Radiative transfer equation and optical depth
Define directional spectral radiation quantities, fix the transfer sign along a ray, and distinguish ray optical depth from a slab's normal optical depth.
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The question for today
What you’ll understand
- Track frequency, area, and solid-angle units consistently.
- Solve pure attenuation with a declared propagation coordinate.
- Convert normal slab depth into inclined-ray depth and distinguish intensity from flux.
A measured spectrum depends both on how matter produces photons and on what happens along their path. Radiative transfer separates these effects without requiring a microscopic calculation at every point. We start with a stationary, nonrelativistic medium and a fixed frequency, defining distance to increase along the direction in which photons propagate. This choice fixes the attenuation sign. Throughout the module, spectral quantities are per unit frequency; a subscript nu never silently changes into a wavelength or frequency-integrated convention.
Intensity is directional, while flux is an angular moment
Specific intensity is power per projected area, frequency interval, and solid angle, with SI units . A ray crossing an area at angle theta to its normal contributes an additional cosine projection. The spectral flux through that oriented surface is the angular integral of intensity times cosine, with units watts per square metre per hertz. It can be zero for an isotropic radiation field even when photons and energy density are present. Intensity itself is not a signed flux.
Define the mean intensity as the full angular average and the spectral energy density as its angular integral divided by the speed of light. For a uniform outward hemisphere, the flux equals pi times its constant intensity, whereas a uniform full sphere gives zero net flux. These factors differ from four pi multiplying an isotropic volume emissivity. Distance dilution of a source's received flux also does not imply dilution of its resolved specific intensity along an empty ray in flat stationary space. Its apparent angular area changes instead.
Transfer along the propagation direction has a loss term
Let s increase along a ray toward its endpoint, often the observer. Define as energy emitted per volume, time, frequency, and solid angle, with units . The extinction coefficient alpha nu has inverse-length units; a mass opacity kappa nu instead has square metres per kilogram and satisfies . Emission adds j times the path increment, while extinction removes alpha times the current intensity times that increment. Extinction may include scattering out of the beam as well as genuine absorption.
Optical depth measured from the incident boundary increases along this same s direction. In pure attenuation the surviving intensity is its boundary value times the exponential of minus optical depth. An optical depth of one therefore leaves e to the minus one of the unscattered original beam, not zero. The inverse extinction coefficient is a local mean free path only when the coefficient can be treated as uniform over that scale. A source function can later add radiation back into the ray, so attenuation alone is not a complete solution for an emitting cloud.
Geometry and boundary conventions cannot be skipped
For a homogeneous plane-parallel slab, a ray with direction cosine magnitude mu relative to the slab normal travels a distance equal to the normal thickness divided by mu. Its optical depth is correspondingly the normal optical depth divided by mu. A grazing ray can therefore be thick even when the normal path is comparatively thin. This geometric relation assumes a laterally extended slab; a finite cloud can instead be exited through a side boundary, so arbitrarily long grazing paths should not be inferred without checking its shape.
Stellar-atmosphere notation often defines optical depth inward from the visible surface while the outgoing coordinate points outward. That depth decreases along an escaping ray, reversing the apparent sign in the differential transfer equation. For outward-normal direction cosine mu and source function , the usual inward-depth form is . It is consistent with our propagation-depth form after applying the chain rule. Always state the depth origin, coordinate direction, and incoming boundary condition before comparing formulas from different geometries.
Work it through
An inclined ray through an absorbing slab
A homogeneous slab has normal thickness 1 m and extinction . With no emission or in-scattering, a ray of direction-cosine magnitude 1/2 enters with . Find its emergent intensity.
The normal optical depth is ln two. The inclined path is , so the ray optical depth is twice ln two, equal to ln four.
The attenuation factor is . Using the normal depth without the inclination correction would incorrectly predict one half.
Multiplying the incident intensity by one quarter gives . No solid-angle factor is inserted because this is one directional intensity, not its angularly integrated flux.
The inclined ray emerges with specific intensity 2 × 10^−9 W m^−2 Hz^−1 sr^−1.
The living notebook
Explore the model
Interactive lab · probability
Independent absorbing layers multiply survival factors
Reset and set BOTH Success probability p = 0.50 and Trials n = 1; interpret success as removal and predict zero-removal probability 0.50. For a fresh comparison, Reset and set p = 0.50 and n = 2; predict zero-removal probability 0.25. Inspect the exact zero-success bars.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forIndependent equal layers illustrate multiplication of survival probabilities. The binomial instrument has no emissivity, ray angle, frequency transport, or radiation-hydrodynamics solver, and does not sample photon histories.
Pause & reason
Check your understanding
Your turn
Try it yourself
At one frequency let a constant intensity be . Compute the outward flux if only an outward hemisphere is filled. Separately, find flux and energy density if the same intensity fills the full sphere, using c = 3 × 10^8 m/s.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The hemisphere cosine integral is pi. The full-sphere energy integral is four pi, while its signed cosine integral vanishes.
Compare with the solution
The outward hemisphere gives . For the separate isotropic full-sphere case, the net flux is zero and , approximately in those units. These are different angular distributions, so their fluxes must not be compared as if one were merely a unit conversion of the other.
Keep the idea
Take it with you
- Specific intensity, spectral flux, emissivity, and opacity carry different angular and geometric meanings.
- With depth increasing along propagation, extinction produces a negative intensity derivative.
- Inclination and boundary conventions determine the optical depth relevant to a ray.
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