I Hubble's law and conservation of energy

[Buzz Bloom]

Hi @George Jones:

In your post #38 I think I now understand everything you are doing except for two topics.
(1) The tensor equation. Although in my youth I could understand a bit about tensor calculus, I have forgotten all of what I once knew about it.
(2) I do not understand what the ε symbol represents and how the ε relates to the Friedmann equation you included, nor the equation with ε_dot (marked "unchanged"), nor other equations involving ε.

ADDED
I now understand that you use ε rather than ρ for the mass density.

Regards,
Buzz

 

[timmdeeg]

I do not understand what form of the Einstein/Friedmann equation you started with, and how you get from that to

ρ_M−2ρ_Λ=0.​

From the acceleration equation. With $p_M=0$ then $\rho+3p$ can be rewritten as $\rho_M+\rho_\Lambda+3p_\Lambda=\rho_M-2\rho_\Lambda$ thereby considering the negative pressure of state $p_V=-\rho_V$. Note $c=1$.

 

[Buzz Bloom]

From the acceleration equation

Hi timmdeeg:

Is the following what you mean by the acceleration equation?
3(dR/dt) = 3Λc²R²(dR/dt)

ADDED
I now am guessing you mean the equation you labeled "acceleration" in your handwritten text. Below I use the prime ' instead of the dot.

ε' = (-4πG/2c2)(ε+3P) + Λ/3​

Is this what you have in mind?

Also, where did you get the Friedmann equation you used? Can you cite a reference?

Also, your differentiating the Friedmann equation doesn't look right to me.

You also seem to be assuming that P and Λ are related. Einstein assumed P=0, and Λ is an independent constant.

Regards,
Buzz

 

[timmdeeg]

Is the following what you mean by the acceleration equation?

No the 2. Friedmann equation which is named acceleration equation sometimes. I guess this clarifies your other questions.

 

[George Jones]

I now understand that you use ε rather than ρ for the mass density.

I have not set $c = 1$. $\varepsilon$ is energy density, which is related to mass density $\rho$ by $\varepsilon = \rho c^2$.

 

[Buzz Bloom]

I have not tried to analyze what you have, but it has been known for almost a century (since at least 1930) that he Einstein static universe is unstable in the following sense.

Hi George:

My mistake was using the word "stable" in my earlier posts when I should have used "metastable".

Metastable

Describes a system which appears to be stable, but which can undergo a rapid change if disturbed.​

Quoted by

https://www.thefreedictionary.com/metastable​

from

Dictionary of Unfamiliar Words by Diagram Group Copyright © 2008 by Diagram Visual Information Limited.​

Regards,
Buzz

 

[kimbyd]

Hi George:

My mistake was using the word "stable" in my earlier posts when I should have used "metastable".

Metastable requires the system be able to remain in that configuration for an extended period of time. I don't think that universe model qualifies, as any inhomogeneities will grow pretty rapidly.

 

[timmdeeg]

Metastable requires the system be able to remain in that configuration for an extended period of time.

Perhaps @metastable would agree. I think "unstable equilibrium" fits better to Einstein's static universe.

 

[Buzz Bloom]

Hi kimbyd and timmdeeg:

I do not want to struggle about vocabulary. I chose "metastable" because I could not find any other word that seemed to be suitable.

To respond clearly I need some notation. I use the apostrophe, (whatever') to represetnt dwhatever/dt. (Using the dot is too awkward for me.) With respect to the equation

let

R₀ = R(t₀)​

be the value of R for which

R'(t0)=R''(t0)=0.​

It will also be useful to let

r(t)=R(t)-R₀.​

Metastable requires the system be able to remain in that configuration for an extended period of time. I don't think that universe model qualifies, as any inhomogeneities will grow pretty rapidly.

The assumptions of the model assumes it is homogeneous, that is, no inhomogeneities. The reason it does not qualify for "stable" is that the sign of R'(t₀+ε) and R''(t₀+ε) will be the same as the sign of (R₀+ε). I believe this condition regarding R'(t₀+ε) and R''(t₀+ε) is true, but to satisfy my curiosity I am currently working on the math to prove it is true. So far I have made many errors, and found them, and I think properly corrected them.
ADDED July 1
And found more errors...

Perhaps @metastable would agree. I think "unstable equilibrium" fits better to Einstein's static universe.

You seem to be correct about what "unstable equilibrium" means.

https://www.merriam-webster.com/dictionary/unstable equilibrium​

Comparing the two definitions, this seems to me to be a synonym of "metastable". There is a subtle difference. The assumptions of the model makes it difficult for something to cause a movement away from metastability.

Regards,
Buzz

 

[kimbyd]

The assumptions of the model assumes it is homogeneous, that is, no inhomogeneities. The reason it does not qualify for "stable" is that the sign of R'(t₀+ε) and R''(t₀+ε) will be the same as the sign of (R₀+ε). I believe this condition regarding R'(t₀+ε) and R''(t₀+ε) is true, but to satisfy my curiosity I am currently working on the math to prove it is true. So far I have made many errors, and found them, and I think properly corrected them.

It doesn't really matter that the model is homogeneous, because our universe manifestly is not. It's very obvious that if this model were to describe anything real, it would need to allow for inhomogeneities. The reason I bring this up is because if you don't have any inhomogeneities, that model can be balanced on the head of a pin, so to speak, such that it remains in the unstable equilibrium for an excessively long period of time.

But if you do have inhomogeneities, then maintaining that balance becomes fundamentally impossible. All you need to break the system is for a baseball to move a little to the left (or whatever direction).

 

[timmdeeg]

https://arxiv.org/pdf/1203.4513.pdf
3 Study of stability
...
Figure 1 ... Notice that the equilibrium at $R = R_E$ is an unstable one. Any small perturbation at $R_E$ makes either the universe to collapse or diverge to R → ∞.

I'm not sure if the question whether or not there are inhomogeneities matters at all.

@Buzz Bloom You might be interested to see how the radius of curvature of Einstein’s static universe $R$ depends on $\rho$, (10) .

 

[kimbyd]

Note the "any small perturbation" part. Those are inhomogeneities.

 

[Buzz Bloom]

@Buzz Bloom You might be interested to see how the radius of curvature of Einstein’s static universe R depends on ρ, (10) .

Hi timmdeeg:

I don't know what this means. I did my own calculation of the relationship between R₀ and ρ, except I calculated
R₀ = 2GM/πc², where
M=ρV, where
V= 2π²R₀³ is volume.

https://www.physicsforums.com/threads/hubbles-law-and-conservation-of-energy.973509​

Post #45​

​

What does "(10)" mean?

ADDED JULY 1
I found equation (10).

Solving for ρ₀G,

ρ₀G = c²/4πR².​

I also found a definition of ρ₀,

This seems to be peculiar for a finite closed universe, making ρ₀ equal to a fraction of the mass M=ρV I used.

ρ₀ = M/2π²​

I only scanned the article, and it seems that the intent is to be more general than Einstein's finite closed model. In a flat or hyperbolic model, the concept of total mass M would not be defined.
Also, the article's equation (2) is different than the one I used for a finite universe with cosmological constant. I need to do some research about the differences.

Thanks for the citation.

Regards,
Buzz

 

[timmdeeg]

Note the "any small perturbation" part. Those are inhomogeneities.

But according to the abstract the article is based on the ideal fluid model, whereas in #60 you talk about the real universe:

It doesn't really matter that the model is homogeneous, because our universe manifestly is not. It's very obvious that if this model were to describe anything real, it would need to allow for inhomogeneities. The reason I bring this up is because if you don't have any inhomogeneities, that model can be balanced on the head of a pin, so to speak, such that it remains in the unstable equilibrium for an excessively long period of time.

But if you do have inhomogeneities, then maintaining that balance becomes fundamentally impossible. All you need to break the system is for a baseball to move a little to the left (or whatever direction).

It seems the universe isn‘t stable even in the ideal case due to perturbations. I wonder if it is less stable in the real case. Or is this consideration flawed insofar as in the real case homogeneity can be assumed if the scales are large enough und thus globally?

 

[kimbyd]

It seems the universe isn‘t stable even in the ideal case due to perturbations. I wonder if it is less stable in the real case. Or is this consideration flawed insofar as in the real case homogeneity can be assumed if the scales are large enough und thus globally?

Again, perturbations are another word for inhomogeneities. Specifically, they're a mechanism to mathematically describe inhomogeneities.

And as long as you have any matter, some level of inhomogeneity is a fundamental requirement. The only way to get away from inhomogeneities entirely is to have every quantum field in its ground state, which means Minkowski or de Sitter space.

Thus whenever you hear the universe described as homogeneous, it must always be considered an approximation.

 

[timmdeeg]

Thanks for your answer. I’m not yet sure if I understand you correctly though.

In #64 you seem to distinguish two cases:

Real universe, inhomogeneities, “ balance becomes fundamentally impossible”.

No inhomogeneities (what I understand to mean the perfect fluid model), “model can be balanced”.

What kind of model do you refer to? As I understand it perturbations are there regardless the model but then no model can be balanced.

 

[kimbyd]

Thanks for your answer. I’m not yet sure if I understand you correctly though.

In #64 you seem to distinguish two cases:

Real universe, inhomogeneities, “ balance becomes fundamentally impossible”.

No inhomogeneities (what I understand to mean the perfect fluid model), “model can be balanced”.

What kind of model do you refer to? As I understand it perturbations are there regardless the model but then no model can be balanced.

My point is that if the universe is perfectly homogeneous, it's technically possible to allow it to sit in the unstable static condition where the cosmological constant perfectly balances matter. It requires infinite fine tuning, but it's technically possible.

But a real universe can't ever be perfectly homogeneous, even in principle, as long as there is some matter (or any quantum field not in its ground state). So even if you set it up so that it is balanced at one point in time, stuff is moving around. All you need is a little motion in one direction and the universe will collapse or start expanding.

 

[timmdeeg]

My point is that if the universe is perfectly homogeneous, it's technically possible to allow it to sit in the unstable static condition where the cosmological constant perfectly balances matter. It requires infinite fine tuning, but it's technically possible.

Do you say that “infinite fine tuning” implies the absence of “any small perturbation” (as mentioned in the article linked in #60)?

 

[PeterDonis]

Do you say that “infinite fine tuning” implies the absence of “any small perturbation” (as mentioned in the article linked in #60)?

Of course it does. "Infinite fine tuning" means the density of matter is exactly the same, to infinite precision, everywhere in the universe (and its value is exactly the value needed to balance the effect of the cosmological constant). That means there can't be any perturbations, no matter how small.

 

[timmdeeg]

So it isn’t correct to think of the perfect fluid model as being infinitely fine tuned, right?

 

[PeterDonis]

So it isn’t correct to think of the perfect fluid model as being infinitely fine tuned, right?

It isn't correct to equate "perfect fluid" with "no inhomogeneities". It's perfectly possible for a perfect fluid to have a density that varies from point to point. Those variations are inhomogeneities. But it's also possible in principle for a perfect fluid to have exactly the same density everywhere, to infinite precision; i.e., to have no inhomogeneities.

In other words, focusing on the "perfect fluid" part is focusing on the wrong thing. The right thing to focus on is whether or not the density is exactly the same everywhere, to infinite precision.

 

[timmdeeg]

Thanks for this helpful answer.

In other words, focusing on the "perfect fluid" part is focusing on the wrong thing. The right thing to focus on is whether or not the density is exactly the same everywhere, to infinite precision.

So the concept of the “perfect fluid” concerns solely the derivation of Friedmann’s Equations.

 

[Michael Price]

The smooth perfect fluid assumption - which is reasonable on the very large scale - is used to derive the Friedmann equations from the Einstein field equations, and from there you can derive the Hubble expansion.
https://en.m.wikipedia.org/wiki/Friedmann_equationsThe first Friedmann equation can be regarded as a total energy conservation equation, where total energy is zero. Despite what some people say, GR has no problem with a local and global conserved energy. You have to use pseudotensors to capture the gravitational energy, but it all works out fine.
https://en.m.wikipedia.org/wiki/Stress–energy–momentum_pseudotensorAs the universe expands energy is lost/transferred from the photons (redshift) and, to a lesser extent, matter as their de Broglie wavelengths are stretched. The energy lost this way is gained by the Hubble expansion. Since the gravitational energy of the expansion is negative this gain of energy slows the expansion.
Hence the expansion in a radiation dominated universes slows more quickly than in a matter-dominated universe.

 

[timmdeeg]

As the universe expands energy is lost/transferred from the photons (redshift) and, to a lesser extent, matter as their de Broglie wavelengths are stretched. The energy lost this way is gained by the Hubble expansion. Since the gravitational energy of the expansion is negative this gain of energy slows the expansion.
Hence the expansion in a radiation dominated universes slows more quickly than in a matter-dominated universe.

However as Sean Carrol put it:

https://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/

There’s nothing incorrect about that way of thinking about it; it’s a choice that one can make or not, as long as you’re clear on what your definitions are. I personally think it’s better to forget about the so-called “energy of the gravitational field” and just admit that energy is not conserved, for two reasons.

That aside I don’t see the context to the OP.

 

[Michael Price]

Sean Carrol concedes it is just a matter of preference. Throwing away the idea of gravitational energy means throwing away the idea that gravitational waves carry energy. Good luck with that!
As for the relevance to the OP - question title says "conservation of energy" and "Hubble's law".

 

[PeterDonis]

Despite what some people say, GR has no problem with a local and global conserved energy

For very specific meanings of those two terms, yes:

Local conservation of energy means the covariant divergence of the stress-energy tensor is zero. This is fairly intuitive, since it says that stress-energy can't be created or destroyed at any point in spacetime. But there are still plenty of counterintuitive aspects to it.

Global conservation of energy means the entire spacetime has zero energy, because "energy" here means the Hamiltonian, which must be zero because of diffeomorphism invariance. This is not intuitive at all.

You have to use pseudotensors to capture the gravitational energy, but it all works out fine.

It depends on what you mean by "works out fine". Gravitational energy is not localizable. Pseudotensors are frame dependent. Both of those things are not "fine" from the viewpoint of many physicists.

Throwing away the idea of gravitational energy means throwing away the idea that gravitational waves carry energy.

This interpretation of "gravitational energy" has nothing whatever to do with pseudotensors or the Hamiltonian. It has to do with the fact that gravitational waves can do work. Something that is frame-dependent can't do work. Something that is identically zero can't do work.

The key point in all this is that "energy" does not have a single meaning. It has multiple possible meanings, and it's easy to confuse yourself and other people by mixing them up.

 

[Michael Price]

It depends on what you mean by "works out fine". Gravitational energy is not localizable. Pseudotensors are frame dependent. Both of those things are not "fine" from the viewpoint of many physicists.
But the ordinary divergence of the (matter tensor plus gravity pseudotensor) is a tensor (and identically zero), as the Wikipedia article points out. So the result (conservation of total energy) is frame independent.

 

[PeterDonis]

the ordinary divergence of the (matter tensor plus gravity pseudotensor) is a tensor

This is just terminology. The covariant divergence of the stress-energy tensor is the ordinary divergence plus extra terms in the connection coefficients. The Wikipedia article is just calling those extra connection coefficient terms "the ordinary divergence of the gravity pseudotensor". That doesn't mean the gravity pseudotensor itself (the thing that the connection coefficient terms are the ordinary divergence of) describes any kind of localizable "energy in the gravitational field". It doesn't.

the result (conservation of total energy) is frame independent

Local conservation, yes. That's all the divergence can tell you about. Global conservation requires doing an integral.

 

[kimbyd]

Do you say that “infinite fine tuning” implies the absence of “any small perturbation” (as mentioned in the article linked in #60)?

No. The assumption of perfect homogeneity does that.

The infinite fine tuning is the perfect balancing of the matter density with the cosmological constant. If you had perfect homogeneity, you could do that balancing. How long the state would last would just be a matter of how precisely the balance was set up.

The point I'm trying to make here is that this balancing can't work, even in principle, for any real universe that contains actual atoms, because it's fundamentally impossible for a real universe to be perfectly homogeneous if it contains atoms. This is the same as stating that it's impossible for a perfect sphere to exist because any real sphere is made of atoms.

Usually the fact that our universe contains atoms is irrelevant at scales much larger than atoms. But there are times when it matters, and I think this is one of them. It's why I say that this is not a metastable equilibrium: the model is fundamentally unstable because it's impossible for any realistic model to be balanced at "no expansion" for any significant amount of time.

Of course, the situation becomes even more stark when we point out that the atoms in our universe are organized into galaxies and galaxy clusters, which are remarkably less homogeneous than a smooth fluid made of real atoms.

 

[Michael Price]

Perhaps the divergence of Landau-Liftshitz pseudotensor is no more than terms in the expansion of the covariant divergence of the matter tensor (I shall try and check) - but it has explanatory power, even so. Where does the energy of a redshifting photon go to? Saying it just sort "disappears" is not satisfactory, in my book anyway.and

The integral of an ordinary divergence is trivial to write down, so the global conservation follows trivally.

And, yes, the gravitational energy is non-localisable. This seems to upset a lot of people. But isn't this the same as the electromagnetic 4-vector being gauge dependent, and can be made to vanish at a chosen point Similarly the gravitational energy is gauge dependent - the gauge transforms in GR are coordinate transforms.

 

[PeterDonis]

Perhaps the divergence of Landau-Liftshitz pseudotensor is no more than terms in the expansion of the covariant divergence of the matter tensor (I shall try and check)

More precisely, the mathematical identity is:

covariant divergence of stress-energy tensor

equals

ordinary divergence of stress-energy tensor + connection coefficient terms

equals

ordinary divergence of stress-energy tensor + ordinary divergence of "energy in gravitational field" pseudotensor

it has explanatory power

Only if you think non-invariant quantities have explanatory power. That is not a common viewpoint among relativity physicists; the common viewpoint is that only invariants have physical meaning.

The integral of an ordinary divergence is trivial to write down

Yes, but picking out the spacelike slice over which to do the integral is not trivial at all. And different spacelike slices will in general give different integrals. Also, integrals of ordinary divergences are coordinate-dependent, since ordinary divergences are not tensors.

isn't this the same as the electromagnetic 4-vector being gauge dependent

No. A 4-vector is a local quantity; it's well-defined at a single spacetime point. "Gravitational field energy" being non-localizable means it is not well-defined at a single spacetime point.

 

[Michael Price]

The gravitational energy, being completely and precisely defined by the metric, is just as well defined as the EM potential. They are both non localisable in the sense that they can be set to zero at a chosen point by an appropriate gauge transformation.

 

[PeterDonis]

The gravitational energy, being completely and precisely defined by the metric

The metric does not define gravitational energy. It defines the spacetime geometry. They're not the same thing.

They are both non localisable in the sense that they can be set to zero at a chosen point by an appropriate gauge transformation.

Huh? You can't make the metric vanish at a point by a gauge transformation.

 

[PeterDonis]

the EM potential. They are both non localisable in the sense that they can be set to zero at a chosen point by an appropriate gauge transformation.

If you are going to make an analogy between gravity and EM in this way, the analogue of the EM potential is not the metric; it's the connection. The connection is what can be made to vanish at any chosen point by a gauge transformation.

 

[Michael Price]

The metric does not define gravitational energy. It defines the spacetime geometry. They're not the same thing.
Huh? You can't make the metric vanish at a point by a gauge transformation.

I didn't say the metric could vanish. But the gravitational energy can be made to.

 

[PeterDonis]

Thread closed for moderation.