Zero energy Universe

1. Apr 25, 2012

johne1618

I have heard people talk about the principle that the total energy of the Universe is zero.

Here is my understanding of the principle.

The positive rest mass energy of each particle is balanced by the negative gravitational potential energy between it and every other particle in the Universe i.e.

$\large m c^2 \propto \frac{G M m}{R}$

where M and R are the mass and radius of our Universe.

I think this relationship provides a better definition of our Universe as a "connected" whole at a particular cosmological time rather than the idea of the observable universe which depends on light rays travelling through time from the Big Bang to us.

Last edited: Apr 25, 2012
2. Apr 26, 2012

johne1618

Here is a another way of looking at it.

The evidence points to the fact that the Universe, for most of its history, has been spatially flat. If we also assume a negligible cosmological constant, the Freidmann equation implies that the density of the Universe, $\rho$, is given by

$\large \rho(t) = \frac{3 H(t)^2}{8 \pi G}$

where $H(t)$ is the Hubble parameter.

Let us define the Hubble radius, $R(t)$, by

$\large R(t) = \frac{c}{H(t)}$

Thus we have

$\large \rho = \frac{3 c^2}{8 \pi G R^2} \ \ \ \ \ \ \ \ (1)$

Now let us imagine a sphere with Hubble radius R centred on our position. The mass of matter in that sphere is given by

$\large M = \frac{4}{3} \pi R^3 \rho$

Rearranging we get

$\large \rho = \frac{3 M}{4 \pi R^3} \ \ \ \ \ \ \ \ \ (2)$

Combining expressions (1) and (2) to get rid of $\rho$ we find

$\large \frac{c^2}{2} = \frac{G M}{R}$

Multiplying both sides by particle mass m we get

$\large \frac{m c^2}{2} = \frac{G M m}{R}$

Thus we find that half the energy of any particle is balanced by the gravitational energy between it and the rest of the Hubble sphere. I would identify the Hubble sphere with our Universe.

Last edited: Apr 26, 2012
3. Apr 26, 2012

RUTA

Why not use the distance R to the particle horizon instead? In that case HR = 2c (using the flat, matter-dominated Einstein-deSitter model as you have).

4. Apr 27, 2012

johne1618

I am considering the size of the Universe at an instant of cosmological time.

The distance to the particle horizon is defined by an integral over cosmological time since the Big Bang.

I don't think I am assuming a matter-dominated Universe (even though I use the term matter I think my argument could include radiation as well). For example I'm not making the assumption that the total number of particles or photons remains constant.

I'm making the following assumptions:

1) The Universe is spatially flat

2) The cosmological constant is negligible

3) Here's the big one. I'm assuming that the Hubble radius is co-moving and thus there is a boundary at the Hubble radius that acts as an edge of the Universe separating what can interact with us in the future from what can't. (The particle horizon is the maximum distance to what has interacted with us in the past). This assumption has the strong implication that the Universe is expanding linearly with time.

Last edited: Apr 27, 2012
5. Apr 27, 2012

Ich

The Hubble radius is not a horizon. In a linearly expanding universe, you can interact with everything in the future.

6. Apr 27, 2012

RUTA

Most people use the particle horizon for that.

7. Apr 27, 2012

Chronos

I agree with Ich in that the Hubble volume is merely a virtual boundary that has no physical meaning.

8. Apr 28, 2012

lpetrich

General relativity has the problem that one cannot define a local gravitational energy. A global one, maybe, but not a local one.

One can find the gravitational energy of a massive object in space-time that is asymptotically flat at large distances from it. One does so by finding its total mass-energy from its gravity, then subtracting out the energy of its material and its kinetic energy.

But one cannot do that in a Friedmann-Lemaitre-Robertson-Walker cosmological model, because it has no large-distance asymptotic flatness. There's been a lot of argument about FLRW total energy, but it's usually considered to be 0.

9. Apr 28, 2012

johne1618

I accept that if the Hubble radius is expanding then eventually every point in space will be in causal contact with us.

But my point is that objects that are presently outside the Hubble radius in a Universe that is expanding linearly (or faster) will never interact with us in the future.

So in that sense there would be a kind of horizon at the Hubble radius.

Last edited: Apr 28, 2012
10. Apr 28, 2012

Ich

In a linearly expanding universe, the Hubble radius is expanding. Linearly, btw. This means that eventually everything will interact with us in the future. We had this issue already settled in the other thread.

11. Apr 28, 2012

Calimero

John, these two sentences are in complete contradiction, and neither of them is true. We live in universe where Hubble radius is increasing, and we think that there exists cosmological event horizon due to the accelerated expansion. In linearly expanding universe, on the other hand, all objects with finite distance will make contact in finite time - there is no event horizon.

You are persistently confusing Hubble radius or sphere with event horizon. Hubble radius marks the distance from where recession velocity, in terms of proper distance per proper cosmological time, exceeds the speed of light. In order for us to receive photons from faster then light receding objects, those photons need to cross from superluminal to subluminal expansion zone, that is they need to cross inside the Hubble sphere. There is only one way in which they can do it - if Hubble sphere is expanding, and thus is able to engulf them.
When expansion of universe is accelerating, expansion of Hubble sphere is decelerating.
At one point in future universe will enter de Sitter expansion, matter density will be negligible, and expansion will be governed with cosmological constant only. Velocities between galaxies will increase linearly, and distances will grow exponentially. That is the point which Ich mentioned to you in the other thread, when H=const, and thus radius of Hubble sphere is also constant. At that point Hubble sphere and cosmological event horizon coincide.

Cosmological event horizon is defined as distance from where photons emitted at tnow will not reach us until t=inf, or as our past light cone at the end of time. It does not have finite value in non-accelerating universe or in universe with non-decelerating Hubble spheres, such as your example. You acknowledged that in other thread, but you keep referring to Hubble sphere as event horizon:

12. Apr 29, 2012

johne1618

Hi,

Sorry everybody!

I now understand that one can only have an event horizon in an accelerating Universe.

John

13. Sep 28, 2012

Ulrich

Something that I cannot understand in this "zero energy universe" is that only gravitational and mass energy is taken into account. What about dark energy that represents over 70% of cosmological density? And isn't there also kinetic energy of all the mass moving? And not to forget the background radiation energy. Even if gravitational and mass energy cancel each other, where do the other form of energies are supposed to come from by the holders of this theory?

14. Sep 28, 2012

Mark M

The cosmological constant doesn't make a difference. For a perfect fluid energy tensor, and a perfect gas equation of state, cosmic pressure and energy density are described by this energy conservation law $$\dot \rho = -3H(\rho + p)$$ For a cosmological constant, $p = - \rho$. So, the contribution of the cosmological constant to the energy density is zero. One way of thinking about this is that negative pressure contributed by the CC is equal to its energy density, so the net contribution is zero.

See: http://arxiv.org/pdf/gr-qc/0605063.pdf

15. Sep 28, 2012

marcus

Hi Ulrich, the cosmological constant is CALLED "dark energy" but no corresponding energy has been measured. It may simply be a small curvature constant, a vacuum curvature rather than a "vacuum energy".

Some people like to think of it as arising from a "dark energy" and this is possible but there is no scientific reason to suppose that. So other people just consider it to be Lambda (a curvature constant in the Einstein GR equation), as in LambdaCDM model, the standard cosmology model.

Either way it would be negligible in the early universe. If it were the result of a tiny constant energy density then in the early universe it would be drowned out by the much higher density of other forms of matter and energy. That is the point. The energy density corresponding to Lambda is only about 0.6 joules per cubic kilometer! Or 0.6 nanojoules per cubic meter.

Such a small constant energy density only matters NOW when the other stuff has thinned out. Back then it would be too small to be noticed, essentially no effect.

What you ask about the other stuff, kinetic, dark matter etc. That is all taken account of when they conjecture about "zero energy". I guess it's good to remember that that is just a conjecture--there is no global energy conservation law that would require it.

16. Sep 29, 2012

Ulrich

Can you explain this further? Here is my understanding: The equation you give is the first law of thermodynamics for cosmology for a general density. It is supposed that the vacuum energy density is independent of time so $\dot \rho = 0$ and correspondingly $p = - \rho$. If now I want to know the total vacuum energy in a volume V, I integrate its density over V. In an infinite universe this would yield an infinite energy. But since according to thermodynamics the pressure $p$ is doing the negative work $dW = - p dV$ in an universe with $dQ=0$, the total vacuum energy cancels this work because of $p = - \rho$. Is this right?

17. Sep 29, 2012

Ulrich

It may be small but small does not signify zero. What the zero-energy-universe-cosmologists are trying to establish is a theory that explains how the universe came out of nothing. However, nothing really signifies zero in the strict mathematical sense. Something around 10^-29 grams per cubic centimeter supposed for dark energy according to http://en.wikipedia.org/wiki/Dark_energy cannot be considered zero, unless it does not make any contribution to the total energy as suggested by Mark.

But in any case, it is supposed that a quantum fluctuation is needed to ignite the expansion of the universe. However, this implies that spacetime already exists before the big bang, which is in contradiction with time that began only with the big bang. So there cannot be any spacetime, which is a condicio sine qua non for quantum fluctuations, before the big bang.

Is this the reason why you call it a conjecture?

Last edited: Sep 29, 2012
18. Sep 29, 2012

twofish-quant

However, the cosmological constant itself has energy which is constant. That's why people call it "dark energy".

I *think* that the zero energy people would argue that the cosmological constant "really isn't energy". This is mathematically justifiable, but it makes the whole idea a lot less convincing (i.e. the universe becomes zero energy because stuff that makes it non-zero isn't energy.)

19. Sep 29, 2012

twofish-quant

One problem here is that "something out of nothing" is a term that makes it sound nice in pop-science books. If you argue that the universe came out of "nothing" that doesn't end the question, because you then have to define the properties of "nothing" and define exactly what you mean by "nothing."

This gets at the essential problem which is that our current models for how the universe works just don't work at event zero.

In the GR view of the world, space time isn't a "thing". Just as an analogy, I take a graph paper and draw a circle on it. You have graph paper lines that define the shape of the circle, but that's graph paper lines are "things." In GR, people think of space time as the same way. It's not a "thing" so when people talk about the cosmological constant, they don't think of it as "a thing", so the energy of the cosmological constant is like lines on a sheet of graph paper. They are just measurement standards and aren't "things."

By contrast, in quantum field theory, everything is a field, and fields are *things*. In the QFT world, you describe things as interactions between fields and fields are things like electrons and photons. One point about QFT, it that there is really no such thing as "nothing." Space always consists of interacting fields, and the vacuum itself is a "thing". For example you start out in a vacuum state. You dig a hole in the vacuum state and you now have stuff and a hole. This corresponds to matter and anti-matter.

These two views of what "things" are fundamentally incompatible, and because they are, it's tough to come up with a theory of everything. You can jokingly but perhaps accurately say that we'll understand everything when we understand nothing.

It's a conjecture because people are thinking out loud, and it's one of probably two dozen ideas that happened at event zero. If I had an hour on the Discovery Channel, what I'd do is to try to give a general survey of the various ideas that have been suggested rather than promote one particular one.

20. Sep 29, 2012

marcus

You are inquiring into the area of quantum cosmology. Not all quantum cosmologists assume what you say (about a quantum fluctuation being needed). If you would like to learn about the directions in current QC research you could have a look at the listings for this conference:
http://www.icra.it/mg/mg13/

It is triennial-- held every 3 years --and this year it was held in Stockholm. Over 1000 physicists participated (several fields dividing into parallel sessions.)

Here are the parallel sessions:
http://www.icra.it/mg/mg13/parallel_sessions.htm
You will see a bunch of "CM" sessions. CM3 is especially about Nonsingular Cosmic Models.
You will see a bunch of "QG" sessions. Some of them are about cosmology with a quantum bounce. These models are also Nonsingular.

Among those people at those sessions concerned with the start of the expansion process I think the majority would not suppose it began from a "quantum fluctuation" but they would be working on models of what conditions led up to it. What kind of time evolution. The "singularity" is widely considered to be unphysical---a failure of the theoretical model. So they work on fixing that---improving the model so that it does not suffer a breakdown.
It is classical (pre-quantum) GR which develops the singularity and it is expected by many researchers that a quantum version of GR will cure the trouble and lead to a NONSINGULAR cosmology.

Maybe some other paths will lead to a nonsingular cosmology, not requiring a quantization of geometry. It is work in progress.

Many of the models being studied have time-evolution extending back before the start of expansion---often with a contraction (which may for example rebound at some extremely high critical density). Quantum effects cause a bounce.

You don't see much about this in popular media but the actual research that is going on, being published and presented at the major conferences tends to involve ways to resolve the classical singularity and extend time-evolution back further. And to search for ways to TEST the predictions of the various nonsingular models.