A recent “Universe” program on the Science Channel said that you will enter the inner event horizon right before reaching the singularity. Nobody knows, of course, what lies thereafter. You may hit a wormhole allowing you to enter another universe.
I heard that it was full of stars.
Well, that just sounds like the well-understood behavior of rainbows. Can we extrapolate to conclude that pots of gold will be found at the event horizon?
It is not a meaningful question. Gravitational compression reaches its measurable maximum at the edge of the event horizon. Once you cross it, you are effectively at the singularity, because spacetime, as we know it, no longer exists. There is no way to measure a “between” inside a black hole.
That’s at the singularity, not at the event horizon. As people keep saying, there’s nothing locally special about the event horizon. It’s even possible to have an event horizon in a region of spacetime that is perfectly flat, and you could cross such an event horizon without even having any evidence at all that you were anywhere near a black hole.
This is pretty far off from my understanding of black holes. The description of black holes that’s in all the textbooks basically says that someone falling into a black hole would have some time to ponder their fate between when they pass the event horizon and when they hit the singularity; the event horizon and the singularity are separated in spacetime. What’s more, in the standard picture of black holes (putting aside any of that “firewall” stuff that the information theorists are going on about), there’s nothing special about the event horizon. One could cross the event horizon of a black hole without even noticing it.
In fact, the singularity of a black hole is separated from the event horizon by time, not space. This is one of the weird properties of black holes: their gravity is so strong and they bend spacetime so much that once you pass the event horizon, “inward” becomes the same direction as “the future”. After you go through the event horizon, you can no more travel back outwards than you can travel back in time.
So, you’re in a spaceship falling toward a singularity. You’re facing directly away from the singularity, and can see a few hundred meters forward within your ship. Your ship happens to be falling at 1 m/s with respect to the singularity as it crosses the horizon. Obviously this would need to be a large enough hole to not kill you with tides.
What do you see, looking toward the part of the ship that’s still outside the horizon? does the light cross the horizon within the ship and return to your eyes, or no?
If no, wouldn’t that seem locally special?
Locally the event horizon is always moving at the speed of light as it is a null (i.e. lightlike) surface (this is true even when there is an apparent horizon inside the event horizon). If you were to shine a torch at the far end of the ship just outside the event horizon, that end of the ship must’ve crossed the event horizon by the time the light from the torch reaches it. Of course that doesn’t mean once in the event horizon that you can’t see light originating from outside the horizon, rather observers outside the horizon can’t see light originating from inside the horizon.
Do you disagree with Chronos above who indicated that there was nothing locally special about the event horizon, or is there some way that your answers are mutually compatible that I’m not seeing?
If it helps, let me further specify that this is quite a large black hole, say 10^10 times the mass of the sun, with a Schwarzschild radius of 400 AU or so. Tides at that distance should be minimal.
There is nothing locally special about the event horizon (excepting the firewall hypothesis): null surfaces intersect every point in spacetime (this can easily be seen from the fact the lightcone of any given observer is a null surface), so the fact that it is a null surface does not make it special. What is special about the event horizon is that it is the boundary of a region that is not causally connected to future null infinity (i.e. it bounds a region where nothing can get arbitrarily far away), but this is very much a global rather than local property.
Locally the event horizon ‘looks’ very much like a Rindler horizon does in flat spacetime. Rindler horizons are in a sense special to their related sets of Rindler observers, but not special to general observers (and in particular not special to inertial/free-falling observers).
We’re both in agreement. No light ever passes out of the event horizon. But this won’t be obvious locally, because you’ll still see light reaching you.
To put my post in a more simplistic way, there is nothing locally special about (abstract/theoretical) surfaces that travel at the speed of light, they ‘exist’ everywhere. What is special about this surface, is that despite locally travelling at the speed of light it cannot get arbitrarily faraway and it is also the boundary between where it is possible to get arbitrarily faraway and where it isn’t. The concept of “arbitrarily faraway” is a global rather than local concept, so that doesn’t make it locally special.
Is it possible for a satellite to send data out if you send it into one?
This seems to imply that light can cross the event horizon (going outward). There should be a region where light has enough velocity to move from a low altitude to a higher altitude, despite the fact that it doesn’t have enough velocity to escape entirely. Is that right? So, a light source inside the event horizon could be seen outside the event horizon but there’d be some distance at which the light source stops being visible? The photons cross the event horizon and then fall back again, like a suborbital rocket which gets up into space but doesn’t have enough speed to stay there?
You’re thinking in terms of escape speed. Light doesn’t work that way. If light starts off outside the event horizon, it will escape, with a greater or lesser degree of redshift depending on how close it was. If it starts off inside the event horizon, it will never exit the event horizon, nor cannot even move in the direction of the event horizon, because that is now a timelike dimension.
Nitpick: once you’re inside the event horizon, the singularity is in a timelike direction from you, while the event horizon is in a spacelike direction.
No in fact it implies the opposite: as the event horizon is moving locally at c outwards, light from within the horizon can never locally catch it up to cross it in the ‘outwards’ direction. For a static black hole light inside the event horizon only loses ‘altitude’ (i.e. the its radial coordinate can only decrease), non-static black holes are a bit more complicated. Certainly though light inside the horizon cannot move outside the horizon.
My God!
Okay, then what’s the answer to Tim’s question:
… although I think he said that backwards. He said “outside the horizon” and I think the relevant question here is INside the horizon.
Suppose you’re in a spaceship with really powerful engines. You ship is pointing away from the singularity and you’re sitting in the nose. The ship crosses the event horizon very slowly (Tim said 1 m/s). Perhaps the engines are running full blast. You twist around in your chair and look towards the tail of the ship. What do you see… (a) right before the tail reaches the event horizon? (b) when the event horizon is at the midpoint of your ship? (c) when the event horizon is just 1 meter behind your captain’s chair? (d) when the entire ship is inside the event horizon?
I think this is what Tim meant to ask. Are you telling me that it’s not possible to cross the event horizon slowly because the event horizon itself is moving at c?
When you are literally at the event horizon you can only travel at c relative to it. For a black hole where the event horizon and apparent horizon coincide, like in a Schwarzschild black hole, to a faraway observer the event horizon appears static and any light moving directly outwards ‘appears’ trapped at the horizon. However the speed of light is always locally c, so to any observer at the event horizon it must appear to travel c.
If we take a region of spacetime which straddles both sides of the event horizon, but which is much smaller than the radius of curvature in the region we can approximate it as flat spacetime. For any free-falling inertial observer the null/ lightlike event horizon means it must always appear to travel at c and so they will soon cross the event horizon, but we can construct non-inertial coordinate systems in this region where the coordinate speed of the event horizon is something other than c. Of particular interest is the coordinate system of an observer who is static relative to the event horizon, in order to stay static they must apply a constant 4-acceleration and so their coordinates will be Rindler coordinates (which will only cover the portion of the region outside the event horizon) and the event horizon will appear as a Rindler horizon. This may make the event horizon seem special, but in fact in our event-horizon-straddling-almost-flat region of spacetime we can construct Rindler coordinates for other similiar null/lightlike surfaces so the event horizon is not locally special.
To start to answer the specific question(s) you’ve posed, firstly to approach the event horizon at 1 m/s from your point of view (lets assume we can construct a sensible coordinate system that allows us to say this) you need to be accelerating (with the acceleration directed away from the BH) and the closer you are to the event horizon the more acceleration you need. At the event horizon the amount of acceleration needed diverges (i.e. goes to infinity). Therefore though it is possible for the nose of the ship, sitting just outside the event horizon to approach the event horizon at 1 m/s, it is not possible for the part of the ship in or at the event horizon to be travelling at 1 m/s (what speed they will be travelling depends on the coordinate system and we may not be even able to assign a coordinate speed to those parts of the ship).
What does the observer at the nose of the ship see? Due to the effects of acceleration, the tail of ship will appear strongly and increasingly red-shifted (i.e. we will see anything happening there in slow motion) in particular:
a) Just before the tails crosses the horizon events there will appear highly red-shifted.
b) Due to the apparent slowing down of time in the tail we are still seeing events at the tail from before it crossed the event horizon. The red-shift (and hence apparent slowing) of events in the tail will be increasing.
c) ditto
d) Depending on the exact details of how all parts of the ship are being accelerated either at the event horizon or before we will realize that it is impossible for the ship to remain rigid whilst the back end (i.e. nose) is approaching the event horizon at 1 m/s as it crosses and so the ship would’ve broken apart. As we cross the event horizon we will be able to see the events in the other parts of the ship that happened at the event horizon. Once inside the event horizon we will be able to see some of the events that have happened inside the horizon.
This might make the event horizon seem locally special, but locally we can attribute all the effects to the acceleration we have applied and as noted above in fact if we tried to approach any similar null surface in such a manner (even if the spacetime is completely flat) then we would get the same effects.