Boundary conditions on Schwarzschild event horizon

Question

Consider the variational problem for a scalar field in Schwarzschild spacetime $M$ with respect to Eddington-Finkelstein coordinates $(v,r, \theta, \varphi)$, i.e.

$$\delta I(\phi) = \int_M dV \ \Big( \text{Euler-Lagrange-Eqs.} \Big) \ \delta \phi + \oint_{\partial M} dS \ n^{\mu} \partial_{\mu} \phi \ \delta \phi \overset{!}{=} 0.$$

The surface integral splits into two parts, one treating the "boundary" at spatial infinity (or rather the appropriate limit), and the other treating the boundary at the event horizon $r_S$. Assuming that the field falls fast enough when approaching spatial infinity, one is left with the requirement

$$\oint_{\partial M} dS \ n^{\mu} \partial_{\mu} \phi \ \delta \phi = 4 \pi \int_{v_1}^{v_2} dv \ r^2 \ n^{\mu} \partial_{\mu} \phi \ \delta \phi \ \Big|^{\infty}_{r_S} = \int_{v_1}^{v_2} dv \ r_S^2 \ \partial_{v} \phi \ \delta \phi \ \Big|_{r_S} \overset{!}{=} 0.$$

In order for this expression to vanish, one must have Dirichlet conditions at the event horizon. But from a physical perspective I see no reason for the need to prescribe ANY boundary condition on the event horizon surface, as the field is purely ingoing at this point and the event horizon cannot influence spacetime points in the exterior region by default. But the term must vanish so that the Euler-Lagrange equations exist.

Am I in error here in assuming the event horizon to be a boundary of the manifold; meaning the surface integral ONLY consists of the term at spatial infinity, since the manifold continues beyond the event horizon? Why do people then call it the "event horizon boundary"?

---

UPDATE:

I consulted with Eric Poisson's "A Relativist's Toolkit", Chapter 3 (Hypersurfaces), where he treats integration over null hypersurfaces. Defining the event horizon surface (EHS) as the set for which $f(r) \equiv r - r_S = 0$, he comes to the conclusion that the directed surface element on the EHS is

$$d\Sigma_{\mu} = - k_{\mu} \, \sqrt{\sigma} \; dv \, d\theta \, d\varphi$$

with $\sigma$ being the determinant of the induced metric on the EHS and

$$ k_{\mu} = - \partial_{\mu} f(r)$$

For the Schwarzschild case this means that

$$(d\Sigma_{\mu}) = (0, 1, 0, 0) \; r_S^2 \; dv \, d\theta \, d\varphi$$

So that the above integral originally in question is

$$\oint_{\partial M} d\Sigma_{\mu} \; g^{\mu \nu} \, \partial_{\nu} \phi \; \delta \phi = 4 \pi \int_{v_1}^{v_2} dv \; r_S^2 \, (g^{rv} \, \partial_{v} \phi + g^{rr} \, \partial_r \phi ) \; \delta \phi = 4 \pi \int_{v_1}^{v_2} dv \; r_S^2 \; \partial_{v} \phi \; \delta \phi$$

which is the result I have already obtained (note that above, I missed a $4\pi$ factor, but this is not relevant to the question). That means even though the fact, that the event horizon hypersurface is null, has been taken into account, this integral still does not vanish in an obvious way (which it must/should, if the Euler-Lagrange-Eqs shall exist).

Answer 1

are you taking a 3+1 decomposition of Schwarzschild first? if so, you should provide the scheme/slicing you're using to do this.

If you are not, then it's very critically important to realize that the horizon is a null surface and conformal infinity is a timelike surface. Your boundary conditions are going to care a lot about whether the $n^{a}$ in some $n^{a}d_{a}\Omega$ term is timelike or null.

There is a bunch of subtle stuff you have to worry about if you are integrating over a null surface. In particular, the metric of a null surface is not an invertible matrix, which means that you lose the natural connection between $g_{ab}$ and $g^{ab}$ when integrating over the horizon, which then means you have no natural way of raising and lowering indices. There are ways around all of these things, but it requires care, and the solutions are not in standard textbooks (Hint: the induced geometry is the important thing here).