---
title: "Put absorption and emission in one equation"
description: "Derive the uniform LTE slab solution, check transparent and isothermal limits, and connect optical depth to the curve of growth."
canonical_url: "https://sandboxphysics.com/stories/stellar-spectroscopy/absorption-emission-and-growth/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/stellar-spectroscopy/absorption-emission-and-growth/"
last_modified: "2026-10-28"
---

[Stellar spectra from dark lines to abundance models](https://sandboxphysics.com/stories/stellar-spectroscopy/)

AC03 · L2

# Put absorption and emission in one equation

Sandbox Physics · 2026-10-28

Gas weakens the incoming beam and adds its own radiation along that same direction. Adding both contributions explains why an absorption line saturates and why the same transition can instead appear bright.

## Before you begin

Read the preceding definitions of intensity and wavelength. A derivative describes the change along a short ray segment; an integral adds those segments. We use only exponential differentiation and integration.

## Choose the ray direction first

Let distance $s$ increase from the background toward the observer, from zero to $L$. The absorption coefficient $\alpha_\lambda$ specifies attenuation per length. The emission coefficient $j_\lambda$ specifies intensity added per length. The net change is emission minus absorption.

Define optical depth by $d\tau_\lambda=\alpha_\lambda ds$. It is dimensionless and increases toward the exit in our convention. The source function $S_\lambda=j_\lambda/\alpha_\lambda$ has intensity units. For a uniform LTE layer it equals $B_\lambda(T)$. LTE ties local atomic populations to local temperature; it does not require an entire star to be isothermal.

$$
\frac{dI_\lambda}{ds}=-\alpha_\lambda I_\lambda+j_\lambda,\qquad \frac{dI_\lambda}{d\tau_\lambda}+I_\lambda=S_\lambda
$$

An optical depth measured inward from the surface would use a different sign convention. Define it before comparing equations.

Reading: Krumholz, ASTR 4003/8003 (2019), Class 1, Radiative transfer I · Condon & Ransom, Essential Radio Astronomy, Appendix E

## Integrate a uniform source

Multiply by $e^\tau$. The left side becomes the derivative of $I_\lambda e^\tau$. Integrate from entrance to exit, pull the constant source function outside the integral, then divide by the integrating factor.

The first resulting term is the attenuated background. The second is radiation supplied by the layer. The outgoing intensity lies between the incident intensity and the source function, approaching the latter as the layer becomes opaque.

$$
\frac{d}{d\tau}(I_\lambda e^{\tau})=S_\lambda e^{\tau}\quad\Longrightarrow\quad I_{\lambda,\rm out}=I_{\lambda,c}e^{-\tau_\lambda}+S_\lambda(1-e^{-\tau_\lambda})
$$

This is an exact solution for a depth-independent source. A complete atmosphere requires temperature and population structure as well.

Reading: Krumholz, ASTR 4003/8003 (2019), Class 1, Radiative transfer I

## Check limits before calculating a spectrum

As $\tau\to0$, the incident background passes through. As $\tau\to\infty$, the output approaches the source function. In the isothermal case $S_\lambda=I_{\lambda,c}$, every optical depth gives the unchanged continuum. This is a particularly useful independent check.

For source-to-background ratio $r=0.2$, optical depths $0.1$, $1$ and $10$ give normalized intensities $0.924$, $0.494$ and $0.200$. Initial attenuation is nearly linear. The opaque line approaches the layer’s twenty-percent brightness.

$$
\frac{I_{\lambda,\rm out}}{I_{\lambda,c}}=r+(1-r)e^{-\tau_\lambda},\qquad r=\frac{B_\lambda(T)}{I_{\lambda,c}}
$$

The thin approximation replaces the exponential by its first-order expansion. Keep the exponential at finite optical depth.

## Specify a consistent artificial atom

Define four levels before calculating the two line opacities. The ground level has zero energy. A common absorbing lower level lies $2\,\mathrm{eV}$ above it. Each line reaches a different upper level, separated from that lower level by its photon energy $hc/\lambda_{0,j}$, about $2.48\,\mathrm{eV}$. All statistical weights are one. The lower level’s excitation energy and the absorbed photon energy are distinct.

Add the four Boltzmann weights to form the partition function, the total weight used to normalize populations. The lower fraction is $p_l(T)$. The ion stage and total ion column remain fixed, while temperature redistributes the four populations. A real atom needs its actual level structure and transition data.

An upper-state atom can also undergo stimulated emission, adding light in the incoming direction. In LTE the upper-to-lower population ratio supplies an exponential, so net absorption includes the factor $f_j(T)$. Including this factor with the lower population makes the opacity consistent with the adopted Planck source function.

Use $T_0=4500\,\mathrm K$ as reference, $N/N_0$ for the total-ion column ratio and strengths $s_1=1$ and $s_2=0.42$. The Doppler parameter is $b(T)$. More net absorbers increase integrated opacity; spreading it over a wider velocity distribution lowers the center. These explicit ratios reproduce both reference and heated spectra.

$$
E_0=0,\quad E_l=2\,\mathrm{eV},\quad E_{u,j}=E_l+hc/\lambda_{0,j},\quad g_i=1
$$

The two vacuum rest wavelengths are five hundred and five hundred point one two nanometres. The atom and ion stage are prescribed.

$$
p_l(T)=\frac{e^{-E_l/(kT)}}{\sum_{i=0,l,u_1,u_2}g_i e^{-E_i/(kT)}},\qquad f_j(T)=1-e^{-hc/(\lambda_{0,j}kT)}
$$

p is a dimensionless population fraction; f removes stimulated emission from the net absorption.

$$
\tau_{0,j}=3s_j\frac{N}{N_0}\frac{p_l(T)}{p_l(T_0)}\frac{f_j(T)}{f_j(T_0)}\frac{b(T_0)}{b(T)}
$$

At reference temperature and column the center optical depths are three and one point two six. Every factor is dimensionless.

$$
\tau_\lambda=\sum_{j=1}^2\tau_{0,j}\exp\left[-\left(\frac{\lambda-\lambda_{c,j}}{\Delta\lambda_{D,j}}\right)^2\right]
$$

Add the two Gaussian optical depths before applying the single finite-depth transfer solution.

Reading: Condon & Ransom, Essential Radio Astronomy, chapter 7.3 · Condon & Ransom, Essential Radio Astronomy, Appendix E

## Integrate the missing light

For small optical depth, expand the exponential to first order and integrate the Gaussian. Equivalent width is proportional to $\tau_0\Delta\lambda_D$. At fixed temperature and velocity distribution, doubling column approximately doubles that area.

Once the center saturates, its deficit approaches $1-r$. Extra opacity mainly widens the opaque part, so equivalent width grows more slowly. This relation is the curve of growth. The model omits pressure-damped wings and therefore does not predict the damping branch of very strong real lines.

$$
W_\lambda=\int(1-I_\lambda/I_{\lambda,c})d\lambda\simeq(1-r)\tau_0\sqrt{\pi}\,\Delta\lambda_D\quad(\tau_0\ll1)
$$

The window must cover the full line and the source ratio is approximately constant across its narrow width. Emission gives negative equivalent width in this convention.

Saved default prediction for the teaching slab

Uniform layer, fixed ion stage and artificial transitions; no random noise. The saved calculation uses 721 wavelengths and a normalized Gaussian instrument response.

| Layer temperature | $4500\,\mathrm K$ |
| --- | --- |
| Column / reference | $N/N_0=1$ |
| Receding velocity | $12\,\mathrm{km\,s^{-1}}$ |
| Resolving power | $R=60000$ |
| Strong-line center | $500.0200\,\mathrm{nm}$ |
| Total equivalent width | $11.952\,\mathrm{pm}$ |
| Layer / background source ratio | $0.201$ |

## [Find where the growth curve bends](https://sandboxphysics.com/labs/stellar-absorption-lines/)

1. Hold the layer cool and start with a small column. Predict what doubling it will do.
2. Increase column and compare center depth with wing width.
3. Set the layer to six thousand kelvin and explain why the entire growth curve reaches zero.

## Exercises with explanations

1. What happens at very large optical depth if the source function is twice the background?

   Read the explanation

   The output approaches twice the background, producing emission. Greater opacity brings the ray closer to the brighter source function.

2. How can thin lines test a growth-curve calculation?

   Read the explanation

   Reduce the column and check that $W_\lambda/N$ approaches a constant agreeing with the Gaussian analytic integral. Halve the wavelength spacing to check convergence as well.

[Previous: Separate line position from line width](https://sandboxphysics.com/stories/stellar-spectroscopy/wavelength-motion-and-width/) [Next: From emergent intensity to a fitted spectrum](https://sandboxphysics.com/stories/stellar-spectroscopy/from-intensity-to-a-fitted-spectrum/)

## Sources and calculation scope

1. [Condon & Ransom, Essential Radio Astronomy, chapter 7.3](https://www.cv.nrao.edu/~sransom/web/Ch7.html)

   Equations 7.38–7.64: transition energy, Einstein detailed balance, stimulated emission as negative absorption, LTE populations and Planck source. Used to define the consistent artificial finite-level atom.

2. [Krumholz, ASTR 4003/8003 (2019), Class 1, Radiative transfer I](https://www.mso.anu.edu.au/~krumholz/teaching/astr8003_s1_2019/notes1.pdf)

   Equations 24–28: integrating factor, formal solution and constant-source slab. Independently derived and checked by numerical transport here.

3. [Condon & Ransom, Essential Radio Astronomy, Appendix E](https://www.cv.nrao.edu/~sransom/web/A5.html)

   Transfer equation 2.27, LTE Kirchhoff relation 2.30 and Planck law 2.86; used at optical wavelengths with the full Planck function.