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The lagrangian density of a classical scalar field is given as $$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi \partial^{\mu}\phi -V(\phi).$$

For a flat and homogenous space time the FRW metric is $g_{\mu \nu}=diag(-1, a(t)^2,a(t)^2,a(t)^2)$. How the Lagrangian density is simplified to $$\mathcal{L}=\frac{1}{2}\dot\phi^2-V(\phi)~?$$

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    $\begingroup$ This is equivalent to asking why $\partial_i\phi=0$. $\endgroup$
    – J.G.
    Aug 18 '21 at 20:04
  • $\begingroup$ Actually I had a typo in my quesstion. I correct it now. as $g^{00}=-1$ I am getting a negative sign in front of the $\frac{1}{2}\dot\phi^2$ term. How to get reid of that negative sign or where have I gone wrong? $\endgroup$
    – Dori
    Aug 18 '21 at 20:19
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There are two common conventions that can be confusing across sources,

  • $+---$ viz. $g_{\mu\nu}=\operatorname{diag}(1,\,-a^2,\,-a^2,\,-a^2)$ and
  • $-+++$ viz. $g_{\mu\nu}=\operatorname{diag}(-1,\,a^2,\,a^2,\,a^2)$.

In both cases, the inverse metric just replaces $a^2$ with $a^{-2}$.

Given $\partial_i\phi=0$ (homogeneity), these conventions respectively write $\dot{\phi}^2$ as $\partial_0\phi\partial^0\phi=\partial_\mu\phi\partial^\mu\phi$ and $-\partial_0\phi\partial^0\phi=-\partial_\mu\phi\partial^\mu\phi$.

Putting relativity aside for the moment, $\dot{\phi}^2$ should have a positive coefficient in $\mathcal{L}$ so a time-dependence in $\phi$ has a kinetic cost. (We're trying to minimize action, after all.) So in the $-+++$ convention,$$\mathcal{L}=-\tfrac12\partial_\mu\phi\partial^\mu\phi-V(\phi).$$

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This is just a matter of the metric sign convention. The Lagrange density for the $(+{} -{} -{} -)$ convention is $$ \mathcal{L} = \partial_\mu \phi \partial^\mu \phi - V(\phi) $$ but for the $(-{} +{} +{} +)$ convention, it is $$ \mathcal{L} = -\partial_\mu \phi \partial^\mu \phi - V(\phi) $$ It is not hard to see that these two densities are the same if you write out the kinetic term in terms of derivatives with respect to time and space.

More generally, given a Lagrangian density defined under one sign convention in a fixed spacetime background, we can switch to the other sign convention by making the substitution $g_{\mu \nu} \to - g_{\mu \nu}$.

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  • $\begingroup$ I am not sure whether this "flip" works for models where the metric is dynamical; I will have to think about it further. $\endgroup$ Aug 18 '21 at 20:28

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