If black holes are 4-dimensional balls of spacetime, they will have a 3-sphere surface with a 3-dimensional volume. Would this allow infalling matter to remain within this surface?
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Spacetime has 3 spatial dimensions and 1 time dimension. The outside of an object like Earth is bounded by a 2D spherical surface. The sphere extends into the timeward direction. If you stand on the Earth, time flows forward without you leaving the surface. This is a 4 dimensional cylinder. Slices perpendicular to the time axis are spheres.
Black holes are like this, but more complex because spacetime is strongly curved near them. Analogies with Earth like spacetime will lead to wrong answers. For example, all paths forward inside a black hole lead to the center. Time ends at the center.
answered Dec 1, 2023 at 2:29
$\begingroup$
$\endgroup$
As correctly stated in the earlier answer, black holes are 4-dimantional cylinders of spacetime. However, the second part of your question regarding the hyper-surfaces is even more interesting.
Imagine an Earth-size planet falling radially to a supermassive black hole, such as Sagittarius A* in the center of Milky Way. The radial Schwarzschild metric is:
$$ \text{d}\tau^2 = \left(1-\dfrac{r_s}{r}\right)\,\text{d}t^2\tag{1}-\left(1- \frac{r_s}{r}\right)^{-1} \,\text{d}r^2 $$
Where the expressions in parenthesis define the (squared) gravitational time dilation and gravitational length contraction accordingly. From here it is very simple to calculate that, once the planet is close to the black hole, in just a few seconds it becomes practically flat in the coordinates of a remote observer. Meanwhile, the observers on the planet don't see this length contraction in their local coordinates, but see the volume of space immediately around them unchanged.
While approaching the horizon, the planet is actually intact (neglecting any tidal forces), but, if remote observers could see it, the planet would appear to them flattened to a subatomic thickness, as if "painted" on the horizon. This is just a perspective of observation, nothing physically actually happens to the planet.
What we "see" as a subatomic-thin layer just outside the horizon appears as a vast 3-dimensional space to the local observers there. We can wait a million years and throw more planets to the black hole with exactly the same result. We would still "see" them "flattened" on the horizon while by then the local observers there would see this "thin layer" extending radially over light years of space.
Thus, your intuition is correct. What we see as a Planck-length-thin 2-sphere layer on the outer surface of the horizon locally represents a 3-dimensional space of an arbitrarily large volume. As mentioned upfront, this layer "looks" like a hyper-surface (of a vanishingly small, but finite thickness) of a 4-dimensional cylinder in the coordinates of a remote observer. (Note that nothing can be actually observed remotely due to the extreme redshift.)
Now let's consider what the spacetime inside the black hole "looks" like. The radial direction on the inside is timelike, as follows from the Schwarzschild metric above. Once you are at the horizon, you move toward the "center" not in space, but in time. This means that the layer of space "painted" on the horizon, as described above, is shrinking over time, its circumference rapidly becomes smaller.
The radial coordinate $ r < r_s $ inside the horizon represents a time-slice of the Schwarzschild spacetime. In the remote coordinates, this time-slice "looks" like a 2-sphere extended over the $t$ dimension, which is spacelike on the inside. Thus, we have the same hyper-surface of a 4-dimensional cylinder as before, except now it is infinitely stretched in space, not in time. In fact, it still is stretched along the very same coordinate $t$. Only the nature of this coordinate is different, it measures time outside, but space inside the horizon.
To the local observers in a free fall, this "thin" hyper-surface still appears as the normal vast 3-dimensional space around them. None of the directions inside points to the singularity, because the singularity inside the Schwarzschild black hole is not a location in space, but a moment of time when time ends everywhere at once. Space does become increasingly tight, because the circumference of the 4-cylinder of spacetime is shrinking to a thin spacelike line along the axis - the line of the Schwarzschild singularity.
