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Sagittarius A*: the riddle of gas cloud G2

  1. May 23, 2015 #1
    So perhaps I'm not the only one here on PF who's heard about this .That said, besides the gas cloud G2 concealing a low-mass star, I cannot help but wonder what azimuthal angle G2 made its closest approach to the supermassive black hole Sagittarius A*. The reason this is relevant is because Sagittarius A* is a black hole with a Kerr metric. So the azimuthal angle I'm talking about is the angular displacement between G2 at its perihelion and the rotation axis of Sagittarius A*. Kerr black holes tend to wobble when infalling matter approaches them above the equatorial plane of rotation but not directly along the axis or rotation and when they do they emit ionizing radiation. For matter to be spaghettified and absorbed into the so-called ring of death the matter must be falling in at an azimuthal angle of ~90° from the rotational axis. So perhaps that explains why G2 wasn't swallowed up by Sagittarius A*. Thoughts?
     
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  3. May 23, 2015 #2

    phyzguy

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    I don't think we know either the angular momentum or the spin axis of Sagittarius A* with any certainty.
     
  4. May 23, 2015 #3
    Oh I certainly agree. However, even if the actual direction of its spin axis is not known with any certainty, I don't see how this precludes the explanation I offered as a possibility.
     
  5. May 24, 2015 #4

    mfb

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    The mass of the cloud, with or without star, is tiny compared to the black hole mass. Its influence on the BH is negligible.
    Source?
    Tidal effects are present in all directions. The interior of the black hole does not matter here, everything relevant happens outside.

    Also, you can assume astronomers know how tidal gravity works.
     
  6. May 24, 2015 #5

    In that case, are you suggesting that Sagittarius A* is not an actual black hole?
     
  7. May 24, 2015 #6

    mfb

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    I am not suggesting this.
    What gave you that impression?
     
  8. May 28, 2015 #7
    Lemme elaborate.....

    You mentioned tidal forces surrounding a (Kerr)black hole are isotropic(as they would be for a Schwarzschild black hole). Now the Kerr metric is the equation:


    e1348994faa6135cc22161e039faff60.png

    Where ##r_{s} = \frac{2GM}{c^{2}}## is the Schwarzschild radius,

    α = ##\frac{J}{Mc}## is the angular momentum

    ρ = r2 + α2cos2θ

    Δ2 = r2-rsr+α2


    Now when θ = 90°, ρ = r2 and sin2θ = 1

    And if you make the substitutions and take the limit we have ## \lim_{r\to 0} c^{2}d\tau^{2} = -\infty ##

    But for 0 < θ < 90°, when we take the limit as r→0 it can be shown by substituting r=0 that the metric tensor neither vanishes nor becomes infinite. So the tidal force F = ##\frac{d(d\tau^{2})}{dr}## is not the same for θ=90° as it is for θ < 90°. Thus the tidal forces surround a black hole with a Kerr metric are not isotropic.
     
  9. May 28, 2015 #8

    mfb

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    No I did not. I said they exist in all directions.
     
  10. May 28, 2015 #9

    Okay...fair enough. The article mentioned that G2 is undergoing spaghettification due to tidal forces but did not get sucked into Sagittarius A*. And my hypothesis was that G2's flyby had an azimuthal angle that is < 90° because at such an angle the Kerr metric tensor does not become infinite as r→0.
     
  11. May 28, 2015 #10

    mfb

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    The cloud was never at r→0. The predicted closest approach was 3000 times the Schwarzschild radius (from here), where relativistic effects are extremely tiny.
     
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