Is energy conserved in cosmology according to the laws of thermodynamics?

[Q-reeus]

If by "violate conservation of energy" you mean a local violation, then that won't happen, because local conservation of energy is built into the Einstein field equations. If you mean a global violation, then this is covered in the FAQ: https://www.physicsforums.com/showthread.php?t=506985 The total energy cannot even be defined in a typical spacetime, so there is no way to discuss whether it changes over time.

Thanks for the link. Well the first consideration is; how local does local have to be? If strictly speaking a point volume, not much use. My brief layman's translation is: In the 'global' context, GR is in general incapable of connecting all the internals of a system to give an unambiguous definition of overall energy-momentum. Hmm...could this fundamentally stem from that lack of gravitational field energy-momentum as source terms in the RHS of the EFE's? Or is it claimed any metric theory would be so burdened with ambiguity?

This sounds like you want to assume GR isn't valid in a discussion of GR. One of the most important and fundamental ideas of GR is that spacetime isn't naturally endowed with coordinates.

That wasn't what was meant. If the assumed global failure stems from this inability to connect the internals from within a curved spacetime, how about assessing a finite system 'from the outside'. Most or at least many cosmologists are happy with the idea of an overall flat universe, so that seems a natural backdrop (the coordinate reference frame) from which to assess the evolution of some given local system - the 'black box'. All I'm saying is, given some initial configuration, this should be able to be followed through to an arbitrary final configuration, and any difference in net energy-momentum unambiguously determined - all referenced to the coordinate values 'at infinity'. So has that been done? If even that is considered an ill-defined problem for GR, then yes I suppose I would be questioning validity of same.

 

[WannabeNewton]

You can't define extrinsic properties of a manifold in GR as it only concerns itself with intrinsic properties so I do not understand what you mean by "from the outside". The ambiguity results from the way parallel transport works on a curved manifold: transporting a vector around a closed loop in one direction will yield a vector with different components than if the vector was transported around the same loop the other way because remember covariant derivatives don't commute in curved space. Can you define the covariant derivative without a metric? It requires an affine connection but you can define an affine connection on a manifold without the use of a metric right?

 

[Q-reeus]

You can't define extrinsic properties of a manifold in GR as it only concerns itself with intrinsic properties so I do not understand what you mean by "from the outside". The ambiguity results from the way parallel transport works on a curved manifold: transporting a vector around a closed loop in one direction will yield a vector with different components than if the vector was transported around the same loop the other way because remember covariant derivatives don't commute in curved space.

I won't argue your comments about parallel transport issues - your '11th grade' education in GR is better than mine. However, let's try and get this into some sort of concrete setting. We have say a binary star system that spirals together and finishes up as a neutron star+exploding gas, light etc. Are you saying there is no way in principle we could have an unambiguous before and after accounting of the net energy-momentum, all referenced from some far vantage point where curvature of the system has negligible effect on the distant observer? If that's the case, this seems tantamount to saying it's impossible in GR to accurately specify any gravitating system to begin with.

 

[WannabeNewton]

Are you saying there is no way in principle we could have an unambiguous before and after accounting of the net energy-momentum, all referenced from some far vantage point where curvature of the system has negligible effect on the distant observer?

No I think in your scenario, where the observer is in an asymptotically flat - space time, the energy - momentum loss can be measured unambiguously through the angular momentum and energy flux carried away by gravitational waves. But I think the point is that in asymptotically flat space - times you can actually define global energy far away from the source but what about in general? You can't even define global energy in curved space - times in the absence of time - like killing fields.

 

[TrickyDicky]

Let's give some context to what is being discussed so we can avoid some gratuitous arguments, what I posted in #22 was in the context of the Noether theorem discussion about symmetries that I introduced previously, so it is obvious that it only works provided a timelike killing vector is introduced. The EFE by themselves tell us nothing about the existence of this Killing vector, that it doesn't exist is an assumption of the current cosmological model.

 

[Q-reeus]

No I think in your scenario, where the observer is in an asymptotically flat - space time, the energy - momentum loss can be measured unambiguously through the angular momentum and energy flux carried away by gravitational waves. But I think the point is that in asymptotically flat space - times you can actually define global energy far away from the source but what about in general? You can't even define global energy in curved space - times in the absence of time - like killing fields.

OK let's agree then that the evolution of a system's energy-momentum can be definitely established wrt asymptotically flat spacetime. I see two issues here. First, can we assume that conservative behavour is always then determined, regardless of the possibly violently energetic and highly curved system local parameters (eg high spinning binary neutron star inspiral)? If so, then shouldn't this logically be extrapolated to say that conservative behavour applies in any arbitrary curved spacetime setting, and it's just that we don't have an unambiguous means of determining it? Which is quite different to saying there can be an actual failure of coem. (hope this doesn't seem gratuitous TrickyDicky! )

 

[bcrowell]

First, can we assume that conservative behavour is always then determined, regardless of the possibly violently energetic and highly curved system local parameters (eg high spinning binary neutron star inspiral)?

In that situation, energy is being radiated away to infinity by gravitational waves.

If so, then shouldn't this logically be extrapolated to say that conservative behavour applies in any arbitrary curved spacetime setting,

No. There are conserved, global, scalar measures of mass-energy for static spacetimes and asymptotically flat spacetimes, but not for arbitrary spacetimes.

 

[Q-reeus]

In that situation, energy is being radiated away to infinity by gravitational waves.

You misunderstand - by conservative it was meant overall conservative, ie conservation of energy-momentum holds, not that there was no dissipative behavour.

No. There are conserved, global, scalar measures of mass-energy for static spacetimes and asymptotically flat spacetimes, but not for arbitrary spacetimes.

Well this is a sticking point for me. Do you agree that say the inspiralling NS scenario will be conservative (net!) wrt asymptotically flat spacetime measure? If not I desist but if you do then this creates a problem. The inspiralling NS system is an interaction of matter in a highly curved local spacetime, so if coem actually fails in GR it should be failing here and thus registered as such at 'infinity'. Conversely, if there is no failure by this reference measure, how can 'failure' in a generally curved background setting be other than a measurement problem, and not intrinsically 'real'?

 

[bcrowell]

Let's give some context to what is being discussed so we can avoid some gratuitous arguments, what I posted in #22 was in the context of the Noether theorem discussion about symmetries that I introduced previously, so it is obvious that it only works provided a timelike killing vector is introduced. The EFE by themselves tell us nothing about the existence of this Killing vector, that it doesn't exist is an assumption of the current cosmological model.

Your #22 was incorrect, for the reasons I gave in #31. Noether's theorem doesn't help, for the reasons given in #31.

Your logic about timelike Killing vectors is incorrect. What's relatively easy to show is that if there's a timelike Killing vector (i.e., the spacetime is stationary), then there is a conserved energy for test particles. That is completely different from being able to define a global, scaler measure of energy for the spacetime as a whole -- that was done by Komar in 1963, and it wasn't just some kind of trivial application of Noether's theorem.

I don't know if you will get around to admitting at some point that your #1 was incorrect. The FAQ gives references -- did you look at them?

 

[TrickyDicky]

Your #22 was incorrect, for the reasons I gave in #31. Noether's theorem doesn't help, for the reasons given in #31.

Your logic about timelike Killing vectors is incorrect. What's relatively easy to show is that if there's a timelike Killing vector (i.e., the spacetime is stationary), then there is a conserved energy for test particles. That is completely different from being able to define a global, scaler measure of energy for the spacetime as a whole

What do you find incorrect about this? It is the mathematical version of #22 and if you want citations you can find it in Penrose Road to reality, where the references to Noether theorem can also be seen:
We define a rate of energy transfer or flux with covector L:
$$L_a=T_{ab}\kappa^{b}$$ with $\kappa$ as the Killing vector
This will make the flux covector divergence vanish when $$\nabla^{a} T_{ab}=0$$
From this It is straight-forward that there is an integral conservation law:
$$\int_{V}L=0$$


Do you get it now?

 

[Ben Niehoff]

What do you find incorrect about this? It is the mathematical version of #22 and if you want citations you can find it in Penrose Road to reality, where the references to Noether theorem can also be seen:
We define a rate of energy transfer or flux with covector L:
$$L_a=T_{ab}\kappa^{b}$$ with $\kappa$ as the Killing vector
This will make the flux covector divergence vanish when $$\nabla^{a} T_{ab}=0$$
From this It is straight-forward that there is an integral conservation law:
$$\int_{V}L=0$$


Do you get it now?

Can you be more specific what you're integrating here? It looks like you're trying to integrate a 1-form over a 3-volume, which makes no sense.

 

[TrickyDicky]

I'd say L is a 3-form.

 

[bcrowell]

What do you find incorrect about this? It is the mathematical version of #22 and if you want citations you can find it in Penrose Road to reality, where the references to Noether theorem can also be seen:

What page in Road to Reality? Ben Niehoff has pointed out that your notation doesn't make sense, which makes me doubt that you're giving an accurate depiction of an argument by Penrose.

Actually I happen to have a relevant quote from the Penrose book handy:

For example, it is not at all a clear-cut matter to apply these ideas to obtain energy-momentum conservation in general relativity, and strictly speaking, the method does not work in this case. The apparent gravitational analogue [of EM gauge symmetry] is 'invariance under general coordinate transformations' [...] but the Noether theorem does not work in this situation, giving something of the nature '0=0'.

--Penrose, Road to Reality, p. 489

Have you still not bothered to look at the references given in the FAQ? Maybe you should write a stern letter to Misner, Thorne, and Wheeler telling them that they're "not even wrong."

 

[TrickyDicky]

What page in Road to Reality? Ben Niehoff has pointed out that your notation doesn't make sense, which makes me doubt that you're giving an accurate depiction of an argument by Penrose.

Actually I happen to have a relevant quote from the Penrose book handy:

--Penrose, Road to Reality, p. 489

Have you still not bothered to look at the references given in the FAQ? Maybe you should write a stern letter to Misner, Thorne, and Wheeler telling them that they're "not even wrong."

Take it easy man, that attitude is not productive.

 

[bcrowell]

Take it easy man, that attitude is not productive.

Still waiting for the page number from Penrose. Still waiting for any indication that you've looked at the references.

 

[WannabeNewton]

I'd say L is a 3-form.

A 3 - form is of the type (0 , 3). Your L defined by contracting the energy - momentum tensor with the killing vector is a one - form.

 

[Ben Niehoff]

OK, so I think what you meant to write is

$$\int_\Sigma \ast L$$

in which case I would agree that this integral gives the same value on any spacelike hypersurface $\Sigma$ by the divergence theorem (given L is constructed from T and a timelike Killing vector as shown above). Hence this is a globally conserved charge (modulo some topological considerations, if the spacetime manifold has spacelike hypersurfaces that cannot be continuously deformed into one another).

So now I think I agree, that if you have a timelike Killing vector, you can define a globally conserved energy.

In the FLRW metric you do not generally have a timelike Killing vector, except in the special case of the inflation period, which is de Sitter space and hence maximally symmetric.

Edited to add: This globally-conserved energy is the energy of stuff in the spacetime. So for vacuum solutions such as Schwarzschild and de Sitter, this energy is zero anyway!

Are there any solutions with a finite density of matter that admit a timelike Killing vector? I'm not very familiar with fluid and dust solutions.

 

[WannabeNewton]

OK, so I think what you meant to write is

$$\int_\Sigma \ast L$$

in which case I would agree that this integral gives the same value on any spacelike hypersurface $\Sigma$ by the divergence theorem (given L is constructed from T and a timelike Killing vector as shown above).

Just to clarify for myself, you are integrating over a space - like hyper surface because the hodge dual of L will give a 3 - form right?

 

[TrickyDicky]

OK, so I think what you meant to write is

$$\int_\Sigma \ast L$$in which case I would agree that this integral gives the same value on any spacelike hypersurface $\Sigma$ by the divergence theorem (given L is constructed from T and a timelike Killing vector as shown above). Hence this is a globally conserved charge (modulo some topological considerations, if the spacetime manifold has spacelike hypersurfaces that cannot be continuously deformed into one another).

So now I think I agree, that if you have a timelike Killing vector, you can define a globally conserved energy.

That's right Ben, I just didn't know how to Latex that asterisk. (Thought the Hodge op. was implicit in the way I wrote it anyway)

In the FLRW metric you do not generally have a timelike Killing vector

As I have noted several times too.

Are there any solutions with a finite density of matter that admit a timelike Killing vector? I'm not very familiar with fluid and dust solutions.

There are, but are not considered mainstream.

 

[PAllen]

There are, but are not considered mainstream.

There are some simple boring ones. A perfect fluid ball (nothing else in the universe) , with its vacuum part coinciding with exterior Schwarzschild is one. It's got everything, static, asymptotic flatness, you name it.

 

[TrickyDicky]

Still waiting for the page number from Penrose.

I don't ave access to the book right now, but I'm quite sure I saw this derivation there.

 

[bcrowell]

I don't ave access to the book right now, but I'm quite sure I saw this derivation there.

You wrote down something and claimed you had a reference to support it. Now other people have pointed out that it was incorrect, and you say you don't have the reference after all.

What is the expression for the global, scalar mass-energy that you claim to have found? You haven't supplied it yet. Once you provide it, it will remain to be shown that it can be interpreted as a measure of mass-energy, e.g., by showing that for a localized distribution of matter, it corresponds properly to the proper acceleration of a distant, static observer. I think it's extremely unlikely that you can demonstrate this, since what you'd presumably be reinventing would be the Komar mass, and I don't think the Komar mass can be reinvented in three lines of math, as you seem to be claiming in #45.

Note that the flux L that you defined in #45 is a vector, not a scalar. That means that any conserved charge you get is going to be a vector, not a scalar. That means you can't define a global conservation law for it, for the reasons given in the first reference in the FAQ -- but you still haven't shown any signs of having looked at any of the references in the FAQ.

Of course none of this has much to do with your statement in #1 that 'I've just read the FAQ about this and IMO it is "not even wrong" to say that energy conservation doesn't apply to Cosmology.' Cosmological spacetimes don't have timelike Killing vectors. I'm still waiting to hear you clearly admit that your assertion in #1 was wrong.

 

[TrickyDicky]

Why should I admit is wrong when it's not just because you say so?
You've been incorrect so many times before without admitting it (like for instance right now even if others with more knowledge than you have clearly pointed out like in post #52 after your #44) that I or anyone can't really take you seriously when you say something is wrong.

Be water, my friend.

 

[TrickyDicky]

BTW, in the MTW cited page, the infamous misleading phrase "conservation of energy doesn't apply to cosmology" (which is basically what I've been saying that needs to be better expressed) is nowhere to be found, actually it only refers to closed universes.

 

[TrickyDicky]

To come back to the OP once some things have been clarified, I must say that it is very hard for me to understand how fundamental physics laws are so easily dismissed by some.

I guess when the claim is made that energy is not conserved in cosmology it is not realized that not only the first law of thermodynamics is rescinded but also the second law since both laws are interconnected.

The measurement of energy is always in relative terms since there is no absolute measure of energy, only the transition of a system from one state into another can be defined, and it takes place through the flow or time-asymmetry of the second law, and the same is true of every measurement process.
Every measurement process also demonstrates the first law as well since the reasoning and relations that hold in the math require something that remains invariant (energy) over those relations (or else one could not get invariant results). The first and second laws are thus automatically entailed in every measurement process.

So I guess according to this, the claim is actually made in the FAQ that Thermodynamics doesn't apply to cosmology which is quite risible since it is used all the time .