advanced29 min studyLesson 882 in Physics

Emission, absorption, and source functions

Solve a uniform emitting slab, separate true absorption from scattering, and connect line coefficients to populations without silently assuming LTE.

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What you’ll understand

  • Interpret the source function as an intensity-valued ratio.
  • Apply the formal slab solution with an explicit incoming boundary.
  • Distinguish local thermal emission, scattering redistribution, and non-LTE line populations.

The source function tells us which intensity an extended uniform medium tends to establish along a ray. It is a ratio of emissivity to extinction, not a temperature by definition. Scattering can make it depend on radiation arriving from other directions, and atomic populations can make it depart from a Planck spectrum. We keep the preceding SI per-frequency and per-solid-angle conventions and measure optical depth along propagation, allowing the boundary solution and its signs to be checked directly.

An emitting slab retains a memory of its boundary

For positive extinction define Sν=jν/ανS_\nu=j_\nu/\alpha_\nu, which has the same units as specific intensity. Dividing the transfer equation by extinction gives a derivative with respect to ray optical depth equal to source function minus intensity. If S is constant, an integrating factor gives an attenuated boundary contribution plus an emitted contribution. The two coefficients sum to one, so a uniform slab produces a weighted interpolation between its incoming intensity and S. This statement concerns a prescribed uniform source function, not every inhomogeneous atmosphere.

In the thin limit the emitted increment is approximately j times path length and the incident beam is only slightly attenuated. In the thick limit the incident contribution is exponentially suppressed and the emerging intensity approaches S. Neither limit identifies S with a blackbody unless the emission and absorption processes justify that identification. A cloud in front of a brighter background can therefore appear in absorption even while emitting photons locally. Whether its line is bright or dark depends on both its source function and the background intensity.

dIνdτν=Sν−Iν,Sν=jναν,Iν,out=Iν,ine−τν+Sν(1−e−τν)\frac{dI_\nu}{d\tau_\nu}=S_\nu-I_\nu,\qquad S_\nu=\frac{j_\nu}{\alpha_\nu},\qquad I_{\nu,\mathrm{out}}=I_{\nu,\mathrm{in}}e^{-\tau_\nu}+S_\nu(1-e^{-\tau_\nu})
This solution assumes a constant source function and positive extinction at the chosen frequency along the ray. I and S have W m to the minus two Hz to the minus one sr to the minus one units, and optical depth is dimensionless.↔ Scroll wide equations horizontally.

Scattering couples the ray to its surroundings

Separate true absorption alpha a from scattering extinction alpha s. For coherent isotropic scattering at the same frequency, the radiation added to a ray is alpha s times the mean intensity J. If the true thermal process has emissivity alpha a times the Planck function B, the total source function is a weighted mean of B and J. The effective thermal weight is ε=αa/(αa+αs)\varepsilon=\alpha_a/(\alpha_a+\alpha_s). This simplified expression assumes isotropic redistribution and no frequency shift; Compton scattering requires a frequency-coupled treatment instead.

A scattering-dominated medium need not establish the gas's thermal spectrum after only one extinction mean free path. Scattering changes directions without necessarily exchanging enough energy to equilibrate photons with matter. Moreover, J is generally determined by the entire transfer problem and its boundaries. Inserting an externally prescribed J into the local formula is a useful algebraic illustration, but does not solve the coupled angular radiation field self-consistently. This nonlocal dependence is why source functions cannot generally be tabulated from local density and temperature alone.

αν=αa+αs,jν=αaBν+αsJν,Sν=εBν+(1−ε)Jν,ε=αaαa+αs\alpha_\nu=\alpha_a+\alpha_s,\qquad j_\nu=\alpha_a B_\nu+\alpha_sJ_\nu,\qquad S_\nu=\varepsilon B_\nu+(1-\varepsilon)J_\nu,\qquad \varepsilon=\frac{\alpha_a}{\alpha_a+\alpha_s}
The expression assumes a thermal true-absorption process plus isotropic coherent scattering. Both alpha coefficients are per length, epsilon is dimensionless, and B, J and S all have intensity units. B is introduced as the Planck function, defined explicitly in the next lesson.↔ Scroll wide equations horizontally.

Line populations determine emission and net absorption

For a narrow isolated transition, normalize its profile ϕν\phi_\nu to unit frequency integral. Spontaneous line emissivity is jν=(hν0/4π)nuAulϕνj_\nu=(h\nu_0/4\pi)n_uA_{ul}\phi_\nu for isotropic emission. Net line absorption includes stimulated emission: αν=(hν0/4π)(nlBlu−nuBul)ϕν\alpha_\nu=(h\nu_0/4\pi)(n_lB_{lu}-n_uB_{ul})\phi_\nu. Here the Einstein B coefficients are defined to multiply profile-averaged mean intensity, not spectral energy density; that choice fixes the four-pi factors and gives Aul/Bul=2hν03/c2A_{ul}/B_{ul}=2h\nu_0^3/c^2. Densities are per cubic metre and the profile has inverse-hertz units.

Using the statistical-weight relation between the two B coefficients gives Sν=(2hν03/c2)/[(gunl)/(glnu)−1]S_\nu=(2h\nu_0^3/c^2)/[(g_un_l)/(g_ln_u)-1] under the common-profile approximation. In LTE the population ratio reduces this to the Planck function, but outside LTE it need not. A population inversion makes net absorption negative and permits amplification. The positive-optical-depth slab interpretations above deliberately exclude that maser regime. Even with positive absorption, a measured source function constrains level populations and radiative transfer together rather than automatically measuring the gas kinetic temperature.

A source function with a prescribed scattering field

At one frequency let αa=1 m−1\alpha_a=1\,\mathrm{m^{-1}}, αs=3 m−1\alpha_s=3\,\mathrm{m^{-1}}, and prescribe Bν=2UB_\nu=2U and Jν=6UJ_\nu=6U, where U=10−9 W m−2 Hz−1 sr−1U=10^{-9}\,\mathrm{W\,m^{-2}\,Hz^{-1}\,sr^{-1}}. For a uniform ray depth ln two and incoming intensity U, find S and the outgoing intensity.

  1. The total extinction is four per metre, so the true-absorption fraction is one quarter and the scattering fraction three quarters. These are interaction weights, not the final fraction of photons escaping.

  2. The source is Sν=(1/4)(2U)+(3/4)(6U)=5US_\nu=(1/4)(2U)+(3/4)(6U)=5U. The prescribed mean intensity is larger than B, so scattering raises S above the local thermal value.

  3. At optical depth ln two the boundary and source weights are each one half. Thus the outgoing intensity is (1/2)U+(1/2)(5U)=3U(1/2)U+(1/2)(5U)=3U.

The result

The prescribed local source is 5U and the outgoing intensity is 3U. This calculation does not determine J self-consistently.

Explore the model

Interactive lab

Interactive lab · probability

Interaction categories have different weights

Reset and set BOTH Success probability p = 0.25 and Trials n = 4; label success as true absorption and predict mean count 1. For a fresh comparison, Reset and set p = 0.75 and n = 4; predict mean count 3. The two count variances are both 0.75.

Interaction categories have different weightsThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

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 forThe exact binomial weights illustrate repeated independent category assignments. The lab computes no scattering directions, mean radiation field, source function, frequency redistribution, or radiation hydrodynamics.

Check your understanding

When does a thick uniform slab necessarily approach the Planck intensity?

Try it yourself

For a positive-absorption line let Q=2hν03/c2=12UQ=2h\nu_0^3/c^2=12U and population-weight ratio (gunl)/(glnu)=3(g_un_l)/(g_ln_u)=3, with U the intensity unit in the worked example. Find the line source function and its thick-slab limit. Can its kinetic temperature be inferred without an LTE assumption?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the population source-function denominator, then distinguish an excitation diagnostic from a kinetic-temperature measurement.

Compare with the solution

The denominator is three minus one, giving Sν=12U/2=6US_\nu=12U/2=6U. A sufficiently thick uniform slab at this frequency tends to six U after its boundary contribution is suppressed. The positive denominator is consistent with positive net absorption. Without LTE, the measured population ratio need not be set by a Boltzmann distribution at the gas kinetic temperature, so this source function alone does not determine that temperature. Collisional rates, pumping, and the surrounding radiation field may also control it.

Take it with you

  • A thick ray approaches its source function, which is not a temperature by definition.
  • Scattering can make the source depend nonlocally on the mean radiation field.
  • Stimulated emission and explicitly defined Einstein coefficients fix line-opacity signs and solid-angle factors.

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