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Assuming a Robertson-Walker metric to describe homogeneous and isotropic cosmological models, Einstein equations with cosmological constant reduce to these 3 non-linear ordinary differential equations for a perfect fluid:

\begin{align} \dot{\rho} &= -3H(\rho + P) \tag{1} \\ \dot{H} &= -H^2-4\pi G(\rho + 3P)/3 + \lambda/3 \tag{2} \\ H^2 &= 8\pi G\rho/3-K/a^2+\lambda/3 \tag{3} \end{align}

Here, dot represent the derivative with respect to time and the function $H=H(t)= \dot{a}/a$ where $a = a(t)$ is the scale factor.

I know that (1) is the conservation of energy equation but i can't understand what (2) and (3) represent?

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If what you want a gut-feeling of what these equations mean, I can share mine.

Equation (2) is a consequence of the two others. Take the time-derivative of (3), remembering that K and $\lambda$ are constants and $H=\dot a/a$, combine this derivative with (1) and (3) and you get (2). So no gut-feeling for that one, just trivial algebra.

Now gut-feeling of (1):

The mass volumic density can change in two ways: either by changing the volume for the same amount of mass (term $-3H\rho$ ) or creating or losing energy (which is the same as mass, of course) because of positive or negative work of the pressure (term $-3HP$). OK, you might not be really convinced, but think it over. If you ignore pressure, then the mass in a comoving volume is strictly constant, and (1) for zero pressure just tells you that the mass volumic density behaves like the inverse cube of $a$.

Finally, equation (3).

Multiply by $a^2$. Then the right-hand-side is $\dot a^2$, kinetic energy per unit mass. Second term, if we had $\rho a^3$ that would be total mass. But we just have $\rho a^2$, so this is mass $M$ divided by radius $a$. $(8\pi GM/3)/a$ opposite of the gravitational potential energy per unit mass. Now $K$ is a constant, total energy. For $\lambda=0$ you have kinetic energy as the total energy minus (negative) potential energy, An intuitive feeling for $\lambda a^2$.... that's another story.

I hope that helps... This is just gut-feeling, it would need a lot of work to make is precise. But his is the way I look at them.

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You can arrive at equations (2) and (3) by plugging the Robertson-Walker metric into the Einstein field equations. The universe is modelled as a perfect fluid so the energy-momentum tensor is $(T_{ab})=\text{Diag}(\rho,P,P,P)$ and the first equation comes from the $T_{00}$ (time) part of the field equations, the second from the spacial part once you have plugged in the metric.

Note that the three equations are not independent, though any choice of two of them are, the continuity equation (1) $\nabla^a T_ab =0 $ may be obtained directly from the field equations by applying $\nabla^a$ and applying the Bianchi identity and metricity property of the Levi-Civita connection.

Hope this helps :)

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  • $\begingroup$ Note that the question asks what those equations physically represent. The method of arriving at them was, from what I can infer, accepted and not in question. $\endgroup$
    – Jim
    May 19, 2015 at 18:28
  • $\begingroup$ Sorry, I thought it was implicit. The Friedmann equations are a manifestation of the Einstein field equations, which describe the relationship between the contents of spacetime and its geometry. Thus the equations (2),(3) are necessary and sufficient constraints on the metric in order that the universe obeys the laws of general relativity. ie. in order that gravity is taken account of. In this case I think the derivation is really all the physical meaning there is... The equations (2),(3) basically are the field equations! Thanks for pointing that out though :) $\endgroup$ May 19, 2015 at 18:58
  • $\begingroup$ The friedmann equations have more physical meaning than that. For instance, (3) describes how the relative expansion of space is directly affected by the different energy densities that inhabit the universe. There are, of course, other interpretations as well. $\endgroup$
    – Jim
    May 19, 2015 at 19:03
  • $\begingroup$ Well, sure. It tells us that the expansion of space is affected by the different energy densities, because we are dealing with a model in which the universe must conform to the field equations, which take into account energy density among other things. But if you understand the field equations, and presumably one should know some GR before studying the Friedmann equations, then you already have the physical meaning. $\endgroup$ May 19, 2015 at 19:12
  • $\begingroup$ I agree with that, but nevertheless, it couldn't hurt to write down the physical meaning of the friedmann equations specifically. Even though anyone that understands their origins should know it already. That is, after all, what the question asks for $\endgroup$
    – Jim
    May 19, 2015 at 19:17

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