Reference: this paper page 15
Let $(M,g)$ be a three-manifold and consider a two-sided compact surface $\Sigma \subset M$.
The mass is defined by
$$m(\Sigma) =\ \bigg( \frac{|\Sigma|}{16\pi} \bigg)^{1/2} \bigg( 1 - \frac{1}{ 16\pi } \int_{\Sigma} H^2 d\sigma - \frac{ \Lambda}{24\pi } |\Sigma| \bigg) \tag{1}$$
where
$\Lambda = \text{inf}_M ~ R$
$R$ is the scalar curvature of $M$.
$H$ is the mean curvature of $\Sigma$.
$K_\Sigma$ is the Gauss curvature of $\Sigma$.
$|\Sigma|$ is the area of $\Sigma$.
The first variation of $m$:
$$\frac{d}{dt}m(\Sigma(t))\bigg|_{t=0} = - \frac{2|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} \varphi \Delta_\Sigma H d\sigma \\ + \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} \bigg[ 2K_\Sigma - \frac{8\pi}{|\Sigma| }+ \bigg( \frac{1}{2|\Sigma|}\int_\Sigma H^2 d\sigma - |A|^2 \bigg) \bigg] H \phi d\sigma \\ + \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_\Sigma (\Lambda - R) H\varphi d\sigma \tag{2}$$
I want to prove (2).
My Attempt:
The mass could also be written as
$$m(\Sigma) =\ \frac{|\Sigma|^{1/2}}{(16\pi)^{1/2}} \frac{1}{16\pi} \int_\Sigma 2\frac{8\pi}{|\Sigma|} d\sigma - \frac{|\Sigma|^{1/2}}{(16\pi)^{1/2}} \frac{1}{ 16\pi } \int_{\Sigma} H^2 d\sigma - \frac{|\Sigma|^{1/2}}{(16\pi)^{1/2}} \frac{ \Lambda}{24\pi } |\Sigma| \bigg) \tag{3}$$
$$m(\Sigma) =\ \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_\Sigma 2\frac{8\pi}{|\Sigma|} d\sigma - \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} H^2 d\sigma - \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_\Sigma \frac{2}{3}\Lambda d\sigma \bigg) \tag{4}$$
Since $\frac{d}{dt}d\sigma = -H\phi d\sigma$, then the first variation becomes
$$\frac{d}{dt}m(\Sigma) =\ - \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_\Sigma 2\frac{8\pi}{|\Sigma|} H\varphi d\sigma - \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} 2H \frac{dH}{dt} d\sigma \\ + \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} H^2 H\varphi d\sigma + \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_\Sigma \frac{2}{3}\Lambda H\varphi d\sigma \bigg) \tag{5}$$
Here
$$\frac{dH}{dt}=\Delta\varphi + \text{Ric}(\nu,\nu)\varphi + |A|^2\varphi = \Delta\varphi + \frac{R}{2}\varphi - K_\Sigma\varphi + \frac{H^2}{2}\varphi + \frac{|A|^2}{2}\varphi$$
so that
$$\frac{d}{dt}m(\Sigma) =\ - \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_\Sigma 2\frac{8\pi}{|\Sigma|} H\varphi d\sigma \\ - \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} 2H \bigg( \Delta\varphi + \frac{R}{2}\varphi - K_\Sigma\varphi + \frac{H^2}{2}\varphi + \frac{|A|^2}{2}\varphi \bigg) d\sigma \\ + \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} H^2 H\varphi d\sigma + \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_\Sigma \frac{2}{3}\Lambda H\varphi d\sigma \bigg) \tag{6}$$
$$\frac{d}{dt}m(\Sigma) =\ - \frac{2|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} H\Delta\varphi d\sigma \\ + \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_\Sigma \bigg( 2K_\Sigma - 2\frac{8\pi}{|\Sigma|}\bigg) H\varphi d\sigma \\ + \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} \bigg( H^2 - H^2 + |A|^2 \bigg)H\varphi d\sigma \\ + \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_\Sigma \bigg( \frac{2}{3}\Lambda - R \bigg) H\varphi d\sigma \tag{7}$$
which is different from (2).
Did I miss something?
The mass could also be written as
\begin{align} m(\Sigma) =\ \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \bigg( \int_{\Sigma} \frac{16\pi}{|\Sigma|}\ d\sigma - \int_{\Sigma} H^2 d\sigma - \int_{\Sigma} \frac{ 2}{3 }\Lambda d\sigma \bigg) \end{align}
Since $\frac{d}{dt}|\Sigma_t|=\frac{d}{dt}\int_{\Sigma}d\sigma = \int_{\Sigma} \frac{d}{dt} d\sigma = - \int_{\Sigma} H\varphi\ d\sigma$, then
\begin{align} \frac{d}{dt}|\Sigma|^{1/2} =\ & \frac{1}{2} |\Sigma_t|^{-1/2} \frac{d}{dt}\bigg( |\Sigma_t| \bigg) = - \frac{1}{2} \frac{1}{|\Sigma_t|^{1/2}} \int_{\Sigma} H\varphi\ d\sigma \end{align}
also $\frac{d}{dt}d\sigma = -H\phi\ d\sigma$, then we get the first variation of mass
\begin{align} \frac{d}{dt}m(\Sigma) =\ &\frac{d}{dt} \bigg( \frac{|\Sigma_t|^{1/2}}{(16\pi)^{3/2}} \bigg) \bigg( \int_{\Sigma} \frac{16\pi}{|\Sigma|} d\sigma - \int_{\Sigma} H^2 d\sigma - \int_{\Sigma} \frac{ 2}{3 }\Lambda d\sigma \bigg) \nonumber\\ & + \frac{|\Sigma_t|^{1/2}}{(16\pi)^{3/2}} \frac{d}{dt} \bigg( 1 - \int_{\Sigma} H^2 d\sigma - \int_{\Sigma} \frac{ 2}{3 }\Lambda d\sigma \bigg) \end{align}
\begin{align} \frac{d}{dt}m(\Sigma) =\ & \bigg( - \frac{1}{(16\pi)^{3/2}|\Sigma_t|^{1/2}} \int_{\Sigma} \frac{1}{2} H\varphi\ d\sigma \bigg) \bigg( \frac{16\pi}{|\Sigma|} |\Sigma| - H^2 |\Sigma| - \frac{ 2}{3 }\Lambda |\Sigma| \bigg) \nonumber\\ & + \frac{|\Sigma_t|^{1/2}}{(16\pi)^{3/2}} \frac{d}{dt} \bigg( 1 - \int_{\Sigma} H^2 d\sigma - \int_{\Sigma} \frac{ 2}{3 }\Lambda d\sigma \bigg) \end{align}
\begin{align} \frac{d}{dt}m(\Sigma) =\ & \bigg( - \frac{|\Sigma|}{(16\pi)^{3/2}|\Sigma_t|^{1/2}} \int_{\Sigma} \frac{1}{2} H\varphi\ d\sigma \bigg) \bigg( \frac{16\pi}{|\Sigma|} - H^2 - \frac{ 2}{3 }\Lambda \bigg) \nonumber\\ & + \frac{|\Sigma_t|^{1/2}}{(16\pi)^{3/2}} \frac{d}{dt} \bigg( 1 - \int_{\Sigma} H^2 d\sigma - \int_{\Sigma} \frac{ 2}{3 }\Lambda d\sigma \bigg) \end{align}
\begin{align} \frac{d}{dt}m(\Sigma) =\ & - \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} \frac{1}{2} \bigg( \frac{16\pi}{|\Sigma|} - H^2 - \frac{ 2}{3 }\Lambda \bigg) H\varphi\ d\sigma \nonumber\\ & + \frac{|\Sigma_t|^{1/2}}{(16\pi)^{3/2}} \frac{d}{dt} \bigg( 1 - \int_{\Sigma} H^2 d\sigma - \int_{\Sigma} \frac{ 2}{3 }\Lambda d\sigma \bigg) \end{align}
Now we have
\begin{align} &\frac{d}{dt} \bigg( - \int_{\Sigma} H^2 d\sigma \bigg) \nonumber\\ = & - \int_{\Sigma} 2H \frac{dH}{dt} d\sigma + \int_{\Sigma} H^2 H\varphi\ d\sigma \nonumber\\ = & - \int_{\Sigma} 2H \bigg(\Delta\varphi + \frac{R}{2}\varphi - K_\Sigma\varphi + \frac{H^2}{2}\varphi + \frac{|A|^2}{2}\varphi\bigg) d\sigma \nonumber\\ & + \int_{\Sigma} H^2 H\varphi\ d\sigma \nonumber\\ % % = & - \int_{\Sigma} 2H \Delta\varphi\ d\sigma - \int_{\Sigma} R H\varphi\ d\sigma + \int_{\Sigma} 2K_\Sigma H\varphi\ d\sigma - \int_{\Sigma} H^2 H\varphi\ d\sigma - \int_{\Sigma} |A|^2 H\varphi\ d\sigma \nonumber\\ % % & + \int_{\Sigma} H^2 H\varphi\ d\sigma \nonumber\\ % % = & - \int_{\Sigma} 2H \Delta\varphi\ d\sigma - \int_{\Sigma} R H\varphi\ d\sigma + \int_{\Sigma} 2K_\Sigma H\varphi\ d\sigma - \int_{\Sigma} |A|^2 H\varphi\ d\sigma \end{align}
and
\begin{align} &\frac{d}{dt} \bigg(- \int_{\Sigma} \frac{2}{3} \Lambda d\sigma\bigg) \nonumber\\ % % = & \int_{\Sigma} \frac{2}{3} \Lambda H\varphi\ d\sigma \end{align}
Thus, the first derivation becomes
\begin{align} \frac{d}{dt}m(\Sigma) =\ & - \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} \frac{1}{2} \bigg( \frac{16\pi}{|\Sigma|} - H^2 - \frac{ 2}{3 }\Lambda \bigg) H\varphi\ d\sigma \nonumber\\ % % & + \frac{|\Sigma_t|^{1/2}}{(16\pi)^{3/2}} \bigg( - \int_{\Sigma} 2H \Delta\varphi\ d\sigma - \int_{\Sigma} R H\varphi\ d\sigma + \int_{\Sigma} 2K_\Sigma H\varphi\ d\sigma - \int_{\Sigma} |A|^2 H\varphi\ d\sigma \nonumber\\ % % & + \int_{\Sigma} \frac{2}{3} \Lambda H\varphi\ d\sigma \bigg) \end{align}
\begin{align} \frac{d}{dt}m(\Sigma) =\ & \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} \bigg( - \frac{8\pi}{|\Sigma|} + \frac{1}{2|\Sigma|} \int_{\Sigma} H^2 d\sigma + \frac{ 1}{3 }\Lambda \bigg) H\varphi\ d\sigma \nonumber\\ % % & + \frac{|\Sigma_t|^{1/2}}{(16\pi)^{3/2}} \bigg( - \int_{\Sigma} 2H \Delta\varphi\ d\sigma - \int_{\Sigma} R H\varphi\ d\sigma + \int_{\Sigma} 2K_\Sigma H\varphi\ d\sigma - \int_{\Sigma} |A|^2 H\varphi\ d\sigma \nonumber\\ % % & + \int_{\Sigma} \frac{2}{3} \Lambda H\varphi\ d\sigma \bigg) \end{align}
\begin{align} \frac{d}{dt}m(\Sigma) =\ & \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} \bigg( 2K_\Sigma - \frac{8\pi}{|\Sigma|} + \frac{1}{2|\Sigma|} \int_{\Sigma} H^2 d\sigma - |A|^2 + \Lambda - R \bigg) H\varphi\ d\sigma \nonumber\\ % % & - \frac{2|\Sigma_t|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} H \Delta\varphi\ d\sigma \end{align}
PROVED.