# Is the law of conservation of energy still valid?

Is the law of conservation of energy still valid or have there been experiments showing that energy could be created or lost?

-
It is absolutely valid in our dimensions. When one goes to big bang models and general relativity, conjectures abound. –  anna v Sep 2 '12 at 4:10
Richart, is there some reason you have to suspect it isn't valid? It will help us give a better answer if we can address that specifically. –  David Z Sep 2 '12 at 4:32
motls.blogspot.de/2010/08/… –  user11151 Sep 2 '12 at 6:10
@annav Well in greece, a lot of things are "valid". –  user11151 Sep 2 '12 at 6:23
There is no global energy conservation in cosmology, if that is what you mean. Locally energy is still conserved ($\nabla_\mu T^{\mu\nu}=0$). –  C.R. Sep 2 '12 at 6:45
show 1 more comment

The energy conservation becomes vacuous or invalid in the general theory of relativity and especially in cosmology. See

http://motls.blogspot.com/2010/08/why-and-how-energy-is-not-conserved-in.html

Why and what does it imply? First of all, Noether's theorem makes the energy conservation law equivalent to the time-translational symmetry. In general backgrounds in GR, the time-translational symmetry is broken (especially in cosmology), so the corresponding energy conservation law is broken, too, despite the fact that the energy conservation law (and the corresponding time-translational symmetry) is an unassailable principle in all of pre-general-relativistic physics.

One example of a possible subtlety we have to be careful about: $\nabla_\mu T^{\mu\nu}=0$ holds in GR but because it contains the covariant derivative $\nabla$, this law can't be brought to the equivalent integral form. The extra Christoffel symbol terms explicitly measure how much the energy conservation law is violated at the given point. There's no way to redefine $T_{\mu\nu}$ so that the conservation law would hold with partial derivatives $\partial_\mu$ but the energy would still retain a coordinate-independent value that actually constrains the final state in any way.

If one views the background as variable and appreciates that the underlying laws as being time-translational-invariant, it doesn't help because the time-translational symmetry is a subgroup of the diffeomorphism group which is a local (gauge) symmetry in GR, and all physical states must therefore be invariant under it. The invariance is the same thing as saying that the generator – the energy itself – identically vanishes. So we may declare that there's a conserved energy in GR but it's zero.

We may see the same point if we try to associate energy to gravitational waves. In general spacetimes, we will fail to find a good formula. It's not hard to see why. The total stress-energy tensor comes from the variation of the action with respect to the metric tensor. The variation of the "matter-field" part of the action gives us the matter part of the energy/momentum density. However, the variation of the gravitational part, the Einstein-Hilbert action, gives us an additional term, the Einstein curvature tensor. Of course, the sum of both vanishes – this condition is nothing else than Einstein's equations – because the metric tensor is a dynamical variable in GR and the action has to be stationary under variations of all dynamical fields.

We may also try to invent other definitions of the total energy in general spacetimes. They will either explicitly refuse to be conserved; or they will be identically zero; or they will depend on the chosen spacetime coordinates (in the latter case, it will actually be the case that the whole "beef" of the energy will be just an artifact of the choice of coordinates and there will be no "meaningful piece" that would actually depend on the matter distribution). There's no way to define "energy" in general (cosmological) situations that would be nonzero, coordinate-choice-independent, and conserved at the same moment.

For asymptotically flat or other asymptotically time-translationally-invariant spacetimes, we may again define the total energy, the ADM mass, but it is not possible to exactly say "where it is located" and the cleanest way to determine the ADM mass is from the asymptotic conditions of the spacetime.

Cosmology

In cosmology, the most explicit example of the text above is the FRW uniform and isotropic cosmology. In that case, the total energy stored in dust which has $p=0$, vanishing pressure, is conserved. However, the total energy stored in radiation is decreasing as $1/a$ where $a$ are the linear dimensions of the Universe simply because each photon (or particle of radiation) sees its wavelength grow as $a$ and energy goes like $1/\lambda$ i.e. $1/a$.

There are other states of matter I could discuss such as cosmic strings and cosmic domain walls which obey different power laws. But the most interesting example I will mention is the cosmological constant. It's an energy density of the vacuum. Because the cosmological constant is "constant", this energy density is always and everywhere the same. So because the density is constant and the volume of spacetime grows as $a^3$ in our spacetime dimension, the total energy stored in the Universe grows as $a^3$, too.

Cosmic inflation is driven by a "temporary cosmological constant" so the total energy of the Universe grows with the volume of the Universe, too. In Alan Guth's words, inflation (or the Universe) is the ultimate free lunch. Inflation explains why the mass/energy of the visible Universe is so much hugely larger than the mass scales of particle physics.

For different mixtures of matter obeying different equations of state (roughly speaking, with different ratios of pressure and energy density), one will see the total energy increase or decrease or be constant. Generally, the total energy of the Universe will tend to increase as the Universe expands if the Universe is filled with matter of increasingly negative pressure; the total energy will decrease if it is filled with matter of increasingly positive pressure.

-
+1 Very clear answer!! I've just learned some nice stuff today! –  J. C. Leitão Sep 2 '12 at 8:32
Shouldn't the total energy of the Universe be infinite ? –  jjcale Sep 2 '12 at 17:07
I've already quoted the link. –  user11151 Sep 3 '12 at 1:29
Thanks, JCL. ... Dear Richart, I realize that. Still, I believe that I also have the right to quote my own text about the very same topic. ;-) @jjcale: the total energy of the visible Universe (whose current radius if 46 billion light years) is finite. Whether or not the total energy of the whole Universe is finite depends on whether the Universe (its volume) is finite. We don't know. We just know that if finite, it is very large - the curvature radius of the whole Universe is at least hundreds of billions of light years, way greater than the radius of the visible Universe. –  Luboš Motl Sep 3 '12 at 6:04
@jjcale see physics.stackexchange.com/questions/24017/… –  raindrop Jan 31 '13 at 22:09

Energy has been shown to be conserved in all circumstances where it is currently possible to test it experimentally. It is also conserved according to theory in any system with time translation invariance. This is the case in all known physics including general relativity.

Some people have tried to argue that energy is not conserved in general relativity, or that conservation is approximate, trivial or meaningless. This is not the case. A variety of falacious arguments are used to support non-conservation, e.g. some theorists say that general relativity does not have time-translation invariance because the gravitational field is not invariant. The obvious solution to this is to include the gravitational field as a dynmaical field with its own time-transaltion invariance. Despite this such incorrect arguments have even made their way into textbooks written by well known cosmologists. This is not something where you should rely on the word of authority. Check the maths and the logic yourself.

This subject has been discussed on physics stackexchange several time before so I wont expand on this answer. Suffice to say that energy is conserved in all established physical laws including general realtivity. It is not approximate, or trivial or true only in special cases. See also the discussions at my blog http://blog.vixra.org/category/energy-conservation/

-
Dear @Phil, concerning the article you linked to – which is probably meant as evidence of your otherwise unjustified and invalid propositions – we have already discussed that but if your "conserved total energy" is always equal to zero, then it is trivial, by the very definition of trivial. In every system, you may invent infinitely many such new "conservation laws" – the number of Phil-Gibbs-owned-platinum-comets is also conserved and zero – and because the value is zero, the knowledge of $E_{\rm initial}$ provides us with no information about the final state. –  Luboš Motl Sep 3 '12 at 6:10
If you need to "authoritatively" learn what a trivial conservation law means, see e.g. page 4 of math.uwaterloo.ca/~karigian/papers/noether.pdf - "... in the sense that their difference is a trivial conservation law, meaning that either P itself vanishes on the solutions, or dP/dt vanishes identically." –  Luboš Motl Sep 3 '12 at 6:18
see also vixra.org/abs/1305.0034 for a discussion of why energy is conserved in GR –  Philip Gibbs May 6 '13 at 12:26

It is still valid.

Only two hypothetical exclusions exist:

1) Quantum uncertainty principle. Energy can be uncertain if time is certain and vice versa. So, virtual particles can violate energy conservation law for a small amounts of time. These violations are averaged in normal scales.

2) General relativity model. Universe are not equal over time. Since energy conservation is a consequence of time uniformity, it is possible that it is violated in cosmological scale.

-

I think Lubos said it very well. Phil Gibbs is, I believe, way off track in thinking energy exists in general relativity. It definitely does not, as there is no background space-time against which one can ask whether the laws are time translationally invariant.

-