Hubble's law and conservation of energy

[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.