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One can consider the 3-sphere of radius a as being embedded in a four-dimensional Euclidean space. One has in this view the condition for any coordinate system with origin at the center of the 3-sphere:

$$a^{2}=g_{\mu\nu}x^{\mu}x^{\nu}$$

Note that the metric here is Riemannian (not pseudoriemannian). In standard Cartesian coordinates for example this reads:

$$a^{2}=x^{2}+y^{2}+z^{2}+w^{2}$$

This is simply the condition that the coordinates lie somewhere on the 3-sphere. For a 3-sphere of changing radius a , one can write this infinitesimally as:

$$da^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}$$

Where it is clear that a will be a function of some external parameter (time or conformal time). Let us now allow a body to move around on the three-sphere while it is expanding. For a coordinate choice with three variables (\chi ) on the three sphere, one has an infinitesimal displacement given by:

$$dl^{2}=g_{\mu\nu}\frac{d\chi^{\mu}}{dt}dt\frac{d\chi^{\nu}}{dt}dt$$

Our total displacement condition then becomes:

$$dL(t,\dot{\chi})^{2}=da(t)^{2}+g_{\mu\nu}d\chi(t)^{\mu}d\chi(t)^{\nu}$$

But $L$ is not invariant; different observers would argue over the value of $L$ . $a$ however is invariant, being agreed upon regardless of position on the three sphere. In this manner it is somewhat more natural to rewrite:

$$-da(t)^{2}=-dL^{2}+g_{\mu\nu}d\chi(t)^{\mu}d\chi(t)^{\nu}$$

It makes sense here to designate $L=\chi^{0}$ and a pseudoriemannian metric such that:

$$-da(t)^{2}=g_{\mu\nu}d\chi^{\mu}d\chi^{\nu}$$

Note this argument also applies for two arbitrary points on $S^{3}$ at two differing values of $a$ . Given the different geometric roles of a and L (as compared to dimensions on $S^{3}$ ), one might choose different units such that $a=c\tau$ and $\chi^{0}=ct$ where c is a constant. Note one then has the natural condition:

$$-c^{2}=g_{\mu\nu}\frac{d\chi^{\mu}}{d\tau}\frac{d\chi^{\nu}}{d\tau}$$ Locally, for large a, this looks a lot like the conditions of Lorentz invariance as in Special relativity. anyway, I thought it was interesting.

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Conformal transformation of a metric $g_{\mu\nu}~\rightarrow~\Omega^2g_{\mu\nu}$ which for the line element $$ ds^2~=~g_{\mu\nu}dx^\mu dx^\nu~\rightarrow~\Omega^2g_{\mu\nu}dx^\mu dx^\nu $$ which for a diagonal metric is $$ ds'^2~=~\Omega^2(d\chi^2~-~d\Sigma^{(3)}), $$ where $\chi$ conformal time and $\Sigma^{(3)}$ the metric of the spatial surface. For a flat surface $d\Sigma^{(3)}~=~dx^2~+~dy^2~+~dz^2$. In general $$ d\Sigma^{(3)}~=~\left(\frac{dr^2}{1 - kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\right). $$ that is mapped by the conformal transformation $d\Sigma^{(3)}~\rightarrow~\Omega^2d\Sigma^{(3)}$.

We now let $$ d\chi~=~\frac{d\chi}{dt}dt~=~\Omega^{-1}dt, $$ which gives the spacetime metric $$ ds^2~=~dt^2~-~\Omega^2\left(\frac{dr^2}{1 - kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\right). $$ The conformal transformation is a time dependent function which we write as $\Omega~=~a(t)$ so this gives the FLRW metric $$ ds^2~=~dt^{2}~-~a(t)^{2}\left(\frac{dr^2}{1 - kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\right). $$

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  • $\begingroup$ Thank you, but I'm not entirely sure this answers my question about the conformal transformation or rather scale factor being an invariant. Wouldn't all observers regardless of frame agree agree on the scale factor? It is tied to the Ricci Scala which is also an invariant. $\endgroup$ – R. Rankin Dec 12 '16 at 20:06
  • $\begingroup$ The scale factor is a sort of time dependent conformal factor that expands the spatial surface. Indeed all observers agree on the same scale factor at a given time as determined by the Hubble surface. $\endgroup$ – Lawrence B. Crowell Dec 13 '16 at 23:00
  • $\begingroup$ @Lawrence_B._Crowell What about a term for the differential of the scale factor here? $d\Omega$ Shouldn't such a quantity be invariant on a Hubble surface as well? $\endgroup$ – R. Rankin Jun 19 '17 at 7:12
  • $\begingroup$ Agreed, which just means the spacetime is isotropic. $\endgroup$ – Lawrence B. Crowell Jun 20 '17 at 11:20

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