If total energy is conserved just transformed and never newly created, is there a sum of all energies that is constant? Why is it probably not that easy?

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4Here is a related question that might be helpful http://physics.stackexchange.com/q/2838/ – Freedom Oct 17 '12 at 01:32
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6The total energy of the universe is not well defined, so we can't even discuss whether it's constant. http://www.physicsforums.com/showthread.php?t=506985 – Oct 17 '12 at 05:36
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5How are people even trying to answer this "yes" or "no" when the definition of energy in GR is a subject of ongoing research? – DanielSank Dec 17 '14 at 06:36
5 Answers
No. The universe is dominated by dark energy, which is consistent with a cosmological constant $\Lambda$. In other words, as the universe expands, the energy density stays roughly the same. So the (energy density)*volume is growing exponentially at late times.
Although the total energy is not well defined (as the volume of the universe may be infinite), the fractional rate of growth is certainly nonzero.
You might wonder how the total energy can grow without violating energy conservation. The answer is that in general relativity, we just need $\boldsymbol{\nabla} \cdot \boldsymbol{T} = 0$, so a cosmological constant is perfectly consistent as $\boldsymbol{\nabla} \cdot \Lambda \boldsymbol{g} = 0$
For a nice explanation by Sean Carroll, see http://blogs.discovermagazine.com/cosmicvariance/2010/02/22/energy-is-not-conserved/

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3The total energy isn't just undefined because of the possibility that the universe is infinite. It's undefined for the reasons given in juanrga's answer. – May 06 '13 at 21:06
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1What about Noether theorem ? If the laws of physics don't depend on time, we should be able to build a conserved quantity, and call it "energy" – agemO Dec 05 '14 at 13:03
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3@agemO Noether's theorem leads to a conserved current. Getting a conserved quantity involves performing a spatial three dimensional integral. This is very subtle in GR. – jwimberley Dec 17 '14 at 14:10
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1And does it eventually lead to a conserve quantity ? What happen when done on a closed universe ? – agemO Dec 17 '14 at 16:09
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1@agemO yes you can use Noether's theorem in this way and get a conserved current even with dark energy. The current can be integrated. There are no special subtleties in doing the integration in GR. The energy in the gravitational field is negative and cancels the increasing dark energy. the total energy in a closed universe is zero, but not in a trivial way. The sum of energies from different fields only adds to zero when the field equations apply. So the correct answer is "yes, energy is conserved." – Philip Gibbs - inactive Dec 19 '14 at 19:30
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In GR, spacetime is dynamical, so in general, it is not time-translational-invariant. Because it's not, one can't apply Noether's theorem to argue that there is a conserved energy. See - http://physics.stackexchange.com/questions/2838/total-energy-of-the-universe – Karthik Sharma Jan 31 '16 at 10:57
Your question is tagged as general-relativity and cosmology, and as textbooks remark (e.g. Peebles [1]) "there is not a general global energy conservation law in general relativity theory.”
Therefore: ”The conclusion, whether we like it or not, is obvious: energy in the universe is not conserved” [2].
[1] Peebles P. J. E., 1993, Principles of Physical Cosmology (Princeton Univ. Press).
[2] Harrison E., 1981, Cosmology ( Cambridge University Press)

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Energy conservation stems from Noether's theorem applied to time (i.e., time-invariance leads to energy conservation, similarly to how spatial-invariance leads to momentum conservation). Since the universe is expanding (and accelerating at that), the state of the universe today is different than it was yesterday and will be tomorrow, hence energy conservation cannot be established for the whole universe.
Locally, however, the stress-energy tensor, $$T^{\mu\nu}=\left(p+\rho\right)u^\mu u^\nu - pg^{\mu\nu},$$ will satisfy the conservation law (of energy and momentum), $$ T^{\mu\nu}{}_{;\nu}=0 $$ (derived through the Bianchi identity, the $;\nu$ subscript denotes the covariant derivatve).
Wald states (Amazon link, emphasis are his) in Chapter 4
The issue of energy in general relativity is a rather delicate one. In general relativity there is no known meaningful notion of local energy density of the gravitational field. The basic reason for this is closely related to the fact that the spacetime metric, $g_{\mu\nu}$, describes both the background spacetime structure and the dynamical aspects of the gravitational field, but no natural way is known to decompose it into its "background" and "dynamical" parts. Since one would expect to attribute energy to the dynamical aspect of gravity but not to the background spacetime structure, it seems unlikely that a notion of local energy density could be obtained without a corresponding decomposition of the spacetime metric. However, for an isolated system, the total energy can be defined by examining the gravitational field at large distances from the system. In addition, for an isolated system the flux of energy carried away from the system by gravitational radiation also is well defined.
Later, in Chapter 11,
...the most likely candidate for the energy density of the gravitational field in general relativity would be an expression quadratic in the first derivatives of the metric. However, since no tensor other than $g_{\mu\nu}$ itself can be constructed locally from only the coordinate basis components of $g_{\mu\nu}$ and their first derivatives, a meaningful expression quadratic in first derivatives of the metric can be obtained only if one has additional structure on spacetime, such as a preferred coordinate system or a decomposition of the spacetime metric into a "background part" and a "dynamical part" (so that, say one could take derivatives of the "dynamical part" of the metric with respect to the derivative operator associated with the background part). Such additional structure would be completely counter to the spirit of general relativity, which views the spacetime metric as fully describing all aspects of spacetime structure and the gravitational field.

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1This is wrong because it treats the gravitational field as a given background field when in fact its evolution is given by dynamical equations which are time invariant and derived from the Einstein-Hilbert action. Noether's theorem therefore does apply. See e.g. Dirac's short book on GR which derived energy conservation in GR this way. – Philip Gibbs - inactive Dec 18 '14 at 22:46
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1The theory of the energy content of the gravitational field is good enough to predict the deceleration of binary pulsars due to gravitational wave radiation. A Nobel prize has been given. MTW, Wald and Peebles are wrong about energy in GR. Einstein, Landau, Lifshitz, Dirac and Weinberg are right. Energy density is just reference frame dependent as you would expect in relativity. That does not make it meaningless. – Philip Gibbs - inactive Dec 19 '14 at 01:16
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1Kyle, What you said is also true of special relativity but nobody is saying there are any problems with energy conervation in SR. Reference frame dependence is not an issue. – Philip Gibbs - inactive Dec 19 '14 at 01:39
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1Kyle, thank you for your advice. My advice to you is that when someone refutes what you say with simple clear logic it does not help to cite vague and irrelevant points from textbooks. In your first quote Wald talks vaguely about the metric being both background and dynamic. There is no sense in that. The metric is dynamic and that is all it is. It is no more a background that any other field is a background. He just ignores the fact that formulations for gravitational energy have been known for decades and tries to argue that they cannot exist. – Philip Gibbs - inactive Dec 19 '14 at 11:35
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1In the second quote he wants the formulation of energy to depend only on first derivatives of the metric. This can be done with pseudotensors but a covariant formulation requires second derivatives because the action has third derivatives. These requirements he wants to impose are artificial and unjustifed. Note that pseudotensors do not require a "preferred" reference frame, they just require someone to choose a reference frame for the purpose of measurement as you do for any other measurment. A good covariant formulation uses the Komar superpotential. – Philip Gibbs - inactive Dec 19 '14 at 11:42
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1In any case, neither of these quotes addresses the problem with the main point of your answer concerning time invariance, which I have refuted. – Philip Gibbs - inactive Dec 19 '14 at 11:45
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1@Kyle Kanos, I also refuted your argument in the comments but you have not answered that. So you have now moved on to the triviality argument which is point 6 refuted in my paper. The Noether current is not given by varying the metric in the action, that only works for the matter part of the current (a special feature of gauge fields that appears in Noether's second theorem). Look up NT in Wikipedia for the correct expression for the current. The current is not zero in general. That is only true in special cases such as a homogeneous cosmology and even then it is never trivially zero. – Philip Gibbs - inactive Dec 21 '14 at 06:14
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1@Kyle Kanos, I will explain again why the argument you use in your answer is wrong. You are treating the gravitational field as a background but it needs to be included in the dynamics to account for the gravitational contribution to energy. The condition for Noether's theorem to apply in the case of energy is that the dynamical equations must be independent of time, not the solution. The full action including gravity does not change in time so NT can be applied. – Philip Gibbs - inactive Dec 21 '14 at 06:23
What we like to call the energy, i.e., the total matter/energy content of space-time, might not be conserved. However, there is a lot of reason to suspect that fundamentally the universe is some big quantum system, and that space-time and particles and fields are emergent from this underlying idea. In that case, we expect there to be a Hamiltonian $H$ and some time evolution rule $i\hbar \partial_t \left|\psi\right\rangle = H \left|\psi\right\rangle$, and unitarity requires that energy be conserved. Papers by Page and Wootters have interesting things to say on the subject.

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2So after criticizing me for saying that energy is conserved you say the same thing but give a more speculative justification based on quantum gravity rather than GR. Don't you think that if enrgy is conserved in a quantum theory there will be a corresponding formulation in the classical limit? – Philip Gibbs - inactive Dec 18 '14 at 11:33
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1I did not mean to give the impression that I was criticizing your answer because I believe that energy was not conserved. I took issue with your stating a controversial viewpoint within GR as a fact, and not letting the reader know that the site you link to is by no means a good place for a beginner to start.
Also, my answer is speculative because it is an open problem. Finally, GR is not expected to be the classical limit to a quantum theory of everything, precisely because of things like singularities, information paradoxes, etc. It is likely an approximation.
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1My answer is only controversial in the sense that there are people here who do not understand how energy works in GR and dispute it despite it having been understood for nearly a hundred years. It sounds like you downvoted me mainly because I linked to viXra without actually finding anything wrong with the paper. GR is expected to be a classical limit of quantum gravity. What else could it be? All classical limits are approximations and are incomplete. – Philip Gibbs - inactive Dec 18 '14 at 22:54
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1"Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics and astronomy" What made you think that answers had to be tailored to suit beginners? – Philip Gibbs - inactive Dec 18 '14 at 23:01
The only thing that prevents us defining a total conserved energy for the entire universe is that if the universe is infinite then the total energy could be infinite or indeterminate.
The statements that say energy is not conserved in general relativity are wrong, irrespective of who says them. You can define energy over any finite volume of space and you can define the flux of energy over the boundary surrounding the volume. The rate at which energy decreases in the volume is equal to the flux of energy across the boundary. This is the the most general way to express energy conservation globally.
All statements to the contrary can be refuted and to avoid arguing around in circles I have done that at length in my write-up at http://vixra.org/abs/1305.0034

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1The way this is worded (and the site you link to) gives the beginning physicist the impression that the only alternative to imperfect peer review is no peer review. – lionelbrits Dec 17 '14 at 21:31
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2No, the alternative is open peer review, but it is not often an available option. – Philip Gibbs - inactive Dec 17 '14 at 21:53
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1viXra accepts submissions from Jesus Christ to show that it does not discriminate even against people making the most outlandish claims such as "I am the son of God" – Philip Gibbs - inactive Dec 17 '14 at 21:56
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2So what was your physics objection to my answer? Do you prefer the appeal to authority citing a text book that is not peer reviewed? – Philip Gibbs - inactive Dec 17 '14 at 22:02
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1I think a textbook like MTW, or Schutz, or Wald, are peer reviewed in the sense that thousands of GR researchers have spent hundreds of thousands of hours poring over minute details of it and submitted errata and decades later, still decide that the material is of sufficient quality to recommend for courses, or base theses on.
Comments on answers need not be limited to concerns about physics. If you want to state a view point that is widely considered incorrect, make that clear to your reader. It's about intellectual honesty.
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2It is not true that my point of view is widely considered incorrect. Conservation of energy in GR was first formulated by Einstein with good alternative but equivalent formulations being given by Landau-Lifshitz, Dirac, Weinberg and others. There are now better methods that dont use pseudotensors. There are of course others who do not understand it, especially people here, but you may notice that I still get more upvotes than downvotes. Tell me a specific fault in my answers and papers instead of appealing to selected authorities or complaining about lack of peer-review. – Philip Gibbs - inactive Dec 18 '14 at 11:25
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1Noether's theorem shows that energy conservation is due to a symmetry in time (time invariance). The universe is accelerating in its expansion so energy cannot be conserved globally as there is no symmetry in time (see my answer). Local conservation is perfectly valid; if you amend your answer to discuss the difference between the local and global aspects, I'd be willing to upvote, but as it stands it is entirely wrong – Kyle Kanos Dec 18 '14 at 13:27
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2Kyle, what you describe is one of the many frequently repeated fallacies that is refuted in my paper that I link to in my answer (point 3). Basically your fault is that you are treating the gravitational field as a given background in which matter and radiation move when in fact it is itself a dynamical field affected by them through equations which are time invariant. When you apply Noether's theorem to the full Lagrangian you get the conserved currents which can be integrated to give global conservation laws for energy including the energy in the gravitational field. – Philip Gibbs - inactive Dec 18 '14 at 22:40
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1@PhilipGibbs: Since your paper has yet to be accepted to any reputable journal, it's basically worthless to the scientific community. Any crank can "publish" (really it should be "post") to vixra, it doesn't have to be real research. Every relevant researcher in GR makes the statement that energy is not conserved in GR, since it was claimed back in 1920!. – Kyle Kanos Dec 19 '14 at 14:09
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2Kyle, it is not a paper of original research so there is no point submitting it to a journal. It seems you cannot defend your answer so you just appeal to authority or critic on the basis that my work is not peer reviewed. Tell me this, do you really think that there is an explicit dependence on time in Einstein's theory of gravity? Do you think that the expansion of the universe is not governed by time-independent equations? That is what you are claiming in your answer and your criticism of my answer. Do you really think that? – Philip Gibbs - inactive Dec 19 '14 at 14:29
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2Kyle, do you understand that the condition required for the application of Noether's theorem for energy is that the dynamical equations are independent of time, not that the solutions are independent of time? Is this the source of your confusion? – Philip Gibbs - inactive Dec 19 '14 at 15:03