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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?

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3 Answers 3

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$\eta$ 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.

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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.

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usually you use this transformation

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

Example

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

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

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

thus

$$dt=a(t)\,d\eta$$

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

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  • $\begingroup$ 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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