# Equation of motion of massless scalar field in F(L)RW spacetime

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.

• Have you checked if there is any hidden total derivative? – ApolloRa May 4 '20 at 12:12

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!
• 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 – martin May 4 '20 at 12:23