# In Big Rip scenario: will there be an equilibrium point?

As the radius of the de Sitter universe shrinks, the de Sitter temperature increases. Also because of the space expansion the content of the universe cools down. Certainly there will be a point where those temperatures will become the same and the space boundary will emit as much energy as it consumes. This supposedly will stop its shrinking (the smaller its radius the more energy it has, so after the consumed photons from the inside area are no longer enough to support the shrinking, the shrinking will slow down).

This possibly will happen at deep sub-atomic level, but at finite scale. So can we say that in Big Rip scenario, an equilibrium will be reached that would prevent reaching the singularity?

• Theory can model whatever you want for you. If you order a new equilibrium point, it can deliver. That's the power of imagination. Aug 3, 2016 at 10:43
• I'm not sure how to answer this question other than "If you want to model a Big Rip that reaches an equilibrium, you can do that but it isn't necessary". Given that a Big Rip is not a likely scenario, going further into detail doesn't seem very useful
– Jim
Aug 3, 2016 at 12:08
• @Jim I am asking, will the equilibrium happen as we understand it now, why the close vote? And whether a cosmological question is "useful" is irrelevant. Aug 3, 2016 at 12:10
• I didn't vote to close, but it seems 2 people have found this question unclear. Again, you can model a big rip with an equilibrium in any way you want to describe it. OR it can be modeled with no equilibrium. The math is surprisingly flexible.
– Jim
Aug 3, 2016 at 12:15
• What @Jim and I are trying to tell you is that there is no on theory in cosmology that can predict anything of the long term future with any precision. We simply don't know how the universe works at this moment, but we can fill that void with any kind of dragon that you like. If it's a green big rip dragon, it's fine, and if somebody else likes a slender blue one, that's just as dandy. Aug 3, 2016 at 12:44

Phantom energy is something that appears pathological in some ways. The Chandra data has w=-0.98+-0.07, and yet From Reiss et al data the value for the important parameter is $w = −1.007 ± 0.081$ . The data is not precise enough to rule out phantom energy from straight dark energy with $w~=~-1$. So one might then give this some serious consideration, for it could turn out to be the case.

The scale factor for the evolution of the spacetime, which I work out in a Newtonian framework here How did the universe shift from "dark matter dominated" to "dark energy dominated"? within a Newtonian context, is $$H^2~=~\left(\frac{\dot a}{a}\right)^2~=~H_0\left[\frac{\Omega_m}{a^3}~+~(1~-~\Omega_m)a^{-3(1+w)}\right]$$ Here $\Omega_m$ holds for matter, which is $.26$ for dark matter and $.04$ for luminous matter so $\Omega_m~=~.3$. Obviously $1~-~\Omega_m$ pertains to dark energy, or phantom energy. Below is a graph of what happens to the scale factor for different values of $w$

The case $w~=~-1/3$ is the coasting cosmology and for smaller $w$ there is exponential expansion. For $w~=~-1$ this is the dark energy condition, and for smaller we have phantom energy.

This has the effect of increasing the vacuum energy or dark energy so that within a finite time the vacuum energy approaches Planck energy density and the universe effectively becomes a singularity. This big rip occurs at a time $t~\simeq~t_0~-~(2/3)H_0^{-1}(1~-~\Omega_m)^{-1/2}$ the Hubble paramter is about $70km/sec-Mpc$. If we assume that the Riess value is real $w~=~-1.007$ then the big rip will occur around $1.6\times 10^{12}$ years. This would be around the time the last stars wink out.

We may think of this as well according to an increase in the cosmological constant. By some mechanism the vacuum energy increases and the cosmological constant increases. The cosmological constant increases by $\Lambda~\simeq~\Lambda_0(1~-~w)t$. For the cosmological constant $H^2~=~\Lambda/3c^2$ $=~8\pi G\rho/3$, it is clear that the vacuum energy increases and the cosmological horizon distance constricts as $r_h~=~\sqrt{3/\Lambda}$. The temperature of the horizon is then $T~=~(2\pi\hbar/k)\sqrt{\Lambda/3}$ or proportional to $1/r_h$ as it constricts.

The question of equilibrium is similar to the problem of quantum black holes. Given a black hole of mass $M$ it has temperature $T~=~1/8\pi M$. Assume this sits in a world with the same background temperature. Now if the black hole absorbs a photon of mass-energy $\delta m$ then by the increase in black hole mass it puts the black hole at a colder temperature and away from this equilibrated (not equilibrium) condition. The converse of course means if the black hole emits a photon it is not hotter and will not preferentially radiate photons into the background. Does something similar happen to a big rip cosmology?

Let us look at a black hole in the de Sitter spacetime. The stationary metric element $g_{tt}~=~1~-~2m/r~-~\lambda r^3/3$ is plotted below for various values of the cosmological constant.

The funny thing is that the horizons of black hole and the cosmology merge and beneath the horizontal black line the metric is effectively spacelike. Before then the horizon of the black hole appears to move out, increasing the horizon area and by extension decreasing its temperature. For $w~=~-1$ there is a nice case of where the energy density $\rho$ and the “work” due to pressure $pc$ are equal in magnitude and opposite in sign. So there is no net energy from dark energy or the vacuum that pumps up black holes to larger mass. However, for $w~<~-1$ we can think of the black hole as gaining mass right from the vacuum until it reaches a type of equilibrium condition with the cosmological horizon. However, if the cosmological horizon is constricting in this is only a temporary condition.

The constriction of the cosmological horizon means it temperature increases, and by extension matter is heated by the frame dragging of extended objects and from the background. The constriction however is not abated by matter reaching some temperature limit, unless that is the string Hagedorn or Planck temperature.

• As I understand, the cosmic horizon shrinks and heats when it consumes matter (photons), unlike a black hole (who grows and cools). So over time its temperature should rise, while the temperature of photons in CMB will drop. Some day the wavelenghth of the CMB photons will reach the dimensions of cosmic horizon, at which point the horizon will be adsorbing as much energy as it emits, so it will stop shrinking. Am I correct? Aug 3, 2016 at 19:28
• In a big rip there will not be enough time for CMB photons to be stretched that far which is $z\sim 10^{28}$. The big rip will likely happen before then. A black in phantom energy will increase in size until the horizons merge. After that the black hole becomes something like a naked singularity. This an other reasons make the phantom energy model problematic. Aug 3, 2016 at 20:31
• Yes, the photons will not have enough time to become that big. But the horizon will shrink as well, so some day the wavelenghth of CMB and radius of the horizon should coincide, should not they? Aug 3, 2016 at 20:51
• I don't think there is anything special about CMB photons being red shifted beyond the cosmological horizon. With dark energy, not phantom energy, in around $10^{40}$ years from now most such regions will contain at most a supermassive black hole. After $10^{100}$ years most horizon bounded regions will be a pure vacuum. There will be some Hawking-Gibbon radiation from the horizon. With phantom energy the constricting horizon means this radiation will be at higher energy. Aug 3, 2016 at 21:16
• "Hawking-Gibbon radiation" - is it the same as de Sitter radiation? "With phantom energy the constricting horizon means this radiation will be at higher energy." - will not the horizon stop constricting when the contents of the universe is empty (no photons left besides those emitted by the horizon)? Is not constriction caused by consuming photons by the horizon? Aug 3, 2016 at 21:54