# Klein-Gordon equation in FRW spacetime

The metric for frw spacetime is $$ds^2=a(n)^2(dn^2 - dx^2)$$ where $$dn$$ is the conformal time differential form. The Klein Gordon equation in curved spacetime is $$\left(\frac{1}{g^{1/2}}\partial_{\mu}(g^{1/2}g^{\mu\nu}\partial_{\nu}) + m^2\right)\phi = 0$$

From this one can obtain the Klein Gordon equation for frw spacetime

$$\ddot{\phi} + 2\frac{\dot{a}}{a}\dot{\phi} - \Delta \phi + m^2a^2\phi = 0$$ (mukhanov 64)

How do we derive this equation from the above equation?

• What have you tried so far? Feb 4 at 15:15
• @JeanbaptisteRoux I’ve calculated it explicitly, but I find that $g^{\mu\nu} = \frac{1}{a(t)^4}n_{\mu\nu}$ so that when the square root of the determinant of the metric (which is $a(t)^2$ acts on the inverse frw metric 1/a(t)^2, they’d cancel out implying that there can be no terms that depend on $a(t)$ Feb 4 at 15:29
• The metric should be written with $d{\vec x}^2$, not $dx^2$. The inverse metric has $a^{-2}$, not $a^{-4}$. The negative determinant is $g=a^8$ so $g^{1/2}=a^4$. There is no cancellation. The final equation follows. Feb 5 at 6:46
• Were you working in $1+1$ rather than $3+1$? Feb 5 at 6:53
• Yes I was working in 2 dimensions Feb 5 at 13:29

According to formula 102 and 107 page 341 with :

$$ds^{2}=a^{2}(n)(dn^{2}-d\chi^{2})=a^{2}(n)(dn^{2}+d(ix)^{2}+d(iy)^{2}+d(iz)^{2})$$, $$H^{2}_{0}=H^{2}_{1}=...=a^{2}(n)$$ $$\Delta_{H}=\frac{1}{H_{0}H_{1}H_{2}H_{3}}\frac{\partial}{\partial q_{i}}\left(\frac{H_{0}H_{1}H_{2}H_{3}}{H^{2}_{i}}\frac{\partial}{\partial q_{i}}\right)$$

$$H_{0}H_{1}H_{2}H_{3}=a^{4}$$ $$\Delta_{H}=\frac{1}{a^{4}}\frac{\partial}{\partial q_{i}}\left(\frac{a^{4}}{H^{2}_{i}}\frac{\partial}{\partial q_{i}}\right)$$ with $$\;q_{0}=n\;,...,\;q_{3}=iz\;\;,$$ we have:$$\Delta_{H}=\frac{2a}{a^{4}}\frac{\partial a(n) }{\partial n}\left(\frac{\partial}{\partial n}\right)+\frac{1}{a^{2}}\frac{\partial }{\partial n}\left(\frac{\partial}{\partial n}\right)+\frac{1}{a^{2}}\frac{\partial }{\partial ix}\left(\frac{\partial}{\partial ix}\right)+\frac{1}{a^{2}}\frac{\partial }{\partial iy}\left(\frac{\partial}{\partial iy}\right)+\frac{1}{a^{2}}\frac{\partial }{\partial iz}\left(\frac{\partial}{\partial iz}\right)$$

$$\Delta_{H}=\frac{1}{a^{2}}\partial^{2}_{nn}+\frac{2a}{a^{4}}\dot{a}(n) \partial_{n}-\frac{1}{a^{2}}(\partial^{2}_{xx}+\partial^{2}_{yy}+\partial^{2}_{zz})$$ we have $$(\Delta_{H}+m^{2})\phi=0 \Leftrightarrow \left(\partial^{2}_{nn}+2\frac{\dot{a}}{a} \partial_{n}-\Delta+m^{2}a^{2}\right)\phi=0$$

• Isn’t there a much simpler way of going about this? Presumably by just carrying out the calculation explicitly by substituting in the metric tensor and determinant? Why can’t we do it that way? Feb 10 at 1:36
• @Obama2020 You can. Feb 10 at 6:13
• This is the simplest method I found (I went back to the source (maths)) after a little research on google scholar* , we put $g=H^{2}$ in formula (107) we have $g^{-1/2}\partial_{u}(g^{1/2}g^{\mu\nu}\partial_{\nu})$. (*)arxiv.org/pdf/1909.01292.pdf Feb 10 at 6:24
• In the link I gave in my comment before, even the general equation from page (11) $\;\;\partial _{\tau}(a^{n-1}\partial\phi_{\tau})=a^{n-1}\Delta \phi\;\;$, n=3=spatial dimension, gives me a factor $\frac{m}{a^{2}}$ , equation I already mentioned in a deleted comment.. Feb 10 at 11:43
• Not the question (I know) but in general shouldn't we go back to the action itself and vary this Klein-Gordon action w.r.t. $a$ and get another whole term? Nov 16 at 1:26