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We can imagine many changes to the laws of physics - you could scrap all of electromagnetism, gravity could be an inverse cubed law, even the first law of thermodynamics could hypothetically be broken - we've all imagined perpetual motion machines at one time or another.

However, the second law of thermodynamics seems somehow more 'emergent'. It just springs out of the nature of our universe - the effectively random movement of physical objects over time. Provided you have a Universe whose state is changing over time according to some set of laws, it seems like the second law must be upheld, things must gradually settle down into the state of greatest disorder.

What I'm particularly wondering is if you can prove in any sense (perhaps using methods from statistical mechanics)? Or is it possible to construct a set of laws (preferably similar to our own) which would give us a universe which could break the second law.

tom
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    An interesting answer to this question can be found in the book 'Time's Arrow and Archimedies' Point' by Huw Price (http://www.amazon.com/Times-Arrow-Archimedes-Point-Directions/dp/0195117980). He points out that if you took our universe and 'ran the tape in reverse' then everything that happened on the microscopic level would be compatible with the laws of physics as we know them. The universe in which entropy always decreases could just as well be our own universe - we'd have no way to tell the difference. (you'd also have to reverse charge and chirality to make it work properly.) – N. Virgo Feb 02 '12 at 17:54
  • I've heard this before, and I guess you're technically right - if we could 'create' a universe in the exact state as our universe and then reverse the motion of all the atoms/charges/forces, etc, then it should run according to relatively familiar laws, except that entropy would constantly be decreasing. Admittedly this is extremely Laplacian, and I'm not sure if quantum mechanics has anything to say on the matter... – tom Feb 02 '12 at 23:12
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    Everything in quantum-mechanics-as-we-know-it respects time-reversal symmetry (technically CPT symmetry), except for the process of observation, in which the wave function "collapses" into a randomly chosen eigenstate. However, to my mind the asymmetry of this is probably the same as the asymmetry involved when you learn that a hidden ball is in my right hand rather than my left. It's not the ball that changes, it's your state of knowledge. There are arguments against this view based on Bell's theorem etc., but space is too limited to discuss them. – N. Virgo Feb 03 '12 at 10:51
  • Possible duplicate: http://physics.stackexchange.com/q/542/2451 and links therein. – Qmechanic Oct 06 '13 at 20:56
  • related: http://physics.stackexchange.com/q/81465/ –  Oct 20 '13 at 19:47
  • Entropy increasing is a necessity in a universe that is always striving for equilibrium of all sort. Basically, one would scrap the second law of thermodynamics by imagining a universe that don’t bother of removing differences of “potentials”. – J. Manuel Jul 12 '17 at 07:53
  • possible duplicate of https://physics.stackexchange.com/questions/10690/is-there-any-proof-for-the-2nd-law-of-thermodynamics –  Oct 16 '17 at 21:51

5 Answers5

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Arnold Neumaier's answer is correct but doesn't seem to have included enough detail to convince people, so here is an answer with a more in-depth explanation.

We have two fundamental theories of physics: general relativity and the standard model of particle physics. The standard model has CPT symmetry, and general relativity has local time-reversal invariance. Although neither of these is technically the same as global time-reversal invariance, for the purposes of the following discussion it's sufficiently accurate to say that the laws of physics are time-reversal invariant. Sometimes you will hear people state this by saying that the "microscopic laws" are time-reversal invariant, the intention presumably being to exclude the second law of thermodynamics, which explicitly distinguishes a forward direction in time. But this is an anachronism, since the second law is no longer considered fundamental but derived.

The question that then arises is, how in the world can you derive a time-asymmetric theorem from time-symmetric assumptions?

Consider the simulation shown below. On the right we have a box that has three areas marked with three colors, and $N=100$ particles that are free to move around in the whole box. (The vertical lines at the boundaries are just visual -- the particles cross them freely.) The simulation was done using this applet. The particles are released at random positions, with random velocity vectors, and their motion is simulated using Newton's laws, which are time-reversal symmetric. The graph on the left shows the number of particles in each area as a function of time.

simulation of particles in a box

Since the particles are initially placed randomly, roughly one third of them are initially in each region. At any randomly chosen time, the number of particles $n$ in, say, the red region has a mean of $\bar{n}=N/3$ and a standard deviation of about $\sqrt{\bar{n}}\approx 6$. Once in a while we get unusually large fluctuations, such as the one marked with a green arrow at $t=19$.

We can now state a derived law L:

(L) If we observe a statistically unlikely value of $n$ at some time $t_0$, there is a high probability that the values of $n$ both before and after $t_0$ (for $t\lesssim t_0-3$ and $t\gtrsim t_0+3$) are closer to the mean.

As $N$ gets larger and larger, L becomes more and more secure; the probability of seeing it violated becomes smaller and smaller. When $N$ gets as big as Avogadro's number, the probability of a violation becomes zero for all practical purposes.

This derived law is still completely time-reversal symmetric, so it doesn't appear to be quite the same as the second law of thermodynamics. But now consider the case where somebody artificially prepares the particles in the box so that they are all initially in the center. (If you run the applet at the link above, this is actually what it does.) The result is shown below.

enter image description here

An observer who doesn't know about the initial preparation of the system, and who only gets to see its behavior during the interval $0\lt t \lesssim 2$, will empirically arrive at a time-asymmetric "law" describing the behavior of the system: the system always evolves from high values of $n_{\text{black}}$ to lower ones. Not knowing the initial preparation of the system, but wishing to believe in a naturalistic theory of the operation of this little "universe," the observer might speculate that the initial, high value of $n$ was an extreme statistical fluctuation. Perhaps at $t\lesssim -2$ the system was in equilibrium. The observer can then explain everything in terms of the time-symmetric law L.

The same analysis applies to the conditions we observe in our universe, with some modifications:

  1. The discussion in terms of $n$ can be replaced with a discussion in terms of the number $\Omega$ of accessible states for a given set of coarse-grained observables -- or we can talk about $\ln\Omega$ or $k\ln\Omega$, i.e., entropy, which is additive.

  2. In the original statement of L we had a constant time 3, which was an estimate of the equilibration time for the toy model. For the real universe, this has to be replaced by some estimate of the equilibration time of the whole universe, which might be very long.

  3. And finally, we have the role played by the mischevious person who secretly initialized the system with all the particles in the center. This naughty trickster was effectively setting a boundary condition. In our universe, this boundary condition consists of the fact that, for reasons unknown to us, our Big Bang had a surprisingly low entropy. If there were some naturalistic principle that the Big Bang should be a typical state rather than a very special one, then our universe should have started out already in a state of maximum entropy.

In the world around us, we see various arrows of time. There is a psychological arrow (we can remember the past, but not the future), a thermodynamic one (candles burn but never unburn), a cosmological one (the Big Bang was in our past, not our future), and various others such as a radiative one (we often observe outgoing spherical radiation patterns but never their time-reversed versions). All of these arrows of time reduce to the cosmological one, which arises from a boundary condition.

The OP asked:

Is a world with constant/decreasing entropy theoretically impossible?

No. In fact, the world that is overwhelmingly the most probable and natural is one in which the entropy is, always has been, and always will be the maximum possible -- but in such a universe there would not be hairless primates tapping on computer keyboards. It is also certainly possible to have a universe in which entropy is always higher in the past and lower in the future. In fact, our own universe is an example, if we simply interchange the arbitrary labels "past" and "future."

A longer discussion of these ideas, with lots of historical context, is given in Callender 2011. Historically, there has been a lot of debate and confusion on these issues, and unfortunately you will hear a lot of this confusion echoing down the halls a hundred years later, perhaps due to the tendency of textbooks to hew to tradition. For example, Ritz and Einstein had a debate in 1909 on the radiative arrow (as discussed in Callender and references therein). Ritz's position, that the radiative arrow is fundamental, is no longer viable.

References

Callender, Craig, "Thermodynamic Asymmetry in Time", The Stanford Encyclopedia of Philosophy (Fall 2011 Edition), Edward N. Zalta (ed.), http://plato.stanford.edu/archives/fall2011/entries/time-thermo

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The short answer is that such a universe cannot be envisaged, not with relevance to our known physics.

Entropy as defined in statistical thermodynamics is proportional to the logarithm of the number of microstates of the closed system, the universe in your question. You would have to devise a universe where the number of microstates diminishes with time.

The great multiplier of microstates in our universe is the photon, which is emitted at every chance it gets, and thus increases the number of microstates. Photons are emitted by electromagnetic interactions and by all bodies consisting of atoms and molecules due to the black body radiation effect. Each emitted (or absorbed, because the state of the atom that absorbed it has changed) photon defines a new microstate to be added to the number of microstates, whose logarithm defines entropy. A universe without electromagnetism would not have atoms.

It is worth noting that all biological systems decrease entropy, as does the crystallization of materials, but this is possible because the systems are open and the energy exchanges create a large number of microstates thus obeying in the closed system the entropy constraint.

anna v
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  • This doesn't make much sense. The sentence "Entropy as defined ..." doesn't connect to what follows. Emission of a photon can be time-reversed, giving absorption of a photon. –  Oct 05 '13 at 04:19
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    @BenCrowell that would count as an extra microstate, increasing the number. and black body radiation is universal too, no reabsorption. – anna v Oct 05 '13 at 04:21
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    The position you seem to be arguing is the one taken by Ritz in a 1909 debate with Einstein. However, it's very hard to tell what your argument is, since your answer is very brief and, at least to me, doesn't seem to have a coherent logical thread. In the year 2013, there is a consensus that Ritz was wrong. See section 2.2 of this paper http://plato.stanford.edu/archives/fall2011/entries/time-thermo/, and references therein. If you want to argue that the modern consensus is incorrect after all, then the burden of proof is on you to put together something more compelling than this. –  Oct 05 '13 at 17:57
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    @BenCrowell My position is more general than the one in par.2.2 of your link. I am sorry if you do not find it clear enough. I am using the statistical mechanics definition of entropy and each photon, radiated (or absorbed) defines a different microstate of the universe and adds to the number of microstates . It is a continually increasing number due to the bosonic nature of photons. – anna v Oct 05 '13 at 19:14
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    I'm sorry, but I just don't see an argument in what you've written. Are you arguing that it's self-evident that the number of microstates must increase over time? That would be wrong, but I'm just guessing whether that's what you have in mind, because you haven't laid out what your position is. It is a continually increasing number due to the bosonic nature of photons. This doesn't make any sense. Ritz's argument is a classical one based on Maxwell's equations. If you have some quantum-mechanical version of Ritz's argument, which you believe fixes it up, you haven't explained what it is. –  Oct 05 '13 at 19:31
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    Anna, I'm quite confused since you seem to be saying that each photon emission & absorption process increases the entropy. Take an isolated, empty black-body cavity. If the radiation field is in thermal equilibrium with the walls of the cavity then the entropy is maximal and constant, by definition. But in this case absorption and emission processes are still taking place all the time, implying by your argument that the entropy is increasing. Could you explain this apparent contradiction? – Mark Mitchison Oct 05 '13 at 21:03
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    @MarkMitchison In the black body setup, concentrating our attention to the inside of a solid sphere one sees the equal sign in the entropy increase law ( there is an equal). The real answer is in my last paragraph. It has to be a closed system for the law of increasing entropy to hold. Out radiation from the black body surface is not included in the entropy of the black body hollow. As I say, crystallization even decreases S in the body as the number of microstates decreases by the ordering of the lattice. But it is not a closed system. A universe is by definition the maximal closed system. – anna v Oct 06 '13 at 04:20
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    @BenCrowell The photons being bosons do not have a limit in their number. They can occupy the same quantum numbers. It is the number of photons. though that is changing the microstate and adds to the counter of microstates. Yes, it is a quantum mechanical argument as photons par excellence exist only in quantum mechanics. I do not have the urge to read the details of Ritz's argument and fix it up or whatever. Quantum mechanics supersedes classical mechanics on all fronts. – anna v Oct 06 '13 at 04:31
  • @MarkMitchison My argument about absorption being another microstate lies in assuming it is a random absorption by any electromagnetic setup. If one can set up a photon gas where the emission and absorption are diagrammatically equal and opposite in sign then the equal sign dominates in the greater than or equal law. It is an idealized thought experiment that cannot hold in a real situation, and again in this thought experiment the system is not closed. The solid surface of the body containin the cavity will be continually radiating to space and thus entropy will increase in total. – anna v Oct 06 '13 at 04:40
  • Right, Anna, but you surely have to be willing to accept the existence in principle of idealised "isolated systems" that are not the entire universe, otherwise what is the point of formulating the second law? You can't measure the entropy of the entire universe so saying it always increases is operationally meaningless. The second law is only useful if we can identify systems which approach equilibrium and then remain there. – Mark Mitchison Oct 06 '13 at 08:55
  • Could you give me an example of a closed system that equilibrates via photon emission and absorption, including what are the thermodynamic variables? I still don't understand how it necessarily holds that any photon absorption or emission process increases the number of microstates that are consistent with the macrostate. Partly because I don't really understand your conception of macrostate. And especially since the "number of photons" in a system is virtually never well defined anyway. I am not just nitpicking, by the way, I am genuinely interested to see if I can make sense of this. – Mark Mitchison Oct 06 '13 at 08:59
  • @MarkMitchison thermodynamics is a limiting case of quantum statistical mechanics. Quantum mechanically no system is really closed because of continuing interactions. Take a black body cavity and suppose it is totally reflecting, maybe metallic with the electrons at the fermi sea. Then one can do a thought experiment where the photons in the cavity are continually reflected or absorbed on the same energy level and radiated back. I think in this case entropy would be constant because the total function describing the cavity would be invariant in time. In reality – anna v Oct 06 '13 at 11:39
  • there would be many energy levels and total absorptions etc. Every single interaction changes the microstate and it can be counted up in the tally. Maybe I should say the density matrix instead of "microstate" but at the limit they are the same http://en.wikipedia.org/wiki/Entropy#Entropy_in_quantum_mechanics . There can be isolated systems within our possibilities of measurement, measurement errors, but when one looks closer they really are not. Up close one has to use quantum mechanics – anna v Oct 06 '13 at 11:45
  • But the statistical density matrix is the macrostate: an ensemble of microstates. If you mean von Neumann entropy as in entanglement entropy of a reduced density matrix, then you are not necessarily talking about thermodynamic entropy. So I still don't understand your argument. – Mark Mitchison Oct 06 '13 at 12:26
  • @annav After thinking a bit I might have an idea what you're getting at. Let me try to rephrase your argument and see if you agree. Quantum mechanics shows us that all subsystems are open, since all forms of matter or energy couple to fluctuations of at least one quantum field, the most prominent being EM. The emission of photons produces correlations between any chosen subsystem and its environment. Since we cannot measure the state of the environment these correlations are lost in the thermodynamic description. The lost information in the correlations is the entropy increase of the subsystem – Mark Mitchison Oct 06 '13 at 12:39
  • Absorption also results in correlations between environment & subsystem in exactly the same way. The "number of microstates" increases in the sense that there are many microstates of the correlated environment + subsystem consistent with our reduced description of the subsystem alone. This argument coincides with my understanding of entropy increase in the canonical ensemble of quantum statistical mechanics. Do you agree? – Mark Mitchison Oct 06 '13 at 12:40
  • @MarkMitchison Yes, that is what I am getting at, obviously not very successfully :). – anna v Oct 06 '13 at 13:01
  • @annav Well, success has now been achieved, at least for our mini-discussion. Thanks for humouring me. And +1 :) – Mark Mitchison Oct 06 '13 at 13:03
  • The long dialog in comments fails to convince me that anything has been clarified. This answer is simply wrong. @annav, if you really believe that there is a valid argument here, which could be expressed clearly, then the responsible thing to do would be to go back and edit the answer to express it clearly with the benefit of Mark Mitchison's comments. All I see is a jumbled mess of non sequiturs, assumptions invoked without justification, and garbled use of terminology that sounds impressive. –  Oct 06 '13 at 15:00
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    @BenCrowell This is not an essay submitted in a class. It is a qualitative answer dependent on the quantum mechanical nature of the underlying reality. The voting decides how clear, unclear or wrong the answers are. Sorry if I cannot make you see my POV but I cannot make it more clear than this. – anna v Oct 06 '13 at 16:54
  • @BenCrowell Do you disagree with my point of view? Seems to be a fairly standard interpretation of what's going on... I appreciate that such a long discussion is hardly ideal but I think the answer + comments is worthy of an upvote. The use of the phrase "number of microstates" is the only slight beef I have, which confused me for a long time. The universe/system is only ever in one microstate. – Mark Mitchison Oct 06 '13 at 18:13
  • @MarkMitchison: Do you disagree with my point of view? I don't know what your point of view is. Would you like to express it in an answer? I appreciate that such a long discussion is hardly ideal but I think the answer + comments is worthy of an upvote. I disagree. I would nominate this answer as one of the worst ever to receive an upvote on physics.SE. –  Oct 06 '13 at 18:22
  • @BenCrowell OK, but it will probably 1) not appear until tomorrow, 2) be very similar to the pair of comments I wrote above, although with less emphasis on the role of photons specifically. I tend to take the Jaynesian viewpoint on the meaning of entropy. In this case Boltzmann's molecular chaos assumption makes plenty of sense, since your inability to measure all degrees of freedom allows the observer to consistently throw away correlations that are there in principle (Joint measurements required to distinguish an uncorrelated mixed state from a correlated pure state are locally inaccessible) – Mark Mitchison Oct 06 '13 at 18:28
  • @BenCrowell " I would nominate this answer as one of the worst ever to receive an upvote on physics.SE" Haha, did I hear someone say hyperbole? Anyway, I'll write up my thoughts in the coming week for you to sink your fangs into :) Although they will answer the different question "Why does entropy increase?" I believe your answer to this question is fairly comprehensively correct. Until then, I should really return to my attempts to not fail my PhD. – Mark Mitchison Oct 06 '13 at 18:29
  • @BenCrowell Also, this answer I just found (by Nathaniel, unsurprisingly) is basically what I would write. Note particularly the discussion in the penultimate paragraph about correlations between relevant and irrelevant d.o.f. – Mark Mitchison Oct 06 '13 at 18:46
  • Also, this answer I just found (by Nathaniel, unsurprisingly) is basically what I would write. Nathaniel's answer is great, and I just gave it a +1. However, I don't see any connection between it and this question, or between it and annav's answer. –  Oct 06 '13 at 19:16
  • The voting decides how clear, unclear or wrong the answers are. No, voting does not decide whether an answer is wrong. Logic and evidence decide whether it's wrong. "If I were wrong, then one would have been enough!" -- http://en.wikipedia.org/wiki/Criticism_of_the_theory_of_relativity#A_Hundred_Authors_Against_Einstein –  Oct 06 '13 at 19:21
  • @BenCrowell Note I said "on this board" It is not an examination and the vote is an opinion mostly driven by intuitive knowledge. It could be dissected and Mark did that, found I am confused with the microstate definition going from classical to quantum mechanical framework( and I will look into it) My answer may be confusing for you but you have not proven it is wrong, and confusion is down-voted, as you did. – anna v Oct 07 '13 at 04:00
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The microscopic laws are reversible in time (if you also change chirality and the sign of all charges). Thus one cannot prove what you'd like to prove.

Statistical mechanics, which is the discipline in which one derives the second law from microphysics, always makes one or the other assumption that induces the direction of time actually observed in our universe: That entropy increases (unless the whole world is in equilibrium, which it currently isn't).

However, you could run the whole universe backward, and it would satisfy precisely the same microscopic laws (if you also change chirality and the sign of all charges). But entropy would decrease rather than increase.

I don't think your friend would like to live in such a world.

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You may or may not be satisfied with this answer, but I found the question fun, and so I gave it a shot.

Take the classic bowls and balls multiplicity set-up. We have four balls, labeled A,B,C,D and two bowls to put them in. Normally, two balls per bowl is the macrostate with the highest multiplicity, and so for the sake of argument let's say that's where you start the system off at.

Now, each microstate has a particular probability of occurring, and that probability was the item I decided to pay attention to for this exercise. The Complex Klein-Gordon lagrangian allows for negative probability densities (from the current 4-vector $j_\alpha =\psi^{*}\overleftrightarrow{\partial_{\alpha}}\psi$), which is one of the reasons it went out of style for electrons. But it also maintains conservation of probability, $\partial_\alpha j^\alpha =0$.

So, consider the bowl example, except let three of the microstates corresponding to the "two balls per bowl" macrostate have negative probability of occurring (for whatever reason $j_0$ for these states leads to negative probability). Then when one sums up the probabilities, the "two balls per bowl" macrostate still has the highest multiplicity, but does not have the greatest probability of occurring, so if the system started out in the "two balls per bowl" state, it is likely that it would evolve into a different state (most likely the "three balls in one bowl" state). Of course, another microstate would have to gain probability to pick up the slack to conserve it.

I don't know if that tickles your fancy but its the first thing that came to mind when I read the question. Someone might have a slightly more reasonable answer.

kηives
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It seems to me that the concept of maximum entropy cannot become a reality. As Cambridge experiments have shown 'zero point energy' occurs before energy reaches absolute zero. Of course this relates to open systems. I'm not sure about closed systems. Although I suspect as long as a closed is able to provide energy to other systems it does not reach M E.Also if a system was able to reach ME it would, by definition cancel itself out and could not exist. It's like saying lets get rid of 'left' and only keep 'right' If left no longer existed right could not exist.

degs
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    Which experiments are you talking about. It seems you have conceptual problem, zero point energy is already the lowest energy results from quantum mechanis. And a system exchange energy with outher is not a closed system – unsym Oct 05 '13 at 01:57