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?

  • $\begingroup$ What have you tried so far? $\endgroup$ Feb 4 at 15:15
  • $\begingroup$ @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 $a(t)$ $\endgroup$
    – Obama2020
    Feb 4 at 15:29
  • $\begingroup$ 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. $\endgroup$
    – Ghoster
    Feb 5 at 6:46
  • 1
    $\begingroup$ Were you working in $1+1$ rather than $3+1$? $\endgroup$
    – Ghoster
    Feb 5 at 6:53
  • $\begingroup$ Yes I was working in 2 dimensions $\endgroup$
    – Obama2020
    Feb 5 at 13:29

1 Answer 1


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

  • $\begingroup$ 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? $\endgroup$
    – Obama2020
    Feb 10 at 1:36
  • $\begingroup$ @Obama2020 You can. $\endgroup$
    – Ghoster
    Feb 10 at 6:13
  • $\begingroup$ 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 $\endgroup$
    – The Tiler
    Feb 10 at 6:24
  • $\begingroup$ 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.. $\endgroup$
    – The Tiler
    Feb 10 at 11:43
  • $\begingroup$ 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? $\endgroup$
    – R. Rankin
    Nov 16 at 1:26

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