We always say due to the negative pressure of dark energy, the acceleration equation shows that dark energy will cause positive acceleration $\ddot a >0$.

For a flat universe with cosmological constant only, the Friedmann equation becomes $(\frac{\dot{a}}{a})^2$ = constant. For convenience let's label the constant $\epsilon_o$.

Then $(\frac{\dot{a}}{a})^2 = {\epsilon_o}$, which leads to $ \frac{\dot{a}}{a} = \pm \sqrt {\epsilon_o}$.

All textbooks talk about the positive solution $a \propto e^{\sqrt {\epsilon_o} t}$, and how cosmological constant makes the universe expands rapidly.

However, isn't the $ \frac{\dot{a}}{a} = - \sqrt {\epsilon_o}$ and thus $a \propto e^{- \sqrt {\epsilon_o} t}$ a valid solution too? Doesn't that mean cosmological constant can make the universe collapse/contract in an exponential manner as well?

  • $\begingroup$ As $t$ goes to infinity your term decays exponentially. I don’t see how it will contribute to universe being collapsed? $\endgroup$
    – seVenVo1d
    Jan 1, 2023 at 2:13
  • $\begingroup$ @seVenVo1d Note that superpositions are not solutions to the Friedmann equation (since it is nonlinear). So ABC's point is that there are two branches of solution, one that grows exponentially with $t$ and one that decays exponentially with $t$. $\endgroup$
    – Andrew
    Jan 1, 2023 at 14:01
  • $\begingroup$ I see. You are right indeed. $\endgroup$
    – seVenVo1d
    Jan 1, 2023 at 15:40

1 Answer 1


de Sitter spacetime (no matter except a positive cosmological constant) is actually a maximally symmetric spacetime with a timelike Killing vector. In some sense, we shouldn't even say it expands or contracts at all.

There is a coordinate system called the "static coordinates" where this is manifest

\begin{equation} ds^2 = -\left(1-\frac{r^2}{\alpha^2}\right) dt^2 + \left(1-\frac{r^2}{\alpha^2}\right)^{-1} dr^2 + r^2 d\Omega^2 \end{equation} where $\alpha^2 \equiv 3/\Lambda$ (in four spacetime dimensions) and $d\Omega^2=d\theta^2+\sin^2\theta d\phi^2$ is the metric on the surface of a unit sphere.

In fact, the coordinates where the Universe grows exponentially or shrinks exponentially do not cover the complete spacetime. The full metric in FRW-like coordinates (this is known as the "closed slicing") has the form \begin{equation} ds^2 = -dt^2 + \alpha^2 \cosh^2\left(\frac{t}{\alpha}\right) d\Omega^2 \end{equation} Here you can see the full metric actually has both an exponentially growing behavior, and an exponentially shrinking behavior, contained in the hyperbolic cosine. To obtain the "usual" FRW solution, one can perform a coordinate transformation, however this only covers half of the full spacetime. Therefore, the exponentially growing and shrinking solutions that you found, cover different halves of the spacetime.

Now, in some sense, this answer has been very mathematical and focused on a special case. There is a broader question you can ask about cosmological solutions with matter, since the Friedmann equation \begin{equation} \frac{3 H^2}{8\pi G } = \rho \end{equation} is time reversal invariant. In other words, if $a(t)$ is a solution of the Friedman equation, then $a(-t)$ is as well -- or in other, other, words, for every growing solution, there is also a contracting solution.

The issue here boils down to entropy. For an expanding Universe, we have a beginning to set initial conditions, and the Universe evolves from here with growing entropy. To arrange for a contracting solution, we would need to settle on some time slice in the past (in general there is no "end" to the Universe), and arrange for the matter fields to be moving exactly opposite to the growing Universe solution. This will lead to a solution of Einstein's equations, in the same sense that a broken egg spontaneously reassembling and rising off the floor into your hand is a solution of Newton's laws. There is nothing mathematically preventing this solution, but it is highly finely tuned and violates the second law of thermodynamics.

I used the following sources to help write this answer:

[1] Wikipedia article on de Sitter space: https://en.wikipedia.org/wiki/De_Sitter_space

[2] "Lecture Notes on Classical de Sitter Space" by Thomas Hartman (Cornell): http://www.hartmanhep.net/GR2017/deSitterLectures-1.pdf

  • $\begingroup$ Huh, thanks for pointing out the behavior of true de sitter space and the references! The point about entropy is interesting -- so from the readings I did it seems lumping universe has higher entropy than smooth universe, and I understand it doesn't make sense to have an universe contracting while having stars and galaxies reverse course as well. The scale factor decreasing doesn't necessarily mean galaxies would reverse course right? So could we have a universe that's contracting, while still getting lumpier with new galaxies forming? That wouldn't violate the thermodynamics law right? $\endgroup$
    – ABC
    Jan 5, 2023 at 5:19
  • $\begingroup$ @ABC Right so, in the case of a closed Universe (positive spatial curvature) that expands and then collapses, I believe there is no problem with the second law. However, there are "cyclic models" of the Universe where the Universe undergoes a series of big bangs and big crunches. I am not an expert on those models and this is stretching my knowledge, so take it with a grain of salt, but I believe there is an issue in those kinds of models with understanding how the entropy can increase when you essentially have to reset the initial conditions every time there is a bounce. $\endgroup$
    – Andrew
    Jan 5, 2023 at 14:19
  • $\begingroup$ Oh right, at the "big bounce" things are supposed to reset, and that'd cause entropy problem. Thanks for the clarification. $\endgroup$
    – ABC
    Jan 7, 2023 at 21:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.