Definition of Energy: Post-Relativity Questions

[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.

 

[PeterDonis]

Under what conditions will the (0,0) element of the Einstein tensor be zero? Can it ever be negative?

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

 

[maline]

What actually has physical implications is the energy density as measured by some particular observer

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

ρρ\rho will only be zero if there is no matter or radiation present at all, and it can never be negative.

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?

 

[PeterDonis]

I thought I had been clear on that.

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.

What I really would like is a definition of rho that does not refer to GR.

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

Energy in the classical sense in only defined in terms of changes

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.

 

[bcrowell]

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.

I am assuming no cosmological constant.

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

 

[PeterDonis]

We do not expect any of the energy conditions to be satisfied under all conditions.

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.

there are many cases where the energy conditions are valid

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.

 

[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.

 

[maline]

"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.

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"?

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

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.

But in relativity, rest mass has energy too; the quantity ρρ\rho includes rest energy, not just kinetic energy, etc. So the statement that (at least for ordinary matter and radiation), ρ=0ρ=0\rho = 0 only if nothing is there has a definite physical meaning; it's not just an arbitrary choice of "zero point".

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".

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

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

 

[bcrowell]

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

This issue doesn't arise with dark energy.

 

[maline]

This issue doesn't arise with dark energy.

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.

 

[PeterDonis]

all coordinate-dependent parameters can be made "invariant" by an appropriate inner product

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.

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.

Sorry, I still don't understand.

I don't see how including rest energy makes the "zero point" more fundamental.

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.

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

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 $g_{ab} u^a u^b = 1$.

I think "dark energy" is more a property of spacetime than something "in" spacetime

You might want to think again. See above.

 

[maline]

Because picking a "zero point" of energy that makes ρρ\rho equal to zero when there is matter or radiation present doesn't make physical sense. Also it violates the Einstein Field Equation

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?

 

[bcrowell]

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?

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?

 

[PeterDonis]

by what criteria, other than the EFE, does it not make sense?

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?

 

[PeterDonis]

how can it be "wrong" without a definition of energy?

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

 

[maline]

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?

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?

 

[bcrowell]

There is another definition of energy: The conserved Noether current due to time translation symmetry.

That doesn't actually work in GR. We have a FAQ about this: https://www.physicsforums.com/threads/what-is-the-total-mass-energy-of-the-universe.506985/

 

[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

That doesn't actually work in GR.

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?

 

[PeterDonis]

I was under the impression that this conservation can also be derived from Noether's theorem. Is this correct?

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

These formulas make no mention of spacetime curvature.

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.

is there a systematic rule for derivation of these formulas based only on properties of the fields?

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.

How is rho determined in this process?

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

 

[maline]

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 gμνgμνg_{\mu \nu}, so it is affected by spacetime curvature.

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 L_matter 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

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.

 

[vanhees71]

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.

They have to do with one of Noether's theorems in this field. Such identities always occur when you have a gauge theory or Hamiltonian systems with constraints. For a nice review, see

http://arxiv.org/abs/hep-th/0009058

 

[PeterDonis]

I am surprised by the fact that the Hilbert SET is gauge- invariant, given that it is constructed from L_matter which includes gauge freedom.

The "gauge freedom" here is just the freedom to choose coordinates. The Lagrangian is a Lorentz scalar, so it is invariant under coordinate transformations.

would this SET still be conserved and gauge- invariant in a hypothetical universe where the spacetime was Miskowski everywhere?

In such a universe the Hilbert SET would be identically zero, because there would be no matter or energy anywhere. I guess that counts as conserved as gauge invariant, but probably not quite in the sense you were thinking.

 

[maline]

The "gauge freedom" here is just the freedom to choose coordinates.

L_matter, being a Lagrangian density, may be arbitrarily changed by any total time derivative. In particular, adding a constant would correspond to changing the zero point of energy. The SET, on the other hand, is uniquely defined. How does this come about from the formula?

In such a universe the Hilbert SET would be identically zero, because there would be no matter or energy anywhere.

That's why I wrote "in a hypothetical universe". Suppose there was no coupling between spacetime & mass-energy. The Hilbert SET would still be well-defined, and useful in SR. That all I was pointing out there (if it's correct).

 

[PeterDonis]

Suppose there was no coupling between spacetime & mass-energy.

Then the laws of physics we know would not apply. It's pointless to ask what the laws of physics say about a scenario in which the laws of physics do not apply.

The Hilbert SET would still be well-defined, and useful in SR.

It can be defined mathematically in SR, by just assuming that spacetime is flat and the metric is the Minkowski metric, yes. But physically, this does not make sense. It is sometimes used as an approximation (when the SET in question has components small enough that the spacetime curvature they produce can be ignored in the particular problem under consideration), but that's all it is--an approximation.

 

[PeterDonis]

L_matter, being a Lagrangian density, may be arbitrarily changed by any total time derivative. In particular, adding a constant would correspond to changing the zero point of energy.

Only if you ignore spacetime curvature. In a curved spacetime, there is a factor of $\sqrt{-g}$ in the integration measure of the action integral. So adding a constant to the Lagrangian density is not just a "zero point" change with no physical effect; it couples to the curvature because of the $\sqrt{-g}$ factor. This is true even in the absence of matter--adding the cosmological constant term to the Einstein-Hilbert Lagrangian changes the field equation.

 

[maline]

So adding a constant to the Lagrangian density is not just a "zero point" change with no physical effect; it couples to the curvature because of the √−g−g\sqrt{-g} factor.

This is exactly my point! The classical definition of the Lagrangian is simply that its variation problem gives the correct dynamic laws. Thus adding a total time derivative produces a Lagrangian that is just as valid as the first. There is no answer to the question "which one is correct?". But now in GR, such a change would apparently produce physical effects! Thus one particular "correct" gauge for the Lagrangian needs to be selected. How is this choice made?

 

[PeterDonis]

The classical definition of the Lagrangian is simply that its variation problem gives the correct dynamic laws. Thus adding a total time derivative produces a Lagrangian that is just as valid as the first.

This is only "classical" in the sense of "non-relativistic". In a relativistic field theory, we work with a Lagrangian density, not a Lagrangian. The Lagrangian density is the Lorentz scalar. The Lagrangian is an integral over "space" of the Lagrangian density--but in a relativistic theory, there is no invariant definition of "space", so the Lagrangian, as opposed to the Lagrangian density, is not an invariant; it's coordinate-dependent.

In other words, the "gauge" problem you are having is a result of trying to use a non-relativistic concept in a relativistic context. The correct fix is to stop doing that. The correct relativistic concept is the Lagrangian density, which is "gauge invariant" in the appropriate sense.

Thus one particular "correct" gauge for the Lagrangian needs to be selected.

No, it doesn't--at least, not if we use the Lagrangian density, as we should (see above). The action integral, as an integral over spacetime of the Lagrangian density, with the $\sqrt{-g}$ factor included in the integration measure, is gauge invariant--it is invariant under any change of coordinates. That is what the $\sqrt{-g}$ factor is for--to ensure that the integral is invariant under any change of coordinates. Without it that would not be true.

 

[maline]

The correct relativistic concept is the Lagrangian density, which is "gauge invariant" in the appropriate sense.

Please elaborate?

 

[maline]

In a relativistic field theory, we work with a Lagrangian density, not a Lagrangian. The Lagrangian density is the Lorentz scalar. The Lagrangian is an integral over "space" of the Lagrangian density--but in a relativistic theory, there is no invariant definition of "space", so the Lagrangian, as opposed to the Lagrangian density, is not an invariant; it's coordinate-dependent.

In other words, the "gauge" problem you are having is a result of trying to use a non-relativistic concept in a relativistic context. The correct fix is to stop doing that. The correct relativistic concept is the Lagrangian density, which is "gauge invariant" in the appropriate sense.

I don't see at all how the fact that we are using Lagrangian densities gets rid of the freedom. If we change the density, say, by some scalar function of spacetime that is independent of the fields, it is clear that when we calculate the action over some compact region, our added function will contribute only a constant and therefore have no effect on the dynamics. What makes this new Lagrangian density incorrect?

Then the laws of physics we know would not apply. It's pointless to ask what the laws of physics say about a scenario in which the laws of physics do not apply.

Is it really pointless to discuss a hypothetical universe where all our laws apply except the EFE, with the spacetime Minkowski everywhere, and of course no gravity? Sounds well-defined to me. Is there something inconsistent about such a picture?

 

[vanhees71]

The action is
$$S=\int \mathrm{d}^4 q \sqrt{-g} \mathcal{L},$$
and if $\mathcal{L}$ is a generally covariant field, the action is a scalar and this implies that the resulting equations of motion are generally covariant. Since $g=\det(g_{\mu \nu})$ is generally not just a constant, changing $\mathcal{L}$ by a constant leads to another model, i.e., other equations of motion. That's why in general relativity the choice of the zero point of the energy density has physical implications. Particularly adding a constant to $\mathcal{L}$ means to add a cosmological constant. Which Lagrangian to choose for a given situation must thus follow other principles, like symmetry principles and finally is subject to empirical justification as any postulated physical law used to derive the equations of motion. See Weinberg, Gravitation and Cosmology or Misner, Thorne, Wheeler for more on this issue.

 

[maline]

Which Lagrangian to choose for a given situation must thus follow other principles, like symmetry principles and finally is subject to empirical justification as any postulated physical law used to derive the equations of motion.

So it's true that GR forces a choice of a particular Lagrangian density, where SR would have allowed an equivalence class of Lagrangian densities? Could you say something about the "symmetry principles" that can be used to make this selection?

 

[vanhees71]

GR doesn't force anything. As I said you need to employ (heuristic) principles to write down the Lagrangian for a given physical situation.

A first guess for generalizing physical laws from special to general relativity is to substitute covariant derivatives, wherever you have a partial derivative in the special theory (using Galileian coordinates). That's however not unique, because covariant derivatives do not commute, while partial derivatives do. That's already a problem for naively translating Maxwell's equations (which are 2nd order equations when written in terms of the four-potential). The Lagrangian, however only involves first-order derivatives, and particularly simple only four-curls, i.e., you have
$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}, \quad \text{with} \quad F_{\mu \nu} = \nabla_{\mu} A_{\nu}-\nabla_{\nu} A_{\mu} = \partial_{\mu} A_{\nu} -\partial_{\nu} A_{\mu}, \quad F^{\mu \nu} = g^{\mu \rho} g^{\nu \sigma} F_{\rho \sigma}.$$
This you plug into the action
$$S=\int \mathrm{d}^4 q \sqrt{-g} \mathcal{L}$$
leading uniquely to the correct (free) Maxwell equations using Hamilton's principle for the variation of the vector field $A_{\mu}$. The variation with respect to $g_{\mu \nu}$ gives you uniquely the correct symmetric (and gauge invariant!) energy-momentum tensor of the electromagnetic field, and this must (within General Relativity) provide the correct local energy and momentum density of the electromagnetic field, while within Special Relativity the energy-momentum tensor is not uniquely defined from Noether's theorem (applied to space-time translation invariance of the Minkowski space), but only total energy and momentum are uniquely defined.

 

[PeterDonis]

So it's true that GR forces a choice of a particular Lagrangian density

No. The point is that, in GR, adding a constant to the Lagrangian density is not just a "change of zero point" with no physical meaning. It is an actual physical change to the model. Physically, it corresponds to adding a cosmological constant (or more generally changing the magnitude of the cosmological constant), which is a physical change that leads to different predictions for physical observables. Mathematically, GR let's you add any constant to the Lagrangian density that you want; but physically, only one value for that constant will give you a model that matches observations.

 

[maline]

GR doesn't force anything. As I said you need to employ (heuristic) principles to write down the Lagrangian for a given physical situation.

A first guess for generalizing physical laws from special to general relativity is to substitute covariant derivatives, wherever you have a partial derivative in the special theory (using Galileian coordinates)

Mathematically, GR let's you add any constant to the Lagrangian density that you want; but physically, only one value for that constant will give you a model that matches observations.

In SR, whenever we write down a Lagrangian density, we are actually referring to the equivalence class of all LD's that differ from this one by a total divergence (or a total time derivative for actual Lagrangians). When generalizing to GR, this equivalence will no longer obtain, so before even beginning to deal with covariant derivatives etc., we must first select one representative of the class as the "correct" one to work with. Is there any general way to make this selection?
It seems that in many cases, the "correct" LD is also already the most familiar one, such as L=−1/4 FμνFμν for classical EM. Perhaps this is because the familiar forms are usually the most simple or symmetric?

 

[vanhees71]

Also in GR the Lagrange density is only determined up to a total four-divergence since
$$\nabla_{\mu} V^{\mu}=\frac{1}{\sqrt{-g}} \partial_{\mu} (\sqrt{-g}V^{\mu}),$$
and thus
$$S'=\int \mathrm{d}^4 q \sqrt{-g} [\mathcal{L}+\nabla_{\mu} V^{\mu}] = \int \mathrm{d}^4 q [\sqrt{-g} \mathcal{L} + \partial_{\mu} (\sqrt{-g}V^{\mu})]= \int \mathrm{d}^4 q \sqrt{-g} \mathcal{L}=S,$$
i.e., the action doesn't change when you change the Lagrangian by a total divergence.