# Does general relativity entail singularities if there's a positive cosmological constant?

I've heard that Hawking and Penrose proved that general relativity entails singularities. But it says in the abstract of what seems to be the paper in which they proved it (The Singularities of Gravitational Collapse and Cosmology) that the theorem applies only if certain assumptions are made, one of which is a zero or negative cosmological constant. Hasn't a positive cosmological constant been favored since the 1998 discovery that the expansion of the universe is accelerating? If so, is it known (i.e. mathematically proven or clearly established from physical evidence) that general relativity entails singularities in that case?

I would say the sign of the cosmological constant would certainly play a factor in determining singularity behaviour of the universe. This can be seen from Raychaudhuri’s equation, which is precisely obtained from Einstein’s field equations, and is given by:

$$\dot{\theta} + \frac{1}{3} \theta^2 + \sigma_{uv}\sigma^{uv} - \omega_{uv} \omega^{uv} + \frac{\kappa}{2} (\mu + 3p) - \Lambda = 0$$

where $\theta$ is the expansion scalar, $\sigma_{uv}$ is the shear tensor, $\omega_{uv}$ is the vorticity, $\mu$ is the energy density, $p$ is the pressure, and $\Lambda$ is the cosmological constant. (This is not the most general form of the Raychaudhuri equation, as I have assumed that the universe model is spatially homogeneous, which has simplified things quite a bit (all partial derivatives are now ordinary time derivatives, however, it will illuminate this discussion slightly). Also, the Raychaudhuri equation was the main motivation behind the Penrose-Hawking singularity theorems.

Now, our universe is understood to be spatially homogeneous and isotropic on the largest scales, and as such, by these symmetries, we must have that the shear and vorticity vanish, such that Raychaudhuri’s equation becomes:

$$\dot{\theta} + \frac{1}{3} \theta^2 + \frac{\kappa}{2} (\mu + 3p) - \Lambda = 0$$

There are many ways to get $\theta(t) \to \infty$, and they depend on the curvature of the universe, the sign of the cosmological constant, the pressure/energy density in the universe, the nature of dark energy,etc… Many models exist in the scientific literature which discuss these issues at length. For example, the recollapse theorems of Barrow and Tipler are actually much more general than the Penrose-Hawking singularity theorems, as Barrow and Tipler use the full Einstein equations, while Penrose-Hawking limit their studies to timelike geodesics.

• Please forgive my stupidity, but I'm unclear whether that's a Yes or a No answer to my question. Can you clarify, in layman's terms? Jul 10 '15 at 7:37
• I would say an answer to your question is Yes. In the case of a positive cosmological constant, one can still have singularities, where for example $\theta(t) \to \infty$. As a special case of a flat FLRW universe above, consider a de Sitter space time, where $\mu = p = 0$ and $\Lambda > 0$. Raychaudhuri's equation simply becomes $\dot{\theta} + \frac{1}{3}\theta^2 - \Lambda = 0$. This ODE will have "exploding" solutions for all $\Lambda > 0$ as long as the initial condition is chosen such that $\Lambda \neq \theta_{0}^2/3$, where $\theta_{0}$ is the initial condition. Jul 10 '15 at 14:24
• That shows that you can have singularities with positive cosmological constant, ok. But what I'm getting at is whether it has been proven that all solutions to Einstein's equations with positive cosmological constant must have singularities. Jul 10 '15 at 17:57
• No such theorem exists. For there are many examples of space times that have no singularities at all. Singularities are not generic, one has to have rather specific conditions to generate them. Jul 10 '15 at 19:10
• Thank you. Would I be correct in assuming that at least some of these space times without singularities are physically plausible? (As plausible as any, I mean. I know that GR isn't generally regarded as the last word on everything) Jul 10 '15 at 21:05

A singularity involves an infinite amount of negative potential energy in a localized volume. A nonzero cosmological constant would only yield a finite amount of positive energy in a localized volume. So the cosmological constant might slow down the rate of singularity production, but it won't stop it.

• Thank you for answering. If it's that straightforwardly Yes, why do Hawking and Penrose make a point of saying they haven't proven it? (They do give an informal argument in favor of Yes in the 7th paragraph) Jul 9 '15 at 23:39
• @Anthony: they assumed it because it wasn't expected and it simplified assumptions. you have manifest singularities in the schwarzschild-de sitter spacetime, so you know that the global statement isn't true. Jul 10 '15 at 4:41
• @Jerry Schirmer: What do you mean by "the global statement" ? Jul 10 '15 at 7:59
• That the cosmological constant can stop singularity formation. There are exact, singular models with a nonzero csomological constant. I know that this isn't quite what you are saying, but it's worth reading for someone coming through this site in the future. Jul 10 '15 at 14:41
• Thanks for clarifying. Do you know if a positive cosmological constant can ever stop (or avoid) singularity formation? Jul 10 '15 at 18:02

Singularities are very likely impossible to create in the real universe.

In other words, as a singularity gets close to forming, the incoming random GR waves and other energy will rip the formation apart, keeping it in a state of almost singularity.

As an example, all Black Holes spin in the real world. The size of the singularity in a spinning Kerr geometry is within a hair of being zero:

Thus we reach the conclusion that at timeline or null geodesic or orbit cannot reach the singularity under any circumstances except in the case where it is confined to the equator, cos() = 0…..Thus as symmetry is progressively reduced, starting from the Schwarchild solution, the extent of the class of geodesics reaching the singularity is steadily reduced likewise, … which suggests that after further reduction in symmetry, incomplete geodesics may cease to exist altogether

Kerr Fields, Brandon Carter 1968.

So while General Relativity in theory has singularities, its not likely that any exist in a real noisy universe. The cosmological constant does not I think enter into the problem.

The wikipedia page on the singularity theorems says as much as well. https://en.wikipedia.org/wiki/Penrose–Hawking_singularity_theorems

It is still an open question whether time-like singularities ever occur…

• Thank you for answering. Eliminating the singularities is often given as one of the motivations for quantum gravity research. Are you saying that simply applying GR to the real universe can accomplish this just as well? Jul 10 '15 at 1:40
• I think you are confusing some topics: Your notion of "random GR waves" and other energy is not well-defined. Second, when you say all Black holes spin in the real world, what do you mean? To have a black hole, all you have to do is satisfy Birkhoff's theorem. In a Kerr metric, there IS a singularity at $r^2 + \alpha^2 \cos^2 0$, it is a ring singularity, and it is irremovable, as the Kretcshmann scalar is singular! Further, the quote you provide from Brandon Carter, is about geodesics in an orbit of the Kerr metric, and is separate from the notion of cosmological singularities. Jul 10 '15 at 3:27
• Continued...The cosmological constant very much enters the problem when it comes to singularities. This can be seen from a simple application of the Raychaudhuri equation, which says that the universe expansion is governed by: $\dot{\theta} = -\frac{\theta^2}{3} - 2 \sigma^2 + 2 \omega^2 + (1/2)(\mu + 3p) + \Lambda$. The terms on the right-hand-side of this equation (including the cosmological constant) may/may not contribute to a singularity. It is NOT an open question whether time-like singularities ever occur, they occur in G.R. all the time. The question exists around their observation. Jul 10 '15 at 3:30
• The question is not about cosmological singularities. Jul 10 '15 at 13:27
• The ring singularity has a set of zero measure of geodesics that run into it. In other words its impossible to hit. Jul 10 '15 at 13:28