Is conformal time observable?

Question

The standard FRW metric with cosmic time is $$ ds^2 = -dt^2 + a^2(t)(\gamma_{ij}dx^i dx^j),$$ and we can measure $t$ as the proper time for comoving observers. Thus it makes sense to talk about the age of the universe $t_0$ in terms of the coordinate $t$.

When using conformal time, the metric becomes $$ds^2 = a^2(\eta)(-d\eta^2+\gamma_{ij}dx^idx^j).$$

Suppose I know the evolution of the scale factor $a(t)$ for all time, in fact lets just assume it's matter dominated, $a(t)\propto t^{2/3}$. Is there any way to make sense of the quantity $\eta_0$, the conformal time today? Is this quantity measurable? If not, and I compute an "observable" where $\eta_0$ appears explicitly, are there any ideas about what could have gone wrong?

Answer 1

It's just a rescaling of the time coordinate. Define $t = f(\eta)$ and ${\bar a}(\eta) = a(f(\eta))$. Then,

$$ds^{2} = -{\dot f}^{2}d\eta^{2} + {\bar a}^{2}d^{3}{x}$$

Thus, if $f(s)$ satisfies $\frac{df}{d\eta} = a(f(\eta))\rightarrow \eta = \int \frac{dt}{a}$, then you have transformed into conformal coordinates, and there is no special meaning for the value of conformal time now. I would probably use the integration constant in the above equation to set the value of the conformal time now to zero.