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.

The Big Bang theory assumes that our universe started from a very/infinitely dense and extremely/infinitely hot state. But on the other side, it is often claimed that our universe must have been started in a state with very low or even zero entropy.

Now the third law of thermodynamic states that if the entropy of a system approaches a minimum, it's temperature approaches absolut zero.

So how can it be that the beginning universe had a high temperature and a low entropy at the same time? Wouldn't such a state be in contradiction to the third law of thermodynamics?

share|improve this question
    
Entropy is the integral of $dQ/t$, so the higher the temperature some energy is, the lower its entropy. Extremely high temperature ~ extremely low entropy, and vice versa. Entropy increases as things cool off. –  Mike Dunlavey Nov 13 '11 at 1:50
    
@Mike Dunlavey "Absolute zero is the theoretical temperature at which entropy reaches its minimum value" is the first sentence in en.wikipedia.org/wiki/Absolute_zero –  asmaier Nov 14 '11 at 14:50
    
You gotta be careful of Wikipedia. This entry gives the classic thermodynamic definition of entropy, in the second law, as Q/T showing how it increases as T decreases, but it decreases as Q decreases. So the real issue is whether we are talking about a system in which Q decreases faster than T. –  Mike Dunlavey Nov 14 '11 at 16:05
    
@Mike Dunlavey Do you want to say that it is possible to reach zero entropy without having zero temperature? –  asmaier Nov 14 '11 at 16:38
1  
@Mike: The correct formula is the first law, TdS=dU+PdV (=dQ). This implies nothing about when S is zero, as dS is only about changes in S. In QM, $S=-k_B \log \Omega$, and the number $\Omega$ of microstates in the ensemble is $\ge 1$ and equals 1 iff the system is in a pure state. Thus $S\ge 0$, and $S=0$ iff the system is in a pure state. Expanding $\rho=e^{-H/k_BT}$ in an eigenbasis of $H$ shows that for a nondegenerate system $T\rightarrow 0$ implies $\rho\rightarrow\psi\psi^*$, where $\psi$ is the ground state. This proves the third law. –  Arnold Neumaier Mar 2 '12 at 22:06
show 3 more comments

6 Answers 6

Here is a case where I think most of the answers in the literature are not very good, because the correct answer was given in 1983 by Paul Davies, and rejected (this later article reviews the idea): http://cosmos.asu.edu/publications/papers/ACosmologicalDissipativeStructure.pdf . Davies's explanation is obviously correct, and it is holographically consistent. It has been rediscovered by others since then, off and on, but it is not popular with Cosmologists, who do not like causal-patch descriptions of cosmology.

When you have an inflating universe, the maximum entropy state is the one which can fit in a deSitter horizon, which has a maximum entropy equal to the cosmological horizon area, which is miniscule. This means that an inflating universe is always low entropy, even in thermal equilibrium. Once inflation ends, the minscule entropy of the initial deSitter phase translates into low entropy initial conditions, smooth homogenous stuff filling the universe.

It is important to note that during inflation, the smooth homogenous stuff is the thermal equilibrium state inside a causal patch. The fluctuations are entirely equilibrated with a miniscule entropy. It is only at the end of inflation, when the cosmological horizon becomes big, that the specialness of the inflating condition is discovered.

The fundamental source of the time-asymmetry is then the expected value of the scalar inflaton field, whatever it is. When you produce a universe with a large value of a scalar producing a cosmological constant, you are starting it in what is effectively a zero entropy state.

share|improve this answer
2  
This downvote reflects the consensus in the literature, but I wish it would come with an explanation. Davies's explanation is absolutely correct. –  Ron Maimon Nov 12 '11 at 23:31
add comment

There is a new arxiv paper that answers this question very well. See arXiv:0907.0659.

share|improve this answer
add comment

I had puzzled about that as well. But while the temperature is high the mass/energy density is extremely uniform (as ilustrated by the uniformity of the cosmic microwave background radiation 380,000 years later in the evolution of the universe). And gravity changes everything. Uniform density is very low for systems dominated by gravity (in the relative sense that clumpy distributions have higher gravitational entropy). So I think this is one way entropy is relatively low.
Another reason may be that the phase space of micro states grows as the universe expands.

share|improve this answer
1  
Is it really the case that the universal expansion contributes significantly to the entropy increase ? The calculation here (around equation 6.12) suggests that expansion doesn't affect the entropy for relativistic particles. The relativistic particles are the biggest entropy contributors apart from gravity. I would have thought that the clumping due to gravity was the reason for gravity's contribution to entropy increase rather than universal expansion ? –  twistor59 Aug 27 '11 at 8:00
    
@twistor59: the particles we are actually concerned with, for which we observe low entropy, are not relativistic. It is patently clear the entropy is low on stars and planets. Even if the entropy of neutrinos and the CMB are high, the stars and planets don't interact strongly enough with them to increase their entropy much. –  Peter Shor Nov 13 '11 at 1:46
    
@PeterShor: Hi Peter, I was responding to the last sentence in Steve C's answer, which I read (maybe incorrectly) as suggesting that the increase in entropy of the whole universe receives a significant contribution from the expansion of the universe. I think the entropy of relativistic particles (together with gravity) vastly outweighs the contribution of the nonrelativistic particles as a contributor to the total entropy of the universe. –  twistor59 Nov 14 '11 at 20:25
add comment

The third law does not say that "if the entropy of a system approaches a minimum, it's temperature approaches absolut zero." It says that if the temperature approaches absolute zero, the entropy does. These are logical converses.

The second law of thermodynamics says that entropy can only increase, so if the early universe had been in a state of maximum entropy, then the cosmos would have experienced its heat death immediately after being born. This contradicts the observation that the present universe contains burning stars, heat engines, and life. These observations imply that the early universe was in a very low-entropy state, which shows that its initial conditions were extremely finely tuned. The reasons for this fine-tuning are not explained by general relativity or the standard model. Adding inflation to the model does not cure this fine-tuning problem.[Penrose 2005]

These ideas are strongly counterintuitive to most people, since we picture the early universe as an undifferentiated soup of hot gas, very much like what we might imagine a heat-dead universe to be like. In what way is the early universe not equilibrated?

We observe that the cosmic microwave background radiation's spectrum is a blackbody curve, which would normally be interpreted as evidence of thermal equilibrium. However, this observation only really tells us that the matter degrees of freedom were in thermal equilibrium. The gravitational degrees of freedom were not. In standard cosmological models, which are constructed to be as simple as possible, there are no gravitational waves. Although the real universe presumably does have gravitational waves in it, they are apparently very weak. In a maximum-entropy universe, the gravitational modes would be equilibrated with the matter degrees of freedom, and they would be very strong, as in models like Misner's mixmaster universe.[Misner 1969]

Even in Newtonian mechanics, gravitating systems violate most people's intuition about entropy. If we psssssht a bunch of helium atoms into a box through an inlet valve, they will quickly reach a maximum-entropy state in which their density is nearly constant everywhere. But in an imaginary Newtonian "box" full of gravitating particles, the maximum-entropy state is one in which the particles have all glommed onto each other in a single blob. This is because of the attractive nature of the gravitational force.

Charles W. Misner, "Mixmaster Universe", Physical Review Letters 22(1969)1071. http://astrophysics.fic.uni.lodz.pl/100yrs/pdf/07/036.pdf

Roger Penrose, 2005 talk at the Isaac Newton Institute, http://www.newton.ac.uk/webseminars/pg+ws/2005/gmr/gmrw04/1107/penrose/

share|improve this answer
1  
Why the downvote? –  Ben Crowell Nov 12 '11 at 19:57
add comment

Well, Entropy of a given system depends on the number of micro states available. This in turn depends on the variation of velocity of particles. At very high temperatures and due to gravitational effect I guess the probability of all the particles having a common velocity increases. Hence the no .of microstates available is quite small. Perhaps this is why it is said to have a low entropy in the beginning.

share|improve this answer
1  
Why would there be less variation in particle velocity with higher temperature? –  AlanSE Nov 12 '11 at 19:55
add comment

The fundamental reason for the increase of entropy in the universe is the increase of the number of particles in the universe.

Because: 1: number of particles does increase, and that does increase entropy 2: I do not see a better explanation here

It may happen that entropy increases without increase of number of particles, for example when heat is convected from a hot object to a cool object, but there is a cool object because at some time particles where created. (the hot object is hot because it was wrapped into a aluminium foil, while the cool object radiated photon particles)

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.