Understanding the Zero Energy Principle of the Universe

[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 traveling through time from the Big Bang to us.

 

[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.

 

[RUTA]

I would identify the Hubble sphere with our Universe.

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).

 

[johne1618]

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).

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.

 

[Ich]

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

 

[RUTA]

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

Most people use the particle horizon for that.

 

[Chronos]

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

 

[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.

 

[johne1618]

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

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.

 

[Ich]

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.

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.

 

[Calimero]

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.

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 t_now 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:

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.

 

[johne1618]

Hi,

Sorry everybody!

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

Thanks for your posts.

John

 

[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?

 

[Mark M]

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?

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

 

[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.

 

[Ulrich]

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.

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?

 

[Ulrich]

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.

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?

 

[twofish-quant]

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.

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.)

 

[twofish-quant]

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.

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."

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.

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.

Is this the reason why you call it a conjecture?

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.

 

[marcus]

... 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...

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.

 

[marcus]

I just did a search of the DESY database for Quantum Cosmology research papers appearing 2009 or later, ranked by number of citations. DESY is a German research institution with an excellent library. The search pulled up 372 papers. I looked over the list of the top 50. The fifty that were most cited in the literature and therefore most representative of research currently in progress. The vast majority involved a quantum bounce from a prior contracting universe. At least 40 were obviously of that type. There were 10 or so that I did not immediately recognize. They could also have been working with bounce-type cosmologies---I did not bother to examine. There were none in the top 50 papers that I could recognize as being of the "quantum fluctuation" sort.

The latter is less fashionable in actual quantum cosmology research now, it seems, than it was say 10 years ago.

 

[Ulrich]

Thanks Marcus for your search. However, I think that we cannot really crasp what happened at the big bang so research on that subject will always stay speculative in my humble opinion.

What is more interesting me at the moment are other questions like how pressures of radiation and vacuum can be handled through thermodynamics, which originally is based on gases? Is this just a heuristic assumption or is there more behind it, that is, has radiation and vacuum pressure some similarity with gas pressure?

Another question is how infinite curvature at the big bang must be considered. According to Hartle, Gravity, 2003, p. 485 this is true for all three standard models (closed, flat and open) because the scale factor a(t) is zero at t=0 and in the denominator of the Einstein cuvature tensor. However, curvature of a flat universe is zero, isn't it? So while infinite curvature at the big bang is evident for closed and open universes why should it be the case for a flat universe?

I have achieved my studies in physics since years and unfortunately did not learn much about cosmology. Now, I can't go to professors and other people anymore to ask my questions. So I hope you guys can answer some of them on here.

 

[Chronos]

There is no infinite curvature if the big bang itself was infinite.

 

[Ulrich]

There is no infinite curvature if the big bang itself was infinite.

Well there is this term $\frac{k c^2+\dot{a}^2}{a^2}$ in the Einstein curvature tensor, where k is -1,0 or 1 for hyperbolic, flat or spherical universe. In all cases this term becomes infinite if $a$ goes to zero, unless $\dot{a}$ also goes to zero, which would yield un undetermined limit. Hartle writes: "It is clear from any of... [these equations]... that the big bang of the FRW model at $a=0$ is a singularity not only in pressure and density, but in the curvature of spacetime as well." Hartle is a known cosmologist (cf. http://en.wikipedia.org/wiki/James_Hartle).

 

[Chronos]

Well there is this term $\frac{k+\dot{a}^2}{a^2}$ in the Einstein curvature tensor, where k is -1,0 or 1 for hyperbolic, flat or spherical universe. In all cases this term becomes infinite if $a$ goes to zero, unless $\dot{a}$ is also infinite at the beginning, which is not realistic. Hartle writes: "It is clear from any of... [these equations]... that the big bang of the FRW model at $a=0$ is a singularity not only in pressure and density, but in the curvature of spacetime as well." Hartle is a known cosmologist (cf. http://en.wikipedia.org/wiki/James_Hartle).

Which leaves open a solution where the BB was infinite.

 

[Ulrich]

Sorry Chronos, I edited my post too late, I meant of course if $\dot{a}$ goes to 0 not to ∞

 

[Mark M]

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 $p = - \rho$.

The fact that $p = - \rho$ is based off of the fact that a cosmological constant has an equation of state of $w = -1$. This then gives you the result that the energy density is independent of time.

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?

In an infinite universe, there is, as you say, no meaningful way to integrate the energy density over all of space. That's why an ideal zero energy universe is closed, as a closed universe is necessarily finite.

In a closed universe, as you explain, the negative pressure of the CC cancels out any contribution to the energy density by the CC.

 

[Ulrich]

The fact that $p = - \rho$ is based off of the fact that a cosmological constant has an equation of state of $w = -1$. This then gives you the result that the energy density is independent of time.

A disagree here because if you say that a cosmological constant has an equation of state of $w = -1$ you still have to explain why this is the case and what is the physical reasoning behind it. The general form of the equation of state is $w=\frac{p}{ρ}$ where p is a pressure and ρ the corresponding density. If we define $p=-ρ$ we in fact obtain $w=-1$. But so far we have just made some mathematical transformation. The physical reasoning on the other hand is that vacuum energy (or the cosmological constant) must not depend upon the expansion of the universe like the other densities, which become thinner and thinner with increasing volume. Because empty space is always the same, whether in a early or late universe. This is physics transformed into mathematics with $\dot{ρ}=0$ leading to $p = - \rho$ through the first law of thermodynamics.

That's why an ideal zero energy universe is closed, as a closed universe is necessarily finite.

I don't think that this is correct neither. A zero energy universe only holds for a flat one with zero curvature when k is taken to be zero (see here), which leads then to what johne1618 posted in the beginning of this thread.

 

[Mark M]

A disagree here because if you say that a cosmological constant has an equation of state of $w = -1$ you still have to explain why this is the case and what is the physical reasoning behind it. The general form of the equation of state is $w=\frac{p}{ρ}$ where p is a pressure and ρ the corresponding density. If we define $p=-ρ$ we in fact obtain $w=-1$. But so far we have just made some mathematical transformation. The physical reasoning on the other hand is that vacuum energy (or the cosmological constant) must not depend upon the expansion of the universe like the other densities, which become thinner and thinner with increasing volume. Because empty space is always the same, whether in a early or late universe. This is physics transformed into mathematics with $\dot{ρ}=0$ leading to $p = - \rho$ through the first law of thermodynamics.

The reason that $w = -1$ for a cosmological constant doesn't have anything to do with the above reasons.

Since it maintains the same value over time, it must not have any dissipative properties such as heat conduction, viscosity, etc. So, it must be a perfect fluid. Therefore $$T_{\mu \nu } = ( \rho + p)U_{\mu} U_{\nu} + pg_{\mu \nu }$$ The first term must be zero in order to ensure that the cosmological constant has no preferred direction, maintaining Lorentz invariance. So, $\rho + p = 0$ and therefore $p = -\rho$. Finally, this means that $w= -1$

I don't think that this is correct neither. A zero energy universe only holds for a flat one with zero curvature when k is taken to be zero (see here), which leads then to what johne1618 posted in the beginning of this thread.

This is false. See the FAQ, here.

 

[audioloop]

I have achieved my studies in physics since and unfortunately not learn much about cosmology. Now, I can't go to professors and other people anymore to ask my questions. So I hope you guys can answer some of them on here.

INTRODUCTORY LECTURES ON QUANTUM COSMOLOGY
Jonathan J.Halliwell
http://arxiv.org/pdf/0909.2566v1.pdf

 

[Ulrich]

Since it maintains the same value over time...

From what do you infer this?

The first term must be zero in order to ensure that the cosmological constant has no preferred direction...

What you are referring to is the general perfect fluid stress-energy tensor. General means that it applies to any density and pressure in a perfect fluid. Also matter and radiation densities have no preferred direction in a FRW-model. So according to your reasonement the first term should be zero for any density, which is not the case.

This is false.

johne1618 has already done this calculation:

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

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

The Friedmann equation is

$H(t)^2-\frac{8 \pi G \rho(t)}{3}=-\frac{k c^2}{a(t)^2}$

So you obtain the density quoted from johne1618 only with a flat model (k=0), so not for a spherical closed model (k=1).

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.

 

[Mark M]

From what do you infer this?

Th definition of the cosmological constant. It takes a constant value $\Lambda$ and does not change. It is equivalent to a vacuum energy with a constant energy density of $$ \rho = \frac{\Lambda} {8 \pi G}$$

What you are referring to is the general perfect fluid stress-energy tensor. General means that it applies to any density and pressure in a perfect fluid. Also matter and radiation densities have no preferred direction in a FRW-model. So according to your reasonement the first term should be zero for any density, which is not the case.

No, it must be the case in order to satisfy Lorentz invariance. Otherwise, you have vacuum energy that is stronger in one area than another, i.e. a preferred direction. So, the first term must be zero.

Once again, this is an issue of general relativity, not Newtonian mechanics. You must use psuedo-tensors to even define a global energy in GR, as Berman does in his calculation. If you read his paper (and our FAQ), you'll see that a closed universe does indeed have zero energy. What John calculated, that the energy of matter and radiation is offset by its gravitational energy, is locally true. However, globally, the issue becomes more complicated because of the issue of defining energy in GR.

 

[Ulrich]

The definition of the cosmological constant.

Exactly, it is a definition. And because it is a definition, one cannot derive it from anything else. $ρ=-p$ follows from that definition and not the other way around. Definitions come first, derivations thereafter.

Once again, this is an issue of general relativity, not Newtonian mechanics.

The Friedmann equation is not Newtonian mechanics. It is derived from the Einstein equation using the FRW metric, although in the flat case it can locally also be derived from Newtonian mechanics.

You must use psuedo-tensors to even define a global energy in GR, as Berman does in his calculation. If you read his paper (and our FAQ), you'll see that a closed universe does indeed have zero energy.

I did read parts of this paper and the FAQ. But apparently pseudo-tensors are coordinate dependend. So getting another value for the energy in spherical coordinates shows that something is flawed with these tensors. This is why there is no consensus about this. A physical quantity must be coordinate invariant like lenghts and angles. So I did not read further.

 

[Mark M]

Exactly, it is a definition. And because it is a definition, one cannot derive it from anything else. $ρ=-p$ follows from that definition and not the other way around. Definitions come first, derivations thereafter.

No, literally, a cosmological constant is a constant energy density. If it wasn't that, it would be something else. For example, phantom energy or quintessence. The data from WMAP points towards dark energy being a cosmological constant. If it did not, then we would not be able to make conclusions about the equation of state. But when I posted the above derivation, I assumed that dark energy is a cosmological constant, an assumption backed up by evidence.

The Friedmann equation is not Newtonian mechanics. It is derived from the Einstein equation using the FRW metric, although in the flat case it can locally also be derived from Newtonian mechanics.

John's derivation is a local one - we can use that explanation within our cosmological horizon. However, because of the numerous ways to define global energy in GR, the issue becomes complex when you ask the question of the total energy of the universe.

The following is from MTW, page 457:

I did read parts of this paper and the FAQ. But apparently pseudo-tensors are coordinate dependend.

Right, as are many things in GR. As long as you stick to the right coordinates (Cartesian coordinates in the case of Berman's derivation), you can avoid the issue. From the paper:

The time-varying result for P0 shows that only Cartesian coordinates must be employed
when applying pseudotensors in General Relativity. In reference (York Jr, 1980) it is stated
that, for closed Universes, the only acceptable result is P₀ = 0 .

So getting another value for the energy in spherical coordinates shows that something is flawed with these tensors.

Why do you conclude this? Can you reference a source from a GR text that backs this claim up?

A physical quantity must be coordinate invariant like lenghts and angles.

Firstly, this isn't true. For example, an observer at asymptotic infinity calculates that an observer outside of a black hole measures a different value for the vacuum than he does. This is Hawking radiation.

Also, lengths and angles aren't invariant. See the Lorentz transformation.

 

[Ulrich]

But when I posted the above derivation, I assumed that dark energy is a cosmological constant, an assumption backed up by evidence.

Definitions are based on such assumptions, so call it what ever you want.

Also, lengths and angles aren't invariant. See the Lorentz transformation.

You are mixing up coordinate invariance with Lorentz invariance. The first implies a change from one coordinate system into another from the point of view of the same observer staying at the same place wheras the second implies a change from one inertial system into another that is moving, which leads to time dilatation, length contraction on so on.

I will read your MTW text tomorrow...