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I try to drive The Klein-Gordon equation for a massless scalar field in case of FRW metric:

$$ ds^2= a^2(t) [-dt^2 + dx^2] $$

So I start by:

$$\left(\frac{1}{g^{1/2}}\partial_{\mu}(g^{1/2}g^{\mu\nu}\partial_{\nu}) \right)\phi = 0$$

The determinant of the metric is $(\sqrt{-g}=a^4)$. So that the equation leads to:

\begin{eqnarray} \nonumber &&\frac{1}{a^4} \left( \partial_0 a^4 g^{0\nu} \partial_\nu + \partial_i a^4 g^{i\nu} \partial_\nu \right) \phi=0, \\ \nonumber && \partial_0 a^4 g^{00} \partial_0 \phi + a^4 g^{ij} \partial_i \partial_j \phi=0,\\ \nonumber && - \partial_0 a^2 \partial_0 \phi + a^4 \nabla^2 \phi=0, \\ \nonumber && - 2 a \dot{a} \dot{\phi} - a^2 \ddot{\phi} + a^4 \nabla^2 \phi =0,\\ && 2 \frac{ \dot{a}}{a^3} \dot{\phi} + \frac{1}{a^2} \ddot{\phi} - \nabla^2 \phi=0. \end{eqnarray}

Please let me know if this derivation is correct.

Qmechanic
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Dr. phy
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    Please let me know if this derivation is correct. Check-my-work questions are off-topic on this site, so your post may get closed. – Ghoster Jan 08 '24 at 17:41
  • You’ve written the metric in a way that makes it look like you are working in 2D spacetime, which I don’t think you’re actually doing. – Ghoster Jan 08 '24 at 17:44
  • Klein-Gordon equation with non-zero mass: https://physics.stackexchange.com/q/748626/ – The Tiler Jan 08 '24 at 17:54
  • @TheTiler. Thanks for the link. The Klein-Gordon equation is given in this thread by $\ddot{\phi} + 2\frac{\dot{a}}{a}\dot{\phi} - \Delta \phi + m^2a^2\phi = 0$ for the same metric, but I wonder why there is a difference a factor $\frac{1}{a^2}$ from my result. – Dr. phy Jan 08 '24 at 19:11
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    The $\Delta$ in the other answer is the Laplacian w.r.t. $\delta_{ij}$. The $\nabla^2$ you are using is w.r.t. the metric $g_{ij} = a^2 \delta_{ij}$. Consequently, $\nabla^2= \frac{1}{a^2} \Delta$ – Prahar Jan 08 '24 at 19:40

1 Answers1

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I soppuse that the answer should be

\begin{eqnarray} \nonumber &&\frac{1}{a^4} \left( \partial_0 a^4 g^{0\nu} \partial_\nu + \partial_i a^4 g^{i\nu} \partial_\nu \right) \phi=0, \\ \nonumber && \partial_0 a^4 g^{00} \partial_0 \phi + a^4 g^{ij} \partial_i \partial_j \phi=0,\\ \nonumber && (\partial_0 a^4) g^{00} \partial_0 \phi+ a^4(\partial_0 g^{00}) \partial_0 \phi +a^4 g^{00} \partial^2_0 \phi+ a^4 \nabla^2 \phi=0, \\ \nonumber && 4 a^3 \dot{a} (-a^{-2}) \partial_0 \phi+ a^4 (2 a^{-3} \dot{a}) \partial_0 \phi +a^4 (-a^{-2}) \ddot{\phi} + a^4 \nabla^2 \phi=0, \\ \nonumber && -4 a \dot{a} \dot {\phi}+2 a \dot{a} \dot{\phi} -a^2 \ddot{\phi} +a^4 \nabla^2 \phi=0, \\ \nonumber && 2 \frac{ \dot{a}}{a^3} \dot{\phi} + \frac{1}{a^2} \ddot{\phi} - \nabla^2 \phi=0. \end{eqnarray} so I believe it is fine

Gabriel Palau
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  • Thanks for your answer @GabrielPalau. – Dr. phy Jan 08 '24 at 19:14
  • I just wonder why in that thread https://physics.stackexchange.com/questions/748626/klein-gordon-equation-in-frw-spacetime. there is a difference a factor $\frac{1}{a^2}$ from my result. – Dr. phy Jan 08 '24 at 19:15
  • In the other question there is a conformal transformation:https://physics.stackexchange.com/questions/612945/is-conformal-transformation-a-coordinate-transformation-or-not – The Tiler Jan 09 '24 at 07:22