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Question about Schwarzschild radius

  1. Dec 5, 2009 #1
    Okay, correct me if I am wrong but Schwarzschild radius of an object if it squeezed to a radius of that point it will collapse into a gravitational singularity and become a black hole. For example, if the sun has a Schwarzschild radius of about 3 kilometers. If we were to be squeezed the sun to that diameter, it would theoretically collapse into a black hole right?

    If that is true, I don't understand. Even if you squeeze the sun into a gravitational singularity, it would still have the same gravity, which is not efficient enough to trap light. So how can this be true?
     
  2. jcsd
  3. Dec 5, 2009 #2

    atyy

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    Roughly, everything will look the same if you stay outside the Schwarzschild radius, but once you go past the Schwarzschild radius, you are doomed. In fact, part of the Schwarzschild solution is appropriate for the region of space outside the sun, even now when it is not a black hole.
     
  4. Dec 5, 2009 #3
    It would have greater gravity. The center of gravity of the sun is a long way away from a grazing light beam. At a much smaller diameter the grazing light beam would be much closer to the center of mass.
     
  5. Dec 5, 2009 #4

    So your saying that, for instance I take my own Schwarzschild radius which would be
    (2)(M)(G)/(C^2)
    (2)((145*1.45)/9.8)(6.674 * 10^-11)/(299972458^2)
    = 3.18628575 * 10^-26 meters

    If I were compressed into a sphere, not even light could escape me if it traveled that close to the center of my mass?
     
  6. Dec 5, 2009 #5

    Nabeshin

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    If you want to think about this in a Newtonian analogy, you can do so as follows (note, in Newtonian gravity, light is not bent by gravity, but I think there is still some worth to the analogy):
    While the sun is as large as it is, a body can only come so close before contacting the sun's surface. Therefore the maximum field produced by the sun is,
    [tex]g_{max}=\frac{G M_{sol}}{r_{sol}^2}[/tex]
    As we compress the sun, the max field increases until at some point this field is so strong not even light escapes, and we have reached the Schwarzschild radius for compression.

    It is just an analogy, but I think it helps to demonstrate why gravity seems "stronger".
     
  7. Dec 5, 2009 #6
    That's the theory, yes.
     
  8. Dec 5, 2009 #7
    It is, if light is assumed to have an equivalent mass due to energy--but this is a side issue.
     
  9. Dec 6, 2009 #8
    So one more verification. Does this mean that every object possesses a black hole, its just the objects we call black holes have a Shcwarzchild's radius that is extremely large compared to other objects? So the center of the sun from 3 kilometers away is completely black and a black hole itself?
     
  10. Dec 6, 2009 #9

    Nabeshin

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    No, not at all. Consider that within a radius of 3km around the sun is only so much mass (I'd do a calculation but I'm lazy). The point is, it is much less than the mass of the entire sun. It is only the enclosed mass that you have to trouble yourself with, so you would see that the schwarzschild radius for this amount of mass would be some much smaller number, 1m or something. But you run into the same problem if you try and say "so there's a black hole at r=1m right?"

    Does this make sense?
     
  11. Dec 6, 2009 #10
    Well, wouldn't that put you into an infinite regression? I don't understand
     
  12. Dec 6, 2009 #11

    Nabeshin

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    Yes, it would. And it serves to show there is no black hole located anywhere inside the sun!

    The calculation of schwarzschild radius assumes that the entire mass of the object is concentrated at a point, ok? That's all well and good, but you can only assume this when you're talking about the area outside of the object (in this case, outside the sun). So when you get a schwarzschild radius inside the sun, you recognize that your results are no longer valid because the assumptions were not met in the first place.
     
  13. Dec 6, 2009 #12
    So, then what is the point of schwarzschilds radius? What do physicsist use it for? Does that mean no object unless the entire mass of an object is concentrated at a point like a singularity of a black hole has a schwarzschilds radius? Does this mean the Schwarszchilds radius for any other object besides black holes are invalid?
     
    Last edited: Dec 6, 2009
  14. Dec 6, 2009 #13

    Nabeshin

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    Sort of. It can be a useful concept (this might sound recursive but I think if you read the post I gave above it'll make more sense) when the object is fully contained within the schwarzschild radius. Although this doesn't a priori mean that it has to be a black hole, it does mean that it eventually does have to become one.

    And yes, it's sort of silly to talk about the schwarzschild radius of the sun or the earth or anything like that, because it has no physical significance (i.e there's nothing special if you went to 3km from the center of the sun). Usually people use it by saying "if you compressed the sun to a 3km sphere, you would form a black hole" which is where you can make the concept useful. It only becomes meaningful when the object is compressed to less than that radius.
     
  15. Dec 6, 2009 #14
    Technically, a golf ball has a Schwartzchild Radius. The radius is of course very small, but it's a real, physical radius nonetheless.
     
  16. Dec 6, 2009 #15

    But that brings me back to my first problem. The sun always has the same gravitational pull though, how can it ever turn into a black hole? It doesn't matter how much you compress it, even if the radius falls below Schwarzschilds radius. It will always have the same pull, am I wrong?
     
  17. Dec 6, 2009 #16

    Nabeshin

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    Refer to my post #5... Umm, I don't know how else to explain it other than how I have so I'll let someone else give it a try and hopefully succeed better.
     
  18. Dec 6, 2009 #17
    Alright man, thanks for helping:)
     
  19. Dec 6, 2009 #18
    As you already know, all objects fall at the same rate on the moon because the moon doesn't have an atmosphere to slow things down. Now to answer your question:

    Suppose you're on the moon, and you have two objects of equal mass: The first object is a lead pellet, and the other object is a huge ball of aerogel that's 1 foot in diameter. Remember, they both weigh the same and are of equal mass.
    Now drop both objects onto a very thin sheet of cellophane that's suspended 1 foot from the lunar surface. Which object caused the most stretching of the sheet of cellophane? That's right: the lead pellet. They are both of equal mass, yet the pellet did more damage! How can that be? They're of equal mass, after all!

    Likewise, consider two interstellar objects: the first object is the Sun, and the second object is exactly as massive as the Sun, only it had been compressed to a diameter of only 3Km. Which interstellar object will have the greater gravitational pull at its surface (important)? The Sun? Or the 3Km object? That's right: Since the Sun has its gravitational field distributed and weakened over a very, very large area equal to its very large radius (minus 3 Km), and the 3Km object has its gravitational pull concentrated and strengthened to a very fine point, the 3Km object will by far exert the greater gravitational pull at its surface (again, important!) even though both objects are of equal mass.
     
    Last edited: Dec 6, 2009
  20. Dec 6, 2009 #19
    From what I have casually read on the subject, Neo_Anderson is technically correct. However, I believe a golf ball's S-Radius (which I assume is EXTREMELY tiny) will display none of the effects that one would typically associate with an S-Radius unless all of the mass of that object (the golf ball, in this case) were to be completely compacted within that S-Radius. And since that never happens to most golf balls I know, the S-Radius of a golf ball has no practical meaning, does it?

    Daisey
     
  21. Dec 6, 2009 #20
    The schwartzchild's golfball would evaporate in like 0.00002 nanoseconds.
     
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