(dimensionally correct, since U is energy/unit mass)
We will choose our scale so that $\kappa $ is an integer: the "curvature" of universe
$\kappa $ = -1 means U > 0, open, negative curvature
$\kappa $ = 0 means U = 0, critical, zero curvature
$\kappa $ = 1 means U < 0, closed, positive curvature
Note some relations:
$$
\color{red}{
v\left( t \right) = \frac{{\partial d\left( t \right)}}{{\partial t}} = H\left( t \right)d\left( t \right) = r \frac{{\partial a\left( t \right)}}{{\partial t}} = r H\left( t \right)a\left( t \right)}
$$
(since r doesn't change with time)
so $$
\color{red}{
\frac{{\partial d\left( t \right)}}{{\partial t}} = H\left( t \right)d\left( t \right)}
$$ so the (almost) general Friedmann equation is
\color{red}{
H\left( t \right)^2 = \left( {\frac{{\dot a}}{a}} \right)^2 = \frac{{8\pi G}}{{3c^2 }}\varepsilon \left( t \right) - \frac{{\kappa c^2 }}{{\left( {R_0 a\left( t \right)} \right)^2 }}}
A better (more fundamental) derivation, due to Berry: metric is
\color{red}{
k'\left( t \right) = - \frac{{\ddot a\left( t \right)}}{{a\left( t \right)}}}
Note this k'(t) is not the usual spatial curvature
Now dimensionally, we expect k(t) to depend on ρ,G,c: so put
\color{red}{
k'\left( t \right) = \alpha \rho \left( t \right)G^l c^m + const}
(const is because empty space can be curved, \color{red}{\rho \left( t \right)}
is only matter)
\color{red}{
\left[ {k'\left( t \right)} \right] = T^{ - 2} \Rightarrow l = 1,m = 0}
Assume $
\rho \left( t \right)
$ depends only on mass
so
\color{red}{
\frac{{\ddot a\left( t \right)}}{{a\left( t \right)}} = - k'\left( t \right) = - \alpha \rho \left( t \right)G + \frac{\Lambda }{3}}
Note the is an acceleration (F=ma) equation.
Note that positive Λ acts like negative density: i.e. drives an acceleration
to get this back to usual form, mult by \color{red}{2\dot a\left( t \right)a\left( t \right)} and put \color{red}{\rho \left( t \right) = \frac{{\rho _0 }}{{a\left( t \right)^3 }}}
gives us
\color{red}{
2\dot a\left( t \right)\ddot a\left( t \right) = - \alpha \frac{{\rho _0 }}{{a\left( t \right)^2 }}2\dot a\left( t \right)G + \frac{\Lambda }{3}2\dot a\left( t \right)a\left( t \right)}
which can be integrated:
\color{red}{
\dot a\left( t \right)^2 = + 2\alpha \frac{{\rho _0 }}{{a\left( t \right)}}G + \frac{\Lambda }{3}a\left( t \right)^2 }
i.e. sign of density flips in the integration from acceleration eqn to energy eqn.
Must agree with Newtonian result in the limit, , so this fixes \color{red}{
\alpha = \frac{{4\pi }}{3}
}
Dimensions: note that \color{red}{\left[ \Lambda \right] = T^{ - 2} } so \color{red}{\left[ {m\Lambda r} \right] = MLT^{ - 2} }
is a force
Vacuum energy density: we can make the Λ term look like an energy density by writing
\color{red}{\frac{{8\pi }}{3}\frac{{\varepsilon _\Lambda G}}{{c^2 }} = \frac{\Lambda }{3}}
so
For adiabatic system \color{red}{dQ = dE - pdV = 0}
. If we have an expanding gas, change in energy = W.D.
$$
\color{red}{
d\left( {\rho V} \right) = dE = - PdV \to \frac{d}{{dt}}\left( {\rho \frac{4}{3}\pi a^3 } \right) = - P\frac{d}{{dt}}\left( {\frac{4}{3}\pi a^3 } \right)}
$$
For cold matter, P = 0, so we get back the old result. (not precisely: really
\color{red}{
P = \frac{{k\rho T}}{\mu }}
so w ∼ 10-10)
For radiation: can repeat old kinetic theory of gases derivation giving P = 1/3 εR.
In general we'll write P = wε. Then
$$
\color{red}{
\frac{{d\left( {\epsilon a^3 } \right)}}{{dt}} = - w\epsilon \frac{{d\left( {a^3 } \right)}}{{dt}}}
$$
Then $$
\color{red}{
\frac{{d\left( {\epsilon a^{3\left( {1 + w} \right)} } \right)}}{{dt}} = 0}
$$
or εa3(1+w) is a constant (to prove it, first show $$
\color{red}{
\frac{{d\left( {a^{3\left( {1 + w} \right)} } \right)}}{{dt}} = \left( {1 + w} \right)a^{3w} \frac{{d\left( {a^3 } \right)}}{{dt}}}
$$
