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 ?

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I won't offer an answer, only the comment that if you see equations that are missing constants, then odds are they have been set to 1 and are working with planck units. This is definitely a source of confusion when authors switch without warning, or don't say so up front – user11547 Oct 17 '12 at 10:15

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, $g_{tt} = c^{2}\,g_{00}$.

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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.

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Now, whether there are the right numbers of $c^2$ floating around is a different issue; I haven't checked the computations on Wiki myself so can't say anything. But because of the identification of space and time, one cannot use dimensional analysis to check whether the number of $c$s are correct, since $c$ is essentially dimension free. – Willie Wong Oct 17 '12 at 7:17

These equations are often done in units where c=1 to make things easier, in this case: $$c^2=1$$

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