# Is conformal time observable?

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?

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.
• I'm more concerned about rescaling ambiguity. I can send $a\rightarrow \lambda a$, $\eta\rightarrow\eta/\lambda$ and $x^i\rightarrow x^i/\lambda$. So the integration constant is not the issue, I'm wondering if the statement "amount of conformal time since a given event" makes any sense. This is what I mean by $\eta_0$. Feb 28, 2014 at 2:43
• e.g. suppose I'm measuring the time since the big bang. For this $t_0=13.9$ billion years. If I set $\eta(t=0)=0$, is $eta_0$ measurable? Feb 28, 2014 at 5:31