Friedmann equation

Question

I've seen in literature

$$\dot{H} + H^2=\ldots$$

Source: https://en.wikipedia.org/wiki/Friedmann_equations

Defining the LHS. Since

$$H = \frac{\dot{a}}{a}$$

And that

$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3}(\rho + 3P)$$

Then replacing gives

$$H^2 = \frac{8\pi G}{3}(\rho + 3P)$$

So my question is how to you arrive at the additive Hubble term

$$\dot{H} + H^2 = \frac{8\pi G}{3}(\rho + 3P)?$$

Answer 1

Note that $H=\frac{\dot{a}}{a}\implies\dot{H}+H^2=\frac{\ddot{a}}{a}$. You're asking about the special case $k=0,\,\Lambda=0$, but seem confused about what results we obtain. In this case, the Friedmann equations are$$H^2=\frac{\dot{a}^2}{a^2}=\frac{8\pi G\rho}{3},\,\dot{H}+H^2=\frac{\ddot{a}}{a}=-\frac{4\pi G(\rho+3p/c^2)}{3}.$$Hence$$\dot{H}=-4\pi G(\rho+p/c^2).$$

Answer 2

The forth equation you wrote, being \begin{equation} H^2=\frac{8\pi G}{3}(\rho+3P) \end{equation} is incorrect. The first Friedmann equation is given by \begin{equation} H^2=\frac{8\pi G}{3}\rho. \end{equation} Now, taking a time derivative in both sides, we find: \begin{equation} 2H\dot{H}=\frac{8\pi G}{3}\dot{\rho}. \end{equation} Knowing that for a bariotropic fluid \begin{equation} \dot{\rho}=-3H(\rho+P), \end{equation} we have \begin{equation} \dot{H}=-4\pi G(\rho+P). \end{equation} Adding the first Friedmann equation to this one we obtain: \begin{equation} H^2+\dot{H}=-\frac{4 \pi G}{3}(\rho+3P), \end{equation} which is the correct form for the equation you are looking for (note that it agrees with the equations in the Wikipedia link you shared.