Spatial dimensions inside the event horizon

Question

Consider a test particle falling radially to a black hole. The radial direction toward the singularity inside the black hole becomes a direction in time while the dimension that used to be time becomes a direction in space.

A massive particle can be stationary in space and just move in time toward the singularity, so no problem there. However, a photon must move locally in space with the speed of light. For example, during one second of time, it must move by one light-second in space in the view of a local observer or by some other distance in the view of a remote observer.

The photon is flying to the black hole radially pointing to the singularity in the center. While it is outside, no problem, for each second of time the photon covers the radial distance of one light second toward the black hole (or some other distance depending on the observer, but still radially toward the black hole).

(I am aware that in the view of a remote observer the photon never crosses the event horizon. This is however beyond the scope of the question that is only concerned the the spacetime geometry on the inside.)

Now, there is a confusion on the inside. The direction in time is now radial toward the singularity in the center. So for each second in time radially toward the singularity the photon must cover one light second (or some other distance) in space... in what direction? It cannot possibly be toward the center, as it is the direction in time, not in space. It also cannot be one of the other two original spatial dimensions, because they do not change while crossing the horizon.

The photon now must move in space in the direction where earlier time was pointing. Well, while time was time, we thought it was pointing to the future, but now, since this dimension is a direction in space, we can no longer say the same. Thus my question is, what is the spatial direction of a radially infalling photon inside a black hole? It cannot be a direction to the center, so where is it moving? Or otherwise what is the flaw in my reasoning?

The null geodesic equation in the Schwarzchild metric for a radially infalling photon ($ds=0$) is:

$$dr=\pm\left( \frac{r_\mathrm{s}}{r_o} - 1 \right) \,dt$$

It solves as $t=f(r)$ where the actual expression is simple enough, but irrelevant to the question. This equation would describe a trajectory of the photon, but only if we knew the direction of $t$. However, it is not evident from the equation where the spatial $t$ dimension is pointing geometrically inside the black hole.

This question can also be generalized as, what is the geometry of space inside a Schwarzchild black hole? At a constant time $r=r_o<r_s$ ($dr=0$), the $(-1,1,1,1)$ metric (where $t$ is a spatial dimension and $c=1$) is:

$$\,{ds}^{2} = \left( \frac{r_\mathrm{s}}{r_o} - 1 \right) \,dt^2 + r_o^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)$$

What kind of space does this formula describe?

Answer 1

A radial null curve inside the event horizon of a Schwarzschild black hole can be explored via the Eddington-Finkelstein coordinates. The metric in such coordinates is: $$ds^2 = -(1 - 2GM/r) dv^2 + (dv dr + dr dv) + r^2 d\Omega^2$$ where:

• $r$ radial coordinate

• $v = t + r^*$

• $t$ time coordinate

• $r^* = r + 2GM \ln (r/(2GM) -1)$ tortoise coordinate

An infalling radial null geodesic is characterized by $v = constant$.

A radial null geodesic is described by setting $ds^2 = 0$, $d\Omega^2 = 0$, thus giving:
$dv/dr = 0$ infalling
or
$dv/dr = 2 (1 - 2GM/r)^{-1}$ outgoing

In this coordinate system there is no problem in tracing the paths of null or timelike particles past the horizon. The light cone tilts over, such that for $r \lt 2GM$ all future directed paths are in the direction of decreasing $r$. At the horizon the outgoing photons keep still at $r = constant$, while the infalling photons point radially to the physical singularity at the center of the black hole. In between (within the light cone) you have the massive particles.

FURTHER
Let us go back to the Schwarzschild coordinates, where the metric is: $$ds^2 = - (1 - 2GM/r) dt^2 + (1 - 2GM/r)^{-1} dr^2 + r^2 d\Omega^2$$

The normal to a surface $r = constant$ is given by $n_\mu = \nabla_\mu f(r) = \partial_\mu f(r)$, where $f(r) = r - constant$. Thus:

• $n_\mu = (0, 1, 0, 0)$ dual vector

• $n^\mu = (1 - 2GM/r) (0, 1, 0, 0)$ vector

• $n^\mu n_\mu = (1 - 2GM/r)$ squared norm

In the exterior region, $r \gt 2GM$, the normal is spacelike, hence a surface $r = constant$ is timelike. In the interior region, $r \lt 2GM$, the normal is timelike, hence a surface $r = constant$ is spacelike.

Therefore in the interior region a surface $r = constant$ is not viable to a particle, whether massive or massless (photon). An infalling particle can only proceed along the radial coordinate towards the center of the black hole.

The flaw in your reasoning is that, even if time and radial coordinates change nature, the coordinate distance to the center of the black hole is still measured by the radial coordinate. The change in nature imposes on the radial coordinate additional constraints peculiar to the interior of a black hole.