Cosmology · Cosmic dark ages, first light and reionisation

Absorption signatures and reionisation evidence

Use a supplied transmission law to understand saturation, finite optical-depth bounds and the difference between mean and effective opacity.

Preparation: ratios, powers of ten and ordinary arithmetic. Physical quantities and notation are introduced as they are needed.

Back to the chapter

After reading, you can

Explain why a saturated resonant absorption feature alone does not establish a fully neutral intergalactic medium.
Convert a supplied transmission upper bound into an optical-depth lower bound under stated assumptions.
Distinguish averaging transmission from averaging optical depth.

A spectrum constrains gas through a model of transmission

A background source can illuminate intervening gas, allowing absorption in its spectrum to probe conditions along the light path. Interpretation depends on the source continuum, atomic transition and spatial or velocity structure sampled.

Hydrogen Lyman-alpha resonance is distinct from photoionization out of the ground state. Resonant absorption or scattering can remove photons from a measured beam without each such event ionizing a hydrogen atom.

Resonant Lyman-alpha opacity can become large while gas is still mostly ionized. A very dark absorption trough therefore does not, by itself, establish that the intergalactic gas is fully neutral or determine a unique global neutral fraction.

The numerical example here supplies optical depths and a simple removed-beam transfer law. It is not a quasar spectrum, a measurement of a reionization date or a conversion of opacity into an empirical neutral fraction.

An exponential law has explicit observational conditions

For each specified resolved spectral or ray element, assume a known intrinsic continuum intensity I_0 and no line emission or scattered light entering the aperture. With no additive foreground or noise in this construction, transmitted intensity obeys F = I/I_0 = exp(−τ).

Both the optical depth τ and continuum-normalized transmission F are dimensionless. Positive finite τ gives positive F; arbitrarily low transmission can be produced by large finite opacity without requiring exactly zero transmitted intensity.

At τ = 0.1, the transmission is about 0.904837. At τ = 5 it is about 0.00673795, and at τ = 10 only about 0.0000453999 remains in the measured beam under the stated exclusions.

The exponential sensitivity makes large optical depths difficult to distinguish using very small transmissions alone. An actual spectrum additionally needs continuum inference, line profiles, instrumental response and a measurement model.

Supplied resolved removed-beam screen with known continuum and no additive signal or noise; these are mathematical inputs, not spectral measurements
Optical depth τ	Transmission F = exp(−τ)	Meaning within this example
0.1	0.904837	Most of the input beam remains
1	0.367879	Transmission reduced by a factor e
5	0.00673795	Less than one percent remains
10	0.0000453999	Very small but nonzero transmission

A dark-looking element can give a finite lower bound

Now supply the mathematical upper bound F ≤ 0.001 for one resolved element in the same known-continuum screen account. This is a chosen teaching threshold, not a measured sensitivity or a confidence limit from a dataset.

Taking the logarithm gives τ ≥ −ln(0.001) = ln(1000) ≈ 6.907755. The inequality reverses appropriately because transmission decreases as opacity increases.

The bound does not select a unique τ. Optical depths of 7, 10 and many larger finite values satisfy the same transmission condition. Infinite opacity is not required.

Applying such reasoning to actual observations requires the uncertainty and resolution account. A bound on a convolved or averaged flux with an uncertain continuum cannot simply be relabelled as the exact optical depth of every resolved ray element.

Low-opacity regions can dominate the transmitted signal

Consider four equal-weight elements with supplied optical depths 0.1, 0.1, 10 and 10. Their arithmetic mean optical depth is (0.1 + 0.1 + 10 + 10)/4 = 5.05.

Their mean transmitted flux is instead [2exp(−0.1) + 2exp(−10)]/4 ≈ 0.452441. Define effective optical depth by τ_eff = −ln⟨F⟩. It is approximately 0.793097 for these elements, far below the mean τ of 5.05.

The low-opacity elements contribute most of the surviving flux. Taking a logarithm after averaging the transmission is a different operation from averaging the four optical depths first.

These four equal weights are known inputs of the construction, rather than a survey estimator. The example establishes a nonlinear averaging effect; it does not supply a probability distribution of intergalactic densities or an observed effective-opacity statistic.

Each supplied optical depth belongs to a resolved spectral or ray element with known intrinsic continuum I_0. The removed-beam screen excludes line emission or scattered light entering the aperture, additive foreground and noise. Its normalized transmission is F = I/I_0 = exp(−τ). Both quantities are dimensionless. Panel A has finite optical-depth abscissa zero to ten and a logarithmic transmission ordinate labelled 1, 0.1, 0.01, 0.001, 0.0001, 0.00001; this logarithmic axis has no zero. The straight-looking blue line is exponential transmission on a log scale. Circles mark τ = 0.1, 1, 5, 10, giving approximately 0.904837, 0.367879, 0.00673795, 0.0000453999. All are positive. An orange dashed guide is the supplied teaching threshold F = 0.001, not a measured sensitivity or confidence bound. A green dotted vertical guide and diamond intersection give τ = ln(1000), approximately 6.907755. F no larger than 0.001 implies τ no smaller than this finite value; larger finite opacities remain possible. Panel B is separate and uses a linear zero-baseline transmission scale for four equally weighted elements with τ = 0.1, 0.1, 10, 10. Their square endpoints show two large and two small positive transmissions. The green dotted mean-flux guide is approximately 0.452441. Their arithmetic mean optical depth is 5.05, whereas effective depth −ln(mean F) is approximately 0.793097. The bar heights near zero are small positive numbers, not zeroed flux. Averaging transmission and then taking a logarithm differs from averaging opacity. These chosen elements are not a survey estimator or fabricated quasar spectrum. Lyman-alpha bound-bound resonance differs from ground-state photoionization, and a saturated trough alone does not determine full neutrality. No empirical neutral fraction, measured flux threshold, current observation or reionization date is inferred. Original Ark teaching model.

A neutral-fraction inference needs more than a trough

Absorption signatures can constrain reionization models when their relation to density, ionization, temperature, velocity and radiation fields is specified. The measurements sample sightlines and spectral regions, rather than directly counting every neutral atom in a cosmic volume.

Saturation limits the information carried by a nearly black region. Remaining transmission, spectral structure and additional probes can be useful, but their interpretation must retain the relevant modelling assumptions and uncertainties.

A global neutral fraction is a defined spatial or density-weighted quantity. It should not be equated with an effective line-of-sight opacity or read uniquely from a simple exponential screen without the intervening physical model.

The transferable lesson is to state what is measured, how it is averaged and what physical conversion is assumed. That separates an observed absorption pattern, a conditional opacity constraint and an inferred ionization history.

Worked reasoning

Convert the supplied flux threshold

Does F ≤ 0.001 require infinite τ in the stated resolved screen?

Use the supplied relation F = exp(−τ).
Take logarithms to obtain τ ≥ −ln(0.001) = ln(1000).
Evaluate the finite lower bound, approximately 6.907755, and retain the inequality rather than assigning one exact τ.

No. It gives τ ≥ 6.907755 in this screen model, with no unique larger optical depth selected.

Try the reasoning

Does a saturated Lyman-alpha trough alone prove that all the intervening gas is neutral?
Show the answer and reasoning
No.
Resonant opacity can be large in mostly ionized gas, and a neutral-fraction inference needs the relevant physical and observation model.
Is τ_eff equal to 5.05 for the four supplied elements?
Show the answer and reasoning
No.
The mean transmission is about 0.452441, so its negative logarithm is approximately 0.793097; the exponential average is nonlinear.
Does finite τ = 10 make exp(−τ) exactly zero?
Show the answer and reasoning
No.
The transmission is about 0.0000453999, positive but small under the stated law.

What to carry forward

Saturated resonant absorption alone does not uniquely determine a global neutral fraction.
A supplied small-flux upper bound can give a finite optical-depth lower bound under explicit conditions.
Mean optical depth and the negative logarithm of mean transmission describe different averages.

References

Sources supporting the explanations, physical distinctions and examples in this lesson.