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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?

Sancol.
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    What have you tried so far? – Jeanbaptiste Roux Feb 04 '23 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 04 '23 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. – Ghoster Feb 05 '23 at 06:46
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    Were you working in $1+1$ rather than $3+1$? – Ghoster Feb 05 '23 at 06:53
  • Yes I was working in 2 dimensions –  Feb 05 '23 at 13:29
  • You’ve offered a bounty, so you must not have understood my comment. What part of it did you not understand? – Ghoster Feb 06 '23 at 17:17
  • You have to do the conformal transformation (3.1) in the link https://books.google.be/books?hl=fr&lr=&id=SEnaUnrqzrUC&oi=fnd&pg=PP11&dq=N.D.+Birrell+and+P.C.W.+Davies,+Quantum+fields+in+curved+space,+Cambridge+University+Press,+Canbridge+U.K.+(1982)&ots=1IUWFfIyq9&sig=Us3nXdQ9MR9W4UH0Rg5MHIj_uP8&redir_esc=y#v=onepage&q&f=false ,, there is another equation (1) in another case https://arxiv.org/pdf/1211.5176.pdf – The Tiler Feb 08 '23 at 10:40
  • You can write $\eta$ as \eta (instead of using $n$ for conformal time). – Ghoster Feb 10 '23 at 00:09
  • @Ghoster when I do that, I still don’t have any $a$ term in the denominator, because $a(t)^8$ multiplying by $a(t)^{-2}$ (inverse metric) it gives $a(t)^6$ which taking the time derivative gives $6a(t)^5 a’(t)$. –  Feb 10 '23 at 02:15
  • No. In your KG equation, what multiplies the inverse metric is not the negative determinant $g=a ^8$ but rather its square root $g^{1/2}=a^4$. Try again. – Ghoster Feb 10 '23 at 06:10
  • @ Ghoster: In the other question there is a conformal transformation:https://physics.stackexchange.com/questions/38138/conformal-transformation-weyl-scaling-are-they-two-different-things-confused – The Tiler Jan 09 '24 at 07:17

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

The Tiler
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  • 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 '23 at 01:36
  • @Obama2020 You can. – Ghoster Feb 10 '23 at 06:13
  • 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 $g^{-1/2}\partial_{u}(g^{1/2}g^{\mu\nu}\partial_{\nu})$. (*)https://arxiv.org/pdf/1909.01292.pdf – The Tiler Feb 10 '23 at 06:24
  • 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.. – The Tiler Feb 10 '23 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? – R. Rankin Nov 16 '23 at 01:26