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Calculate the Gravity of a Black Hole

  1. Oct 25, 2015 #1
    Can I bring an old thread back to life? The last post in Gravity for a stellar black hole
    was in Mar-2008, but I have been looking for That Thread since 2000!

    Unfortunately for me, it raised more questions than answers.

    Woops. The Username "SamT" is already taken, but it's been my handle for decades.
  2. jcsd
  3. Oct 25, 2015 #2

    Simon Bridge

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    Welcome to PF;
    You say that post creates more questions than answers ... what were your questions?
  4. Oct 25, 2015 #3
    Hello, Simon,

    I have very little formal Science education, and in the 4 decades since I was last in school, I have forgotten the trigonometry and calculus I learned then. I have been pondering Black holes Since Hawking made them famous.

    It seems to me that from the perspective of a photon, that the definition of the gravitational acceleration (g) at an Event Horizon is C/s. IOW, Space/Time is warped, (in the classical definition,) 45* towards the Singularity.

    However, in the referenced thread, Steve says, "For example, for a 3 sol black hole, the gravity at the event horizon, using GM/r^2, works out at 5.0845x10^12 m/s^2." (roughly 1600C/s)

    Using MS Excel, when I plug Steve's numbers into George Jones formula, I get an answer of 1.77334E+11 m/s^2. (roughly 500 C/s)

    Those two values indicate a S/T warpage of ~90*

    I know that from the Photon's viewpoint, there is no acceleration, it is just that space/time warps and the photon just keeps traveling in the same old (but very warped) straight line terminating at the singularity.

    Steve also said, "Also, working backwards, you can work out at what point the gravity exceeds 299,792,500 m/s^2, r = (GM/c)^0.5, for a 3 sol black hole this would be at 1152.2 km radius."

    I suppose this means that the photon "sees" the Singularity 1152.2Km away.

    Ohhh, my head hurts

    The Question: Why are both Steve's (Newtonian) and George's (GR) acceleration numbers so much higher than C/s?

    Follow up Question: What would a distant observer measure this SR as? 8.8 km or 1100km?
  5. Oct 25, 2015 #4

    Simon Bridge

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    Don't try to do relativity from the pov of a photon, it does not work. Photons cannot be observers.
    Try again, but leave photons out of it.
  6. Oct 25, 2015 #5
    I keep remembering the story of Albert sitting in a bus and imagining he was a photon.

    thank you for your interest.
  7. Oct 25, 2015 #6

    Simon Bridge

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    Do you have a reference to that because it produces nonsense.
    Einstein's most famous thought experiment involved chasing a beam of light - is this what you meant?

    An object moving fast in the frame of, say, the center of mass of a black hole, will observe the black-hole frame to be length contracted in the direction of travel ... so distances there are shorter in proportion to the speed. The closer to the speed of light the object travels, the shorter the distances ... at the speed of light, the math tells us all distances in the direction of travel are contracted to zero ... so talk of how far away the photon sees the black hole is nonsense: it sees the black hole as right there!

    So don't use a photon as an observer: you get nonsense.
    There are a lot of other problems with what you wrote, but this is a good place to start.
    In relativity you have to be very careful about how you set stuff up.

    No responsibility for advise not taken.
  8. Oct 25, 2015 #7

    Simon Bridge

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    Because they are different calculations. Usually the acceleration of gravity due to a body is what an accelerometer that is (radially) stationary in the frame of reference of the center of mass of the body would read. However, that is not the only acceleration of gravity ... if you are in freefall, then your accelerometer would read zero.

    The correct Swartzchilde radius is easy to look up ... it is ##R_s=2GM/c^2##. It is a good exercize to do the calculation yourself.
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