# What is the Lagrangian for FRW universe?

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?

• Which authors ? – Qmechanic Jul 25 '19 at 9:54
• i.e. this paper equation 2.2 – Ammar Qasim Jul 25 '19 at 10:21
• Hint: The term $a^3\frac{\ddot{a}}{a}$ is equal to $-2a^3(\frac{\dot{a}}{a})^2$ modulo a total time derivative. – Qmechanic Jul 25 '19 at 10:34
• 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. – Ammar Qasim Jul 25 '19 at 10:54