Monday, 30 November 2015

general relativity - What is the 'apparent horizon' of a black hole?



The Wikipedia article is full of general relativistic and differential geometry jargon, and its accuracy is questionable. So, without such (or explaining said jargon without more jargon), what is the 'apparent horizon' of a black hole?



Answer



The apparent horizon is a technical term, so you're going to have to have SOME jargon to explain it--the distinctions between trapping, apparent, dynamical and event horizons does just require some knowledge of differential geometry and general relativity, even if you can explain them using common words, as below.


Freeze a moment in time in some particular coordinate system. Then, if that coordinate system contains an apparent horizon, it will be closed two-dimensional surface (think a ball, but not necessarily undistorted) where all light rays but one are forced to move into the surface when evolved in time, and the last light ray will be 'frozen' on the surface at that moment of time (think of the outgoing light ray at $r=2M$ in the Schwarzschild solution--it tries to get away, but is just stuck on the surface).


It is called apparent because it:



  1. depends on the notion of constant time on that one surface

  2. is therefore coordinate dependent, and in some cases, can be eliminated entirely by choice of coordinate

  3. can have strange, unpredictable behaviour

  4. is generally inside the proper event horizon, but


  5. if the ''stack'' of apparent horizons forms a timelike surface, the surface will be two-way transversible, and therefore, doesn't correspond with what we normally think of as a 'horizon'.


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