grNotesVarPrin

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Notes on the Lagrangian formulation of General

Relativity

These notes are not a substitute in any manner for class lectures. Thesenotes are now, essentially, complete. Please let me know if you find errors.Thanks, Steve Detweiler

I. LAGRANGIAN FORMULATION OF A CLASSICAL FIELD THEORY IN

CURVED SPACETIME

The Lagrangian density L of a classical, massless scalar field is given by

L(ψ, aψ, gab) ≡ L√−g = −1

2

√−ggabaψbψ, (1)

while the action is

S = 4-vol

L(ψ, aψ, gab)√−g d4x. (2)

The assumption that the action is an extremum under arbitrary variations of ψ leads to theEuler-Lagrange equation,

δS ψ =

4-vol

δψ

δL

δψ+ δ(aψ)

δL

δ(aψ)

√−g d4x

=

4-vol

a

δψ

δL

δaψ

√−g d4x +

4-vol

δψ

δL

δψ− a

δL

δaψ

√−g d4x,

= 3−

surf

naδψδL

δaψ |

h

|d3x +

4-vol

δψ δL

δψ − a

δL

δaψ

√

−g d4x, (3)

where for the second equality we have commuted a and δ and integrated by parts, and thethird equality follows by writing the volume integral of a divergence as a surface integral. na

is the unit normal of the boundary of the 4-volume, and hab is the metric of the boundary.And the Euler-Lagrange equation

0 =δL

δψ− a

δL

δaψ

= a

gabbψ

= 0

= aaψ = 0 (4)

is the field equation for the massless scalar field. Note, however, that Eq. (1) is the appro-priate Lagrangian density only if the boundary conditions specify either the value of ψ onthe boundary or that the normal derivative of ψ is zero on the boundary; otherwise, theboundary integrals would not vanish.

We also expect that the action should be invariant under a change in the metric thatresults only from a change in the coordinate system. Thus we evaluate

δS g =

4-vol

δgab δL

δgab+

1

2

δg

gL

√−g d4x

= 4-vol δgab

δL

δgab− 1

2gabL

√−g d4x, (5)

1



where we have used δg/g = gabδgab = −gabδgab, which follows from elementary analysis of the determinant of a matrix. Now, δS g is to vanish for arbitrary δgab which results from a co-ordinate transfortmation. In particular, for xa → xa +ξa, where ξa is an infinitesimal changein the coordinates, δgab = −£ξgab = −2(aξb). Thus, for such a change in coordinates

δS g = 4-vol

−2aξb

δLδgab

− 12

gabL√−g d4x

=

4-vol

2ξba

δL

δgab− 1

2gabL

√−g d4x (6)

where we have integrated by parts, and dropped the surface term with the assumptionthat ξb = 0 on the boundary of the 4-volume. Thus the invariance of the action under aninfinitesimal change in the coordinate system requires that

−1

2

√−gT ab ≡ δLδgab

=√−g

δL

δgab− 1

2gabL (7)

T ab be divergence-free. T ab is called the “stress-energy tensor” of the classical scalar field,

T ab = aψbψ − 1

2gab(cψ)cψ. (8)

II. LAGRANGIAN FORMULATION OF GENERAL RELATIVITY

The Lagrangian density of the gravitational field should be derived from a scalar whichdescribes the geometry of spacetime, and we let

Lg = √−gLg = 116π

(R − 2Λ)√−g (9)

where R is the scalar curvature of spacetime, and Λ is a constant which is usually called the“cosmological constant.” The corresponding action is

16πS =

4-vol

(R − 2Λ)√−g d4x. (10)

This is not the only possible choice for a Lagrangian density, other scalars could be con-structed from combinations of products of the Riemann and Ricci tensors. But, this choice

leads to general relativity. With R = gab

Racb

c

,

16πδLg = δgabRacbc + gabδRacb

c. (11)

In Eq. (27) at the start of §V we show that this δRacbc term reduces to a boundary integral,

and the first term on the right hand side is δgabRacbc = δgabRab It easily follows that the

stress-energy tensor of the gravitational field is

−8π√−gT gravab ≡ δLg

δgab=

√−g

Rab − 1

2gabR + Λgab

(12)

2



And the action is an extremum under arbitrary variations of the the metric if and only if the Einstein equations with a cosmological constant

Rab − 1

2gabR + Λgab = 8πT ab (13)

are satisfied.Exercise 1: Fill in this last step to show that the Einstein equations, Eq. (13), really dofollow from an action principle. Assume that the δRacb

c term, alluded to above, really doesreduce to a boundary integral and can be ignored with an appropriate choice of boundaryconditions.

III. DIFFERENTIAL GEOMETRY WITH TWO METRICS

We begin with an n dimensional manifold with two different metrics, gab and gab, alongwith respective metric-compatible derivative operators,

agbc = 0 and

agbc = 0. First we

see that a − a defines a connection tensor γ cab. Given two arbitrary covariant vectorfields, ξa and λa along with a scalar field α, consider

(a − a)(αξb + λb) = a(αξb + λb) − a(αξb − λb)

= (aα)ξb + αaξb + aλb − (aα)ξb − αaξb − aλb

= αaξb + aλb − αaξb − aλb

= α(a − a)ξb + (a − a)λb. (14)

Thus, we see that a − a corresponds to a linear operator from covariant vectors ontotwo-indexed covariant tensors. As we discussed last semester a linear operator from tensor

fields to tensor fields is, in fact, a tensor field. And, we define the tensor γ c

ab by

aξb = aξb − γ cabξc (15)

for an arbitrary ξb. The rule for the difference of a and a when operating on tensorsof other types follows the same rule as the covariant derivative, partial derivative and theChristoffel symbol.

A constructive expression for γ cab is easily found by following a procedure that is verysimilar to that which we used to find the Christoffel symbols in the first set of notes fromlast semester. From the definition of a metric-compatible derivative operator

agbc = 0 =

agbc

−γ dabgdc

−γ dacgbd, (16)

so that

agbc = γ dabgdc + γ dacgbd,

bgca = γ dbcgda + γ dbagcd,

cgab = γ dcagdb + γ dcbgad. (17)

The second two of these follow from the first with cyclic permutations of the indices. Addthe first two and subtract the third to obtain

agbc +

bgca

− cgab = 2γ dabgdc, (18)

3



where the symmetry of the Christoffel symbols is used. Now, raising the c index results in

γ dab =1

2gdc (agbc + bgca − cgab) . (19)

The relationship between the Riemann tensors for the two metrics also follows the deriva-

tion of the Riemann tensor in terms of the derivatives of the Christoffel symbols in the firstset of notes. Let ξa be an arbitrary vector field, and consider

abξc − baξc = a

bξc − γ dbcξd− γ eab

eξc − γ decξd− γ eac

bξe − γ dbeξd

− b

aξc − γ dacξd

+ γ ebaeξc − γ decξd

+ γ ebc

aξe − γ daeξd

= Rabcdξd − a

γ dbcξd

− γ eabeξc − γ decξd

− γ eacbξe − γ dbeξd

+ b

γ dacξd

+ γ eba

eξc − γ decξd

+ γ ebcaξe − γ daeξd

= Rabc

dξd − a

γ dbcξd

− γ eacbξe − γ dbeξd

+

b γ dacξd + γ ebc aξe

−γ daeξd

= Rabcdξd − a

γ dbc

ξd + γ eacγ dbeξd

+ b

γ dac

ξd − γ ebcγ daeξd

= Rabcdξd − a

γ dbc

ξd + b

γ dac

ξd − γ ebcγ daeξd + γ eacγ dbeξd, (20)

where the first equality follows from the description of the difference of the covariant deriv-atives in terms of γ abc, the second from the Ricci identity for the two different derivativeoperators, the third from the symmetry of γ abc, the fourth from the Leibnitz rule for differ-entiation, and the fifth by rearranging terms. From the Ricci identity we also have

a

bξc

− b

aξc = Rabc

dξd. (21)

With ξa being arbitrary, it is necessary that

Rabcd = Rabc

d − aγ dbc + bγ dac − γ deaγ ebc + γ debγ eac, (22)

after a rearrangement of terms and indices.Eq. (19) and (22) have many different applications such as with conformal transformations

or, of immediate use, for perturbation analysis.

IV. PERTURBATION ANALYSIS

If the two metrics from the preceding section are only infinitesimally different, thenδgab ≡ gab − gab is assumed to be small, and we work only through first order in δgab

throughout this section. From Eq. (19)

γ dab =1

2gdc (aδgbc + bδgca − cδgab) , (23)

note that γ dab is first order in δ. And Eq. (22) becomes

δRabcd ≡ Rabc

d − Rabcd = −aγ dbc + bγ dac, (24)

4



where the terms quadratic in γ abc are second order and are dropped. The contraction of thislast equation on b and d yields

δRac ≡ Rac − Rac = −aγ bbc + bγ bac, (25)

After substitution from Eq. (23), this implies

−2δRac = ddδgac + acδg − abδgbc − cbδgba + 2Rbadcδgbd − 2Rb

(aδgc)b (26)

where δg ≡ gabδgab. This is equivalent to Eq.(35.58a) of MTW[1].

Exercise 2: Show that Eq. (26) follows from Eq. (25).

Solution: From Eq. (25)

−2δRac = a[gbd(cδgdb + dδgcb − bδgcd)] − b[gbd(aδgcd + cδgad − dδgac)]

= ac(gbdδgdb) − b[aδgcb + cδgab − bδgac]

=

a

c(gbdδgdb)

− b

aδgcb

− b

cδgab +

b

bδgac

= bbδgac + ac(gbdδgdb) − baδgcb − bcδgab

= bbδgac + ac(gbdδgdb)

−abδgcb − Rbac

dδgdb − Rbab

dδgdc

−cbδgab − Rbca

dδgdb − Rbcbdδgda

= bbδgac + acδg − abδgcb − cbδgab + 2Rbac

dδgdb − 2R(adδgc)d.

The fifth equality follows from the Ricci identity, and the sixth from symmetries of theRiemann tensor.

Exercise 3: Show that the contracted Bianchi identities follow automatically if the grav-itational action is invariant under an infinitesimal coordinate transformation. HINT: followa similar analysis as that leading up to Eq. (7).

V. CONCLUSION OF THE LAGRANGIAN FORMULATION OF GENERAL

RELATIVITY

Now, we revisit the boundary integrals in the variation of the action of the gravitationalfield which is

part of δS g = 4-vol

gabδRab√−g d4x

=

4-vol

−aγ bba + aγ ab

b√−g d4x

= − 4-vol

∂ a√−gγ bb

a − √−gγ abb

d4x

= − 3-surf

na

γ bb

a − γ abb

|h| d3x. (27)

We use Eq. (25) for the second equality, and versions of Gauss’ law (cf. Eq. (75)) for thirdand fourth equalities. Thus the term which was dropped earlier leading up to Eq. (12) is,

5



indeed a surface integral. Let’s look at this term a little more closely. γ dab defined in Eq. (23)is an infinitesimal quantity, and the surface term is

3-surf

−na

γ bb

a − γ abb

|h| d3x = − 3-surf

nabδgab − gbcnaaδgbc

|h| d3x (28)

after γ dab is removed in favor of δgab. A more interesting form for this surface term followsfrom a tedious analysis which is very similar to that between Eqs. (87) and (95). ( To derive

this expression from Eq. (95) make the following substitutions: N → 1, r → na, σab → hab,D → , hab → gab, and δhab → δgab.) The result is that the final surface term

− 3-surf

nahbc (bδgca − aδgbc)

|h| d3x = − 3-surf

δ(hab)anb + δ(hab)habcnc

|h| d3x

+ 2δ

3-surf

habK ab

|h| d3x

. (29)

Recall that = ±1, depending upon whether na is spacelike or timelike, and that theextrinsic curvature of the boundary K ab = ha

chbd

cnd. This expression shows directly how

the boundary conditions and the action are related. If the action is modified to be

16πS =

(R − Λ)

√g d4x − 2

3-surf

habK ab

|h| d3x, (30)

then the variation of the additional boundary integral ultimately cancels out the “total”variation in Eq. (29). Only the first surface integral on the right hand side of Eq. (29) wouldremain. This modified action would then be appropriate for boundary conditions whichspecify just the 3-metric hab on the boundary, which would make the remaining surfaceterm vanish. This choice is a version of Dirichlet boundary conditions. Note that thisis a weaker condition than specifying the entire 4-metric on the boundary, which is oftenerroneously said to be required.

Also, recall that in a Lagrangian formulation of classical mechanics, the coordinates qi areheld fixed at the end points while the action is being extremized. The dynamical equationsfollow as a consequence. In that same spirit, it is apparent that the choice of Eq. (30) forthe action implies that hab should be held fixed at the “end point”, i.e. on the initial andfinal spacelike hypersurfaces. Thus, out of the entire metric, only hab is seen to be playingthe role of the “canonical coordinates” qi.

As an alternative to Eq. (30), we might have chosen

16πS =

(R − Λ)

√g d4x −

3-surf

habK ab

|h| d3x (31)

(Note that there is no factor of 2 in front of the boundary integral.) for the action, then theboundary integrals remaining after the variation are

−

3-surf

δ(hab)K ab + δ(hab)habK c

c

|h| d3x + δ

3-surf

habK ab

|h| d3x

=

3-surf

−δ(hab)K ab − δ(hab)habK c

c + δ(hab)(K ab − 1

2habK c

c) + habδK ab

|h| d3x

=

3-surf

−3

2δ(hab)habK c

c + habδK ab

|h| d3x

= 3-surf habδ |h|3/2K ab |h|−1 d3x (32)

6





The action principle now becomes

δS =

t2t1

δ( pq − H )dt

= t2t1

pδq + qδp − δp

δH

δp − δq

δH

δq

dt

=

t2t1

d

dt( pδq) − ˙ pδq + qδp − δp

δH

δp− δq

δH

δq

dt

= [ pδq]t2t1 +

t2t1

−δq

˙ p +

δH

δq

+ δp

q − δH

δp

dt (38)

The action S is an extremum under an arbitrary change in path through phase space (withfixed endpoints) if and only if Hamilton’s equations,

˙ p = −δH

δq and q =

δH

δp (39)

are satisfied. These are the dynamical equations of classical mechanics.A Hamiltonian approach to classical mechanics relies upon the use of a “time” function.

This goes against the grain for general relativity, but the 3+1 formalism does an adequate job of treating gravity in a manner which is suitable for the Hamiltonian formalism. To findthe Hamiltonian for gravity we start with (We are dropping the cosmological constant formy convenience in trying to get these notes correct.)

16πL = R (40)

and rewrite it in terms of the 3+1 variables. One immediate complication is that R involvessecond order derivatives of the metric; hence, we use a more convenient description of R,

R = 2nanb

Rab − 1

2gabR

− 2nanbRab (41)

From the Hamiltonian constraint

2nanb

Rab − 1

2gabR

= R − K abK ab + K 2, (42)

and from the Ricci identity

naRcac

bnb = na(canc − acnc)

= c(naanc) − (cna)anc − a(nacnc) + (ana)cnc

= c(naanc) − K acK ac − a(nacnc) + K 2; (43)

also√

g = N √

h. Altogether,

16πL = R = R − K abK ab + K 2 − 2−K acK ac + K 2 − a(nacnc) + c(naanc)

= R − K abK ab + K 2 + 2K acK ac − 2K 2 + 2a(nacnc) − 2c(naanc)

=

R+ K abK ab

−K 2 + 2

a(na

cnc)

−2

c(na

anc); (44)

8



here R and Rab are the scalar curvature and the Ricci tensor of constant time hypersurfacewhose metric is hab.

For a Hamiltonian treatment of General Relativity we must first identify the “coor-dinates.” From the Lagrangian formulation and with the machinery of the initial valueformalism behind us, as a reasonable first try, we let the “coordinates” be the 3-metric hab

of a spacelike hypersurface. hab along with the lapse N and shift vector β a provide all tencomponents of the spacetime metric gab. In addition, we recall that the extrinsic curvatureis related to the time derivative of hab by (note that = −1)

hab ≡ £thab = −2NK ab + £β hab = −2N K ab + Daβ b + Dbβ a (45)

so that

K ab =1

2N

−hab + Daβ b + Dbβ a

. (46)

Next, First we must find the “momenta” conjugate to the “coordinates.” Note that thetime derivatives of the lapse, N , and shift vector β a don’t appear in the Lagrangian, and

these quantities have no conjugate momenta. This is consistent with the implication fromthe Lagrangian formulation that the “coordinates” should be hab and not include the lapseand shift. Below we see that the lapse N and shift β a are Lagrange multipliers which enforcethe constraint equations.

We are now prepared to find the “momenta” P ab ≡ πab/16π which are canonically conjugate to the “coordinates” hab.

P ab =δL

δhab

=δL

δK cd

δK cd

δhab

= − δLδK cd

1

2N δac δb

d = − 1

2N

δLδK ab

. (47)

We ignore the boundary terms in determining the canonical momenta, as those are controlled

by boundary conditions, not by dynamical equations.Exercise 4: Perform the last step just described and show that

πab = 16πP ab = −√

h(K ab − habK ). (48)

Note that πab is not a tensor, but rather a tensor density because of the factor of √

h. Inthe following I find it easiest to perform the usual tensor manipulations on the particularcombination πab/

√h which is a tensor.

It is now convenient to rewrite many of the 3+1 equations in terms of πab rather than interms of K ab. For starters:

K =

π

2√h, (49)

K ab = − 1√h

πab − 1

2habπ

(50)

and

K abK ab =1

h

πabπab − π2 +

3

4π2

=

1

h

πabπab − 1

4π2

. (51)

The Hamiltonian constraint is now

R − K abK ab + K 2 = R − πabπab

h+

π2

2h≡ N , (52)

9



and the momentum constraint is

Db(K ab − habK ) = −Db(πab/√

h) ≡ −Ba. (53)

A dynamical equation is

£thab = −2NK ab + Daβ b + Dbβ a

=2N √

h

πab − 1

2habπ

+ Daβ b + Dbβ a ≡ Gab. (54)

After a lengthy calculation, it may be verified that the “evolution equation,” from theEinstein equations is

£tπab = −N h1/2

Rab − 1

2habR

+

1

2N h−1/2hab

πcdπcd − 1

2πc

cπdd

− 2Nh

−1/2π

ac

πc

b

−1

2πc

c

π

ab+ h

1/2 D

a

D

b

N − h

ab

D

c

DcN

+ h1/2Dc

h−1/2πabβ c

− πacDcβ b − πbcDcβ a + 8πN h1/2S ab

≡ P ab + 8πN h1/2S ab (55)

for the time derivative of πab; this also defines P ab for future use.At this stage, the Lagrangian density along with the “coordinates” hab and their “mo-

menta” πab have been identified. The next step is to determine the Hamiltonian from

16π

Σ

Hd3x = 16π

Σ

P abhab − L

d3x. (56)

The right hand side here is Σ

πab 2N √

h

πab − 1

2πhab

+ 2πabDaβ b − R + K abK ab − K 2

N

√h

d3x

=

Σ

πab 2N √

h

πab − 1

2πhab

+ 2πabDaβ b −

R +

πabπab

h− π2

2h

N

√h

d3x

=

Σ

−R +

πabπab

h− π2

2h

N

√h + 2πabDaβ b

d3x (57)

where we have used Eq. (44) (without the divergence terms) and Eqs. (49)-(53). Now weintegrate the β a term by parts

Σ

2πabDaβ bd3x =

Σ

2Da

πab

√h

β b

√hd3x −

Σ

2β bDa

πab

√h

√hd3x. (58)

The first term on the right is a divergence term, may be written as a surface integral, andwe will discard it for the time being—we consider all of the surface terms later. Finally wehave for the Hamiltonian H

16πH = 16π Σ Hd3x = Σ −R +πabπab

h− π2

2hN √

h − 2β bDaπab

√h d3x, (59)

10



up to possible surface terms which we might wish to include.Note that

16πδH

δN =

Σ

−R +

πabπab

h− π2

2h

δN

√hd3x, (60)

so that the Hamiltonian is an extremum with respect to arbitrary variations of the lapse

δN , if and only if the Hamiltonian constraint is satisfied.Note that

16πδH

δβ a=

Σ

−2δβ bDa

πab

√h

d3x, (61)

so that the Hamiltonian is an extremum with respect to arbitrary variations of the shift δβ a,if and only if the momentum constraint is satisfied.

VIII. SUMMARY OF CURRENT STATE

Our goal is to formulate a Hamiltonian approach to general relativity, which pays properattention to “boundary terms”. At this point what is lacking is the treatment of these veryboundary terms. It is most convenient, now, to define a first guess for H as in Eq. (59),then to derive the equations of motion while carefully tracking all boundary terms. At theend, we will be able to redefine H by consideration of these same boundary terms.

Before proceeding on this task, it is useful to consider how the geometry on a submanifold(a boundary, in our application) changes under a perturbation of the metric of the fullmanifold. This is covered in the following section.

IX. VARIATIONS OF GEOMETRICAL QUANTITIES IN 3+1 LANGAUGE

The boundary of Σ is assumed to be a two-dimensional surface upon which some givenscalar field r is constant. The unit normal to the boundary is ra, and the 2-metric of theboundary is σab with determinant σ. The function ρ (defined below) plays the role of the“lapse function” for this boundary two dimensional boundary as embedded in the threedimensional Σ.

The variation of the Hamiltonian is manipulated using the following conventions:

δhab ≡ δ(hab), (62)

δhab ≡ hachbdδhcd = −δ(hab), (63)

h ≡ det(hab), (64)

and δh/h = hab δhab = δhaa

. (65)

The unit normal to a surface defined by a scalar field r = constant is

ra ≡ ρDar, (66)

whereρ−2 ≡ habDar Dbr. (67)

Here r is not necessarily related to the radial coordinate of flat space. The metric of theboundary, as well as the projection operator onto the boundary is

σab

≡hab

−rarb (68)

11



and the determinant of σab is σ. Also,

Darb = σ(acσb)

dDcrd − ρ−1raσbd Ddρ. (69)

Variations of the geometry at the boundary, with the location of the boundary fixed,

obey δ(ra) = δρDar = raδρ/ρ, (70)

andraδ(ra) = −δ(ra)ra = δρ/ρ (71)

so thatδra ≡ δ(ra) = σb

aδrb − raδρ/ρ. (72)

Now from δ(σabrb) = 0 and Eq. (70) it follows that

δ(σab)rb = 0. (73)

And the variation in hab may be related to the variation of σab via Eq. (68),

δ(hab) = δ(σab) + raσcbδrc + rbσc

aδrc − 2rarbδρ/ρ. (74)

Before starting on the variation of the Hamiltonian it is useful to note a variety of formsof Gauss’ law:

DbAb

h1/2 d3x =

∂

∂xb(h1/2Ab) d3x =

(Dbr) Abh1/2 d2x

= rbAb

ρh1/2 d2x = rbAbσ1/2 d2x (75)

where we use Eq. (66) andh = σρ2. (76)

X. THE VARIATION FINALLY BEGINS

The starting point of the Hamiltonian formalism is the definition

16πH 0 ≡ −

N

R + h−1

1

2πa

aπbb − πabπab

+ 2β aDb(h−1/2πa

b)

h1/2 d3x. (77)

With R = habRab,

δR = δ(hab)Rab + habδRab = −δhabRab + habδRab

= −δhabRab + hab(DcD(aδhb)c − 1

2DaDbδhc

c − 1

2DcDcδhab)

= −δhabRab + DaDbδhab − DaDaδhcc (78)

the second equality follows from an intermediate step in the derivation of Eq. (26), above,which you came across in your solution to Exercise 2.

12



The variations of the different parts of the volume integral are

h−1/2δ

N Rh1/2

= δN R − N δhab(Rab − 1

2habR) + δhab(DaDbN − habDcDcN )

+ Da(N Dbδhab) − Da(NDaδhbb)

− Da(δhabDbN ) + Da(δhbbDaN ), (79)

h−1/2δ

N

h

1

2πa

aπbb − πabπab

h1/2

=

δN

h

1

2πa

aπbb − πabπab

− N δhc

c

2h

1

2πa

aπbb − πabπab

+N

h

πc

cπabδhab − 2πacπcbδhab

+

N

h

δπabhabπc

c − 2δπabπab

, (80)

and

h−1/2δ

2β aDb

πa

b/h1/2

h1/2

= 2δβ aDb(πab/h1/2) − 2δπabDaβ b/h1/2

− δhab

2πbcDcβ a/h1/2 − Dc

β cπab/h1/2

− Da

β aπbcδhbc/h1/2

+ Da

2β bδπab/h1/2

+ Da

2β bπacδhbc/h1/2

. (81)

After the substitution of these expressions into δH 0

16πδH 0 = − δN N h1/2 + 2δβ aBah1/2 + δhabP ab − δπabGab d3x

+

Da

β aπbcδhbc/h1/2 − 2β bδπab/h1/2 − 2β bπacδhbc/h1/2

h1/2 d3x

+

−Da(NDbδhab) + Da(NDaδhbb) + Da(δhabDbN ) − Da(δhb

bDaN )

h1/2 d3x; (82)

recall the definitions of N , Ba, P ab and Gab in Eqs. (52)-(55). After the application of Gauss’law, the variation of H 0 becomes

16πδH 0 = −

δN N h1/2 + 2δβ aBah1/2 + δhabP ab − δπabGab

d3x

+

raβ aπbcδhbc/h1/2 − 2raβ bδπab/h1/2 − 2raβ bπacδhbc/h1/2

σ1/2 d2x

+

−N raDbδhab + N raDaδhbb + raδhabDbN − raδhb

bDaN

σ1/2 d2x.(83)

From the Lagrangian formulation of general relativity, we expect that δH 0 should beexpressible in a form where each surface integral is either a total variation or consists of terms containing only the variations of the 3-metric of the boundary. Such terms containδN , δβ a or δσab, but do not contain derivatives of these quantities. We proceed to manipulatethe surface integrals of Eq. (83) into precisely this form.

13



The terms in the surface integrals on the boundary which involve πab are easily rewrittenas

16πδH π∂ 0 ≡

raβ aπbcδhbc/h1/2 − 2raβ bδπab/h1/2 − 2raβ bπacδhbc/h1/2

σ1/2 d2x

=

[raβ aπbcδhbc/h1/2 + 2raδβ bπba/h1/2]σ1/2 d2x

− 2δ

raβ bπb

a/h1/2

σ1/2 d2x

. (84)

This follows from

2δ

(raβ bπb

a/h1/2)σ1/2 d2x = 2δ

β bπachbcDar d2x

=

2raδβ bπb

a/h1/2 + 2raβ bδπab/h1/2 + 2raβ bπacδhbc/h1/2

σ1/2 d2x, (85)

where the first equality depends upon Eq. (76).The terms involving δhab present more of a challenge.

16πδH h∂ 0 ≡ −N raDbδhab + N raDaδhb

b + raδhabDbN − raδhbbDaN

σ1/2 d2x. (86)

The first two terms of the integral in Eq. (86) are [−N rahbcDbδhac + N rahbcDaδhbc]σ1/2 d2x

=

[N raσcbDbδ(hac) − N raσbcDaδ(hbc)]σ1/2 d2x. (87)

The first term on the right hand side of Eq. (87) gives N raσc

bDbδ(hac)σ1/2 d2x =

N raσc

bDb[δ(σac) + raσdcδrd + rcσd

aδrd − 2rarcδρ/ρ]σ1/2 d2x

=

N raσc

bDb[δ(σac) + raσdcδrd − 2rarcδρ/ρ]σ1/2 d2x

= N [−σcbδ(σac)Dbra + Dc(σc

dδrd) − 2ρ−1δρσbcDbrc]σ1/2 d2x(88)

where Db is the two dimensional derivative operator on the boundary, and we made useof Eq. (74). Now, NDc(σc

dδrd) = Dc(N σcdδrd) − σc

dδrdDcN and the divergence herecontributes to an integral of a two dimensional divergence over the two dimensional boundarywhich has no boundary, itself. The result is zero from Stoke’s theorem. Finally, the firstterm in the integral on the right hand side of Eq. (87) is reduced to

N raσcbDbδ(hac)σ1/2 d2x =

[−Nδ(σab)Darb − σa

bδrbDaN − 2N ρ−1δρ σbcDbrc]σ1/2 d2x.

(89)

14



The second term of the right hand side of Eq. (87) is

−

N raσbcDaδ(hbc)σ1/2 d2x = −

N raσbcDa[δ(σbc) + rcσdbδrd + rbσd

cδrd − 2rbrcδρ/ρ]σ1/2 d2x

=− N raσbcDa[δ(σbc) + 2rcσd

bδrd]σ1/2 d2x

= −

N raDa[σbcδ(σbc)] − raδ(σbc)Da(rbrc) + 2raσbcDa(rcσd

bδrd)σ1/2 d2x

= −

N raDa[σbcδ(σbc)] + 2raσdcδrdDarcσ1/2 d2x. (90)

Now focus on

2δ(σabDanb) = 2δ

1

h1/2

∂

h1/2ra

∂xa

= −δhh

Dara + Da

δhh

ra

+ 2Daδra

= −δhbbDara + Da(δhb

bra + 2δra)

= −δhbbDara + Da(δhb

bra + 2σbaδrb − 2raδρ/ρ)

= −δhbbDara + Da

δ (h/ρ2)

h/ρ2ra + 2σb

aδrb

= −δhbbDara + Da

δσ

σra + 2σb

aδrb

= −h−1δhDara − Da σbcδ σbc ra − 2σb

aδrb . (91)

Substitute this into the result of Eq. (90) to obtain

−

N raσbcDaδ(hbc)σ1/2 d2x

= −

N raDa[σbcδ(σbc)] + 2raσdcδrdDarc

− 2δ(σabDanb) − h−1δhDcrc − Da[σbcδ(σbc)ra] + 2Da(σbaδrb)σ1/2 d2x

=

N [2δ(σabDarb) + h−1δhDcrc + σbcδ(σbc)Dara − 2σc

aDa(σcbδrb)]σ1/2 d2x

=

N [2δ(σabDarb) + h−1δhDcrc − σδσDara − 2σcaDa(σcbδrb)]σ1/2 d2x

=

N [2δ(σabDarb) + 2ρ−1δρDara − 2σc

aDa(σcbδrb)]σ1/2 d2x. (92)

But, N [−2σc

aDa(σcbδrb)]σ1/2 d2x =

[−2Da(N σa

bδrb) + 2σabδrbDaN ]σ1/2 d2x

=

[2σa

bδrbDaN ]σ1/2 d2x (93)

15



from Stoke’s Theorem again using the fact that a boundary has no boundary, itself. Finally

−

N raσbcDaδ(hbc)σ1/2 d2x

= [2N δ(σabDarb) + 2Nρ−1δρDara + 2σabδrbDaN ]σ1/2 d2x

= 2δ

N Daraσ1/2 d2x

+

[(−N σ−1δσ − 2δN + 2Nρ−1δρ)Dara + 2σa

bδrbDaN ]σ1/2 d2x. (94)

Now substituting Eqs. (89) and (94) into Eq. (87) while using Eq. (76), we have N [−raDbδhab + raDaδhb

b]σ1/2 d2x =

[−Nδ(σab)Darb − N σ−1δσDara

+ σabδrbDaN

−2δN Dara] + 2δ NDara d2x (95)

Also using Eq. (74) note that (σa

bδrbDaN +raδhabDbN − h−1δhraDaN )σ1/2 d2x

=

σa

bδrbDaN − ra[δ(σab) + raσbcδrc + rbσa

cδrc

−2rarbρ−1δρ]DbN − (2ρ−1δρ + σ−1δσ)raDaN σ1/2 d2x

= −

σ−1δσraDaNσ1/2 d2x. (96)

After substitution Eqs. (89) and (94) into Eq. (87) and use of Eqs. (76) and (96), the δhsurface terms of 16πδH 0 in Eq. (83) reduce to

16πδH h∂ 0 = 2δ

NDaraσ1/2 d2x

−

N δ(σab)Darb + (2δN + Nσ−1δσ)Dara + σ−1δσraDaN

h1/2 d3x.(97)

Using Eq. (83) with Eqs. (84) and (97) we define

16πH 1 = 16πH 0 + 2raβ bh−1/2πabσ1/2 d2x − 2NDaraσ1/2 d2x, (98)

and the variation of H 1 results in Eq. (101), below.

XI. CONCLUSIONS

At the end of the previous section we defined

16πH 1 ≡ −

(N N + 2β aBa) h1/2 d3x

+ 2naβ bh−1/2πabσ1/2 d2x − 2NDanaσ1/2 d2x (99)

16



where the volume integral is over Σt. The vector na is the outward pointing unit normalto the bounding two-surface, σab is both the metric of this two-surface and the projectionoperator onto the two-surface,

σab ≡ hab − nanb, (100)

σ is its determinant and−

Dana is the trace of the extrinsic curvature of the boundary asembedded in Σt.

An arbitrary, infinitesimal variation of N , β a, hab and πab, which holds the location of the boundary fixed, results in an infinitesimal change in H 1

16πδH 1 = −

δN N h1/2 + 2δβ aBah1/2 + δhabP ab − δπabGab

d3x

−

δσab [NDanb − σabDc (N nc)] + 2δN Dana

d2x

+

naβ ah−1/2πbcδhbc + 2naδβ bh−1/2πab

σ1/2 d2x. (101)

In our analysis, at least one connected component of the boundary is usually in a weak-field zone. And we assume that this component is spherically symmetric with constantCartesian radius r. The outward-pointing unit (flat space) vector, normal to a surface of constant r, is n0

i ≡ 0i r and should not be confused with na which is orthogonal to the

2-surface that bounds Σt and is normalized with hab. The metric of the constant-r 2-sphereembedded in flat space is σ0

ij where

σ0ij = r0

i n0 j (102)

and the determinant of σ0ij is σ0, the 0 distinguishes this from the determinant of σab, the

actual metric of the 2-surface that bounds Σt. The symbol r used as a tensor index on oneof the Cartesian tensors denotes the implied contraction of the index with ni

0 and ∂ r = ni00

i

We assume that at a large distance the shift vector might have an axial component andmay be written as

β i = ΩΦi + S i, (103)

where S j falls off sufficiently fast, and where Φi∂/∂xi ≡ ∂/∂φ.The first surface integral of Eq. (101) may be rewritten as

− r0

δσab [N Danb − σabDc (Nnc)] + 2δN Dana

d2x

= −δ

r0

N 20i ni

0σ1/2 d2x

− r0

δ(σ1/2σab)N (Danb − 1

2σabDcnc) + δ(σ1/2N 2)(N −1Dana − 0

i ni0)

+ 2δ(σ1/2)naDaN

d2x. (104)

The total variation of the surface integral on the right hand side will be absorbed into yetanother definition for H .

17



The term involving naδβ bπab in Eq. (101) is r0

2naδβ bh−1/2πabσ1/2 d2x = δ

2Ω

r0

naΦbh−1/2πabσ1/2 d2x

+ 16πΩ δJ

+ r0 2n

a

δS

b

h

−1/2

πabσ

1/2

d

2

x, (105)

where

16πJ ≡ − r0

2naΦbh−1/2πabσ1/2 d2x; (106)

with the boundary in the weak-field region J is the angular momentum. Again, the surfaceintegral of the total variation on the right hand side here will be absorbed in a new definitionfor H .

The new version of H including the total variations from Eqs. (104) and (105 is

16πH 2 ≡ − (N N + 2β

a

Ba) h1/2

d3

x

− r0

N

2Dana − N 0i ni

0

σ1/2 d2x +

r0

0i ni

0σ1/20 d2x

+

r0

2naS bh−1/2πabσ1/2 d2x. (107)

The second surface integral is a constant 8πr0 that does not change under a variation but isincluded so that the surface integrals of H 2 evaluate to M . The last surface integral vanishesas r0 → ∞.

An arbitrary, infinitesimal variation of N, β a, hab and πab results in an infinitesimal change

of H 2, and

16πδH 2 = −

[E eqns] d3x + 16πΩ δJ

− r0

δ

σ1/2σab

N

Danb − 1

2σabDcnc

+ δ

σ1/2N 2

N −1Dana − 0

i ni0

+ 2δ

σ1/2

naDaN

d2x

+

r0

naβ ah−1/2πbcδhbc + 2naδS bh−1/2πab

σ1/2 d2x. (108)

The symbol [E eqns] is an abbreviation for the first integrand in Eq. (101) which givesboth the constraint equations and the dynamical Einstein equations. Each of the remainingsurface integrals vanish as r0 → 0 for an asymptotically flat metric.

The “value” of H when the field equations are satisfied gives the mass of the system, asmeasured at infinity, in terms of boundary integrals:

16πM ≡ − r0

N

2Dana − N 0i ni

0

σ1/2 d2x +

r0

0i ni

0σ1/20 d2x (109)

and if N is approximately a constant N 0 = 1 in the weak field region then N should bereplaced by N/N 0 in the two places it occurs.

18



You might note that the form for H 2 is rather more complicated than that given by, say,Wald [2], or Poisson [3]. Our form for the Hamiltonian allows for evaluation under modestlyweaker asymptotic requirements than that of the others. But, this is not a substantialimprovement.

[1] C. W. Misner, K. S. Thorne, and J. A. Wheeler, Gravitation (Freeman, San Fransisco, 1973).

[2] R. M. Wald, General Relativity (University of Chicago, Chicago, 1984).

[3] E. Poisson, A Relativist’s Toolkit (Cambridge University Press, Cambridge, 2004).

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