Mass/energy in general relativity

[Umaxo]

Hi,

i got little confused after this conversation

It is best to adopt the modern terminology and refer to it simply as the mass. There is no other kind of mass used in the modern lexicon so there is no need to preface it with "rest" to distinguish it. The history of relativistic physics gave rise to terms that we no longer use: relativistic mass, transverse mass (which is identical to relativistic mass), and longitudinal mass.

Yes i know, i am not using the other masses and neither did i come in (serious) contact with them in college (i was just confused how the concept comes into being, without realising it since it doesn't influence any of the calculations i have seen or did). But in my university everyone says "rest mass" - i guess it is just an habbit, or to make it clearer what is meant. I mean, in GR there are more definitions of mass, so i guess it is not that bad idea to use "full name" for them.

In the standard terminology of physics there is only one kind of mass, the ordinary mass. It just causes confusion to call it the rest mass.

in https://www.physicsforums.com/threads/some-questions-about-light-and-relativity.918094/page-2

There was also this conversation:

Hi,

I don't want to spam the topic, so i write directly to you.

You wrote: "In the standard terminology of physics there is only one kind of mass, the ordinary mass. It just causes confusion to call it the rest mass.".

Isnt this valid only for test particles (and som other special cases)? If you want to speak about mass of macroscopic objects, in general you should run into dificulties as i understand.

No difficulties. When you measure the mass of a macroscopic object it includes the energies of the particles that make up that object. In other words, the mass of that object is not equal to the sum of the masses of those constituent particles. This is the lesson of the Einstein mass-energy equivalence.

So i want to clear the confusion.

As was pointed out to me, in STR mass is defined as m²=E² - p², which can be done for all objects. But I was taught, that in GR there are, in general, some problems.

I know it can be done by computation/measurement of 4-momentum in asymptoticly flat region and then using STR definition of mass, but not all solutions of einstein equations are asymptoticly flat.

So are there really no dificulties with mass in GR as is Mister T saying? He says, that "the mass of a macroscopic object it includes the energies of the particles that make up that object", but again i was taught there is problem with this approach, because there is no gravitational energy in GR, so there is no contribution from gravity to overall mass if one tries to compute it this way. However, gravitation has its inprint on the mass of the object.

And about the question wheter there is indeed only one mass in modern physics - one could imagine observer who doesn't have acces to informations from asymptoticly flat region and has only acces to information near some object where there is "wild" curvature. But he still wants to somehow measure properties of the object, so different definition of mass would be needed for him. So are there some different kinds of masses in GR (i guess they should reduce to same number if p^u can be appropriately defined), or indeed there is only definition m²=p_up^u?

Thanks for all the future replies:)

 

[Vitro]

@Umaxo, I don't think Mister T's comment was in the context of GR for an object of large dimensions in a curved spacetime where there is significant tidal gravity across the object, or an object that's both large and very massive and itself causes significant spacetime curvature on its various parts. I don't know if to such an object one could assign a naive scalar mass, I would think not.

 

[phinds]

... there is no contribution from gravity to overall mass

I'm no expert on this stuff but I believe that is exactly correct. Mass is affected by gravity, but gravity does not contribute to mass. Binding energy in atoms/molecules does contribute to mass.

 

[Mister T]

So are there really no dificulties with mass in GR as is Mister T saying?

What I'm saying is that no difficulty arises from referring to the rest mass as simply the mass.

He says, that "the mass of a macroscopic object it includes the energies of the particles that make up that object", but again i was taught there is problem with this approach, because there is no gravitational energy in GR, so there is no contribution from gravity to overall mass if one tries to compute it this way. However, gravitation has its inprint on the mass of the object.

That is something that someone with a knowledge of GR would have to answer for us. But calling the mass the rest mass won't remove any difficulty in the physics. In fact, my point is that it introduces difficulty for people trying to learn physics.

 

[pervect]

So are there really no dificulties with mass in GR as is Mister T saying? He says, that "the mass of a macroscopic object it includes the energies of the particles that make up that object", but again i was taught there is problem with this approach, because there is no gravitational energy in GR, so there is no contribution from gravity to overall mass if one tries to compute it this way. However, gravitation has its inprint on the mass of the object.

And about the question wheter there is indeed only one mass in modern physics - one could imagine observer who doesn't have acces to informations from asymptoticly flat region and has only acces to information near some object where there is "wild" curvature. But he still wants to somehow measure properties of the object, so different definition of mass would be needed for him. So are there some different kinds of masses in GR (i guess they should reduce to same number if p^u can be appropriately defined), or indeed there is only definition m²=p_up^u?

Thanks for all the future replies:)

What sort of mass are we talking about? The ADM, Bondi, and Komar masses are conserved quantities in an asymptotically flat (for ADM and Bondi masses) or stationary (for the Komar mass) spacetime. And they behave as much as Umaxo describes (small details may vary, I haven't read the thread that closely).

Energy density is one of the components of the stress energy tensor. There is a conservation law for the stress energy tensor as a whole, $\nabla_a T^{ab} = 0$, but the energy density itself can and does vary with time. These seem to behave as the other posters are talking about.

So I think there is some talking past each other here.

 

[PAllen]

... The ADM, Bondi, and Komar masses are conserved quantities in an asymptotically flat (for ADM and Bondi masses) or stationary (for the Komar mass) spacetime. ...

Bondi mass, by design, is not necessarily conserved. It excludes radiation reaching (null) infinity, so it decreases over time for a radiating system. In fact, a definition of the energy of radiation including GW is the difference between ADM mass (which is conserved) and Bondi mass.

 

[pervect]

Bondi mass, by design, is not necessarily conserved. It excludes radiation reaching (null) infinity, so it decreases over time for a radiating system. In fact, a definition of the energy of radiation including GW is the difference between ADM mass (which is conserved) and Bondi mass.

That's a good point, though the correct view then gets rather vague. I suppose the old sci.physics.faq does one of the best jobs, http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

"Is energy conserved in General Relativity" is the question and title of the FAQ, and the answer starts out as "In special cases, yes. In general — it depends on what you mean by "energy", and what you mean by "conserved".

This has the advantage of actually being correct, but has the disadvantage of having little if any predictive power without more explanation.

 

[Umaxo]

ADM, Bondi, and Komar masses

I need to check the definitions of these, but due to my exams i don't have enaugh time right now, so please be patient with me:)

But i want ask this, since i think it could be answered without knowing much about the above masses:

I'm no expert on this stuff but I believe that is exactly correct. Mass is affected by gravity, but gravity does not contribute to mass. Binding energy in atoms/molecules does contribute to mass.

Is there anyone who can verify this?

- In this case, mass could be always defined simply from integrating energy-momentum tensor, right? Then the above masses, as i understood, are defined only because we don't know much about details of matter distribution and interactions inside the objects in question, so we need to come to conlusion by indirect methods, like the inprit matter leaves on spacetime geometry. And the reason why there are at least those 3 masses mentioned is because it is not that easy to analyze the inprints. Is that correct?

 

[PAllen]

Is there anyone who can verify this?

- In this case, mass could be always defined simply from integrating energy-momentum tensor, right? Then the above masses, as i understood, are defined only because we don't know much about details of matter distribution and interactions inside the objects in question, so we need to come to conlusion by indirect methods, like the inprit matter leaves on spacetime geometry. And the reason why there are at least those 3 masses mentioned is because it is not that easy to analyze the inprints. Is that correct?

All 3 most common special cases of system mass/energy in GR (Komar, ADM, and Bondi) are affected by gravitational potential energy. The same components each locally measured, at the same temperature, will have lower total mass if they are closer together. The reflects the decrease in potential energy.

Also, ADM mass effectively 'weighs' gravitational radiation of a total system.

So, generally, I would disagree with that statement. What I am guessing it refers to is that the stress-energy tensor T does not include any explicit gravitational contribution. Thus, gravity does not explicitly appear in the source term of the GR field equations. However, the non-linearity of the GR field equations is considered by many to imply 'gravity gravitates' to some extent.

 

[Umaxo]

All 3 most common special cases of system mass/energy in GR (Komar, ADM, and Bondi) are affected by gravitational potential energy. The same components each locally measured, at the same temperature, will have lower total mass if they are closer together. The reflects the decrease in potential energy.

So this is wrong then:

When you measure the mass of a macroscopic object it includes the energies of the particles that make up that object. In other words, the mass of that object is not equal to the sum of the masses of those constituent particles. This is the lesson of the Einstein mass-energy equivalence.

right? Since, in general we don't have gravitational potential energy.

 

[jbriggs444]

So this is wrong then:

I do not know how you see a conflict in those two passages. @PAllen points out that mass lowers as potential energy decreases. @Mister T points out that mass is not additive because energy contributes to it.

 

[Umaxo]

I do not know how you see a conflict in those two passages. @PAllen points out that mass lowers as potential energy decreases. @Mister T points out that mass is not additive because energy contributes to it.

I know there is problem with defining conserved energy on macroscopic scales, because it cannot be made invariant.

Since in general energy is not conserved, i believe this implies you cannot in general define potential energy. Thus you cannot always compute mass as sum of its consituents energies (which i now realize @Mister T didnt wrote). Is this right? Or problems with invariance of conserved macroscopic energy doesn't mean that you cannot make computations?
So, I agree the way @Mister T formulated it (and i guess meant) doesn't contradict what @PAllen said. I guess I just read more than was written.

 

[jbriggs444]

The problem with not computing mass as the sum of the masses of the constituents is far more basic. No mumbling about curved space or global definitions is required.

Take two steel balls. One kilogram each. Individually. This is an invariant measure. No matter how fast or how slow these balls are moving according to any frame of reference you choose to adopt, their individual masses are always 1 kg each.

Now accelerate the balls to some significant fraction of the speed of light relative to one another. Consider the two balls as one combined system. What is the mass of the two balls considered together? It is greater than 2 kg. No matter what frame of reference you choose, at least one of the two balls has kinetic energy that contributes significantly to the combined mass. It turns out that the mass of the combined system computed this way by including the kinetic energy is always the same no matter what frame of reference you use to evaluate it. That is to say that it is "invariant".

Mathematically, $m^2 = E^2 - p^2$. If you have the two kilogram masses and a frame of reference in which their center of mass is at rest (p=0) then $E^2$ is greater than 2 kg because it includes 2 kg from the balls' rest energies plus the contribution from their kinetic energies in this frame. So the combined mass is greater than 2 kg.

Take two protons, two neutrons and two electrons. Weigh them. Put them together into a helium atom and weigh them again. The mass is not the same. The atom weighs less. Mass is not an additive quantity. [Of course, you had to allow a good quantity of energy to radiate away when you assembled that atom -- that's what balances the energy books]

 

[Mister T]

I know there is problem with defining conserved energy on macroscopic scales, because it cannot be made invariant.

Conserved means amount stays the same. Invariant means the same in all reference frames. They are two different things.

This conversation started with a discussion of rest mass. Simply calling it the mass implies nothing about the physics. It's equivalent to rest energy regardless of what you call it. It's a quantity that's invariant but not conserved. Total energy is not invariant but it is conserved.

 

[PeterDonis]

mass could be always defined simply from integrating energy-momentum tensor, right?

No. That definition is the Komar mass; the other two PAllen mentioned (ADM mass and Bondi mass) do not require any knowledge of the stress-energy tensor, only the asymptotic spacetime geometry.

 

[Umaxo]

Total energy is not invariant but it is conserved.

Yes, i wrote nonsense.

I need to think about it a little:) I will get back after my exams in tuesday.

 

[Umaxo]

The problem with not computing mass as the sum of the masses of the constituents is far more basic.

I didnt try to suggest that mass can be computed from masses of its constituens.

If you have the two kilogram masses and a frame of reference in which their center of mass is at rest (p=0) then $E^2$ is greater than 2 kg because it includes 2 kg from the balls' rest energies plus the contribution from their kinetic energies in this frame. So the combined mass is greater than 2 kg.

I was asking, wheter this can be done in general in GR - to compute mass of composite object through energies (not masses) of its constituents. For two particles, the computation in STR should go as $m_{total}^2=p_{total}^2= (p_u+p´_u)(p^u+p´^u)=m^2+m´^2+2p_u p´^u$ in natural units, where m is mass, p is 4momentum. But you cannot directly do this in GR, because every 4-momentum lives on its own tangent space and you cannot simply add them up for macroscopic objects.
You could perhaps parallel transport them to same location and add them up there and compute mass from it. But parallel transport gives different results for different paths, so in this way you wouldn't get invariant mass.

 

[Umaxo]

No. That definition is the Komar mass; the other two PAllen mentioned (ADM mass and Bondi mass) do not require any knowledge of the stress-energy tensor, only the asymptotic spacetime geometry.

But the fact that ADM mass and Bondi mass do not require knowledge of stress-energy tensor doesn't by itself imply that mass cannot be computed from some kind of integral of stress-energy tensor.

I guess i went wrong by talking about stress-energy tensor and integrals. See #17, I think i made it little clearer what my question is.

Or at least if someone could answer this weaker question:

Is it in general possible to define invariant mass, if we know all the information of interrior of macroscopic object (i.e. all the fields, state equations, geometry etc. inside) without referring to asymptotic behaviour, i.e. just by some computation inside the star? - all the masses mentioned require special spacetimes, but if answer to this question is yes, then one could in principle compute invariant mass for any spacetime one can imagine.

 

[PeterDonis]

the fact that ADM mass and Bondi mass do not require knowledge of stress-energy tensor doesn't by itself imply that mass cannot be computed from some kind of integral of stress-energy tensor

Yes, it does, because it means that "mass", without specifying what definition of mass you are using, is not a well-defined term. On one definition of "mass"--Komar mass--you do compute it from an integral of the stress-energy tensor. On other definitions--ADM mass or Bondi mass--you don't. So your question as it stands is not well-defined, since you didn't specify what definition of mass you are using.

And, btw, you also cannot fall back on a general statement that the different definitions will all give the same answer, because in general that is not the case. The different definitions apply in different domains: Komar mass applies to any stationary spacetime, while ADM and Bondi mass apply to any asymptotically flat spacetime. But there are asymptotically flat spacetimes which are not stationary, and there are stationary spacetimes which are not asymptotically flat.

In fact, even in cases where both definitions apply--spacetimes which are both stationary and asymptotically flat--the different definitions don't always give the same answer. The simplest counterexample is Schwarzschild spacetime. It is a vacuum solution, so the stress-energy tensor is zero everywhere, which means the Komar mass is zero. But the ADM mass (and Bondi mass, which for any stationary spacetime is equal to the ADM mass) is $M$, the "mass" that appears in the metric.

Is it in general possible to define invariant mass, if we know all the information of interrior of macroscopic object (i.e. all the fields, state equations, geometry etc. inside) without referring to asymptotic behaviour, i.e. just by some computation inside the star?

Unfortunately, no, it isn't.

 

[Umaxo]

Thanks, that actually answered all of my questions:)

and i also want to thank for all of the replies, they were very helpfull.

 

[Umaxo]

actually, i have one more question:

It is a vacuum solution, so the stress-energy tensor is zero everywhere, which means the Komar mass is zero.

Isn´t there problem with spacetime not being stationary under event horizon? Since you mentioned that is the requirement for Komar mass to be applicable.

 

[PeterDonis]

Isn´t there problem with spacetime not being stationary under event horizon?

You can choose a spacelike slice that does not contain any portion of the region beneath the horizon, and evaluate the Komar mass on that slice (which gives the answer zero). Good question though.

 

[PAllen]

In fact, even in cases where both definitions apply--spacetimes which are both stationary and asymptotically flat--the different definitions don't always give the same answer. The simplest counterexample is Schwarzschild spacetime. It is a vacuum solution, so the stress-energy tensor is zero everywhere, which means the Komar mass is zero. But the ADM mass (and Bondi mass, which for any stationary spacetime is equal to the ADM mass) is $M$, the "mass" that appears in the metric.

Actually, the volume integral formulation of Komar for SC geometry is undefined because of the singularity. There is a surface integral formulation of Komar mass which is equivalent to the volume integral when both are defined. The surface integral is well defined for SC geometry and yields the metric mass parameter.

 

[PeterDonis]

the volume integral formulation of Komar for SC geometry is undefined because of the singularity

Not if you evaluate it on a spacelike surface that does not contain the singularity. The only requirement is that the surface in question must be a surface of constant "time", where "time" is the coordinate whose corresponding integral curves are also the integral curves of the timelike Killing vector field of the spacetime. A surface of constant Schwarzschild coordinate time $t$ that covers the region outside the horizon meets this requirement. (Strictly speaking, for the pure vacuum case, you have to extend the surface from region I into region III of the maximally extended spacetime, i.e., into the second "exterior" region, in order for it to be a Cauchy surface for the spacetime; but this is easily done.)

 

[PAllen]

Not if you evaluate it on a spacelike surface that does not contain the singularity. The only requirement is that the surface in question must be a surface of constant "time", where "time" is the coordinate whose corresponding integral curves are also the integral curves of the timelike Killing vector field of the spacetime. A surface of constant Schwarzschild coordinate time $t$ that covers the region outside the horizon meets this requirement. (Strictly speaking, for the pure vacuum case, you have to extend the surface from region I into region III of the maximally extended spacetime, i.e., into the second "exterior" region, in order for it to be a Cauchy surface for the spacetime; but this is easily done.)

One must be careful to construct a space like surface that is complete and doesn't include either the singularity or pass through a nonsationar region. Specifically, the surface must pass through the origin in Kruskal coordinates, and this surface is not simply connected. But admittedly, such surfaces exist. This seems to lead to a paradoxical situation - the surface integral formulation of Komar mass differs from the volume integral. This would seem to disprove some generalization of Stokes theorem. I wonder if it is physically valid to talk about Komar mass if 'almost all' spacelike slices include either nonstationary region or singularity.

 

[PeterDonis]

One must be careful to construct a space like surface that is complete and doesn't include either the singularity or pass through a nonsationar region.

I think the surface I described meets this requirement; it doesn't pass through either of the non-stationary interior regions of the maximally extended spacetime. It does pass through the origin in Kruskal coordinates, which is a single $\mathbb{S}^2$ on which the vector field $\partial / \partial t$ is, strictly speaking, undefined, but I think that can be dealt with by taking appropriate limits, since it's only a single $\mathbb{S}^2$. (If the surface passed through a finite portion of an interior region, this trick would not work.)

the surface must pass through the origin in Kruskal coordinates, and this surface is not simply connected. But admittedly, such surfaces exist. This seems to lead to a paradoxical situation - the surface integral formulation of Komar mass differs from the volume integral

I don't think it would in this particular case. Yes, the surface in question is not simply connected, because it has topology $\mathbb{S}^2 \times \mathbb{R}$ instead of $\mathbb{R}^3$. But I think the surface integral over any of the $\mathbb{S}^2$ will still give the same answer that the volume integral does. I think the key point is that it is still clear which "side" of the surface is "closer to infinity", so to speak (this is heuristic but I think it can be made rigorous). It would be different if the topology were $\mathbb{S}^3$, since then there would be no natural way to distinguish one side of a given $\mathbb{S}^2$ surface from the other, which is the usual condition under which Gauss's Law-type surface integrals break down.

 

[PAllen]

I think the surface I described meets this requirement; it doesn't pass through either of the non-stationary interior regions of the maximally extended spacetime. It does pass through the origin in Kruskal coordinates, which is a single $\mathbb{S}^2$ on which the vector field $\partial / \partial t$ is, strictly speaking, undefined, but I think that can be dealt with by taking appropriate limits, since it's only a single $\mathbb{S}^2$. (If the surface passed through a finite portion of an interior region, this trick would not work.)
I don't think it would in this particular case. Yes, the surface in question is not simply connected, because it has topology $\mathbb{S}^2 \times \mathbb{R}$ instead of $\mathbb{R}^3$. But I think the surface integral over any of the $\mathbb{S}^2$ will still give the same answer that the volume integral does. I think the key point is that it is still clear which "side" of the surface is "closer to infinity", so to speak (this is heuristic but I think it can be made rigorous). It would be different if the topology were $\mathbb{S}^3$, since then there would be no natural way to distinguish one side of a given $\mathbb{S}^2$ surface from the other, which is the usual condition under which Gauss's Law-type surface integrals break down.

The surface integral form of Komar mass always leads to the mass parameter of an SC BH. It is easy to find discussions of this in the literature, e.g.

http://www.aei.mpg.de/~gielen/black.pdf , page 47

The implication I have always drawn is that the volume integral being zero is simply an invalid application of the volume integral

 

[PeterDonis]

It is easy to find discussions of this in the literature, e.g.

The argument given here appears to be using the volume integral version to start with; it defines the current $J_a = - T_{ab} k^b$, where $k^b$ is the timelike KVF, and then defines the Komar energy as the integral over a spacelike 3-surface of $* J$. That is a volume integral. It then appears to say that this volume integral as it stands cannot be converted into a surface integral, and goes through some manipulations involving the properties of the KVF to define a different current $J'_a = - 2 R_{ab} k^b$, which gives rise to a volume integral that can be converted into a surface integral. The fact that this surface integral is nonzero (it is $M$ for the Schwarzschild case, as noted) appears to indicate that the corresponding volume integral is also nonzero. (The equality here is given in the equation just above definition 6.6.)

I admit this looks fishy on the surface, because the current $J'$, whose volume integral is being taken, is equal by the EFE to an expression involving the stress-energy tensor which should be identically zero in any vacuum spacetime, which would seem to indicate that the volume integral itself should be zero. And yet the surface integral is obviously nonzero. So there would appear to be some missing steps in the argument. I'll take a more detailed look at this source (which I have not seen before) when I get a chance, to see if there's something else I'm missing. (One possibility is that taking the dual via the $*$ operation, which is done in the volume integrals, makes a difference that I haven't taken into account.)

 

[PAllen]

@PeterDonis
Let's get back to basics. The Komar mass surface integral is based on the notion of integrating force around a surface and relating it to mass. In a stationary spacetime, the timelike killing field defines a class of static observers whose proper acceleration may be taken as force, but with discounting relative to normalization at infinity - which again is well defined because of the ability to define a potential for a stationary spacetime. The volume integral coming out the same then flows from stokes theorem or suitable generalizations. A key point is that the surface integral formulation of Komar mass is unaffected, in principle, by the interior - keep the metric on the surface and out to infinity the same, changing the interior at will, and the Komar mass is unchanged. This fits with Birkhoff's theorem in the spherically symmetric case. Thus, replacing a body inside an integrating surface with a BH with the same external geometry at and outside the integrating surface cannot change anything. The outlier here, which I have alway taken to be simply invalid, is the volume integral komar mass for a BH.

Your own analysis can be taken to support this - there is really no valid integration volume for the komar mass. You can avoid the singularity, and with care, avoid the interior, but you cannot avoid the horizon, which is does not have a timelike killing vector field. You try to gloss over this suggesting limits, but I think this defect - absence of a valid volume, helps explain that the result is not meaningful.

 

[PAllen]

@PeterDonis
Let's get back to basics. The Komar mass surface integral is based on the notion of integrating force around a surface and relating it to mass. In a stationary spacetime, the timelike killing field defines a class of static observers whose proper acceleration may be taken as force, but with discounting relative to normalization at infinity - which again is well defined because of the ability to define a potential for a stationary spacetime. The volume integral coming out the same then flows from stokes theorem or suitable generalizations. A key point is that the surface integral formulation of Komar mass is unaffected, in principle, by the interior - keep the metric on the surface and out to infinity the same, changing the interior at will, and the Komar mass is unchanged. This fits with Birkhoff's theorem in the spherically symmetric case. Thus, replacing a body inside an integrating surface with a BH with the same external geometry at and outside the integrating surface cannot change anything. The outlier here, which I have alway taken to be simply invalid, is the volume integral komar mass for a BH.

Your own analysis can be taken to support this - there is really no valid integration volume for the komar mass. You can avoid the singularity, and with care, avoid the interior, but you cannot avoid the horizon, which is does not have a timelike killing vector field. You try to gloss over this suggesting limits, but I think this defect - absence of a valid volume, helps explain that the result is not meaningful.

Here is a funny thought - in the eternal BH spacetime (rather than a post collapse BH), one is required to perform the Komar surface integral for both exteriors in opposite sense, getting M and -M, for total of zero. Never heard of such a thing, but if it could be justified, it would restore consistency as well as provide sort of an explanation.

 

[PeterDonis]

the surface integral formulation of Komar mass is unaffected, in principle, by the interior - keep the metric on the surface and out to infinity the same, changing the interior at will, and the Komar mass is unchanged.

Yes, I agree with this.

Your own analysis can be taken to support this - there is really no valid integration volume for the komar mass.

I agree that just looking at the volume integral, without trying to work out if/how it is related to the surface integral, makes it seem obvious that, formally, the volume integral should be zero for any vacuum spacetime--and since this clearly doesn't make sense for Schwarzschild spacetime, that would seem to indicate that the volume integral is invalid for this spacetime. I also agree that trying to finesse this for the maximally extended spacetime by taking limits in order to deal with the fact that the Killing vector field is null at the origin of Kruskal coordinates might not be valid.

What confuses me somewhat is the fact that the source you linked to, which is specifically about black holes, doesn't appear to mention any of this when it writes down the Komar volume and surface integrals, and then just blithely says that the Komar energy of Schwarzschild spacetime is $M$. That makes me wonder whether there is some subtle point that I'm missing, that makes it OK to take the volume integral in Schwarzschild spacetime and somehow has it work out to be $M$ instead of zero.

in the eternal BH spacetime (rather than a post collapse BH), one is required to perform the Komar surface integral for both exteriors in opposite sense, getting M and -M, for total of zero.

I had thought of this as well--in fact it would actually make a kind of sense if the two exterior regions were connected, since the "direction" of the timelike KVF is opposite in the two exterior regions.