What exactly is the relationship between Omega and omega? My understanding is that Omega, the solid angle, is A/r^2, therefore dA is an 'infinitesimal area' and therefore d(omega) is an 'infinitesimal' solid angle. But omega is also a direction vector, so how does d(omega) become a infinitesimal solid angle?
woojinko
While I don't have a great enough understanding to answer the part about d(direction vector)=infinitesimal solid angle, I believe that there is no distinction between capital Omega and lowercase omega here. https://en.wikipedia.org/wiki/Solid_angle and the fact that we directly substitute Omega = A/r^2 in for omega in d(omega) shows this fact.
kingdish
Here we use the definition of spherical coordinates to get the definition of dA and then use the definition of solid angle to get dω
bbtong
We're using Theta (latitude) and Phi (longitude) as measurements to get some point on the sphere.. I'm still a bit confused as to what putting this into sin(theta) * d(phi) does, to a ELI5 level?
sirejdua
dA is the area of the little rectangle that is drawn on the surface of the sphere. To find this area, we need the width and height. To find the width, at the top "rsin theta" refers to the distance between the y axis and the rectangle. There is a circle drawn along the y axis with that radius, so it looks like the width is the arclength of that circle with angle dphi, so the width is rsin theta * dphi. The height is a bit easier, because the height is the arclength along the circle that passes through the y axis with radius r with arc d theta, so the height is r * d theta.
The point of computing this width and height is so that we can figure out the area of the infinitesimal rectangle, so can integrate over a surface by breaking it up into infinitesimal rectangles.
What exactly is the relationship between Omega and omega? My understanding is that Omega, the solid angle, is A/r^2, therefore dA is an 'infinitesimal area' and therefore d(omega) is an 'infinitesimal' solid angle. But omega is also a direction vector, so how does d(omega) become a infinitesimal solid angle?
While I don't have a great enough understanding to answer the part about d(direction vector)=infinitesimal solid angle, I believe that there is no distinction between capital Omega and lowercase omega here. https://en.wikipedia.org/wiki/Solid_angle and the fact that we directly substitute Omega = A/r^2 in for omega in d(omega) shows this fact.
Here we use the definition of spherical coordinates to get the definition of dA and then use the definition of solid angle to get dω
We're using Theta (latitude) and Phi (longitude) as measurements to get some point on the sphere.. I'm still a bit confused as to what putting this into sin(theta) * d(phi) does, to a ELI5 level?
dA is the area of the little rectangle that is drawn on the surface of the sphere. To find this area, we need the width and height. To find the width, at the top "rsin theta" refers to the distance between the y axis and the rectangle. There is a circle drawn along the y axis with that radius, so it looks like the width is the arclength of that circle with angle dphi, so the width is rsin theta * dphi. The height is a bit easier, because the height is the arclength along the circle that passes through the y axis with radius r with arc d theta, so the height is r * d theta.
The point of computing this width and height is so that we can figure out the area of the infinitesimal rectangle, so can integrate over a surface by breaking it up into infinitesimal rectangles.