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--> --> --> $ S = \frac{1}{16\pi G_0}\int {\rm{d}} x^4\sqrt{-g}\left( \varphi R-\frac{\omega}{\varphi}\nabla_a\varphi\nabla^a\varphi \right), $ | (1) |
$\tag{2a} R_{ab}-\frac{1}{2}g_{ab}R = \frac{8\pi G_0}{\varphi}{\cal{T}}_{ab}, $ |
$\tag{2b} \nabla_a\nabla^a\varphi = 0, $ |
$\begin{aligned}[b] {\cal{T}}_{ab} =& \frac{1}{8\pi G_0}\left[\frac{\omega}{\varphi}\left(\nabla_a\varphi\nabla_b\varphi-\frac{1}{2}g_{ab}\nabla_c\varphi\nabla^c\varphi\right)\right.\\&+\nabla_a\nabla_b\varphi-g_{ab}\nabla_c\nabla^c\varphi\Bigg].\end{aligned} $ | (3) |
From a previous study [34], it is known that
$ S = \frac{1}{16\pi\tilde G}\int\sqrt{-\tilde{g}}\left( \tilde R-\frac{2\omega+3}{2} \tilde{\nabla}_a \tilde{\varphi} \tilde{\nabla}^a \tilde{\varphi} \right), $ | (4) |
$\tag{5a} \tilde R_{ab}-\frac{1}{2}\tilde g_{ab}\tilde R = 8\pi G_0\tilde{{\cal{T}}}_{ab}, $ |
$\tag{5b} \tilde\nabla_a\tilde\nabla^a\tilde\varphi = 0, $ |
$ \tilde{{\cal{T}}}_{ab} = \frac{2\omega+3}{16\pi G_0}\left( \tilde\nabla_a \tilde\varphi\tilde\nabla_b\tilde\varphi-\frac{1}{2}\tilde g_{ab}\tilde\nabla^c\tilde\varphi\tilde\nabla_c\tilde\varphi\right). $ | (6) |
As discussed in Ref. [34], Eqs. (2) can be solved using the generalized Bondi-Sachs coordinates
$\begin{aligned}[b] {\rm{d}} s^2 =& e^{2\beta}\frac{V}{r} {\rm{d}} u^2-2e^{2\beta} {\rm{d}} u {\rm{d}} r+h_{AB}( {\rm{d}} x^A\\&-U^A {\rm{d}} u)( {\rm{d}} x^B-U^B {\rm{d}} u), \end{aligned}$ | (7) |
$\begin{aligned}[b] \beta =& {\cal{O}}\left({r^{-1}}\right),\quad V =& -r+{\cal{O}}\left({r^0}\right),\quad\\ U^A =& {\cal{O}}\left({r^{-2}}\right), \end{aligned}$ | (8) |
$ \det(h_{AB}) = r^4\left( \frac{\varphi_0}{\varphi}\right)^2\sin^2\theta. $ | (9) |
$\tag{10a} \varphi = \varphi_0+\frac{\varphi_1}{r}+\frac{\varphi_2}{r^2}+{\cal{O}}\left({\frac{1}{r^3}}\right), $ |
$\tag{10b} g_{uu} = -1+\frac{2m+\varphi_1/\varphi_0}{r}+{\cal{O}}\left({\frac{1}{r^2}}\right), $ |
$\tag{10c}\begin{aligned}[b] g_{ur} =& -1+\frac{\varphi_1}{\varphi_0r}+\frac{1}{r^2}\left[ \frac{1}{16}\hat c_{A}^{B}\hat c^{A}_{B}+\frac{2\omega-5}{8}\left( \frac{\varphi_1}{\varphi_0} \right)^2\right.\\&\left.+\frac{\varphi_2}{\varphi_0} \right]+{\cal{O}}\left({\frac{1}{r^3}}\right), \end{aligned}$ |
$\tag{10d}\begin{aligned}[b] g_{uA} =& \frac{{\cal{D}}_B\hat c^B_A}{2}+\frac{2}{3r}\left[ N_A+\frac{1}{4}\hat c_{AB}{\cal{D}}_C\hat c^{BC}\right.\\&\left.-\frac{\varphi_1}{12\varphi_0}{\cal{D}}_B\hat c^B_A \right] +{\cal{O}}\left({\frac{1}{r^2}}\right),\end{aligned} $ |
$\tag{10e}\begin{aligned}[b] g_{AB} =& r^2\gamma_{AB}+r\left( \hat c_{AB}-\gamma_{AB}\frac{\varphi_1}{\varphi_0} \right)+\hat d_{AB}\\&+\gamma_{AB}\left( \frac{1}{4}\hat c_{C}^{D}\hat c^{C}_{D}+\frac{\varphi_1^2}{\varphi_0^2}-\frac{\varphi_2}{\varphi_0} \right) +{\cal{O}}\left({\frac{1}{r}}\right).\end{aligned} $ |
$ \tag{11a}\dot {m} = -\frac{1}{4}{\cal{D}}_A{\cal{D}}_BN^{AB}-\frac{1}{8}N_{AB}N^{AB}-\frac{2\omega+3}{4}\left( \frac{N}{\varphi_0} \right)^2, $ |
$\tag{11b} \begin{aligned}[b] \dot {N}_A =& {\cal{D}}_Am +\frac{1}{4}({\cal{D}}_B{\cal{D}}_A{\cal{D}}_C\hat c^{BC}-{\cal{D}}_B{\cal{D}}^B{\cal{D}}_C\hat c_A^C)\\ & -\frac{1}{16}{\cal{D}}_A(N^{B}_{C}\hat c_{B}^{C})+\frac{1}{4}N^{B}_{C}{\cal{D}}_A\hat c_{B}^{C}+\frac{1}{4}{\cal{D}}_B(N_{A}^{C}\hat c^{B}_{C}\\ &-\hat c_{A}^{C}N^{B}_{C}) +\frac{2\omega+3}{8\varphi_0^2}(\varphi_1{\cal{D}}_AN-3N{\cal{D}}_A\varphi_1), \end{aligned} $ |
$\tag{11c} \dot\varphi_2 = \frac{\varphi_1N}{\varphi_0}-\frac{1}{2}{\cal{D}}^2\varphi_1, $ |
As in GR, the asymptotically flat spacetime in BD also exhibits BMS symmetries. An infinitesimal BMS transformation,
$\tag{12a} \delta_\xi\varphi_1 = fN+\frac{\psi}{2}\varphi_1+Y^A{\cal{D}}_A\varphi_1, $ |
$\tag{12b} \delta_\xi\hat c_{AB} = -fN_{AB}-2{\cal{D}}_A{\cal{D}}_Bf+\gamma_{AB}{\cal{D}}^2f+ {\cal{L}}_Y\hat c_{AB}-\frac{\psi}{2}\hat c_{AB}, $ |
$\tag{12c} \delta_\xi N_{AB} = f\dot N_{AB}+ {\cal{L}}_YN_{AB}, $ |
$\tag{12d} \delta_\xi N = f\dot N+\psi N+Y^A{\cal{D}}_AN, $ |
In Ref. [34], we did not calculate the “conserved charges” of the asymptotically flat spacetime in the BD. In the present study, we computed them using the covariant phase space formulism devised by Wald and Zoupas [39]. For that purpose, we started with the asymptotic structure of BD, which will be described in the next section.
A.Radiative modes
Consequently, in Einstein frame, we can effectively perform the conformal completion for GR with a canonical scalar field. Many results obtained in GR can be carried over directly. For example, the conformal transformation of $ \Omega\bar S_{ab}+2 \bar{\nabla}_a \bar{n}_b-\bar f \bar{g}_{ab} = \Omega^{-1}\bar L_{ab}, $ | (13) |
$ \bar L_{ab} = \frac{2\omega+3}{2}\Omega^2\left(\bar{{\cal{T}}}_{ab}-\frac{1}{6} \bar{g}_{ab}\bar{{\cal{T}}}\right), $ | (14) |
$ \Omega \bar{\nabla}_a \bar{\nabla}^a \bar{\varphi}+ \bar{\varphi} \bar{\nabla}_a \bar{n}^a-2\bar f \bar{\varphi} = 0. $ | (15) |
In addition, the conformal factor can be freely chosen. A new conformal factor,
$\tag{16a} \bar{g}'_{ab} = \varpi^2 \bar{g}_{ab},\quad \bar{\varphi}' = \varpi^{-1} \bar{\varphi}, $ |
$\tag{16b} \bar{n}'_a = \varpi \bar{n}_a+\Omega \bar{\nabla}_a\varpi, $ |
$\tag{16c} \bar f' = \varpi^{-1}\bar f+2\varpi^{-2} \bar{n}^a \bar{\nabla}_a\varpi+\varpi^{-3}\Omega( \bar{\nabla}^a\varpi) \bar{\nabla}_a\varpi. $ |
$\bar f\ddot = 0,\quad \bar{\nabla}_a \bar{n}_b\ddot = 0. $ | (17) |
$ {\cal{L}}_{ \bar{n}} \bar{g}_{ab}\ddot = 0, $ | (18) |
According to the above discussion, the structure of
The first-order structure is covariant derivative
$ {\cal{D}}_a\gamma_{bc} = 0,\quad {\cal{D}}_a \bar{n}^b = 0. $ | (19) |
$ {\cal{D}}_{[a}{\cal{D}}_{b]}\nu_c = \frac{1}{2}{\cal{R}}_{abc}{}^d\nu_d. $ | (20) |
$ {\cal{S}}_b{}^a \bar{n}^b = ({\cal{S}}_b{}^b-{\cal{R}}) \bar{n}^a, $ | (21) |
$ {\cal{R}}_{abc}{}^d = \gamma_{c[a}{\cal{S}}_{b]}{}^d+{\cal{S}}_{c[a}\delta_{b]}{}^d, $ | (22) |
Owing to the topology of
$ \rho_{ab} \bar{n}^b = 0,\quad \gamma^{ab}\rho_{ab} = {\cal{R}}, \quad {\cal{D}}_{[a}\rho_{b]c} = 0. $ | (23) |
$ N_{ab} = {\cal{S}}_{ab}-\rho_{ab}. $ | (24) |
Finally, the third-order structure can be introduced. According to [52],
$ \bar R_{abcd} = \bar C_{abcd}+ \bar{g}_{a[c}\bar S_{d]b}- \bar{g}_{b[c}\bar S_{d]a}. $ | (25) |
$ K^{ab} = -4\Omega^{-1}\bar C^{acbd} \bar{n}_c \bar{n}_d,\quad {}^*K^{ab} = -4\Omega^{-1}{}^*\bar C^{acbd} \bar{n}_c \bar{n}_d, $ | (26) |
$ \gamma_{ac}K^{cb} = -\bar\epsilon_{acd} \bar{n}^d{}^*K^{cb},\quad \gamma_{ac}{}^*K^{cb} = \bar\epsilon_{acd} \bar{n}^dK^{cb}, $ | (27) |
$\tag{28a} {\cal{D}}_{[a}{\cal{S}}_{b]}{}^c = \frac{1}{4}\bar\epsilon_{abd}{}^*K^{dc}, $ |
$\tag{28b} {\cal{D}}_bK^{ab} = \frac{2(2\omega+3)}{3}\left[ \bar{\varphi} {\cal{L}}_{ \bar{n}}\bar N-2\bar N^2\right] \bar{n}^a, $ |
$\tag{28c} {\cal{D}}_b{}^*K^{ab} = 0. $ |
The gauge transformations of the above structures are given by [50]
$\tag{29a} \gamma'_{ab} = \varpi^2\gamma_{ab},\quad \bar{n}'^a = \varpi^{-1} \bar{n}^a, $ |
$\tag{29b} {\cal{D}}'_a\nu_b = {\cal{D}}_a\nu_b-2\varpi^{-1}\nu_{(a}{\cal{D}}_{b)}\varpi+\varpi^{-1}\gamma_{ab}\varpi^c\nu_c, $ |
$\tag{29c} N'_{ab} = N_{ab},\quad\bar N' = \varpi^{-2}\bar N, $ |
$\tag{29d} K'^{ab} = \varpi^{-5}K^{ab},\quad {}^*K'^{ab} = \varpi^{-5}{}^*K^{ab}, $ |
$ {\cal{D}}'_a\nu_b = {\cal{D}}_a\nu_b+\kappa \gamma_{ab} \bar{n}^c\nu_c, $ | (30) |
$ \Sigma_{ab} = ({\cal{D}}'_a-{\cal{D}}_a)\ell_b, $ | (31) |
$ N_{ab} = -2 {\cal{L}}_{ \bar{n}}\sigma_{ab}. $ | (32) |
The metric solution presented in the previous section is actually in the Bondi gauge. To demonstrate this, first, the solution is transformed to the one in Einstein frame, and then, a conformal transformation is performed with
$ \begin{aligned}[b] {\rm{d}} \bar s^2 =& \left[-\Omega^2+2\Omega^3m+{\cal{O}}\left({\Omega^4}\right)\right] {\rm{d}} u^2+2\left[1+{\cal{O}}\left({\Omega^2}\right)\right] {\rm{d}} u {\rm{d}}\Omega\\&+\left[ \Omega^2{\cal{D}}_B\hat c_A^B+{\cal{O}}\left({\Omega^3}\right) \right] {\rm{d}} u {\rm{d}} x^A+\Bigg[\gamma_{AB}+\Omega\hat c_{AB}\\&\left.+\Omega^2\left( \hat d_{AB}+\frac{\varphi_1}{\varphi_0}\hat c_{AB}+\frac{\gamma_{AB}}{4}\hat c_C^D\hat c^C_D \right)+{\cal{O}}\left({\Omega^3}\right)\right] {\rm{d}} x^A {\rm{d}} x^B,\end{aligned} $ | (33) |
$ \bar\varphi = \frac{\varphi_1}{\varphi_0}+\Omega\left( \frac{\varphi_2}{\varphi_0}-\frac{\varphi_1^2}{2\varphi_0^2}\right)+{\cal{O}}\left({\Omega^2}\right). $ | (34) |
2
B.BMS generators
As discussed in Ref. [56], an infinitesimal asymptotic symmetry, $\begin{aligned}[b] \Omega^2\delta_\xi \tilde{g}_{ab} =& \Omega^2 {\cal{L}}_{\xi} \tilde{g}_{ab}\\=& {\cal{L}}_\xi \bar{g}_{ab}-2 \bar{K} \bar{g}_{ab} = 2\Omega \bar{X}_{ab}, \end{aligned}$ | (35) |
$ {\cal{L}}_\xi \bar{g}_{ab} = 2( \bar{K} \bar{g}_{ab}+\Omega \bar{X}_{ab}). $ | (36) |
$ \begin{aligned}[b] &- \bar{\nabla}_a \bar{\nabla}_b \bar{K}+4 \bar{n}_{(a} \bar{X}_{b)}+2\Omega \bar{\nabla}_{(a} \bar{X}_{b)}- \bar{g}_{ab} \bar{n}_c \bar{X}^c \\& \quad -\frac{1}{2} {\cal{L}}_\xi( \bar S_{ab}-\Omega^{-2}\bar L_{ab})- {\cal{L}}_{ \bar{n}} \bar{X}_{ab} = 0, \end{aligned} $ | (37) |
$ {\cal{L}}_\xi \bar{n}^a = - \bar{K} \bar{n}^a+\Omega \bar{\nabla}^a \bar{K}-2\Omega^2 \bar{X}^a. $ | (38) |
$ {\cal{L}}_{ \bar{n}} \bar{K} = \frac{1}{2}( {\cal{L}}_\xi\bar f- \bar{K}\bar f), $ | (39) |
$ \delta_\xi \bar{\varphi} = {\cal{L}}_\xi \bar{\varphi}+ \bar{K} \bar{\varphi}, $ | (40) |
Now, we know how a BMS generator acts on
$\begin{aligned}[b] {\cal{L}}_\xi \gamma_{ab} = 2\bar K\gamma_{ab},\quad{\cal{L}}_\xi \bar{n}^a = -\bar K \bar{n}^a. \end{aligned}$ | (41) |
$ \xi^a = \alpha \bar{n}^a-\Omega \bar{\nabla}^a\alpha+\Omega^2u^a, $ | (42) |
$\tag{43a} \bar{K} = \Omega(\alpha\vartheta-\varsigma_\alpha+\varrho), $ |
$\tag{43b}\begin{aligned}[b] \bar{X}_{ab} =& -\bar\nabla_a\bar\nabla_b\alpha-\frac{1}{2}(\alpha\vartheta-2\varsigma_\alpha+2\varrho)\bar g_{ab}\\&-\frac{\alpha}{2} (\bar S_{ab}-\Omega^{-2}\bar L_{ab})+2 \bar{n}_{(a}u_{b)}+\Omega \bar{\nabla}_{(a}u_{b)}, \end{aligned}$ |
$ \tag{43c}\begin{aligned}[b] \bar{X}_a =& \frac{1}{2}\bar\nabla_a(\alpha\vartheta-2\varsigma_\alpha+\varrho)-\frac{1}{2}(\bar S_{ab}-\Omega^{-2}\bar L_{ab})\bar\nabla^b\alpha\\&+\frac{1}{2} \bar{n}^b \bar{\nabla}_bu_a+\frac{\Omega}{4}[3\vartheta u_a+(\bar S_{ab}-\Omega^{-2}\bar L_{ab})u^b], \end{aligned} $ |
$\tag{43d}\begin{aligned}[b] \bar{X} =& -\bar\nabla^2\alpha-2(\alpha\vartheta-2\varsigma_\alpha+\varrho)-\alpha\left(\frac{\bar R}{3}-\Omega^{-2}\bar L \right)\\&+\Omega \bar{\nabla}_au^a, \end{aligned}$ |
For a generic BMS generator
$\begin{aligned}[b]& \bar{n}^a\xi_a = 0,\\& {\cal{D}}_{(a}\xi_{b)} = \bar{K} \gamma_{ab}, \\& {\cal{L}}_{ \bar{n}}\xi_a = 0, \end{aligned}$ | (44) |
Once a foliation of
$ \xi^a\ddot = \left( \alpha+\frac{u}{2}{\cal{D}}\cdot Y \right) \bar{n}^a+Y^a, $ | (45) |
One should also know how a BMS generator transforms
$ ( {\cal{L}}_\xi{\cal{D}}_a-{\cal{D}}_a {\cal{L}}_\xi)\nu_b = (\xi^d{\cal{R}}_{dab}{}^c-{\cal{D}}_a{\cal{D}}_b\xi^c)\nu_c. $ | (46) |
$ \begin{aligned}[b]\delta_{\alpha \bar{n}}\Sigma_{ab} =& ( {\cal{L}}_{\alpha \bar{n}}{\cal{D}}_a-{\cal{D}}_a {\cal{L}}_{\alpha \bar{n}})\ell_b \\=& -{\cal{D}}_a{\cal{D}}_b\alpha-\frac{\alpha}{2}N_{ab}+\kappa'\gamma_{ab}, \end{aligned}$ | (47) |
$ \begin{aligned}[b] \delta_Y\Sigma_{ab} =& ( {\cal{L}}_\xi{\cal{D}}_a-{\cal{D}}_a {\cal{L}}_\xi)\ell_b \\=& -\frac{u}{2}{\cal{D}}_a{\cal{D}}_b({\cal{D}}\cdot Y)-\frac{1}{2}\sigma_{ab}{\cal{D}}\cdot Y\\&+ {\cal{L}}_Y\sigma_{ab}-\frac{u}{4}({\cal{D}}\cdot Y)N_{ab}-\ell_{(a}{\cal{D}}_{b)}({\cal{D}}\cdot Y)\\&+\frac{1}{4}\gamma_{ab}\ell{\cal{D}}\cdot Y+\frac{1}{2}\gamma_{ab}Y^c{\cal{D}}_c\ell, \end{aligned} $ | (48) |
$ \begin{aligned}[b] ( {\cal{L}}_\xi \bar{\nabla}_a- \bar{\nabla}_a {\cal{L}}_\xi)\nu_b =& -\nu_c\delta_\xi\bar\Gamma^c_{ab} = \nu_c[ \bar{n}^c \bar{X}_{ab}-2 \bar{n}_{(a} \bar{X}_{b)}{}^c\\&-2\delta^c_{(a} \bar{\nabla}_{b)} \bar{K}+ \bar{g}_{ab} \bar{\nabla}^c \bar{K}+{\cal{O}}\left({\Omega}\right)]. \end{aligned} $ | (49) |
$ \bar{X}_{ab} = ( {\cal{L}}_\xi{\cal{D}}_a-{\cal{D}}_a {\cal{L}}_\xi)\ell_b+2\ell_{(a}{\cal{D}}_{b)} \bar{K}-\gamma_{ab}\ell_c \bar{\nabla}^c \bar{K}, $ | (50) |
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A.(Pre)symplectic currents
Following Ref. [39], we can start with the variation of the action described by Eq. (1), $\begin{aligned}[b] \delta S =& \frac{1}{16\pi G_0}\int {\rm{d}}^4\sqrt{-g}(E_{ab}\delta g^{ab}+E_\varphi\delta\varphi)\\&+\int {\rm{d}}^4x\sqrt{-g}\nabla_a\theta^a, \end{aligned}$ | (51) |
$ \begin{aligned}[b] \theta_{abc}(\delta g,\delta\varphi) =& \frac{1}{16\pi G_0}\epsilon_{dabc}\bigg[ \varphi g^{de}g^{fh}(\nabla_f\delta g_{eh}-\nabla_e\delta g_{fh})\\ & +g^{de}g^{fh}(\delta g_{fh}\nabla_e\varphi-\delta g_{eh}\nabla_f\varphi)-\frac{2\omega}{\varphi}\delta \varphi\nabla^d\varphi\bigg].\end{aligned} $ | (52) |
$ \begin{aligned}[b] \omega_{abc} =& \delta\theta_{abc}(\delta' g,\delta'\varphi)-\delta'\theta_{abc}(\delta g,\delta\varphi) \\=& \frac{1}{16\pi G_0}\epsilon_{dabc}\varphi w^d+\frac{1}{16\pi G_0}\epsilon_{dabc}\\&\times \bigg[2g^{d[e}g^{f]h}(\delta\varphi\nabla_f\delta' g_{eh}\!-\!\delta'g_{eh}\nabla_f\delta\varphi)\\ &+(g^{dp}g^{eq}g^{fh}+g^{de}g^{fp}g^{qh})\delta g_{pq}\delta'g_{eh}\nabla_{f}\varphi\\&+\frac{1}{2}g^{fh}\delta'g_{fh}\delta g^{de}\nabla_e\varphi-\frac{2\omega}{\varphi}\delta'\varphi\bigg(\delta g^{de}\nabla_e\varphi+\nabla^d\delta\varphi\\&+\frac{1}{2}g^{ef}\delta g_{ef}\delta'\varphi\nabla^d\varphi\bigg)-\langle\delta\leftrightarrow\delta'\rangle\bigg], \end{aligned} $ | (53) |
$\begin{aligned}[b] w^a =& (g^{a[e}g^{d]c}g^{bf}+g^{ae}g^{b[f}g^{c]d}+g^{a[d}g^{b]c}g^{ef})\\&\times(\delta'g_{bc}\nabla_d\delta g_{ef}-\delta g_{bc}\nabla_d\delta'g_{ef}). \end{aligned}$ | (54) |
To resolve the complication, all quantities in Eqs. (52) and (53) should be replaced by the corresponding ones in Einstein frame. One may also directly calculate the presymplectic potential current and the symplectic current using the action described in Eq. (4) in Einstein frame, which are
$\tag{55a}\begin{aligned}[b] \tilde\theta_{abc} =& \frac{1}{16\pi\tilde G} \tilde{\epsilon}_{dabc}[ \tilde{g}^{de} \tilde{g}^{fh}( \tilde{\nabla}_f\delta \tilde{g}_{eh}- \tilde{\nabla}_e\delta \tilde{g}_{fh})\\&-(2\omega+3)\delta \tilde{\varphi} \tilde{\nabla}^d \tilde{\varphi}], \end{aligned}$ |
$ \tag{55b}\begin{aligned}[b] \tilde\omega_{abc} =& \frac{1}{16\pi \tilde{G}} \tilde{\epsilon}_{dabc}\tilde w^d -\frac{2\omega+3}{16\pi \tilde{G}} \tilde{\epsilon}_{dabc}\bigg[ \delta' \tilde{\varphi} \tilde{\nabla}^d\delta \tilde{\varphi}\\&+\delta' \tilde{\varphi}\delta \tilde{g}^{de} \tilde{\nabla}_e \tilde{\varphi}+\frac{1}{2} \tilde{g}^{ef}\delta \tilde{g}_{ef} \delta' \tilde{\varphi} \tilde{\nabla}^d \tilde{\varphi}-\langle\delta\leftrightarrow\delta'\rangle\bigg], \end{aligned} $ |
$ \tag{56a}\theta_{abc}(\delta g,\delta\varphi) = \tilde\theta_{abc}(\delta \tilde{g},\delta \tilde{\varphi})+\Delta_{abc}, $ |
$\tag{56b} \Delta_{abc}\equiv\frac{3}{16\pi \tilde{G}} \tilde{\epsilon}_{dabc}\left( \delta\tilde g^{de} \tilde{\nabla}_e \tilde{\varphi}+\frac{1}{2} \tilde{g}^{ef}\delta\tilde g_{ef} \tilde{\nabla}^d \tilde{\varphi}+ \tilde{\nabla}^d\delta \tilde{\varphi} \right). $ |
$ \begin{aligned}[b] \tilde{\epsilon}^{abcd} \tilde{\nabla}_a\Delta_{bcd} =& \frac{9}{8\pi \tilde{G}}\bigg[ \frac{1}{2} \tilde{g}^{ab}\delta \tilde{g}_{ab} \tilde{\nabla}^2 \tilde{\varphi}+\bigg( \delta \tilde{g}^{ab} \tilde{\nabla}_a \tilde{\nabla}_b \tilde{\varphi}\\&+ \tilde{\nabla}_a\delta \tilde{g}^{ab} \tilde{\nabla}_b \tilde{\varphi}+\frac{1}{2} \tilde{g}^{ab} \tilde{\nabla}_c\delta \tilde{g}_{ab} \tilde{\nabla}^c \tilde{\varphi}+ \tilde{\nabla}^2\delta \tilde{\varphi} \bigg) \bigg], \end{aligned} $ | (57) |
Now, let us choose an arbitrary, closed, embedded 3-dimensional hypersurface,
$ \Xi_\Sigma(\delta \tilde{g},\delta \tilde{\varphi};\delta' \tilde{g},\delta' \tilde{\varphi}) = \int_\Sigma\tilde\omega_{abc}(\delta \tilde{g},\delta \tilde{\varphi};\delta' \tilde{g},\delta' \tilde{\varphi}). $ | (58) |
$ {\not\! \delta} Q_\xi[\Sigma] = \int_\Sigma\tilde\omega_{abc}(\delta \tilde{g},\delta \tilde{\varphi}; {\cal{L}}_\xi \tilde{g}, {\cal{L}}_\xi \tilde{\varphi}). $ | (59) |
$ {\not\!\delta} Q_\xi[ \partial\Sigma] = \int_{ \partial\Sigma}[\delta\tilde Q_{ab}-\xi^c\tilde\theta_{cab}(\delta \tilde{g},\delta \tilde{\varphi})], $ | (60) |
$ \tilde Q_{ab} = -\frac{1}{16\pi \tilde{G}} \tilde{\epsilon}_{abcd} \tilde{\nabla}^c\xi^d, $ | (61) |
$ \int_{ \partial\Sigma}\xi^c\tilde\omega_{cab}(\delta \tilde{g},\delta \tilde{\varphi};\delta' \tilde{g},\delta' \tilde{\varphi}) = 0. $ | (62) |
Equations (55) and (61) are the most important ones for calculating the “conserved charges” at
$ \delta \bar{g}_{ab} = \Omega^2\delta \tilde{g}_{ab}\ddot = 0, $ | (63) |
$ \delta \bar{g}_{ab} = \Omega\tau_{ab},\quad\delta \tilde{g}_{ab} = \Omega^{-1}\tau_{ab}. $ | (64) |
$ \tau_{ab} \bar{n}^b = \Omega\tau_a. $ | (65) |
Now, it is straightforward to reexpress Eqs. (55) and (61) in the unphysical spacetime
$\begin{aligned}[b] \tilde\theta_{abc} =& \frac{1}{16\pi\tilde G_0}\bar\epsilon_{abcd}\left\{ \Omega^{-1}\left[ \bar{\nabla}_e\tau^{de}- \bar{\nabla}^d\tau-3\tau^d\right.\right.\\&\left.\left.-(2\omega+3)\chi \bar{\varphi} \bar{n}^d \right]-(2\omega+3)\chi \bar{\nabla}^d \bar{\varphi}\right\}, \end{aligned}$ | (66) |
$ \delta\bar S_{ab}\ddot = 4 \bar{n}_{(a}\tau_{b)}- \bar{n}^c \bar{\nabla}_c\tau_{ab}- \bar{g}_{ab} \bar{n}^c\tau_c+(2\omega+3) \bar{n}_a \bar{n}_b\chi \bar{\varphi}. $ | (67) |
$\begin{aligned}[b] \delta \bar S_{ab}\ddot =& - \bar{n}_{(a} \bar{\nabla}_{b)}\tau- \bar{n}^c \bar{\nabla}_c\tau_{ab}+ \bar{n}_{(a} \bar{\nabla}^c\tau_{b)c}+ \bar{n}_{(a}\tau_{b)}\\&-\frac{1}{3} \bar{g}_{ab}( \bar{n}^c\tau_c- \bar{n}^c \bar{\nabla}_c\tau). \end{aligned}$ | (68) |
$\tag{69a} \bar{\nabla}^b\tau_{ab}- \bar{\nabla}_a\tau-3\tau_a-(2\omega+3)\chi \bar{\varphi} \bar{n}_a\ddot = 0, $ |
$\tag{69b} \bar{n}^a \bar{\nabla}_a\tau+2 \bar{n}^a\tau_a\ddot = 0. $ |
$ \tilde Q_{ab}(\xi) = -\frac{1}{16\pi \tilde{G}}\bar\epsilon_{abcd} \bar{\nabla}^c(\Omega^{-2}\xi^d), $ | (70) |
$\begin{aligned}[b] \tilde\omega_{abc} = &-\frac{1}{32\pi \tilde{G}}\bar\epsilon_{abc}(\tau'^{de}\delta N_{de}-\tau^{de}\delta N'_{de})\\&+\frac{2\omega+3}{16\pi \tilde{G}}\bar\epsilon_{abc}(\chi'\delta \bar N-\chi\delta\bar N'), \end{aligned}$ | (71) |
$ \tilde\Theta_{abc}(\delta \tilde{g},\delta \tilde{\varphi}) = -\frac{1}{32\pi \tilde{G}}\bar\epsilon_{abc}\tau^{de} N_{de}+\frac{2\omega+3}{16\pi \tilde{G}}\bar\epsilon_{abc}\chi \bar N, $ | (72) |
2
B.Fluxes
Once $ \begin{aligned}[b] F_{\xi,{\cal{B}}} =& \int_{{\cal{B}}}\tilde\Theta_{abc}( {\cal{L}}_\xi \tilde{g}, {\cal{L}}_\xi \tilde{\varphi}) \\=&-\frac{1}{16\pi \tilde{G}}\int_{{\cal{B}}}\bar\epsilon_{abc}\Big\{ N_{de}[( {\cal{L}}_\xi{\cal{D}}_p-{\cal{D}}_p {\cal{L}}_\xi)\ell_q\\ &+2\ell_{(p}{\cal{D}}_{q)} \bar{K}]\gamma^{dp}\gamma^{eq}-(2\omega+3)( {\cal{L}}_\xi \bar{\varphi}+ \bar{K} \bar{\varphi})\bar N\Big\},\end{aligned} $ | (73) |
$ F_{\xi,{\cal{B}}} = -(Q_\xi[{\cal{C}}_2]-Q_\xi[{\cal{C}}_1]). $ | (74) |
If
$\tag{75a}\begin{aligned}[b] {\cal{H}}_\alpha \!=& \frac{1}{16\pi \tilde{G}}\int_{{\cal{I}}}\bar\epsilon_{abc}\bigg[N_{de}\left( {\cal{D}}_p{\cal{D}}_q\alpha\!+\!\frac{\alpha}{2}N_{pq} \right)\gamma^{dp}\gamma^{eq}\\&+\alpha(2\omega+3)\bar N^2\bigg], \end{aligned}$ |
$\tag{75b} \begin{aligned}[b] {\cal{H}}_Y =& \frac{1}{16\pi \tilde{G}}\int_{{\cal{I}}}\bar\epsilon_{abc} \bigg\{ N_{de}\bigg[ \frac{u}{2}{\cal{D}}_p{\cal{D}}_q({\cal{D}}\cdot Y)\\&+\frac{1}{2}\sigma_{pq} {\cal{D}}\cdot Y- {\cal{L}}_Y\sigma_{pq}+\frac{u}{4}N_{pq} {\cal{D}}\cdot Y \bigg]\gamma^{dp}\gamma^{eq} \\ & +(2\omega+3)\bar N\left[ \frac{1}{2}(u\bar N+ \bar{\varphi}) {\cal{D}}\cdot Y + {\cal{L}}_Y \bar{\varphi} \right]\bigg\}, \end{aligned} $ |
$\tag{76a}\begin{aligned}[b] {\cal{H}}_\alpha =& \frac{\varphi_0}{16\pi G_0}\int\alpha\bigg[ {\cal{D}}_A{\cal{D}}_BN^{AB}+\frac{1}{2}N_A^BN^A_B\\&+(2\omega+3)\left( \frac{N}{\varphi_0} \right)^2 \bigg] {\rm{d}} u {\rm{d}}^2{\bf{\Omega}}, \end{aligned}$ |
$\tag{76b} \begin{aligned}[b] {\cal{H}}_Y =& {\cal{H}}_{\alpha'}+\frac{\varphi_0}{32\pi G_0}\int Y^A \left[ \frac{1}{2}(\hat c_B^C{\cal{D}}_AN^B_C-N_B^C{\cal{D}}_A\hat c^B_C)\right.\\&+{\cal{D}}^B(N^C_B\hat c_{AC}-\hat c^C_B N_{AC})\\& \left.+\frac{2\omega+3}{\varphi_0^2}(N{\cal{D}}_A\varphi_1-\varphi_1{\cal{D}}_AN)\right] {\rm{d}} u {\rm{d}}^2{\bf{\Omega}}, \end{aligned} $ |
2
C.“Conserved charges”
Now, we can calculate the “conserved charges”. According to the decomposition expressed in Eq. (45), any BMS generator $\tag{77a} \xi^a_1\ddot = \frac{u-u_0}{2}\psi( \partial_u)^a+Y^A( \partial_A)^a, $ |
$\tag{77b} \xi^a_2\ddot = \left( \alpha+\frac{u_0}{2}\psi \right)( \partial_u)^a, $ |
For the Lorentz generator
$ Q_{\xi_1}[{\cal{C}}_0] = \oint_{{\cal{C}}_0}\tilde Q_{ab}(\xi_1), $ | (78) |
$\begin{aligned}[b] Q_{\xi_1}[{\cal{C}}_0] =& \frac{1}{16\pi \tilde{G}}\oint_{{\cal{C}}_0}Y^A\Biggr[ 2N_A+\frac{1}{16}{\cal{D}}_A(\hat c_{BC}\hat c^{BC})\\&+\frac{2\omega+3}{4}\frac{\varphi_1{\cal{D}}_A\varphi_1}{\varphi_0^2} \Biggr] {\rm{d}}^2{\bf{\Omega}}. \end{aligned}$ | (79) |
$ \delta Q_{\xi_2}[{\cal{C}}_0] = \oint_{{\cal{C}}_0}[\tilde Q_{ab}(\alpha \bar{n})-\alpha \bar{n}^c\tilde\theta_{cab}+\alpha \bar{n}^c\tilde\Theta_{cab}]. $ | (80) |
$\begin{aligned}[b]F_{\alpha \bar{n},{\cal{B}}} =& -\frac{1}{16\pi \tilde{G}}\int_{{\cal{B}}}\bar\epsilon_{abc}\Big[\gamma^{df}\gamma^{eh} N_{fh}\left( {\cal{L}}_{\alpha \bar{n}}{\cal{D}}_f\right.\\&\left.-{\cal{D}}_f {\cal{L}}_{\alpha \bar{n}}\right)\ell_h+(2\omega+3)\alpha \bar N^2\Big], \end{aligned}$ | (81) |
$ Q_\alpha[{\cal{C}}] = \frac{1}{8\pi \tilde{G}}\oint_{{\cal{C}}}P^d\ell_d \bar{n}^c\bar\epsilon_{cab}, $ | (82) |
$\begin{aligned}[b] P^a =& \frac{\alpha}{4}K^{ab}\ell_b+ N_{cd}\gamma^{bd}\gamma^{c[e} \bar{n}^{a]}(\alpha{\cal{D}}_e\ell_b+\ell_e{\cal{D}}_b\alpha)\\&-\frac{2\omega+3}{6}\alpha \bar{n}^a \bar{\varphi} \bar N. \end{aligned}$ | (83) |
$ Q_{\xi_2}[{\cal{C}}_0] = \frac{1}{8\pi \tilde{G}}\oint_{{\cal{C}}_0}\left( 2\alpha m-u_0Y^A{\cal{D}}_A m\right) {\rm{d}}^2{\bf{\Omega}}. $ | (84) |
The total “conserved charge” is the sum of Eqs. (79) and (84):
$\begin{aligned}[b] Q_\xi[{\cal{C}}] =& \frac{\varphi_0}{8\pi G_0}\oint_{{\cal{C}}}\left[ 2\alpha m-u {\cal{L}}_Ym +Y^AN_A+\frac{1}{32} {\cal{L}}_Y(\hat c_{B}^A\hat c^{B}_A)\right.\\&\left.+\frac{2\omega+3}{8}\frac{\varphi_1 {\cal{L}}_Y\varphi_1}{\varphi_0^2}\right] {\rm{d}}^2{\bf{\Omega}}, \end{aligned}$ | (85) |
$\begin{aligned}[b]& \int_{ \partial\Sigma}\left\{\eta^c\tilde\theta_{cab}( {\cal{L}}_\xi \tilde{g}, {\cal{L}}_\xi \tilde{\varphi})-\xi^c\tilde\theta_{cab}( {\cal{L}}_\eta \tilde{g}, {\cal{L}}_\eta \tilde{\varphi})\right.\\&\left.\quad+\tilde{{\cal{L}}} \tilde{\epsilon}_{abcd}\eta^c\xi^d-\tilde Q_{ab}[ {\cal{L}}_\eta\xi]\right\} = 0, \end{aligned}$ | (86) |
Now, let us work out the “conserved charges” for some specific BMS generators. First, consider a generic supertranslation generator
$ {\cal{P}}_\alpha[{\cal{C}}] = \frac{\varphi_0}{4\pi G_0}\oint_{{\cal{C}}}\alpha m {\rm{d}}^2{\bf{\Omega}}. $ | (87) |
$ M = \frac{\varphi_0}{4\pi G_0}\oint_{{\cal{C}}} m {\rm{d}}^2{\bf{\Omega}}, $ | (88) |
$ Y^A = {\cal{D}}^A\mu+\epsilon^{AB}{\cal{D}}_B\upsilon, $ | (89) |
$ {\cal{K}}_\mu[{\cal{C}}]\! =\! -\frac{\varphi_0}{8\pi G_0}\oint_{{\cal{C}}}\mu\left( {\cal{D}}^AN_A\!+\!2um-\frac{\hat c_A^{B}\hat c^A_B}{16}-\frac{2\omega+3}{8}\frac{\varphi_1^2}{\varphi_0^2} \right) {\rm{d}}^2{\bf{\Omega}}, $ | (90) |
$ {\cal{J}}_\upsilon[{\cal{C}}] = -\frac{\varphi_0}{8\pi G_0}\oint_{{\cal{C}}}\upsilon\epsilon^{AB}{\cal{D}}_AN_B {\rm{d}}^2{\bf{\Omega}}, $ | (91) |
We also consider the memory effects between vacuum states in the tensor and scalar sectors. Following Ref. [34], a vacuum state in the scalar sector is simply given by
$ \hat c_{AB} = \left({\cal{D}}_A{\cal{D}}_B-\frac{1}{2}\gamma_{AB}{\cal{D}}^2\right)\Phi+\epsilon_{C(A}{\cal{D}}_{B)}{\cal{D}}^C\Upsilon, $ | (92) |
2
A.Displacement memory effects
Let us start with the displacement memory effect in the tensor sector. First, we rewrite the flux-balance law associated with supertranslation $ \begin{aligned}[b] F_{\alpha \bar{n},{\cal{B}}} =& \frac{\varphi_0}{16\pi G_0}\int_{{\cal{B}}}\alpha\bigg[ {\cal{D}}^A{\cal{D}}^BN_{AB}+\frac{N_{AB}N^{AB}}{2}\\&+(2\omega+3)\left( \frac{N}{\varphi_0} \right)^2 \bigg] {\rm{d}} u {\rm{d}}^2{\bf{\Omega}} = -\Delta{\cal{P}}_\alpha, \end{aligned} $ | (93) |
$ \oint_{{\cal{C}}}\alpha{\cal{D}}^2({\cal{D}}^2+2)\Delta\Phi {\rm{d}}^2{\bf{\Omega}} = \frac{32\pi G_0}{\varphi_0}({\cal{E}}_\alpha+\Delta{\cal{P}}_\alpha), $ | (94) |
Now, let us consider the displacement memory effect in the scalar sector. Following the above argument, one may want to consider the flux-balance law for Lorentz generator
$ \begin{aligned}[b] F_{Y,{\cal{B}}} =& \Delta{\cal{P}}_{\alpha'}-\frac{\varphi_0}{16\pi G_0}\int_{{\cal{B}}}\bigg[Y^AJ_A +\upsilon\epsilon_{AB}N^{CA}\hat c_C^B\bigg] {\rm{d}} u {\rm{d}}^2{\bf{\Omega}}\\ & +\frac{\varphi_0}{64\pi G_0}\oint_{{\cal{C}}}\mu\Delta\left( \frac{2\omega+3}{\varphi_0^2}\varphi_1^2+\frac{\hat c_A^B\hat c^A_B}{2} \right) {\rm{d}}^2{\bf{\Omega}}\\ =& -\Delta{\cal{K}}_\mu-\Delta{\cal{J}}_\upsilon, \end{aligned} $ | (95) |
$ J_A = \frac{1}{2} N^B_C{\cal{D}}_A\hat c^C_B-\frac{2\omega+3}{\varphi_0^2}N{\cal{D}}_A\varphi_1. $ | (96) |
$ \oint\mu\Delta\left[ \frac{2\omega+3}{\varphi_0^2}\varphi_1^2+\frac{\hat c_A^B\hat c^A_B}{2} \right] {\rm{d}}^2{\bf{\Omega}} = -\frac{16\pi G_0}{\varphi_0}(\Delta{\cal{K}}_\mu+\Delta{\cal{P}}_{\alpha'}+{\cal{J}}_\mu), $ | (97) |
$ {\cal{J}}_\mu = \frac{\varphi_0}{16\pi G_0}\int_{{\cal{B}}}\mu {\cal{D}}^AJ_A {\rm{d}} u {\rm{d}}^2{\bf{\Omega}}. $ | (98) |
In fact, the equation of motion gives a constraint on
$\begin{aligned}[b] \Delta\varphi_1^2 =& \frac{16\varphi_0^2}{2\omega+3}\bigg\{\frac{1}{32}\Delta(\hat c_A^B\hat c_B^A)+ {\cal{D}}^{-2}{\cal{D}}^A\Delta N_A \\&-\int_{u_i}^{u_f} {\rm{d}} u\left[ m+\frac{1}{2}{\cal{D}}^{-2}{\cal{D}}^AJ_A \right]\bigg\},\end{aligned} $ | (99) |
2
B.Spin memory effect
Spin memory effect exists only in the tensor sector, as it depends on the leading order term in $ \begin{aligned}[b] {\cal{F}}_{Y,{\cal{B}}} =& \frac{\varphi_0}{32\pi G_0}\int_{{\cal{B}}}Y^A{\cal{D}}^B({\cal{D}}_A{\cal{D}}_C\hat c^C_B-{\cal{D}}_B{\cal{D}}_C\hat c^C_A) {\rm{d}} u {\rm{d}}^2{\bf{\Omega}}\\ =& \frac{\varphi_0}{64\pi G_0}\int_{{\cal{B}}}\epsilon_{AB}Y^A{\cal{D}}^B{\cal{D}}^2({\cal{D}}^2+2)\Upsilon {\rm{d}} u {\rm{d}}^2{\bf{\Omega}}. \end{aligned} $ | (100) |
$ \oint_{{\cal{C}}}\upsilon{\cal{D}}^2{\cal{D}}^2({\cal{D}}^2+2)\Delta{\cal{R}} {\rm{d}}^2{\bf{\Omega}} = -\frac{32\pi G_0}{\varphi_0}(\Delta{\cal{J}}_\upsilon+{\cal{Q}}_\upsilon+\bar{\cal{J}}_\upsilon), $ | (101) |
$\tag{102a} {\cal{Q}}_\upsilon = -\frac{\varphi_0}{16\pi G_0}\int_{{\cal{B}}}\upsilon\epsilon_{AB}N^{AC}\hat c^B_C {\rm{d}} u {\rm{d}}^2{\bf{\Omega}}, $ |
$\tag{102b} \bar{\cal{J}}_\upsilon = \frac{\varphi_0}{16\pi G_0}\int_{{\cal{B}}}\upsilon\epsilon^{AB}{\cal{D}}_AJ_B {\rm{d}} u {\rm{d}}^2{\bf{\Omega}}. $ |
2
C.Center-of-mass memory effect
Now, consider the CM memory. Given that $\tag{103a} \oint_{{\cal{C}}}\alpha{\cal{D}}^2({\cal{D}}^2+2)\Delta\Phi_n {\rm{d}}^2{\bf{\Omega}} = \frac{32\pi G_0}{\varphi_0}{\cal{E}}_\alpha, $ |
$\tag{103b} \oint_{{\cal{C}}}\alpha{\cal{D}}^2({\cal{D}}^2+2)\Delta\Phi_o {\rm{d}}^2{\bf{\Omega}} = \frac{32\pi G_0}{\varphi_0}\Delta{\cal{P}}_\alpha. $ |
$\setcounter{equation}{104} \Delta{\cal{K}} = \int_{u_i}^{u_f}u \partial_u\Phi_o {\rm{d}} u. $ | (104) |
$ \Delta{\cal{P}}_{\alpha'} = -\frac{\varphi_0}{64\pi G_0}\oint_{{\cal{C}}}\mu{\cal{D}}^2{\cal{D}}^2({\cal{D}}^2+2)\Delta{\cal{K}} {\rm{d}}^2{\bf{\Omega}}. $ | (105) |
$ \oint_{{\cal{C}}}\mu{\cal{D}}^2{\cal{D}}^2({\cal{D}}^2+2)\Delta{\cal{K}} {\rm{d}}^2{\bf{\Omega}} = \frac{64\pi G_0}{\varphi_0}({\cal{J}}_\mu-\Delta{\cal{K}}'_\mu), $ | (106) |
$ \Delta{\cal{K}}'_\mu = -\frac{\varphi_0}{8\pi G_0}\oint_{{\cal{C}}}\mu\Delta({\cal{D}}^AN_A+2um) {\rm{d}}^2{\bf{\Omega}}. $ | (107) |
$\tag{A1} \int_{{\cal{C}}}\tilde Q_{ab} = \int_{S_0}\tilde Q_{ab}+\int_{\Sigma'} {\rm{d}}_a\tilde Q_{bc}, $ |
The integrand of the third integral can be contracted with
$\tag{A2}\begin{aligned}[b] \bar{\nabla}^b \bar{\nabla}_{[a}(\Omega^{-2}\xi_{b]}) =& \Omega^{-2}\bar R_{ab}\xi^b+ \bar{\nabla}_a \bar{\nabla}_b(\Omega^{-2}\xi^b)\\&- \bar{\nabla}^b \bar{\nabla}_{(a}(\Omega^{-2}\xi_{b)}),\end{aligned} $ |
$\tag{A3} \begin{aligned}[b] \bar{\nabla}_a \bar{\nabla}_b(\Omega^{-2}\xi^b)- \bar{\nabla}^b \bar{\nabla}_{(a}(\Omega^{-2}\xi_{b)}) =& \Omega^{-1}(3X_a+ \bar{\nabla}_aX- \bar{\nabla}^bX_{ab})\\ & +\Omega^{-3}\xi^b(2 \bar{\nabla}_a \bar{n}_b+ \bar{g}_{ab} \bar{\nabla}_c \bar{n}^c\\&-3\Omega^{-1} \bar{g}_{ab} \bar{n}_c \bar{n}^c). \end{aligned} $ |
$\tag{A4}\begin{aligned}[b] \Omega\bar R_{ab}&+2 \bar{\nabla}_a \bar{n}_b+ \bar{g}_{ab} \bar{\nabla}_c \bar{n}^c-2\Omega^{-1} \bar{g}_{ab} \bar{n}_c \bar{n}^c\\&-\frac{1}{2}\Omega^{-1} \bar{g}_{ab}\bar L = \Omega^{-1}\bar L_{ab}, \end{aligned}$ |
$\tag{A5}\begin{aligned}[b] \bar{\nabla}^b \bar{\nabla}_{[a}(\Omega^{-2}\xi_{b]}) =& \Omega^{-1}(3X_a+ \bar{\nabla}_aX- \bar{\nabla}^bX_{ab})\\&+\Omega^{-4}\xi^b\left( \bar L_{ab}+\frac{1}{2} \bar{g}_{ab}\bar L \right). \end{aligned}$ |
$\tag{A6} \begin{aligned}[b] \bar{\nabla}^b \bar{\nabla}_{[a}(\Omega^{-2}\xi_{b]}) =& \Omega^{-1}\bigg[ 3X_a+ \bar{\nabla}_aX- \bar{\nabla}^bX_{ab}\\&+\frac{2\omega+3}{2}( \bar{K} \bar{\varphi}^2+ \bar{\varphi} {\cal{L}}_\xi \bar{\varphi}) \bar{n}_a\bigg]\\ &+ \bar{K} \bar{\varphi} \bar{\nabla}_a \bar{\varphi}+ \bar{\nabla}_a \bar{\varphi} {\cal{L}}_\xi \bar{\varphi}. \end{aligned} $ |
$ \tag{A7a}\begin{aligned}[b] {\cal{L}}_\xi\bar R_{ab} =& -2 \bar{\nabla}_a \bar{\nabla}_b \bar{K}+2 \bar{n}_{(a}(X_{b)}- \bar{\nabla}_{b)}X+ \bar{\nabla}^cX_{b)c})\\&+4X_{c(a} \bar{\nabla}_{b)} \bar{n}^c-X_{ab} \bar{\nabla}_c \bar{n}^c-X \bar{\nabla}_a \bar{n}_b-2 {\cal{L}}_{ \bar{n}}X_{ab}\\&+\Omega(2 \bar{\nabla}_c \bar{\nabla}_{(a}X^c_{b)}+2 \bar{\nabla}_{(a}X_{b)}- \bar{\nabla}_c \bar{\nabla}^cX_{ab}- \bar{\nabla}_a \bar{\nabla}_bX),\end{aligned} $ |
$\tag{A7b} \begin{aligned}[b] {\cal{L}}_\xi\bar R =& -2 \bar{\nabla}_a \bar{\nabla}^a \bar{K}+4 \bar{n}^aX_a-2X_{ab} \bar{\nabla}^a \bar{n}^b\\&-2X \bar{\nabla}_a \bar{n}^a-4 {\cal{L}}_{ \bar{n}}X-2 \bar{K}\bar R+2\Omega( \bar{\nabla}_a \bar{\nabla}_bX^{ab}\\&- \bar{\nabla}_a \bar{\nabla}^aX+2 \bar{\nabla}_aX^a-X^{ab}\bar R_{ab}). \end{aligned} $ |
$\tag{A8} \begin{aligned}[b] &\bar{n}_{(a}\left[3X_{b)}+ \bar{\nabla}_{b)}- \bar{\nabla}^cX_{b)c}+\frac{2\omega+3}{2} \bar{\varphi}( {\cal{L}}_\xi \bar{\varphi}+ \bar{K} \bar{\varphi}) \bar{n}_{b)}\right]\\&\quad -\frac{1}{6} \bar{g}_{ab}( \bar{\nabla}_c \bar{\nabla}^c \bar{K}+4 \bar{n}^cX_c+2 {\cal{L}}_{ \bar{n}}X)\ddot = 0. \end{aligned} $ |
$\tag{A9a} 3X_a+ \bar{\nabla}_aX- \bar{\nabla}^bX_{ab}+\frac{2\omega+3}{2} \bar{\varphi}( {\cal{L}}_\xi \bar{\varphi}+ \bar{K} \bar{\varphi}) \bar{n}_a\ddot = 0, $ |
$\tag{A9b} 2 \bar{n}^aX_a+ {\cal{L}}_{ \bar{n}}X\ddot = 0. $ |
$\tag{B1} \int_{ \partial\Sigma}[\eta^c\tilde\theta_{cab}( {\cal{L}}_\xi \tilde{g}, {\cal{L}}_\xi \tilde{\varphi})-\xi^c\tilde\theta_{cab}( {\cal{L}}_\eta \tilde{g}, {\cal{L}}_\eta \tilde{\varphi})] = 0, $ |
$\tag{B2} \eta^c\tilde\theta_{cab} = \frac{1}{16\pi \tilde{G}}\beta F(\alpha) \bar{n}^c\bar\epsilon_{cab}, $ |
$\tag{B3} \begin{aligned}[b] F(\alpha) =& - \bar{n}_a\Omega^{-1}[ \bar{\nabla}_b\lambda^{ab}- \bar{\nabla}^a\lambda-3\lambda^a-(2\omega+3)\chi \bar{\varphi} \bar{n}^a]\\&+(2\omega+3)\chi\bar N \ddot = - \bar{\nabla}_a \bar{\nabla}_b\lambda^{ab}+ \bar{\nabla}^2\lambda\\&+3 \bar{\nabla}_a\lambda^a+(2\omega+3)\chi\bar N, \end{aligned} $ |
$\tag{B4} F(\alpha)\ddot = - \bar{\nabla}_a \bar{\nabla}_b\lambda^{ab}+3 \bar{\nabla}_a\lambda^a+\frac{3}{4} \bar{\nabla}^2\lambda+\frac{1}{24}\bar R\lambda+(2\omega+3)\chi\bar N. $ |
$\tag{B5} \bar{K} = \Omega(\alpha\vartheta-\sigma_\alpha), $ |
$\tag{B6} \lambda_{ab} = -2\bar\nabla_a\bar\nabla_b\alpha-(\alpha\vartheta-2\sigma_\alpha)\bar g_{ab}-\alpha (\bar S_{ab}-\Omega^{-2}\bar L_{ab}), $ |
$\tag{B7} \lambda_a = \bar\nabla_a(\alpha\vartheta-2\sigma_\alpha)-(\bar S_{ab}-\Omega^{-2}\bar L_{ab})\bar\nabla^b\alpha, $ |
$\tag{B8} \lambda = -2\bar\nabla^2\alpha-4(\alpha\vartheta-2\sigma_\alpha)-\alpha\left(\frac{\bar R}{3}-\Omega^{-2}\bar L \right). $ |
$\tag{B9}\begin{aligned}[b] F(\alpha)\ddot = & -\frac{\alpha(2\omega+3)}{3}\left[2\bar N^2- \bar{\varphi} {\cal{L}}_{ \bar{n}}^2 \bar{\varphi}\right]-\frac{1}{4}\left( \bar{\nabla}^2-\frac{\bar R}{6}\right)\\&\times\left( \bar{\nabla}^2\alpha+2\alpha\vartheta-4\sigma_\alpha+\frac{\alpha \bar R}{6}\right), \end{aligned}$ | (B9) |
$\tag{B10} \bar{\nabla}^2\alpha\ddot = {\cal{D}}^2\alpha+2\sigma_\alpha. $ | (B10) |
$\tag{B11}\begin{aligned}[b] {\cal{L}}_{ \bar{n}} \bar{\nabla}^2\alpha =& 2 {\cal{L}}_{ \bar{n}}\sigma_\alpha+\Omega\bigg( \bar\nabla^2\sigma_\alpha+2\vartheta\sigma_\alpha-\frac{\bar R\sigma_\alpha}{3}\\&+\frac{1}{6}\bar\nabla_a\alpha\bar\nabla^a\bar R-\vartheta\bar{\nabla}^2\alpha+\bar S^{ab}\bar\nabla_a \bar\nabla_b\alpha\bigg)+{\cal{O}}\bigg({\Omega^2}\bigg),\end{aligned} $ | (B11) |
$\tag{B12} \begin{aligned}[b]\bar{\nabla}^2 \bar{\nabla}^2\alpha =& {\cal{D}}^2{\cal{D}}^2\alpha+4\bar\nabla^2\sigma_\alpha+4\vartheta\sigma_\alpha-\frac{2}{3}\bar R\sigma_\alpha\\&+\frac{1}{3}\bar\nabla_a\alpha\bar\nabla^a\bar R-2\vartheta\bar\nabla^2\alpha+2\bar S^{ab}\bar\nabla_a\bar\nabla_b\alpha+{\cal{O}}\left({\Omega}\right).\end{aligned}$ |
$\tag{B13} X^{ \bar{n}}_{ab} = -\frac{\vartheta}{2} \bar{g}_{ab}-\frac{1}{2}(\bar S_{ab}-\Omega^{-2}\bar L_{ab}),\quad X^{ \bar{n}}_a = \frac{ \bar{\nabla}_a\vartheta}{2}, $ |
$\tag{B14} X^{ \bar{n}} = -2\vartheta-\frac{\bar R}{6}+\frac{\Omega^{-2}\bar L}{2}, \quad \bar{K}^{ \bar{n}} = \Omega\vartheta. $ |
$\tag{B15} \bar{\nabla}^2 \bar{K}^{ \bar{n}} = 2 {\cal{L}}_{ \bar{n}}\vartheta+\Omega\left( \bar{\nabla}^2\vartheta+ 2\vartheta^2-\frac{\vartheta\bar R}{6} \right)+{\cal{O}}\left({\Omega^2}\right), $ |
$\tag{B16} {\cal{L}}_{ \bar{n}}\vartheta = -\Omega\left(\frac{1}{2} \bar{\nabla}^2\vartheta+ \vartheta^2-\frac{\vartheta\bar R}{12} \right)+{\cal{O}}\left({\Omega^2}\right). $ |
$\tag{B17} \bar{\nabla}^2\vartheta = -\vartheta^2+\frac{\vartheta\bar R}{12}+\frac{1}{2}{\cal{D}}^2\vartheta+{\cal{O}}\left({\Omega}\right). $ |
$\tag{B18}\begin{aligned}[b]{\cal{L}}_{ \bar{n}}\bar R =& \Omega\bigg\{-\frac{\vartheta\bar R}{2}+\frac{3}{2}\bar S^{ab}\bar S_{ab}+(2\omega+3)\\&\times\left[ 4\bar N^2+ \bar{\varphi} {\cal{L}}_{ \bar{n}}^2 \bar{\varphi} \right]\bigg\}+{\cal{O}}\left({\Omega^2}\right),\end{aligned} $ |
$\tag{B19} \bar{\nabla}^2\bar R = -\vartheta\bar R+3\bar S^{ab}\bar S_{ab}+2(2\omega+3)\left[ 4\bar N^2+ \bar{\varphi} {\cal{L}}_{ \bar{n}}^2 \bar{\varphi} \right]+{\cal{O}}\left({\Omega}\right). $ |
$\tag{B20}\begin{aligned}[b] F(\alpha) =& \frac{1}{2}\bigg({\cal{D}}^2{\cal{D}}^2\alpha+2{\cal{D}}^2\alpha+2 N^{ab}{\cal{D}}_a{\cal{D}}_b\alpha\\&+\frac{\alpha}{2} N^{ab} N_{ab} \bigg)+\frac{2(2\omega+3)}{3}\alpha {\cal{L}}_{ \bar{n}}( \bar{\varphi}\bar N). \end{aligned}$ |
$\tag{B21} F(\alpha) = \frac{1}{2}\left({\cal{D}}^2{\cal{D}}^2\alpha+2{\cal{D}}^2\alpha \right), $ |