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?

  • 8
    $\begingroup$ 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 $\phi$. $\endgroup$ – 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 $$ \Box\phi=-\frac{\partial V\left(\phi\right)}{\partial\phi}. $$

| cite | improve this answer | |
  • 1
    $\begingroup$ Isn't it more generally written as $$\dfrac{\partial(\sqrt{-g}\mathcal{L})}{\partial\phi}]-\partial^{\mu}\left[\dfrac{\partial(\sqrt{-g}\mathcal{L})}{\partial(\partial^{\mu}\phi)}\right]=0$$? $\endgroup$ – Souradeep Jul 4 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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