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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.

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  • $\begingroup$ vector fields are also described as "derivations" $\endgroup$
    – janmarqz
    Apr 14, 2014 at 16:11
  • $\begingroup$ @janmarqz: Thanks, I wasn't familiar with that viewpoint of a tensor field. Do you know how I can actually find the normals systematically, given a metric? $\endgroup$
    – JPhy
    Apr 14, 2014 at 16:28

1 Answer 1

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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.

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