How can an expanding Universe look homogeneous?

[Vincentius]

Observation shows that the Universe is homogeneous (and isotropic) at the large scale, while one expects to see inhomogeneity (increasing density at greater distances) on the past light cone due to expansion. This seems inconsistent. Am I misunderstanding something here?

 

[phinds]

Observation shows that the Universe is homogeneous (and isotropic) at the large scale, while one expects to see inhomogeneity (increasing density at greater distances) on the past light cone due to expansion. This seems inconsistent. Am I misunderstanding something here?

Yes, you are comparing apples (the universe then) to oranges (the universe now).

 

[Vincentius]

I understand the difference between light cone and constant time slices. The latter are considered homogeneous by the cosmological principle. Looks fair. The constant time slices are spacelike however, so one can not observe anything within the slice. One can observe on the light cone, which I expect shows inhomogeneity due to expansion. Is that correct? So the question still stands how one infers homogeneity on the large scale through observation.

 

[Grinkle]

I think the evolution of the universe is taken into account when reaching that conclusion.

The universe looks the same no matter which direction we look in. Looking further out, the universe looks different - but it looks different in the same way no matter which direction we look in.

I may be missing your point, though.

 

[Orodruin]

One can observe on the light cone, which I expect shows inhomogeneity due to expansion. Is that correct?

Yes.

So the question still stands how one infers homogeneity on the large scale through observation.

Easy question, you cannot.

However, that was not your original question. Your original question was whether it was inconsistent to have inhomogeneity on the past lightcone and homogeneity on spatial slices (it isn't).

 

[Orodruin]

The universe looks the same no matter which direction we look in. Looking further out, the universe looks different - but it looks different in the same way no matter which direction we look in.

That's isotropy, not homogeneity.

 

[rootone]

The faster you go, the rounder you get.

 

[Vincentius]

Yes.

Easy question, you cannot.

However, that was not your original question. Your original question was whether it was inconsistent to have inhomogeneity on the past lightcone and homogeneity on spatial slices (it isn't).

Perhaps the original question was confusing. It did not mention spatial slices, though. The question is purely about which statement is right: A) observation shows that the Universe is homogeneous on the large scale, or B) Observation shows inhomogeneity due to expansion. The first statement I do not understand but I come across frequently, while the second statement hardly ever. Perhaps both are true but need different interpretation?

 

[PeroK]

Perhaps the original question was confusing. It did not mention spatial slices, though. The question is purely about which statement is right: A) observation shows that the Universe is homogeneous on the large scale, or B) Observation shows inhomogeneity due to expansion. The first statement I do not understand but I come across frequently, while the second statement hardly ever. Perhaps both are true but need different interpretation?

The universal expansion is the same everywhere, hence homogeneous. To be inhomogenous, the universal expansion would have to vary from one part of the universe to another.

 

[PeterDonis]

The question is purely about which statement is right: A) observation shows that the Universe is homogeneous on the large scale, or B) Observation shows inhomogeneity due to expansion.

Both statements are correct, although "inhomogeneity" is a somewhat misleading way of describing observation B). What we observe as we go further and further along our past light cone is better described as "times further in the past" rather than "increasing distances", and the fact that the observed density increases as we go further along our past light cone just means the universe was denser in the past.

With that caveat, observation B) is obvious (you stated it yourself in the OP). A is not quite so obvious, but it follows from the observed isotropy, which is part of observation B) (we see the same behavior of density in all directions). If the universe were not homogeneous at a given "instant" of cosmological time (more technically, on a given spacelike slice of constant cosmological time), we would not expect to see isotropic behavior of the density when we look back along our past light cone.

 

[Orodruin]

A) observation shows that the Universe is homogeneous on the large scale, ... The first statement I do not understand but I come across frequently,

Please provide a reference.
Homogeneity is an assumption, albeit a reasonable one. You cannot show that the Universe is homogeneous by direct observation. (You could infer that what you see is going to evolve essentially to what you see locally at the present by observing your past lightcone and applying your knowledge of GR and how matter behaves).

If the universe were not homogeneous at a given "instant" of cosmological time (more technically, on a given spacelike slice of constant cosmological time), we would not expect to see isotropic behavior of the density when we look back along our past light cone.

Unless we are in a special place where the Universe appears isotropic although it is not homogeneous. You cannot draw the conclusion of homogeneity based on isotropy alone. Isotropy and homogeneity are different things. I think it is also unclear what we mean by "cosmological time" unless the Universe is homogeneous as our usual definition of cosmological time involves a RW spacetime - which is homogeneous by definition.

 

[PeterDonis]

You cannot draw the conclusion of homogeneity based on isotropy alone.

Yes, I agree it's not a certain conclusion. But I think it is a plausible inference.

I think it is also unclear what we mean by "cosmological time" unless the Universe is homogeneous

Yes, agreed.

 

[TEFLing]

Quoting one particle physicist,

"Based on observations and the second law of thermodynamics, we know that at the beginning of the universe, a quantity called entropy was extremely low, and has been increasing ever since. Entropy can be thought of as a measure of disorder or randomness. A low-entropy beginning to our universe means that the big bang origin was not a chaotic event but was highly ordered"

If the tire universe began from a singular highly ordered, ultra low entropy state... Then everything everywhere began in a very similar initial condition. And if everything everywhere started in extremely similar states. Wouldn't we expect everything to evolve towards similar states?

Is it possible that the uniformity homogeneity and isotropy of the cosmic microwave background radiation? merely reflects the initial. Low entropy state?

 

[Grinkle]

Then everything everywhere began in a very similar initial condition.

I think this was answered here -

Unless we are in a special place

Wouldn't we expect everything to evolve towards similar states?

My understanding is that without an initial rapid inflation one would expect differences in the evolution of the temperature profile; it requires more than just a uniform initial state (which is a part of the explanation also) to explain what we observe today. One needs to hypothesize inflation also to explain it.

 

[PeterDonis]

Wouldn't we expect everything to evolve towards similar states?

No, because entropy increases as the universe evolves, and in the presence of gravity, things becoming more and more clumped is higher entropy.

 

[TEFLing]

No, because entropy increases as the universe evolves, and in the presence of gravity, things becoming more and more clumped is higher entropy.

So I understand that going forward in time towards a hypothetical big crunch is not analogous to going backwards in time towards The Big Bang.Going forward to in time towards a big crunch, densities increase, energy densities increase, temperatures increase and entropy also increases. Like the compression stroke of a car engine.

Whereas going backwards in time towards The Big Bang. Although matter and energy densities and temperatures increase, entropy decreases.

What does that even mean? How can you have high temperature low entropy?

The only thing I can think of is some kind of Fermi Fluid at zero Kelvin. Which even though it has no temperature? Still, by the exclusion principle. The Fermi "surface" of the Fermi "sea" Maintains occupied states have very high energy??

What exactly is ultra high temperature yet ultra low entropy??

 

[PeterDonis]

How can you have high temperature low entropy?

Because the universe gets more and more uniform--less and less clumped--as you go back towards the Big Bang. The decrease in entropy due to that extreme uniformity more than compensates for the increase in entropy due to the higher density and temperature.

If the universe were to collapse into a Big Crunch, it would not be the same; the universe would still be getting more clumped gravitationally as it collapsed.

 

[Grinkle]

the universe would still be getting more clumped gravitationally as it collapsed.

Is there potentially a point during a Crunch where density starts to be come more uniform due to how high the average density has become? To illustrate my mental image - if the entire universe attains the average density of the average density of a neutron star, would that be a very uniform density, or still clumped (I don't know if neutron stars have a very uniform density but I imagine they do)? Or is that so much mass as such a high density that it can't be described without a quantum-gravity or other new theory?

 

[PeterDonis]

Is there potentially a point during a Crunch where density starts to be come more uniform due to how high the average density has become?

No. The increasing density would actually increase clumping, if anything.

 

[phinds]

@Grinkle, to expand just slightly on what Peter said, keep in mind that one of the reasons that cosmic inflation has been posited for the first tiny fraction of time after the singularity is to explain how it is that we have such relative homogeneity and isotropy now compared to the very early universe, where, it is believed, things were less so.

 

[Grinkle]

@phinds @PeterDonis Thanks, both. My intuitions for this are clearly way out of whack.

I do recall from a different thread Peter explaining to me that in the presence of gravity clumping is a high entropy state because there are many more likely clumped states than un-clumped states, which made perfect sense to me at the time (and still does).

 

[TEFLing]

Because the universe gets more and more uniform--less and less clumped--as you go back towards the Big Bang. The decrease in entropy due to that extreme uniformity more than compensates for the increase in entropy due to the higher density and temperature.

If the universe were to collapse into a Big Crunch, it would not be the same; the universe would still be getting more clumped gravitationally as it collapsed.

thanks for your reply :)

so, you're discussing the collapses of stars??

dS = dQ/T

as stars gravitationally collapse, their interiors heat up to millions & billions of degrees (T), even as they lose heat?

dQ < 0
dS* = dQ/T < 0

whereas the surrounding environment is much cooler (t), so when it absorbs all that heat, its entropy increase is greater?

dS = dQ/t >> 0

dS_net = dS + dS* = dQ(1/t -1/T) > 0

 

[phinds]

so, you're discussing the collapses of stars??

No, he is discussing the collapse of the universe. That is, galaxies get closer together then merge. This is not a gravitational collapse of individual stars, which is a different thing.

 

[Grinkle]

I am not seeing how dS = dQ/T can describe a system where entropy is strongly influenced by gravity.

If I have my B level concepts straight, considering a system containing a neutron star and nothing else, with a relatively large boundary (say one light year) that system has a very high entropy because it is in a very probable state given that gravity is clumping all the particles together as one would expect. If there were no gravity in this one-light year diameter system, then the entropy would be, I think, very low, because only considering thermal interactions and no gravity it is very unlikely (probably no mechanism to even have such a configuration) that all the particles would end up grouped together as tightly as the particles of a neutron star.

So trying to make sense of dS = dQ/T in a system where entropy is very influenced by gravity may not work - it seems like some needed modelling is missing. I tried a couple Google searches but can't find anything that looks like what I am picturing.

 

[PeterDonis]

so, you're discussing the collapses of stars??

I'm talking about the effects of gravitational clumping on the scale of the universe as a whole, since we are discussing the universe as a whole. That includes collapses of galactic clusters, galaxies, and stars (and of course white dwarfs, neutron stars, and black holes).

dS = dQ/T

Not in the presence of gravity. In the presence of gravity dQ/T is not the only contribution to the entropy. We don't know exactly how to derive the gravitational contribution from microstates of the system because we don't have a good theory of quantum gravity; but there are many cases in which there are other terms in the entropy besides dQ/T, so we know the general pattern, which is that dS = dQ/T will not be valid whenever there are other significant changes that can happen to the system besides heat transfer.

 

[TEFLing]

I am not seeing how dS = dQ/T can describe a system where entropy is strongly influenced by gravity.

If I have my B level concepts straight, considering a system containing a neutron star and nothing else, with a relatively large boundary (say one light year) that system has a very high entropy because it is in a very probable state given that gravity is clumping all the particles together as one would expect. If there were no gravity in this one-light year diameter system, then the entropy would be, I think, very low, because only considering thermal interactions and no gravity it is very unlikely (probably no mechanism to even have such a configuration) that all the particles would end up grouped together as tightly as the particles of a neutron star.

So trying to make sense of dS = dQ/T in a system where entropy is very influenced by gravity may not work - it seems like some needed modelling is missing. I tried a couple Google searches but can't find anything that looks like what I am picturing.

well, if gravity is an attractive interaction, how would one model the entropy of (say) a gas of pure H which would naturally tend to "clump" and bond into H₂ ? How could you compare the entropy of pure H to the favored pure H₂ state ?

 

[Grinkle]

How could you compare the entropy of pure H to the favored pure H2 state ?

This is chemistry and the change in entropy of the chemical reaction is modeled by the heat tranfer of the reaction. To make it about gravity, ask how one models a cloud of $H_2$ with and without taking gravity into consideration. @PeterDonis points out in post 25 there isn't an established way to model it.

 

[PeterDonis]

how would one model the entropy of (say) a gas of pure H which would naturally tend to "clump" and bond into H2 ?

The same way you model the entropy of any system that contains multiple particle species that can change into each other: using a chemical potential. But there is no useful analogy here with systems in which gravity is present.