Scalar field lagrangian in curved spacetime

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SUMMARY

The discussion focuses on deriving the Euler-Lagrange equations for a scalar field \(\phi\) in curved spacetime, specifically within the context of inflation theory. The action is defined as \(I[\phi] = \int \left[\frac{1}{2}g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi + V(\phi)\right]\sqrt{-g} d^4x\). Key errors identified include a typo in the derivative notation and the omission of a factor of 1/2 in the derivative calculation. The correct resulting equation is \(\frac{1}{\sqrt{-g}}\partial_\mu\left(\sqrt{-g}g^{\mu\nu}\partial_\nu\phi\right) = \frac{\partial V(\phi)}{\partial \phi}\).

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Researchers in theoretical physics, particularly those focusing on cosmology, inflation theory, and field theory in curved spacetime.

resaypi
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Homework Statement


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

Homework Equations


Euler-Lagrange equations for a scalar field is given by
[itex]\partial_\mu \frac{\partial L}{\partial\left(\partial_\mu\phi\right)} - \frac{\partial L}{\partial \phi} = 0[/itex]

The Attempt at a Solution


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

But according to the book the resulting equation is
[itex]\frac{1}{\sqrt{-g}}\partial_\mu\left(\sqrt{-g}g^{\mu\nu}\partial_\nu\phi\right) = \frac{\partial V\left(\phi\right)}{\partial \phi}[/itex]

What am I doing wrong?
 
Last edited:
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Hi resaypi!

Looks right to me except for:

1. typo with [itex]\partial_nu[/itex] instead of [itex]\partial_{\nu}[/itex]

2. no factor of 1/2 when you take the [itex]\frac{\partial}{\partial (\partial_{\mu} \phi)}[/itex] derivative

3. [itex]\frac{\partial \left[\sqrt{-g}V\left(\phi\right)\right]}{\partial \phi} = \sqrt{-g}\frac{\partial V\left(\phi\right)}{\partial \phi}[/itex]
 

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