Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
4  
Here is a related question that might be helpful physics.stackexchange.com/q/2838 –  Hal Swyers Oct 17 '12 at 1:32
5  
The total energy of the universe is not well defined, so we can't even discuss whether it's constant. physicsforums.com/showthread.php?t=506985 –  Ben Crowell Oct 17 '12 at 5:36
2  
How 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 at 6:36

5 Answers 5

up vote 13 down vote accepted
+100

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/

share|improve this answer
2  
The 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. –  Ben Crowell May 6 '13 at 21:06
1  
What 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 5 at 13:03
1  
@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 at 14:10
1  
And does it eventually lead to a conserve quantity ? What happen when done on a closed universe ? –  agemO Dec 17 at 16:09
    
@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 at 19:30

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.

share|improve this answer
    
Why is there a "-" in the stress-energy for a perfect fluid? –  MBN Dec 17 at 16:56
    
@MBN: The metric has a signature of (+,-,-,-) here. –  Kyle Kanos Dec 17 at 17:00
    
This 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 at 22:46
    
@PhilipGibbs: Actually, there is no meaningful definition of "energy density" in general relativity, so there cannot be such a conservation law. I'm updating my answer with a relevant quote from Wald (I presume a similar quote appears in MTW). –  Kyle Kanos Dec 19 at 0:58
    
The 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 at 1:16

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)

share|improve this answer

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

share|improve this answer
    
The 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 at 21:31
    
Also, vixra.org/author/jesus_christ ... Really? –  lionelbrits Dec 17 at 21:52
    
No, the alternative is open peer review, but it is not often an available option. –  Philip Gibbs - inactive Dec 17 at 21:53
    
viXra 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 at 21:56
    
So 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 at 22:02

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.

share|improve this answer
2  
So 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 at 11:33
1  
I 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. –  lionelbrits Dec 18 at 13:31
1  
My 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 at 22:54
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 at 23:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.