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Blackholes and time

  1. Dec 23, 2003 #1
    Hi all, this is my first thread so be nice.

    Ok, so I remember reading something which said that if an astronaut falls into the event horizon, his vantage point of time will slow down (due to the massive gravitational field) and will eventually halt at the event horizon. He will therefore see the fate of the universe--if he lives through this of course. My problem is the following: this seems to imply that matter will never ever fall into the event horizon (which i know is not true), and if not, how do black holes consume matter and increase in size (such as the ones in the center of the galaxy). Since that can not be the case, what have i misinterpreted here? Please help.

    Another problem, sorry for the length! I seem to be suffering from the same confusion as this poor man: http://superstringtheory.com/forum/basicboard/messages3/63.html His point is, how do blackbodies radiate a continuous sprectrum of energy when matter, when excited, radiates only spectral bands peculiar to the matter. What is the mechanism of blackbody radiation?-- as opposed to the mechanism of emmission (electron excitation).
     
    Last edited: Dec 23, 2003
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  3. Dec 23, 2003 #2

    chroot

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    The astronaut falling into the black hole will do so in finite time according to his watch. The outside observer, however, will never see the astronaut cross the event horizon.

    - Warren
     
  4. Dec 23, 2003 #3

    LURCH

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    The time dilation effect to which you refer is generally used to show that the subject will never reach the center of the BH. From the inf-falling observer's point of view the Event Horizon is erally a sort of "Non-event" Horizon; nothing especially noticable happens when he crosses it.
     
  5. Dec 23, 2003 #4
    ahhh gotcha Lurch. thanks for the clarification guys. Technically, passage past the EH can be very tranquil so long as there is no violent accretion disk you would have to pass through.
     
  6. Dec 23, 2003 #5

    mathman

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    Just to get my two cents in! chroot is right and lurch is wrong. The outside observer will see the infall never quite reach the event horizon. Also it will appear to get redder to the point of invisibility. A good book to read is "Black Holes and Time Warps" by Kip Thorne.
     
  7. Dec 24, 2003 #6
    Quote:What is the mechanism of blackbody radiation?-- as opposed to the mechanism of emmission (electron excitation).

    Think about it for a while, here is a clue 'Once Upon A Time'?
     
    Last edited: May 2, 2004
  8. Dec 30, 2003 #7


    Schwarzschild radius:
    [tex]r_s = \frac{ 2 G M_s}{c^2}[/tex]

    Gravitational Acceleration:
    [tex]g = \frac{ G M_s}{r_g^2}[/tex]

    [tex]r_s = r_g[/tex]

    [tex]\frac{ 2 G M_s}{c^2} = \sqrt{ \frac{ G M_s}{g}}[/tex]

    Gravitational acceleration for Schwarzschild Black Hole:
    [tex]g = \frac{ c^4}{4 G M_s}[/tex]

    Fatal Acceleration:
    [tex]g_f = 200g_e[/tex]

    [tex]M_s = \frac{ c^4}{4 G g_f}[/tex]

    Minimum survivable Schwarzschild Black Hole Mass for event horizon crossing:
    [tex]M_s \geq 1.543E+40 kg[/tex]
    M_s >= 1.543*10^40 kg

    Tidal Acceleration:
    The tidal acceleration is the differential acceleration between two points due to the difference in gravitational acceleration caused by their differing distances from a body. Taking the differential of the gravitational acceleration due to a body of mass M with radius R:

    [tex]g_t = \frac{ 2 G M_s}{r_t^3} dr_1[/tex]

    [tex]r_t = \sqrt[3]{ \frac{ 2 G M_s}{g_t} dr_1}[/tex]

    [tex]r_s = r_t[/tex]

    [tex]\frac{ 2 G M_s}{c^2} = \sqrt[3]{ \frac{ 2 G M_s}{g_t} dr_1}[/tex]

    [tex]dr_1 = 2 m[/tex]
    [tex]g_t = 200g_e[/tex]

    [tex]M_s = \frac{ c^3}{2 G} \sqrt{ \frac{dr_1}{g_t}}[/tex]

    Minimum survivable Schwarzschild Black Hole Tidal Mass for event horizon crossing:
    [tex]M_s \geq 6.448E+33 kg[/tex]
    M_s >= 6.448*10^33 kg

    ---

    Schwarzschild Black Hole Tidal Acceleration:
    [tex]g_t = \left( \frac{ c^3}{2 G M_s} \right)^2 dr_1[/tex]

    Confirmed:
    M_s = 1.991*10^36 kg, g_t = 2.056*10^-2 m*s^-2
    M_s = 1.991*10^39 kg, g_t = 2.056*10^-8 m*s^-2

     
    Last edited: Dec 31, 2003
  9. Dec 31, 2003 #8
    Shouldn't we be more concerned with the maximum tidal acceleration that we can withstand? If we are in free fall, does the acceleration of gravity really matter? Shouldn't we calculate the minimum mass a black hole needs to be based on the tidal force an averaged size being could withstand at the event horizon?

    A person 6 ft in height at the Event horizon of a million solar mass black hole will experience a maximum tidal acceleration of approximately .04 m/s^2. At the event horizon of A billion solar mass black hole, the tidal acceleration would equal approximately 40 * 10^-9 m/s^2.

    Could someone double check those figures?
     
  10. Jan 8, 2004 #9
    Re: Re: Blackholes and time

    Update of a Good overview by no other than G T Hooft:http://uk.arxiv.org/abs/gr-qc/0401027

    There is a god!:wink:
     
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