“And black holes also don’t have infinite density. When we talk about a black hole, we generally mean the entire region inside the event horizon, the surface of no return–to put it another way, the region from which no light escapes. This horizon has a radius called the Schwarzschild radius, which is directly proportional to the mass of the black hole (specifically, Rs = 2GM/c², where G is the gravitational constant, M is the mass of the object, and c is the speed of light). You can calculate the Schwarzschild radius for any mass, whether it’s a black hole or not. For instance, the sun has a Schwarzschild radius of about 3 kilometers.”
If this is true, with the sun’s Schwarzchild radius being about 3 kilometers, how do we get light and radiation on Earth, if all light is sucked back in 3 Km above the sun?
As discussed in this thread: hawkings radiation, the three km radius is from the center of the sun, not from the surface. The surface of the sun is way way beyond the 3 km radius.
And welcome to the Straight Dope Message Boards, moonphrogg, glad to have you with us!
We can blame the author of the article for the confusion. Strictly speaking, if an object is not a black hole, it doesn’t have a Schwartzchild radius. The article should have said, in part, “…if the sun were a black hole, [f]or instance, [it would have] a Schwarzschild radius of about 3 kilometers.”
What I said in the Report stands. Every object has a Schwarzschild radius, but it’s only significant if the object is a black hole.
And yes, the aforementioned radius is equal to double the mass, in the appropriate units, and in those same appropriate units, the mass of the Sun is about 1.4 km. “Appropriate units” here means that G (Newton’s constant) and c (Einstein’s constant, the speed of light) are given a value of 1, a trick commonly used by relativists to make the equations simpler. You’ll note that if you take the formula in the report, R[sub]S[/sub] = 2GM/c[sup]2[/sup], and set c and G equal to 1, you’ll be left with just R[sub]S[/sub] = 2M . Like RM said.
You know, my degree is in physics and I have never understood this whole Schwarzchild radius thing as it applies to light. Rs is calculated to be the situation where escape velocity is equal to C. But light is not affected the same way by gravity as objects with rest mass are. Escape velocity calculations assume that the object is slowed by gravity and will stop and return to the point of origin if the initial velocity is less than escape velocity. When light is directed away from center of mass, it doesn’t slow down, only reduce its frequency (increase wavelength) - no change in speed at all.
Furthermore, if I hurl a rock directly away from earth at exactly escape velocity (under ideal conditions - thought experiment), it will travel an infinite distance. When hurled at a small enough speed less than escape velocity, the rock can travel any arbitrary distance I choose.
So, just because an object (or even light) can’t technically escape from a black hole, it can travel any arbitrary distance away before returning.
Finally, all of these calculations assume initial impulse with no further forces applied other than gravity. Escape velocity is meaningless in the situation of powered flight. I can theoretically climb a ladder any distance I choose from either the Earth or a black hole (ignoring tidal effects which can be quite large in the latter situation) without even getting close to escape velocity.
As I understand it, light is affected by gravity not like a baseball or rocket is affected by gravity… Gravity curves space; light travels in a straight line through space, but if the space is curved, it follows that. Where escape velocity equals C the space just curves enough that light follows it back to where it came from- in a circle sort of. Keep in mind this is from 1st semeter 11th grade physics (and I’m in 2nd semester 11th grade physics right now), so I’m much less qualified than those who have a degree in physics.
You are absolutely correct, but where my problem lies is when the light is directed radially away from the center of mass. I can’t see how curvature affects it in that situation. e.g. When a rock if thrown in any direction but straight up, its path is curved into a parabola by gravity. Throw it straight up and no curve.
The reality is General Relativity is not this simple and thus black holes are not very comprehensible (by me) until I have have a better understanding of it (though this is not likely to happen in my lifetime).
The whole “escape velocity is equal to c” thing is a convenient Newtonian shortcut to explaining an event horizon, and it happens to give the right answer. But really, it’s just a coincidence that it does so: In many other cases where one uses a Newtonian shortcut in GR, one ends up being off by a factor of 2 or so. So you shouldn’t really rely on such shortcuts. A better explanation, in the context of GR, is that there is no path from the inside of a black hole to the outside. From any point inside a black hole, the center is always in the future and the exterior is always in the past. So any particle travelling from the past to the future (which, so far as we know, is all of them) must sooner or later (probably sooner) end up at the center, once it crosses the horizon.
Yes GR is very complicated, heck we only describe them mathematically about what we think is going on beyond that shroud of darkness. But the problem with firing light radially away from the center of a blackhole is that there isn’t a radially away direction. Mathematically speaking space-time gets so curved that all pathways lead toward the center, so while you might “see” light coming into the blackhole (if you were somehow inside the event horizon) that pathway wouldn’t lead “out”. Its very difficult to understand, and the whole “bowling ball on a matress” analogy can only go so far (where here it fails).
