The question is:

What is the mathematical and/or physical basis for saying that a (static) spacetime manifold with more than one asymptotically flat region at infinity ("end") has a distinct ADM mass for each end?

Whilst there are related/subsidiary questions below, this is what I would most like to understand.

Background and Commentary

Multiplicity of ADM masses is uniformly accepted as true in the physics community but after much research I am unable to understand why it is true: the boundary integral in the ADM formalism seems to me to suppose that a boundary at infinity encompasses everything, and thus that the metric at infinity is determined by everything contained within that boundary... but the mass "closer to another end" is surely causally connected and mathematically contained by the first boundary. [NB, if the manifold is separated into components, one per asymptotically flat end, would those components not be geodesically incomplete, thus singular and thus unsuited to the ADM formalism??]

I can see that in some circumstances, there might be a reason for distinct ADM masses, such as when the components of a spacetime manifold are joined at a singularity (such as the central singularity of a Schwarzschild black hole sitting between the "normal" spacetime and the other half of the maximal analytic extension), but even then it does not seem obvious.

On a related point, Garcıa-Compean and Manko comment, in the context of Kerr/Kerr-Newman spacetimes,

The procedure of the continuation of r into negative values is well described in the classical books on general relativity and leads to appearance of the second asymptotically flat region of negative ADM mass, provided the mass of the first asymptotically flat region is positive definite. The very fact that the same singularity may look as having positive or negative mass depending on a particular asymptotically flat region in which an observer is situated seems rather unphysical, but quite surprisingly this has never been questioned or objected in the literature. [3]

As a further conundrum, consider a spacetime with two ends - such as two empty universes (two "copies" of the same thing to ensure symmetry) joined by a wormhole. Now a wormhole may, in principle, have negative mass, and by symmetry even if distinct ADM masses are allowed, aren't both ADM masses are going to be negative? If so, how is this to be squared with the various positive mass results (e.g. Witten...)

Some Sources

Robert Bartnik has proven that "the mass of an asymptotically flat n-manifold is a geometric invariant" [1]. Although the paper only includes the argumentation for a manifold with a single end he also says that generalisation to multiple ends is straightforward (if you are sufficiently mathematical, which, obviously, I am not).

I took this to mean that a manifold with multiple ends nonetheless has a single mass: if the mass of the manifold is a geometric invariant, surely a manifold can only have one, no matter how many asymptotic ends it has.

However, Hubert Bray, using Bartnik's result, explicitly says in another paper [2]

The above definition for an asymptotically flat manifold is easily generalized to allow for more than one asymptotically flat end, where each connected component of M\K, referred to as an end, is required to satisfy the stated asymptotically flat condition. For example, each Schwarzschild metric with m > 0 has two asymptotically flat ends.... The definition of total mass then applies to each end separately, with each end having its own total mass.

[1] @article{bartnik_mass_1986, title = {The mass of an asymptotically flat manifold}, volume = {39}, url = {http://onlinelibrary.wiley.com/doi/10.1002/cpa.3160390505/abstract}, number = {5}, urldate = {2014-11-07}, journal = {Communications on pure and applied mathematics}, author = {Bartnik, Robert}, year = {1986}, pages = {661--693} }

[2] @article{bray_positive_2011, title = {On the positive mass, Penrose, and {ZAS} inequalities in general dimension}, url = {http://arxiv.org/abs/1101.2230}, urldate = {2015-01-29}, journal = {{arXiv} preprint {arXiv}:1101.2230}, author = {Bray, Hubert L.}, year = {2011} }

[3] @article{garcia-compean_are_2012, title = {Are known maximal extensions of the Kerr and Kerr-Newman spacetimes physically meaningful and analytic?}, url = {http://arxiv.org/abs/1205.5848}, urldate = {2014-11-07}, journal = {{arXiv} preprint {arXiv}:1205.5848}, author = {Garci­a-Compean, H. and Manko, V. S.}, year = {2012} }

  • $\begingroup$ Why would the Hamiltonian evolution conserve anything other than the sum of the ADM masses across all connected subcomponents of the boundary? I also don't see why it's necessarily true that different boundaries will be causally connected. $\endgroup$ Commented Jan 29, 2015 at 16:38
  • $\begingroup$ @JerrySchirmer Thx for the interest. If I understand that correctly, I'm not sure we would disagree: the Hamiltonian would conserve the mass. The question is why that mass is the sum. The integration can be performed over any boundary so why are they not all equal? I didn't say it that boundaries had to be causally connected (the MAE Schwarzschild example is a counter example), the question is really about those cases in which they are causally connected and there seems to be no physical justification for ADM differences. Interesting that no one has jumped in with an obvious "answer" yet $\endgroup$ Commented Jan 30, 2015 at 20:19

3 Answers 3


Firstly, the ADM mass of an end is in no way the sum of the masses of the things in any portion of (or entirety of) the spacetime. Even the ADM mass of a black hole or even a regular star is in no way the sum of the masses of the parts. For instance if you put 10^20 cold iron atoms into a spherical asteroid in an otherwise empty universe, it will have an ADM mass that is less than 10^20 times the mass of an individual iron atom. The ADM mass is not equal to the sum of "other" masses of any kind.

The ADM mass of an end is just a function of the metric in one of the asymptotically flat regions. So you can choose a region of the manifold that topologically looks like $\mathbb{R}^n-B^n$ and where the metric is close enough to the flat metric in spherical coordinates and then compute the metric out there, and then compute the mass (of the end) from that. It isn't related to any other end, or to the compact region $K$.

If you have a manifold with multiple ends, colloquially you can compactify all the ends except one (in a way that doesn't change the mass of that end) and compute the mass of that remaining end. So when Robert Bartnik says that "the mass of an asymptotically flat n-manifold is a geometric invariant" I read that as that the mass didn't depend on a choice of coordinates in which the end is asymptotically flat. And that the generalization to multiple ends and/or geodesically incomplete manifolds just means that the mass of each end is a separate geometric invariant (in the sense that each end has the mass it has regardless of which coordinate system you use of the potentially many coordinate systems in which it looks asymptotically flat).

As for the wormhole and its relationship to positive mass theorems, in my experience only very few proofs of positive mass theorems explicitly state the assumptions of various energy conditions (and sadly even fewer statements of the theorems themselves explicitly mention these assumptions), but to my knowledge every positive mass theorem assumes them. But more importantly don't let the word mass confuse you, the mass of an end is not the sum of some type of primitive masses added up over a region.

Different ends of the same spacetime can have different masses. The mass of an end is not the sum of any basic mass of stuff inside the/a bounding surface of an end.

And I want to make another thing clear about geodesic (in)completeness and the ADM formalism. The ADM mass is simply a definition about the asymptotic behavior of the metric on an end. The metric has the property (or not) regardless of whether it is geodesically complete. If being geodesically incomplete interferes with your proof or your computation, then simply adjust the manifold to make it complete in a way that doesn't change whether it has a mass or what mass it has, and then move on with your life. As a definition it just depends on the end. And it is a property of the spatial metric, it really has nothing to do with evolution or the ADM formalism (again don't let the name confuse you), the definition is just about the (in space) asymptotic (spatial) metric of an end. You don't need a lapse or a shift to talk about the ADM mass of an end.

Edited to include explicit examples with ends with different masses

Robert Bartnik 100% definitely means what I say rather than what you say, because otherwise he'd be wrong because there are examples of manifolds with multiple ends where the ends have different ADM masses. This simply isn't a matter of opinion. A specific metric with two differently massed ends is at the end of this answer.

I will construct a manifold with two ends, each of which has a different ADM mass. First I will develope background techniques, since a reader might need to remedy many misconceptions before I get to my counterexample.

The metrics in question are spatial metrics (so 3d manifolds when you have a 4d spacetime such as in general relativity). But I'm going to start with some 4d manifolds so you can see the exact connection between the theory and general relativity.

Our first 4d spacetime is the maximal time symmetric Schwarzschild black hole, let's use Kruskal-Szekeres coordinates with metric:

$$\frac{4(2M)^3}{r}e^{-r/(2M)}\left(dR^2-dT^2\right)+r^2\left(d\theta^2+\sin^2\theta d\phi^2\right),$$

with $T^2-R^2=\left(1-\frac{r}{2M}\right)e^{r/(2M)}.$ This has a white hole singularity at $T=-\sqrt{1+R^2}$ and a black hole singularity at $T=+\sqrt{1+R^2}$. Event horizons at $T=R$ and $T=-R$. It is geodesically incomplete. A reader is probably familiar with it (or should become familiar). Note that the region with $R>+\sqrt{T^2}$ is causally unrelated to the region with $R<-\sqrt{T^2}$

Now, take the surface where $T=0$, this is a 3d manifold with metric:

$$\frac{4(2M)^3}{r}e^{-r/(2M)}dR^2+r^2\left(d\theta^2+\sin^2\theta d\phi^2\right),$$

and note that it is geodesically complete, and that the region with $R<0$ is still most definitely causally unrelated to the region with $R>0$. Geodesic completeness is unrelated to causal influence when you talk about spatial sections, and even being geodesically complete in a spatial slice tells you nothing about whether you are geodesically complete as a 4d manifold (after you put in a lapse and a shift if even possible).

So let's keep that example in mind, but let's now develop another basic tool. Imagine a flat region of minkowski space, say $\{(\tau,x,y,z): x^2+y^2+z^2<4\}$ and an asymptotically flat Schwarzschild solution $$-(1-\frac{1}{r})dt^2+(1-\frac{1}{r})^{-1}dr^2+r^2\left(d\theta^2+\sin^2\theta d\phi^2\right) \text{ with } r>2.$$

These two solutions can be sewed together, corresponding to to single solution where there is a source term at $r=2$ of positive energy density. If you don't like the singularness of the source term, smooth it out from $r=2-\epsilon$ to $r=2+\epsilon$. The point is that if you place actual positive energy on a spherical shell, then the effective mass "outside" the shell looks bigger (while the inside doesn't notice except for the time dilation between inside and outside the shell). This is very common, every single star in the entire universe, now or ever, has this property. Every planet, every asteroid, etc. Yu can imagine that spherical shell moving in time, but for any fixed Schwarzschild/Minkowksi time it is at some radius, so the spatial metric always looks like a Schwarzschild glued to a Minkowski, or a smoothed version. Smoothing a region that has a Schwarzschild of lesser mass on the inside and larger Schwarzschild mass on the outside is nothing difficult and results in a positive energy density for joining, this is exactly what every normal planet or star does.

So now imagine the solution before $$\frac{4(2M)^3}{r}e^{-r/(2M)}dR^2+r^2\left(d\theta^2+\sin^2\theta d\phi^2\right),$$

(recall that $0^2-R^2=\left(1-\frac{r}{2M}\right)e^{r/(2M)}$).

If we place the right amount of mass or energy density at $r=8M$, then we can set up two coordinate system, the first allows $R_1$ to be positive or negative, has $\theta$ and $\phi$ as normal, and has metric:

$$\frac{4(2M)^3}{r_1}e^{-r_1/(2M)}dR_1^2+r_1^2\left(d\theta^2+\sin^2\theta d\phi^2\right).$$

It holds for every negative $R_1$ and for all $R_1$ up until the first place where $r_1=8M$ when $r_1$ is implicitly defined through $0^2-R_1^2=\left(1-\frac{r_1}{2M}\right)e^{r_1/(2M)}$, i.e. for any $R_1<+\sqrt{3e^4}$. For our other coordinate system we have $R_1$ is positive, has $\theta$ and $\phi$ as normal, and has metric:

$$\frac{4(4M)^3}{r_2}e^{-r_2/(4M)}dR_2^2+r_2^2\left(d\theta^2+\sin^2\theta d\phi^2\right).$$

It holds for large positive $R_2$ up until the first place where $r_2=8M$ when $r_2$ is implicitly defined through $0^2-R_2^2=\left(1-\frac{r_2}{4M}\right)e^{r_2/(4M)}$, i.e. for any $R_2>+e^1$.

If you don't like the discontinuity you can instead add a shell of positive energy density near $r=8M$ so that when $r>8M+\epsilon$ you are exterior Schwarzschild with mass $2M$ and such that when $3M<r<8M-\epsilon$ you are exterior Schwarzschild with mass $M$, and just continue that mass $M$ solution all the way tot he other end. If you disagree about this (all in the region $r>3M>2M=r_s$) then you basically disagree that you can make more massive stars by adding positive energy density material in a shell about the star. It has nothing to do with singularities, geodesic completeness or anything, since the exterior to an actual (uncharged non rotating static stationary) star is Schwarzschild.

So the mass $M$ solution goes all the way to the other side but on this side it transitions to a mass $2M$ solution, one ends has mass $M$, the other has mass $2M$. It is a geometric invariant of each end.

You could also imagine adding some mass, $m$, in one shell, then half that mass, $m/2$, at a larger surface area shell and then adding yet another mass, $m/4$, at an even larger surface area shell and so on so that the mass of the end is increased by $2m$. Each time you only affect the outside, since the you can adjust the end without changing the interior at all. So it is definitely the mass of an asymptotic end, not the mass of some volume.

You fixation of thinking it is about a volume when it is about a limit of surface integrals is a problem even with just one end in the almost simplest situation a regular situation of a normal star with layers of normal positive energy density. And I don't know where you get that idea, the definition itself says it is about a limit of surface integrals. Geodesic completeness is unrelated to anything, we can remove the region where we transition from a star of one mass to a solution of a star of the other mass, so the smoothness of the transition doesn't matter. So all that remains is that ends of different masses can indeed be joined together, exactly as I said.

  • $\begingroup$ I appreciate the time but... ADM mass is supposed to be an integral over a surface that encloses a volume - that's its utility; geodesic completeness just affects what can causally count as inside that volume and hence what can affect the surface values (NB I didn't say anything about trying to add masses of atoms as such). With multiple ends this enclosure could be thought of as enclosing other ends, possibly "compactified" (I thought eg by mapping down to a point). I can't agree with your reading of Bartnik, which seems explicit - if he had meant "of an end" I think he would have said so. $\endgroup$ Commented Mar 17, 2015 at 14:25
  • $\begingroup$ @JulianMoore I thought you were asking a question, so I answered it, and you disbelieve me. So I've edited the question. If you think geodesic incompleteness means causal influence, I give a counterexample. If you think that two ends have to have the same ADM mass, again I give a counterexample. You say that an ADM mass is "supposed" to have something to do with a volume, and you are just 100% wrong. The definition itself is about a limit of surface integrals. I don't want to have to make a sequence of counterexamples by guessing your beliefs, so further conjectures should be explicit. $\endgroup$
    – Timaeus
    Commented Mar 17, 2015 at 19:19
  • $\begingroup$ Re: surface vs volume. From scholarpedia (easier to copy TeX) $ P^0 \equiv E = -\int d^3r \nabla^2 g^T =- \oint dS_i g^T \!_{,i} =\oint dS_i (g_{ij,j} - g_{jj,i}) $ Then in ADM's original paper [1] see 1st Eq 2.1 (surface integral) and then Eq 3.2 on p1002 where it says "all these expressions then reduce to Eq. (2.1). The latter can be converted to a volume integral [Eq. 3.2] ... and the last member is correctly the energy [i.e a volume integral]" [1]R. Arnowitt, S. Deser, and C. W. Misner, Phys. Rev. 122, 997 (1961). Seems explicit to me. Can you explain why am I "100% wrong"? $\endgroup$ Commented Mar 18, 2015 at 22:04
  • $\begingroup$ "@Timaeus" (how do you make that actually appear?) forgot to namecheck you on the last comment... PS Re: "If you think geodesic incompleteness means causal influence" did you mean "geodesic completeness means causal influence" or "geodesic incompleteness doesn't mean lack of causal influence". I'm afraid the original wording doesn't make sense to me. Typo? $\endgroup$ Commented Mar 18, 2015 at 22:31
  • $\begingroup$ @JulianMoore The answerer is automatically notified when you comment to their answer, so the stackexchange software doesn't allow you to waste bytes putting a namecheck (which means we are in irony mode right now). I meant "If you think geodesic completeness means causal influence". For your other comment I will look up the reference when I can get a copy of the paper, but you are 100% wrong because I provided a counterexample. Likely the paper says that in the special case where there is a compact interior to the surface (i.e. just one end of the space) it reduces to the volume integral. $\endgroup$
    – Timaeus
    Commented Mar 19, 2015 at 15:32

OK, here we go:

When I'm doing Hamiltonian analysis of GR, I break the metric $g_{ab}$ into a three metric $\gamma_{ij}$ and a lapse $\alpha$ and a shift 3-vector $\beta_{i}$. Then, if the three-manifold that I am evolving has an outer boundary, then the ADM mass is given by $\int H - H_{o}$, where $H$ is the extrinsic curvature of the outer boundary relative to the 3-surface, and $H_{o}$ is this extrinsic curvature measured relative to some flat space. This term, along with the ADM momentum and angular momentum, is the only non-zero part of the hamiltonian evolution of GR, and it is therefore conserved.

This question doesn't make sense to me, though, because:

1) If the boundary has disconnected subcomponents, we expect the conservation to only hold across the sum of these subcomponents

2) It doesn't make sense to talk about the causal connection of two disconnected subcomponents of boundary within a spacelike surface -- everything is disconnected

3) if the 3-surface suddenly acquires a new disconnected subcomponent to its boundary, then it has changed topology, and the Hamiltonian evolution has ceased to be smooth.

  • $\begingroup$ Thks Jerry; think I can advance things now. Agreed, Hamiltonian~energy/mass & is conserved. Maybe my use of "components" (Schwarzschild MAE as contra-ex) confused things? Think of 2 universes, each with an asymptotically flat infinity; as two disconnected "components" obviously the ADM sum is conserved. Now excise a spatial 3sphere from each and identify the 2sphere surfaces; there is now one manifold with two "ends". Each infinity seems to contain the other, so surely integration over either should give the same result for the total content? Consensus says no, my Q is why? $\endgroup$ Commented Jan 31, 2015 at 9:44
  • $\begingroup$ (continued) How is this reconciled with Bartnik who says ADM mass is a manifold invariant? If it is, then it shouldn't matter over which infinity the ADM is calculated. The joined universe is a perfectly good manifold... isn't it? $\endgroup$ Commented Jan 31, 2015 at 9:46
  • $\begingroup$ @JulianMoore: the joined universe is certainly a perfectly good manifold, though I am wondering if you will run into problems of smoothness or constraint violation at the seam. I need to think about this. $\endgroup$ Commented Feb 1, 2015 at 0:22
  • $\begingroup$ Thanks for taking extra time to think about it. I don't think standard wormhole spacetimes have any such problems; I do think however that some people construct joins and deliberately build in discontinuities (e.g. Visser, a wormhole from joining two Schwarzschild geometries) but their existence doesn't in general seem to me necessary (& makes me wonder if their discontinuities are legit). I look forward to hearing you thoughts on the matter... and I have to say it is reassuring that a PhD relativist needs to think about it - no wonder I've been struggling. $\endgroup$ Commented Feb 1, 2015 at 13:46
  • $\begingroup$ @JulianMoore: Here's my issue with this -- The Hamiltonian and Momentum constraints are a set of elliptic equations on the three-manifold. Therefore, a complete set of $\gamma_{ij}$, $K_{ij}$ and their derivatives on a 2-surface should yield the whole 3-manifold. This should preclude discontinuous things like gluing. But I don't see an obvious reason why, other than that. $\endgroup$ Commented Feb 8, 2015 at 21:18

After further research and thought, I think I can now offer an answer to my own question. I first restate the key question, then state the conclusions and finally comment on them in the context of clarifications of the original “Background and Commentary” arising from the responses of Timaeus and JerrySchirmer, who I thank for their input.

The answer I am about to give is tentative insofar as I believe it to be coherent and well supported, but further input is welcomed. I have avoided using technical terminology except where it is essential and I am confident of its usage – which is not to say that my confidence may not, occasionally, be misplaced.

Further comment is welcome.

The original question was:

What is the mathematical and/or physical basis for saying that a (static) spacetime manifold with more than one asymptotically flat region at infinity ("end") has a distinct ADM Mass for each end?

Notation, Terminology, Key Facts, etc.

Geometric units c= G = 1 are used.

For accessibility, I use “mass” as a loose shorthand for the stress-energy tensor T in Einstein’s field equations; curvature loosely attributed to mass is in fact due to energy, momentum density, shear stress and pressure, without cosmological constant:

$ R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R = {8 \pi } T_{\mu \nu} $

The ADM Mass of a Schwarzschild spacetime equals the Schwarzschild mass. The integral of a sum is the same as the sum of the integrals of the summands, thus the sum of two surface integrals is the same as the integral over the sum (union) of two disjoint surfaces.

In the case under discussion, surface and volume integrals are related by Stokes theorem.

$ \int_{\partial \Omega}\omega=\int_\Omega \mathrm {d}\omega $


There is no such thing as the “ADM Mass” of an individual end in a spacetime with multiple asymptotically flat regions. The limit of the surface integral (as defined by Arnowitt et al 1) that defines “ADM Mass” as the surface of integration tends to spatial infinity at one of several ends may exist, but it does not correspond to the “mass” of anything identifiable.

Phrases such as “the ADM Mass of end X” are informal references to the value of the appropriate surface integral, it being understood – or should be understood – that only the sum has physical meaning.

Loosely speaking – though in a way that can be made precise (see below) – this is because, in a spacetime with multiple ends, the boundary of that spacetime is the union of the distinct spatial surfaces “at infinity.” No individual such surface bounds anything – a non-boundary cannot define a volume – and if there is no volume the surface integral is just a surface integral without the required correspondence to the volume integral. The value of the ADM formalism is specifically that it allows the determination of “mass” from a tractable surface integral (rather than what is probably a very messy volume integral).

The ADM Mass of a manifold is the sum over all ends of the limits of the integral over each end since the union of all spatial surfaces of integration is a boundary. The ADM Mass of an (asymptotically) flat manifold is, per Bartnik 2 an invariant of the manifold (obviously including the metric). The value of the surface integral given by Arnowitt et al cannot be an invariant of any individual end since (generally) mass may be moved around and (specifically) transformations may be applied to compactify individual ends so as to render the value of the integral zero, thereby allowing The ADM Mass to be calculated from one remaining uncompactified end. This is nicely set out and illustrated by Bray [3] (see Theorems 8 & 9). It is important to note that the compactification produces a sphere less a point, but that this singularity is then overcome; in other words, the compactification does not strictly map spatial infinity to a point, as would happen with the full spatial inversion required to give a compact interior for naïve application of Birkhoff’s theorem.

Note - being precise: the ADM Mass is defined as the limit of a particular integral over a closed surface as that surface expands out to the region where the metric is asymptotically flat (subject to certain additional requirements on convergence). In a simply connected spacetime, any closed surface, such as the surface of a sphere, is the boundary of a volume (in the case of a spherical surface, the interior of that sphere; technically an “open ball”). In a multiply connected spacetime this may no longer be true; this was the basis for Wheeler’s idea of “charge without charge.”

As an analogy in good topological style, imagine that the material of a drinking mug (topological torus) contains an electric field, and that there is no field source in the handle itself. If the field is integrated over a surface enclosing the region where one end of the handle joins the rest of the mug, a divergence will be obtained, incorrectly suggesting the presence of a source within that surface (in the handle). However, if the integration is performed over two such surfaces, at either end of the handle, no divergence is found because taken together those two surfaces bound a region that does not in fact contain a field source. Note that this is an analogy; gravity is not a vector field like the electric field.

Similarly, in a space with electric field sources, integration of the field over a non-bounding surface will yield a non-zero result that indicates the presence of charges but it can say nothing about the quantity of charge in any particular place.

In the connected universe scenario of the question, the wormhole joining them is like the handle in the analogy above: no single surface enclosing either end is a boundary of the whole but the union of two such is a boundary. Take the union of two such surfaces and take the sum of the limits of the integrals as each component tends to infinity.

NB For simplicity, most calculations and derivations are now done nowadays with respect to harmonically flat spacetimes and the result applied to asymptotically flat spacetimes on the basis that the latter can be converted to the former with arbitrarily small error.


I was previously confused by the idea of independent masses for individual ends for the following reason. Consider a spherical shell of mass around the junction between two asymptotically flat universes. According to Birkhoff’s theorem the solution to the Einstein Field Equations in a spherically symmetric spacetime has Schwarzschild geometry outside the inner surface of the mass shell and within the mass shell spacetime is Minkowski (actually, the metric is apparently (-k, 1, 1, 1) where k is a constant that can be scaled away).

Le the two universes be A and B. Consider now the normals to the sphere; the net convergence/divergence of normals on a closed surface in a simply connected space distinguishes inside from out. Starting at spatial infinity in A, the normals are seen to converge, A is therefore largely outside the sphere, and by Birkhoff the geometry of A is Schwarzschild and within the sphere the geometry must be Minkowski. However, follow the normals through the shell and they will be found to diverge, indicating that B is outside the sphere, must be Schwarzschild, etc. – a contradiction.

Since there is mass in the A + B system it seemed that A & B could not both be flat, nor by symmetry could they be different, so they should both be Schwarzschild and with the same Schwarzschild mass the ADM Mass calculated by integration over either end should have been the same. However, if the ADM Mass were the sum of the surface integrals this approach would have made ADM = 2 x Schwarzschild – except that it is now clear that the individual integrals do not each contribute 1x Schwarzschild (unless one end is suitably compactified as described by Bray).

I also note that the spaces either side of the mass shell are not compact because they are not bounded and the interior aspect of Birkhoff’s theorem probably does not apply, and the stated scenario also seems to lack the required overall spherical symmetry.


Largely a repeat: per Bartnik, a given manifold and metric has an invariant mass equal to the ADM Mass. It can be calculated even for a manifold with multiple ends, particularly by compactifying all but one (see Bray); if the ends are not compactified, the sum of the surface integrals is the ADM Mass. The value of each integral alone - which need not be equal, the “ADM Mass of an end” (sic), is not meaningful because only the union of the surfaces constitutes a boundary.


  1. R. Arnowitt, S. Deser, and C. W. Misner, Phys. Rev. 122, 997 (1961).
  2. R. Bartnik, Communications on Pure and Applied Mathematics 39, 661 (1986).
  3. H. L. Bray, Journal of Differential Geometry 59, 177 (2001).
  • $\begingroup$ Throughout the entire answer I see references to a spacetime, for instance whether a spacetime is simply connected. But the ADM mass is defined for spatial submanifolds, and you can cut out a region in spacetime that leaves the spacetime simply connected but leaves some its spatial submanifolds not simply connected, for example imagine removing a ball in spacetime that is bounded in time. Simple connectedness of a larger spacetime manifold in which your spatial manifold is embedded is a red herring. $\endgroup$
    – Timaeus
    Commented Mar 30, 2015 at 23:41
  • $\begingroup$ I would not personally go so far as to say that the ADM mass of an end is without physical meaning. For instance you can compare the mass of an end to the result of some of the bounded surface integrals, and those differences are as meaningful as the Schwarzschild mass, it isn't the sum of the masses of anything, but it still means something. When the number changes it signifies something. I also don't know why you want to talk about moving mass around, again, the ADM mass is for a spatial submanifold. $\endgroup$
    – Timaeus
    Commented Mar 30, 2015 at 23:45
  • $\begingroup$ @Timaeus Physical meaning - I don't follow the argument. Can you elaborate? As I read it, the meaningfulness of "mass of an end" seems to be assumed, and if you meant "compare the integrated value of one end with..." I don't see the meaning of the difference. $\endgroup$ Commented Mar 31, 2015 at 8:02
  • $\begingroup$ @Timaeus. Disjoint surfaces - agreed & fixed, thanks $\endgroup$ Commented Mar 31, 2015 at 8:02
  • $\begingroup$ @Timaeus. Re connectedness. Limiting discussion to purely Riemannian spatial (sub)manifolds would be of little value when the larger context is the application of ADM mass in issues of GR, such as in the possibility of creating closed timelike curves by joining two such spacetimes as described at their asymptotic infinities; in which context absence of spatial topology change in time is important, per Geroch's theorem. The larger context had been unstated to avoid confusing the issue. HTH $\endgroup$ Commented Mar 31, 2015 at 8:08

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