Comparing FLRW and Scale Factor Metrics

[tartaneto]

The generic FLRW metric is dS^2 = a^2.(dx^2 + dy^2) - (c.dt)^2. Is it equivalent to the metric dS^2 = (dx^2 + dy^2) - (c.dt/a)^2 with the scale factor in the denominator of the time dimension? (I suppressed one dimension just for simplicity).

Thanks for the help.

 

[PAllen]

That's not the generic FLRW metric. That's only for the metric for the spatially flat case. Please specify whether you want to restrict attention to this special case.

 

[tartaneto]

Yes, it can be only for the flat case. My point is: in order to get the metric growing either I can keep the space component growing or the time component decreasing and that is the reason of my question if the are equivalent or not.

 

[fzero]

A metric of the form

$$ds^2 = - \left( \frac{c dt}{A(t)}\right)^2 + ds_3^2 ~~~~(*)$$

is a flat metric, which you can see by defining a new variable

$$ t' = \int^{t'} \frac{c dt}{A(t)},$$

so that

$$ds^2 = - (dt')^2 + ds_3^2. $$

Therefore there is no choice of $A(t)$ such that this metric is equivalent to the FLRW metric.

$$ ds^2 = - c^2 dt^2 + a(t)^2 ds_3^2 .$$

If we were dealing with the open or closed versions of $ds_3^2$ in (*), we'd find that the curvature of the 4-metric was independent of $A(t)$ and was just given by the spatial curvature.

An equivalent form to the FLRW metric is

$$ ds^2 = a(\tau)^2 \left( - d\tau^2 + ds_3^2 \right),$$

which can be obtained by defining the so-called conformal time

$$ \tau = \int^{\tau} \frac{c dt}{a(t)}.$$

The problem with the form that you postulate is that the coordinate transformation you have to make to remove the $a(t)$ from the spatial part involves mixing the time variable with the radial variable:

$$ r' = a(t) r.$$

However, now $dr'$ involves a term with $dt$ and you'll inevitably find cross terms in the transformed metric of the form $dt' dr'$ that don't vanish (these terms preserve the curvature). There doesn't appear to be an appropriate choice of $t'$ such that you end up with the simple expression that you're hoping for, even if we were willing to let the function $A=A(r)$.