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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 $-c^2$?

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  • $\begingroup$ Images of mathematical expressions are very strongly discouraged here. Please use Mathjax instead. $\endgroup$ Dec 20, 2022 at 13:43
  • $\begingroup$ $g^{\mu\nu}$ being the inverse of $g_{\mu\nu}$ does not mean that $g_{00}= 1/g^{00}$. It is a matrix inverse. $\endgroup$
    – mike stone
    Dec 20, 2022 at 14:13
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    $\begingroup$ 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 $\mathsf{L}^{-2}$. $\endgroup$
    – J.G.
    Dec 20, 2022 at 14:24
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    $\begingroup$ 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. $\endgroup$
    – hwan
    Dec 20, 2022 at 15:22
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    $\begingroup$ @mikestone Not in general, but if the metric is diagonal (like the FLRW is), then it is correct. $\endgroup$
    – Javier
    Dec 20, 2022 at 15:39

1 Answer 1

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I don't have the required 50 reputation to write this as a comment so this will be an answer instead.

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 $ds^2=-dt^2+a^2(t)\gamma_{ij}(x^\mu)dx^idx^j$.

It is worth noting that, under this convention, time now has the unit of length.

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 $c=1$.

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