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I Visualizing Solid Angle of a 3d Object (say a Sphere)

  1. Jul 18, 2017 #1
    Hello Everybody!
    Concept of Solid Angle was pretty much straight forward until they were on surface patches were taken into account which were visualized as base of cone.
    I am having difficult when 3d Objects like Sphere/Cylinder .
    We can very easily calculate the respective area and plugin the value to find answer but what baffles me is
    That only a part of the the sphere 'surface area is capture by the cone(say 2pi*(R)^2 instead of 4pi*(R)^2 )
    In real life, you can see that: if a ball is at some distance you can only see the part facing you not on the opposite.
    I google and found one wolfram demonstration which starts with a small patch goes then from there as solid angle covers half of hemisphere,pretty much straightforward
    And, after that point it starts covering the other half of the sphere , I want to know why we are coniderering a part we are not facing at all (or can't perceive it until we turn our heads around)?
  2. jcsd
  3. Jul 18, 2017 #2


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    The part we don't see is still part of the sphere surface.

    If you have a question about a specific explanation, it would help to reference this.
  4. Jul 18, 2017 #3
    But Ain't we should consider only the part we face only as per the definition?
    I asked this question in reference to proof of Gauss Theorem , everything makes sense except the part we don't see?

    Also lets suppose we consider part we 'don't see

    then lets draw a simple cone(with no extras like sphere,etc..), obviousuly the part 'we see is' pi*R^2 so solid angle will be pi but if we consider surface we don't see then can't we claim there is infinite plane beyond that surface...the part we don't see?
  5. Jul 18, 2017 #4


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    Consider it where?
    See above, please explain what exactly you are asking questions about.
  6. Jul 19, 2017 #5


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    @Adjax : Are you confusing the actual solid angle of the sphere with the solid angle of the cone which bounds all the rays of light coming from the sphere to an external point of observation ?
  7. Jul 20, 2017 #6
    Well the defination of solid angle I know is that Area subtended by base of cone at a point.

    So if I put sphere into the picture (Imagine fitting the sphere into the cone) ,then the maximum surface are captured by the base of the cone is 2pi*r^2 ,not the 4*pi*r^2.
    The other part certainly not captured( The part we don't see, for example we can only say a part of moon's surface not the one behind it , so ain't the solid angle should be the the surface visible divided by the distant)

    But everywhere I see the Gauss laws derivation , they include the not visible portion/portion-of-view also

    This is my source of Confusion!

    @Nidum: Is there any alternative/ more general definition too, which I might not be knowing/ flossed off in texts?
  8. Jul 20, 2017 #7


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    That's the largest part of the surface you can see from a given point on the outside. It is not a solid angle, and it has nothing to do with solid angles. The solid angle the sphere occupies from a point outside is always smaller than 2 pi.

    If you are interested in finding the full solid angle, you need the view from inside the sphere. The "cone" is not a single well-defined cone there, you'll need an integral.
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