Definition of isotropy of radiation and its consequences

Question

I have encountered different notions of isotropy of radiation and I would like to know if they are the same and what the exact definition of isotropy is, if one exists.

Let's take black body radiation inside a cavity at thermal equilibrium for an example. It's a fact, that in this case the radiation is isotropic, but what does that mean exactly? Let's consider a point in the cavity and two small surfaces $d\sigma_1$ and $d\sigma_2$ located at said point, but with different orientations in space. Let $L_1(\theta_1,\phi_1,\nu)$ be the spectral radiance of $d\sigma_1$ where $\theta_1$ and $\phi_1$ are the angles of the solid angle $d\Omega_1$ and $\nu$ is a certain frequency and $L_2(\theta_2,\phi_2,\nu)$ be the corresponding spectral radiance of $d\sigma_2$. The angles $\theta_i$, $\phi_i$ are measured with respect to the normal to the surface $d\sigma_i$.

The first notion of isotropy that I encountered is: $L_1(\theta_1,\phi_1,\nu)$ = $L_1(\tilde{\theta}_1,\tilde\phi_1,\nu)$ for all possible angles $\theta_1$, $\phi_1$, $\tilde\theta_1$, $\tilde\phi_1$. That means, if the surface is chosen, it does not matter in which direction you look.

The second notion is: Let $\theta := \theta_1=\theta_2$ and $\phi := \phi_1 = \phi_2$. Then $L_1(\theta,\phi,\nu) = L_2(\theta,\phi,\nu)$ (Planck used this in his book "Vorlesungen über die Theorie der Wärmestrahlung" in the derivation of equation (21)). That means that the orientation does not matter.

So which one is the definition of isotropy of radiation? Maybe none or both?

Answer 1

I think I have found an answer to my question. The answer is that the first notion is the correct definition and the second notion of isotropy follows from the first notion. This goes as follows:

If $d\sigma_1$ and $\theta, \phi$ are given, where $\theta, \phi$ are measured with respect to the normal vector of $d\sigma_1$, then you can choose an small surface $d\sigma_2$ orientated such that its normal vector points towards $d\Omega(\theta,\phi)$. Then $L_1(\theta,\phi,\nu) = L_2(0,0,\nu)$. And by isotropy it follows that $L_2(0,0,\nu) = L_2(\theta,\phi,\nu)$.