Evolution of Angular Momentum and Center of Mass at Null Infinity

Abstract

We study how conserved quantities such as angular momentum and center of mass evolve with respect to the retarded time at null infinity, which is described in terms of a Bondi–Sachs coordinate system. These evolution formulae complement the classical Bondi mass loss formula for gravitational radiation. They are further expressed in terms of the potentials of the shear and news tensors. The consequences that follow from these formulae are (1) Supertranslation invariance of the fluxes of the CWY conserved quantities. (2) A conservation law of angular momentum à la Christodoulou. (3) A duality paradigm for null infinity. In particular, the supertranslation invariance distinguishes the CWY angular momentum and center of mass from the classical definitions.

Appendices

Appendix A. Christodoulou–Klainerman Connection Coefficients and Curvature Components in Bondi–Sachs Formalism

We write the limit of connection coefficients and curvature components defined in [7,8,9] in terms of the Bondi–Sachs metric coefficients.

We choose the null vector fields $L = \frac{\partial }{\partial r}$ and ${\underline{L}}= \frac{2}{U}\left( \partial _u - W^D \partial _D - \frac{V}{2}\partial _r \right)$, which satisfy $\langle L,{\underline{L}}\rangle = -2$.

Definition A.1

The second fundamental forms and torsion are defined by

$$\begin{aligned} \chi _{AB}&= \langle D_A L, \partial _B \rangle = \frac{1}{2} \text{ tr }\chi g_{AB} + {\widehat{\chi }}_{AB}\\ \underline{\chi }_{AB}&= \langle D_A {\underline{L}}, \partial _B \rangle = \frac{1}{2} \text{ tr }\underline{\chi }g_{AB} + {\widehat{\underline{\chi }}}_{AB}\\ \zeta _A&= \frac{1}{2} \langle D_A L, {\underline{L}}\rangle \end{aligned}$$

Their limit as $r \rightarrow \infty$ are defined by

$$\begin{aligned} \Sigma&= \lim _{r\rightarrow \infty } {\widehat{\chi }}\\ \Xi&= \lim _{r\rightarrow \infty } r^{-1} {\widehat{\underline{\chi }}} \\ Z&= \lim _{r\rightarrow \infty } r\zeta . \end{aligned}$$

They are related to the metric coefficients in the corresponding Bondi–Sachs coordinate system as follows:

Proposition A.2

$$\begin{aligned} \Sigma _{AB}&= -\frac{1}{2} C_{AB}\\ \Xi _{AB}&= N_{AB}\\ Z_A&= -\frac{1}{2} \nabla ^B C_{AB} \end{aligned}$$

Proof

Starting with $g_{AB} = r^2 \sigma _{AB} + rC_{AB} + O(1)$, the determinant condition gives $\text{ tr }\chi = \frac{2}{r}$ and we compute

$$\begin{aligned} \chi _{AB} = r\sigma _{AB} + \frac{1}{2}C_{AB} + O(r^{-1}) \end{aligned}$$

to get $\Sigma _{AB} = -\frac{1}{2}C_{AB}$. Direct computation gives

$$\begin{aligned} \underline{\chi }_{AB} = r(-\sigma _{AB} + \partial _u C_{AB} ) + O(1) \end{aligned}$$

and hence $\text{ tr }\underline{\chi }= -\frac{2}{r} + O(r^{-2})$ and ${\widehat{\underline{\chi }}}_{AB} = r \partial _u C_{AB} + O(1)$. The limit of torsion follows from $\zeta _A = -\frac{1}{r} W_A^{(-2)} + O(r^{-2})$. $\quad \square$

Definition A.3

The mass aspect and conjugate mass aspect function of Christodoulou–Klainerman are defined by

$$\begin{aligned} \mu = K + \frac{1}{4}\text{ tr }\chi \text{ tr }\underline{\chi }- \text{ div }\zeta \qquad \underline{\mu } = K + \frac{1}{4}\text{ tr }\chi \text{ tr }\underline{\chi }+ \text{ div }\zeta . \end{aligned}$$

Here K denotes the Gauss curvature of the two-sphere $r=$const. Their limits are defined by

$$\begin{aligned} N&= \lim _{r\rightarrow \infty } r^3 \mu \\ {\underline{N}}&= \lim _{r\rightarrow \infty } r^3 \underline{\mu }. \end{aligned}$$

We express them in terms of the Bondi–Sachs metric coefficients as follows:

Proposition A.4

$$\begin{aligned} N = 2m + \frac{1}{2}\nabla ^A\nabla ^B C_{AB}, \qquad {\underline{N}} = 2m - \frac{1}{2}\nabla ^A\nabla ^B C_{AB} \end{aligned}$$

Proof

We compute $K = \frac{1}{r^2} + \frac{1}{2r^3} \nabla ^A\nabla ^B C_{AB} + O(r^{-4})$ and $\frac{1}{4}\text{ tr }\chi \text{ tr }\underline{\chi }= -\frac{1}{r^2} + \frac{1}{r^3}(2m-\frac{1}{2}\nabla ^A\nabla ^B C_{AB})$ and the assertion follows. $\quad \square$

We turn to curvature components. The convention of Riemann curvature tensor is

$$\begin{aligned} R(X,Y)Z&= (D_X D_Y -D_Y D_X - D_{[X,Y]} )Z\\ R(X,Y,W,Z)&= \langle R(X,Y)Z,W \rangle . \end{aligned}$$

Definition A.5

Define the curvature components

Here is the area form of the two-sphere with respect to $g_{AB}$. Their limits are defined by

$$\begin{aligned} {\underline{A}}_{AB}&= \lim _{r\rightarrow \infty } r^{-1}\underline{\alpha }_{AB}\\ {\underline{B}}_A&= \lim _{r\rightarrow \infty } r\underline{\beta }_A\\ P&= \lim _{r\rightarrow \infty } r^3\rho \\ Q&= \lim _{r\rightarrow \infty }r^3 \sigma \\ B_A&= \lim _{r\rightarrow \infty } r^3 \beta _A \end{aligned}$$

Note that $({\underline{A}},{\underline{B}})$ were denoted by (A, B) in [7].

We express them in terms of the Bondi–Sachs metric coefficients as follows:

Proposition A.6

$$\begin{aligned} {\underline{A}}_{AB}&= -2 \partial _u N_{AB}\\ {\underline{B}}_A&= \nabla ^B N_{AB}\\ P&= -2m - \frac{1}{4}C_{AB}N^{AB}\\ Q&= \epsilon ^{AB} \left( -\frac{1}{4}C_A^D N_{DB} - \frac{1}{2} \nabla _A \nabla ^D C_{DB} \right) \\ B_A&= -N_A \end{aligned}$$

Proof

The formula for ${\underline{A}}$ is obtained from (6) of [7], $2 \frac{\partial \Xi }{\partial u} = -{\underline{A}}$, which is the rescaled limit of the propagation equation $\widehat{{\underline{D}}} {\widehat{\underline{\chi }}} = -\underline{\alpha }$.

The formula for ${\underline{B}}$ is obtained from (2) of [7], $\nabla ^B \Xi _{AB} = {\underline{B}}_A$, which is the rescaled limit of the Codazzi equation .

The formula for P and Q are obtained from (3) of [7],

$$\begin{aligned} \epsilon ^{AB} \nabla _A Z_B = Q - \frac{1}{2}\Sigma \wedge \Xi , \qquad \nabla ^A Z_A = {\underline{N}} + P - \frac{1}{2} \Sigma \cdot \Xi , \end{aligned}$$

which is the rescaled limit of the Hodge system

Finally, we consider the Codazzi equation

Its leading order at $O(r^{-2})$ leads to (1) of [7] and its subleading order at $O(r^{-3})$ leads to

$$\begin{aligned} \begin{aligned}&\left( -\frac{1}{4} \partial _A |C|^2 + \frac{1}{2} C^{BD} \nabla _D C_{AB} + \frac{1}{4} \nabla _A C_D^E C^D_E + \frac{1}{2} \nabla _D C^{DE}C_{AE} \right) + \frac{1}{4}C_{AB} \nabla _D C^{BD}\\&\quad = \zeta _A^{(-2)} - B_A. \end{aligned} \end{aligned}$$

We simplify the second term in the parentheses by the identity $\nabla _{(D}C_{B)A} = \nabla _A C_{BD} + \nabla ^E C_{AE} \sigma _{BD} - \nabla ^E C_{E(D} C_{B)A}$ and the left-hand side becomes $\frac{1}{8} \partial _A |C|^2 + \frac{1}{4}C_{AB}\nabla _D C^{BD}$. Direct computation yields $\zeta _A^{(-2)} = -N_A + \frac{1}{8} \partial _A |C|^2 + \frac{1}{4}C_{AB}\nabla _D C^{BD}$ and the formula for B follows. $\quad \square$

Appendix B. Integration by Part Formula

In this section, we prove two integration formula that will be used to compute angular momentum and center of mass in spacetime with vanishing news.

Theorem B.1

Let $Y^A = \epsilon ^{AB} \nabla _B \tilde{X}^k, k = 1,2,3$. Let $F_{AB} = 2 \nabla _A\nabla _B f - \Delta f \sigma _{AB}$ and $P_{BA} = \nabla _B\nabla ^D C_{DA} - \nabla _A\nabla ^D C_{DB}$. Then

$$\begin{aligned} \int _{S^2} Y^A \left( \frac{1}{4} C_{AB} \nabla _D F^{DB} + \frac{1}{4} F_{AB} \nabla _D C^{DB} -\frac{3}{4} P_{BA} \nabla ^B f -\frac{1}{4} \nabla ^B P_{BA} f \right) =0. \end{aligned}$$

Proof

We integrate by parts the last two terms to get

$$\begin{aligned}&\int _{S^2} -\frac{1}{2} Y^A (\nabla _B\nabla ^D C_{AD} - \nabla _A\nabla ^D C_{BD}) \nabla ^B f + \frac{1}{2}\nabla ^B Y^A \nabla _B\nabla ^D C_{AD} \cdot f.\\&\quad = \int _{S^2} \frac{1}{2} \nabla _B Y^A \nabla ^D C_{AD} \nabla ^B f + \frac{1}{2} Y^A \nabla ^D C_{AD} \Delta f - \frac{1}{2} Y^A \nabla ^D C^B_D \nabla _A\nabla _B f\\&\qquad + \int _{S^2} \frac{1}{2} Y^A \nabla ^D C_{AD} f - \frac{1}{2} \nabla ^B Y^A \nabla ^D C_{AD} \nabla _B f\\&\quad = \int _{S^2} -\frac{1}{2} Y^A \nabla ^D C_D^B (\nabla _A\nabla _B f - \frac{1}{2}\Delta f \sigma _{AB}) + \frac{1}{4} Y^A \nabla ^D C_{AD} (\Delta + 2)f \end{aligned}$$

$\quad \square$

Theorem B.2

Let $F_{AB} = 2 \nabla _A\nabla _B f - \Delta f \sigma _{AB}$ and $P_{BA} = \nabla _B\nabla ^D C_{DA} - \nabla _A\nabla ^D C_{DB}$. Then

$$\begin{aligned} \int _{S^2} \nabla ^A {{\tilde{X}}}^k \left( C_{AB} \nabla _D F^{BD} + F_{AB}\nabla _D C^{BD} + \frac{1}{2} \nabla _A (C_{BD}F^{BD} ) - f\nabla ^B P_{BA} - 3 P_{BA} \nabla ^B f \right) =0. \end{aligned}$$

Proof

We integrate by parts the last two terms to get

$$\begin{aligned}&\int _{S^2} \nabla ^A {{\tilde{X}}}^k (-2P_{BA})\nabla ^B f\\&\quad = \int _{S^2} -2\nabla ^A {{\tilde{X}}}^k (\nabla _B\nabla ^D C_{DA} - \nabla _A\nabla ^D C_{DB})\nabla ^B f\\&\quad = \int _{S^2} -2{{\tilde{X}}}^k \nabla ^D C_{DA}\nabla ^A f + 2 \nabla ^A {{\tilde{X}}}^k \nabla ^D C_{DA}\Delta f + 4 {{\tilde{X}}}^k \nabla ^D C_{DB}\nabla ^B f \\&\qquad - 2\nabla ^A {{\tilde{X}}}^k \nabla ^D C_{DB} \nabla _A\nabla ^B f\\&\quad = \int _{S^2} 2 {{\tilde{X}}}^k \nabla ^D C_{DA}\nabla ^A f -\nabla ^A {{\tilde{X}}}^k \nabla ^D C_{DB} F_{AB} + \nabla ^A {{\tilde{X}}}^k \nabla ^D C_{DA}\Delta f \\&\quad = \int _{S^2} -2 \nabla ^D {{\tilde{X}}}^k C_{DA}\nabla ^A f -2 {{\tilde{X}}}^k C_{DA} \nabla ^D\nabla ^A f \\&\qquad -\nabla ^A {{\tilde{X}}}^k \nabla _D C^{DB} F_{AB} + \nabla ^A {{\tilde{X}}}^k \nabla ^D C_{DA}\Delta f \\&\quad = \int _{S^2} 2\nabla ^A {{\tilde{X}}}^k \nabla ^D C_{DA} \nabla ^A f + \frac{1}{2}\Delta {{\tilde{X}}}^k C_{DA}F^{DA} \\&\qquad - \nabla ^A {{\tilde{X}}}^k \nabla _D C^{DB} F_{AB} + \nabla ^A {{\tilde{X}}}^k \nabla ^D C_{DA}\Delta f\\&\quad = \int _{S^2} -\nabla ^A {{\tilde{X}}}^k C_{DA}\nabla ^D(\Delta + 2)f -\frac{1}{2}\nabla ^A {{\tilde{X}}}^k \nabla _A (C_{DB}F^{DB}) \\&\qquad - \nabla ^A {{\tilde{X}}}^k \nabla _D C^{DB} F_{AB}. \end{aligned}$$

$\quad \square$