This is
the basic equation of cosmology, and its solution tells how universe will
evolve on large scales.
We can
derive it using Newtonian approximation, if we accept two results from General
Relativity:
First,
there is, so called, Birkhoff's ,theorem, which says that, for a spherically
symmetric system, the force due to gravity at some radius is determined only by the mass interior to this radius.
And also,
the energy contributes to gravitational mass density. So we need to add them up
together.
Imagine a
sphere with a mean density of row, radius of R, acceleration to a particle in
surface, which is the second derivative of the radius, is given by formula.
Ȑ= - 4л G ϱ R
3
Now, if
the density changes, it changes according to the inverse cube of the radius. R^-3
ϱ = ϱo R^-3
If we,
use subscript zero for present moment, then we can derive the density at any
given time through the cube of the radius divided the present one, which we can
assume to be unity for simplicity.
Ṙ= - 4л
G ϱ0
3 R^2
This tells
us that if the density is not zero, the universe must be either expanding or
contracting, because the second derivative is finite. This sign depends,
obviously, on the first derivative. This also tells us that relativistic
universe cannot be static.
We can
integrate by multiply the whole thing with first derivative of Ṙ.
Ȑ Ṙ + 4л
G ϱ0 Ṙ =0
3 R^2
From
calculus, derivative of the square of the first derivative is given by this
expression.
d (Ṙ^2)/dt = 2 Ṙ Ȑ
Plug it
in, and the equation now looks a little more complicated.
1 d (Ṙ^2) + 4л
G ϱ0 dṘ =0
2 dt 3 R^2
dt
But there
is a reason for this we also remember that
1
dR = - d (1/R)
R^2 dt dt
We plug
that in and we have a simple differential equation, which says that derivative
time derivative of this thing in brackets is actually zero.
d [Ṙ^2 - 8лG ϱ0/3]
= 0
dt R
The thing
in brackets must be constant, so it's called K. Turns out that's curvature
constant.
Ṙ^2 – (8лG ϱ0/3) = k
R
Now, if
we replace present day density row zero with general density at any given time,
rho, scaling by, appropriate scale factor, and then divide by R squared, we get
the following equation.
Ṙ^2 – 8лG ϱ = - k
R 3
R^2
That is
the Friedmann equation in the absence of the cosmological constant.
So,
finally we get this result,
Ṙ^2 – 8лG ϱ =
- k
R 3
R^2
Let’s
take a, just, inspection of this.
1. If k=0 as it could be, then R dot must be
always positive, because you move the gravity turn to the right side and the
expansion goes forever. But it slows down, because the density goes down.
This is
so called critical or flat universe.
2. If curvature constant k>0, then
initially R dot will be positive, and that means the universe will be
expanding, but then at some point, we'll flip the sign and start contracting,
so that closed the universe.
3. If curvature constant k<0 then r dot
must be always positive. It's never zero and universe expands forever, so
that's the open universe.
This is
the Friedmann equation, and we've neglected details including what happens with
cosmological constant. But if cosmological constant is included, that is just
yet another term.
Ṙ^2 – 8лG ϱ - 1Ϡ
= - kc^2
R 3
3 R^2
We can
think of it as another density term, although it behaves differently from other
kinds of density.
So
Friedmann equation is the equation of motion for a universe, for homogeneous,
isotropic, relativistic universe. And solving this equation for a given parameters
gives good cosmological models. But in order to do this we need one more
equation, which is the equation of state, which tells us how the density is of
whatever is filling up the universe changing as the universe expands.
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