Question about $3p/c^2$ in Friedmann equation The Friedmann Equation:

$$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3}$$

I want to know, where does the $3p/c^2$ come from. I know it is something like density of energy or something like that but I don't know where it comes from.
There are two Friedman equations. 


One related to the energy-density component in the Einstein field equations.
One related to the pressure component in the Einstein field equations.


The Einstein field equations are the following ten equations (16 equations for 4x4 tensor components but only ten of them are independent due to symmetry)

$$G_{\mu\nu} + \Lambda g_{\mu\nu} = \kappa T_{\mu\nu} = \frac{8 \pi G}{c^4} T_{\mu\nu}$$

where $G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} g_{\mu\nu}R$ is the Einstein tensor, $g_{\mu\nu}$ the metric tensor, $\Lambda$ is the cosmological constant, $\kappa = \frac{8 \pi G}{c^4}$ is the Einstein constant, $T_{\mu\nu}$ is the energy-momentum tensor



Friedman was actually assuming that only the matter/energy density component is non-zero $T_{44}= \epsilon = \rho c^2$. He assumed "Die Materie ist inkohärent" (ie non interacting dust with $p=0$) by which the components $T_{ii}$ are non-zero for $i \neq 4$. The other components (energy flux and momentum) are set equal to zero based on the assumption that the motion of energy and matter is considered to be negligibly small.

LeMaître kept the pressure in place arguing that the pressure associated with the kinetic energy of matter might be negligible but this may not be the case for radiation pressure. So he used also $T_{11} = T_{22} = T_{33} = -p$



Using the Robertson Walker metric metric $d s^2 =  dt^2 - a(t)^2 \left[ \frac{dr^2}{1-k r^2}+r^2d\Omega^2\right]$ the equations become expressions in terms or the scale factor $a(t)$:

$$\begin{array}{rcrcl} G_{00} &amp;=&amp; c^{-2}\frac{3k}{a^2} + 3c^{-2} \frac{\dot a^2}{a^2} &amp;=&amp; \frac{8 \pi G}{c^4} \rho c^2 - \Lambda  \\ -G_{11} = -G_{22} = -G_{33} &amp;=&amp; c^{-2}\frac{k}{a^2} + c^{-2}\frac{\dot a^2}{a^2} + 2 c^{-2} \frac{\ddot a}{a} &amp;=&amp; -\frac{8 \pi G}{c^4} p + \Lambda \end{array} $$



If you subtract the second equations from the first equations

$$G_{00}+G_{11}+G_{22}+G_{33} = -6 c^{-2} \frac{\ddot a}{a} =
  \frac{8 \pi G}{c^4} (3p+ \rho c^2) + 2\Lambda   $$

and rearrange some terms, then you could also write 

$$
 \frac{\ddot a}{a} = -\frac{4 \pi G}{3}  \left( \rho + \frac{3p}{c^2} \right) + \frac{ \Lambda c^2}{3}   $$