2
$\begingroup$

De sitter universe is dominated by a cosmological constant (devoid of matter and radiation), corresponding to dark energy in the far future or the inflaton field in the early universe, leading to an exponential expansion of space.

What would theoretically happen to de Sitter universe if the cosmological constant suddenly disappears? $~~$ Would it continue to expand at the same rate, or expand at a slower rate, or simply stop expanding?

$\endgroup$
1
$\begingroup$

Einstein's equations in the presence of a cosmological constant $\Lambda$ (and no matter) are

$$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} + \Lambda g_{\mu \nu} = 0$$

which gives maximally symmetric de Sitter ($\Lambda > 0$) or Anti de Sitter ($\Lambda < 0$) solutions. In the absence of the cosmological constant, $\Lambda =0$, and Einstein's equations reduce to

$$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = 0$$

which have Ricci flat solutions: $R_{\mu \nu} = R = 0$. This means that spacetime curvature, if it is at all present, must reside entirely in the Weyl tensor $C_{\mu \nu \rho \sigma}$. Schwarzschild solution is an example of a Ricci flat solution.

But this is an idealized picture, since $\Lambda$ doesn't 'vanish' in an instant out of nowhere, and neither are vacuum solutions practical from the point of view of cosmology. For a typical cosmological timeline, one has to solve Friedmann's equations and deduce how the universe goes from a matter-dominated era to a $\Lambda$-dominated era.

$\endgroup$
  • $\begingroup$ So, what would theoretically happen to the expansion rate of de Sitter universe if the cosmological constant vanishes? $~$Would it continue to expand at the same rate or not? $\endgroup$ – Forge Mar 1 at 13:20
  • $\begingroup$ @Forge I don't think that you can ask that question. Not even theoretically. $\endgroup$ – Avantgarde Mar 2 at 14:51
1
$\begingroup$

This question can't be answered by general relativity, because its assumptions are not consistent with GR. It's not consistent with the Einstein field equations for the cosmological constant to suddenly vanish. This is similar to questions about what happens to the solar system if the sun suddenly vanishes.

$\endgroup$
  • 1
    $\begingroup$ What about the decay of the hypothetical inflaton field in the early universe? $~$ Isn’t it actually the same thing with the cosmological constant vanishing? $\endgroup$ – Forge Mar 1 at 14:54
  • 1
    $\begingroup$ @Forge: The cosmological constant term in the Einstein field equations is $\Lambda g^{ab}$. If you take the divergence of this, you get $\nabla ^b \Lambda$. Since the other terms in the field equations have zero divergence, any variation in $\Lambda$ violates the field equations. What about the decay of the hypothetical inflaton field in the early universe? Isn’t it actually the same thing with the cosmological constant vanishing? No, it's not the same thing. The cosmological constant has a specific definition: it's a term $\Lambda g^{ab}$ in the field equations. $\endgroup$ – Ben Crowell Mar 1 at 17:21
  • $\begingroup$ There is no physical difference between a cosmological constant and a field with a non-zero vacuum energy. Nature does not care whether you write the term on the left or righthand side of the Einstein equation. $\endgroup$ – mmeent Mar 4 at 7:09
  • $\begingroup$ @mmeent: I agree. Is that relevant somehow? $\endgroup$ – Ben Crowell Mar 4 at 19:35
1
$\begingroup$

The short answer: The expansion of the universe would continue, but its acceleration would stop. (You also need to figure out what happend to all the energy.)

The longer answer: For a spacial homogeneous an isotropic universe filled with a perfect fluid the Einstein equations reduce to the Friedman equations: $$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3}\rho-\frac{\kappa}{a^2}$$ and $$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho+3p)$$, where $\rho$ is the energy density $p$ the pressure $G$ the gravitational constant, $a$ the scale factor of the universe, and $\kappa$ the curvature of the spacial slices.

deSitter space corresponds to a solution of these equations with $\kappa > 0$ and $p=-\rho$. Vanishing of the cosmological constant would mean that $p$ changes, while $\kappa$ and $\rho$ must stay continuous (anything else would violate conservation of energy).

The first Friedmann equation thus says that -- regardless of what happens to the energy released by the vanishing of the cosmological constant -- the rate of expansion $\dot{a}/a$ will stay continuous through the "vanishing" of the cosmological constant. (The first part of the short answer).

For what happens next, we need to know what happend to the vacuum energy represented by the cosmological constant. This boils down to specifying what happens to the pressure $p$. If all the energy would decay into radiation we would see $p$ jump to $p = \rho/3$. If all the energy decay into (non-relativitic) matter, we would get $p=0$. Some combination of the two would and up somewhere in between. In both cases however, the sign of the RHS of the second Friedmann equation changes from positive to negative, which tells us that the expansion of the universe has stopped accelerating and has start decelerating. (The second part of the short answer.)

$\endgroup$
  • $\begingroup$ You also need to figure out what happend to all the energy. This is kind of the main point -- we can't conclude anything since by assumption this violates GR. $\endgroup$ – Ben Crowell Mar 1 at 17:16
  • $\begingroup$ @BenCrowell This is exactly what happens at the end of most inflation models. The inflaton decays causing the "cosmological constant" to disappear. The energy in this case is dumped in other particles and fields causing the universe to heat up. $\endgroup$ – mmeent Mar 3 at 23:12
0
$\begingroup$

In the de Sitter model,

$$\lambda=3H^2/c^2.$$

If $\lambda$ vanishes, then so does the de Sitter model

$$ds^2=c^2dt^2-e^{2Ht}[dr^2+r^2(d\theta^2+sin^2(\theta)d\phi^2]$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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