# The Ricci scalar in FRW, where am I getting wrong?

I'm trying to derive Ricci scalar with FRW metric, but additional $$c^2$$ makes me confused.

The book by D. Baumann says \begin{align} R &= g^{\mu\nu}R_{\mu\nu} \\ &= -R_{00}+\frac{1}{a^2}R_{ii}=\frac{6}{c^2}\left[\frac{\ddot{a}}{a}+\left(\frac{\dot{a}}{a}\right)^2+\frac{kc^2}{a^2R_0^2}\right],\tag{2.133} \end{align} where $$R_{00}=-\frac{3}{c^2}\,\frac{\ddot{a}}{a}$$

And for the calculation of the Ricci scalar, since $$g^{\mu \nu}$$ is inverse of $$g_{\mu\nu}$$ so I thought if $$g_{00} = -c^2$$ then inverse of it should be $$g^{00}=-1/c^2$$ because the FRW metric says $$ds^2 = -c^2dt^2 + a^2(t)\gamma^{ij}dx^idx^j$$.

My assumption on Ricci scalar was $$R = g^{00}R_{00} + g^{ij}R_{ij} = -1/c^2 R_{00} + 1/a^2 R_{ii}$$(in cartesian coordinate with $$x=0$$.)

But I don't understand why the additional $$1/c^2$$ factor is missing on the text book.

Or is it conventional not to right speed of light $$c$$ in $$g_{00}$$ component of the metric? So that even it is in SI unit, not natural unit, I should right down $$g^{00} = -1$$, not $$-c^2$$?

• Images of mathematical expressions are very strongly discouraged here. Please use Mathjax instead. Dec 20, 2022 at 13:43
• $g^{\mu\nu}$ being the inverse of $g_{\mu\nu}$ does not mean that $g_{00}= 1/g^{00}$. It is a matrix inverse. Dec 20, 2022 at 14:13
• Since $ds^2=g_{\mu\nu}dx^\mu dx^\nu$ it's usually convenient to choose coordinates for which $[x^\mu]=\mathsf{L}$ for all $\mu$, so $[ds^2]=\mathsf{L}^2$ while $g_{\mu\nu}$ is dimensionless. For example, in this convention we take $x^0=ct$ rather than $x^0=t$. (However, it gets awkward if the space coordinates might require e.g. $x^2=R\theta,\,x^3=R\phi$ for a suitable length $R$, rather than $x^2=\theta,\,x^3=\phi$; one might even take $R=r=x^1$.) Then $[\Gamma^\mu_{\nu\rho}]=\mathsf{L}^{-1}$, while Riemann/Ricci terms are $\mathsf{L}^{-2}$.
– J.G.
Dec 20, 2022 at 14:24
• I'm sorry for uploading image instead of equations. I'll make sure not to upload image next time Thank you for letting me know. And now i understand what was wrong with my assumption. Thank you.
– hwan
Dec 20, 2022 at 15:22
• @mikestone Not in general, but if the metric is diagonal (like the FLRW is), then it is correct. Dec 20, 2022 at 15:39

It is a common convention to write $$c=1$$ (and often $$\hbar=1$$) since we can use dimensional analysis at the end to redimensionalise our equations (there will be a unique way of writing the powers of $$c$$ and $$\hbar$$). The lecture notes by Baumann use exactly this convention (so I assume the book does too). As such, the metric would read $$ds^2=-dt^2+a^2(t)\gamma_{ij}(x^\mu)dx^idx^j$$.
A note on metric inversion: When calculating the inverse, as others have noted, this will be a matrix inverse, however often the metric is already diagonal and of the form (in FRW)) $$ds^2 = -dt^2 + a^2(t)f^i(x^\mu)dx_i^2$$. Then the components of the inverse metric are indeed just the reciprocals of the coefficients of $$dt^2$$, $$dx_i^2$$ etc.
Punchline: Yes, this should just be a matter of convention with the author writing $$c=1$$.