2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$g=-a^6\Rightarrow \sqrt{-g}=a^3$$

$\endgroup$
  • $\begingroup$ I did not get. What about $r^4\sin^2\theta$? $\endgroup$ – mithusengupta123 Dec 27 '19 at 13:31
  • $\begingroup$ You're using Cartesian coordinates, so there is no $r$ values... $\endgroup$ – Kyle Kanos Dec 27 '19 at 13:35
  • $\begingroup$ What if one chooses spherical coordinates? $d^4x$ can also stand for $dt (r^2\sin\theta)drd\theta d\phi$ $\endgroup$ – mithusengupta123 Dec 27 '19 at 13:44
  • $\begingroup$ 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. $\endgroup$ – Kyle Kanos Dec 27 '19 at 13:58
  • $\begingroup$ The way you did this applies only to flat geometry ($k=0$), which is not the general case. Am I correct? $\endgroup$ – mithusengupta123 Dec 27 '19 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.