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The Einstein-Hilbert action for spacetime with only a cosmological constant is given by $$S=\int\Big[\frac{R}{2}-\Lambda\Big]\sqrt{-g}\ d^4x.$$ I assume a spatially flat FRW metric in Cartesian coordinates with line element $$ds^2=-dt^2+a^2(t)(dx^2+dy^2+dz^2).$$ The Ricci scalar is given by $$R=6\left\{\frac{\ddot{a}}{a}+\Big(\frac{\dot{a}}{a}\Big)^2\right\}$$ and the determinant of the metric is given by $$g=-a^6.$$ I want to use the calculus of variations to find the form of $a(t)$ for a de-Sitter Universe.

The Einstein-Hilbert action is \begin{eqnarray} S &=& \int\Bigg[3\left\{\frac{\ddot{a}}{a}+\Big(\frac{\dot{a}}{a}\Big)^2\right\}-\Lambda\Bigg]a^3\ d^4x \\ &=& \int\Big[3\ a^2\ \ddot{a}+3\ a\ \dot{a}^2-\Lambda\ a^3\Big]d^4x \end{eqnarray} The extremum of the action is given by the Euler-Lagrange equations $$\frac{\delta S}{\delta a}=\frac{\partial L}{\partial a}-\frac{d}{dt}\frac{\partial L}{\partial \dot{a}}+\frac{d^2}{dt^2}\frac{\partial L}{\partial \ddot{a}}=0.$$ We have $$\frac{\partial L}{\partial a}=6\ a\ \ddot{a}+3\ \dot{a}^2-3\ \Lambda\ a^2$$ $$\frac{d}{dt}\frac{\partial L}{\partial \dot{a}}=\frac{d}{dt}\Big[6\ a\ \dot{a}\Big]=6\ \dot{a}^2+6\ a\ \ddot{a}$$ $$\frac{d^2}{dt^2}\frac{\partial L}{\partial \ddot{a}}=\frac{d^2}{dt^2}\Big[3\ a^2\Big]=\frac{d}{dt}\Big[6\ a\ \dot{a}\Big]=6\ \dot{a}^2+6\ a\ \ddot{a}$$ Therefore the Euler-Lagrange equation is $$6\ a\ \ddot{a}+3\ \dot{a}^2-3\ \Lambda\ a^2=0.$$ I was hoping this equation would have the exponential solution for a de-Sitter Universe given by $$a(t)=e^{\sqrt{\Lambda/3}\ t}.$$ But this expression does not solve the Euler-Lagrange equation.

What's wrong?

Addendum

Oops - Sorry it does solve the EL equation! I must have just slipped up when I plugged in the expression.

$$a=e^{\sqrt{\Lambda/3}t}$$ $$\dot{a}=\sqrt{\frac{\Lambda}{3}}\ e^{\sqrt{\Lambda/3}t}$$ $$\ddot{a}=\frac{\Lambda}{3}\ e^{\sqrt{\Lambda/3}t}$$

Substituting into EL $$6\ a\ \ddot{a}+3\ \dot{a}^2-3\ \Lambda\ a^2=0$$ gives $$6\times e^{\sqrt{\Lambda/3}t}\times \frac{\Lambda}{3}\ e^{\sqrt{\Lambda/3}t}+3\times \frac{\Lambda}{3}\times e^{2\sqrt{\Lambda/3}t} - 3\times \Lambda\times e^{2\sqrt{\Lambda/3}t}=0.$$

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  • $\begingroup$ Sorry, it does seem to solve the EL equation. What's the problem? $\endgroup$ – user2309840 Feb 24 '17 at 2:54
  • $\begingroup$ Oops - you're right, sorry! Must have just slipped up when I tried to verify the solution to the EL equation last night. $\endgroup$ – John Eastmond Feb 24 '17 at 7:25

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