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)##.

 

[sardar]

It is important to understand that the two metrics you have provided are not equivalent. The FLRW metric, also known as the Friedmann-Lemaitre-Robertson-Walker metric, is a solution to Einstein's field equations and describes the geometry of the universe in the context of general relativity. It is a spacetime metric that takes into account the expansion of the universe over time, represented by the scale factor "a".

On the other hand, the second metric you have provided is a simple Euclidean metric with the scale factor "a" in the denominator of the time dimension. This metric does not take into account the expanding universe and does not accurately describe the geometry of the universe as we know it.

In fact, the FLRW metric is a special case of the more general metric known as the Robertson-Walker metric, which allows for a more accurate description of the universe by including additional parameters such as the curvature of space and the density of matter.

Therefore, it is important to use the correct metric when studying the universe and its evolution. The FLRW metric, with its inclusion of the scale factor, is a crucial component in understanding the dynamics of the expanding universe and is widely used in cosmology and astrophysics.