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Consider the equation of motion of a scalar field $\phi(t,x^i)$, $$\nabla^\mu\nabla_\mu\phi=\frac{dV(\phi)}{d\phi}$$ where $V(\phi)$ is the potential. Specialise to a massless field ($V(\phi)=0$) in the FRW/FLRW metric (coordinates $\{t,r,\varphi,\theta\}$, with $\phi=\phi(t)$. I need to show that $\phi(t)-\phi(t_0)=k\int_{t_0}^tdt'a(t')^{-3}$ for some constant of integration $k$.

My attempt is as follows:

$$\nabla^\mu\nabla_\mu\phi=g^{\alpha\mu}(\partial_\alpha\nabla_\mu\phi-\Gamma^\lambda_{\alpha\mu}\nabla_\lambda\phi)=g^{tt}(\partial_t\partial_t\phi-\Gamma^\lambda_{tt}\partial_t\phi)+g^{ij}(\partial_i\partial_j\phi-\Gamma^\lambda_{ij}\partial_\lambda\phi)$$ where I used that $\nabla=\partial$ for scalars, and I carried out the implicit (Einstein) summation. Next, I use the following information

  • the only relevant non-zero Christoffel symbol is $\Gamma^t_{ij}=a\dot{a}\delta_{ij}$ (where $\dot{}$ denotes differentiation wrt $t$)
  • $\phi$ is a function of $t$ only, so $\partial_i\phi=0$
  • $g^{tt}=-1$ and $g^{ij}=\delta^{ij}/a^2$

to find

$$\nabla^\mu\nabla_\mu\phi=-\partial^2_t\phi-\frac{\delta^{ij}\delta_{ij}\dot{a}}{a}\partial_t\phi=-\partial^2_t\phi-\frac{3\dot{a}}{a}\partial_t\phi$$

Therefore, I am left with

$$-\partial^2_t\phi-\frac{3\dot{a}}{a}\partial_t\phi=0$$

No clue how to proceed. Any hints? Thanks.

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  • $\begingroup$ Have you checked if there is any hidden total derivative? $\endgroup$ – ApolloRa May 4 at 12:12
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Set $\Phi= \dot \phi$ then your equation is $$ \frac{\dot \Phi}{\Phi} =- 3 \frac{\dot a}{a} $$ or $$ \frac {d\ln \Phi}{dt}= -3 \frac{d\ln a}{dt} $$ so $\ln[\Phi(t)a^3(t)]=constant$, or equivalently $$ \Phi(t)= k a^{-3}(t). $$ Therefore $$ \phi(t)-\phi(0)= k\int^t_0 a^{-3}(t)dt. $$ TaDa!

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  • $\begingroup$ Thanks! I had realised that $a^{-1}\dot{a}=d(\ln a)/dt$ but for some reason I did not make any progress with it. Thanks again for your answer $\endgroup$ – martin May 4 at 12:23

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