I'm trying to understand the classical derivation of Friedmann equation but I'm missing one step.

So, I start with accelerations, where $a$ is a scale factor


$\ddot{a}=-\frac{4\pi G}{3}\rho a$

Now multiply both sides by $\dot{a}$, and here is the question: Why LHS is



2 Answers 2


Not sure I see the problem.

$\frac{1}{2}\frac{{\rm d}}{{\rm d}t}\big((\dot{a})^{2}\big) = \frac{1}{2}*2*\dot{a}*\frac{{\rm d}}{{\rm d}t}\big(\dot{a}\big) = \dot{a}\ddot{a}$. Maybe you don't recognize the notation - a dot is often used to denote a time derivative. Thus two dots indicates a double time derivative.

  • $\begingroup$ ok I see... the product rule ;p I tried to do it LHS->RHS, what I should do is just the product rule on the RHS, silly me... $\endgroup$
    – Luke
    Apr 25, 2013 at 1:59

I should point out that third derivative has an important defintion. It is representative of non-conserved models of expansion, so long as the equation os state is not assumed zero.

Since the equation of state [are] the derivatives on

$\dot{\rho} + 3\dot{P} = 0$

Which is usually the case for conservation, but since we know we have a theory outside of our comfortable conserved physics, I can say the equation of state need not be zero at all. As pointed out by Lloyd Motz, conservation in cosmology is a completely unfounded assumption. So we now say

$\dot{\rho} + 3\dot{P} \ne 0$

Let's show how to do that calculation. Starting with a simple Friedmann equation,

$(\frac{\dot{R}}{R})^2 = \frac{8 \pi G}{3}\rho$


$\dot{R}^2= \frac{8 \pi GR^2}{3}\rho$

$m\dot{R}^2= \frac{8 \pi GmR^2}{3}\rho$

Our Friedmann equation now has units of energy. We could construct a Langrangian by taking both sides away from each other, but we don't need to make the Langrangian, not part of the general point. Taking the third derivative through I get

$2m\ddot{R}\dot{R}= \frac{8 \pi GmR^2}{3}\dot{\rho}$

Ref: https://adsabs.harvard.edu/full/1989ComAp..13...67M

Here, Lloyd explains how the third derivative represents nonconservation.


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