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Solid Angle Woes

  1. May 11, 2006 #1

    eep

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    Perhaps this should be moved to the math forum, but I've never quite been able to understand solid angles. For example, in terms of thermal radiation and the planck distribution (black bodies), we can show that the energy flux denisty [itex]J_u[/itex] (the rate of energy emission per unit area) is equal to

    [tex]
    J_u = \frac{cU({\tau})}{V} * \frac{1}{4}
    [/tex]

    Where [itex]U({\tau})[/itex] is the total energy in the cavity.The geometrical factor [itex]\frac{1}{4}[/itex] can be derived by solving the following problem:

    Show that the spectral desntiy of the radiant energy flux that arrives in the solid angle [itex]d\Omega[/itex] is [itex]cu_{\omega}\cos\theta{\cdot}\frac{d\Omega}{4\pi}[/itex] where [itex]\theta[/itex] is the angle the normal to the unit area makes with the incident ray, and [itex]u_\omega[/itex] is the energy density per unit frequency range. The sum of this quantity over all incident rays is [itex]\frac{1}{4}cu_\omega[/itex].

    Now, my understanding is that the solid angle is a projection of your object (in this case a small hole, or perhaps the beam emerging from the hole) onto a sphere of radius 1. I just really have no idea of what's going on, can anyone point me in the right direction?
     
  2. jcsd
  3. May 11, 2006 #2

    Andrew Mason

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    A solid angle at point A subtended by an object is: the ratio of the area of the projection of the object from point A onto a sphere - any sphere - whose centre is A, divided by R^2 (R is the radius of the sphere). The analogy is to an ordinary angle: projection of object onto a circle divided by R. They are hard to visualize because congruent solid angles can have quite different shapes.

    AM
     
  4. May 16, 2006 #3

    CarlB

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    Solid angles are areas of the surface of a sphere. The unit sphere has area 4 pi, so 4 pi acts for solid angles the way that 2 pi acts for regular angles.

    Any three vectors that are neither parallel nor antiparallel define three points on the sphere that are the vertices of a spherical triangle. This is similar to how three points on the plane define a regular triangle. The formula for the area (or solid angle) of a spherical triangle is kind of complicated.

    A right spherical triangle is one that has three 90 degree angles. Such a spherical triangle defines an octant of the sphere and thus has an area of 1/8 of 4 pi or pi/2. Another special type of spherical triangle is one that has three 180 degree angles. Such a spherical triangle is a half sphere and has a solid angle of 2 pi.

    An interesting fact of quantum mechanics is that if one is given three unit vectors, say r, g and b, that are neither parallel nor antiparallel, then the projection operators for spin-1/2 in the directions r, g and b, say R, G and B multiply together to give a complex phase related to the area of the spherical triangle defined by r, g and b. That is, if

    [tex]R G B R = \eta e^{i\delta} R[/tex]

    where \eta and \delta are real numbers, then [tex]\delta = area(r,g,b)/2[/tex]. The formula for [tex]\eta[/tex] is the square root of the product of the three functions [tex](1+\cos(\theta))/2[/tex] where \theta is the angle between r and g, r and b, and between g and b.

    Carl
     
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