Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

After watching the first episode of wonders of the solar system, one question came up which is not explained.

Bryan Cox says that ultimately the universe will be devoid of matter, so not even a single particle will actually exist. This however, seems to be a state where entropy is zero. There is no entropy since there's no disorder.

On the other hand the 2nd law of thermodynamics says that the disorder increases.

So, I am pretty sure that I am thinking wrong about this in someway.

I can imagine that the answer might be that an empty universe has an infinite amount of entropy but it feels pretty counter-intuitive for me.

share|improve this question
add comment

2 Answers 2

up vote 10 down vote accepted

Dear Paulo, the space will be empty in the sense that the density of particles will be tiny (the particles will include diluted gas as well as the Hawking radiation from the black holes that will have evaporated).

However, the Universe will also be huge because its expansion will continue - and it will become the exponential expansion of the de Sitter space (because of the dark energy, also known as the cosmological constant, that has a constant positive energy density).

It's only the total entropy of the system that is increasing and the growth of the total volume of the Universe beats the decrease of the entropy density.

There's another method to look at the entropy in the future. One may only look inside the "cosmic horizon" of the de Sitter space that our Universe is already becoming. This cosmic horizon - the maximally distant place where you can still see, if you wish - has a fixed radius and indeed, the entropy inside the cosmic horizon will go zero.

In this description, however, the cosmic horizon itself carries a huge entropy equal to $$ S = A / 4A_0$$ where $A_0=\hbar G / c^3$ is the Planck area, about $10^{-70}$ squared meters. The formula is identical to the Bekenstein-Hawking formula for a black hole. You may imagine that this entropy $S$ of the event horizon remembers the entropy of everything that may exist behind the cosmic horizon.

For the value of the radius of the cosmic horizon which is determined by the cosmological constant in our Universe, $S$ is approximately $10^{120}$ in natural units (essentially bits of information, also equal to $10^{97}$ J/K or so) that is about 20 orders of magnitude larger than the current entropy $10^{100}$ of the visible Universe - the latter figure is dominated by the entropy of the large black holes in the middle of galaxies.

(Before people knew about the large black holes and their entropy, they thought that most of the entropy of the current Universe was carried by the cosmic microwave background - but the CMB entropy is roughly 10-20 orders of magnitude lower than the entropy of the black holes.)

The entropy of the Universe will continue to increase but in some proper language, it will converge to $10^{120}$ which is the maximum allowed entropy of our Universe. The latter statement - that the entropy of a de Sitter can't exceed this bound equal to the inverse cosmological constant in the Planck units - is a bit of the lore of quantum gravity. I suppose that Brian Cox didn't want to explain this point because the TV program would become really confusing for his typical audiences, much like this comment inevitably will. Whoever is confused by this bound should keep on assuming that $10^{120}$ is infinity and the entropy will continue to grow indefinitely.

share|improve this answer
    
I can know understand why he didn't explain it further. Thanks! :) –  Paulo J. Matos Mar 14 '11 at 12:38
add comment

The problem with volume answers to this question is that if the universe is ultimately going to be de-Sitter, the volume that one is talking about is outside the horizon, and so of dubious status considering logical positivism.

A good answer to this can be obtained by considering the entropy stored in the cosmological horizon. In a deSitter space, the maximum area of the cosmological horizon is in an empty universe, and the gain in area of the horizon swamps the entropy of anything that fell through. That this is always true is the content of the Bekenstein bound on entropy. So as the universe dilutes, the entropy of the cosmological horizon always increases by more than the entropy of whatever is inside goes down.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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