A question related to radiometry:
Irradiance $E$ at a point $x$ can be written as:
$$E = \int_\Omega L(x, \omega) cos(\theta) d\omega$$
I understand this formula and where it comes from. The equation for radiance can be written as:
$$L = {d^2\Phi \over {d\omega dA^\perp cos(\theta)}}$$
What I don't understand is if we substitute this last equation in the first one for irradiance, don't we get:
$$E = \int_\Omega {d^2\Phi \over {d\omega dA^\perp cos(\theta)}} cos(\theta) d\omega \rightarrow \int_\Omega {d^2\Phi \over {dA^\perp}} \rightarrow \int_\Omega d({d\Phi \over dA^\perp}) \rightarrow \int_\Omega dE$$
which doesn't make sense to me? What am I missing? Is irradiance the integral of differential irradiance over the hemisphere?
Best Answer
Radiance is the correct term for what we loosely describe as "brightness."
Its units are Watts per square meter, per steradian.
Irradiance is simply Watts per square meter, and describes the power per unit of area impinging on a surface.
Radiance is invariant under all optical transformations (reflection, refraction, etc.)
The only way to change the radiance is to introduce statistical processes (scattering by diffuse materials).
This always involves energy losses, due to large angle and back scattering, and absorption in the diffusing medium.
The constancy of radiance is a consequence of the second law of thermodynamics. If it were possible to losslessly change radiance, we could increase the radiance of an emission from a black body source, and impinge it on a second black body, which would drive that body to a higher temperature than the source. Because of bi-directionality, if you can't increase radiance, you can't decrease it either (losslessly).