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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.
  • 960

1 Answers1

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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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