# Scalar field equation of motion in FRW metric

Consider a scalar field $$\phi$$ with the following Lagrangian density:

$$\mathscr{L}=-\frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi-V(\phi),$$ and consider a FRW metric, whose line element is given by $$\mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left[\frac{\mathrm{d} r^{2}}{1-k r^{2}} + r^{2} \mathrm{d}\theta^{2} + r^{2} \sin^{2} \theta \mathrm{d}\phi^{2}\right],$$ with $$a(t)$$ being the FRW scale factor. According to e.g. Turner 1983, the equation for motion for $$\phi$$ in this setting turns out to be $$\ddot{\phi}+3 H \dot{\phi}+V^{\prime}(\phi)=0.$$

How do I derive this? I have varied the action of the scalar field and obtained the scalar field equation of motion for a generic metric $$g_{\mu\nu}$$: $$g_{\mu\nu} \partial^\mu \partial^\nu \phi - \frac{\delta V(\phi)}{\delta \phi}=0.$$ Now, I suppose that the $$\ddot{\phi}$$ term in the equation of motion is sourced by the $$g_{00}$$ component of the metric tensor, but what is the origin of the term $$3 H \dot{\phi}$$ given the metric I have written above?

• Are you sure the $g_{\mu\nu} \partial^\mu \partial^\nu \phi$ term shouldn't read $g_{\mu\nu} \nabla^\mu \nabla^\nu \phi$? Commented Jun 13, 2022 at 16:09
• @scaphys I have varied the action with respect to the field, hence the term depending on $\delta \sqrt{-g} / \delta \phi$ should be null Commented Jun 13, 2022 at 16:16
• @NíckolasAlves I thought covariant derivatives of a scalar coincided with its gradients Commented Jun 13, 2022 at 16:17
• @gangio not the double covariant derivative: $\nabla \partial \neq \partial \partial$ Commented Jun 13, 2022 at 16:32

You made a mistake when you varied the action. Explicitly, the Lagrangian density is: $$\mathcal L = (-\frac{1}{2}g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi-V(\phi))\sqrt{-g}$$ so the Euler-Lagrange equations actually give you: $$-\partial_\mu (\sqrt{-g}g^{\mu\nu}\partial_\nu\phi)+\sqrt{-g}V'(\phi) = 0$$ which you usually rearrange as: $$-\frac{1}{\sqrt{-g}}\partial_\mu (\sqrt{-g}g^{\mu\nu}\partial_\nu\phi)+V'(\phi) = 0$$ and you recognize in the first term the Laplace-Beltrami operator, which is a covariant quantity that you can rewrite covariantly as $$\nabla_\mu\nabla^\mu\phi$$.

The author was considering spatially homogeneous solutions, ie $$\phi(t)$$, so calculating: $$g = -a^6\frac{r^4\sin^2\theta}{1-kr^2}$$ the equation simplifies to: $$-\frac{1}{\sqrt{-g}}\frac{d}{dt} (-\sqrt{-g}\dot\phi)+V'(\phi) = 0 \\ \ddot\phi+3\frac{\dot a}{a}\phi+V'(\phi) = 0 \\$$ and if you set the Hubble constant to $$H = \frac{\dot a}{a}$$, you obtain the advertised equation (you'll notice that the factor $$3$$ comes from the $$3$$ spatial dimensions).

Hope this helps and tell me if something is not clear.

• That was super helpful, thank you! Only thing I am struggling with is how to vary the lagrangian by $\delta \partial_\mu \phi$. E.g. is it right if I say $\frac{\delta \mathrm{L}}{\delta \partial_\mu \phi} = -\frac{1}{2}g^{\mu\nu}\frac{\delta \partial_\mu \phi}{\delta \partial_\mu \phi}\partial_\nu \phi -\frac{1}{2}g^{\mu\nu}\partial_\mu \phi \frac{\delta \partial_\nu \phi}{\delta \partial_\mu \phi} = -\frac{1}{2}g^{\mu\nu}\partial_\nu \phi - \frac{1}{2} g^{\mu\nu}\partial_\mu \phi \delta_\mu^\nu$? Commented Jun 13, 2022 at 20:13
• No problem! You got the idea but the formalism is a bit off. Remember, the repeated indices in the Einstein convention are bound. You’d rather get: $$\frac{\delta L}{\delta \partial_\sigma\phi} = -\frac{1}{2}g^{\mu\nu}(\delta^\sigma_\mu\partial_\nu\phi+ \delta^\sigma_\nu\partial_\mu\phi)$$ using $$\frac{\delta \partial_\alpha\phi }{\delta \partial_\beta\phi}=\delta^\beta_\alpha$$
– LPZ
Commented Jun 13, 2022 at 20:36
• You are right of course, many thanks for your time :) Commented Jun 14, 2022 at 7:42
• sorry to bother you, but I have further (very basic) doubts. The term $V'(\phi)$ appearing in the Euler-Lagrange equation is sourced by the term $\frac{\delta \mathrm{L}}{\delta \phi}$ right? If this is true, does this imply that $\frac{\delta \partial_\nu \phi}{\delta \phi}=0$? Thank you Commented Jun 14, 2022 at 13:58
• $\mathcal L$ is a function of $\phi, \partial_\mu\phi$, so they are treated as independent variables (coordinates of the tangent bundle). Therefore, you have $\frac{\partial (\partial_\mu\phi)}{\partial \phi} = 0$ as well as $\frac{\partial \phi}{\partial (\partial_\mu\phi)} = 0$. Remember that partial derivatives mean you are keeping other variables constant, most subtleties involve in defining precisely which other variables are you keeping constant.
– LPZ
Commented Jun 14, 2022 at 14:53

You do not obtain the $$\frac{\dot{a}}{a}$$ term probably because you have not used covariant derivative. I did the same mistake once...

The field is a scalar so when acting on the field, the covariant derivative reduces to the standard partial one. But the derivatives of the scalar field are the component of a vector field. Thus, to compute the second order derivatives you must use the covariant one. The equation of motion is then $$$$\nabla_{\mu}\left(\frac{\partial L}{\partial (\partial_{\mu}\phi)}\right)-\frac{\partial L}{\partial \phi}.$$$$

The covariant derivative act on the component of the $$v=v^{\nu}\boldsymbol{e}_{\nu}$$ where $$\boldsymbol{e}_{\nu}$$ is a basis vector, as: $$$$\nabla_{\mu}v^{\nu}=\partial_{\mu}v^{\nu}+v^{\alpha}\Gamma^{\nu}_{\mu\alpha}.$$$$ If you apply this to the case in which $$v^{\nu}=\partial^{\nu}\phi$$ and substitute the christoffel symbols for the FLRW metric, you will obtain the correct equation of motion.

I should mention that there is an abuse of notation. The covariant derivative acts on vector not on components. What i have written is the $$\nu$$ component of the covariant derivative of the $$v$$ vector, which is usually indicated as $$v^{\nu}_{;\mu}$$ instead of $$\nabla_{\mu}v^{\nu}$$. Written out explicitly is $$\nabla_{\mu}v=v^{\nu}_{;\mu}\boldsymbol{e}_{\nu}$$.

Hope this helps.