Is our locally measured time actually conformal time?

1. Apr 13, 2015

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 $T$ ?

2. Apr 13, 2015

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.

3. Apr 13, 2015

Staff: Mentor

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.

4. Apr 14, 2015

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 $T(t)= \ln a(t) \Rightarrow a(t)= e^{T}$

5. Apr 14, 2015

wabbit

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

6. Apr 14, 2015

ChrisVer

Oops I'm sorry!

7. Apr 14, 2015

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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook