The Euclidean Schwarzschild metric describing a manifold (a black hole, though this is not relevant to the question) is given by,
$$\mathrm{d}s^2 = \left( 1-\frac{2GM}{r}\right)\mathrm{d}\tau^2 + \left( 1-\frac{2GM}{r}\right)^{-1} \mathrm{d}r^2 + r^2 \mathrm{d}\theta^2 + r^2 \sin^2 \theta \, \mathrm{d}\phi^2$$
where $\tau$ is periodic, hence the manifold is topologically equivalent to $\mathbb{R}^2 \times S^2$. The boundary of the manifold is described by a metric,
$$(\mathrm{boundary}):\mathrm{d}s^2=\left( 1-\frac{2GM}{R}\right)\mathrm{d}\tau^2 + R^2 \mathrm{d}\theta^2 + R^2 \sin^2 \theta \, \mathrm{d}\phi^2$$
where we introduce a radial 'cut off' or regulator $R > GM$. The lecturer then stated that the inward pointing unit normal vector was given by,
$$n^a = -\delta^a_r \sqrt{1-\frac{2GM}{R}}$$
How does one obtain this normal? In addition, how does one, in general for any metric, find the normals to the boundary? I also saw another lecturer write that the relevant normal was,
$$n=-\sqrt{1-\frac{2GM}{R}} \frac{\partial}{\partial r}$$
How is this a normal vector? It has no indices, and the $\partial / \partial r$ suggests it is an operator, rather than a vector.
Best Answer
Since the boundary is the level set $r = 2m$, the (unnormalized) normal can be found as the gradient of $r$: $$ \nu^a = \nabla^a r = g^{ab} \nabla_b r = g^{ab} \delta_b^r = g^{ar}.$$
To get the unit normal, just divide by the length:
$$ n^a = (g(\nu,\nu))^{-1/2} \nu^a = \left(\left( 1-\frac{2GM}{r}\right)^{-1}\right)^{-1/2} g^{ar}. $$
The minus sign shows up because we want the inward pointing normal.