.. because, how would all the current matter and energy manifest then? Isn't everything coming to complete stop, minimal temperature, zero interaction state impossible? Wouldn't there still be some rare minor events occuring? And if so, given infinite amount of time couldn't these result in events resulting in some kind of chain-reaction?
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$\begingroup$ AFAIK, the total energy of the universe is zero, because it also combines potential energies of various interactions, so it should violate this law. $\endgroup$– TosicCommented May 3, 2019 at 11:53
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$\begingroup$ The expansion of space is relevant here -- the space between particles could be growing fast enough that they don't meet. $\endgroup$– Steve LintonCommented May 3, 2019 at 12:14
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1$\begingroup$ @Tosic IS zero-energy universe an accepted theory? I haven't heard of gravity being counted as negative energy, as it suggests. $\endgroup$– marko-36Commented May 3, 2019 at 12:53
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2$\begingroup$ Energy conservation is not a law that applies to the evolution of the universe as a whole. $\endgroup$– ProfRobCommented May 3, 2019 at 13:41
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1$\begingroup$ It shouldn't be, no, actually, I made a mistake there (I even confused the Big Freeze with the Big Crunch)... I'm sure the provided links will help you much more than I can. $\endgroup$– TosicCommented May 3, 2019 at 15:42
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1 Answer
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The current model ($\Lambda$CDM) predicts that space will expand exponentially, and keep on doing it indefinitely far into the future. The end result, after the stars burn out, galaxies disperse, protons decay and black holes evaporate is a very thin soup of stable particles (see Adams & Laughlin 1997 or their popular book The Five Ages of the Universe).
Since each stable particle finds itself separated by exponentially growing distances due to the expanding space they will not be able to interact. Their density scales as $\rho(t)\sim \rho_0/a(t)^3 = \rho_0 e^{-3t/H_0}$. The probability of a collision over time $dt$ for a particle is roughly $\sigma v \rho(t) dt$, which if we integrate to infinity becomes $\Pr[\text{any collision}]=\sigma v \rho_0 \int_0^\infty e^{-3t/H_0} dt = \sigma v \rho_0 H_0/3 $ which is finite - there will only be a finite number of interactions for each particle, ever.
At this point temperatures are extremely low since all background radiation has been redshifted to undetectable levels, and only the very cold horizon radiation remains ($\approx 10^{-30}$ K). Particles end up in a equilibrium state with this heat bath, and this is the real heat death of the universe.
This low temperature is still enough to rarely excite the particles, or generate new particles... or Boltzmann brains or entire worlds. Maybe. Eternity is a very long time, so low-probability processes will happen if they are allowed. They will however be so rare that the effects of one will typically dissipate long before anything else happens.
answered May 3, 2019 at 21:59
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$\begingroup$ Presuming protons do decay (I'm sceptical). $\endgroup$– PM 2RingCommented May 3, 2019 at 22:20
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$\begingroup$ @PM2Ring - There are quite a lot of possible baryon-nonconserving mechanisms theorized. But even if there are none, over very long spans free energy minimization might cause dissolution of matter since the entropy gain can be large; see the discussion in math.ucr.edu/home/baez/end.html $\endgroup$ Commented May 3, 2019 at 23:28
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$\begingroup$ Thanks, I'll read it shortly. I'm familiar with en.wikipedia.org/wiki/Timeline_of_the_far_future and have often linked it, but I love the way John Baez writes. $\endgroup$– PM 2RingCommented May 3, 2019 at 23:43