I Metric of the Universe and dependence on Cosmological P

[Arman777]

Let us suppose we have a metric in the form of,
$$ds^2=-c^2dt^2+[(a^2(t)+b(r)e^{-lt})(dr^2+r^2d\Omega^2)]$$
Where scale factor is defined as $(a^2(t)+b(r)e^{-lt})$

Is this metric describes homogeneity and isotropy or not ? I think it cannot since there's an $r$ dependence, and there are only 3 allowed spatial geometries which satisfy Cosmological Principle. So r dependence on scale factor will not work, and as time goes to $-∞$ the scale factor goes to infinite. So I guess this metric cannot describe our universe ?

 

[kimbyd]

It's isotropic at $r=0$ but not homogeneous.

 

[Arman777]

It's isotropic at $r=0$ but not homogeneous.

What it means to say $r=0$ ? Isnt it the location of observer..?

 

[Ibix]

What it means to say $r=0$ ? Isnt it the location of the observer..?

$r=0$ is arbitrary in a homogeneous universe, but yours isn't homogeneous. It's a special point in yours - the only one at which the universe is the same in all directions.

 

[kimbyd]

What it means to say $r=0$ ? Isnt it the location of observer..?

It's the origin of your coordinate system, which yes, would be the location of a specific observer. If you translated those coordinates to the location of some other observer at a different location, then they wouldn't see an isotropic universe at all.

 

[Arman777]

It's the origin of your coordinate system, which yes, would be the location of a specific observer. If you translated those coordinates to the location of some other observer at a different location, then they wouldn't see an isotropic universe at all.

$r=0$ is arbitrary in a homogeneous universe, but yours isn't homogeneous. It's a special point in yours - the only one at which the universe is the same in all directions.

I see thanks, Hence we can say that this metric cannot represent the Universe, if we consider the universe obeys CP.

 

[kimbyd]

BTW, doing this in full is beyond the scope of the post, but it is possible to write down a metric which looks like it might depend upon location, namely by including spatial curvature:

$$ds^2 = dt^2 - a(t)^2\left({dr^2 \over 1 - kr^2} + r^2d\Omega^2\right)$$

It really looks like the curvature term $1/(1-kr^2)$ might make it depend upon location in the same way as the metric in the OP. After all, if you translate coordinates, won't that denominator be different?

One way to see why it doesn't work is going through the rather daunting challenge of calculating the Ricci scalar $R$ for this metric. There's a lot of math involved in doing that, but when you do you find a result that is purely a function of $a(t)$ and its derivatives (Wikipedia has the answer here). The Ricci scalar is a measure of how much curvature there is at every point in space and time. A space-time that has a Ricci scalar that is independent of spatial coordinates is by definition homogeneous and isotropic.

Furthermore, I'm pretty sure that it's been proven that this is the only kind of metric which can have this property (that is, if you ever find a metric which has this property, you can use a coordinate transformation to turn it into the FLRW metric above).

That said, the cosmological principle does not necessarily apply to our universe, and it's not completely absurd to consider metrics that do not obey the cosmological principle. It does seem that a universe that doesn't obey the cosmological principle is a bit contrived, but it can't be rejected out of hand as impossible.

 

[Arman777]

That was really nice thanks.