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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?

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

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  • 2
    $\begingroup$ 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

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