I never quite understood what H^2 meant here- what is a hemisphere in this picture? Also what's the difference between dw and dA?
mpotoole
$H^2$ and $A$ refer to the domains of integration.
In the first integral, we're integrating all possible incident light rays as a function of direction $\omega'$ (i.e., we choose a light ray in the direction $\omega'$ and measure the contribution of the ray). The set of all possible incident light rays are given by sampling a hemisphere centered around point $p$, represented by $H^2$. The measure $d\omega'$ weighs the contribution of this ray by the differential solid angle (the field of view that this ray covers).
In the second integral, we're integrating all possible incident rays as a function of points $p'$ on the aperture (i.e., we choose a light ray by sampling a point $p'$ on the aperture, define a ray passing through points $p$ and $p'$, and measure the contribution of the ray). The set of all possible incident light rays are given by the sampling all points on the aperture, and the domain of integration in this case is $A$---representing the area of the aperture. The measure $dA$ weighs the contribution of the ray by the differential area on the aperture.
Note that the (differential) solid angle is related to the (differential) area on the aperture by the following function: $d\omega' = \frac{cos(\theta)}{|p'-p|^2}dA$. That is, if you were to rotate a differential area by $\theta$, the solid angle subtended from point $p$ would get smaller by $cos(\theta)$. If you were to move the differential area further away from the sensor, then the solid angle would decrease according to $\frac{1}{|p'-p|^2}$.
So, in short, when changing the domain of integration (e.g., from a hemisphere $H^2$ to the surface of an aperture $A$), we need to change the measure ($d\omega'$ or $dA$) accordingly.
I never quite understood what H^2 meant here- what is a hemisphere in this picture? Also what's the difference between dw and dA?
$H^2$ and $A$ refer to the domains of integration.
In the first integral, we're integrating all possible incident light rays as a function of direction $\omega'$ (i.e., we choose a light ray in the direction $\omega'$ and measure the contribution of the ray). The set of all possible incident light rays are given by sampling a hemisphere centered around point $p$, represented by $H^2$. The measure $d\omega'$ weighs the contribution of this ray by the differential solid angle (the field of view that this ray covers).
In the second integral, we're integrating all possible incident rays as a function of points $p'$ on the aperture (i.e., we choose a light ray by sampling a point $p'$ on the aperture, define a ray passing through points $p$ and $p'$, and measure the contribution of the ray). The set of all possible incident light rays are given by the sampling all points on the aperture, and the domain of integration in this case is $A$---representing the area of the aperture. The measure $dA$ weighs the contribution of the ray by the differential area on the aperture.
Note that the (differential) solid angle is related to the (differential) area on the aperture by the following function: $d\omega' = \frac{cos(\theta)}{|p'-p|^2}dA$. That is, if you were to rotate a differential area by $\theta$, the solid angle subtended from point $p$ would get smaller by $cos(\theta)$. If you were to move the differential area further away from the sensor, then the solid angle would decrease according to $\frac{1}{|p'-p|^2}$.
So, in short, when changing the domain of integration (e.g., from a hemisphere $H^2$ to the surface of an aperture $A$), we need to change the measure ($d\omega'$ or $dA$) accordingly.