates, but if it is time dependend you can't $\endgroup$
    – Yukterez
    Commented Jun 24, 2022 at 19:00
  • $\begingroup$ You don't need staticity for conformal flatness $\endgroup$
    – Eletie
    Commented Jun 25, 2022 at 8:08
  • $\begingroup$ @Yukterez, of course, but the statement of Florides is about the opposite: if expressible in static coordinates (the FLRW metric is conformal to Minkowski) then the FLRW metric can only be one of these 6. $\endgroup$
    – Octaaf
    Commented Jun 25, 2022 at 9:40
  • $\begingroup$ @Eletie: indeed. Florides shows that, given Minkowski (or any other static representation of FLRW), the solution set of expanding FLRW spacetimes is restricted to Milne and (A)dS. The implicit statement seems the FLRW metric is a (energy) conservation equation, therefore, restricts what you can put into the stress energy tensor if space expands. $\endgroup$
    – Octaaf
    Commented Jun 25, 2022 at 10:31
  • 1
    $\begingroup$ The scale factor $\rm a$ can be time dependend in a static metric, but $\rm \dot{a}/a=H$ can not $\endgroup$
    – Yukterez
    Commented Jun 25, 2022 at 11:58

1 Answer 1

0
$\begingroup$

The flat FLRW metric in static spherical Schwarzschild/Droste-style coordinates is

$$g_{\mu \nu}=\left( \begin{array}{cccc} \rm c^2-H^2 r^2 & 0 & 0 & 0 \\ 0 & \frac{1}{\rm H^2 r^2/c^2-1} & 0 & 0 \\ 0 & 0 & \rm -r^2 & 0 \\ 0 & 0 & 0 & \rm -r^2 \sin ^2 \theta \\ \end{array} \right)$$

or in Gullstrand/Painlevé-style proper distance raindrop coordinates

$$g_{\mu \nu}=\left( \begin{array}{cccc} \rm c^2-H^2 r^2 & \rm H \ r & 0 & 0 \\ \rm H \ r & -1 & 0 & 0 \\ 0 & 0 & \rm -r^2 & 0 \\ 0 & 0 & 0 & \rm -r^2 \sin ^2 \theta \\ \end{array} \right)$$

or the cartesian form with $\rm \{ t,x,y,z\}$ instead of $\rm \{ t, r, \theta, \phi \}$

$$g_{\mu \nu}=\left( \begin{array}{cccc} \rm c^2-H^2 (x^2+y^2+z^2) & \rm H \ x & \rm H \ y & \rm H \ z \\ \rm H \ x & -1 & 0 & 0 \\ \rm H \ y & 0 & \rm -1 & 0 \\ \rm H \ z & 0 & 0 & \rm -1 \\ \end{array} \right)$$

but if $\rm H$ is not constant and you have a time dependend Hubbleparameter the whole metric becomes time dependend, so that is only static when the other densities are neglible compared to the dark energy density. Then we have basically De Sitter space where $\rm H=c \sqrt{\Lambda/3}$ is constant, or Minkowski if you also set $\Lambda=0$.

In our universe this will be the case in a few billion years when the Hubble radius becomes asymptotically constant, see here at around $\rm t > 35 \ Gyr$ where $\rm H$ doesn't really change anymore. During inflation the Hubbleradius $\rm c/H$ is also expected to be constant, so that era can also be described in static coordinates.

For $\rm H$ to change you need $\rho$ to change, and then your curvature

$$ \rm K=12 (2 H^4+2 H^2 \dot{H}+\dot{H}^2)/c^4$$

is no longer constant (since the $\rm \dot{H}=dH/dt\neq 0$) and the metric is no longer static, but rather $\rm t$-dependend. With $\rm \dot{H}=0$ the curvature invariant reduces to $\rm K=24H^4/c^4$ and the requirement of constant curvature that your reference presumes for static coordinates is fulfilled. I haven't read it though, but since you quoted it yourself:

Octaaf quoted Florides: "the only FLRW spacetimes that are expressible in static form are the 6 FLRW spacetimes of constant curvature"

I don't expect it to be different there.

$\endgroup$
3
  • $\begingroup$ That still leaves the question: all FLRW spacetimes satisfy the FLRW metric, so can be conformally represented by static Minkowski. But then, according to Florides, can only be of constant curvature. Hence the question. $\endgroup$
    – Octaaf
    Commented Jun 25, 2022 at 13:22
  • $\begingroup$ Take a look at en.wikipedia.org/wiki/…. This article makes clear that the FLRW spacetimes satisfying the FLRW metric can include alternative geometries in addition to having flatness or constant curvature. $\endgroup$
    – Buzz
    Commented Jul 6, 2022 at 23:10
  • $\begingroup$ Yes, but any FLRW geometry is still conformally flat, so that does not change the question, which by Florides’ result suggests that only those of constant curvature are admissible. $\endgroup$
    – Octaaf
    Commented Jul 8, 2022 at 11:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.