Deriving the correct expression for the action of the scalar field at the time of inflation

Question

In its matrix form, the FRW metric is $$g^{\rm FRW}_{\mu\nu}=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & -\frac{a^2}{1-kr^2} & 0 & 0\\0 & 0 & -a^2r^2 & 0\\0 & 0 & 0 & -a^2r^2\sin^2\theta\end{pmatrix}.$$ It's determinant is therefore, $$\det g=g=-\frac{a^6r^4\sin^2\theta}{1-kr^2}\Rightarrow \sqrt{-g}\neq a^3.$$

Assuming homogeneity $\nabla\phi=0$, the action of scalar field during inflation is $$S=\int d^4x\sqrt{-g}(\frac{1}{2}g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi-V(\phi))=\int d^4x\sqrt{-g}(\frac{1}{2}\dot\phi^2-V(\phi)).$$ But since $\sqrt{-g}\neq a^3$ my expression does not match with D. Tong's expression (1.81, NB: PDF to a preprint of Tong's text). How is this obtained starting from the Lagrangian above? Please help me spot my mistake.

Answer 1

Tong says he is using $k=0$, so one can easily use $$ \mathrm d\Sigma^2=\mathrm dx^2+\mathrm dy^2+\mathrm dz^2$$ instead of using radial coordinates. And hence your FLRW metric is of the form, $$g_{\mu\nu}=\text{diag}(1,\,-a^2,\,-a^2,\,-a^2).$$ This clearly gives the desired form of $$g=-a^6\Rightarrow \sqrt{-g}=a^3$$