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 ?

  • 4
    $\begingroup$ 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 $\endgroup$
    – 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, $g_{tt} = c^{2}\,g_{00}$.


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.

  • 1
    $\begingroup$ 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. $\endgroup$ Oct 17 '12 at 7:17

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.