Mathematically deriving Hubble's constant

Assuming homogeneity and isotropy, the spacetime metric is of the form $g = -d\tau^2 + a^2(\tau)\gamma$ where $a$ is the scale factor and $\gamma$ is the spatial metric corresponding to a flat, spherical or hyperbolic universe. Given any two isotropic observers at time $\tau$, let $R$ be the spatial distance between them (so $R = a^2(\tau)\gamma$). Wald says that $\frac{dR}{d\tau} = \frac{R}{a}\frac{da}{d\tau}$. I calculate something different though: \begin{align*} \frac{dR}{d\tau} &= \lim\limits_{h \rightarrow 0} \frac{R(\tau + h) - R(\tau)}{h} \\ &= \lim\limits_{h \rightarrow 0} \frac{a^2(\tau + h)\gamma - a^2(\tau)\gamma}{h} \\ &= \gamma \lim\limits_{h \rightarrow 0} \frac{a^2(\tau + h) - a^2(\tau)}{h} \\ &= \gamma \frac{d(a^2)}{d\tau} \\ &= \gamma \cdot 2a \frac{da}{d\tau} \\ &= \frac{R}{a^2} \cdot 2a \frac{da}{d\tau} \\ &= 2\frac{R}{a}\frac{da}{d\tau} \end{align*} Where did I go wrong?

You statement $R = a^2 \gamma$ is wrong at least because of the dimensions. The correct one would be the following: if the 2 observers had initial coordinate distance $r$, then the physical distance between them is $R = a r$. Then, follows:
$$R(\tau) = a(\tau) r$$
$$\frac{d R}{d \tau} = \frac{d a}{d \tau} r = \frac{d a}{d \tau} \frac{R}{a}$$
On the physical distance. By physical here I mean the one calculated in the local Minkowski metric — and coordinate transformation for it will be precisely the definition of $R$
• Ah, $a^2(\tau)\gamma$ gives the squared distance. When I correct this, I get the correct calculation. Commented Jun 19, 2016 at 22:47