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?

  • $\begingroup$ Are you sure the $g_{\mu\nu} \partial^\mu \partial^\nu \phi$ term shouldn't read $g_{\mu\nu} \nabla^\mu \nabla^\nu \phi$? $\endgroup$ Commented Jun 13, 2022 at 16:09
  • $\begingroup$ @scaphys I have varied the action with respect to the field, hence the term depending on $\delta \sqrt{-g} / \delta \phi$ should be null $\endgroup$
    – gangio
    Commented Jun 13, 2022 at 16:16
  • $\begingroup$ @NíckolasAlves I thought covariant derivatives of a scalar coincided with its gradients $\endgroup$
    – gangio
    Commented Jun 13, 2022 at 16:17
  • 3
    $\begingroup$ @gangio not the double covariant derivative: $\nabla \partial \neq \partial \partial$ $\endgroup$
    – Eletie
    Commented Jun 13, 2022 at 16:32

2 Answers 2


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.

  • $\begingroup$ 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 $\frac{\delta \mathrm{L}}{\delta \partial_\mu \phi} = -\frac{1}{2}g^{\mu\nu}\frac{\delta \partial_\mu \phi}{\delta \partial_\mu \phi}\partial_\nu \phi -\frac{1}{2}g^{\mu\nu}\partial_\mu \phi \frac{\delta \partial_\nu \phi}{\delta \partial_\mu \phi} = -\frac{1}{2}g^{\mu\nu}\partial_\nu \phi - \frac{1}{2} g^{\mu\nu}\partial_\mu \phi \delta_\mu^\nu$? $\endgroup$
    – gangio
    Commented Jun 13, 2022 at 20:13
  • $\begingroup$ 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 $$\frac{\delta \partial_\alpha\phi }{\delta \partial_\beta\phi}=\delta^\beta_\alpha$$ $\endgroup$
    – LPZ
    Commented Jun 13, 2022 at 20:36
  • $\begingroup$ You are right of course, many thanks for your time :) $\endgroup$
    – gangio
    Commented Jun 14, 2022 at 7:42
  • $\begingroup$ 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 $\endgroup$
    – gangio
    Commented Jun 14, 2022 at 13:58
  • $\begingroup$ $\mathcal L$ 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. $\endgroup$
    – 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 \begin{equation} \nabla_{\mu}\left(\frac{\partial L}{\partial (\partial_{\mu}\phi)}\right)-\frac{\partial L}{\partial \phi}. \end{equation}

The covariant derivative act on the component of the $v=v^{\nu}\boldsymbol{e}_{\nu}$ where $\boldsymbol{e}_{\nu}$ is a basis vector, as: \begin{equation} \nabla_{\mu}v^{\nu}=\partial_{\mu}v^{\nu}+v^{\alpha}\Gamma^{\nu}_{\mu\alpha}. \end{equation} 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.