# Is the universe an adiabatic system?

1. Dec 31, 2014

### Daemon

Hello everyone, I wanted to ask you if the universe is an adiabatic system where the energy it cant be transmited outside itself. If is an adiabatic system, what happend with the information and energy inside a blackhole? How the universe started with a 0 entropy state? Could be the hyperinflation produceed by the ''dark energy''a sympton of an isothermic process in the other way? I would like to hear your reflexions about it.

Thanks a lot

2. Dec 31, 2014

### PWiz

Firstly, we don't have a proper definition for "edge of the universe", and we can only define the boundaries based on what we can currently confirm, hence we define the basic physical dimensions of the "observable universe". How can you think of something "outside" the universe when you can't know where its limits end? If something exists out of what we consider the universe, then since it exists and everything that exists is part of the universe, it is included as well. There are many "theories" over here but no definite answer for this question, so defining the universe as a "closed" system is subject to speculation.
Secondly, a black hole has something that we call the "event horizon". If you consider spacetime to be a grid sheet and any mass in the universe as a ball placed on it, you will see that they that it will cause a depression depending on how heavy it is, and other smaller balls naturally roll towards the depression. Hence, all objects with invariant mass cause "depressions" in spacetime. But a black hole is a depression so deep that it's basically a bottomless well on the grid - the event horizon is the point beyond which you can't climb out of the well. No matter which direction you try to take after crossing the event horizon, you won't be able to get away from the black hole: all directions will lead back to it. Hence, light or any other form of radiation carrying information is forever trapped in the black hole. To an outside observer, it vanishes from spacetime after crossing the event horizon. Consider it entering a locker from which it never comes out (talk about one-way tickets).
Finally, the universe was in the state of singularity before the Big Bang, and hence its entropy was 0. As it exploded into so many more physically present entities, there was the "spread" of energy and a gradual distribution, more or less random, among the ejected materials, so as the individual state of each entity began to vary, the entropy value increased. A room has to have some stuff lying about so that you can call it "ordered" because it's neatly stacked or "messy" because it's all spread around, but you can't say anything about it when there's nothing in the room(which was the case before the Big Bang, so no disorderliness = 0 entropy). And yes, dark energy is supposedly theorized to be causing hyperinflation, since nothing else can be held responsible with any credible evidence for causing massive galaxies with God knows what mass to move away at greater than light speed. (Try considering the relativistic mass at that separation velocity)

3. Dec 31, 2014

### Staff: Mentor

Do you mean the entire universe or just the observable universe? The latter is finite in extent and there is more universe outside it (and its boundary changes with time). The former has no "outside", by definition.

The energy is still there, visible as the hole's mass. What happens to information that falls in is an unresolved question at this point.

4. Dec 31, 2014

### Staff: Mentor

Yes, we do: it's that there is no "edge of the universe". The universe has no spatial boundary; either it's finite but unbounded (unlikely based on our current observations but still possible), or it's infinite, hence unbounded.

This model has a lot of limitations, and in particular it doesn't generalize well to black holes. I would not advise using it.

No, it doesn't; it just enters a region of spacetime that the outside observer can't see. Spacetime itself is not observer-dependent, but what region of spacetime is visible can be observer-dependent.

No, it wasn't. In the original Big Bang model, there is no such thing as "before" the Big Bang; there is an initial singularity, but it is not, strictly speaking, part of spacetime. However, that is a highly idealized model, and AFAIK no cosmologists think it's realistic.

Our best current model only goes back to the inflationary era; the question of what started inflation, or what, if anything, existed before it, is left open, until we have enough of a theory (presumably a theory of quantum gravity) to be able to model states of the universe with very large spacetime curvature and so have something non-singular with which to replace the unphysical singularity in the original, classical GR model.

5. Dec 31, 2014

### Staff: Mentor

We don't know. There are a number of speculations in this area, but AFAIK nobody really has a good explanation at this point. In our current models we just take this as given since it's consistent with observations and it's the best we can do.

6. Dec 31, 2014

### Staff: Mentor

The galaxies in question are not actually moving faster than light, so this calculation is meaningless. The "relative velocity" in question is a coordinate velocity and does not have a direct physical meaning; and the coordinates in question are certainly not special relativistic coordinates, so a coordinate velocity in these coordinates can't be plugged in to a special relativistic "relativistic mass" formula anyway.

7. Dec 31, 2014

### Chronos

Assuming the universe includes all of reality implies there is nothing else with which it can 'share' anything. The question becomes irrelevant in this context.

8. Dec 31, 2014

### PWiz

@PeterDonis You've quoted quite a lot of my text, so I'll just go in a chronological order of numbering
1)If you read the last line of my first para, it says the same thing: the universe is not closed. When I say that the physical boundaries of the universe are not defined, does it not mean that the universe has no boundaries? (eg: $\frac {1}{0}$ is not defined, so it does not have any value)
2) Agreed, but I don't think it hurts to use it for basic understanding.
3) Pardon my laziness. Rephrasing it, from the frame of reference of an outside observer, any form of radiation crossing the event horizon of the black hole enters a region of spacetime where the presence of the radiation cannot be confirmed by the observer, since all suggestive information never reaches the observer.
4) I'm no qualified cosmologist (I'm still in high school, but I make sincere attempts to find credible information before making ignorant proclamations), but wasn't the Big Bang a supposed event that is said to be the origin of the current universe(I don't think implausibility makes a theory void when there is no proven and accepted model in it's place)? And if it is an event, then surely there was a "time" before it consisting of "initial singularity", since you yourself mention that spacetime is independent from it?
5)I'm not 100% sure about this so I'd like your input - surely there isn't any stationary frame of reference in the universe? If hypothetically speaking there was one, then aren't all velocities in the universe "relative" to it(upon which we carry out all our calculations)? Velocities of which frame of reference are used in Einstein's relativistic mass formula?
P.S. Thanks for making me reconsider what I earlier said, it always helps to learn along the way.:)

9. Dec 31, 2014

### Staff: Mentor

No. "No boundaries" is a defined condition; it's not the same as an operation like dividing by zero. The surface of a sphere is a finite manifold with no boundary, but there's nothing "undefined" anywhere.

No, although this misconception appears in many pop science discussions of the topic. In the actual theory, the term "Big Bang" simply refers to the fact that the early universe was in a very hot, dense state and was expanding very fast. It makes no claim beyond that.

What does "stationary frame of reference" mean? You may be getting at a true statement here, but I'm not sure. See further comments below.

Any inertial frame you like. But in curved spacetime (such as the universe), there are no global inertial frames; inertial frames are only local, in both space and time. So any "velocity" you would plug into this formula has to be measured locally. There's no way to obtain any "velocity" for a galaxy a billion light years away that will have any meaning in the relativistic mass formula, because there is no inertial frame that covers both that galaxy and the Earth.

10. Dec 31, 2014

### PWiz

@PeterDonis I see, so relativistic mass is only a locally observable property. That definitely explains a lot, I'll really think over it. And by "stationary frame of reference", I mean a frame which is "truly at rest" , and from which all relative velocities are the "true" velocities, a frame which is the "zero frame".(Globally considering spacetime). I don't know if I'm making myself clear here, but since every quantity is measured from a "zero" reference (eg: electric and gravitational fields have zero potentials at a hypothetical point that is infinitely far away) , it seems only natural that frames of reference for physical observations should also have a zero point(hypothetical or real). I think I'll make new thread about this for more in-depth discussion so that I don't go off topic here.

11. Dec 31, 2014

### PWiz

12. Jan 1, 2015

### Chronos

You missed the point. Your question is philosophical, not scientific.

13. Jan 1, 2015

### Torbjorn_L

The universe has a very low entropy density and a 0 energy density (according to some, if not all, cosmologists).

When you start to think about the universe as a system, you get problems with thermodynamics as it becomes ill defined. I don't think there is any consensus on any of that. And then thermodynamics is ill defined, you know you are knee deep in bantha poodoo.

Possibly you can extend thermodynamic energy to describe the entire universe, because there are indicators (using the same GR Lagrangian that are used for quantizing it to get gravitons, another non-classical extension). They tend to agree with an integration of the energy density, i.e. for all practical purposes the universe has no energy to "transmit", nor would we expect any other sinks/sources that would do so. (Say, if a universe tunnels out of another, as some speculate. It would be constrained by the uncertainty principle, so nothing appreciable is "transmitted". Et cetera.)

So the idea is not only meaningless as is, if it would work it would still be meaningless, and it has a hard time working since the context is ill defined. How much "ill defined" do we have to iterate before we give up? [rhetorical question ]

In the same way, you need to extend thermodynamic entropy to include BH horizons to get a "generalized second law of thermodynamics" to even try to estimate the total entropy of the observable universe.

Here is the best effort I [a layman] know of, by Egan & Lineweaver: http://arxiv.org/pdf/0909.3983v3.pdf . Even they end up with two different schemes consistent with such a generalization, one without a holographic boundary (akin to a BH horizon) and one with. However, both versions give about the same result. The universe behaves pretty much as if general relativity rules, i.e. classically.

As a context, here is notes from a seminar they held on the subject around the time they published their papers. [ http://www.univers2009.obspm.fr/fic...e/R7_Inflation_and_Dark_Energy/Lineweaver.pdf ] If you go to slide 8 you can see that the typical models used to describe total entropy of the universe (including their own earlier one) are called "Semi-popular treatments".

I think the moral is that we can extend physics from where we are, and that it happens to be easy to derive a dynamic (inflation) cosmology. But same as thermodynamics gets little informative on entire system behavior outside of equilibrium it is little informative on entire universe behavior. I can speculate that these things are somewhat correlated, but I don't really know. (Especially after decoupling, when the universe is explicitly in a non-equilibrium thermodynamic state due to an inner phase change.*)

*Or at least, that is how I would try to describe it, not to be confused with the earlier phase changes as standard particles freeze out et cetera. Those can happen, same as a liquid/crystal phase change can be without much latent heat, and again it is more the dynamic than the thermodynamic that rules.

Last edited: Jan 1, 2015
14. Jan 1, 2015

### Staff: Mentor

Reference, please? Whether the entropy density is "low" depends on what you define as "low" (e.g., are we comparing to a far future "heat death" state or to the state of the early universe?). However, the claim that there is 0 energy density seems obviously false, not only because the universe obviously contains galaxies, but because current cosmological models have nonzero density of all main types of stress-energy (ordinary matter, dark matter, and dark energy).

15. Jan 4, 2015

### Torbjorn_L

[Seems I don't get PF's alert system. I saw a quote the same day, but there wasn't any response when I looked. Now there is a response...]

My first link was intended as an entropy reference. As can be seen from their figs 6 & 7, whether one include the cosmological event horizon the total entropy is effectively constant and so the density goes down with the expansion.

But I was thinking of comparing it to a black hole as in the OP, which I think has maximum entropy density.

The reference I was thinking of when I wrote "there are indicators (using the same GR Lagrangian that are used for quantizing it to get gravitons, another non-classical extension)" is Faraoni's and Cooperstocks's "ON THE TOTAL ENERGY OF OPEN FRIEDMANN-ROBERTSON-WALKER UNIVERSES", The Astrophysical Journal, 587:483–486, 2003 April 20 ; http://iopscience.iop.org/0004-637X/587/2/483/pdf/56020.web.pdf ]

"The idea that the universe has zero total energy when one includes the contribution from the gravitational field is reconsidered. A Hamiltonian is proposed as an energy for the exact equations of FRW cosmology: it is then shown that this energy is constant. Thus, open and critically open FRW universes have the energy of their asymptotic state at infinite dilution, which is Minkowski space with zero energy. It is then shown that de Sitter space, the inflationary attractor, also has zero energy, and the argument is generalized to Bianchi models converging to this attractor."

I think that is effectively the gravitational Lagrangian (applied to a FRW universe) that is used to get gravitons at low energies and large scales, but of course I haven't checked. [ https://golem.ph.utexas.edu/~distler/blog/archives/000639.html ]

That energy is balanced and conserved in cosmology may not be such an astonishing idea. I've seen this attempt (which I lack the mathematical chops to verify - no tensors, only the most basic differential geometry), but I don't know if it has been published outside of arxiv:

"These are questions that surfaced relatively recently. As I mentioned in my history post, the original dispute over energy conservation in general relativity began between Klein, Hilbert and Einstein in about 1916. It was finally settled by about 1957 after the work of Landau, Lifshitz, Bondi, Wheeler and others who sided with Einstein. After that it was mostly discussed only among science historians and philosophers. However, the discovery of cosmic microwave background and then dark energy have brought the discussion back, with some physicists once again doubting that the law of energy conservation can be correct.

Energy in the real universe has contributions from all physical fields and radiation including gravity and dark energy. It is constantly changing from one form to another, it also flows from one place to another. It can travel in the form of radiation such as light or gravitational waves. Even the energy loss of binary pulsars in the form of gravitational waves has been observed indirectly and it agrees with experiment. None of these processes is trivial and energy is conserved in all cases. But what about energy on a truely [sic!] universal scale, how does that work?

On scales larger than the biggest galactic clusters, the universe has been observed to be very close to homogeneous and isotropic. ...

In a previous post I gave the equation for the Noether current in terms of the fields and an auxiliary vector field that specifies the time translation diffeomorphisms. The Noether current has a term called the Komar superpotential but for the standard cosmology this is zero. The remaining terms in the zero component of the current density come from the matter fields and the spacetime curvature and are given by

J^0 = \rho + \frac{\gamma}{a} + \frac{\Lambda c^2}{\kappa} - 3 \frac{\dot{a}^2}{\kappa a^2}

The first term is the mass-energy from cold matter, (including dark matter) at density \rho . The second term is the energy density from radiation. The third term is dark matter energy density and the last term is the energy in the gravitational field. Notice that the gravitational energy is negative. By the field equations we know that the value of the energy will be zero. This equation is in fact one of the Freidmann equations that is used in standard cosmology.

If you prefer to think of total energy in an expanding region of spacetime rather than energy density, you should multiply each term of the equation by a volume factor a^3.

It should now be clear how energy manages to be conserved in cosmology on large scales even with a cosmological constant. The dark energy in an expanding region increases with the volume of the region that contains it, but at the same time the expansion of space accelerates exponentially so that the negative contribution from the gravitational field also increases in magnitude rapidly. The total value of energy in an expanding region remains zero, and therefore constant. This is not a trivial result because it is equivalent to the Friedmann equation that captures the dynamics of the expanding universe.

So there you have it; the cosmological energy conservation equation that everybody has been asking about is just this

E = M c^2 + \frac{\Gamma}{a} + \frac{\Lambda c^2}{\kappa} a^3 – \frac{3}{\kappa}\dot{a}^2 a = 0

It is not very complicated or mysterious, and it’s not trivial because it describes gravtational dynamics on the scale of the observable universe."

[ http://blog.vixra.org/2010/08/17/energy-is-conserved-in-cosmology/ and its references. Among other things the arxiv paper claims that there is a slicing that covers black holes, so that aspect of GR is good too, assuming it is correctly done.]

Last edited: Jan 4, 2015
16. Jan 4, 2015

### Staff: Mentor

It's not, but you didn't say zero energy, you said zero energy density, which is not the same thing. The "zero energy" idea is basically finding an expression for "gravitational potential energy" that is negative and exactly balances total energy from matter, radiation, etc. But there is no expression for "gravitational potential energy density"; the whole concept of gravitational potential energy only works globally, not locally. Locally, the energy density of the universe is positive.

Sure, the Hamiltonian approach that underlies the "zero energy" idea works for any spacetime that is a solution of the Einstein Field Equation. But it's still zero energy, not zero energy density. It's an alternative way of looking at the global dynamics of the universe; it doesn't change the local fact that there is nonzero energy density present.

17. Jan 8, 2015

### Torbjorn_L

The universe is very flat, so the energy density is zero when seen over enough volume and locally approximatively zero in the LCDM model, isn't it?

Interesting question if it wasn't, is the clumpiness of matter under gravitation the issue and so the LCDM somehow smooth that out?

18. Jan 8, 2015

### George Jones

Staff Emeritus
No. The energy density is small, but certainly non-zero. I think that you are confusing "the universe is flat" with "spacetime is flat". "The universe is flat" means that 3-dimensional spatial sections of constant (cosmic) time are flat, it does not mean that spacetime is flat. In a flat universe, the spatial curvature tensor specific to a 3-dimensional spatial section is zero, but the spacetime curvature tensor is non-zero.