The energy associated with the liberation of these electrons is the radiative energy we seek to measure. This energy interacts with the semiconductor substrate and liberates photoelectrons at a rate assumed to be linear with respect to the integration-time.

$$E = \frac{hc}{\lambda}$$

Energy carries the normal units Joules ($[E]=~ J$)¹⁾, but be careful! These counts are proportional to the received radiant energy (in the form of photons for this discussion) incident on a single pixel. This single pixel receives photons from a specific angular area of the sky, the pixel has an area over which it received photons, an orientation of that area, a time during which it was 'allowed' to receive photons, and a sensitivity to photons over a specific range of wavelengths.

So these counts are proportional not to the total energy of a target observed by the telescope, but to the differential amount of that energy called (in the broadest sense) the spectral radiance.

The differential amount of energy shown in the numerator of the right hand side tends to vanish as the factors in the denominator tend to zero. The spectral radiance (the left hand side; $I(\lambda, \theta, \phi)$) is the limiting value this ratio approaches. Is can be thought of as a “ray” of light or “pencil” of radiation. The spectral radiance is a convenient theoretical quantity since its dependence on differential solid angle leave it conserved over its optical path (barring any sources or sinks). Therefore, the spectral radiance is the basis of all radiative transfer theory.

$$I(\lambda, \theta, \phi) = \frac{d^6 E}{d\lambda\cdot dt\cdot d^2A \cos{\theta}\cdot d^2\Omega(\theta,\phi)}$$

The spectral radiance²⁾ has units $[I(\lambda, \theta, \phi)]=J\cdot (m\cdot t\cdot m^2\cdot sr^2)^{-1} = W\cdot (m\cdot m^2\cdot sr^2)^{-1}$

Said slightly differently, the spectral radiance is the amount of energy received by an area ($d^2A$) with some orientation relative to incidence ($\cos{\theta}$) and some a sensitivity to photons over a specific range of wavelengths ($d\lambda$) from a specific angular area of the sky ($d^2\Omega(\theta,\phi)$) during some time ($dt$).

Each pixel effectively integrates this spectral radiance with respect to wavelength, time, perpendicular area, and solid angle. The resulting measure of radiant energy is what is responsible for liberating photoelectrons in the CMOS semiconductor substrate which are eventually readout proportional to a voltage they generate as counts.

For angularly unresolved sources, we are forced to assume that the energy we received in each pixel is the integral of all the spectral radiance that emerged from the source parallel to our line of sight. Since we only can hope to get components of radiance from one hemisphere of the radiance field generated by the source, we can integrate over half of the sphere ($\theta\in[0,\frac{\pi}{2}], \phi\in[0,2\pi]$).

$$\int_{2\pi}I(\lambda, \theta, \phi)\cos{\theta}\cdot d^2\Omega(\theta,\phi) = \int_{2\pi}\frac{d^6 E}{d\lambda\cdot dt\cdot d^2A \cos{\theta}\cdot d^2\Omega(\theta,\phi)}\cos{\theta}\cdot d^2\Omega(\theta,\phi)$$

Note that this is the first angular moment of the spectral radiance field. And it is defined to be the spectral flux.

$$F(\lambda) = \frac{d^4 E}{d\lambda\cdot dt\cdot d^2A_{\perp}}$$

The radiative, spectral flux has units $[F(\lambda)]=J\cdot (m\cdot t\cdot m^2)^{-1}=W\cdot (m\cdot m^2)^{-1}$.

BLAH THIS PAGE IS NOT DONE. I need to explain how to photometrically calibrate still with this background out of the way … maybe this is another page.