I Why does an Empty Universe have to obey Negative Curvature?

[Arman777]

Its stated that empty universe should have a hyperbolic geometry (Milne Universe) but I don't understand how its possible.
$$H^2=\frac {8\pi G\epsilon} {3c^2}-\frac {\kappa c^2} {R^2a^2(t)}$$For an empty universe when we set $\epsilon=0$ we get
$$H^2=\frac {-\kappa c^2} {a^2(t)}$$

$$\ddot{a}(t)=-\kappa c^2$$
However, $$\frac {\ddot{a}(t)} {a(t)}=-\frac {4\pi G} {3c^2}(\epsilon+3P)$$

and for the acceleration equation, we get $\frac {\ddot{a}(t)} {a(t)}=0$,

for $\epsilon=0$ non-trivial solution happens only $\ddot{a}(t)=0$ for $\kappa=0$

So what's the problem here ?

 

[Jorrie]

Its stated that empty universe should have a hyperbolic geometry (Milne Universe) but I don't understand how its possible.

Without going into the technical stuff, as I understand it, a Milne universe is spatially flat, but has negative spacetime curvature, because the (empty) space is expanding at a constant rate ($\ddot a = 0)$.

 

[Arman777]

Without going into the technical stuff, as I understand it, a Milne universe is spatially flat, but has negative spacetime curvature, because the (empty) space is expanding at a constant rate ($\ddot a = 0)$.

Hmm, so that's a different metric then the FLRW since FLRW rerpesents only spatial metrics. But why do we need a negative spacetime curvature ? Why it can't be just flat ?

So $\kappa=0$ is true but space-time in negatively curved ?

 

[George Jones]

Milne universe is spatially flat, but has negative spacetime curvature

Actually, it is the other way 'round: the Milne universe has zero spacetime curvature, and negative spatial curvature.

For an empty universe when we set $\epsilon=0$ we get
$$H^2=\frac {-\kappa c^2} {a^2(t)}$$

$$\ddot{a}(t)=-\kappa c^2$$

How did you get the second equation? $H^2 = \left( \dot a /a \right)^2$, which gives
$$\left(\frac{\dot a}{a} \right)^2 = \frac{-\kappa c^2} {a^2}.$$
Consequently, $\kappa$ must be negative.

 

[Jorrie]

Hmm, so that's a different metric then the FLRW since FLRW rerpesents only spatial metrics. But why do we need a negative spacetime curvature ? Why it can't be just flat ?

So $\kappa=0$ is true but space-time in negatively curved ?

Nope, FLRW is a spacetime metric, because H has time in it: $H = \frac{\dot {a}}{a}$.

You must distinguish between curved space and curved spacetime. Minkowski spacetime is flat, because it does not expand: $\kappa=0$ refers to zero spatial curvature, but you must also have $\dot {a} = 0$ to get flat spacetime.

 

[Arman777]

oh yes there's also $-c^2dt^2$

Nope, FLRW is a spacetime metric, because H has time in it: $H = \frac{\dot {a}}{a}$.

You must distinguish between curved space and curved spacetime. Minkowski spacetime is flat, because it does not expand: $\kappa=0$ refers to zero spatial curvature, but you must also have $\dot {a} = 0$ to get flat spacetime.

I see now. I don't know why I said its not a space-time metric..

Actually, it is the other way 'round: the Milne universe has zero spacetime curvature, and negative spatial curvature.

How did you get the second equation? $H^2 = \left( \dot a /a \right)^2$, which gives
$$\left(\frac{\dot a}{a} \right)^2 = \frac{-\kappa c^2} {a^2}.$$
Consequently, $\kappa$ must be negative.

oh I see cause there's square. I didnt notice that. So for a non trivial solution $\kappa=-1$ but it can be also $\kappa=0$ right ?

 

[Arman777]

If we say $\kappa=0$ then $a(t)=C$

and for
$\kappa=-1$

$\dot{a}(t)=c/R$ or $a(t)=tc/Rt_0$ ?

Whats the unit of $a(t)$ its unitless right ? From $s_p=a(t)r$, but in $V=HD$, $H$has a unit of 1/s so, if $a(t)$ is unitless its derivative "gains" unit ??

Are above equations true ?

I also noticed that I missed $R^2$ term in the Friedmann Equation

 

[Jorrie]

Actually, it is the other way 'round: the Milne universe has zero spacetime curvature, and negative spatial curvature.

Oops yea, I goofed the words. Thanks for correction.

 

[Arman777]

I edited my post..It seems I made a mistake

 

[Orodruin]

Nope, FLRW is a spacetime metric, because H has time in it: $H = \frac{\dot {a}}{a}$.

You must distinguish between curved space and curved spacetime. Minkowski spacetime is flat, because it does not expand: $\kappa=0$ refers to zero spatial curvature, but you must also have $\dot {a} = 0$ to get flat spacetime.

The Milne universe is Minkowski space, just with different coordinates. It is similar to using Rindler coordinates, but in the future light-cone of the Mikwoski space origin rather than the spatially separated region. You certainly do not need $\dot a = 0$ to get flat spacetime, as the Milne coordinates clearly show. It is similar to how Euclidean space is still flat regardless of whether you use spherical coordinates or not.

 

[kimbyd]

Without going into the technical stuff, as I understand it, a Milne universe is spatially flat, but has negative spacetime curvature, because the (empty) space is expanding at a constant rate ($\ddot a = 0)$.

Nit: negative spatial curvature, but that's an artifact of the coordinate system used. The total space-time curvature is identically zero.

Basically, there's positive curvature from the expansion that is matched by the negative spatial curvature.

And as Orodruin mentioned, it's the same exact space-time as the non-expanding Minkowski space-time. It's just a different coordinate system that makes it look curved.