Now use the useful approximation that the atmosphere is very thin compared with the underlying radius of the object, so \(\delta r \ll r\). Then the radial direction hardly changes across the layer and the ray angle changes very little. In other words, \(\theta_1 \approx \theta_2\). In that case we neglect the angular derivative term and locally treat the atmosphere as plane-parallel.
\[
\mu\,\frac{dI_\nu}{dz} = \eta_\nu - \alpha_\nu I_\nu.
\]
Here \(z\) is the local vertical coordinate. The equation has exactly the same algebraic form as the steady one-dimensional Cartesian equation. The only extra ingredient is the direction cosine \(\mu=\cos\theta\), which tells us how tilted the ray is relative to the vertical.
A simple example of this approximation is the atmosphere of the Earth. The atmosphere is very thin compared with the Earth's radius, so over a local patch the geometry behaves almost like a plane-parallel slab.
Next define the optical depth along the vertical direction, not along the ray:
\[
d\tau_\nu^{z} = -\alpha_\nu\,dz.
\]
The minus sign is important. We usually set \(\tau_\nu^{z}=0\) at the outer surface and let optical depth increase inward. So if \(z\) increases upward, then moving inward makes \(z\) decrease and \(\tau_\nu^{z}\) increase.
\[
\tau_\nu^{z}(z_2) - \tau_\nu^{z}(z_1)
=
-\int_{z_1}^{z_2}\alpha_\nu\,dz.
\]
Using \(S_\nu = \eta_\nu/\alpha_\nu\), the plane-parallel transfer equation becomes
\[
\mu\,\frac{dI_\nu}{dz}
=
\alpha_\nu(S_\nu - I_\nu).
\]
Now change variables from \(z\) to \(\tau_\nu^z\). Since \(d\tau_\nu^z/dz = -\alpha_\nu\), we have
\[
\frac{dI_\nu}{dz}
=
\frac{dI_\nu}{d\tau_\nu^{z}}
\frac{d\tau_\nu^{z}}{dz}
=
-\alpha_\nu\,\frac{dI_\nu}{d\tau_\nu^{z}}.
\]
Substitute this into the transfer equation:
\[
-\mu\,\alpha_\nu\,\frac{dI_\nu}{d\tau_\nu^{z}}
=
\alpha_\nu(S_\nu - I_\nu).
\]
After dividing by \(\alpha_\nu\), we get
\[
\mu\,\frac{dI_\nu}{d\tau_\nu^{z}} = I_\nu - S_\nu.
\]
This optical depth is not measured along the ray. It is measured along the vertical direction, opposite to the outward \(z\)-axis. Along a ray tilted by angle \(\theta\), the corresponding optical depth is larger by a factor \(1/\mu\):
\[
\tau_\nu^{\rm ray} = \frac{\tau_\nu^{z}}{\mu}.
\]
Now solve the equation. Rewrite it in linear form:
\[
\frac{dI_\nu}{d\tau_\nu^{z}} - \frac{1}{\mu}I_\nu = -\frac{1}{\mu}S_\nu.
\]
The integrating factor is
\[
e^{-\tau_\nu^{z}/\mu}.
\]
Multiply the equation by this factor:
\[
e^{-\tau_\nu^{z}/\mu}\frac{dI_\nu}{d\tau_\nu^{z}}
-
\frac{1}{\mu}e^{-\tau_\nu^{z}/\mu}I_\nu
=
-\frac{1}{\mu}S_\nu\,e^{-\tau_\nu^{z}/\mu}.
\]
The left-hand side is now an exact derivative:
\[
\frac{d}{d\tau_\nu^{z}}
\left(
I_\nu e^{-\tau_\nu^{z}/\mu}
\right)
=
-\frac{1}{\mu}S_\nu\,e^{-\tau_\nu^{z}/\mu}.
\]
Integrate between two vertical optical depths \(\tau_{\nu,1}^{z}\) and \(\tau_{\nu,2}^{z}\), with \(\tau_{\nu,2}^{z} > \tau_{\nu,1}^{z}\):
\[
I_\nu(\tau_{\nu,2}^{z},\mu)\,e^{-\tau_{\nu,2}^{z}/\mu}
-
I_\nu(\tau_{\nu,1}^{z},\mu)\,e^{-\tau_{\nu,1}^{z}/\mu}
=
-\int_{\tau_{\nu,1}^{z}}^{\tau_{\nu,2}^{z}}
S_\nu(t)\,e^{-t/\mu}\,\frac{dt}{\mu}.
\]
Rearrange it to solve for the shallower point \(\tau_{\nu,1}^{z}\):
\[
I_\nu(\tau_{\nu,1}^{z},\mu)
=
I_\nu(\tau_{\nu,2}^{z},\mu)
e^{-(\tau_{\nu,2}^{z}-\tau_{\nu,1}^{z})/\mu}
+
\int_{\tau_{\nu,1}^{z}}^{\tau_{\nu,2}^{z}}
S_\nu(t)\,
e^{-(t-\tau_{\nu,1}^{z})/\mu}\,
\frac{dt}{\mu}.
\]
Now read this formula directly. The first term is just the exponential attenuation of the incoming beam. The factor depends on the optical-depth difference \(\Delta\tau_\nu^{z}=\tau_{\nu,2}^{z}-\tau_{\nu,1}^{z}\), but because the ray is tilted by \(\theta\), the path length introduces the extra \(1/\mu\).
The second term is different. The source is not added at one big point. Each small point along the ray adds a small contribution to the beam. Then that small contribution has to travel the rest of the way to the final point, and during that travel it is attenuated. That is why the source term has to be integrated point by point along the whole ray.
Now take a semi-infinite medium. This simply means the object is so thick that, for radiative-transfer purposes, it behaves like an atmosphere extending to very large optical depth. At the outer boundary, \(\tau_\nu^{z}=0\). Deep inside, \(\tau_\nu^{z}\to\infty\). For an emergent ray we are interested in \(I_\nu(0,\mu)\), so the formal solution becomes
\[
I_\nu(0,\mu)
=
I_\nu(\infty,\mu)e^{-\infty/\mu}
+
\int_0^\infty S_\nu(t)\,e^{-t/\mu}\,\frac{dt}{\mu}.
\]
The exponential kills the deep boundary term, so
\[
I_\nu(0,\mu)
=
\int_0^\infty S_\nu(t)\,e^{-t/\mu}\,\frac{dt}{\mu}.
\]
If the source function is constant, \(S_\nu(t)=S_\nu\), then
\[
I_\nu(0,\mu)
=
S_\nu\int_0^\infty e^{-t/\mu}\,\frac{dt}{\mu}
=
S_\nu.
\]
This matches the optically thick result we already derived for the homogeneous slab, so this result is exactly what we should expect.
As a direct check, the outward flux in this constant-source case becomes
\[
F_\nu^{z}(0)
=
2\pi\int_0^1 I_\nu(0,\mu)\,\mu\,d\mu
=
2\pi\int_0^1 S_\nu\,\mu\,d\mu
=
\pi S_\nu.
\]
Now take the next simple case, where the source function varies linearly with vertical optical depth:
\[
S_\nu(\tau_\nu^{z}) = a_\nu + b_\nu \tau_\nu^{z}.
\]
Insert this into the surface solution:
\[
I_\nu(0,\mu)
=
\int_0^\infty
\left(a_\nu + b_\nu t\right)
e^{-t/\mu}\,\frac{dt}{\mu}.
\]
Separate the two pieces:
\[
I_\nu(0,\mu)
=
a_\nu\int_0^\infty e^{-t/\mu}\,\frac{dt}{\mu}
+
b_\nu\int_0^\infty t\,e^{-t/\mu}\,\frac{dt}{\mu}.
\]
For the first integral, let \(x=t/\mu\), so \(dt=\mu\,dx\):
\[
\int_0^\infty e^{-t/\mu}\,\frac{dt}{\mu}
=
\int_0^\infty e^{-x}\,dx
= 1.
\]
For the second integral, use the same substitution:
\[
\int_0^\infty t\,e^{-t/\mu}\,\frac{dt}{\mu}
=
\mu\int_0^\infty x\,e^{-x}\,dx
=
\mu.
\]
Therefore
\[
I_\nu(0,\mu) = a_\nu + b_\nu \mu.
\]
Since \(S_\nu(\tau_\nu^{z}) = a_\nu + b_\nu \tau_\nu^{z}\), this means
\[
I_\nu(0,\mu) = S_\nu(\tau_\nu^{z} = \mu).
\]
This is the Eddington-Barbier relation. It says that the emergent intensity in the direction \(\mu\) is approximately the source function evaluated at the vertical optical depth \(\tau_\nu^{z}=\mu\). This approximation is extremely useful for understanding some very basic properties of radiative transfer.
Now ask what we actually see. We see the source function, but not from every depth equally. Along a given line of sight we effectively see to optical depth of order unity along the ray. In vertical coordinates that means \(\tau_\nu^{z}\approx \mu\).
So if we look straight out, \(\mu=1\), then ray optical-depth unity corresponds roughly to \(\tau_\nu^{z}=1\), and we see deeper. If we move toward the limb, \(\mu\) becomes small. Then the condition \(\tau_\nu^{\rm ray}=1\) is reached earlier in the vertical scale, so we see a higher layer.
If the source function increases inward, then deeper layers are brighter. Therefore the centre of the disc (\(\mu=1\)) appears brighter than the limb (\(\mu\approx 0\)). This is the basic explanation of limb darkening.
One can say it in the simplest possible way: optical depth of order unity tells us how far we can see into the medium.
Now compute the emergent flux at the surface. The outward flux is
\[
F_\nu^{z}(0)
=
2\pi\int_0^1 I_\nu(0,\mu)\,\mu\,d\mu.
\]
Insert the linear-source result \(I_\nu(0,\mu)=a_\nu+b_\nu\mu\):
\[
F_\nu^{z}(0)
=
2\pi\int_0^1 (a_\nu+b_\nu\mu)\,\mu\,d\mu.
\]
\[
F_\nu^{z}(0)
=
2\pi
\left(
a_\nu\int_0^1 \mu\,d\mu
+
b_\nu\int_0^1 \mu^2\,d\mu
\right).
\]
\[
F_\nu^{z}(0)
=
2\pi
\left(
\frac{a_\nu}{2}
+
\frac{b_\nu}{3}
\right)
=
\pi\left(a_\nu+\frac{2b_\nu}{3}\right).
\]
But the linear source function is \(S_\nu(\tau_\nu^{z})=a_\nu+b_\nu\tau_\nu^{z}\). So
\[
F_\nu^{z}(0)
=
\pi\,S_\nu\!\left(\tau_\nu^{z}=\frac{2}{3}\right).
\]
This result says something very important. Along a single ray, the intensity effectively emerges from around optical depth unity along that ray. But once we take that quantity and average it over outward angles to form the flux, the characteristic vertical optical depth becomes \(2/3\).
That is why we usually define the photosphere, or the effective visible surface of a star or any other optically thick object, near optical depth \(2/3\). In that sense, the photosphere of a star, the visible surface of an accretion disc, or the apparent surface of any optically thick atmosphere is the layer around \(\tau \sim 2/3\).
