Conservation of Energy in the Universe at Large

1. May 17, 2014

Islam Hassan

Does the conservation of energy law apply to:

i) The initial impetus that drove the Big Bang;
ii) The inflationary phase of the early universe;
iii) The present expansion and acceleration of the universe?

If not, under what conditions/hypotheses do cosmologists admit exceptions to the conservation of energy.

IH

2. May 17, 2014

Simon Bridge

Last edited by a moderator: Sep 25, 2014
3. May 18, 2014

abitslow

Your question does not have a satisfactory simple answer. Unfortunately, the answer is (as far as we now know):"It depends." I know that is far from helpful, and I wish it were better. So, lets say you accept that answer (to all 3 of your questions), ok? You're going to say (I predict):"It depends on what??". That I can answer: it depends on the cosmological model you choose.
Simon provided you a link to the quote:"The theory of general relativity [GR] leaves open the question of whether there is a conservation of energy for the entire universe." GR isn't just "a theory" it is a theoretical framework (meaning there are some parameters you can pick which will affect the results in profound ways). That's why after we found out that expansion was accelerating we just said "oh, that is interesting!" rather than saying:"OMG! This changes everything!! Toss out our (old) theories, we need a NEW idea!!". There are plenty of posts here about the gravitational potential energy between two cosmologically separated objects (surely it must increase as their distance apart increases!) as well as the more obvious question of where a photon's energy "goes" when it experiences cosmological redshift. (a red photon has less energy than a blue photon).
There are two answers to this question: 1. We must assume that this energy is transformed into something else, let's call it "Dark Energy". OR 2. Yeah, we don't know where it goes, or if it goes anywhere, so lets not assume it "goes" anywhere (it is lost/not conserved). {note that assuming #2 does NOT (necessarily) mean that we can't also invoke something like "dark energy"; but if we do, it will be for other reasons.} Much of theoretical Cosmology is about trying to set up models of our Universe which are self-consistent as well as consistent with our local understanding of Quantum Mechanics, GR, and Thermodynamics. Often we get a theoretical framework and then we play with the adjustable parameters. There are such a huge number of possible Theories of Everything, that I'm pretty sure no one knows how many there are at any one time. My (cynical) guess is there are enough so that each Cosmologist can work on one that no one else is working on and continue to do so for his/her entire career - as one theory goes down in flame, another can be chosen - Star Trek opens with "Space: the final frontier..." We now know that in Physics it's "Theories: the infinite frontier...

4. May 18, 2014

bapowell

There is no conservation of energy principle operative in general relativity (due, in part, to the lack of any globally applicable sense of time evolution). But, there is a conservation of stress-energy, i.e. the covariant 4-divergence of the energy-momentum tensor is zero. This broader conservation principle does not apply to the big bang itself, because this process is not described by or understood within the context of modern observational cosmology. It does, however, apply to your ii) and iii) as these are processes consistently understood within general relativity and quantum field theory.

5. May 19, 2014

Chronos

To put it in simple terms, energy has no global definition under GR - which is a complication. The stress energy tensor is as close as it gets, as noted by bapowell.

6. May 21, 2014

johne1618

Warning: A heretical view about energy conservation in an expanding Universe.

Imagine at the present time $t_0$ we have a pair of co-moving mirrors at a proper distance $L_0$ apart.

As I understand it according to quantum field theory there exist zero-point electromagnetic modes between the plates with energy:
$$E_0 = \frac{1}{2} \frac{h c}{ L_0}.$$
At time $t_1$ the co-moving mirrors are now at a proper distance $L_1 = a(t_1) L_0$.

The energy $E_1$, at time $t_1$, is now given by:
$$E_1 = \frac{1}{2}\frac{h c}{L_1} \\ E_1 = \frac{1}{2}\frac{h c}{a(t_1) L_0} \\ E_1 = \frac{E_0}{a(t_1)}$$
Now one could argue that no work has been done by the electromagnetic modes so that their energy should not decrease.

Perhaps the energy scale itself must increase by a factor $a(t_1)$ in order to ensure that the energy of the modes does not change?

The energy scale is ultimately given by the Planck mass so maybe the Planck mass itself increases by a factor of $a(t_1)$?

7. May 21, 2014

Simon Bridge

That discussion contains a number of potentially misleading statements ... i.e. we do not usually use the term "zero point energy" to describe the energy eigenvalue of the ground state for a single particle in a box (your photon between two mirrors), and you have confused the energy eigenstate with the energy of the particle. i.e. a particle does not have to occupy an energy eigenstate of the system it is in. Further, you have neglected the role of the observer in the description.

Just as in relativity you have to think about who is doing the measuring (i.e. what is the reference frame) in quantum mechanics you have to think in terms of the measurement itself. QM is silent about what happens between measurements.

The situation as described is a standard exercize for introductry quantum physics students - you start with a particle in a box, prepared in the ground state. At some time, the width of the box has increased - this means the particle is no longer in an energy eigenstate: the energy has become uncertain - and the possible outcomes of a subsequent measurement is determined by resolving the old wavefunction against the new eigenstates.

There are wrinkles due to the way that the width increases... in this case, gradually, over time, as a metric expansion of space itself. If you are really interested in that, you should start a new thread. The short answer is that the mechanism for the change is important for figuring where the energy comes from.

Diatribe on what "expanding spacetime" means
The idea of an expanding universe can easily lead to confusion, and this note tries to counter
some of the more tenacious misconceptions. The worst of these is the ‘expanding space’ fallacy.

http://www.roe.ac.uk/~jap/book/expandspace.pdf

Particle in a box with moving walls:
We analyze the non-relativistic problem of a quantum particle that bounces back and forth between two moving walls. http://iopscience.iop.org/1751-8121/46/36/365301

Last edited: May 21, 2014
8. May 22, 2014

Staff: Mentor

There is no need for mirrors or even quantum mechanics at all. We can just consider the cosmic microwave background. Neglecting absorption, the number of photons stays constant, but their wavelength increases and therefore their energy is decreasing.
This is accounted for in the state equations of cosmology, and it is indeed a relevant contribution to the early evolution of the universe.

9. May 26, 2014

Islam Hassan

Thank you everyone for your insightful input. Some questions though:

- I understand that GR is an extremely successful theory in measuring/explaining all types of phenomenon outside of the consideration of black holes. How much of an impediment is the fact that it cannot precisely accomodate the conservation of energy principle? Are there any observations that GR cannot make/predict due to this 'problem'?

- Is the conservation of momentum likewise problematic in GR?

IH

10. May 26, 2014

Staff: Mentor

Excuse me, but there is a simple answer that does not need relativity or quantum physics. Only Newtonian physics is needed.

Conservation of energy is a special case where time translation invariance applies. Actually, you don't need any physics at all for this Noether's Theorum is pure mathematics.

In the general case, the Hamiltonian (or any other expression for the total energy of a system) may have a term that explicitly depends on time. In other words, the answer changes if we choose a different time to label t=0. In the case of cosmology, the scale factor a(t) is the villain. In that case, there is no energy conservation.

It is only the special case (where there is no time term) that conservation of energy applies. Our everyday life is one of those special cases. For example Newton's second law, F=ma, gives identical answers now or one hour from now. Hence, conservation of energy does apply.

11. May 26, 2014

Staff: Mentor

This is not an issue. Why would you need global energy conservation? Locally, energy/momentum is conserved.

How do you define a global momentum? You cannot study the conservation of something you cannot even define properly.

12. May 27, 2014

Islam Hassan

The following statement may be total and utter rubbish...but perhaps naively and at the global level...to ascertain whether (and these are heretical questions):

i) Cosmic energy may be (or appear to be) sourced from 'nothingness'; or
ii) Energy may 'leak' (or appear to leak) off into nothingness from the cosmos.

IH

13. May 27, 2014

Staff: Mentor

I guess you can write that, yes. Once you consider it in terms of the Friedmann equations, for example, everything looks fine.

14. May 30, 2014

George Jones

Staff Emeritus
In general relativity, conservation of energy is not a global concept expressed in terms of integrals, it is a local concept expressed in terms of derivatives, i.e., the stree-energy tensor is divergenceless,
$$0 = \nabla_\mu T^{\mu \nu}.$$
In standard FLRW cosmology, this leads to a very striking equation, reminiscent of conservation of energy and the first law of thermodynamics.

The above equation, together any FLRW metric and Einsitein's equation, leads to
$$0 = \dot{\rho} + 3\left(\rho + p \right) \frac{\dot{a}}{a},$$
where $\rho$ is density, $p$ is pressure, $a$ is the scale factor of the universe.

Multiplying by volume $V = a^3$ gives
\begin{align} 0 &= a^3 \dot{\rho} + 3 a^2 \dot{a} \rho + 3 a^2 \dot{a} p \\ &= \frac{d}{dt} \left( a^3 \rho \right) + p \frac{d}{dt} \left( a^3 \right) \\ 0 &= dE + p dV. \end{align}