Derivation of the solar resource at Earth's surface

Question

I am reading Jenny Nelson's The Physics of Solar Cells and am a bit confused with the derivation of the solar irradiance at the Earth's atmosphere she supplies. I'll outline it here, and I provide also pictures in case that's not clear.

She begins with the statement of the spectral photon flux $\beta(E, s, \theta, \phi)$; no problems here, I've seen this derivation in Reif's statistical mechanics text. Importantly, we emphasize that this is the flux (per unit energy and per unit solid angle of photon direction) of photons per unit area of the blackbody's surface area. I'll come back to this point later as it's the source of my confusion.

As regards the solar resource at the Earth, Nelson then goes on to integrate this flux over the angles $\theta < \theta_{sun}$ in order to obtain the photon flux, $b$, at the Earth. Now the trouble for me is that this sort of geometric argument seems to have two troubling premises:

1. That we are considering some surface element (I'll call it $dS'$ to make it special) of the blackbody (our Sun here) which is directly "in line" with the Earth, so that the only photons being captured by the Earth are those photons which were emitted from this $dS'$, and nothing else, i.e. all the photons from any other $dS \neq dS'$ (please forgive the abuse of notation in treating the $dS$ as actual pieces of surface).

2. Throughout the derivation, we never integrate over the surface area element $dS$. It seems that, by the end, we are treating the flux $b$ (and irradiance $L$) on a per unit of Earth's surface area basis, even though $dS$ was defined on the surface of the blackbody.

Am I mistaken in having the two concerns above and, if not, can someone explain how these concerns are waved away/justified?

Derivation page 1

Derivation page 2

Answer 1

Here, spectral photon flux $\beta$ refers to the number of photons that leave a unit area of a black body (the Sun) $dS'$ and reach a unit area of a target (the Earth) $dS$ with a certain frequency $\omega$. So to get the total number of photons that arrive to the Earth you must sum over all $dS'$, $dS$ and $\omega$. Sum in this case is integrating because we are in the continuum. So to the total number of photons $N_s$ are:

\begin{equation} N_s = \int_0^\infty\int_{S_{Earth}}\int_{S_{sun}} \beta() dS' dS d\omega \end{equation}

But as you can follow in the derivation you mention, from equation 2.1 to equation 2.2 he is only integrating over $dS'$, so they are calculating the numbre of photons that reach a part of the Earth $dS$ (maybe here is your problem, $dS$ is on the Earth) coming from any part of the Sun. To integrate over all the Sun surface you can perform a coordinates change and integrate over solid angle $d\Omega$.

So about your questions:

1. Yes, only parts of the black body that are in-line with the surface of Earth contribute to the radiation, but if you think on that, you can make a straight line between any part part of the Earth and the half of the Sun. Hope that the image helps you with that.

2. As I said above, dS is defined on the Earth surface, $d\Omega$ is what represent the Sun surface, in the notes you facilitate $dS$ is not defined, so take into account that all the problem is reversible, you can consider $dS$ over the Sun, but then $d\Omega$ represents the Earth surface.