# Definition of energy

1. Feb 15, 2016

### maline

This thread was triggered by @Anonymous Vegetable 's question re. nuclear fusion.

In GR, energy density (in some coordinate system) is a parameter with physical implications- it is the (0,0) element in the stress energy tensor. This is in contrast with the situation in classical mechanics, where only energy differences have physical implications.
In classical mechanics, the zero point of energy was defined more or less for convenience in calculation: for gravity it was when all bodies are at infinite distance (implying negative energy for finite distances), for electromagnetism it was zero field strength (so that electric potential energy was always positive), and in chemistry, "zero enthalpy" was defined arbitrarily for each element.
Post- relativity, the question of how to define the zero point becomes meaningful. Under what conditions will the (0,0) element of the Einstein tensor be zero? Can it ever be negative? I am assuming no cosmological constant.

Last edited: Feb 15, 2016
2. Feb 15, 2016

### Staff: Mentor

This isn't really the correct question, because picking out a particular component of a tensor is a coordinate-dependent operation. What actually has physical implications is the energy density as measured by some particular observer. If we denote that observer's 4-velocity by $u^a$, then the energy density that observer will observe is $\rho = T_{ab} u^a u^b$. In a local inertial frame in which the observer is at rest, we have $u^a = (1, 0, 0, 0)$, so we get $\rho = T_{00}$. But this is only true in that particular frame; in other frames $\rho$ will involve other components of $T$ besides the 0-0 one. But the numerical value of $\rho$, given a particular observer 4-velocity $u$, will be invariant, independent of coordinates.

So the physical question is, under what conditions could $\rho$ be zero, and could it ever be negative? If we restrict ourselves to ordinary matter and radiation, then the answer is that $\rho$ will only be zero if there is no matter or radiation present at all, and it can never be negative. However, this is really just a definition of what we mean by "ordinary matter and radiation"; it doesn't prove that there cannot be some kind of substance for which $\rho$ could be zero or negative when that substance is present. Such a hypothetical substance is often called "exotic matter", and most physicists believe that it cannot exist. One key reason for that is that, if exotic matter existed, it could lead to violations of causality; this is discussed briefly in the Wikipedia page on "energy conditions" (which are conditions that express mathematically what I have been saying in words above):

https://en.wikipedia.org/wiki/Energy_condition

3. Feb 15, 2016

### maline

Yes, that's what I am referring to. I thought I had been clear on that.

What I really would like is a definition of rho that does not refer to GR. Energy in the classical sense in only defined in terms of changes, so what does the quantity really refer to?

4. Feb 15, 2016

### Staff: Mentor

No, you weren't, because "the 0-0 component of the stress-energy tensor" does not mean the same thing as "the energy density measured by a particular observer". The former is a coordinate-dependent quantity; the latter is an invariant, independent of coordinates.

I'm not sure I understand. If you're not using GR as your physical theory, what are you using?

No, energy in the pre-relativistic sense is only defined in terms of changes. But in relativity, rest mass has energy too; the quantity $\rho$ includes rest energy, not just kinetic energy, etc. So the statement that (at least for ordinary matter and radiation), $\rho = 0$ only if nothing is there has a definite physical meaning; it's not just an arbitrary choice of "zero point".

You might be confused by the fact that observers in different states of motion will measure different values of $\rho$ for the same object. This does not change the meaning of $\rho = 0$; a value of zero for one observer will still be a value of zero for every observer.

You might also be confused by the fact that "gravitational potential energy" does not appear in $\rho$. So you can't change the "zero point" of $\rho$ by redefining the "zero point" of gravitational potential energy. Gravitational potential energy, in GR, is really just a convenience; in spacetimes in which it is defined (note that it is not defined in all spacetimes, only in a special class of them), it can help in analyzing free-fall motions such as orbits, and it can also help to heuristically understand the externally measured mass of isolated static systems. But all of these things are global, not local; they don't affect the definition or the physical meaning of $\rho$, which is a locally measured quantity.

5. Feb 15, 2016

### bcrowell

Staff Emeritus
We do not expect any of the energy conditions to be satisfied under all conditions. See "Twilight for the energy conditions?," Barcelo and Visser, http://arxiv.org/abs/gr-qc/0205066 . However, there are many cases where the energy conditions are valid. For example, they are probably valid at all points in stellar evolution.

Why do you want to make an assumption contrary to reality?

6. Feb 15, 2016

### Staff: Mentor

Yes, this is a good point; as a matter of fundamental theory, there is probably no way around having quantum corrections lead to energy condition violations in some regimes.

Yes, including all cases that only involve ordinary matter and radiation in weak gravitational fields. I would expect this to include all cases that the OP has in mind.

7. Feb 15, 2016

### PAllen

Also, a good reason to believe that macroscopically, the energy conditions remain useful, is that (if GR is assumed to be true), energy conditions are a necessary condition for geodesic timelike motion to follow from the field equations. In fact, assuming energy conditions to be false allows demonstration that bodies may move on paths that are spacelike and non-geodesic.

Last edited: Feb 16, 2016
8. Feb 15, 2016

### maline

By the same token, all coordinate-dependent parameters can be made "invariant" by an appropriate inner product. Is this really different from just saying "in some coordinate system"?

What I would like to see is a definition of energy in terms of what's going on within the spacetime, rather than referencing the effect the energy has on the spacetime.

I don't see how including rest energy makes the "zero point" more fundamental. The fact that rest energies tend to be rather high-and positive- makes it easier for a positive energy condition to apply, but in classical terms we could theoretically choose some arbitrary mass density as "zero rest energy".

I am trying to avoid the thorny issues of describing "gravitational energy". First I need to decide what ordinary, nice, local energy is!

9. Feb 15, 2016

### bcrowell

Staff Emeritus
This issue doesn't arise with dark energy.

10. Feb 15, 2016

### maline

Thank you, I didn't know that. Nevertheless, I think "dark energy" is more a property of spacetime than something "in" spacetime, so it's not my focus here.

11. Feb 15, 2016

### Staff: Mentor

Sure, this is one way of giving a physical interpretation to tensor components. But you should not put "invariant" in quotes. The invariants are the actual physical quantities; any correspondence to particular components of particular vectors, tensors, etc. is coordinate-dependent and doesn't contain any physics.

Sorry, I still don't understand.

Because picking a "zero point" of energy that makes $\rho$ equal to zero when there is matter or radiation present doesn't make physical sense. Also it violates the Einstein Field Equation; the condition "no matter or radiation present" corresponds to the RHS of the EFE being zero, which implies $\rho = 0$ for any observer.

As bcrowell said, dark energy is not "gravitational energy"; it has a well-defined local energy density. More precisely, it has a well-defined stress-energy tensor at each event in spacetime; the SET is just a constant (the cosmological constant) times the metric tensor at that event. So the energy density $\rho$ for dark energy measured by an observer with an arbitrary 4-velocity $u$ is just the cosmological constant $\Lambda$, since for any 4-velocity $u$ we have $g_{ab} u^a u^b = 1$.

You might want to think again. See above.

12. Feb 15, 2016

### maline

This is what I'm trying to get at: by what criteria, other than the EFE, does it not make sense? It may be counterintuitive, useless, and just annoying, but how can it be "wrong" without a definition of energy?

13. Feb 15, 2016

### bcrowell

Staff Emeritus
There are in principle three types of mass-energy you can define: active gravitational mass, passive gravitational mass, and inertial mass. Active and passive gravitational mass are equal, as has been verified to extremely high precision in experiments that search for nonconservation of momentum. Inertial mass is equal to gravitational mass, as has been verified to extremely high precision by Eotvos experiments.

If you want to define active gravitational mass, the only definition that works is the Einstein field equation. (Or, in the limit of weak fields and low velocities, you can define it in Newtonian terms, but that amounts to the same thing.)

Did you want to define mass-energy by one of the other definitions?

14. Feb 15, 2016

### Staff: Mentor

Let me turn the question around: by what criteria would you argue that making $\rho = 0$ mean anything other than "no matter or radiation present" does make sense?

15. Feb 15, 2016

### Staff: Mentor

Why do you think there must be a theory-independent definition of "energy"? What physical phenomenon requires that?

16. Feb 15, 2016

### maline

There is another definition of energy: The conserved Noether current due to time translation symmetry. In the LCMF of the system, this is equal to the other three- up to a constant. Is there a way to get rid of the constant before balancing the equation?

17. Feb 15, 2016

### bcrowell

Staff Emeritus
18. Feb 16, 2016

### maline

I am not referring to global energy, only to local conservation as per the Bianchi identities. I was under the impression that this conservation can also be derived from Noether's theorem. Is this correct?
Anyhow, in practice, it is possible to calculate the stress/energy tensor given the various field values, using the appropriate formulas for each field. These formulas make no mention of spacetime curvature. They seem to simply describe "what's there", in a way that would be correct even in Minkowski spacetime if gravity did not exist. Now is there a systematic rule for derivation of these formulas based only on properties of the fields? How is rho determined in this process?

19. Feb 16, 2016

### Staff: Mentor

No. The Bianchi identities have nothing to do with Noether's theorem; they are general geometric identities that hold in any spacetime geometry, not just one with time translation symmetry.

In spacetimes with time translation symmetry, you can in fact use Noether's theorem to define a local "energy tensor"--but it will in general not be the same as the stress-energy tensor that appears in the EFE. See the Wikipedia page on the SET for a brief discussion:

https://en.wikipedia.org/wiki/Stress–energy_tensor#Variant_definitions_of_stress.E2.80.93energy

The formulas in flat spacetime make no mention of spacetime curvature. But in curved spacetime, they certainly do, because (a) they involve the metric, and in curved spacetime you have to use the curved spacetime metric, not the flat Minkowski metric, and (b) they involve derivatives, and in curved spacetime (actually, even in flat spacetime in non-Cartesian coordinates) you have to use covariant derivatives, not partial derivatives.

The Hilbert stress-energy tensor, as described in the Wikipedia article I linked to above, is the systematic rule that gives the SET that appears in the EFE. It works for any field whose field equation can be derived from a Lagrangian. The article gives some examples. Note that the rule involves the metric $g_{\mu \nu}$, so it is affected by spacetime curvature.

The same way I described before, once you have the stress-energy tensor.

20. Feb 17, 2016

### maline

Okay, hopefully now I can finally ask something concrete! I am surprised by the fact that the Hilbert SET is gauge- invariant, given that it is constructed from Lmatter which includes gauge freedom. Is there a simple explanation of what makes this work out? If not, I'd still like to see the derivation.
Also, would this SET still be conserved and gauge- invariant in a hypothetical universe where the spacetime was Miskowski everywhere? That's what I intended with