My high school calculus has totally lost the plot here. According to the book, you integrate$$2\frac{\ddot{a}}{a}+\frac{\dot{a}^{2}}{a^{2}}=\varLambda$$to get$$a\dot{a}^{2}=\frac{\varLambda}{3}a^{3}+K,$$where the dots are derivatives wrt time and $K$ is a constant of integration. (It's a cosmology equation, by the way – $a$ is the scale factor, $\varLambda$ is the cosmological constant.) Well, I've no idea how the author did this, so I thought I'd try to differentiate the second equation to get to the first. This works, but only if I differentiate the left term wrt time, and the right term wrt to $a$, which doesn't seem legal. Is this to do with implicit differentiation? Advice much appreciated. Thanks.


1 Answer 1


Multiply through by $a^2\dot{a}$ to get

$$ 2a\ddot{a}\dot{a} + \dot{a}^3 = \Lambda a^2\dot{a} $$

$$ a \frac{d}{dt}(\dot{a}^2) + \dot{a}\cdot\dot{a}^2 = \Lambda a^2 \dot{a} $$

By the product rule, we have

$$ \frac{d}{dt}(a\dot{a}^2) = \Lambda a^2 \dot{a} $$

Finally, integrate

$$ a\dot{a}^2 = \Lambda \frac{a^3}{3} + K $$

  • $\begingroup$ How did you see that? $\endgroup$
    – Peter4075
    May 28, 2018 at 10:46
  • $\begingroup$ Given the answer, one could always work backwards. $\endgroup$
    – mickep
    May 28, 2018 at 10:48

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