My professor for study FRLW metric respect to conformal time started from $$ ds^2=c^2dt^2-a^2(t) f(r,\Omega) $$ and, after that, define the conformal time $\eta$ as $$ cdt=a(\eta) d\eta $$ but Wikipedia, for example, define conformal time as $$ cdt=a(t) d \eta $$ can you help me?


3 Answers 3


$\eta$ can be considered a function of $t$, and $t$ can be considered a function of $\eta$, so $a$ can be considered a function of either one. Which parameterization is more useful depends on what you are doing.

As a simple analogy, sometimes in 1D classical mechanics one is interested in $v(t)$ and sometimes (such as when using an energy-conservation argument) in $v(x)$.

Mathematicians tend not to do this kind of thing, since $v(t)$ and $v(x)$ are different functions of their arguments! Physicists, on the other hand, are comfortable naming the two different functions “$v$” because they are both velocity functions; they seldom even write the argument.

Cosmologists only care that $a$ is the Friedmann scale factor. Whether it is expressed as a function of cosmological time, conformal time, or something else like temperature is less important.


One of the main advantages of using the conformal time is that the action of the scalar and non-massive vector fields are reduced to a form similar to that which occurs in a plane-Minkowski spacetime.

The scale factor a (t) has the meaning of the radius of curvature of space at time t (either in the closed space / 3-sphere / model or in the open / 3-hyperboloid model. In the flat space model a (t) has no physical meaning. In the flat space model, the relationship of the scale factor at different instants of time a (t1) / a (t2) has physical significance.


usually you use this transformation

$$dt=a(\eta)\,d\eta\tag 1$$


$$a(\eta)=\eta$$ $\Rightarrow$ $$t=\int \eta\,d\eta=\frac{\eta^2}{2}+c~\,\text{with}~c=0$$ thus $$\eta=\sqrt{2}\,t$$

is $~dt=a(t)\,d\eta\,?$

$$dt=a(\eta)\,d\eta=\eta\,d\eta=\sqrt{2}\,t\,\sqrt{2}\,dt\ne dt$$



is only fulfill if $\eta=t$, but not for a general case

  • $\begingroup$ but in general we don't have $\eta=t$, so what is the right transformation? $\endgroup$
    – Tony Stack
    Commented Dec 13, 2020 at 9:53
  • $\begingroup$ so i think the transformation is not valid for general case $\eta=\eta(t)$ $\endgroup$
    – Eli
    Commented Dec 13, 2020 at 10:05

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