A straightforward calculation to the Lagrangian for FRW as $\mathscr{L}=\sqrt{-g}R$ gets me the following result:

$$\mathscr{L}\sim a^3\big(\frac{\ddot{a}}{a}+(\frac{\dot{a}}{a})^2+\frac{k}{a^2} \big)$$

However, in the context of the Hamiltonian formulation, I notice authors citing the Lagrangian for FRW as

$$\mathscr{L}\sim a^3\big((\frac{\dot{a}}{a})^2+\frac{k}{a^2} \big)$$

Why are they dropping the $\ddot{a}$ term? or is there a certain gauge for which this term is irrelevant?

  • $\begingroup$ Which authors ? $\endgroup$ – Qmechanic Jul 25 '19 at 9:54
  • $\begingroup$ i.e. this paper equation 2.2 $\endgroup$ – Ammar Qasim Jul 25 '19 at 10:21
  • 2
    $\begingroup$ Hint: The term $a^3\frac{\ddot{a}}{a}$ is equal to $-2a^3(\frac{\dot{a}}{a})^2$ modulo a total time derivative. $\endgroup$ – Qmechanic Jul 25 '19 at 10:34
  • $\begingroup$ I see, $a^3\frac{\ddot{a}}{a}=-2a^3\big(\frac{\dot{a}}{a}\big)^2+\frac{d}{dt}(a^2\dot{a})$ and the total derivative is disregarded as a surface term. $\endgroup$ – Ammar Qasim Jul 25 '19 at 10:54

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