Density $\rho$ in the Friedmann equations In the Friedmann equations:
$$\ddot{a}=-\frac{4}{3}\pi G(\rho+\frac{3p}{c^2})$$
$$\dot a^2+Kc^2=\frac{8}{3}\pi G\rho a^2$$
I didn't understand if $\rho$ is the mass density deriving from $m_0$ (the rest mass) or from $\gamma m_0$. In other words $\rho c^2$ is the energy density due only to the rest energy $E=m_0 c^2$ or due to the total energy $E=\gamma m c^2$ (rest energy, Kinetic energy, internal energy...)?
I think that $\rho$ is $m_0\over V_0$ (where $V_0$ is the proper volume) so that I could write $\rho=\frac{m_0}{V_0}=\frac{E}{V_0 c^2}=\frac{\epsilon}{c^2}$, with $\epsilon$ density (rest) energy but I'm not sure.
Someone could make clear my ideas please?
Summary
We can summarize the point thanks to the help of Ben Crowell and Elio Fabri:
$\rho$ is in general the energy density of my cosmological fluid, but we are in the comoving frame so $\rho$ is related to rest energy ($Mc^2$) of the entire fluid (with mass $M$) because we see the fluid still. All the particles (galaxies) of the cosmological fluid contribute to this rest energy through their own energy (rest energy, kinetic energy, interaction energies), as in the example by Ben Crowell of the proton in a boby.
 A: The $\rho$ comes from the component $T^{tt}$ of the stress-energy tensor, which is the density of mass-energy $E$, not the density of mass. We never have any way of knowing or defining the density of mass. For example, I could say that a proton in my body has some mass which contributes to my mass, but its mass may actually be in forms such as the kinetic energy of its quarks. Also, mass is not additive in relativity, but the stress-energy tensor is a tensor, which means we want to be able to talk about adding stress-energy tensors.
BTW, do yourself a favor and stop writing factors of $c$ when you do GR. $c=1$ in any system of units that is sensible for GR.
A: $\def\qy#1#2{#1\,{\rm#2}} \def\10#1#2{#1\cdot10^{#2}}$
Let's start from Robertson-Walker metric:
$$ds^2 = dt^2 - a^2(t) \left(\!{dr^2 \over 1 - k\,r^2} + 
r^2 d\theta^2 + r^2 \sin^2\!\theta\,d\phi^2\!\right)$$
Two things must be noted with this choice of coordinates:
1) Time coordinate $t$ has coefficient $g_{tt}=1$. This means that $t$
is proper time for objects "standing still" at some $r,\theta,\phi$
space location.
2) For the same reason a component $u^t=u_t$ of a 4-vector, as well as
a component $T^{tt}=T_t^t=T_{tt}$ of a tensor are proper components,
i.e. they measure a physical quantity in local comoving frame.
This is relevant in order to understand the meaning of $\rho$ in
Friedmann equations. $\rho$ is another name for $T^{tt}$ i.e. energy
density. (BTW, I wouldn't use the term "mass-energy" as @BenCrowell
does. IMO it's apt to cause confusion.). Every energy contribution is
included, both for "cold" and for "hot" matter.
Maybe it could be useful to dwell on this point. By "cold" matter is
generally meant baryons, as they appear in stars, in galactic dust, in
sparse hydrogen atoms. It may look peculiar to call "cold" matter in the
centre of a star, but let's try to put some numbers.
Consider Sun's centre, where temperature reaches $\qy{10^7}K$. A
proton there has a kinetic energy of the order
$$kT \simeq \10{1.4}{-23}\cdot10^7 = \qy{\10{1.4}{-16}}J$$
to be compared with its "rest energy"
$$m\,c^2 \simeq \qy{\10{1.6}{-27}}{kg} \cdot \qy{10^{17}}{m^2/s^2} =
\qy{\10{1.6}{-10}}J.$$
So kinetic energy is 6 orders of magnitude smaller than rest energy
and is entirely negligible. Even for electrons there are 3 orders and
electrons energy is already unimportant in itself.
The opposite is true for photons, whose energy is entirely kinetic, and
for neutrinos, whose mass is extremely small. This is considered "hot"
(i.e. extreme relativistic) matter, also named "radiation". The contribution of radiation to $\rho$ is negligible for
$a(t)>0.01$. The ratio $\rho_{\rm hot}/\rho_{\rm cold}$ scales as
$1/a(t)$.
