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).
Best Answer
$L_r = \frac{d^2\Phi}{d\omega \space dA\cos{\theta}}$
I can separate out the derivatives in the above equation to get:
$L_r = \frac{d}{d\omega}(\frac{d\Phi}{ \space dA\cos{\theta}})$ [note, alternatively I could have done $L_r = \frac{d}{dA}(\frac{d\Phi}{ \space d\omega\cos{\theta}})$, but I believe the former will allow for a more intuitive explanation of the formula.]
Let us examine the portion in brackets, that is, $\frac{d\Phi}{ \space dA\cos{\theta}}$. Firstly, note that $\theta$ represents the angle between the surface normal and the specified direction from which $L_r$ is measured, so we can think of $A cos\theta$ as being the area perpendicular to the direction L is measured.
This term is thus calculating how much power there is in the light, per the component of unit area that is perpendicular to the light, at a particular point.
The reason for the derivative, rather than simply a "divide by" is because we must consider each "infintisimally small" surface element separately, because each surface element could have a different orientation, and thus a different value of $\cos{\theta}$, since $\theta$ depends upon the orientation of the surface element. A simple "divide by" over a large area would thus not work, as there would be no value of $\theta$ applicable to the entire area.
Let us now consider the $\frac{d}{d\omega}$ component. This operator acts on the quantity "Power per perpendicular unit area", and thus evaluates to "Power per perpendicular unit area, per unit frequency". The power per unit area could be due to a contribution from a variety of frequencies, and this quantity $L_r$ can thus tell us how the power contributions change as frequency changes. Ultimately the quantity could be used to evaluate the power contributions in a certain frequency range, by computing the integral $\int_{\omega_0}^{\omega_0 + \delta} L_r d\omega$.