Understanding the Zero Energy Principle of the Universe

[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...

 

[Mark M]

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

Right, but considering it's generally accepted that dark energy is a cosmological constant (See here), we arrive at the conclusion that it doesn't make a contribution to the energy density.

 

[twofish-quant]

Right, but considering it's generally accepted that dark energy is a cosmological constant (See here), we arrive at the conclusion that it doesn't make a contribution to the energy density.

It's not anywhere near that certain.

The problem with the current WMAP measurements is that there is a degeneracy between the curvature of the universe and evolution of the dark energy equation of state. The current results are consistent with zero curvature and zero evolution of the cosmological constant, but if both are non-zero, their effects will cancel each other out.

Now we *will* be able to establish that both are effectively zero pretty soon (just take more measurements to beat down the error bars in the next few years) but it's too early to declare this yet.

 

[twofish-quant]

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?

Nope. It's all hot gas. That's why people are pretty certain about the interpretation of things like big bang nucleosynthesis and the cosmic microwave background. They just involve gas and radiation for things that we can do Earth based experiments for.

One other good thing is that the "weird physics" for the most part does not effect the basic calculations that happen at between say 10 seconds and 500,000 years after event zero. The thing about dark energy is that it's a "constant" energy field which means that in the early universe, any pressure and energy that comes out of it is going be much less than "normal" gas and radiation pressure.

One weird thing about dark energy is that it doesn't seem to matter until very recently. This bothers a lot of people.

Another question is how infinite curvature at the big bang must be considered.

Some people take a look at the problem, throw up their hands and then look at some other part of the problem.

 

[twofish-quant]

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.

The cool thing is that this is *NOT* the situation, and we are likely to have a very good idea of what happened at "event zero" in the next decade. What you basically do is to take the data that is coming from PLANCK and WMAP, and "peel away the onion." You remove effects that happened at various times in the early universe, and what you have left is pretty close to event zero.

see http://www.cmu.edu/cosmology/events/cosmic-acceleration/will_kinney.pdf

The reason theorists are busy writing these papers is that we've got a lot of data that's coming.

 

[twofish-quant]

One other thing. It's tempting to say "the numbers look close to zero so they must be zero" but in this situation there are in fact good reasons to think that the both the flatness of the universe and the "constantness" of the cosmological constant could be close to zero, but not zero.

Getting a "flat" universe is a pretty generic feature of an inflationary universe. You take a curved object, blow it up, and it looks flat. So if inflation is right, we ought to expect a flatness of the universe near zero, but fluctuations in the early universe would set things up so that it's not zero.

Similarly, there are theoretical reasons why the cosmological constant may turn out to be non-constant. Either the dark energy is increasing, constant, or decreasing. If dark energy were decreasing then we wouldn't be here. The universe would have blown itself apart before stars formed and so we wouldn't see it.

So that means that dark energy could must be either increasing or constant. There are limits on how much dark energy could increase. At has to have started at zero, 13 billion years ago, and then gradually increased to it's current tiny amount, and that puts limits on how quickly dark energy has increased, and it turns out that distinguishing between dark energy that grows from zero to the current amount, and one that was always at the current amount is difficult.