Solid angles in integral? Could somebody explain why we have to integrate over the solid angle when calculating total luminosity in the following:

The integral over the wavelength makes sense, that a=way we have a total power from the entire spectrum, and multiplying by it's area makes sense, because the Planck function is per area. But why solid angle?
 A: It's part of the definition of $B_\lambda$. Inherent in the definition of $B_\lambda$ is the treatment of photons as though they are particles that have a position and momentum (in this case, they're a low density of wave packets). I actually find it easier to understand what's going on by considering the phase space density of photons
$$\mathcal{N}_\gamma\left(\vec{x},\vec{p}\right) \equiv \frac{\partial^2N}{\partial x^3 \partial p^3}. \tag1$$
In (1) you naturally have two three dimensional spaces, and $\mathcal{N}_\gamma$ is the number of photons per unit phase space volume. Because you have two 3-dimensional volumes involved, you have two solid angles – one for real space, and one for momentum space. You can work out that this is related to spectral radiance by
$$I_\nu\left(\vec{x},\nu,\hat{k}\right) = hcp^3 \mathcal{N}_\gamma\left(\vec{x},\vec{p}\right),$$
with $\vec{p} = h\nu \hat{k} / c$.
What I like about using $\mathcal{N}_\gamma$ is that


*

*the blackbody photon density is given by $\frac{2}{h^3\left[e^{h\nu / kT}-1\right]}$, which is very simple, and

*$\mathcal{N}_\gamma$ is invariant under the expansion of the universe, making cosmology/GR calculations easier.


So, where does the $\cos\theta$ come from in the solid angle integral in the question? Well, to understand that you need to do a little bit of work. For photons just moving through space, photon number is conserved. So, $\mathcal{N}_\gamma$ participates in a continuity equation with a phase space current density, just like charge density and electric current density, but with more dimensions. The phase space velocity is given by
\begin{align}
   \dot{\vec{x}} & = c \hat{k}\\
   \dot{\vec{p}} & = 0
\end{align}
leading to the phace space current density, $\vec{\mathcal{J}}_\gamma$,
\begin{align}
  \vec{\mathcal{J}}_\gamma\left(\vec{x},\vec{p}\right) = \left[\begin{array}{c}
    c\hat{k} \mathcal{N}_\gamma \\
    0
   \end{array}\right].\tag2
\end{align}
So, to find the rate at which photons are traveling through some surface in phase space you would calculate
$$\int \vec{\mathcal{J}}_\gamma \cdot \hat{n}\,\mathrm{d}A, \tag3$$
where $\hat{n}$ is perpendicular to the surface. The dot product in (3) is where the $\cos\theta$ comes from.
A: As Wikipedia explains, the spectral radiance $B_\lambda$ of a blackbody is the power emitted per unit area of the body, per unit solid angle of emission, per unit wavelength. To get the power you have to integrate over the area, the solid angle, and the wavelength.
