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Introduction to Black Hole Physics$

Valeri P. Frolov and Andrei Zelnikov

Print publication date: 2011

Print ISBN-13: 9780199692293

Published to Oxford Scholarship Online: January 2012

DOI: 10.1093/acprof:oso/9780199692293.001.0001

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(p.431) Appendix E Boundary Term for the Einstein—Hilbert Action

(p.431) Appendix E Boundary Term for the Einstein—Hilbert Action

Introduction to Black Hole Physics
Oxford University Press

(p.431) Appendix E

Boundary Term for the Einstein—Hilbert Action

E.1 An Example Illustrating the Problem

Let us consider a massless scalar field ϕ in a D‐dimensional region V of the spacetime with the boundary ∂V ≡ Σ. The action that describes the dynamics of the field ϕ is of the form

S [ ϕ ] = 1 2 V ( ϕ ) 2 g d D x .

Its variation is

δ S [ ϕ ] = V δ ϕ ϕ g d D x V ( δ ϕ ϕ ) g d D x = V δ ϕ ϕ g d D x Σ δ ϕ ϕ ; μ d σ μ .

If we fix the value of ϕ on the boundary, i.e. δϕǀΣ =0, then the variational problem is well defined because the boundary term in Eq. (E.1.2) vanishes. Since the variation δϕ in the bulk is arbitrary we obtain the equation for the scalar field in the bulk

ϕ = 0.

However, it is possible to modify the action by adding an extra boundary term to it. This does not change the variation of the action in the bulk. The difference will only be in the boundary variations. To give an example, let us consider the following action

S α [ ϕ ] = 1 2 V ( ϕ ) 2 g d D x + α Σ ϕ ϕ ; μ d σ μ .

The second term in the modified action can be written as a bulk integral of a total derivative. This gives another (equivalent) form of the action

S α [ ϕ ] = 1 2 V [ ( 1 2 α ) ( ϕ ) 2 2 α ϕ ϕ ] g d D x .

(p.432) The variation of Eq. (E.1.4) gives

δ S α [ ϕ ] = V δ ϕ ϕ g d D x + ( α 1 ) Σ δ ϕ ϕ ; μ d σ μ + α Σ ( δ ϕ ) ; μ ϕ d σ μ .

For α = 0 the action Eq. (E.1.4) reproduces Eq. (E.1.1) and the variational problem is well defined for the Dirichlet boundary condition δϕǀΣ =0. For α = 1 the variational problem is also well defined, but for the Neumann boundary condition n μ(δϕ);μǀΣ = 0. In the general case, in order to make the variation procedure self‐consistent one needs to impose the following condition

[ ( α 1 ) δ ϕ + α n μ ( δ ϕ ) ; μ ] Σ = 0.

The lesson of this example is that the boundary term in the action and the boundary conditions are related to each other and have to be consistent. Natural boundary conditions are those that imply the vanishing of boundary terms in the first variation of the action functional for a free problem, that is when functions are not subject to constraints on the boundary. As stated by Courant (1943) “Now the dominant fact is: appropriate boundary conditions for differential equations are obtained as natural boundary conditions of corresponding variational problems.”

E.2 Boundary Term for the Einstein–Hilbert action

The Einstein–Hilbert action contains second derivatives of the metric, and thus it is similar to the action (E.1.5) for the scalar field. It is more involved but the algorithm for treatment of the boundary terms is quite similar to the scalar case. In most physical problems the metric on the boundary of the manifold is assumed to be given, i.e. the Dirichlet problem for the metric is usually considered. In this case the variation of the metric at the boundary is set to be zero δg αβ |Σ = 0. To be more accurate it is the intrinsic (D −1)‐metric on the boundary that is to be fixed rather than all the components of g αβ. In order for the Dirichlet boundary conditions to be natural boundary conditions one has to determine the form of the boundary term in the action. This means that the normal derivatives of the metric variations at the boundary should not appear in the variation of the total action, including the surface contribution, because otherwise the variational problem will not be consistent.

In the variation of the Einstein action Eq. (5.2.11) only the term g g α β δ R α β contains second derivatives of the metric variation. After integration they will be boiled down to a surface integral that contains not higher than the first derivative variations. Written explicitly these surface terms are

δ S [ g ] = c 3 16 π G Σ g α β ( δ g μ α ; β δ g α β ; μ ) d σ μ + .

Here,… denote terms proportional to δg αβ in the bulk and on the boundary. We have to add to the Einstein action Eq. (5.1.6) a surface integral, variation of which cancels the first term in Eq. (E.2.1). Let us determine this integral. On the surface Σ we have at hand the metric itself, the unit normal vector n μ, and their first derivatives. Variations of other possible terms are of higher order in derivatives of δg αβ, so we do not need to include them.

(p.433) The obvious candidate for the surface integral, which is linear in derivatives, has the form

Σ n α ; α n μ d σ μ .

The orientation is defined by the requirement that n μ be the unit outward‐pointing vector field at the boundary. The surface element is

d σ μ = ε ( n ) n μ | h | d D 1 y .

Here, y i are the coordinates on the boundary, h ij is the induced (D − 1) metric on the boundary, and h = det(h ij). We exclude a special case when the boundary surface is null, so that the normal vector n μ can be either space‐like n α n α = +1 or time‐like n α n α = − 1

ε ( n ) = n α n α = ± 1.

The variations of n μ and μ are

δ n μ = ( 1 2 ε ( n ) n μ n α g μ α ) n β δ g α β , δ ( d σ μ ) = 1 2 g α β δ g α β d σ μ .

The first relation here trivially follows from the expression

n μ = g μ ν f , ν / ε ( n ) g α β f , α f , β ,

provided the equation f (x) = const defines the boundary surface. Now consider the variation of n α ; α

δ ( n ; α α ) = ( δ n α ) ; α + δ ( Γ α β α ) n β .

Taking into account that μ is proportional to n μ one can show that

Σ δ ( n α n μ d σ μ ) = 1 2 ε ( n ) Σ [ g α β δ g α μ ; β + ( g α β ε ( n ) n α n β ) δ g α μ ; β ] d σ μ + ε ( n ) Σ δ ( Γ α β α ) d σ β + = 1 2 ε ( n ) Σ ( g α β ε ( n ) n α n β ) δ g α μ ; β d σ μ + 1 2 ε ( n ) Σ g α β ( δ g μ α ; β δ g α β ; μ ) d σ μ .

One can see that the last term in this expression, when taken with the proper coefficient, is exactly the same as the variation Eq. (E.2.1). Thus, by subtracting it one can ‘kill’ the dangerous terms containing normal derivatives of the metric. The other term in this expression, which accompanies it, survives on the boundary, and its vanishing requires that tangential derivatives of the metric variation have to be zero. This is evident, since the combination

p α β = g α β ε ( n ) n α n β

(p.434) is the projection operator onto the boundary

p α ϵ n ϵ = 0 , p α ϵ p ϵ = p α β .

We can rewrite Eq. (E.2.2) in terms of the extrinsic curvature using the relation

n α ; α = K .

Here, K is the trace of the extrinsic curvature

K α β = p μ p ν n ν ; μ , K = p α β K α β .

Combining all these formulas together we finally obtain the proper Einstein action with boundary terms (York 1972, 1986; Gibbons and Hawking 1977), corresponding to the variational problem when the metric variations and the tangential derivatives of the metric variations are zero on the boundary, while normal derivatives of the metric variations are not fixed

S [ g ] = c 3 16 π G ( V d D x   g   [ R 2 Λ ] 2 B ε B ( n ) Σ K n μ d σ μ ) = c 3 16 π G ( V d D x   g   [ R 2 Λ ] 2 B ε B ( n ) Σ d D 1 y | h | K ) .

Here, ΣB means the sum over all time‐like and space‐like boundaries of the region V with the appropriate ε B (n) = ± 1. The surface integral in this action is called the Gibbons–Hawking– York boundary term.

Note that in the literature one can find different signs for the boundary term in Eq. (E.2.13). This apparent discrepancy comes from different sign conventions in the definitions of the extrinsic curvature (see Eqs. (3.8.9), and (E.2.11)). We accepted here the (Misner et al. 1973) sign conventions.

E.3 Boundary Term for the Euclidean Einstein–Hilbert Action

In the Euclidean spacetime the calculations are practically the same. The only differences are:

  • The normal to Σ is always space‐like and hence ε(n) = 1.

  • The total sign in the Euclidean bulk action is opposite.

The Euclidean Einstein‐Hilbert action with the boundary term is

S E [ g ] = c 3 16 π G ( V d D x   g   [ R 2 Λ ] 2 Σ K n μ d σ μ ) = c 3 16 π G ( V d D x   g   [ R 2 Λ ] 2 Σ d D 1 y h K ) .