Introduction to Quantum Gravity

Abstract

All particles, whether massive or massless, experience the gravitational interaction due to their energy content.

Appendices

Appendix A: Variation of a Determinant

In varying $\sqrt{-g}$ with respect to g _{μ ν} where g(x) = det [ g _{μ ν} (x) ], we need to know how to vary a determinant. This is explicitly used in Problem 1.3, and in obtaining the field equation in (1.2.2). This appendix shows you how to do that.

Given two matrices N, M, integrate the following from λ = 0 to λ = 1:

$$\displaystyle{ \frac{\mathrm{d}} {\mathrm{d}\lambda }\,\mathrm{e}^{\lambda \,N}\,\mathrm{e}^{-\lambda \,M}\, =\,\mathrm{ e}^{\lambda \,N}(N - M)\,\mathrm{e}^{-\lambda \,M}, }$$

(A-1.1)

to obtain

$$\displaystyle{ \mathrm{e}^{N}\,\mathrm{e}^{-M}\, =\, 1\, +\,\int _{ 0}^{1}\!\mathrm{d}\lambda \,\mathrm{e}^{\lambda \,N}(N - M)\,\mathrm{e}^{-\lambda \,M}, }$$

(A-1.2)

or for N = M + δM

$$\displaystyle{ \mathrm{e}^{N} -\mathrm{ e}^{M}\, =\,\int _{ 0}^{1}\!\mathrm{d}\lambda \,\mathrm{e}^{\lambda \,N}(N - M)\,\mathrm{e}^{-\lambda \,M}\,\mathrm{e}^{M}. }$$

(A-1.3)

$$\displaystyle{ \updelta \,\mathrm{e}^{M} =\int _{ 0}^{1}\mathrm{d}\lambda \,\,\mathrm{e}^{\lambda \,M}\,(\updelta M)\,\mathrm{e}^{-\lambda M}\,\mathrm{e}^{M}. }$$

(A-1.4)

We multiply the latter equation from the right by e^−M, and take the trace to obtain

$$\displaystyle{ \mathrm{Tr}\,\big[\big(\updelta \,\mathrm{e}^{M}\big)\,\mathrm{e}^{-M}\,\big]\, =\,\mathrm{ Tr}\big[\,\updelta \:\!M\,]\, =\,\mathrm{ Tr}\,\big[\,\mathrm{e}^{-M}\,\updelta \:\!\mathrm{e}^{M}\,\big], }$$

(A-1.5)

where the second equality follows from the fact that

$$\displaystyle{ \mathrm{Tr}\,[\,\mathrm{e}^{\lambda \,M}\,(\updelta M)\,\mathrm{e}^{-\lambda \,M}\,] =\mathrm{ Tr}\,[(\updelta M)\,\mathrm{e}^{-\lambda \,M}\mathrm{e}^{\lambda \,M}\,] =\mathrm{ Tr}\,[\,\updelta M\,]. }$$

(A-1.6)

For a given matrix A, let M = ln [ A ]. (A-1.5) then gives

$$\displaystyle{ \mathrm{Tr}\,[\,A^{-1}\,\updelta \,A\,]\, =\,\updelta \, [\,\mathrm{Tr}\,(\mathrm{ln}\,[\,A\,])\,]. }$$

(A-1.7)

On the other hand for a c-number a: δ e^a = e^a δ a, and for a  = Tr (ln[ A ]), e^{Tr ln[ A ]}  = det[ A ]. This c-number equation then gives

$$\displaystyle{ \updelta \,\mathrm{det}\,[\,A\,]\, =\,\mathrm{ det}\,[\,A\,]\,\updelta \,[(\mathrm{Tr}\,\mathrm{ln}\,[\,A\,])], }$$

(A-1.8)

which upon comparison with (A-1.7) leads to

$$\displaystyle{ \updelta \,\mathrm{det}\,[\,A\,]\, =\,\mathrm{ det}\,[\,A\,]\,\mathrm{Tr}\,[A^{-1}\,\updelta \,A]. }$$

(A-1.9)

Appendix B: Parametric Integral Representation of the Logarithm of a Matrix

For a given matrix M into consideration, one may conveniently introduce another matrix N commuting with M, such as the unit matrix, and express

$$\displaystyle{ \mathrm{ln}\,M -\mathrm{ ln}\,N =\int _{ 0}^{1} \frac{(M - N)\,\,\mathrm{d}\lambda } {N + (M - N)\:\!\lambda } =\mathrm{ i}(M - N)\int _{0}^{\infty }\!\mathrm{d}s\,\mathrm{e}^{-\mathrm{i}s\:\!(N-\mathrm{i}\,\epsilon )}\int _{ 0}^{1}\mathrm{d}\lambda \,\mathrm{e}^{-\mathrm{i}\,s\,\lambda (M-N)}. }$$

(B-1.1)

An integration of the right-hand side over λ gives the following parametric integral representation

$$\displaystyle{ \mathrm{ln}\,M = -\int _{0}^{\infty }\frac{\mathrm{d}s} {s} \,\mathrm{e}^{-\mathrm{i}\,s\,(M-\mathrm{i}\,\epsilon )} +\Big [\,\mathrm{ln}\,N +\int _{ 0}^{\infty }\frac{\mathrm{d}s} {s} \,\mathrm{e}^{-\mathrm{i}\,s\,(N-\mathrm{i}\,\epsilon )}\Big], }$$

(B-1.2)

where we note that the second term, within the square brackets, is independent of the matrix M in consideration. Of particular interest is the trace of lnM given by

$$\displaystyle{ \mathrm{Tr}\:\![\,\mathrm{ln}\,M\,] = -\int _{0}^{\infty }\frac{\mathrm{d}s} {s} \,\mathrm{Tr}\:\!\big[\,\mathrm{e}^{-\mathrm{i}\,s\,(M-\mathrm{i}\,\epsilon )}\big] +\mathrm{ Tr}\:\!\Big[\,\mathrm{ln}N +\int _{ 0}^{\infty }\frac{\mathrm{d}s} {s} \,\mathrm{e}^{-\mathrm{i}\,s\,(N-\mathrm{i}\,\epsilon )}\Big], }$$

(B-1.3)

where, again the second term on the right-hand side is independent of the matrix M. In a path integral representation, through exponentiation of the expression in (B-1.3), such a second term just gives rise to an overall multiplicative constant factor to the generating functional 〈 0₊ | 0₋〉 independent of the matrix M in consideration.

Appendix C: Content of the Euler-Poincaré Characteristic

The purpose of this Appendix is to establish the following fundamental result in four dimensional spacetime that

$$\displaystyle{ \updelta \Big(\sqrt{-g}\big[\,R_{\mu \nu \sigma \lambda }\,R^{\mu \nu \sigma \lambda } - 4\,R_{\mu \nu }R^{\mu \nu } + R^{2}\,\big]\Big), }$$

(C-1.1)

is a total derivative. ​⁸⁸ This is the content of the so-called Euler-Poincaré characteristic which holds true in four dimensional spacetime. It is also known as the Gauss-Bonnet Theorem in four dimensions. According to Problem 1.16, this is established by considering equivalently the far simpler expression

$$\displaystyle{ \updelta \big(\sqrt{-g}\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\kappa }R_{\mu \nu \sigma \lambda }\,R_{\alpha \beta \gamma \,\kappa }\big). }$$

To establish that this is a total derivative, we make use of the following results: ​⁸⁹

$$\displaystyle{ \updelta \sqrt{-g} = \frac{1} {2}\sqrt{-g}\,g\,^{\rho \rho '}\updelta g_{\rho \rho '}, }$$

(C-1.2)

$$\displaystyle{ \updelta \eta ^{\mu \nu \sigma \rho } = -\frac{1} {2}\,\eta ^{\mu \nu \sigma \rho }\,g\,^{\rho \rho '}\updelta g_{\rho \rho '}, }$$

(C-1.3)

$$\displaystyle{ \updelta R_{\mu \nu \sigma \lambda } =\updelta g_{\mu \rho }\,R\,^{\rho }\:\!_{\nu \sigma \lambda } + g_{\mu \rho }\big(\nabla _{\sigma }\,\,\updelta \varGamma _{\nu \lambda }\:\!^{\rho }-\nabla _{\lambda }\,\updelta \varGamma _{\nu \sigma }\:\!^{\rho }\big), }$$

(C-1.4)

$$\displaystyle{ \eta ^{\sigma \lambda \gamma \,\kappa }\,\nabla _{\sigma }\,R_{\alpha \beta \gamma \,\kappa } = 0, }$$

(C-1.5)

$$\displaystyle{ \nabla _{\rho }\eta ^{\mu \nu \alpha \beta } = 0, }$$

(C-1.6)

established, respectively, in Problems 1.3, (i), Problem 1.15, Problem 1.7, (ii), Problem 1.17, Problem 1.18.

From (C-1.2)–(C-1.4), the variation in question is given by

$$\displaystyle\begin{array}{rcl} & \updelta \Big(\!\sqrt{-g}\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\kappa }R_{\mu \nu \sigma \lambda }R_{\alpha \beta \!\gamma \,\kappa }\Big)\! =\! \frac{1} {2}\:\!g\,^{\rho \rho '}\updelta g_{\rho \rho '}\!\Big(\!\sqrt{-g}\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\kappa }R_{\mu \nu \sigma \lambda }R_{\alpha \beta \!\gamma \,\kappa }\!\Big)& \\ & \qquad \qquad \qquad +\, 2\,\Big(\!-\frac{1} {2}\Big)g\,^{\rho \rho '}\updelta g_{\rho \rho '}\Big(\sqrt{-g}\,\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\kappa }R_{\mu \nu \sigma \lambda }R_{\alpha \beta \gamma \,\kappa }\Big) & \\ & \qquad \qquad \qquad +\, 2\,\updelta g_{\mu \rho }\sqrt{-g}\,\,\eta ^{\mu \nu \alpha \beta }\,\eta ^{\sigma \lambda \gamma \,\kappa }\,R\,^{\rho }\:\!_{\nu \sigma \lambda }R_{\alpha \beta \!\gamma \,\,\kappa } & \\ & \qquad \qquad \qquad +\, 4\,\sqrt{-g}\,\,\eta ^{\mu \nu \alpha \beta }\,\eta ^{\sigma \lambda \gamma \,\kappa }\,\nabla _{\sigma }\,\big(\updelta \varGamma _{\nu \lambda }\:\!^{\rho }\big)\,g_{\mu \rho }\,R_{\alpha \beta \!\gamma \,\,\kappa }\,. &{} \\ \end{array}$$

(C-1.7)

The first two expressions on the right-hand side of the above equation may be combined and give

$$\displaystyle{ -\frac{1} {2}\,g\,^{\rho \rho '}\updelta g_{\rho \rho '}\Big(\sqrt{-g}\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\:\!\kappa }R_{\mu \nu \sigma \lambda }R_{\alpha \beta \gamma \,\:\!\kappa }\Big). }$$

From (C-1.5), (C-1.6) and the fact that ∇ _σ g _{μ ρ}  = 0, the last term in (C-1.7) may be equivalently rewritten as

$$\displaystyle{ 4\,\sqrt{-g}\,\,\nabla _{\sigma }\Big(\eta ^{\mu \nu \alpha \beta }\,\eta ^{\sigma \lambda \!\gamma \,\:\!\kappa }\,\updelta \varGamma _{\nu \lambda }\:\!^{\rho }g_{\mu \:\!\rho }\,R_{\alpha \beta \!\gamma \,\kappa }\Big), }$$

which from Problem 1.5 (i) is a total derivative.

Therefore as an intermediary step we have

$$\displaystyle\begin{array}{rcl} & & \quad \;\quad \;\updelta \Big(\sqrt{-g}\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\kappa }R_{\mu \nu \sigma \lambda }R_{\alpha \beta \gamma \,\kappa }\Big) \\ & =& -\frac{1} {2}\,g\,^{\rho \rho '}\updelta g_{\rho \rho '}\Big(\sqrt{-g}\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\kappa }R_{\mu \nu \sigma \lambda }R_{\alpha \beta \gamma \,\kappa }\Big) \\ & & +\,2\,\updelta g_{\mu \:\!\rho }\sqrt{-g}\,\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\kappa }R\,^{\rho }\:\!_{\nu \sigma \lambda }\,R_{\alpha \beta \gamma \,\kappa } \\ & & +\,4\,\sqrt{-g}\,\,\nabla _{\sigma }\Big(\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \kappa }\,\updelta \varGamma _{\nu \lambda }\:\!^{\rho }\,g_{\mu \rho }\,R_{\alpha \beta \gamma \,\kappa }\Big).{} \\ \end{array}$$

(C-1.8)

We rewrite the second term on the right-hand side of the above equation as

$$\displaystyle{ 2\,\updelta g_{\mu \:\!\rho }\sqrt{-g}\,\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\:\!\kappa }R\:\!_{\rho \:\!'\!\nu \sigma \lambda }\,g\,^{\rho \rho '}R_{\alpha \beta \gamma \,\:\!\kappa }. }$$

(C-1.9)

To investigate the nature of this second term on the right-hand side of (C-1.8), we proceed as follows.

A totally anti-symmetric tensor of rank 5, i.e., having five indices, in four dimensional spacetime, is necessarily zero due to the simple fact that at least two of its indices must be equal. Now consider the part

$$\displaystyle{ \updelta g_{\mu \rho }\,R_{\rho \:\!'\!\nu \sigma \lambda }R_{\alpha \beta \gamma \,\:\!\kappa }, }$$

in this expression, concentrating on the following five indices (μ, ρ ′, ν, α, β). Suppressing the other indices in the just given part, we may, in turn, consider the expression:

$$\displaystyle{ \Big(\updelta g\:\!_{\mu \,\centerdot }\,R\:\!_{\rho '\nu \,\centerdot \,\centerdot }\,R\:\!_{\alpha \beta \,\centerdot \,\centerdot }\Big)\eta ^{\mu \nu \alpha \beta }. }$$

(C-1.10)

By interchanging in turn the index ρ ​′ with the other indices within the round brackets, we may generate the expression

$$\displaystyle\begin{array}{rcl} & & \Big(\updelta g\:\!_{\mu \,\centerdot }\,R\:\!_{\rho '\nu \,\centerdot \,\centerdot }\,R\:\!_{\alpha \beta \,\centerdot \,\centerdot }\,-\,\updelta g\:\!_{\rho '\,\centerdot }\,R\:\!_{\mu \nu \,\centerdot \,\centerdot }\,R\:\!_{\alpha \beta \,\centerdot \,\centerdot }\,-\,\updelta g\:\!_{\mu \,\centerdot }\,R\:\!_{\nu \rho '\,\centerdot \,\centerdot }\,R\:\!_{\alpha \beta \,\centerdot \,\centerdot } \\ & &\;\;\;\;\,\,-\,\updelta g\:\!_{\mu \,\centerdot }\,R\:\!_{\alpha \nu \,\centerdot \,\centerdot }\,R\:\!_{\rho '\beta \,\centerdot \,\centerdot }-\updelta g\:\!_{\mu \,\centerdot }\,R\:\!_{\beta \nu \,\centerdot \,\centerdot }\,R\:\!_{\alpha \rho '\,\centerdot \,\centerdot }\Big)\eta ^{\mu \nu \alpha \beta } = 0,{} \\ \end{array}$$

(C-1.11)

where due to the anti-symmetry property of the multiplicative factor η ^{μ ν α β}, the above must be equal to zero as indicated, since an antisymmetric tensor of rank 5 is zero as mentioned above. By re-inserting the suppressed fixed indices, the above equation leads, upon multiplying by g ^{ρ ρ ​′} η ^{σ λ​γ  ​κ}, to

$$\displaystyle\begin{array}{rcl} \updelta g\:\!_{\mu \:\!\rho }\,R\,^{\rho }\,\!\!_{\nu \sigma \lambda }\,R\:\!_{\alpha \beta \gamma \,\kappa }\,\eta ^{\mu \nu \alpha \beta }\,\eta ^{\sigma \lambda \gamma \,\kappa }& =& g\,^{\rho \rho \:\!'}\updelta g\:\!_{\rho \:\!'\rho }\,R\:\!_{\mu \nu \sigma \lambda }\,R\:\!_{\alpha \beta \gamma \,\kappa }\,\eta ^{\mu \nu \alpha \beta }\,\eta ^{\sigma \lambda \gamma \,\kappa } \\ & +& \quad \;\,\,\updelta g\:\!_{\mu \:\!\rho }\,R\:\!_{\nu }\,\!\,^{\rho }\:\!\!_{\sigma \lambda }\,R\:\!_{\alpha \beta \gamma \,\kappa }\,\eta ^{\mu \nu \alpha \beta }\,\eta ^{\sigma \lambda \gamma \,\kappa } \\ & +& \quad \;\,\,\updelta g\:\!_{\mu \:\!\rho }\,R\:\!_{\alpha \nu \sigma \lambda }\,R\,^{\rho }\:\!_{\beta \gamma \,\:\!\kappa }\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\:\!\kappa } \\ & +& \quad \;\,\,\updelta g\:\!_{\mu \:\!\rho }\,R\:\!_{\beta \nu \sigma \lambda }\,R\:\!_{\alpha }\:\!^{\rho }\:\!\!_{\gamma \,\:\!\kappa }\,\eta ^{\mu \nu \alpha \beta }\,\eta ^{\sigma \lambda \gamma \,\:\!\kappa }.{} \\ \end{array}$$

(C-1.12)

In the second line simply write R ​ _ν  ​^ρ ​​ _{σ λ}  = −R  ​^ρ ​​ _{ν σ λ} . In the third line exchange (σ λ) ↔ (γ  ​κ), and α ↔ β. In the fourth line first write R ​ _α  ​^ρ ​ _γ κ  = − R ​ ^ρ ​ _{α γ κ} then exchange (σ λ) ↔ (γ  ​κ), and α ↔ ν, and use, in the process, the anti-symmetry of η ^{σ λ γ  ​κ} and, in particular, the anti-symmetry of R ​ _{β α σ λ} in its first two indices. These allow us to rewrite (C-1.12) as

$$\displaystyle\begin{array}{rcl} \updelta g\:\!_{\mu \:\!\rho }\,R\:\!\,^{\rho }\:\!\!_{\nu \sigma \lambda }\,R\:\!_{\alpha \beta \gamma \,\,\kappa }\,\eta ^{\mu \nu \alpha \beta }\,\eta ^{\sigma \lambda \gamma \,\kappa }\,& =& g\,^{\rho \rho '}\updelta g_{\rho '\rho }\,R_{\mu \nu \sigma \lambda }\,R_{\alpha \beta \gamma \,\,\kappa }\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\,\kappa } \\ & -& \,3\,\,\updelta g_{\mu \:\!\rho }\,R\:\!\,^{\rho }\:\!_{\nu \sigma \lambda }\,R_{\alpha \beta \gamma \,\,\kappa }\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\,\kappa }.{} \\ \end{array}$$

(C-1.13)

We may now solve for δg ​ _μ ​ρ  R ​ ^ρ ​ _{ν σ λ}  R ​ _{α β γ κ}  η ^{μ ν α β} η ^{σ λ γ κ}, to obtain

$$\displaystyle{ \updelta g_{\mu \:\!\rho }\,R\;\!\,^{\rho }\:\!\!_{\nu \sigma \lambda }\,R_{\alpha \beta \gamma \,\kappa }\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\kappa } = \frac{1} {4}\,g\,^{\rho \rho '}\,\updelta g_{\rho '\rho }\,R_{\mu \nu \sigma \lambda }\,R_{\alpha \beta \gamma \,\kappa }\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\kappa }. }$$

(C-1.14)

This equation allows us to rewrite the term displayed in (C-1.9) as

$$\displaystyle\begin{array}{rcl} & & \;\;2\,\updelta g_{\mu \rho }\sqrt{-g}\,\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\kappa }R\:\!_{\rho '\nu \sigma \lambda }\,g\,^{\rho \rho '}R_{\alpha \beta \gamma \,\kappa } \\ & =& \frac{1} {2}\,g^{\rho \rho '}\,\updelta g_{\rho \rho '}\Big(\sqrt{-g}\,\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\kappa }\,R_{\mu \nu \sigma \lambda }R_{\alpha \beta \gamma \,\kappa }\Big).{} \\ \end{array}$$

(C-1.15)

Thus the first and second term on the right-hand side of the intermediary step in (C-1.8) cancel out. All told, we finally obtain

$$\displaystyle{ \updelta \Big(\sqrt{-g}\,\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\kappa }R_{\mu \nu \sigma \lambda }R_{\alpha \beta \gamma \,\kappa }\Big) = 4\sqrt{-g}\,\nabla _{\sigma }\Big(\eta ^{\mu \nu \alpha \beta }\eta ^{\sigma \lambda \gamma \,\kappa }\updelta \varGamma _{\nu \lambda }\:\!^{\rho }\,g_{\mu \rho }R_{\alpha \beta \gamma \,\kappa }\Big), }$$

(C-1.16)

which from Problem 1.5 (i), may be rewritten in the form $4\partial _{\sigma }(\sqrt{-g}\,\xi ^{\sigma })$, and is a total derivative.

Appendix D: Invariant Products of Three Riemann Tensors

In this appendix we show that the invariant products of three Riemans tensors on the mass shell, i.e., when R _{μ ν}  = 0 and hence also R = 0, are necessarily proportional to the invariant 

$$\displaystyle{ R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }\,R\:\!^{\sigma \lambda }\:\!\!_{\kappa \rho }\,R\:\!^{\kappa \rho }\:\!\!_{\mu \nu }. }$$

(D-1.1)

At the very outset, we note the following:

• (1) The condition R _{μ ν}  = 0 (R = 0), and the anti-symmetry conditions of R _{μ ν σ λ} in the interchange μ ↔ ν or of σ ↔ λ, imply that in order to generate a non-trivial, i.e., non-vanishing, invariant out a product of three Riemann, no two indices within each of the Riemann tensors may be set equal.

• (2) Exactly two indices must be common between any two of Riemann tensors in the product. If four indices are common between two of the Riemann tensors are common, no non-trivial invariant may be formed out of this scalar and the third Riemann fourth rank tensor. If three indices are common between two Riemann tensors, no non-trivial invariant may be constructed out of this second rank tensor and the third Riemann fourth rank tensor. Similarly if only one index is common between two of the Riemann tensors, no non-trivial invariant may be constructed of this sixth rank tensor and the third Riemann fourth rank tensor. Again we use the mass shell condition R _{μ ν}  = 0 (R = 0).

A direct application of the cyclic relation of the permutation of the indices of the Riemann tensor in the first equation in (1.1.41), which holds irrespective of the mass shell condition, gives

$$\displaystyle{ R\,^{\lambda \sigma }\:\!_{\kappa \,\rho } =\, -\,R\,^{\lambda }\:\!\!_{\kappa \,\rho }\:\!^{\sigma }\, -\, R\,^{\lambda }\:\!\!_{\rho }\:\!^{\sigma }\:\!\!_{\kappa }, }$$

(D-1.2)

from which

$$\displaystyle{ R\:\!^{\mu \nu }\:\!_{\sigma \lambda }\,R\,^{\lambda \sigma }\:\!_{\kappa \,\rho } =\, -\,R\:\!^{\mu \nu }\:\!_{\sigma \lambda }\,R\,^{\lambda }\:\!_{\kappa \,\rho }\:\!^{\sigma } -\, R\:\!^{\mu \nu }\:\!_{\sigma \lambda }\,R\,^{\lambda }\:\!_{\rho }\:\!^{\sigma }\:\!_{\kappa }. }$$

(D-1.3)

Upon relabeling the indices on the right-hand side of the above equation, and using the elementary properties in (1.1.42), the above equation leads to

$$\displaystyle{ R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }R\,^{\lambda \sigma }\:\!\!_{\kappa \,\rho } = 2\,R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }\,R\,^{\lambda }\:\!_{\kappa }\:\!^{\sigma }\:\!\!_{\rho }. }$$

(D-1.4)

Upon multiplying the latter by R ​^κ ρ ​​ _{μ ν} gives the identity

$$\displaystyle{ R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }\,R\,^{\lambda }\:\!\!_{\kappa }\:\!^{\sigma }\:\!_{\rho }\,R\,^{\kappa \rho }\:\!\!_{\mu \nu } = -\,\frac{1} {2}\,R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }\,R\:\!^{\sigma \lambda }\:\!\!_{\kappa \rho }\,R\,^{\kappa \rho }\:\!\!_{\mu \nu }, }$$

(D-1.5)

where we have used the relation R ​^{λ σ} ​​ _{κ ρ}  = −R ​^σ λ ​​ _{κ ρ} on the right-hand side of the above equation. With relabeled indices, we may apply the identity in (D-1.3) to R ^{κ ρ} ​​ _{μ ν}  R ^{μ ν} ​​ _{σ λ} in the above equation to obtain

$$\displaystyle{ R\,^{\kappa \,\rho }\:\!\!_{\mu \nu }\,R\:\!^{\nu }\:\!\!_{\sigma }\:\!^{\mu }\:\!_{\lambda }\,R\,^{\lambda }\:\!\!_{\kappa }\:\!^{\sigma }\:\!\!_{\rho } = \frac{1} {4}\,R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }\,R\:\!^{\sigma \lambda }\:\!\!_{\kappa \rho }\,R\,^{\kappa \rho }\:\!\!_{\mu \nu }. }$$

(D-1.6)

More appropriately, by relabeling indices, this may be rewritten as

$$\displaystyle{ R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }\,R\,^{\lambda }\:\!_{\rho }\:\!^{\sigma }\:\!_{\kappa }\,R\,^{\kappa }\:\!_{\mu }\,^{\rho }\:\!_{\nu } = \frac{1} {4}\,R\:\!^{\mu \nu }\:\!_{\sigma \lambda }\,R\:\!^{\sigma \lambda }\:\!\!_{\kappa \,\rho }\,R\,^{\kappa \rho }\:\!\!_{\mu \nu }. }$$

(D-1.7)

We denote the left-hand side of the above equation by

$$\displaystyle{ R\:\!^{\mu \nu }\:\!_{\sigma \lambda }\,R\,^{\lambda }\:\!_{\rho }\:\!^{\sigma }\,_{\kappa }\,R\,^{\kappa }\:\!_{\mu }\:\!^{\rho }\:\!_{\nu }\, \equiv \, (\,\mu \nu \sigma \lambda )\,(\lambda \rho \:\!\sigma \kappa )\,(\kappa \mu \rho \nu ). }$$

(D-1.8)

Due to the identity in (D-1.4), and the fact that R _{μ ν σ λ} is odd in the interchange μ ↔ ν, or of σ ↔ λ, and even in the interchange (μ, ν) ↔ (σ, λ), it is sufficient to establish that the following two invariants

$$\displaystyle{ (\,\mu \nu \sigma \lambda )\,(\,\sigma \rho \mu \kappa )\,(\,\lambda \kappa \nu \rho ),\quad (\,\mu \nu \sigma \lambda )\,(\,\mu \rho \:\!\sigma \kappa )(\,\lambda \kappa \nu \rho ), }$$

(D-1.9)

are, in turn, also proportional to the invariant in (D-1.1). [ ​Note that in the products of the Riemann tensors corresponding to the ones in (D-1.9) and the one in (D-1.8) any two of the Riemann tensors, in a given product, have exactly two indices in common.]

To the above end, consider the invariant R ^{μ ν} ​ _{σ λ}  R ^{σ λ} ​ _{κ ρ}  R ^{κ ρ} ​ _{μ ν} , and concentrate on the indices in R ^{μ ν} ​_{▪  ​ ▪} R ^{σ λ} ​_{▪  ​ ▪} R ^κ ▪ ​_{▪  ​ ▪} , suppressing the other indices. Using the fact that a totally anti-symmetric tensor with five indices must be zero in four dimensional spacetime, ​⁹⁰ we can construct the following totally anti-symmetric tensor in four dimensions:

$$\displaystyle\begin{array}{rcl} 0\,& =& \Big(R\:\!^{\mu \nu }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\sigma \lambda }\:\!_{\centerdot \:\!\centerdot }- R\:\!^{\sigma \nu }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\mu \lambda }\:\!_{\centerdot \:\!\centerdot }- R\,^{\lambda \nu }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\sigma \mu }\:\!_{\centerdot \:\!\centerdot }- R\:\!^{\mu \sigma }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\nu \lambda }\:\!_{\centerdot \:\!\centerdot }- R\:\!^{\mu \lambda }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\sigma \nu }\:\!_{\centerdot \:\!\centerdot }\Big)R\:\!^{\kappa \,\centerdot }\:\!_{ \centerdot \:\!\centerdot } \\ &-\,& \Big(R\:\!^{\kappa \nu }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\sigma \lambda }\:\!_{\centerdot \:\!\centerdot }- R\:\!^{\sigma \nu }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\kappa \lambda }\:\!_{\centerdot \:\!\centerdot }- R\,^{\lambda \nu }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\sigma \kappa }\:\!_{\centerdot \:\!\centerdot }-\:\! R\:\!^{\kappa \sigma }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\nu \lambda }\:\!_{\centerdot \:\!\centerdot }-\, R\:\!^{\kappa \lambda }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\sigma \nu }\:\!_{\centerdot \:\!\centerdot }\:\!\Big)R\:\!^{\mu \,\centerdot }\:\!_{ \centerdot \:\!\centerdot } \\ &-\,& \Big(R\:\!^{\mu \kappa }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\sigma \lambda }\:\!_{\centerdot \:\!\centerdot }- R\:\!^{\sigma \kappa }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\mu \lambda }\:\!_{\centerdot \:\!\centerdot }- R\,^{\lambda \kappa }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\sigma \mu }\:\!_{\centerdot \:\!\centerdot }- R\:\!^{\mu \sigma }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\kappa \lambda }\:\!_{\centerdot \:\!\centerdot }-\:\! R\:\!^{\mu \lambda }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\sigma \kappa }\:\!_{\centerdot \:\!\centerdot }\!\Big)R\:\!^{\nu \,\centerdot }\:\!_{ \centerdot \:\!\centerdot } \\ &-\,& \Big(R\:\!^{\mu \nu }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\kappa \lambda }\:\!_{\centerdot \:\!\centerdot }- R\:\!^{\kappa \nu }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\mu \lambda }\:\!_{\centerdot \:\!\centerdot }- R\,^{\lambda \nu }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\kappa \mu }\:\!_{\centerdot \:\!\centerdot }\,- R\:\!^{\mu \kappa }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\nu \lambda }\:\!_{\centerdot \:\!\centerdot }\,\:\!-\:\! R\:\!^{\mu \lambda }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\kappa \nu }\:\!_{\centerdot \:\!\centerdot }\Big)R\:\!^{\sigma \,\centerdot }\:\!_{ \centerdot \:\!\centerdot } \\ &-\,& \Big(R\:\!^{\mu \nu }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\sigma \kappa }\:\!_{\centerdot \:\!\centerdot }- R\:\!^{\sigma \nu }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\mu \kappa }\:\!_{\centerdot \:\!\centerdot }- R\:\!^{\kappa \nu }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\sigma \mu }\:\!_{\centerdot \:\!\centerdot }\,- R\:\!^{\mu \sigma }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\nu \kappa }\:\!_{\centerdot \:\!\centerdot }\,-\:\! R\:\!^{\mu \kappa }\:\!_{\centerdot \:\!\centerdot }\,R\:\!^{\sigma \nu }\:\!_{\centerdot \:\!\centerdot }\Big)R\:\!^{\lambda \,\centerdot }\:\!_{ \centerdot \:\!\centerdot }\,\,.{} \\ \end{array}$$

(D-1.10)

Inserting the remaining fixed indices, for example, in R ^μ ▪ ​_{▪  ​ ▪} , R ^ν ▪ ​_{▪  ​ ▪} , we obtain from the original expression R ​^μ ​ρ ​ _{μ ν} , R ​^ν ​ρ ​ _{μ ν} , which vanish on the mass shell. This means that the second and third lines in the above equation do not contribute on the mass shell. Similarly, within the first pair of round brackets only the first term gives a non-zero contribution. Within the fourth pair of round brackets, only the second and fourth terms are non-vanishing, and in the fifth pair of round brackets, only the third and the fifth terms give non-zero contributions. All told, (D-1.10) implies that

$$\displaystyle\begin{array}{rcl} R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }\,R\:\!^{\sigma \lambda }\:\!\!_{\kappa \rho }\,R\:\!^{\kappa \rho }\:\!\!_{\mu \nu } =& -& R\:\!^{\kappa \nu }\:\!\!_{\sigma \lambda }\,R\:\!^{\mu \lambda }\:\!\!_{\kappa \rho }\,R\:\!^{\sigma \rho }\:\!\!_{\mu \nu } - R\:\!^{\mu \kappa }\:\!\!_{\sigma \lambda }\,R\:\!^{\nu \lambda }\:\!\!_{\kappa \rho }\,R\:\!^{\sigma \rho }\:\!\!_{\mu \nu } \\ & -& \,R\:\!^{\kappa \nu }\:\!\!_{\sigma \lambda }\,R\:\!^{\sigma \mu }\:\!\!_{\kappa \rho }\,R\,^{\lambda \rho }\:\!\!_{\mu \nu } - R\:\!^{\mu \kappa }\:\!\!_{\sigma \lambda }\,R\:\!^{\sigma \nu }\:\!\!_{\kappa \rho }\,R\,^{\lambda \rho }\:\!\!_{\mu \nu },{} \\ \end{array}$$

(D-1.11)

which upon relabeling the indices on the right-hand side of the equation, and using the elementary properties of the Riemann curvature tensor in (1.1.42), we obtain the following identity

$$\displaystyle{ R\:\!^{\mu \nu }\:\!_{\sigma \lambda }\,R\:\!^{\sigma \lambda }\:\!_{\kappa \rho }\,R\:\!^{\kappa \rho }\:\!_{\mu \nu } = 4\,R\:\!^{\mu \nu }\:\!_{\sigma \lambda }\,R\:\!^{\sigma \rho }\:\!_{\mu \kappa }\,R\:\!^{\lambda \kappa }\:\!_{\nu \rho }. }$$

(D-1.12)

Or more appropriately,

$$\displaystyle{ R\:\!^{\mu \nu }\:\!_{\sigma \lambda }\,R\:\!^{\sigma \rho }\:\!_{\mu \kappa }\,R\:\!^{\lambda \kappa }\:\!_{\nu \rho } = \frac{1} {4}\,R\:\!^{\mu \nu }\:\!_{\sigma \lambda }\,R\:\!^{\sigma \lambda }\:\!_{\kappa \rho }\,R\:\!^{\kappa \rho }\:\!_{\mu \nu }, }$$

(D-1.13)

establishing that the first invariant in (D-1.9) is proportional to the invariant in question on the right-hand of the above equation.

On the other hand, we may rewrite the basic identity in (D-1.13) as

$$\displaystyle\begin{array}{rcl} & & \frac{1} {4}\:\!R\:\!^{\mu \nu }\:\!_{\sigma \lambda }\,R\:\!^{\sigma \lambda }\:\!_{\kappa \rho }\,R\:\!^{\kappa \rho }\:\!_{\mu \nu }\! =\! R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }\,R\:\!^{\sigma \rho }\:\!\!_{\mu \kappa }\,R\,^{\lambda \kappa }\:\!\!_{\nu \rho }\! =\! R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }\Big(\!\!-R\:\!_{\mu }\:\!^{\sigma \,\rho }\:\!_{\kappa } - R\:\!_{\mu }\:\!^{\rho }\:\!_{\kappa }\:\!^{\sigma }\!\Big)R\,^{\lambda \kappa }\:\!\!_{\nu \rho } \\ & & \quad \;\; = -R\:\!_{\mu }\:\!^{\sigma \,\rho }\:\!_{\kappa }\,R\,^{\lambda \kappa }\:\!\!_{\nu \rho }\,R\:\!^{\mu \nu }\:\!_{\sigma \lambda } + R\:\!^{\mu \nu }\:\!_{\sigma \lambda }\,R\:\!_{\mu }\:\!^{\rho }\:\!^{\sigma }\:\!\!_{\kappa }\,R\,^{\lambda \kappa }\:\!\!_{\nu \rho }\! = -R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }\,R\,^{\lambda }\:\!\!_{\rho }\:\!^{\sigma }\:\!\!_{\kappa }\,R\:\!^{\kappa }\:\!_{\mu }\:\!^{\rho }\:\!\!_{\nu } \\ & & \quad \;\, +\:\! R\:\!^{\mu \nu }\:\!_{\sigma \lambda }\,R\:\!_{\mu }\:\!^{\rho \sigma }\:\!\!_{\kappa }\,R\,^{\lambda \kappa }\:\!\!_{\nu \rho } =\! -\frac{1} {4}\,R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }\,R\:\!^{\sigma \lambda }\:\!\!_{\kappa \rho }\,R\:\!^{\kappa \rho }\:\!_{\mu \nu } + R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }\,R\:\!_{\mu }\:\!^{\rho \sigma }\:\!\!_{\kappa }\,R\,^{\lambda \kappa }\:\!\!_{\nu \rho }\,,{} \\ \end{array}$$

(D-1.14)

where in writing the fourth equality, we have relabeled the indices of the first term, and in writing the fifth equality we have used the identity in (D-1.7) for the term just mentioned. The above equation gives for the very last on its extreme right-hand side the identity

$$\displaystyle{ R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }\,R\:\!_{\mu }\:\!^{\rho }\:\!^{\sigma }\:\!\!_{\kappa }\,R\,^{\lambda \kappa }\:\!\!_{\nu \rho } = \frac{1} {2}\,R\:\!^{\mu \nu }\:\!\!_{\sigma \lambda }\,R\:\!^{\sigma \lambda }\:\!\!_{\kappa \rho }\,R\:\!^{\kappa \rho }\:\!\!_{\mu \nu }, }$$

(D-1.15)

establishing that the second invariant in (D-1.9) is proportional to the invariant in question as well.

The remarkable property of invariant products of three Riemann curvature tensors, on the mass shell, being proportional to the invariant product in (D-1.1) is a key identity in investigating the divergent part of quantum general relativity in the two-loop contribution as discussed above within the text. ​⁹¹

Appendix E: Bekenstein-Hawking Entropy Formula of a Black Hole

The Bekenstein-Hawking Entropy formula ⁹² was briefly discussed in the introduction to this chapter. It relates the entropy of a Black Hole (BH) to the surface area of the event horizon. It arises as a consequence of investigations by Hawking⁹³ that a BH is a thermodynamic object, it radiates and has a temperature associated with it. ⁹⁴ To find its temperature of a BH, note that the so-called thermodynamic partition function e^{−β H}, in performing averages (by taking a trace), where H is the Hamiltonian of a system,⁹⁵ may be obtained from the time development unitary operator e^{− i t H∕ℏ}, by the substitution: t → −iℏ β. The trace operation gives rise to a periodicity condition on β.

The infinitesimal distance squared in Minkowski space, i.e., in flat spacetime is given by ds ² = −η _{μ ν}  dx ​^μdx ^ν, with metric η _{μ ν}  = diag[ −1, 1, 1, 1]. For a spherically symmetric (uncharged, non-rotating) body of mass M, for example, the gravitational effect of the body causes a curvature of spacetime around it, which amounts in replacing the line element squared by ds ² = −g _{μ ν}  dx ​^μdx ^ν, where g _{μ ν} is the corresponding metric which takes this distortion into account. The line element squared outside the body now takes the form

$$\displaystyle{ \mathrm{d}s^{2} =\Big (\,\Big[1 -\frac{2\,\mathrm{G_{N}}M} {\mathrm{c}^{2}\,r} \Big]^{1/2}\mathrm{c}\,\mathrm{d}t\Big)^{2} -\Big (\,\Big[1 -\frac{2\,\mathrm{G_{N}}M} {\mathrm{c}^{2}\,r} \Big]^{-1/2}\mathrm{d}r\Big)^{2} - r^{2}\Big[(\mathrm{d}\theta )^{2} + (\sin \theta \,\mathrm{d}\phi )^{2}\Big], }$$

(E-1.1)

where the point r = R _BH = 2 G_N M∕c² at which the coefficient of (dr)² blows out defines the radius of the BH, specifying the location of the horizon.⁹⁶

The above metric is referred to as the Schwarzschild metric, and R _BH is also referred to as the Schwarzschild radius.

We note that near the horizon $r\gtrsim 2\,\mathrm{G_{N}}M/\mathrm{c}^{2}$

$$\displaystyle{ \mathrm{d}\,\Big[1 -\frac{2\,\mathrm{G_{N}}\,M} {\mathrm{c}^{2}\,r} \Big]^{1/2} = \frac{\mathrm{G_{N}}M} {\mathrm{c}^{2}\,r^{2}} \Big[1 -\frac{2\,\mathrm{G_{N}}M} {\mathrm{c}^{2}\,r} \Big]^{-1/2}\mathrm{d}r\, \simeq \frac{\mathrm{c}^{2}} {4\,\mathrm{G_{N}}\,M}\Big[1 -\frac{2\,\mathrm{G_{N}}M} {\mathrm{c}^{2}\,r} \Big]^{-1/2}\mathrm{d}r. }$$

(E-1.2)

This suggests that near the horizon, we may define the variable

$$\displaystyle{ \rho = \frac{4\,\mathrm{G_{N}}\,M} {\mathrm{c}^{2}} \Big[1 -\frac{2\,\mathrm{G_{N}}M} {\mathrm{c}^{2}\,r} \Big]^{1/2}, }$$

(E-1.3)

carry out the transformation t → −iℏ β, as discussed earlier, to obtain

$$\displaystyle{ -\,\mathrm{d}s^{2} \rightarrow \Big (\rho \,\mathrm{d}\Big[\hslash \frac{\mathrm{c}^{3}} {4\,\mathrm{G_{N}}\,M}\,\beta \Big]\Big)^{2} +\big (\mathrm{d}\rho \big)^{2} + \frac{4\,\mathrm{G_{N}}^{2}\,M\,^{2}} {\mathrm{c}^{4}} \Big[(\mathrm{d}\theta )^{2} + (\sin \theta \,\mathrm{d}\phi )^{2}\Big]. }$$

(E-1.4)

The first two terms represent the line element squared in a two dimensional space (a disc), with [ ℏc³∕4G_N M ]β representing an angle. For [ ℏc³∕4G_N M ] β = 2π and β = 1∕k_B T, where k_B is the Boltzmann constant, the temperature of the black hole is thus given by⁹⁷

$$\displaystyle{ T_{\mathrm{BH}} = \frac{\hslash \mathrm{c}^{3}} {8\,\pi \,\mathrm{G_{N}}\,M\,\mathrm{k_{B}}}, }$$

(E-1.5)

where note that a very massive black hole is cold.

Recall that entropy S represents a measure of the amount of disorder with associate encoded information, and invoking the thermodynamic interpretation of a BH, we may write

$$\displaystyle{ \frac{\partial S} {\partial (M\,\mathrm{c}^{2})} = \frac{1} {T}, }$$

(E-1.6)

which upon integration with boundary condition that for M → 0, S → 0, gives the celebrated result

$$\displaystyle{ S_{\mathrm{BH}} = \frac{\mathrm{c}^{3}\mathrm{k_{B}}} {4\,\mathrm{G_{N}}\hslash }\,A,\quad \quad A = 4\,\pi \bigg(\frac{2\mathrm{G_{N}}\,M} {\mathrm{c}^{2}} \bigg)^{2}, }$$

(E-1.7)

referred to as the Bekenstein-Hawking Entropy formula of a BH, where A is the (surface) area of the horizon. Using the fact that G_N ℏ∕c³ = ℓ _P ² denotes the Plank length squared, we may appropriately rewrite (E-1.7) as

$$\displaystyle{ S_{\mathrm{BH}} =\mathrm{ k_{B}} \frac{A} {4\,\ell\,_{\mathrm{P}}^{2}}. }$$

(E-1.8)

As mentioned earlier, the proportionality of entropy to the area rather than to the volume of a BH horizon should be noted.

Problems

1.1. Establish the anti-symmetry property:

e _{β μ} (∂ _ν e _α  ​^μ +Γ _{ν σ}  ​^μ e _α  ​^σ)  = − e _{α μ} (∂ _ν e _β  ​^μ +Γ _{ν σ}  ​^μ e _β  ​^σ),

obtained by simply interchanging the Lorentz indices α, β.

1.2. Show that:

$$\displaystyle\begin{array}{rcl} \mathrm{(i)}\;\;[\nabla _{\alpha },\nabla _{\beta }]\,\xi ^{\alpha }\,& \,=& \,R\,^{\alpha }\,\!_{\sigma \alpha \beta }\,\xi ^{\sigma }\, =\, R_{\sigma \beta }\,\xi ^{\alpha }\;\text{for a vector field}\;\;\xi ^{\alpha }. {}\\ \mathrm{(ii)}[\nabla _{\alpha },\nabla _{\nu }]\,h^{\alpha \beta }\,& \,=& \,R_{\sigma \nu }\,h^{\sigma \beta }\, +\, R^{\beta }\:\!_{\sigma \alpha \:\!\nu }\,h^{\sigma \alpha }\;\text{for a symmetric tensor field}\;h^{\alpha \beta }. {}\\ \end{array}$$

1.3. Show that:

$$\displaystyle\begin{array}{rcl} \mathrm{(i)}\;\;\updelta \sqrt{-g}\,& \,=& \,\frac{1} {2}\,\sqrt{-g}\,g^{\alpha \beta }\,\updelta g_{\alpha \beta }\, = -\frac{1} {2}\,\sqrt{-g}\,g_{\alpha \beta }\,\updelta g^{\alpha \beta },\,\mathrm{where}\,g\, =\,\mathrm{ det}[g_{\mu \nu }]. {}\\ \mathrm{(ii)}\,\partial _{\mu }\sqrt{-g}\,& \,=& \,\frac{1} {2}\,\sqrt{-g}\,g^{\alpha \beta }\partial _{\mu }g_{\alpha \beta }\, =\, \sqrt{-g}\,\varGamma _{\mu \sigma }\,^{\sigma }. {}\\ \end{array}$$

1.4. Establish the equalities:

$$\displaystyle\begin{array}{rcl} \mathrm{(i)}\,g^{\mu \sigma }\,\varGamma _{\mu \sigma }\,^{\alpha }\,& \,=& \,- \frac{1} {\sqrt{-g}}\,\partial _{\mu }(\sqrt{-g}\,g^{\alpha \mu }). {}\\ \mathrm{(ii)}\,\delta ^{\mu }\,_{\nu }\,\varGamma _{\mu \sigma }^{\nu }\,& \,=& \,\frac{\partial _{\sigma }\sqrt{-g}} {\sqrt{-g}} = \partial _{\sigma }\,\mathrm{ln}[\sqrt{-g}\:]. {}\\ \end{array}$$

1.5.

• (i) For a vector field ξ ^μ, show that $\sqrt{-g}\,\nabla _{\mu }\xi ^{\mu }\, =\, \partial _{\mu }(\sqrt{-g}\,\xi ^{\mu })$.

• (ii) For a scalar field ϕ, show that $\sqrt{-g}\,g^{\mu \nu }\partial _{\mu }\phi \partial _{\nu }\phi \, = -\sqrt{-g}\,\phi (g^{\mu \nu }\nabla _{\mu }\!\nabla _{\nu }\phi )$, up to a total derivative.

• (iii) For a fixed real scalar field ϕ, show that $\updelta [\sqrt{-g}\,(g^{\mu \nu }\partial _{\mu }\phi \,\partial _{\nu }\phi + m^{2}\phi ^{2})]$ is given by

  $$\displaystyle{ \sqrt{-g}\,[\partial _{\mu }\phi \,\partial _{\nu }\phi -\frac{1} {2}\,g_{\mu \nu }(\partial ^{\rho }\phi \,\partial _{\rho }\phi + m^{2}\phi ^{2})]\updelta g^{\mu \nu }. }$$

1.6. Use (1.1.14) to infer that although Γ _ρ ν  ​^γ is not a tensor, the variation δΓ _ρ ν  ​^γ is a tensor.

1.7. Show that the infinitesimal variations of the Riemann tensor, the Ricci tensor and the connection may be written as:

$$\displaystyle\begin{array}{rcl} \mathrm{(i)}\;\updelta R^{\mu }\,_{\nu \rho \sigma }\,& \,=& \,\nabla _{\rho }\,\updelta \varGamma _{\nu \sigma }\,^{\mu }\,-\,\nabla _{\sigma }\,\updelta \varGamma _{\nu \rho }\,^{\mu }. {}\\ \mathrm{(ii)}\;\,\updelta R_{\mu \nu \rho \sigma }\,& =& \,\updelta g_{\mu \lambda }\,R^{\lambda }\:\!_{\nu \rho \sigma } + g_{\mu \lambda }\big(\nabla _{\rho }\,\updelta \varGamma _{\nu \sigma }\,^{\lambda }\,-\,\nabla _{\sigma }\,\updelta \varGamma _{\nu \rho }\,^{\lambda }\big). {}\\ \mathrm{(iii)}\;\;\;\;\updelta R_{\nu \sigma }\,& \,=& \,\nabla _{\rho }\,\updelta \varGamma _{\nu \sigma }\,^{\rho }\,-\,\nabla _{\sigma }\,\updelta \varGamma _{\nu \mu }\,^{\mu }. {}\\ \mathrm{(iv})\;\;\updelta \varGamma _{\mu \nu }\,^{\rho }\,& \,=& \,\frac{1} {2}g^{\rho \sigma }(\nabla _{\mu }\updelta g_{\nu \sigma } + \nabla _{\nu }\updelta g_{\mu \sigma } -\nabla _{\sigma }\updelta g_{\mu \nu }). {}\\ \end{array}$$

1.8. Establish that $\sqrt{-g}\,g^{\mu \nu }\,\updelta R_{\mu \nu }\, =\, 0$, up to a total derivative.

1.9. Verify that $\updelta (\sqrt{-g}g^{\mu \nu }R_{\mu \nu })\, =\, \sqrt{-g}(R_{\mu \nu } -\frac{1} {2}\,g_{\mu \nu }R)\updelta g^{\mu \nu }$.

1.10. Derive the following fundamental equalities which hold true up to total derivatives. These are generalization of results to be used in partial integrations in curved spacetime.

$$\displaystyle\begin{array}{rcl} \mathrm{(i)}\,\quad & & \,\sqrt{-g}\,(\nabla _{\mu }T^{\mu \,_{1}\ldots \mu \,_{k}\mu \,_{k+1}\ldots \mu \,_{n}})\,S_{\mu \,_{ 1}\ldots \mu \,_{k}}\,^{\mu }\,_{\mu \,_{k+1}\ldots \mu \,_{n}} {}\\ \,=\,& -& \,\sqrt{-g}\,T^{\mu \,_{1}\ldots \mu \,_{k}\mu \,_{k+1}\ldots \mu \,_{n}}\,\nabla _{\mu }\,S_{\mu _{ 1}\ldots \mu _{k}}\,^{\mu }\,_{\mu _{k+1}\ldots \mu \,_{n}}, {}\\ \end{array}$$

for n rank and n + 1 rank tensors $T^{\mu \,_{1}\ldots \mu \,_{k}\mu \,_{k+1}\ldots \mu \,_{n}}$, $S_{\mu \,_{1}\ldots \mu \,_{k}}\,^{\mu }\,_{ \mu \,_{k+1}\ldots \mu \,_{n}}$, respectively,

$$\displaystyle\begin{array}{rcl} & & \mathrm{(ii)}\,\,\sqrt{-g}\,(\nabla _{\mu }T^{\mu \,_{1}\ldots \mu \,_{k}\mu \mu \,_{k+1}\ldots \mu \,_{n}})\,S_{\mu \,_{ 1}\ldots \mu \,_{k}\mu \,_{k+1}\ldots \mu \,_{n}} {}\\ & & \quad \, =\, -\,\sqrt{-g}\,T^{\mu \,_{1}\ldots \mu \,_{k}\mu \mu \,_{k+1}\ldots \mu \,_{n}}\,\nabla _{\mu }\,S_{\mu \,_{ 1}\ldots \mu \,_{k}\mu \,_{k+1}\ldots \mu \,_{n}}, {}\\ \end{array}$$

for n + 1 rank and n rank tensors $T^{\mu \,_{1}\ldots \mu \,_{k}\mu \mu \,_{k+1}\ldots \mu \,_{n}}$, $S_{\mu \,_{1}\ldots \mu \,_{k}\mu \,_{k+1}\ldots \mu \,_{n}}$, respectively. In particular, for n = 0, i.e., for a scalar field ϕ, (i) reads

$$\displaystyle{ \mathrm{(iii)}\,\;\;\;\;\qquad \qquad \sqrt{-g}\,(\nabla _{\mu }\phi )\,S^{\mu }\, =\, -\,\sqrt{-g}\,\phi \,\nabla _{\mu }\,S^{\mu }. }$$

1.11. The action of a massive real scalar field is defined by:

$$\displaystyle{ W_{\mathrm{matter}} = -\frac{1} {2}\int (\mathrm{d}x)\sqrt{-g}(g^{\mu \nu }\partial _{\mu }\phi \,\partial _{\nu }\phi + m^{2}\phi ^{2}). }$$

• (i) Derive the field equation for the field ϕ.

• (ii) The energy-momentum tensor is defined in (1.2.22) by

  $$\displaystyle{ T_{\mu \nu } = - \frac{2} {\sqrt{-g}}\, \frac{\updelta } {\updelta g^{\mu \nu }}W_{\mathrm{matter}}.\;\;\;\;\;\mathrm{Find}\;\;T_{\mu \nu }. }$$

• (iii) Show that ∇ _μ T ^{μ ν} = 0, ∇ _ν T ^{μ ν} = 0.

1.12. Show that the action in (1.3.36) of first order formulation of linearized gravity is invariant under the infinitesimal gauge transformations:

$$\displaystyle{ \updelta \,\xi ^{\mu \nu }\, =\, \partial ^{\mu }\varLambda ^{\nu } + \partial ^{\nu }\varLambda ^{\mu } -\eta ^{\mu \nu }\,\partial _{\alpha }\varLambda ^{\alpha },\quad \updelta \,\varGamma _{\alpha \beta \gamma }\, =\, \partial _{\alpha }\partial _{\beta }\varLambda _{\gamma }. }$$

1.13. Show that the expressions in (1.4.3) and (1.4.4) are the same.

1.14. Find δΓ _{μ ν}  ​^σ up to second order in h _{μ ν} , with the latter describing a fluctuation about a background metric g _{μ ν} .

1.15. Given the tensor $\eta ^{\mu \nu \lambda \sigma } =\varepsilon ^{\mu \nu \lambda \sigma }/\sqrt{-g}$, with ɛ ^{μ ν λ σ} totally anti-symmetric, and ɛ ⁰¹²³ = +1. Show that, under an infinitesimal variation δg ^{ρ κ}, δη ^{μ ν λ σ} = (1∕2)η ^{μ ν λ σ} g _{ρ κ} δg ^{ρ κ} = − (1∕2)η ^{μ ν λ σ} g ^{ρ κ}δg _{ρ κ} .

1.16. Show that R _{μ ν ρ σ} R _{λ κ γ ε}  η ​^{μ ν λ κ} η ​^{ρ σ γ ε} $= 4\big(4R_{\mu \nu }R^{\mu \nu } - R_{\mu \nu \rho \sigma }R^{\mu \nu \rho \sigma } - R^{2}\big)$.

1.17. Show that η ^{ρ σ α λ}∇ _λ R _{μ ν ρ σ}  = 0.

1.18. Show that ∇ _λ η ^{μ ν α β} = 0.

1.19. Show that the solution of the equation in (1.9.10) is given by the one in (1.9.11), thus leading to the explicit expression for the holonomy in (1.9.12).