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Shapes of black holes (energy talk)

  1. May 17, 2015 #1
    ok, hear me out on a lymph, because i'm going to either be talking crap or something that sounds crazy
    either way....
    my question is:
    why does the phenomena of a black hole look so perfectly circular/spherical?
    now here's my reasoning

    spheres are the shape that takes the least energy to be sustained due to the ration of squares numbers and cube numbers

    now if you start heating a bubble which had little energy, then the surface starts to wobble, and then it bursts, because objectively speaking, more energy makes things go out of the state the least energy, from electron shells to bubbles,

    p.s., i know the sun is quite spherical to look at from far, far away, but the energy density of it is still nothing compared to a black hole which is suspected to interfere with time due to it amount in energy

    but then why is that, a black hole which is said to have the highest energy density, appear to be so perfectly circular (ie, there is suppose to be a disk of gravity limit) it may be because we are very far, but still too perfectly circular in my perspective....

    what do you think about my ridiculous thought?
     
  2. jcsd
  3. May 17, 2015 #2

    DaveC426913

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    A black hole is actually a point (or so we think. It's a singularity - which means we don't really know what it is). When we talk about the shape of a black hole, we are usually talking about the event horizon. The event horizon is not a physical object; it is simply the boundary beyond which light cannot escape.

    Naturally, the boundary will be equidistant from the central point in all directions.

    Make sense?

    P.S. What this mean???
     
  4. May 18, 2015 #3
    hear me out on a lymph ... = don't take it too seriously

    but yeah, why is it equidistant? i get that when you draw a shape with constant distance from center point, you'd get a circle

    but normal (objectly speaking) the higher the energy, the further away from pi it'd tend to be, so it's wouldn't it be either : event horizon is not really circular, or, there is a even bigger counter force that makes it low enough energy to have a good relationship with pi
     
  5. May 18, 2015 #4

    DaveC426913

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    Lymph is a bodily fluid...o_O

    A circle is two dimensional. In 3 dimensional space, you'd get a sphere.

    Why?

    No, it's spherical.

    What?
     
  6. May 18, 2015 #5

    ShayanJ

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    Only Schwarzchild BH is spherically symmetric. More complicated BHs aren't.
     
  7. May 18, 2015 #6

    Nugatory

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    The first step in the derivation of the Schwarzchild solution to the Einstein Field Equations is "assume that the gravitating body is spherical". It is, therefore, somewhat less than surprising that the resulting solution describes a spherically symmetric black hole.

    It doesn't follow from this that an actual physical black hole is a perfect sphere - it's not!. The spherical solution is a very good approximation so we use it, just as we treat the earth as a sphere instead of the slightly lumpy and pear-like shape that it actually is, when calculating orbits.
     
  8. May 18, 2015 #7

    Orodruin

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    To add to the above, equidistant is a misnomer here. The r coordinate is time-like inside the horizon and there is no "distance" to the singularity in that sense. There is also nothing strange about the spherical symmetry - it only holds for black holes without angular momentum. Black holes with angular momentum do not exhibit spherical symmetry.
     
  9. May 18, 2015 #8

    BiGyElLoWhAt

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    Even if they have a spherically symmetric density? So spherical fronts comprising the black hole, and the density at any point on this shell is uniform in density. I'm having trouble picturing how that would come out.
     
  10. May 18, 2015 #9

    Orodruin

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    Solutions like the Schwarzschild solution are vacuum solutions, there is no mass density distribution.
     
  11. May 18, 2015 #10

    BiGyElLoWhAt

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    If there is no density, then how can it have angular momentum about its center of mass?
     
  12. May 18, 2015 #11

    Orodruin

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    There are some different ways of defining mass in GR. You should not be surprised to find the Schwarzschild solution a vacuum solution. After all, it describes the solution outside of a spherically symmetric distribution. (There is also an internal solution, but that is not what I am discussing)
     
  13. May 18, 2015 #12

    BiGyElLoWhAt

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    Yes, but I don't see how rotation affects the symmetry. Does it not still have the spacial symmetry? I understand that there is 0 matter distributed outside the horizon, but the symmetry of spacetime outside should be dependant on the symmetry of matter within. Conceptually speaking, of course, I have never even attempted to derive a solution to EFE.
     
  14. May 18, 2015 #13

    ChrisVer

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    Indeed the Schwarchzild black hole is a vacuum solution, because the energy-momentum tensor is zero everywhere in your spacetime (except for the r=0 point, where you have the singularity, but oh well that's outside the spacetime). Choosing appropriate coordinates (coordinate invariance of EFEs allow it) you can take that there is no density either.
    A Schwarchzild black hole is not rotating.

    Then why rotating objects break the spherical symmetry? That's even known from Newtonian mechanics, and a great example of it is the non-spherical shape of the Earth (due to its rotation). In particular in classical mechanics, instead of just the gravitational potential, you will also get some potential due to the rotation, which will change the spherically symmetric solutions.
     
  15. May 18, 2015 #14

    Orodruin

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    Rotation breaks the full rotational symmetry by introducing a preferred direction, the axis of rotation. You should not expect the solution to be spherically symmetric (and indeed it is not, see http://en.wikipedia.org/wiki/Kerr_metric).
     
  16. May 18, 2015 #15

    Nugatory

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    The Einstein field equations include not just the distribution of matter but also the distribution of stress and energy. Even if you have a spherically symmetric matter distribution, it is clear that if your spherically symmetric mass is rotating the distribution of stress and energy is not going to be spherically symmetric (it will be different at the poles and the equator) so the spacetime that solves the EFE will also not be spherically symmetric.

    And that's before your initially spherically symmetric mass distribution starts moving around because it's not in a spherically symmetric spacetime...
     
  17. May 19, 2015 #16

    BiGyElLoWhAt

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    Yes, yes, I overlooked that. Thanks for pointing that out to me, guys.
     
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