Does an observer on an expanding three-sphere naturally have a hyperbolic sense of time?

Question

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.

Answer 1

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 $$ ds^2~=~dt^{2}~-~a(t)^{2}\left(\frac{dr^2}{1 - kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\right). $$