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Physics
Classical Physics
Radiance and energy density of a black body
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[QUOTE="Charles Link, post: 6063936, member: 583509"] The calculation is simply computing what goes out the aperture. As viewed from a location at ## (r, \theta, \phi) ##, the aperture looks like it has area ## A \cos{\theta} ##. (The circular aperture will look like an oval or ellipse when viewed from angle ## \theta ##). Meanwhile from that point at ## (r,\theta, \phi) ##, the particles will emanate uniformly in all directions, and the computation basically computes what fraction passes through the aperture of area ## A ##, (projected area of ## A \cos{\theta} ##), that cross the surface at radius ## r ## with surface area ## 4 \pi r^2 ##. That fraction is ## \frac{A \cos{\theta}}{4 \pi r^2} ##. ## \\ ## The contributions of all of the various locations in the entire hemisphere are then summed in an integral. [/QUOTE]
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Classical Physics
Radiance and energy density of a black body
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