Calculation of the angular momentum of the deformed BTZ black hole

Question

Consider the following deformed black string solution of low energy supergravity (describe single trace $T\bar{T}$ deformed BTZ black hole) given in eq.(2.17-2.18) here

$$ds^2 = -\frac{N^2}{1+\frac{\rho^2}{R^2}}d\tau^2+\frac{d\rho^2}{N^2}+\frac{\rho^2}{1+\frac{\rho^2}{R^2}}\left(d\theta-N_{\theta}d\tau\right)^2,\quad \theta \sim \theta +2\pi$$ $$B = \frac{\rho^{2}}{r_{5}}\sqrt{\left(1+\frac{\rho_{-}^{2}}{R^{2}}\right)\left(1+\frac{\rho_{+}^{2}}{R^{2}}\right)}\frac{1}{1+\frac{\rho^{2}}{R^{2}}}d\tau\wedge d\theta$$ $$e^{2\Phi} = \frac{kv}{p}\sqrt{\left(1+\frac{\rho_{-}^{2}}{R^{2}}\right)\left(1+\frac{\rho_{+}^{2}}{R^{2}}\right)}\frac{1}{1+\frac{\rho^{2}}{R^{2}}}$$ with: $$N^{2}=\frac{\left(\rho^{2}-\rho_{-}^{2}\right)\left(\rho^{2}-\rho_{+}^{2}\right)}{\rho^{2}r_{5}^{2}},\quad N_{\theta}=\frac{\rho_{-}\rho_{+}}{r_{5}\rho^{2}},\quad r_{5}=\sqrt{\alpha'k}$$ Where $\rho_{\pm}$ are the inner and outer horizons, $k,p,v$ are constants of the theory $\alpha'$ is the string tension, and $R$ is deformation parameter and also the asymptotic radius of the $\theta$ circle, in the $R\to\infty$ limit this solution reduces to the (undeformed) BTZ black string solution, and in the limit $\rho\to\infty$ the asymptotic regime is a flat spacetime with linear dilaton. The canonical coordinates $(t,x,\phi)$ of the flat spacetime with linear dilaton is obtained after making the following transformation: $$t=\frac{R}{r_5}\tau,\quad x=R\theta,\quad \phi =r_5\log\left(\frac{\rho}{R}\right)$$

According to eq.(21) in here the ADM mass of dilaton gravity systems, which tends to flat spacetimes in the asymptotics (given in the canonical normalized coordinates) is independent of the Kalb-Ramond (B) field and given by the formula: $$M_{ADM}=\frac{1}{2\kappa_0^2}\oint_{S^\infty} e^{-2\Phi}\left[-\partial_i\gamma_{tt}+\partial^{\mu}\gamma_{\mu i}-\partial_i\gamma-2\gamma_{i\mu}\partial^\mu\Phi\right]dS^i$$ Where $\gamma_{\mu\nu}=g_{\mu\nu}-\eta_{\mu\nu}$ with $g_{\mu\nu}$ the original metric, $\gamma = det(\gamma_{\mu\nu})$, and the integral is taken on the circle in radial infinity with normal to the sphere such that $dS^i=n^idS$ In our case according to eq.(2.23) here: $$\kappa_0^2=8\pi G_3=\frac{2\pi\sqrt{\alpha'}}{k^{3/2}v}$$ preforming the integral will yield to: $$M_{ADM}=\frac{\pi R}{\kappa_{0}^2}\lim_{\phi\to\infty}\left[\sqrt{g_{xx}}e^{-2\Phi}\left[-\partial_{\phi}\gamma_{tt}+g^{\phi\phi}\left(\partial_{\phi}\gamma_{\phi\phi}-2\gamma_{\phi\phi}\partial_{\phi}\Phi\right)-\partial_{\phi}\gamma\right]\right]$$ $$M_{ADM}=\frac{Rp}{\alpha'}\frac{\left(1+\frac{\rho_+^2+\rho_-^2}{R^2}\right)}{\sqrt{\left(1+\frac{\rho_-^2}{R^2}\right)\left(1+\frac{\rho_+^2}{R^2}\right)}}=\frac{Rp}{\alpha'}\sqrt{\left(1+\frac{\tilde{\rho}_-^2}{R^2}\right)\left(1+\frac{\tilde{\rho}_+^2}{R^2}\right)}$$ with: $$\tilde{\rho}_\pm=\frac{\rho_\pm}{\sqrt{1+\frac{\rho_{\mp}^2}{R^2}}}$$ which is indeed compatible with eq.(2.24) and eq.(2.34) in here

My question is what is the analogous formula for computing the ADM angular momentum of a general dilaton-gravity system, and how to apply it on the 2+1 dimensional system above. According to eq.(2.32) in here the result should be: $$J_{ADM}=\frac{p}{\alpha'}\frac{\rho_+\rho_-}{\sqrt{\left(1+\frac{\rho_+^2}{R^2}\right)\left(1+\frac{\rho_-^2}{R^2}\right)}}=\frac{p}{\alpha'}\tilde{\rho}_+\tilde{\rho}_-$$

This post imported from StackExchange Physics at 2024-12-29 21:31 (UTC), posted by SE-user Daniel Vainshtein