I would say free falling, as in falling, or being accelerated along by gravity/curved space-time with out opposing forces is the correct term. Nothing in the term implies anything about direction.
The laws of physics should come up with one result. We are describing one universe. Either the guy fell into the blackhole or he didn't. Yes I know quantum mechanics is weird, uncertantity principle, things can exist at different places at the same time/probability waves and what not, but it doesn't apply here. Your explanation would lead some to think that in one reference frame the guy crossed the event horizon and all information about him was lost for ever to the external universe, and another reference frame where he does not cross the horizon and information about him is accessible to the outside universe forever. How do you solve this apparent paradox?
You neither need a "ground" nor a "down" towards which to fall: "free falling" just means "moving according to gravitational pull without resisting it". The ISS and the many satellites in orbit around Earth are in a free fall around Earth. The Moon is in a free fall as well. Earth is in a free fall due to the Sun's gravity, and all the planets as well. Correct, that's a good point. [DISCLAIMER]: I'm neither an astrophysicist nor a cosmologist. My field is particle physics, so I could be not 100% accurate. But here's my answer. We could say that the event "the man crosses the event horizon" does happen also for the external observer, just that... it happens after the end of time (after it has passed an infinite amount of time in the external reference frame). This from a purely classical viewpoint. However we do have to take into account QM, because things could actually go in a very different way. If blackhole evaporation holds true (and it makes perfect sense), then for an external observer the blackhole is evaporating, and therefore within a finite amount of time there would be no event horizon anymore: the man will be able to reach the origin, not because he will get through the event horizon, but because the event horizon will shrink as he (slowly) gets closer and closer, up to the point when there's no event horizon anymore (for astrophysical blackholes, this would take gazillion years, but still less than infinite). But still, the man never crosses the event horizon, in the external reference frame. As you said, what happens in one valid reference frame happens in every valid reference frame. So what's the version of the story according to the falling man? According to him, while he gets closer and closer to the event horizon, he is "welcomed" by a huge amount of Hawking radiation while the event horizon rapidly shrinks to non existence. So it really happens exactly the same thing, just in a different amount of time: for the falling man it all happens in a matter of seconds (or slightly more, but not gazillions years for sure), but after this all ends, when he looks around, he sees a long since dead universe, because according to any external observer it has taken gazillions years for all of this to happen and the universe has reached heat death meanwhile.
Yeah, I don't really keep up with particle physics, I apologize for killing part of you. Also, zeno's paradox is pretty interesting, as I am a huge fan of infinite series and such. But looking at the math it makes sense. Im pretty sure the concept of infinite series in general is conceptually hard to grasp. ∞ lim ∑ 1/2^x =2 x->∞ i=0 ahh that was hard to type. Furthermore, it is noteworthy to mention in general that event horizons aren't a physical object, it comes from the calculation of where vesc ≥ c. Thus the logic that velocity v where 0<v<c for all v must conclude that no object with velocity v can exit such an event. The explanation provided above is probably better than I could have explained it. Either way, whether you crossed the event horizon or you were close enough to the black hole when it completely evaporated away, you would die. However, it is more probable that you would never reach the event horizon due to imaginary particles creating hawking radiation/time dilation, yes.
You are both wrong. Of course stuff can fall into black holes. Its quite obvious really. There is stuff, lots actually, inside black holes, inside the event horizon. Stuff therefore has to be able to collapse inside an event horizon to get there to create a black hole in the first place. What happens is that light travelling from the falling objects gets redshifted out of existence by time dilation, so the person observing sees the object get redder and redder as it approaches the event horizon, and then dimmer and dimmer until it is no longer visible using any equipment or possible technology. Any information about the object is lost to the outside universe. The object sees outside light increasingly warped until it sees darkness and a small circle of light. That circle will signify passing the event horizon. The circle gets smaller and bluer until its blue shifted all the way to gamma radiation. Then we don't know what because general relativity breaks in the vacinity of a singularity. Quite possibly, something called quantum gravity will be needed to solve this puzzle. Hawking radiation is chaotic, non-information bearing. Its caused by virtual particles popping into existence near a black hole. Normal the pairs like to stick together. But tidal forces pull them apart. That energy causes the particles to become real particles. One falls into the black hole, the other rockets away. Energy for this to occur comes from mass loss in the blackhole. The bigger the event horizon, the higher the amount of hawking radiation, the higher the amount of entropy contained with in. So the event horizon is really just a measure of entropy, or disorder.
For what I recall of the General relativity course, according to the Schwarzschild metric (which is the metric of a far away observer), the falling man never reaches the event horizon. I'm reasonably sure of this. You don't need stuff to be actually inside the event horizon in order to have a blackhole, because outside of the event horizon the metric of a blackhole is indistinguishable from the metric of a star: if you have all the matter of a star tightly packed just an inch outside of where the event horizon would be (but still inside of where any falling observer would be, and with a spherically symmetric distribution), then falling objects will behave as if they were falling towards a truly singular blackhole. In a way, this means that you cannot create a true blackhole starting from scratch (at least, classically), but just something that for most practical purposes is indistinguishable from a blackhole.
Everything is in a free fall to everything because everything attracts due to gravity. "moving according to gravitational pull without resisting it". If everything is attracted to everything then there is always resistance. Free fall is a confusing term especially when used to refer to planetary motion or anything with gravity.
I'm sorry, but no, you're just wrong. I didn't invent this use of the term, it's just standard use, and for a very good reason. If you stand up right on the ground, you're not in a free fall, despite being attracted by Earth: because you are resisting the gravitational pull due to the upward force the ground applies to your feet (which is not a gravitational force, but rather an electromagnetic one). On the other hand, if you were shot by a cannon, during your ballistic journey, you would be subject only to gravitational forces (neglecting the aerodynamic resistance), and therefore you would be in a free fall. And the ISS in orbit around Earth is in a free fall, because it is subject only to gravitational force: the only difference with a bullet is that ISS' trajectory never crosses the ground while the bullet's does, but this is not essential. This picture shows why "ISS orbiting Earth" and "bullet in a ballistic trajectory" are actually the same thing: http://i.stack.imgur.com/wzgCq.png
Sorry but the connotation of the term "free fall" is falling in a straight line path directly towards an object, solely due to gravity. There are quite a few other terms to use other than "free fall" to convey a similar meaning.
Obviously word are only labels and you can redefine a word in any way it pleases you, but usually the point of using words is to communicate with other people and therefore it would be quite foolish to use a different meaning than the one used by the majority of people talking about that stuff. In this case, "free fall" belongs to Physics and therefore, if you don't like the use I made of this expression, you have the option to personally send an email to the million of physicists in the world asking them to change the use the expression in a way that pleases you. However, if you chose this course of action, I can't guarantee they won't consider you a lunatic. https://en.wikipedia.org/wiki/Free_fall
Damn i know little about such stuff i will have to re-watch Star Gate again the answer will be there somewhere, also i believe i have a physics paper somewhere here in my room, i would have to check it out as i am very fascinated in quantum physics but also do not know much about this as well. Really good thing to get into here very hard to learn though.
While Wikipedia is only a tertiary source, this doesn't mean that what you find there is automatically bullshit: I linked to Wikipedia only to show you that the usage of the term is very broad and not restricted to a very few physicists. If you want a secondary source, the first book that comes to my mind is: "Lev D. Landau, Evgen M. Lifshitz: The Classical Theory of Fields, Vol. 2", in particular the beginning of Chapter X.
I'll still try to explain one last time. You say that "free fall" means "when something goes straight down due to gravity", and that everything else is not free fall. Even if you insist that the globally accepted definition of "free fall" is counter-intuitive, I'm going to show you that a ballistic trajectory is exactly the same thing of what you naively call "free fall down in a straight line". If you have a basic understanding of high school Physics, you should know of inertial reference frames: you can change reference frame with a transformation of coordinates: Code: x' = x + V_x·t + x_0 y' = y + V_y·t + y_0 This is not the most general transformation, but it's enough for our purposes. Say, you have a body initially at height h_0 and you just open your hand at t=0. It will move according to the law: Code: y(t) = h_0 - 1/2·g·t^2 This is what you call "free fall down in a straight line". I omitted x(t) because it's just constant, but a complete description would include x(t)=something constant. Now change to a new reference frame using the first block of code with non-null V_x>0 and V_y>0 (x_0 and y_0 don't really matter) and see what you get. Do it yourself with paper and pencil, then open the following spoiler. Spoiler YOU JUST GET THE BALLISTIC TRAJECTORY! Indeed, you have a motion that can be decomposed into: an horizontal uniform motion; a vertical uniformly accelerated motion, where: initially the body has a positive velocity and goes up, slowing down; after the peak, the body starts going down, accelerating downwards. So the ballistic trajectory is just "a body falling straight down" seen from another (still inertial!) reference frame. And, as the picture I posted a few comments above explained, the ballistic trajectory is no different than the ISS in orbit. Therefore: "falling down on a straight line" = "any ballistic trajectory (with no force above gravity)" = "body in orbit around another massive body (with no force above gravity)". And the bottom line of all this is "any body subject only to gravitational force" = "free fall". After this post, I'm really done. If you want to understand and learn something, good for you, but if you persist arguing that the worldwide common usage of the term "free fall" is stupid, there's nothing any mortal on Earth can do to help you.
I took AP Physics in highschool last year. I have yet to insist on anything except that the connotation of free fall is to be moving towards a definite floor/ground, only being pulled by gravity. I said connotation for a reason.
What's wrong with saying "uh, turns out I was wrong, but better that I learned this sooner than later"? I showed you that "fall straight down" and "going through a ballistic trajectory" are exactly the same thing and there's no point pretending that there's a meaningful difference. I'm not expecting you to believe me on authority, but did you even read what I wrote? What I took the effort and time to write for your benefit? I said this, @0TheRedHerring0 confirmed this (and we both have a university formation on Physics), Wikipedia confirms this and wherever you're going to look you'll find this very usage of the term "free fall", but you still insist that you're right and the rest of the world is wrong, just about something conventional like language? Geez, you're not going anywhere with this attitude, man.
Again, I specifically said connotation and not denotation. Look up the definition of both and you will understand.
