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I've seen in literature

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

Source: https://en.m.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

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

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

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

I've seen in literature

$\dot{H} + H^2$

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

Defining the LHS. Since

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

And that

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

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:

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

?

I've seen in literature

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

Source: https://en.m.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

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