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I don't understand something in transport theory:

$$P(x,\vec{w})=p(x,\vec{w}) \cos\theta \, dw \, dA$$

This is the number of particles flowing across a differential surface element in the direction $\vec{w}$. I understand that I need something like

$$\text{total number of particles} = \text{densityofparticles} \times \text{volumeconsidered}$$

so the density towards a direction $* \cos\theta * dA$ makes sense to me.. but I don't understand why isn't there a 1/3 term in the equation.. aren't we interested in the area where the solid angle subtends so it should be a pyramid?

enter image description here

Since my comprehension of solid angles isn't perfect either, I need someone to explain this to me please

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You have the wrong model, I believe. Solid angle isn't needed here. You want to think about a parallelepiped (box) of material that flows across the patch of surface. So the height of the box is $\cos\theta dw$ (if I guess what your notation means) and its base is $dA$.

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  • $\begingroup$ I suppose you're right, your statement makes sense. But then why is my book citing the solid angle there? When is it needed? The book was stating that when we want to generalize this concept to take into account any direction w in every point on the space, we need a solid angle in the equation as I wrote $\endgroup$ – Marco A. Sep 5 '13 at 14:39
  • $\begingroup$ If they wanted to do solid angle, then the surface integral should be over the sphere, not over an arbitrary surface. Perhaps you'd better show us the whole passage from the book. $\endgroup$ – Ted Shifrin Sep 5 '13 at 20:26
  • $\begingroup$ Here's the passage: "The total number of particles crossing the surface is $P(x) = p(x)dV = p(x) v dt cos\theta dA$. More generally all the particles through a point will not be flowing with the same speed and in the same direction. Fortunately the above calculation is fairly easy to generalize to account for a distribution of particles with different velocities moving in different directions. The particle density is now a function of two independent variables, position x and direction w. Then the number of particles flowing across a differential surface element in the direction w is: [1] $\endgroup$ – Marco A. Sep 6 '13 at 9:26
  • $\begingroup$ This is very strange. They seem to be imagining that at a given point $x$ there are infinitely many particles instantaneously moving in all various directions. So they're imagining the density function on the "sphere bundle" of points in the region together with all directions (a unit sphere) at that point. Although this makes no physical sense to me whatsoever, then for a given patch $dA$ of surface, we have to imagine all different points and directions from those points that would have landed us in the patch. And I assume the $dt$ is missing from your original formula? $\endgroup$ – Ted Shifrin Sep 6 '13 at 12:40
  • $\begingroup$ yes, no dt at all (I suppose the solid angle vector already does the trick). It may make sense since this is a computer-graphics oriented book so I guess they're thinking about light particles. I still don't understand this "solid angle".. what does this have to do with a surface "patch" ? If you can explain this to me in simple terms (perhaps with a small drawing) I'll be very grateful since this is the main issue I'm having figuring out where's the volume to multiply with the points density $\endgroup$ – Marco A. Sep 6 '13 at 14:55

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