# Scalar field lagrangian in curved spacetime

I am studying inflation theory for a scalar field $\phi$ in curved spacetime. I want to obtain Euler-Lagrange equations for the action:

$$I\left[\phi\right] = \int \left[\frac{1}{2}g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi + V\left(\phi\right) \right]\sqrt{-g} d^4x$$

Euler-Lagrange equations for a scalar field is given by

$$\partial_\mu \frac{\partial L}{\partial\left(\partial_\mu\phi\right)} - \frac{\partial L}{\partial \phi} = 0$$

$$\partial_\mu \frac{\partial L}{\partial\left(\partial_\mu\phi\right)} = \frac{1}{2}\partial_\mu\left(\sqrt{-g}g^{\mu\nu}\partial_\nu\phi \right)$$

$$\frac{\partial L}{\partial \phi} = \frac{\partial \left[\sqrt{-g}V\left(\phi\right)\right]}{\partial \phi}$$

But according to the book the resulting equation is

$$\frac{1}{\sqrt{-g}}\partial_\mu\left(\sqrt{-g}g^{\mu\nu}\partial_\nu\phi\right) = \frac{\partial V\left(\phi\right)}{\partial \phi}$$

What am I doing wrong?

• First, You forgot a $2$ factor, because the kinetic term is quadratic in first derivatives of $\phi$, and secondly, $\sqrt{-g}$ does not depend on $\phi$. – Trimok Aug 21 '13 at 9:18

The correct Euler-Lagrangian equation for scalar in curved spacetime is $$\frac{\partial\mathcal{L}}{\partial\phi}=\frac{1}{\sqrt{-g}}\partial_{\mu}\left[\sqrt{-g}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi\right)}\right],$$ where the Lagrangian density should be $$\mathcal{L}=\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi-V\left(\phi\right)$$ and it doesn't contain the $\sqrt{-g}$ factor. Note this is the same as $$\frac{\partial\mathcal{L}}{\partial\phi}=\nabla_{\mu}\left[\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi\right)}\right],$$ in terms of covariant derivative, $\nabla_{\mu}$.
The right-hand side is $$\nabla_{\mu}\left[\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi\right)}\right] = \nabla_{\mu}\left(g^{\mu\nu}\partial_{\nu}\phi\right) = g^{\mu\nu}\nabla_{\mu}\left(\partial_{\nu}\phi\right) \equiv \Box\phi,$$ where the second equality is true because the covariant derivative, $\nabla_{\mu}$, commutes with the metric tensor, $g^{\mu\nu}$. The left-hand side is $$\frac{\partial\mathcal{L}}{\partial\phi}=-\frac{\partial V\left(\phi\right)}{\partial\phi}.$$ So the equation of motion for a scalar field $\phi$ in curved spacetime is $$\Box\phi=-\frac{\partial V\left(\phi\right)}{\partial\phi}.$$
• Isn't it more generally written as $$\dfrac{\partial(\sqrt{-g}\mathcal{L})}{\partial\phi}]-\partial^{\mu}\left[\dfrac{\partial(\sqrt{-g}\mathcal{L})}{\partial(\partial^{\mu}\phi)}\right]=0$$? – Souradeep Jul 4 at 8:04