Another thread about BHs made me remember a question I’ve been wanting to ask about the event horizon.
Let’s say that my buddy and I are falling straight into a BH, one all isolated so there isn’t a bunch of hot plasma to obscure our view. My buddy has head start, so that he’ll get there first. As we approach it, does the event horizon stay the same radius? Will I watch him suddenly disappear when he gets there?
If I think of the event horizon as the surface where something going even the speed of light can no longer escape, and I try to relate that to throwing a ball up and having it come back down, as I get closer, some of the light that won’t escape will still reach my eyes. So as I approach the BH and sink into the gravity well, I’ll see the event horizon shrink ahead of me.
Is this the way it works, or does the time dilation somehow keep the event horizon in the same place?
You won’t see your friend disappear as he crosses the event horizon. The event horizon has no local significance to inertial (free-falling) observers: if you consider a “small” (compared to the horizon curvature) box with you and your friend inside, no experiment you can do inside the box can detect your crossing of the event horizon. In particular, you won’t see any change in your friend’s appearance, as long as he’s relatively close to you. So this:
is a good way of thinking about it. But this:
is not right; the event horizon is not something that you can sense with local experiments when freely falling.
(In fact the position of the event horizon is not even causally determined: its current position depends on future events. For example, consider a ray of light very close to, but just outside of, the apparent event horizon of an isolated spherical black hole. Eventually it will escape. But if I add some mass to the black hole soon enough, the “real” event horizon will grow and envelop the ray of light; so the real event horizon at a given time depends on whether I decide to add mass to the black hole at a later time.)
The event horizon’s position will change a little as you fall in; you and your friend have added some mass to the black hole, so the horizon will expand slightly.
You see by light entering your eye that was reflected off of an object. Your friend (being the object in question here) has passed behind the event horizon of a black hole. The event horizon is defined as the boundary where nothing, not even light, can escape.
Therefore, there will never be a light ray (or photon) that will bounce off of your friend and reach your eye (assuming you are outside of the event horizon).
However, if you are both inside the event horizon I believe you could see each other. Light can still move around inside a black hole…theoonly difference is that there isn’t a path it can take to escape the black hole. The light ray or photon will forever remain inside the event horizon. For sufficiently large black holes I believe the event horizon can be so broad that it would be possible to cross the boundary and not know it happened (for smaller ones the gravitational differences between your head and your feet could see you stretched like a piece of spaghetti so you probably would notice).
Interesting…
So what’s “relatively” close? If we were looking at ,say, Mars with a high-powered telescope and it (Mars) and it fell into a black hole what, then, would we observe?
The event horizon is the boundary beyond which light cannot escape to infinity. If you consider the boundary beyond which light cannot escape to you, you would get a different shape. Once you are inside the event horizon, light from the event horizon could reach you. I’m not sure about this, but I think that once you pass the horizon, you could not see anything closer to the singularity than your position, but you could see outward.
The problem with this reasoning is that you’re implicitly assuming that you are stationary, not infalling. A stationary observer hovering just outside the event horizon cannot see inside of the horizon. (He would see his friend, redshifted more and more and asymptotically approaching the horizon, but never crossing it.) But an infalling observer can “catch up” to light beams emitted from within the horizon. Even though these light beams can’t escape the horizon either, they can last long enough for you to meet them. (For example, a light beam emitted directly outward from just inside the horizon can stay close to the horizon for an arbitrarily long time while you fall in to meet it.)
More formally, what’s happening is that the Schwarzschild black hole solution has no physical singularity except at zero radius. This means that local inertial coordinate frames can be defined, even at the event horizon; so physics ought to behave “normally” in small, inertially-moving frames even as they fall through the event horizon.
Hombre:
This depends on the size of the black hole, relative to the distance to Mars. The “local” coordinates are only a good approximation to flat space over a distance of order the spacetime curvature. Near the event horizon this curvature is of the same order as the horizon circumference. If the black hole is, say, a 1-billion-solar-mass Schwarzschild black hole, then the Schwarzschild radius is r = 2GM/c[sup]2[/sup] ~ 20 AU (20 times the distance from Sun to Earth)–if I did the math right–much farther than the distance to Mars. So if Mars is falling in, we’re probably falling in with it, and we won’t see anything terribly unusual when looking at Mars as it falls through the event horizon. (We’ll start seeing unusual things as it nears the singularity, though.) But if the black hole is much smaller than an AU then Mars might fall in while Earth does not. Then our situation would be more like that of the stationary observer above: we would see the red planet get redder and redder.
If I am hearing you right you are both saying that the event horizon is only the boundary at which light from behind it must return to the BH. However, said light can pass the boundary and go some arbitrary distance away before it gets turned around. Presumably that photon can go as far as it’d like as long as the BH lies in its future (unless it is absorbed or interferred with such as striking your retina). Kinda like a long period comet.
If that is the case then I gues it depends on how far away the observer is from the event horizon and his friend. Just outside of the event horizon there would presumably be lots of photons that could still make it to you. The further away you go the fewer that will have the energy to reach you till the point that none can reach you. At that point I would suppose your friend would disappear from your perspective.
However, I always assumed the event horizon kept everything in…there being no path that will take the photon past the event horizon. Every path would lead either nealry perpendicular to the singularity (at best) or head towards the singularity.
[sub]NOTE: I realize that there will be a point further from a BH that our friend could not escape than the event horizon itself where light itself can’t escape. Light being the ultimate arbiter that if it can’t do it then nothing can at that point.[/sub]
Actually you won’t see your friend disappear even if you’re not falling in after them. To a stationary observer far away, objects falling into the black hole appear to slow down as they approach the event horizon. Indeed to this observer, the object (in this case, your unlucky friend) will appear to take an infinite amount of (your) time to reach the horizon. Your friend unfortunately measures the same interval as finite. As a conseqence of this mismatch, they only reflect/emit a finite amount of photons in reaching the horizon. You thus see them taking forever to reach there, but getting fainter and fainter as they do so. The same effect is why black holes sometimes used to be called “frozen stars”: in principle, you can still see the original surface of the star just above the horizon, eternally falling in.
If you’re falling as well, then, as already noted, you don’t see an horizon at all.
For what it’s worth, here’s my prediction: Your buddy will look redder and redder and then disappear.
First, I believe that light simply cannot cross the event horizon (going outwards), as demonstrated by the following thought experiment:
If light could cross outwards, only to fall back in, you could set up a system of “relays” to get information out of the black hole. i.e. if light can make it 1 foot out of the black hole, you set up an antenna one foot out of the black hole to detect the signal. That signal is then broadcast to infinity.
A second thought experiment: You are shining a light on a friend who is falling into a black hole. Whether you are moving or not, your friend is accelerating away from you. Thus, you will observe a red shift in your friend’s appearance, i.e. the signal will get weaker and weaker. As your friend crosses the event horizon, that signal will disappear since the light bouncing off your friend finally hits an insurmountable barrier.
No, light from within the event horizon can’t get out. (If the light could get beyond the event horizon, it could eventually make it to infinity (–And Beyond!).) But you can still catch up to it. Think about the light leaving your friend just as he crosses the event horizon. The light that leaves from right on the horizon will “hover” there forever. The light that leaves from just outside will hover there for a long time, moving outward slowly and eventually escaping. The light that leaves from just inside will also hover there for a long time, moving inward slowly and eventually hitting the singularity. But while it’s hovering near the horizon, you’re still falling into the black hole. If you’re close enough to your friend, you will fall through the horizon before it has a chance to leave the neighborhood of the horizon.
(I say the outwardly-directed light near the horizon moves “slowly”; this is only when viewed by a stationary observer far away from the black hole, because of the time dilation near the horizon. A freely-falling observer nearby, like you, will see it moving at the speed of light, as usual. This dependence on observer’s frame is probably what’s causing most of the confusion.)