Einstein tensor in Friedmann equations : where is the missing $c^2$? I would like to demonstrate the several forms of the Friedmann equations WITH the $c^2$ factors. Everything is fine ... apart that I have a missing $c^2$ factor somewhere.
In all the following $\rho$ is the mass density and not the energy density $\rho_{E}=\rho c^2$
If we look at the wikipedia French page concerning the Friedmann equations, according to the demonstration of the last paragraph we have :
The Einstein field equation : $G_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu\nu}$
The Einstein tensor : 
$G_{\mu\nu} = 
\begin{pmatrix} 
G_{00}&0&0&0 \\
0&G_{ij}&0&0 \\
0&0&G_{ij}&0 \\
0&0&0&G_{ij}
\end{pmatrix}$
The Energy-Momentum tensor :
$T_{\mu\nu} = 
\begin{pmatrix} 
T_{00}&0&0&0 \\
0&T_{ij}&0&0 \\
0&0&T_{ij}&0 \\
0&0&0&T_{ij}
\end{pmatrix}$
with :
$G_{00} = 3H^2+3\frac{k}{a^2}c^2$
$G_{ij} = -\left(3\frac{H^2}{c^2}+2\frac{\dot{H}}{c^2}+\frac{k}{a^2}\right)$
$T_{00} = \rho c^2$
$T_{ij} = -P$
But : $T_{00}$ and $T_{ij}$ have the same physical unit ($P$ and $\rho c^2$ are in $kg.m^{-1}.s^{-2}$) whereas $G_{00}$ and $G_{ij}$ does not have the same unit : in the first one we have $H^2$ and in the second one we have $\frac{H^2}{c^2}$ for example.
My question are : is there a mistake in the french wikipedia demonstration ? Where is the missing $c^2$ ? Where can I find a good demonstration with the $c^2$ factors ?
EDIT :
Maybe I've found something. At the beginning of the demonstration, the author say that the metric is of the form :
$ds^2=c^2dt-a^2\gamma_{ij} dx^i dx^j$
where $\gamma_{ij}$ depends on the coordinates choice. This formula seems ok to me. 
But then he writes that :
$g_{00} = c^2$
$g_{ij} = -a^2\gamma_{ij} $
I have a doubt on $g_{00}$ : is it equal to $c^2$ or to $1$ ? In fact, if we choose to write $g_{00} = c^2$, then $T_{00} = \rho c^4$ isn't it ?
 A: I recently re-derived these equations with all the dimensionful constants in place. Your last statement in the "Edit" is correct: $T_{00} = \rho_{E}\,c^{2} = \rho\,c^{4}$. It's easy to lose track of factors of $c$ in calculations like this; the usual culprit is mixing up $t$ and $x^{0} = c\,t$, and $\partial_t$ and $\partial_0 = c^{-1}\,\partial_{t}$. For instance, $g_{tt} = c^{2}\,g_{00}$. 
A: Actually, in the context of general relativity, $c$ has no (physical) unit. 
More precisely, $c$ is meter per second. Meter is a measure of length. Second is a measure of time. In GR we unified space and time, and hence a meter and a second are different units of measurement for the "same thing". The number $c$ is a pure scalar that is just a conversion factor. 
In terms of more everyday situations, consider the unit "millimeters per meter". This has no physical unit, and is a pure scalar that equals to 1000. It represents a conversion factor between two different ways of measuring the same physical unit. $c$ is like that in relativity. 
A: These equations are often done in units where c=1 to make things easier, in this case:
$$
c^2=1
$$
A: I don't understand Roberts answer based on dimensional analysis.
$T_{00}$ has dimensions of energy density. He established this by writing $\rho_E$ which is an energy density. Further he then attached a factor of the speed of light squared to it as
$$T_{00} = \rho_E c^2$$
To me this is wrong. Rather:
$$T_{00} = \rho_E = \rho c^2$$
Is correct.
They also justified
$$T_{00} = \rho c^4$$
By faulty consequence. The dimensions are simply wrong concerning the answers given to you. The energy density is usually defined by writing it as a mass density with a coefficient of $c^2$ making the appropriate dimensions balance.
Where the latter refers to a mass density. So no, the previous comments are wrong and the wiki article is correct.
Also, the energy density and pressure have the same units, so there is no contradiction in the wiki article, as suggested by the OP.
Edit: Also, $g_{00} =c^2$ is generally wrong. The units are almost right (but usually we write it as $\phi$ weighted by $c^2$), but the metric is usually written as
$g_{00} = -(1 + \frac{\phi}{c^2})$
Robert is right concerning tracking the speed of light in context of the partial derivatives. When dealing with an object that has space dimensions on one side, and time on another, will require that it is fixed by coefficient of the speed of light on the space component. However, it is more conventional to just put an inverse of the speed of light on the temporal part so that it has spacelike units, just like so:
$\frac{1}{c^2}\frac{\partial^2}{\partial t^2} = \frac{\partial^2}{\partial x^2}$
