# How did Friedmann and Lemaitre based on Einstein's “original” GR theory find an expanding solution?

Wikipedia says,

Einstein included the cosmological constant as a term in his field equations for general relativity because he was dissatisfied that otherwise his equations did not allow, apparently, for a static universe: gravity would cause a universe that was initially at dynamic equilibrium to contract. To counteract this possibility, Einstein added the cosmological constant.[3] However, soon after Einstein developed his static theory, observations by Edwin Hubble indicated that the universe appears to be expanding; this was consistent with a cosmological solution to the original general relativity equations that had been found by the mathematician Friedmann, working on the Einstein equations of general relativity.

Ordinary matter and energy would always cause the Universe to contract and the rate of contraction would be increasing with time. How did then Friedman and Lemaitre get an expanding Universe solution based on Einstein's original theory of general relativity (with a zero cosmological constant but with ordinary matter and energy)?

Einstein's equations with $$\Lambda=0$$ are perfectly capable of describing an expanding spacetime. The cosmological principle leads to a metric of the form

$$ds^2 = -c^2 dt^2 + a(t)^2 d\Sigma^2$$

where $$a(t)$$ is the so-called scale factor which is conventionally set to $$1$$ at the present time, and $$d\Sigma^2$$ is a spatial 3-metric with constant curvature $$\frac{k}{a^2} = \frac{1}{R_0^2}$$ with $$R_0$$ the curvature radius. The application of Einstein's equations to this metric leads to the Friedmann equations, which govern the time-evolution of the scale factor:

$$\frac{\dot a^2 + kc^2}{a^2} = \frac{8\pi G\rho}{3} \qquad (1)$$ $$\frac{\ddot a}{a} = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right)\qquad (2)$$ where $$k\in\{0, \pm 1\}$$ denotes the curvature of the spacelike hypersurfaces $$\Sigma_t$$, $$\rho$$ is the energy density (the $$00$$ component of the stress-energy tensor) and $$p$$ is the corresponding pressure. In order for the universe to be static, we would need to have that $$\dot a =0$$; equation (1) then implies that

$$a^2 = \frac{3kc^2}{8\pi G\rho} \implies \frac{1}{R_0^2} = \frac{k}{a^2} = \frac{8\pi G\rho}{3c^2}$$

and therefore that $$k>0$$, meaning that the universe is spatially closed - a sphere with radius $$R_0$$. However, this is not a steady-state configuration; if $$\rho + \frac{3p}{c^2} \neq 0$$, then $$\ddot a < 0$$ and the scale factor would begin to decrease. For ordinary (cold) matter and electromagnetic radiation, $$p = 0$$ and $$p=\frac{\rho c^2}{3}$$, respectively, so this would seem to be the case.

The addition of a cosmological constant $$\Lambda$$ solves this problem. The Friedmann equations become

$$\frac{\dot a^2 + kc^2}{a^2} = \frac{8\pi G\rho+\Lambda c^2}{3} \qquad (3)$$ $$\frac{\ddot a}{a} = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right)+\frac{\Lambda c^2}{3}\qquad (4)$$

Choosing $$\frac{\Lambda c^2}{3}=\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right)$$ makes $$\ddot a=0$$; setting $$\dot a=0$$ in equation (3) then yields

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

Assuming cold normal matter ($$p=0$$), this can be written $$\frac{1}{R_0^2}=\frac{k}{a^2} = \frac{4\pi G \rho}{c^2} = \Lambda$$

This solution is brittle, however; note that if $$a\rightarrow a+\delta a$$, then $$\rho \rightarrow \rho + \delta \rho$$ where $$\delta \rho = \frac{k}{a^2}\left(1-2\frac{\delta a}{a}\right)$$ which means via equation (4) $$\ddot{\delta a} \propto \delta a$$ and so the equilibrium is unstable. Small perturbations will cause runaway expansion $$(\delta a > 0 )$$ or contraction $$(\delta a < 0 )$$.

The confirmation (Hubble, 1929) that the universe was not static - that the scale factor was indeed evolving with $$\dot a > 0$$ - meant that this workaround using the seemingly arbitrary $$\Lambda$$ was unnecessary, and so Einstein abandoned it. It wasn't until 1998 that it was discovered that the expansion of the universe is accelerating, meaning that $$\ddot a>0$$.

This is a different beast. It either requires a cosmological constant which is sufficiently large to make the right hand side of equation (4) positive, or it requires a new kind of matter with equation of state $$p < -\frac{1}{3}\rho c^2$$ (or possibly some combination of the two). The basic form of the $$\Lambda_{CDM}$$ model considers only the cosmological constant; extensions or modifications of the model allow for different possibilities (see e.g. quintessence).

Ordinary matter and energy would always cause the Universe to contract and the rate of contraction would be increasing with time.

This is not true. If we assume a flat universe containing only cold baryonic matter (i.e. dust, with $$p=0$$), the Friedmann equations yield

$$a(t) \propto t^{2/3}$$

which increases forever without bound. Of course the real constitution of the universe is more interesting than this, but the point is that if $$\dot a>0$$ at some initial time, there is no cosmological constant needed to describe a universe which expands forever.

• Let me summarize your answer which has 3 parts. In the 1st part, you argue that if at some point in time $\dot{a}=0$ but $\ddot{a}<0$ (as with ordinary matter and radiation), the scale factor will decrease with time and it give a contracting solution. The 2nd part argues that by adding a cosmological constant, we can make $\dot{a}=\ddot{a}=0$. Though this leads to a static solution, that is not stable against perturbations. In the 3rd part, you have given an example to show it is possible to have an expanding Universe with dust if it started with $\dot{a}>0$. Is that a fair summary? – mithusengupta123 Jun 26 at 12:44
• One confusion. If Einstein's addition of CC ensures $\ddot{a}=0$, and if $\dot{a}=0$, doesn't that prevent both expansion and contraction (because a becomes time-independent)? – mithusengupta123 Jun 26 at 12:53
• @mithusengupta123 Yes, that summary is fair. I'm not sure I understand your second question - yes, if we choose that specific value for $\Lambda$, then we can make $a$ static, as you said in your first comment. – J. Murray Jun 26 at 14:05
• I mean, the CC can be chosen in such a way that $\ddot{a}=0$, and if $\dot{a}=0$, there is no expansion or contraction. So Einstein's inclusion of CC together with $\dot{a}=0$, implies that by introducing CC, he could avoid both the expansion as well as the contraction (barring perturbations). So Einstein's CC, even if motivated for preventing contraction, could also prevent expansion. Is that wrong? – mithusengupta123 Jun 26 at 14:26
• @mithusengupta123 Einstein's original motivation was to prevent both contraction and expansion. He wanted a static solution, and needed $\Lambda$ for that, since otherwise a static solution is not possible. – J. Murray Jun 26 at 14:31