I Friedmann Equation - Positive Curvature

[Arman777]

Friedmann Equation without the cosmological constant can be written as;
$$H^2=\frac {8πρG} {3}-\frac {1} {a^2(t)}$$
for $Ω>1$ (which in this case we get a positive curvature universe), and for a matter dominant universe;
$$ρ=\frac {1} {a^3(t)}$$
so in the simplest form the Friedmann equation becomes;
$$H^2=\frac {1} {a^3(t)}-\frac {1} {a^2(t)}$$

For a(t) larger than 1. The equations becomes negative but I remember that

Proper distance is proportional to a(t), $D=a(t)Δx$ so when $a(t)$ is smaller then $1$ it means universe collapses. In this case how it expends in the first place ?

$a(t)$ is a function of $t$ so in this sense even its get smaller its possible that to get expand and then collapse I think but how can I find it ?

 

[mfb]

Why did you skip the matter term for the collapsing universe? It is still there, and larger than the other term.

a(t) is always smaller than 1. You used this already to determine that the matter term is larger in the expansion phase.

 

[Arman777]

Why did you skip the matter term for the collapsing universe? It is still there, and larger than the other term.

a(t) is always smaller than 1. You used this already to determine that the matter term is larger in the expansion phase.

I understabd that my condition happenz for a(t) is larger than 1. But in early times universe was expanding so hows that possible that a(t) is smaller than 1 and universe still expands ?

 

[Arman777]

I didnt quite understand the general solution for $k=1$.

We have;

$$H^2=\frac {8πGρ} {3}-\frac {1} {a^2(t)}$$
And
$$\frac {dρ} {dt}+3Hρ=0$$

Is it enough to find a solution ?

 

[timmdeeg]

See here, Matter Only (k = 1).

 

[Arman777]

See here, Matter Only (k = 1).

$a(θ)$ comes from the FLRW metric ?

 

[George Jones]

Proper distance is proportional to a(t), $D=a(t)Δx$ so when $a(t)$ is smaller then $1$ it means universe collapses. In this case how it expends in the first place ?

What does "expand" mean?

 

[Arman777]

What does "expand" mean?

It can still expand but the speed of expansion can slow down or derivative of it

 

[kimbyd]

Friedmann Equation without the cosmological constant can be written as;
$$H^2=\frac {8πρG} {3}-\frac {1} {a^2(t)}$$
for $Ω>1$ (which in this case we get a positive curvature universe), and for a matter dominant universe;
$$ρ=\frac {1} {a^3(t)}$$
so in the simplest form the Friedmann equation becomes;
$$H^2=\frac {1} {a^3(t)}-\frac {1} {a^2(t)}$$

For a(t) larger than 1. The equations becomes negative but I remember that

Proper distance is proportional to a(t), $D=a(t)Δx$ so when $a(t)$ is smaller then $1$ it means universe collapses. In this case how it expends in the first place ?

$a(t)$ is a function of $t$ so in this sense even its get smaller its possible that to get expand and then collapse I think but how can I find it ?

Your statement that when $a(t)$ is smaller than $1$ the universe collapses is incorrect. In this model, $a(t) \le 1$ always. As long as $a(t) < 1$, then $1/a^3 > 1/a^2$, so that the right hand side is positive. Early on, $H > 0$. As the universe expands, it reaches a point where the matter term and the curvature term are equal, so that $H=0$. Then the universe recollapses after that point. $H^2 \ge 0$ always. It has to: otherwise the rate of expansion would be an imaginary number. You can verify this behavior by looking at the second Friedmann equation and considering appropriate initial conditions.

 

[Arman777]

Your statement that when $a(t)$ is smaller than $1$ the universe collapses is incorrect. In this model, $a(t) \le 1$ always. As long as $a(t) < 1$, then $1/a^3 > 1/a^2$, so that the right hand side is positive. Early on, $H > 0$. As the universe expands, it reaches a point where the matter term and the curvature term are equal, so that $H=0$. Then the universe recollapses after that point. $H^2 \ge 0$ always. It has to: otherwise the rate of expansion would be an imaginary number. You can verify this behavior by looking at the second Friedmann equation and considering appropriate initial conditions.

So the inital conditions are

$\ddot a(t)<0$ and $\dot a(t)>0$ and $H=\frac {\dot a(t)} {a(t)}>0$

at some point it will be,

$\ddot a(t)<0$ and $\dot a(t)=0$. So does $H=\frac {\dot a(t)} {a(t)}=0$

Then since $\ddot a(t)<0$;

$\dot a(t)<0$, So does $H=\frac {\dot a(t)} {a(t)}<0$

But in all of these cases $H^2≥0$

Is this true ?

If its true then what will be the value of $a(t)$ when $\dot a(t)=0$ ?

And Inital conditons are like this casue of the acceleration equation ?

 

[mfb]

Is this true ?

Sure.

If its true then what will be the value of $a(t)$ when $\dot a(t)=0$ ?

Some maximum value that you can calculate from the condition of H=0. But keep in mind that this scale is arbitrary. Multiplying the scale factor by 2 everywhere doesn't change the universe.

And Inital conditons are like this casue of the acceleration equation ?

We don't know if the initial conditions have some deeper reason, but using them we can calculate the evolution of the universe.

 

[Arman777]

Sure.Some maximum value that you can calculate from the condition of H=0. But keep in mind that this scale is arbitrary. Multiplying the scale factor by 2 everywhere doesn't change the universe.We don't know if the initial conditions have some deeper reason, but using them we can calculate the evolution of the universe.

You are right yes, scale factor itself don't mean anything..

 

[Arman777]

Thats nice then.

So I want to ask just in case.

For $k=0$

$\ddot a(t)=0$ and $\dot a(t)>0$

For $k=-1$

$\ddot a(t)>0$ and $\dot a(t)>0$.