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Solid angle formula

  1. Apr 28, 2009 #1
    Hello.

    How does one derive the first formula on this page(the integral Omega = ...)

    http://en.wikipedia.org/wiki/Solid_angle" [Broken]

    ?

    I guess the problem is to project some surface onto the unit sphere, so the formula makes sense with the dot product and all, but I just cannot derive it in detail. I have tried very hard but not had any luck :(
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 28, 2009 #2

    tiny-tim

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    Hi daudaudaudau! :smile:

    The formula is ∫∫S.(r/r) (ndS/r2),

    and dS/r2 is the area of the projection of dS onto the unit sphere. :wink:
     
  4. Apr 29, 2009 #3
    That is true if dS is "parallel" to the sphere with respect to a given point (say the center). But what about the case when dS is at an angle wrt. the sphere?
     
  5. Apr 29, 2009 #4

    sylas

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    That's handled by the dot product in the formula you've asked about.

    The solid angle subtended by any surface is the sum (integral) of the solid angle from all the little patches of surface dS. For each patch of surface, there is a normal unit vector, given as "n" in the formula, which captures exactly the angle of dS wrt line of sight. The vector dot product r.n effectively scales to account for that angle. The division by r^3 is effectively a division by r^2, since it has to divide out the magnitude of the r vector in the dot product, and this scales to account for the distance of that patch dS.

    Cheers -- Sylas
     
  6. Apr 29, 2009 #5
    If I draw two straight lines on a piece of paper, I can see that the projection of one line onto the other has to do with the dot product. I just cannot see this for the case of surface elements, no matter what I draw.
     
  7. Apr 29, 2009 #6

    tiny-tim

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    Hi daudaudaudau! :wink:

    As sylas :smile: says …
    Imagine a cone cut by a plane … obviously the solid angle (from the vertex) is the same for any plane.

    For a "horizontal" plane, the surface is a circle, and for any other plane it is an ellipse, and the area of the ellipse is greater (for the same average distance), but the dot-product brings it back to size. :smile:
     
  8. Apr 29, 2009 #7
    I bet you are right, but how can I prove it beyond handwaving?
     
  9. Apr 30, 2009 #8
    I think I figured this one out. Imagine we have a surface element dS which we wish to project onto another surface. The surface element dS is a rectangle with side lengths du and dv, and along each side there is a unit vector u and v, respectively, that tells us the orientation of that particular side. Of cause the area of dS is du*dv.

    To project this surface element onto another surface with normal vector R, all we have to do is project the two unit vectors u and v and then take the cross product of these two projections to get the area. This gives us a result which can be manipulated(using the BAC-CAB rule) into the dot product that I was looking for...

    Any comments on this?
     
  10. Apr 30, 2009 #9

    tiny-tim

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    Hi daudaudaudau! :smile:
    Yes, I think you've got it …

    but it's a bit difficult to tell without seeing any equations …

    and you could refine it by saying that the "directed area" of the element (that's the vector normal to the element, with magnitude equal to its area) is the cross product du x dv :wink:
     
  11. Apr 30, 2009 #10
    I'll be a little more specific then.

    Say we want to project some surface S onto the unit sphere wrt. the center of the sphere. So we divide the sphere into a grid of parallelograms such that the area of S is the sum of the areas of the parallelograms. Each parallelogram is described by two vectors, du and dv. Now to project this parallelogram onto the sphere with unit normal vector r, I find for instance the component of du along r and subtract this from du to get a component only perpendicular to r

    [tex]
    U = \mathbf{du}-(\mathbf r\cdot\mathbf {du})\mathbf r
    [/tex]

    [tex]
    V = \mathbf{dv}-(\mathbf r\cdot\mathbf {dv})\mathbf r
    [/tex]

    And now both of these vectors have to be divided by R, the distance between sphere and surface, to account for the distance in the projection. So in total the projected area is

    [tex]
    dS' = \frac{\mathbf U\times\mathbf V}{R^2}=\frac{\mathbf {du}\times\mathbf {dv} + (\mathbf r\times\mathbf {du})(\mathbf r\cdot\mathbf {dv})-(\mathbf r\times\mathbf {dv})(\mathbf r\cdot\mathbf {du})}{R^2}=\frac{\mathbf {du}\times\mathbf {dv} + \mathbf r\times \left[\mathbf{du}(\mathbf r\cdot\mathbf {dv})-\mathbf{dv}(\mathbf r\cdot\mathbf{du})\right]}{R^2}
    [/tex]

    Now comes the BAC CAB rule (eek, twice!)

    [tex]
    dS' = \frac{\mathbf {du}\times\mathbf {dv} + \mathbf r\times \left[\mathbf r\times(\mathbf{du}\times\mathbf{dv})\right]}{R^2}=\frac{\mathbf {du}\times\mathbf {dv} + \mathbf r(\mathbf r\cdot(\mathbf{du}\times\mathbf{dv}))-(\mathbf{du}\times\mathbf{dv})(\mathbf r\cdot\mathbf r)}{R^2}=\frac{\mathbf r(\mathbf r\cdot(\mathbf{du}\times\mathbf{dv}))}{R^2}
    [/tex]

    So this is almost the formula from wikipedia, except my surface element is still a vector. I ought to take the magnitude of it, but then it will always be positive, and the one in wikipedia can be both positive and negative.
     
  12. Apr 30, 2009 #11

    sylas

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    It goes negative when the surface wraps around and you are looking at the back instead of the front. In some contexts this is very useful. If you just want to calculate the area of a shadow, however, you have to do what amounts to "hidden surface removal", and make sure you only integrate over those dS elements not obscured by another part of the same surface which is closer to you. For a closed surface, you can omit segments where dot product goes negative. But if you just take an absolute value, you end up double counting when there are two parts of the surface projected to the same point on the sphere.

    Cheers -- sylas
     
  13. Apr 30, 2009 #12
    Very interesting. I was just thinking of another thing. Say we want to project just a flat plane onto the sphere. The plane has no holes. Surely it should not be necessary to integrate over the whole plane, but only along the edges, because the edges determine the size of the solid angle.
     
  14. Apr 30, 2009 #13

    sylas

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    Uh... yeah, in special cases, you can solve the integral into a simpler analytic expression.
     
  15. May 1, 2009 #14

    tiny-tim

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    Hi daudaudaudau! :smile:

    You've made it very complicated …
    It's much simpler to say the projected area is (U x V).r/r3

    'cos then all the "r x d…" terms get squished by the ".r" :wink:
     
  16. May 2, 2009 #15
    Yes that is a nice trick :)
     
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