If the universe is infinite, does that mean that everything exists somewhere?

A. Neumaier

This is undecidable.

Suppose you'd record every detail about the history of the universe, wait till it has died, and then replay it in a perfect simulation (where of course, everything is already there). The physical laws would be exactly the same - without the slightest detectable difference.

TrickyDicky

Nicely put Marcus. I'd like to add to your insightful phrase in my bold that maybe sometimes we get hung upon abstractions about Spacetime, but it's good to remember ourselves once in a while that spacetime is just a geometrical abstraction to describe the relations within matter in its broad meaning of mass-energy continuum. In this sense matter is all there is and surely what makes the difference.

About the stationary observers, they illustrate the way the GR equations were designed in a general covariant way to have 6 independent differential equations with 6 unknown quantities and another 4 unknown quantities that are arbitrarily fixed with the choice of coordinates.
This condition allows us to stablish the rest frame or stationary observers as we set the coordinate space and the coordinate time for a particular metric, and therefore we can determine a rest state wrt these coordinates so in this sense the fundamental observers appear not only in the "Friedmann model" but in any metric we might build from the GR equations.

In our cosmological model this rest frame is embodied by the CMB like you say, we measure our motion with respect to this radiation that fills the vacuum thru the universe.

This is for a very practical reason, the CMB are photons and we are able to detect them, quite easily (from 1965 at least), we could say the CMB is the "visible" part of the energy density of the vacuum, which is indirectly observe or "felt" as dark energy (and also as dark matter according to some models with inhomogeneities such as those of T. Buchert et al., but these models are not mainstream).

Chalnoth

Yes, very true. So to do this properly, you'd have to integrate over all macrostates. That result, also, will have to be finite.

The specific superposition of states is just a representational issue and thus cannot be a physical effect. That is to say, a particle that is in an eigenstate of energy is in a superposition of states in position. So you can recast any particle that we "see" as being in a superposition of states as being in a particular eigenstate by constructing your operator appropriately.

A. Neumaier

Nothing in quantum mechanics allows you to deduce this!

But entropy only counts the eigenstates of the energy. On the other hand, most states in nature are not eigenstates (only stationary states are). Thus the vast majority of observable states is not counted by entropy.

Chalnoth

This stems from the exact same arguments as in quantum field theory: there has to be some high-energy cutoff.

Now you're mixing different descriptions of the same system. But it isn't true in any event. The computation of entropy has to count the full set of microstates, which for real particles also includes things like spin and angular momentum, as well as energy.

Deuterium2H

Coming full circle, and getting back to the original question/post...I think that my arguments using mathematically-based Set Theory, and Chalnoth's physics-based arguments (Thermodynamics, Statistical and Quantum Mechanics) have both converged on an answer that is rather non-intuitive. Certainly, it goes against popular "opinion". But if mathematics can teach us anything, it is that transfinite Set Theory is itself counter intuitive. This just so happens to be very much the case, as well, with Quantum Theory.

The answer to the the original post is quite simply this...

Given an infinite Universe, it is does NOT necessarily follow that "everything exists somewhere". Or, in other words, as previously argued...the Universe being infinite is a necessary condition, but not a sufficient condition to ensure that any/every event that has a finite probability must occur/exist somewhere in the Universe.

A. Neumaier

Can you show why it should follow from that???

If you look at the books of statistical mechanics, you'll find that microstates means only ''energy eigenstate'', and not ''arbitrary state''.

Chalnoth

If the integration over macrostates is limited at some high energy, and every component of that integration is finite (that is, if the function is well-defined everywhere), then it will have to be finite, because it will be a representation of a sum over a finite (but large) number of states.

I don't think this is true at all. The basis you do your sums in is completely irrelevant. It has to be, by nature of the underlying mathematics. The only reason why the sums are done in the energy basis is because:
1. Most introductory statistical mechanics books neglect complications like spin, angular momentum, and any other potential quantum numbers that are different from energy.
2. It is much easier to do the sums in terms of energy because the total energy of the system is one of the macroscopic variables we use.

In principle you could always change to some other basis, and if it's done right you have to come up with the exact same answer, but it's going to be much more difficult to connect the other basis to the macroscopic variables.

That said, this is an off-topic argument, because it simply has no application to my original statement, which had nothing whatsoever to do with entropy. Remember, I was making two separate points when talking about the finite number of potential states. The entropy argument was one argument, and is a separate one from the purely quantum-mechanical one.

The purely quantum-mechanical argument is that as long as you cut off your states at some high energy, there are a finite (though large) number of states. You came back and stated that you can also have superpositions of those states, and since there can be an infinite number of superpositions, this finite number of quantum states leads to an infinite number of possibilities.

Not so, I said, because the superpositions are merely a representational issue: any superposition of states can be represented as an eigenstate of the right operator. You'll still always get the exact same number of states, no matter the representation you use, as long as you do the counting correctly. This second argument for the finite number of states has nothing to do with statistical mechanics.

A. Neumaier

But this can be argued only locally. The energy cutoff of QFT is something at the level of individual scattering events, while the integration over macrostates in statistical mechanics never had such a cutoff.


No. The only reason why the sums are done in the energy basis is because the canonical ensemble involves the Hamiltonian, and the trace defining the entropy reduces to a sum _only_ in a representation where the basis states are energy eigenstates.


And I pointed out that both your hypothesis and your conclusion are flawed.

Chalnoth

Typically you don't do any integration over macrostates in statistical mechanics. The integrations are over microstates. And you don't need any cutoff there because we are generally considering systems that are at such low temperatures that any cutoff that would come in from high-energy physics is irrelevant.

But when considering all possible states of the system, you have to integrate the number of states over the ensemble of all possible macrostates. As long as the number of states for any given macrostate is finite, and as long as you have to cut off your integral at some energy (so that the integral doesn't go to infinite), the result also has to be finite.

And the reason why the canonical ensemble includes the Hamiltonian is because energy is one of the macroscopic variables. It is the only operator used because in the classical treatment, energy is the only thing that is allowed to be mixed (the particle number and volume tend to be fixed). When considering more complicated systems, such as a quantum system including spin or one where the particle number is allowed to vary, you have to make the ensemble a bit more complicated, so that it incorporates these added degrees of freedom.

It doesn't really matter, though. You can still transform to another basis if you like. The results will necessarily come out the same. It's just that the math will be horribly difficult, and thus it's much easier to just remain in the eigenbasis of your ensemble.

No, because you changed topics and started talking about statistical mechanics in an argument that had nothing to do with statistical mechanics.

A. Neumaier

As if entropy and counting quantum states could be done without statistical mechanics.

Chalnoth

Huh? Counting states is a component of statistical mechanics, but hardly requires it. Entropy doesn't even need to come into the argument when all you're interested in is the total number of possible states.

Terraqueous

I'm not completely sure on this, but I think the answer is 10^-6 seconds.

GODISMYSHADOW

Are you suggesting the universe is a simulation?

A. Neumaier

No, only that we couldn't distinguish it experimentally from a simulation. The physical laws would be exactly the same if the simulation was perfect.

Tanelorn

"No, only that we couldn't distinguish it experimentally from a simulation. The physical laws would be exactly the same if the simulation was perfect."


I am not sure I understand what you are actually saying there but I can say that there is a great difference between a mathematical simulation on a computer with a cpu executing single arithmetic instructions one bit at a time and the space time reality we are part of. In a similar way it is highly unlikely that life like intelligence can ever be created on such a simple calculating device.

A. Neumaier

Of course. If our universe were a simulation, it would have been simulated on one of God's hyper-computers with a very different physics and technology.

The point is, we couldn't see the difference in the results.