# The spatial Schwarzschild metric

The Schwarzschild spacetime is defined by the following line element \begin{equation*} ds^2 = - \left( 1 - \frac{2m}{r} \right)dt^2 + \frac{1}{1-\frac{2m}{r}}dr^2 + r^2 d\theta^2 + r^2\sin \theta^2 d\phi^2. \end{equation*}

The spatial Schwarzschild manifold is expressed on $\mathbb{R}^3 \setminus \{0\}$ by \begin{equation*} (1+\frac{m}{2r})^4 \delta_{ij}, \end{equation*} with $\delta_{ij}$ the flat metric, $r=|x| > 0$ and $m>0$.

How do we derive the second expression from the Schwarzschild manifold. I read somewhere that it is the $\{t=0\}$-slice of the Schwarzschild manifold but I don't see how to obtain the metric form. Thanks

• $\uparrow$ Read where? – Qmechanic Dec 24 '15 at 16:14
• General Relativity and Gravitation by Bishop and Maharaj, p.164 – Sasha Dec 24 '15 at 16:20

This is from taking the metric in isotropic coordinates, then setting $t=\textrm{const}$. However you have mixed two different radial coordinates in calling them both "r". See answer here