Cosmology · The cosmic web, voids and intergalactic matter
Absorption forests and line-of-sight sampling
Follow supplied optical-depth components through transmitted flux, instrumental convolution and a deliberately wrong continuum normalization.
Preparation: ratios, powers of ten and ordinary arithmetic. Physical quantities and notation are introduced as they are needed.
Back to the chapterAfter reading, you can
- Explain how a background spectrum samples intervening gas along a line of sight.
- Distinguish adding optical depths from smoothing transmitted flux.
- Keep continuum, coordinate-area and physical equivalent-width assumptions explicit.
Many absorption features need a physical mapping
A background source can illuminate intervening gas whose resonant absorption appears at different observed wavelengths. A collection of such features is often called an absorption forest.
Relating a feature to gas location and composition requires a transition identification, redshift and velocity model, ionization conditions and an instrument response. A fitted component need not represent one distinct uniform cloud.
Construct a synthetic example with a background source assigned redshift 2.5 and features near redshift 2. These are chosen labels, not an actual quasar spectrum or observational redshift record.
The example supplies optical-depth profiles directly. It infers no gas density, neutral column, temperature or physical Doppler width from them.
Optical depths add before transmission is calculated
Let λ_α be the positive rest wavelength of an identified transition and choose λ_ref = 3λ_α. Define u = ln(λ/λ_ref)/w with w = 0.001, so λ/λ_ref = exp(wu). All wavelengths remain positive.
Ignoring peculiar velocities for these labels gives 1+z = 3exp(wu). Centres u = −1, 0 and 1.2 then correspond to redshifts about 1.9970015, 2 and 2.0036022, all in front of the supplied source.
Supply τ(u) = Σ_j τ₀,j exp[−((u−u_j)/b_j)²] with triples (u_j,b_j,τ₀,j) equal to (−1,0.35,0.8), (0,0.5,1.5) and (1.2,0.3,0.6). The widths b_j are dimensionless in u, not velocities or measured thermal widths.
With unit continuum, complete coverage and no added line emission into the beam, transmitted flux is F = exp(−τ). The analytic profiles extend over real u for the mathematical integrals; that extension is not an absorber catalogue beyond the source. Display only −3 ≤ u ≤ 3.
A response averages light rather than optical depth
Choose a normalized Gaussian response R_σ(t) = exp[−t²/(2σ²)]/(√(2π)σ) with σ = 0.25 in u units. Its convolved normalized flux is F_R = R_σ * F.
Exponentiation is nonlinear, so convolving optical depth and then exponentiating generally gives a different result. At u = 0, F is about 0.223079, F_R about 0.303943 and exp[−R_σ*τ] about 0.290799.
A response can raise flux near one trough and lower it near an adjacent high point. At u = −0.5, for example, the supplied intrinsic flux is about 0.519049 while the convolved value is about 0.449535.
The table contains exact-profile numerical evaluations, not noisy pixels, a fitted spectrograph or a likelihood. Gaussian optical-depth components also omit natural damping wings and a full gas-dynamical line calculation.
| u | Total τ | Intrinsic F | Correct R_σ * F | Incorrect exp(−R_σ * τ) |
|---|---|---|---|---|
| −3 | ≈0 | 1.000000 | 1.000000 | 1.000000 |
| −1 | 0.827473 | 0.437152 | 0.533930 | 0.523132 |
| −0.5 | 0.655757 | 0.519049 | 0.449535 | 0.434406 |
| 0 | 1.500228 | 0.223079 | 0.303943 | 0.290799 |
| 0.5 | 0.554411 | 0.574410 | 0.550052 | 0.512419 |
| 1.2 | 0.604727 | 0.546224 | 0.670074 | 0.660656 |
| 2 | 0.000489762 | 0.999510 | 0.982115 | 0.980384 |
| 3 | ≈0 | 1.000000 | 1.000000 | 1.000000 |
A conserved log-coordinate area is not a wavelength equivalent width
Define A_u = ∫(1−F)du over the complete real coordinate line. A normalized translation-invariant response preserves this area because the response redistributes the integrable flux deficit without changing its total.
For these supplied profiles, both full intrinsic and full convolved A_u are about 1.44711072. The finite displayed response window has area about 1.4471107103; its excluded tails are small but not erased or renormalized.
Physical wavelength equivalent width instead uses W_λ = λ_ref w∫exp(wu)(1−F)du, since dλ = λ_ref wexp(wu)du. A_u is dimensionless, while W_λ has wavelength units.
Conservation of A_u under this response in log wavelength does not assert conservation of the differently weighted wavelength integral or of every finite-window area. The integration variable and boundaries must travel with an area claim.
A continuum error can imitate a broad deficit
Now deliberately divide the convolved flux by a continuum estimate 1.05 instead of the true value one. Far from the features, normalized flux tends to 1/1.05 ≈ 0.952381 instead of one.
Across the finite window of width six in u, the apparent deficit area becomes A_app = 6(1−1/1.05) + A_R,window/1.05 ≈ 1.663915. The five percent offset is a chosen sensitivity test, not a measured continuum uncertainty.
Actual absorption inference must handle continuum placement, velocity blending, thermal structure, instrumental response and noise together. A deeper feature does not by itself identify a unique separate cloud or total baryon column.
This synthetic sequence separates optical-depth addition, nonlinear transmission, instrumental averaging and normalization. Keeping those stages explicit makes line-of-sight evidence useful without mistaking a convenient profile model for a direct three-dimensional gas map.
Worked reasoning
Calculate the effect of a supplied continuum offset
What apparent area results on −3 ≤ u ≤ 3 when F_R is divided by 1.05?
- The true convolved finite-window area is A_R,window ≈ 1.4471107103.
- Integrate 1−F_R/1.05: the constant offset contributes 6(1−1/1.05), and the true deficit contributes A_R,window/1.05.
- The result is about 1.66391496, rather than 1.44711071.
The chosen high continuum creates an extra finite-window deficit; it does not reveal additional gas in this construction.
Try the reasoning
Should the supplied instrument response act on F or on τ before exponentiation?
Show the answer and reasoning
On transmitted flux F.
The response averages arriving light; nonlinear exponentiation does not commute with that averaging.
Is the conserved A_u the usual wavelength equivalent width?
Show the answer and reasoning
No.
Wavelength integration includes λ_ref wexp(wu) and has wavelength units; A_u is a dimensionless log-coordinate area.
Do three supplied optical-depth components prove three distinct physical clouds?
Show the answer and reasoning
No.
Components describe a chosen profile decomposition; gas structure, velocity and ionization require further inference.
What to carry forward
- Absorption forests record line-of-sight transmission under transition, velocity and ionization assumptions.
- Optical depths add, transmission is exponential and instrumental convolution acts on flux.
- Continuum choice, coordinate Jacobians and finite windows matter before spectral areas become physical claims.
References
Sources supporting the explanations, physical distinctions and examples in this lesson.