Is our locally measured time actually conformal time?

[jcap]

The FRW metric at the origin $r=0$, with $c=1$, is given by:
$$ds^2=-dt^2+a(t)^2dr^2$$
Now one can change variables so that near the origin the FRW metric is approximated by the Minkowski metric describing flat spacetime:
$$dS^2=-dT^2+dR^2$$
where:
$$dT=\frac{dt}{a(t)}$$
$$dS=\frac{ds}{a(t)}$$
$$dR=dr$$
All the physics experiments that we perform locally are assumed to occur in flat spacetime as described above.

Surely therefore our locally measured time is not the cosmological time $t$ but rather the conformal time $T$ ?

 

[Chalnoth]

The local measured time, for any observer, is the proper time. The proper time is, in this case, almost identical to the cosmological time.

 

[PeterDonis]

one can change variables so that near the origin the FRW metric is approximated by the Minkowski metric describing flat spacetime:

​

$$
dS^2=-dT^2+dR^2
$$

where:

$$​

dT=\frac{dt}{a(t)}
$$

$$
dS=\frac{ds}{a(t)}
$$

$$
dR = dr
$$
​

This isn't a correct transformation. You can't transform $ds$ to $dS$; $ds$ is the actual, physical line element, and it has to be left invariant, otherwise you're comparing apples and oranges. The correct transformation gives:

$$
ds^2 = a^2(T) \left( - dT^2 + dR^2 \right)
$$

In order to make the metric Minkowski in our local vicinity, you have to define $a(T_0) = 1$, where $T_0$ is our current value of $T$. But that only works locally, and we can make observations that go beyond that local region of spacetime. And in any case, proper time for an observer at constant $R$ is given by $a(T) dT$, not $dT$; again, you can try to obscure this by defining $a(T_0) = 1$, but we can make observations covering a larger region of spacetime than the local region covered by that definition.

 

[ChrisVer]

You can make the following:

$ds^2= dt^2 - a^2(t) dr^2 = a^2(t) \big( \frac{dt^2}{a^2(t)} - dr^2 \big)$

now you can write the first term in the parenthesis as a single variable by changing $\frac{dt}{a(t)}= d( \ln a(t) ) \equiv dT$.

So:

$ds^2= a^2(t) \big[ dT^2 - dr^2 \big]$ or to have everything in the same coordinates: $ds^2= e^{2T} \big[ dT^2 - dr^2 \big]$
since $T(t)= \ln a(t) \Rightarrow a(t)= e^{T}$

 

[wabbit]

You can make the following:

$ds^2= dt^2 - a^2(t) dr^2 = a^2(t) \big( \frac{dt^2}{a^2(t)} - dr^2 \big)$

now you can write the first term in the parenthesis as a single variable by changing $\frac{dt}{a(t)}= d( \ln a(t) ) \equiv dT$.

So:

$ds^2= a^2(t) \big[ dT^2 - dr^2 \big]$ or to have everything in the same coordinates: $ds^2= e^{2T} \big[ dT^2 - dr^2 \big]$
since $T(t)= \ln a(t) \Rightarrow a(t)= e^{T}$

But $d(\ln a(t))=\frac{\dot a(t)}{a(t)}dt\neq\frac{dt}{a(t)}$
You need $T=\int\frac{dt}{a(t)}$ instead

 

[ChrisVer]

Oops I'm sorry!

 

[wabbit]

No it's an interesting point, I have been using this $T=\int\frac{dt}{a(t)}$ but I didn't know it was called conformal time - never really got to checking what that name was referring to, seemed a bit exotic... So now I know, and I can also see it's a useful thing - thanks.