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

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.

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 $$g=-a^6\Rightarrow \sqrt{-g}=a^3$$
• I did not get. What about $r^4\sin^2\theta$? – mithusengupta123 Dec 27 '19 at 13:31
• You're using Cartesian coordinates, so there is no $r$ values... – Kyle Kanos Dec 27 '19 at 13:35
• What if one chooses spherical coordinates? $d^4x$ can also stand for $dt (r^2\sin\theta)drd\theta d\phi$ – mithusengupta123 Dec 27 '19 at 13:44
• Then you clearly get those pesky sine functions and $r$ terms appearing that make your life more difficult (because you'd have to cancel them out with the gradient operators, I imagine). Or you could make life super easy and just user Cartesian coordinates. – Kyle Kanos Dec 27 '19 at 13:58
• The way you did this applies only to flat geometry ($k=0$), which is not the general case. Am I correct? – mithusengupta123 Dec 27 '19 at 14:02