Chapter 2

Radiative Transfer Introduction

This chapter starts from the first differential equation of radiative transfer and then studies the special case of equilibrium radiation. From there it develops the blackbody spectrum, its low- and high-frequency limits, integrated blackbody quantities, the meaning of effective temperature, and finally the formal solution of radiative transfer in practical slabs, atmospheres, and clouds.

Scope of this part: the slab derivation of the transfer equation, equilibrium radiation, Planck's law, Rayleigh-Jeans and Wien limits, wavelength form, Wien's displacement law, integrated blackbody intensity, energy density, flux, pressure, effective temperature, optical depth, source function, and the formal radiative-transfer solution in optically thick and optically thin media.

Chapter Contents

A short guide to the structure before the detailed derivations.

  1. 1Radiative Transfer Equation
  2. 1.1Slab Picture and Basic Equation
  3. 1.2Vacuum Limit
  4. 2Equilibrium Radiation and Planck Spectrum
  5. 2.1Planck Function
  6. 2.2Low-Frequency Limit
  7. 2.3High-Frequency Limit
  8. 2.4Wavelength Form and Wien's Law
  9. 3Integrated Blackbody Quantities
  10. 3.1Intensity, Energy Density, and Ultraviolet Catastrophe
  11. 3.2Flux and Pressure of an Isotropic Blackbody Field
  12. 4Effective Temperature
  13. 5Formal Solution of the Transfer Equation
  14. 5.1Optical Depth and Source Function
  15. 5.2Pure Absorption / Pure Extinction
  16. 5.3General Solution and Homogeneous Slab
  17. 5.4Optically Thick and Optically Thin Limits
  18. 5.5Line Formation and Variable Source Function
  19. Ref.References

1. Radiative Transfer Equation

The question here is simple: if a beam with specific intensity \(I_\nu^{\mathrm{in}}\) enters a small piece of material, what will the outgoing intensity \(I_\nu^{\mathrm{out}}\) be after it travels a short distance \(\Delta s\) along the ray direction \(\hat n\)?

1.1 Slab Picture and the Basic Equation

Take a thin slab of material of thickness \(\Delta s\). Radiation enters from the left with specific intensity \(I_\nu^{\mathrm{in}}\) and leaves from the right with specific intensity \(I_\nu^{\mathrm{out}}\). Over such a short path, the change in intensity comes from two competing effects: the medium can add radiation to the beam, and it can remove radiation from the beam.

The amount added over the short distance \(\Delta s\) is written as \(\eta_\nu \Delta s\). Here \(\eta_\nu\) is the emission coefficient. It measures how much intensity per unit path length is inserted into the beam at frequency \(\nu\).

The amount removed is proportional to both the path length and the amount of radiation already present in the beam. That is why the loss term is written as \(\alpha_\nu I_\nu \Delta s\). Here \(\alpha_\nu\) is the absorption coefficient.

\[ \Delta I_\nu = I_\nu^{\mathrm{out}} - I_\nu^{\mathrm{in}} = \eta_\nu \Delta s - \alpha_\nu I_\nu \Delta s. \]

Now divide by \(\Delta s\). In the limit of a very thin slab, \(\Delta s \to 0\), the finite difference becomes a derivative along the ray:

\[ \frac{\Delta I_\nu}{\Delta s} = \eta_\nu - \alpha_\nu I_\nu \quad \Longrightarrow \quad \frac{dI_\nu}{ds} = \eta_\nu - \alpha_\nu I_\nu. \]

This is the basic equation of radiative transfer. Along a ray, intensity increases because emission adds radiation to the beam and decreases because absorption removes radiation from it.

Interpretation. The term \(\eta_\nu\) is a source term. The term \(\alpha_\nu I_\nu\) is a sink term. The balance between them determines whether the beam brightens, dims, or stays unchanged along the path.

It is also useful to understand why these two coefficients appear in different forms. The emission term is written as \(\eta_\nu \Delta s\) because the amount added depends mainly on how much emitting material lies in the small path length \(\Delta s\). If the slab is twice as thick, there is twice as much material available to emit, so the added intensity is twice as large.

The absorption term has the form \(\alpha_\nu I_\nu \Delta s\) because absorption removes a fraction of the beam that is already present. If the incoming beam is brighter, more intensity can be removed. If the path is longer, there is more opportunity to absorb. That is why the loss term must be proportional to both \(I_\nu\) and \(\Delta s\).

A useful microscopic picture is to think of \(\alpha_\nu\) as behaving like number density times cross-section, \(\alpha_\nu \sim n\sigma_\nu\). Then \(n\sigma_\nu \Delta s\) is the small probability that radiation is removed while crossing the slab. This is why \(\alpha_\nu\) multiplies the incoming intensity instead of appearing by itself.

With this intuition, the transfer equation becomes very natural: the medium creates radiation through \(\eta_\nu\), removes radiation through \(\alpha_\nu I_\nu\), and the competition between these two effects controls the evolution of the beam.

Transfer equation slab diagram Incoming specific intensity enters a slab of thickness delta s and exits with a modified outgoing intensity along the ray direction n hat. incoming Iν Δs outgoing Iν n direction path length Δs added by emission: ην Δs removed by absorption: αν Iν Δs
The slab picture behind the transfer equation. Over a short distance \(\Delta s\), the outgoing beam differs from the incoming beam by the competition between emission and absorption.
Symbol Meaning in this chapter
\(I_\nu\) Specific intensity at frequency \(\nu\)
\(\eta_\nu\) Emission coefficient: intensity added per unit path length
\(\alpha_\nu\) Absorption coefficient: fractional removal per unit path length
\(s\) Distance measured along the ray

1.2 Vacuum Limit

If there is no interaction with matter, nothing can be added to or removed from the beam. In that case

\[ \eta_\nu = 0, \qquad \alpha_\nu = 0. \]
\[ \frac{dI_\nu}{ds} = 0. \]
\[ I_\nu = \text{constant along the ray}. \]

This is the same statement already met in Chapter 1: in vacuum, specific intensity is conserved along a ray.

2. Equilibrium Radiation and the Blackbody Spectrum

The special case of equilibrium radiation is the point where the transfer equation stops changing the beam. That stationary solution is the blackbody spectrum.

2.1 Equilibrium Radiation and Planck Function

In equilibrium, the intensity no longer changes along the path. That means the gain and loss terms balance each other:

\[ \frac{dI_\nu}{ds} = 0. \]
\[ \eta_\nu = \alpha_\nu I_\nu. \]

So the equilibrium solution is a frequency-dependent intensity that stays constant through the medium. In thermal equilibrium, that special spectrum is the Planck function:

\[ I_\nu = B_\nu(T) = \frac{2h\nu^3}{c^2} \frac{1}{e^{h\nu/kT} - 1}. \]

This is Planck's law of radiation. It tells us the specific intensity of blackbody radiation at temperature \(T\).

The shape is worth reading physically. At low frequency the spectrum rises as a power law. It reaches a peak, and at high frequency it falls rapidly because of an exponential suppression.

Schematic blackbody spectral shape A schematic plot of the Planck function against logarithmic frequency showing power law rise, peak, and exponential cutoff. Bν(T) log ν power-law rise peak exponential cut-off
The Planck spectrum shows a low-frequency rise, a peak, and a high-frequency exponential fall.

2.2 Low-Frequency Limit: Rayleigh-Jeans Form

At low frequency, the exponential argument is small:

\[ x = \frac{h\nu}{kT} \ll 1. \]

Use the first-order expansion \(e^x \approx 1 + x\). Then

\[ B_\nu(T) = \frac{2h\nu^3}{c^2} \frac{1}{e^{h\nu/kT}-1} \approx \frac{2h\nu^3}{c^2} \frac{1}{1 + h\nu/kT - 1} \]
\[ B_\nu^{\mathrm{RJ}}(T) = \frac{2kT}{c^2}\nu^2. \]

This is the Rayleigh-Jeans limit. It gives the classical low-frequency approximation to the full Planck spectrum. In this regime \(B_\nu \propto \nu^2\) and \(B_\nu \propto T\).

2.3 High-Frequency Limit: Wien Form

At high frequency, the exponential argument is large:

\[ \frac{h\nu}{kT} \gg 1. \]

Now \(e^{h\nu/kT} - 1 \approx e^{h\nu/kT}\), so

\[ B_\nu(T) \approx \frac{2h\nu^3}{c^2} e^{-h\nu/kT}. \]

This is the Wien tail. The factor \(\nu^3\) still appears, but the exponential suppression dominates and drives the spectrum down rapidly at large frequency.

2.4 Wavelength Form and Wien's Displacement Law

Your notes next move from the frequency form \(B_\nu\) to the wavelength form \(B_\lambda\). The two descriptions must represent the same energy in the same small spectral interval, so

\[ B_\nu(T)\,d\nu = B_\lambda(T)\,d\lambda. \]

Therefore

\[ B_\lambda(T) = B_\nu(T)\left|\frac{d\nu}{d\lambda}\right|. \]

Since \(c = \lambda \nu\), we have \(\nu = c/\lambda\) and

\[ \left|\frac{d\nu}{d\lambda}\right| = \frac{c}{\lambda^2}. \]

Now substitute \(\nu=c/\lambda\) into the Planck function:

\[ B_\nu(T) = \frac{2h(c/\lambda)^3}{c^2} \frac{1}{e^{hc/\lambda kT}-1} = \frac{2hc}{\lambda^3} \frac{1}{e^{hc/\lambda kT}-1}. \]

Multiplying by \(c/\lambda^2\) gives the wavelength version:

\[ B_\lambda(T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda kT}-1}. \]

To find the wavelength of peak emission, differentiate \(B_\lambda\) with respect to \(\lambda\) and set the result to zero. This leads to a transcendental equation which is usually written in terms of \(x = hc/(\lambda kT)\):

\[ 5(1-e^{-x}) = x, \qquad x \approx 4.965. \]

From that result one obtains Wien's displacement law:

\[ \lambda_{\max} T = 2897.8\ \mu\mathrm{m\,K}. \]

If a stellar spectrum peaks near \(\lambda_{\max} \approx 0.5\,\mu\mathrm{m}\), then

\[ T \approx \frac{2897.8\ \mu\mathrm{m\,K}}{0.5\ \mu\mathrm{m}} \approx 5.8\times 10^3\ \mathrm{K}. \]

So the characteristic temperature is about \(5800\ \mathrm{K}\), which is exactly the familiar solar-scale example.

3. Integrated Blackbody Quantities

Once the spectrum is known, we can integrate over frequency and then connect blackbody radiation with the energy density, flux, and pressure discussed in Chapter 1.

3.1 Integrated Intensity, Energy Density, and the Ultraviolet Catastrophe

Define the frequency-integrated blackbody intensity by

\[ B(T) = \int_0^\infty B_\nu(T)\,d\nu. \]

Insert the Planck function and change variables to \(x = h\nu/kT\):

\[ B(T) = \int_0^\infty \frac{2h\nu^3}{c^2} \frac{1}{e^{h\nu/kT}-1}\,d\nu \]
\[ \nu = \frac{xkT}{h}, \qquad d\nu = \frac{kT}{h}\,dx. \]
\[ B(T) = \frac{2(kT)^4}{h^3 c^2} \int_0^\infty \frac{x^3}{e^x-1}\,dx. \]

The remaining definite integral is standard:

\[ \int_0^\infty \frac{x^3}{e^x-1}\,dx = \frac{\pi^4}{15}. \]

So the integrated blackbody intensity becomes

\[ B(T) = \frac{\sigma T^4}{\pi}. \]

This is the Stefan-Boltzmann scaling in intensity form: \(B(T)\propto T^4\).

Now use the Chapter 1 relation between intensity and radiation energy density. For isotropic blackbody radiation, \(B_\nu(T)\) is independent of direction, so

\[ E = \oint_\Omega \int_0^\infty \frac{B_\nu(T)}{c}\,d\nu\,d\Omega = \frac{4\pi}{c}\int_0^\infty B_\nu(T)\,d\nu. \]
\[ E = \frac{4\pi}{c}B(T) = \frac{4\pi}{c}\frac{\sigma T^4}{\pi} = \frac{4\sigma T^4}{c} \equiv aT^4. \]

This is the standard blackbody radiation energy density.

Your notes then point out what goes wrong if we replace the full Planck function by the Rayleigh-Jeans approximation all the way to infinite frequency. In that case

\[ E^{\mathrm{RJ}} = \frac{4\pi}{c}\int_0^\infty B_\nu^{\mathrm{RJ}}(T)\,d\nu = \frac{4\pi}{c}\int_0^\infty \frac{2kT}{c^2}\nu^2\,d\nu. \]
\[ E^{\mathrm{RJ}} \to \infty. \]

That divergence is the ultraviolet catastrophe. It is precisely the failure that Planck's law fixes.

3.2 Flux and Pressure of an Isotropic Blackbody Field

For an isotropic blackbody radiation field, the specific intensity is independent of \(\theta\) and \(\phi\). So in the flux integral the angular dependence only comes from the directional cosine \(\mu = \cos\theta\).

\[ F_z = \oint_\Omega \int_0^\infty B_\nu \mu\,d\nu\,d\Omega = 2\pi \left(\int_0^\infty B_\nu\,d\nu\right)\int_{-1}^{1}\mu\,d\mu. \]
\[ F_z = 0. \]

An isotropic blackbody field has zero net flux because every direction has an opposite partner that cancels it.

Now look at the pressure along the \(z\)-direction. From Chapter 1,

\[ P_{zz} = \frac{1}{c} \oint_\Omega \int_0^\infty B_\nu \mu^2\,d\nu\,d\Omega. \]
\[ P = \frac{2\pi}{c} \left(\int_0^\infty B_\nu\,d\nu\right) \int_{-1}^{1}\mu^2\,d\mu. \]
\[ P = \frac{2\pi}{c}\frac{\sigma T^4}{\pi}\left(\frac{2}{3}\right) = \frac{4\sigma T^4}{3c}. \]
\[ P = \frac{E}{3}. \]

So isotropic blackbody radiation has zero net flux but a non-zero isotropic radiation pressure. The pressure tensor becomes

\[ \mathbf{P} = \begin{pmatrix} P & 0 & 0 \\ 0 & P & 0 \\ 0 & 0 & P \end{pmatrix}. \]
Isotropic blackbody field Radiation arriving equally from all directions around a point, illustrating zero net flux and nonzero pressure. isotropic field opposite directions cancel in flux but add in pressure F = 0, P = E/3
In an isotropic field, the first angular moment vanishes, but the second angular moment survives. That is why the flux is zero while the pressure is non-zero.

4. Effective Temperature

The notes now move from an isotropic radiation field in the interior to the free-streaming surface of a star or emitting object. That is where effective temperature is defined.

4.1 Why the Surface is Different

At first sight one might worry that a blackbody has zero flux, because the isotropic blackbody field in the previous subsection gave \(F=0\). But that result applies to radiation inside an isotropic field where every direction is balanced by an opposite direction.

At the surface of a star, the situation is different. The outward hemisphere contributes radiation, but there is no matching incoming radiation from outside. So the field is no longer isotropic there.

If \(\mu = \cos\theta\), then only the outward directions with \(0 \le \mu \le 1\) contribute at the surface. The inward hemisphere \(-1 \le \mu < 0\) contributes nothing in the idealized no-incoming-radiation picture.

Surface flux geometry A surface element with outward normal, outgoing directions only, and no incoming radiation from outside. normal θ μ = cos θ surface of the source outgoing hemisphere contributes no incoming radiation from outside
At a free surface, only outward directions contribute. This is why a radiating surface has non-zero net flux even though an isotropic interior field does not.

4.2 Blackbody Surface Flux

Start from the flux expression, but restrict the angular integral to the outward hemisphere:

\[ F_z = 2\pi \int_0^1 \int_0^\infty I_\nu\,\mu\,d\nu\,d\mu. \]

At an ideal blackbody surface, the outward intensity is the Planck function, \(I_\nu = B_\nu(T)\). So

\[ F_z = 2\pi \int_0^1 \int_0^\infty B_\nu(T)\,\mu\,d\nu\,d\mu. \]

The angular and frequency parts separate:

\[ F_z = 2\pi \left(\int_0^\infty B_\nu(T)\,d\nu\right) \left(\int_0^1 \mu\,d\mu\right). \]
\[ F_z = 2\pi \left(\frac{\sigma T^4}{\pi}\right) \left(\frac{1}{2}\right). \]
\[ F_z = \sigma T^4. \]

This is the Stefan-Boltzmann law in flux form for a blackbody surface.

4.3 General Definition of Effective Temperature

For a real object, the emerging intensity need not be exactly Planckian. Still, we define the effective temperature by asking: what blackbody temperature would produce the same total emergent flux?

\[ 2\pi \int_0^1 \int_0^\infty I_\nu\,\mu\,d\nu\,d\mu = \sigma T_{\mathrm{eff}}^4. \]

The left-hand side is the actual emergent flux from the surface. The right-hand side defines \(T_{\mathrm{eff}}\) as the blackbody temperature that would emit the same flux.

So in general \(T_{\mathrm{eff}}\) is a flux-based temperature, not automatically the same as the local material temperature. How close they are depends on how well the emitting surface can be approximated as a blackbody.

5. Formal Solution of the Transfer Equation

We now solve the radiative-transfer equation. The guiding example is a stellar atmosphere or a gas slab placed along the line of sight. The question is always the same: if some radiation enters the material, what intensity comes out after extinction and emission both act throughout the path?

5.1 Reference Cases, Units, Optical Depth, and Source Function

Before solving the general equation, it is useful to recall the two simplest limiting situations already seen earlier in the chapter.

\[ \frac{dI_\nu}{ds} = \eta_\nu - \alpha_\nu I_\nu \]
\[ \text{vacuum:}\qquad \eta_\nu = 0,\ \alpha_\nu = 0 \quad\Longrightarrow\quad \frac{dI_\nu}{ds}=0,\ \ I_\nu=\text{constant} \]
\[ \text{equilibrium:}\qquad \eta_\nu = \alpha_\nu I_\nu \quad\Longrightarrow\quad \frac{dI_\nu}{ds}=0,\ \ I_\nu=B_\nu \]

Now we move to the more interesting case where the beam is actually modified while passing through matter.

The absorption coefficient has units of inverse length:

\[ [\alpha_\nu] = \mathrm{cm^{-1}}. \]

This makes sense physically. The combination \(\alpha_\nu ds\) must be dimensionless, because it measures the small fractional attenuation produced by a tiny segment of path.

The left-hand side of the transfer equation has the same units as the emission coefficient:

\[ \frac{dI_\nu}{ds} = \frac{dE}{dA\,dt\,d\nu\,d\Omega\,ds} = \left(\frac{dE}{dV}\right)\frac{1}{dt\,d\nu\,d\Omega}, \]

because \(dV=dA\,ds\). So \(dI_\nu/ds\) is energy added to the beam per unit volume, per unit time, per unit frequency interval, and per unit solid angle. This is why \(\eta_\nu\) carries the same units as \(dI_\nu/ds\).

The next useful step is to absorb the extinction coefficient into the path coordinate by defining the optical depth:

\[ d\tau_\nu = \alpha_\nu\,ds, \qquad \tau_\nu = \int \alpha_\nu\,ds. \]

Optical depth is the extinction accumulated along the beam. It is not a geometric distance. It tells you how opaque the medium is along that line of sight. If \(\alpha_\nu\) is spatially constant over a path of length \(L\), then

\[ \tau_\nu = \alpha_\nu L. \]

Using \(d\tau_\nu = \alpha_\nu ds\), divide the transfer equation by \(\alpha_\nu\):

\[ \frac{dI_\nu}{\alpha_\nu ds} = \frac{\eta_\nu}{\alpha_\nu} - I_\nu. \]
\[ \frac{dI_\nu}{d\tau_\nu} = S_\nu - I_\nu, \qquad S_\nu \equiv \frac{\eta_\nu}{\alpha_\nu}. \]

The quantity \(S_\nu\) is called the source function. It is the ratio of emission to extinction. It has the same units as specific intensity, which is why it can be compared directly with \(I_\nu\).

Important notation. The path coordinate is \(s\), while the source function is \(S_\nu\). They are completely different quantities even though both are written with the same letter in Roman script.

5.2 Pure Extinction or Pure Absorption

Now solve the simplest nontrivial case: no radiation is added to the beam. That means

\[ \eta_\nu = 0 \qquad\Longrightarrow\qquad S_\nu = 0. \]

So the transfer equation in optical-depth form becomes

\[ \frac{dI_\nu}{d\tau_\nu} = -I_\nu. \]

Separate variables:

\[ \frac{dI_\nu}{I_\nu} = -d\tau_\nu. \]

Integrate between an input point and an output point along the ray:

\[ \int_{I_\nu^{\rm in}}^{I_\nu^{\rm out}} \frac{dI_\nu}{I_\nu} = -\int_{\tau_\nu^{\rm in}}^{\tau_\nu^{\rm out}} d\tau_\nu. \]
\[ \ln\!\left(\frac{I_\nu^{\rm out}}{I_\nu^{\rm in}}\right) = -\left(\tau_\nu^{\rm out}-\tau_\nu^{\rm in}\right). \]
\[ I_\nu^{\rm out} = I_\nu^{\rm in} \exp\!\left[-\left(\tau_\nu^{\rm out}-\tau_\nu^{\rm in}\right)\right]. \]

This is the exponential attenuation law for light in a purely absorbing or purely extinguishing medium.

For an atmosphere, it is common to measure optical depth from the top boundary. Then \(\tau_\nu^{\rm in}=0\) at the entrance, and if the total optical depth of the atmosphere is \(\tau_\nu\), the emerging beam is simply

\[ I_\nu^{\rm out} = I_\nu^{\rm in} e^{-\tau_\nu}. \]

If \(\tau_\nu\) is large, the radiation is strongly attenuated. This is exactly why radiation in some frequency ranges, such as parts of the ultraviolet, may fail to pass efficiently through an atmosphere if the corresponding absorption is strong.

The same idea appears in many astrophysical situations: starlight can be extinguished by a dusty interstellar or intergalactic medium, and the amount of suppression is controlled by the optical depth along the path.

Pure absorption through an atmosphere Incoming light enters an atmosphere at optical depth zero and emerges at the observer after passing through total optical depth tau nu. source of light stellar atmosphere incoming Iν outgoing Iν τin = 0 τν
A simple atmosphere example. Radiation enters at the top and is attenuated as it crosses the total optical depth \(\tau_\nu\) before reaching the observer.

5.3 General Solution and the Homogeneous Slab

Now return to the full optical-depth form

\[ \frac{dI_\nu}{d\tau_\nu} = S_\nu - I_\nu. \]

Rearrange it into the standard first-order linear form:

\[ \frac{dI_\nu}{d\tau_\nu} + I_\nu = S_\nu. \]

The integrating factor is \(e^{\tau_\nu}\). Multiply both sides:

\[ e^{\tau_\nu}\frac{dI_\nu}{d\tau_\nu} + e^{\tau_\nu} I_\nu = e^{\tau_\nu} S_\nu. \]

The left-hand side is now an exact derivative:

\[ \frac{d}{d\tau_\nu}\!\left(e^{\tau_\nu} I_\nu\right) = e^{\tau_\nu} S_\nu. \]

Integrate this between two positions on the ray, labelled “in” and “out”:

\[ \int_{\tau_\nu^{\rm in}}^{\tau_\nu^{\rm out}} \frac{d}{d\tau_\nu}\!\left(e^{\tau_\nu}I_\nu\right)\,d\tau_\nu = \int_{\tau_\nu^{\rm in}}^{\tau_\nu^{\rm out}} e^{\tau_\nu} S_\nu(\tau_\nu)\,d\tau_\nu. \]
\[ I_\nu^{\rm out} e^{\tau_\nu^{\rm out}} - I_\nu^{\rm in} e^{\tau_\nu^{\rm in}} = \int_{\tau_\nu^{\rm in}}^{\tau_\nu^{\rm out}} e^{t} S_\nu(t)\,dt. \]

This is already the formal solution. Solving for the emergent intensity gives

\[ I_\nu^{\rm out} = I_\nu^{\rm in} e^{-(\tau_\nu^{\rm out}-\tau_\nu^{\rm in})} + \int_{\tau_\nu^{\rm in}}^{\tau_\nu^{\rm out}} S_\nu(t)\,e^{-(\tau_\nu^{\rm out}-t)}\,dt. \]

The meaning is very clear. The first term is the attenuated background intensity. The second term is the sum of all locally emitted contributions, each of which is weighted by how much attenuation it suffers before reaching the output point.

To get a closed expression, assume a homogeneous slab: the source function is constant throughout the material, \(S_\nu=\text{constant}\). Then we can take it out of the integral:

\[ I_\nu^{\rm out} e^{\tau_\nu^{\rm out}} - I_\nu^{\rm in} e^{\tau_\nu^{\rm in}} = S_\nu \left(e^{\tau_\nu^{\rm out}} - e^{\tau_\nu^{\rm in}}\right). \]

If we choose the incoming boundary to have zero optical depth, \(\tau_\nu^{\rm in}=0\), and call the slab optical depth \(\tau_\nu^{\rm out}=\tau_\nu\), then

\[ I_\nu^{\rm out} e^{\tau_\nu} - I_\nu^{\rm in} = S_\nu\left(e^{\tau_\nu}-1\right). \]
\[ I_\nu^{\rm out} = e^{-\tau_\nu} I_\nu^{\rm in} + S_\nu \left(1-e^{-\tau_\nu}\right). \]

This is the standard homogeneous-slab solution. It is one of the most useful formulas in introductory radiative transfer because every important physical case comes out of it as a limit.

Homogeneous slab solution geometry Background intensity enters a slab of constant source function and leaves as the sum of attenuated input and internally produced radiation. Iν in Sν = constant homogeneous slab Iν out τin τout
The emergent intensity from a slab is the sum of the attenuated incoming beam and the slab's own internally generated contribution.

5.4 Optically Thick and Optically Thin Limits

The homogeneous-slab solution contains all the familiar cases.

\[ I_\nu^{\rm out} = e^{-\tau_\nu} I_\nu^{\rm in} + S_\nu \left(1-e^{-\tau_\nu}\right) \]

Vacuum

\[ \tau_\nu = 0 \qquad\Longrightarrow\qquad I_\nu^{\rm out} = I_\nu^{\rm in}. \]

Pure Absorption

\[ \eta_\nu = 0 \quad\Longrightarrow\quad S_\nu = 0 \qquad\Longrightarrow\qquad I_\nu^{\rm out} = I_\nu^{\rm in} e^{-\tau_\nu}. \]

Optically Thick Medium

If \(\tau_\nu \gg 1\), then \(e^{-\tau_\nu}\) is negligibly small. The background beam is forgotten, and the emergent intensity becomes

\[ I_\nu^{\rm out} \approx S_\nu. \]

This is why \(S_\nu\) is called the source function: in an optically thick medium, the observer mainly sees the radiation field created by the medium itself, not the background entering from behind it.

A foggy medium gives the right intuition. If the fog is thick enough, you do not see the building behind it; you mostly see the fog itself.

Optically thick and optically thin intuition A two-panel sketch showing a thick foggy slab that hides the background and a thin cloud that emits weakly over a transparent line of sight. optically thick: τν ≫ 1 background hidden observer sees only the slab / fog optically thin: 0 < τν ≪ 1 thin cloud small correction to the incoming beam
In the thick limit, the medium hides the background and the observer sees mainly \(S_\nu\). In the thin limit, the medium only perturbs the incoming beam slightly.

Optically Thin Medium

If \(0 < \tau_\nu \ll 1\), expand the exponential to first order:

\[ e^{-\tau_\nu} \approx 1-\tau_\nu. \]

Substitute this into the slab solution:

\[ I_\nu^{\rm out} \approx I_\nu^{\rm in}(1-\tau_\nu) + S_\nu \tau_\nu. \]
\[ I_\nu^{\rm out} \approx I_\nu^{\rm in} + \tau_\nu\left(S_\nu - I_\nu^{\rm in}\right). \]

This form is extremely useful because it tells you immediately whether the medium increases or decreases the intensity relative to the background.

5.5 Cloud Emission, Line Formation, and Variable Source Function

Now take a cloud with no background source behind it. Then \(I_\nu^{\rm in}=0\). In the optically thin limit the previous result reduces to

\[ I_\nu^{\rm out} \approx S_\nu \tau_\nu. \]

If the absorption coefficient is roughly constant along the cloud, then \(\tau_\nu = \alpha_\nu L\), where \(L\) is the thickness of the cloud. Since \(S_\nu=\eta_\nu/\alpha_\nu\), we obtain

\[ I_\nu^{\rm out} \approx \frac{\eta_\nu}{\alpha_\nu}\,\alpha_\nu L. \]
\[ I_\nu^{\rm out} \approx \eta_\nu L. \]

This result is very intuitive. If there is no background beam, the cloud itself supplies all the radiation you observe in that direction. So the observed intensity is the emission added per unit length, multiplied by the distance travelled through the cloud.

Optically thin cloud with no background A cloud emits toward the observer with no background source behind it, illustrating the optically thin approximation I out about S tau about eta L. cloud no background source Iν in = 0 Iν out optically thin: Iν out ≈ Sντν ≈ ηνL
When there is no background source, an optically thin cloud contributes only its own small emissive signal along the line of sight.

The same optically thin formula also gives a direct criterion for whether a spectral feature appears in emission or absorption. From

\[ I_\nu^{\rm out} \approx I_\nu^{\rm in} + \tau_\nu\left(S_\nu - I_\nu^{\rm in}\right), \]

we see that:

  1. If \(S_\nu > I_\nu^{\rm in}\), the medium adds more than it removes, so the feature appears in emission.
  2. If \(S_\nu < I_\nu^{\rm in}\), the medium removes more than it adds, so the feature appears in absorption.
  3. If \(S_\nu = I_\nu^{\rm in}\), there is no contrast and the feature disappears into the background.

If the intervening material is hotter, or more precisely has a larger source function than the background, the line tends to appear in emission. If it is cooler than the background, the line tends to appear in absorption. Nebulae often give emission lines, while stellar atmospheres commonly produce absorption lines.

Now consider the more general situation. In reality the source function often varies throughout the medium. Then the integral in the formal solution cannot be simplified by pulling \(S_\nu\) outside:

\[ I_\nu^{\rm out} = I_\nu^{\rm in} e^{-(\tau_\nu^{\rm out}-\tau_\nu^{\rm in})} + \int_{\tau_\nu^{\rm in}}^{\tau_\nu^{\rm out}} S_\nu(t)\,e^{-(\tau_\nu^{\rm out}-t)}\,dt. \]

That is the genuine formal solution. When \(S_\nu\) varies, we either make useful approximations or solve the problem numerically. One important approximation we will meet later is the Eddington-Barbier relation, which classically explains effects such as limb darkening.

References

These references support the transfer equation, optical depth, source function, the blackbody spectrum, formal slab solutions, and the Stefan-Boltzmann relations used in this chapter.