Absorption & Emission Lines

Set a source temperature, a gas temperature and an optical depth to see the same spectral line go dark, go bright, or disappear.

Worked example — Absorption & Emission Lines

Continuum radiance	2.24 × 10¹³ W·m⁻²·sr⁻¹·m⁻¹
Gas's own radiance	7.55 × 10¹² W·m⁻²·sr⁻¹·m⁻¹
Transmitted through the gas	8.25 × 10¹² W·m⁻²·sr⁻¹·m⁻¹
Emitted by the gas	4.77 × 10¹² W·m⁻²·sr⁻¹·m⁻¹
Observed line radiance	1.30 × 10¹³ W·m⁻²·sr⁻¹·m⁻¹
Radiance difference	-1.49 × 10¹³ W·m⁻²·sr⁻¹·m⁻¹
Contrast	-9.40 × 10¹² W·m⁻²·sr⁻¹·m⁻¹
Contrast, relative to continuum	-41.9%

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Absorption & Emission Lines

A spectral line is one wavelength where a gas is far better at absorbing and emitting than at its neighbours. Put that gas in front of something and two things happen at once: it removes light coming from behind, by a factor e^−τ, and it adds light of its own. What you see is the sum, I = I_sourcee^−τ + B(T_gas)(1 − e^−τ) — the same equation whether the line comes out dark or bright.

What each symbol means

τ: optical depth: how much of the light behind survives the gas, as e to the minus τ
B(T): the brightness a body at temperature T emits at this wavelength

Assumes: One line, one slab of gas at a single temperature, and local thermodynamic equilibrium — that the gas emits as a body at its own temperature. A real stellar atmosphere has a temperature that changes with depth, which is what gives real lines their shape rather than a single depth.

Dark or bright is decided by what sits behind 🖖

The page opens on the Sun: a 5772 K source, a cloud of hydrogen at 4500 K in front of it, optical depth 1, H-alpha at 656.28 nm. The percentage readout says -41.9%, and the line is dark. Drag the gas slider up to 8000 K. The readout says +127.0%, and the same line is now brighter than the continuum on either side of it. Nothing about the hydrogen changed. Fraunhofer mapped the Sun's dark lines in 1814 and could not say what they were; Kirchhoff could, in 1859, and the rows below are his answer.

The Sun's dark lines say this much and no more. The gas drawing them sits near 4500 K, hot enough to melt tungsten, the most stubborn metal we have, and it reads dark because the photosphere beneath it is hotter. Put the gas back at 4500 K and pull the source slider down to 4000 K instead: +53.4%. The cloud you were calling an absorber is glowing, and nobody touched it.

At one temperature the line vanishes exactly 🖖

There is one gas temperature at which the line disappears completely, and it is the source's own. Drag the gas slider up to meet it at 5772 K, or click the vanishing-point sample. The contrast row reads 0, and the relative-contrast row under it reads 0.0%. Not a residue in the fifteenth decimal, not a value too small to plot: zero.

It stays zero wherever else you go. Push τ to 5, thick enough that the cloud replaces almost all the light behind it, then walk the wavelength across the whole range from 100 nm to 2000 nm. Nothing appears.

Kirchhoff's law of thermal radiation is why. A gas in local thermodynamic equilibrium, which is the assumption this model rests on, emits at each wavelength in exact proportion to how strongly it absorbs there, and one temperature governs both. Match that temperature to the source and the cloud returns precisely what it takes. It is absorbing furiously and emitting furiously, and the spectrum reaching you holds no evidence it is there.

Optical depth sets how deep, never which way 🖖

Put the gas back at 4500 K under the 5772 K source and take the τ slider for a walk. At its floor, 0.01, the percentage readout is -0.7%. At its ceiling, 5, it is -65.9%, the line centre having lost two-thirds of the continuum's brightness. Every setting between them is negative. The line deepens the whole way and never crosses zero.

Optical depth enters the contrast only as the factor (1 − e^(−τ)), the fraction of the source's light the cloud intercepts and replaces with its own. That fraction climbs from nearly nothing towards 1 and can never go negative, so it scales the contrast and cannot touch its sign. The temperatures own the sign.

Watch the number rather than the picture while you drag. The spectrum panel rescales to whatever line it is drawing, so the shallowest notch and the deepest one are drawn exactly the same size.

Problem solved in full

How deep the line goes, and how cold the gas has to be 7 steps
Light leaves a photosphere at 5772 K and passes through a layer of that star's own atmosphere at 4500 K. The layer is one optical depth thick at H-alpha, 656.28 nm in air. Work out how far below the continuum the line core sits. Then read the same physics backwards: a second star with the same 5772 K surface and the same optical depth shows its H-alpha core 25.0% below its continuum. How hot is the gas doing the absorbing?

Open this problem in the tool
1. Planck's law wants an exponent before it wants anything else, and at a fixed wavelength that exponent is one constant divided by a temperature. At 656.28 nm, hc/λk works out to 21923.2 K, so the source's exponent is 21923.2/5772 = 3.7982. Put it in and the continuum radiance is 2.24 × 10¹³ W·m⁻²·sr⁻¹·m⁻¹, which is what the Continuum radiance row prints.
  
  $I_{0} = \frac{2 h c ^{2}}{λ ^{5} ( e ^{h c / λk T_{0}} - 1 )} = \frac{1.19104 \times 1 0 ^{- 16}}{( 6.5628 \times 1 0 ^{- 7} ) ^{5} ( e ^{3.7982} - 1 )} = 2.24 \times 1 0^{13}$
2. The gas obeys the same law at its own temperature. 21923.2/4500 = 4.87183, a bigger exponent, so a bigger denominator and less radiance: 7.55 × 10¹². The Gas's own radiance row agrees. The line is already decided here, before optical depth enters. A cooler surface radiates less at every wavelength, so the core can only end up darker than the continuum.
  
  $B = \frac{2 h c ^{2}}{λ ^{5} ( e ^{h c / λk T} - 1 )} = \frac{1.19104 \times 1 0 ^{- 16}}{( 6.5628 \times 1 0 ^{- 7} ) ^{5} ( e ^{4.87183} - 1 )} = 7.55 \times 1 0^{12}$
3. A slab of optical depth τ does two things at once. It passes a fraction e^−τ of the light behind it, which at τ = 1 is 0.36788, and it replaces the rest with its own emission, 0.63212 of B. Transmitted 8.25 × 10¹² plus emitted 4.77 × 10¹² gives 1.30 × 10¹³ in the Observed line radiance row. Those two numbers are the whole of radiative transfer through a slab.
  
  $I = I_{0} e^{- τ} + B (1 - e^{- τ}) = 8.25074 \times 1 0^{12} + 4.77324 \times 1 0^{12} = 1.30 \times 1 0^{13}$
4. Contrast is what the observer sees minus what they would have seen with no gas there. Write it the way the Contrast row does, as (B − I₀)(1 − e^−τ), because the subtraction then happens before the scaling and two equal temperatures cancel to a clean zero. −1.48767 × 10¹³ times 0.63212 is −9.40 × 10¹².
  
  $Δ I = (B - I_{0}) (1 - e^{- τ}) = - 1.48767 \times 1 0^{13} \cdot 0.63212 = - 9.40 \times 1 0^{12}$
5. Divide by the continuum and the depth is 41.9%. The row beneath, Line vanishes at, reads 5772 K, and that figure is exact rather than nearly right: at the source's own temperature B − I₀ is a number minus itself, so the contrast is zero at every wavelength and every optical depth. A thick cloud at the right temperature is still invisible, which is why depth and direction are two separate questions.
  
  $\frac{Δ I}{I _{0}} = \frac{- 9.40386 \times 1 0 ^{12}}{2.24278 \times 1 0 ^{13}} = - 0.419, Δ I = 0 ⟺ T = T_{0} = 5772 K$
6. Now the measurement you actually get from a telescope, which is a depth and nothing else. Call the fractional depth d, so d = −ΔI/I₀ = (1 − B/I₀)(1 − e^−τ), and solve for the gas's radiance instead of for the depth: B = I₀(1 − d/(1 − e^−τ)). With d = 0.250 and τ = 1, the bracket is 1 − 0.395495 = 0.604505, and B = 1.3558 × 10¹³. The tool will not take this step for you.
  
  $d = (1 - \frac{B}{I _{0}}) (1 - e^{- τ}) \Rightarrow B = I_{0} (1 - \frac{d}{1 - e ^{- τ}}) = 2.24278 \times 1 0^{13} \times 0.604505 = 1.3558 \times 1 0^{13}$
7. One more inversion and the temperature falls out. Turn B = 2hc²/(λ⁵(eˣ − 1)) around into eˣ = 1 + 2hc²/(λ⁵B), which comes to 73.158, so x = ln 73.158 = 4.2926. The exponent was 21923.2/T all along, so T = 21923.2/4.2926 = 5107 K. Set the gas temperature slider there and the Contrast, relative to continuum row reads −25.0%.
  
  $e^{x} = 1 + \frac{2 h c ^{2}}{λ ^{5} B} = 73.158 \Rightarrow x = 4.2926, T = \frac{h c}{λk x} = \frac{21923.2}{4.2926} = 5107 K$
Answer

The 4500 K layer takes the H-alpha core 41.9% below the continuum, at 1.30 × 10¹³ against a continuum of 2.24 × 10¹³. Read the second star backwards and its 25.0% core needs gas at 5107 K, a mere 665 K cooler than the surface behind it. The inversion also says when to stop trusting it. Let the gas fall towards absolute zero and B goes to zero with it, so the deepest core an optical depth of 1 can produce is 1 − e⁻¹, or 63.2% of the continuum. Measure a line 70% deep and no gas temperature fits it; the cloud has to be thicker instead.

$T = \frac{h c / λk}{ln ( 1 + \frac{2 h c ^{2}}{λ ^{5} I _{0} ( 1 - d / ( 1 - e ^{- τ} ) )} )} = 5107 K$

References (1)

Who drew the Sun's dark lines, and who explained them: J. Fraunhofer, "Bestimmung des Brechungs- und des Farbenzerstreuungs-Vermögens verschiedener Glasarten." Denkschriften der Königlichen Akademie der Wissenschaften zu München 5, 1814–15, pp. 193–226. G. Kirchhoff, "Über den Zusammenhang zwischen Emission und Absorption von Licht und Wärme." Monatsberichte der Königlich Preussischen Akademie der Wissenschaften zu Berlin, 1859, pp. 783–787.

By the 42Math team · Reviewed against published sources. Corrections welcome.