- #1
etotheipi
To prove (4.4.58) from Wald$$P = \frac{1}{45} \sum_{\mu, \nu = 1}^{n} \left(\frac{d^3 Q_{\mu \nu}}{dt^3} \right)^2$$from starting by writing$$P = - \frac{d}{dt} \int_S t_{a0} dS^a = - \frac{d}{dt} \int_S - \frac{1}{8 \pi} G_{a0}^{(2)} [ \gamma_{cd} ] dS^a = \frac{1}{8\pi} \frac{d}{dt} \int_S \left(R_{a0}^{(2)} - \frac{1}{2} \eta_{a0} R^{(2)} \right) dS^a$$we know that
$$\begin{align*}
R_{a0}^{(2)} = \frac{1}{2} \gamma^{cd} \partial_a \partial_0 \gamma_{cd} &- \gamma^{cd} \partial_c \partial_{(a} \gamma_{0)d} + \frac{1}{4}(\partial_a \gamma_{cd}) \partial_0 \gamma^{cd} + (\partial^d {\gamma^c}_0) \partial_{[d} \gamma_{c]a} \\
&+ \frac{1}{2} \partial_d(\gamma^{dc} \partial_c \gamma_{a0}) - \frac{1}{4}(\partial^c \gamma) \partial_c \gamma_{a0} - (\partial_{d} \gamma^{cd} - \frac{1}{2} \partial^c \gamma) \partial_{(a} \gamma_{0)c}
\end{align*}$$from which it also follows we can write the ##- \frac{1}{2} \eta_{a0} R^{(2)}## term as
$$\begin{align*}
- \frac{1}{2} \eta_{a0} R^{(2)} =
-\frac{1}{4} \eta_{a0} \gamma^{cd} \partial_e \partial^e \gamma_{cd} & + \frac{1}{2} \eta_{a0} \gamma^{cd} \partial_c \partial_{e} {\gamma^{e}}_{d} - \frac{1}{8} \eta_{a0} (\partial_e \gamma_{cd}) \partial^e \gamma^{cd} - \frac{1}{2} \eta_{a0} (\partial^d {\gamma^c}^e) \partial_{[d} \gamma_{c]e} \\
&- \frac{1}{4} \eta_{a0} \partial_d(\gamma^{dc} \partial_c \gamma) + \frac{1}{8} \eta_{a0} (\partial^c \gamma) (\partial_c \gamma) + \frac{1}{2} \eta_{a0} (\partial_{d} \gamma^{cd} - \frac{1}{2} \partial^c \gamma) \partial_{e} {\gamma^{e}}_{c}
\end{align*}
$$where$$\bar{\gamma}_{\mu \nu} = \frac{2}{3R} \frac{d^2 q_{\mu \nu}}{dt^2} = \gamma_{\mu \nu} - \frac{1}{2} \eta_{\mu \nu} \gamma$$How on Earth are you supposed to do that integral?!? Apparently some terms disappear but which ones?
$$\begin{align*}
R_{a0}^{(2)} = \frac{1}{2} \gamma^{cd} \partial_a \partial_0 \gamma_{cd} &- \gamma^{cd} \partial_c \partial_{(a} \gamma_{0)d} + \frac{1}{4}(\partial_a \gamma_{cd}) \partial_0 \gamma^{cd} + (\partial^d {\gamma^c}_0) \partial_{[d} \gamma_{c]a} \\
&+ \frac{1}{2} \partial_d(\gamma^{dc} \partial_c \gamma_{a0}) - \frac{1}{4}(\partial^c \gamma) \partial_c \gamma_{a0} - (\partial_{d} \gamma^{cd} - \frac{1}{2} \partial^c \gamma) \partial_{(a} \gamma_{0)c}
\end{align*}$$from which it also follows we can write the ##- \frac{1}{2} \eta_{a0} R^{(2)}## term as
$$\begin{align*}
- \frac{1}{2} \eta_{a0} R^{(2)} =
-\frac{1}{4} \eta_{a0} \gamma^{cd} \partial_e \partial^e \gamma_{cd} & + \frac{1}{2} \eta_{a0} \gamma^{cd} \partial_c \partial_{e} {\gamma^{e}}_{d} - \frac{1}{8} \eta_{a0} (\partial_e \gamma_{cd}) \partial^e \gamma^{cd} - \frac{1}{2} \eta_{a0} (\partial^d {\gamma^c}^e) \partial_{[d} \gamma_{c]e} \\
&- \frac{1}{4} \eta_{a0} \partial_d(\gamma^{dc} \partial_c \gamma) + \frac{1}{8} \eta_{a0} (\partial^c \gamma) (\partial_c \gamma) + \frac{1}{2} \eta_{a0} (\partial_{d} \gamma^{cd} - \frac{1}{2} \partial^c \gamma) \partial_{e} {\gamma^{e}}_{c}
\end{align*}
$$where$$\bar{\gamma}_{\mu \nu} = \frac{2}{3R} \frac{d^2 q_{\mu \nu}}{dt^2} = \gamma_{\mu \nu} - \frac{1}{2} \eta_{\mu \nu} \gamma$$How on Earth are you supposed to do that integral?!? Apparently some terms disappear but which ones?
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