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I always thought the uniformity in the temperature of the CMB was supposed to be expected, since it's a much more probable initial condition for the universe. I finally found someone explaining what I mean in much better words (link):

Horizon problem isn't really a problem

If we examine from statistical mechanics principles what thermal equilibrium really means, we see that it is the most probable macrostate for a system (in other words, the state with highest entropy). Systems evolve towards thermal equilibrium not because nature has any sort of preference for evening out energy among all degrees of freedom, but simply because having a roughly equal partition of energy among degrees of freedom is OVERWHELMINGLY probable.

For exactly the same reason why it is overwhelmingly probable for a closed system to move toward thermal equilibrium, it is overwhelmingly probable for a completely randomly selected initial condition to be in thermal equilibrium. No causal contact is necessary.

The only "counter-argument" I could find for that, ironically enough, comes from Jason Lisle (link):

(...) in the early universe, the temperature of the CMB would have been very different at different places in space due to the random nature of the initial conditions.

But if that "random nature of the initial conditions" is of the same order of magnitude as quantum fluctuations, wouldn't that apply to the early instants of inflation too? If so, how would thermal equilibrium be even possible under such quantum fluctuations during inflation?

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Wood
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  • The problem is that the early universe is not in thermodynamic equilibrium. Assuming the predictions of ordinary GR, there is not enough time for it to settle into an equilibrium state before the expansion has "frozen" the temperature difference. The problem with this can be solved in a number of ways, one of which is inflation, which basically stretches "the lumps in the dough" so thin that they look homogeneous, or one can assume that the initial expansion was much slower than GR predicts, or that the universe was much larger and pre-mixed etc.. – CuriousOne Dec 16 '14 at 04:14
  • @CuriousOne You're just stating the basic definition of the Horizon Problem. That's not really my question. – Wood Dec 16 '14 at 07:00
  • I gave you three or four possible solutions... besides pointing out that the problem only occurs IF GR is assumed to be valid at that early era, which I personally would not assume. If anything, this could be the first clear hint that GR is not the correct theory. If I am not mistaken, the problem is not present in Einstein-Cartan theory (i.e. GR with torsion), and that's a totally straight forward extension of GR. – CuriousOne Dec 16 '14 at 07:23
  • @CuriousOne I'm not asking for the solution to the Horizon Problem. Read my question again. I'm arguing that the traditional explanation (the one you gave) is flawed. No causal contact is necessary for thermodynamic equilibrium. So my question is: how is the Horizon Problem really a problem? To me, it looks like GR alone already predicts the uniformity of the temperature of the CMB. No inflation would be necessary to explain it. – Wood Dec 16 '14 at 18:26
  • I have no idea where you are getting the idea from that an interaction free system can ever attain thermal equilibrium. That is just not so, not even for the ideal gas, for which thermal equilibrium is a naive postulate to give first term students a system to work with that they can actually understand (even though it is 100% flawed). The only interaction in GR are gravitational waves, and they are just not strong enough to thermalize a conventional big bang model. That, it seems, is not true in Einstein-Cartan any longer. – CuriousOne Dec 17 '14 at 02:02
  • @CuriousOne Please, read the question first. http://physics.stackexchange.com/q/153491/67685 You're trying to answer my question before even reading it. – Wood Dec 17 '14 at 12:28
  • Systems only evolve towards thermal equilibrium it they interact. Thermal equilibrium is not something that just happens. It's something that happens because all parts of the system are in contact. A "random choice" is a physical process which does not happen without interaction, unless the semantics of "random choice" is merely meant to hide one problem behind another (in this case the question "What caused the choice to be random and who expended the energy needed to throw so many dice?"). – CuriousOne Dec 17 '14 at 12:55
  • @CuriousOne I'm not saying that it moved towards thermal equilibrium without interaction, I'm arguing that it might have already started in equilibrium. – Wood Dec 23 '14 at 10:47
  • How did it get into that equilibrium? You are basically replacing one miraculous assumption with another. – CuriousOne Dec 23 '14 at 10:49
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    @CuriousOne It didn't "get into equilibrium", it always was in equilibrium. A "miraculous assumption" would be to think otherwise, since that would be far more unlikely. – Wood Dec 23 '14 at 10:53
  • How exactly is it far more likely? Imagine I had say a deck of cards, and the early universe was defined by randomly drawing 4 cards. Given enough time, perhaps the cards will go to 4 aces (equilibrium) but the chance that you start with 4 aces randomly picked is like very slim. Personally I don't think trusting that paragraph is good, I can't find any other instances of this argument. – QCD_IS_GOOD Dec 25 '14 at 23:09
  • @JoshuaLin That's not what equilibrium means. A better analogy is to think of you randomly selecting a thousand numbers from 0 to 10 and calculating the average. It's much more likely that the result will be closer to 5 than 10 (or 0). The average corresponds to the macrostate and the specific collection of numbers picked correspond to the microstate. Please read this excellent answer that has a lot to do with my question. – Wood Dec 26 '14 at 05:05
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    @Wood Don't worry, I at least understand your concern. But I've searched for years without finding a satisfactory answer, instead just seeing the same unjustified dogma repeated over and over. If you don't get a good answer over the next few days, feel free to ping me, since I would probably put my own bounty on the question then. –  Dec 26 '14 at 23:42
  • @CuriousOne has not answered the basic question here. At $t=0$, the entire universe is in causal contact and thermal equilibrium (every point in the universe has the same temperature as every other point). Please tell us what physical principle causes the universe to drop out of thermal equilibrium at $t=0+\delta$. That is, what process of Thermodynamics allows one spot to have a different temperature to the spot immediately to the right or to the left as the universe uniformly cools adiabatically. –  Mar 22 '20 at 13:34

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I’m not sure I fully understand the question but I’ll try. The “curious one” says that if the initial condition is picked at random then it should be a state of thermal equilibrium (maximum entropy) since the overwhelming majority of states look equilibrium-like. Now there has to be something wrong with that in the cosmological context. States of thermal equilibrium are static and the world is manifestly not static.

What statistical mechanics actually says is that for a fixed total energy most states look like thermal equilibrium at a temperature that depends on the total energy. In other words, for a given energy most states maximize the entropy. If we don’t constrain the energy then the state of maximum entropy has infinite energy and is not really a state. Even more to the point, the idea of energy is very confusing in general relativity. In a cosmological setting the total energy is always zero. So the whole framework of statistical mechanics and maximum-entropy states is not well defined. To put it bluntly there is no theory of the initial state. That’s the problem.

What inflation does is it makes the later history very insensitive to the initial state. Whatever theory for the initial state is put on the table, inflation will wipe out its memory. The result of inflation is a “fixed point” or “attractor” which means a particular behavior that a very wide class of initial conditions will result in.

Leonard Susskind

Leonard Susskind
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  • I'll repeat my very basic issue with this line of reasoning. At $t=0$, the universe was causally connected. Every point in the universe had the same temperature as every other point. At $t=0+\delta$, what physical process causes those points to have a different temperature from the point immediately adjacent to it? –  Mar 22 '20 at 13:28
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The problem stemmed from having to deal with how such a vast region of space had such a fine tuned uniformity. Without inflation that same volume could not have maintained the same uniformity once you consider the mean free path between the particles. Thermal equilibrium requires not only a high temperature. It also requires a sufficient density to allow the reverse reactions to occur.

Prior to inflation the temperature and density is sufficient. Then inflation occurs. That sudden volume change would normally cause a sudden cooling. If inflation has multiple waves or perturbations there would have been anistropies crop up. However the slow roll process at the end of inflation caused a significant reheating effectively wiping slate clean of any previous anistropies and previous particles that were not in equilibrium. This makes determing which inflation out of the 70+ inflation models more difficult.

The latest Planck dataset favors an inflation model with a single scalar and low kinetic term. However this does not rule out multiscalar models. http://arxiv.org/pdf/hep-th/0503203.pdf"Particle Physics and Inflationary Cosmology" by Andrei Linde http://www.wiese.itp.unibe.ch/lectures/universe.pdf:"Particle Physics of the Early universe" by Uwe-Jens Wiese Thermodynamics, Big bang Nucleosynthesis These articles will help the finer details see chapter 3 of the second one. http://arxiv.org/abs/1303.3787

Hraish kumar
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    That doesn't address the main point that the original conditions of the universe would favor uniformity, which already solves the Horizon Problem. No causal contact is necessary to maintain equilibrium if the system is already in equilibrium. – Wood Dec 26 '14 at 05:16