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Molecular line cubes: space, velocity and hidden depth

Molecular line cubes · L2

When gas obscures its own line

Sandbox Physics ·

Adding molecules can make a line harder to see through. The extra gas emits, but also obscures radiation behind it. A thick line peak approaches a temperature-dependent ceiling even while the molecular column continues to rise.

Follow gain and loss through a layer

Let the path coordinate ss increase toward the observer. A short segment adds emissivity jνj_\nu and removes a fraction of the intensity IνI_\nu through absorption coefficient αν\alpha_\nu, measured in inverse metres. Its path integral is dimensionless optical depth.

Writing Sν=jν/ανS_\nu=j_\nu/\alpha_\nu defines the source function. For a constant source, change variables to optical depth, multiply the transfer equation by its integrating factor and integrate. The result separates attenuated incident light from the layer’s own surviving emission.

For several layers, feed each output into the next layer in physical order. A photon emitted in the front layer crosses less intervening material than one emitted behind it. When temperatures differ, swapping the layers generally changes the spectrum.

dIνds=jν−ανIν,dτν=ανds,Sν=jναν\frac{dI_\nu}{ds}=j_\nu-\alpha_\nu I_\nu,\quad d\tau_\nu=\alpha_\nu ds,\quad S_\nu=\frac{j_\nu}{\alpha_\nu}
Optical depth accumulates along the direction of propagation.
Iν,out=Iν,ine−τν+Sν(1−e−τν)I_{\nu,\rm out}=I_{\nu,\rm in}e^{-\tau_\nu}+S_\nu(1-e^{-\tau_\nu})
This is the exact constant-source layer solution. Distinct sources require ordered recursion.

Reading: Mangum & Shirley, How to Calculate Molecular Column Density, arXiv 1501.01703v4 (2017)

Kelvin on a brightness axis has a definition

LTE makes the source function equal to the Planck function. Radio data often express intensity in RJ-equivalent kelvin: a linear rescaling of intensity. That choice of units does not require replacing the Planck spectrum by its low-frequency approximation.

The function Jν(T)J_\nu(T) writes the full Planck intensity in those units. With a 2.725-kelvin incident background, the saved line excess follows the expression below. At 345 gigahertz, hν0/k=16.557 Kh\nu_0/k=16.557\,\mathrm K, so cold-gas calculations need the exponential factor.

For a complete numerical example, take gas at 20 kelvin. Then Jν(T)=12.8507 KJ_\nu(T)=12.8507\,\mathrm K and the background term is 0.03812 K0.03812\,\mathrm K. A line-centre optical depth of one gives 8.099 K8.099\,\mathrm K of excess brightness. At very large optical depth the limit is 12.8126 K12.8126\,\mathrm K. None of these brightness values equals the supplied 20-kelvin thermodynamic temperature.

Jν(T)=hν0/kehν0/(kT)−1,ΔTRJ=[Jν(T)−Jν(Tbg)](1−e−τν)J_\nu(T)=\frac{h\nu_0/k}{e^{h\nu_0/(kT)}-1},\qquad \Delta T_{\rm RJ}=[J_\nu(T)-J_\nu(T_{\rm bg})](1-e^{-\tau_\nu})
RJ-equivalent brightness is the unit convention; the source on the right retains the full Planck law.

Reading: Mangum & Shirley, How to Calculate Molecular Column Density, arXiv 1501.01703v4 (2017) · Sandbox Physics independent molecular-line science producer and reader package

Saturation changes width as well as height

At small optical depth, expanding the exponential makes brightness nearly proportional to column. At large optical depth the centre is already close to the source function. More column mainly makes the wings opaque over a wider velocity interval.

The figure holds temperature and microscopic dispersion fixed. Its increasing optical depths produce a flatter, broader line even though the molecular velocity distribution is unchanged. A measured half-maximum width can therefore include a transfer effect.

A retained offline stress case increases an already thick column by a factor of ten. Its peak ratio is nearly one while its integrated-brightness ratio is about 1.178. That case lies outside the main Lab column range. It illustrates different saturation of peak and area, without turning either into a unique real-column estimate.

Uniform-slab profiles with increasing central optical depth flatten and broaden at fixed temperature and velocity dispersion.
Synthetic analytic uniform layers, with the full Planck source. The independent calculation plots recession velocity against RJ-equivalent line excess.

Reading: Sandbox Physics independent molecular-line science producer and reader package · Mangum & Shirley, How to Calculate Molecular Column Density, arXiv 1501.01703v4 (2017)

A static foreground can make two peaks

Place a cooler layer in front of a warmer emitting layer and give both zero mean velocity. The cool foreground absorbs most strongly near line centre. The centre can fall below the neighbouring wings, leaving two apparent peaks.

There are no oppositely moving parcels in this example. Assigning every double peak to two velocity components would miss a possible transfer explanation. Lines with different opacity or excitation conditions can help test that interpretation.

The interactive branch is isothermal rigid rotation. On each sightline, its temperature and mean recession speed are constant with depth. Every segment therefore shares the source function and normalized profile. Integrating density into a column and applying one slab solution is exactly equivalent to ordered transfer under those conditions.

Temperature gradients, inflow and a distinct cool foreground break that equivalence. The independent package keeps general sightline calculations for such offline examples. The fast branch is a mathematically justified special case; it does not solve arbitrary cloud excitation or dynamical evolution.

Reading: Sandbox Physics independent molecular-line science producer and reader package · Mangum & Shirley, How to Calculate Molecular Column Density, arXiv 1501.01703v4 (2017)

Exercises and explanations

  1. Which optical-depth change comes closer to doubling the peak: 0.01→0.020.01\to0.02 or 10→2010\to20?

    Read the explanation

    The first. Its exponential is close to a linear term. Both thick cases already approach the same source function, so additional column mainly affects the wings.

  2. Why not add the optical depths of a warm and a cool layer and use an arbitrary average temperature?

    Read the explanation

    Their source functions differ. Front-layer emission and rear-layer emission encounter different subsequent absorption. An average discards these ordered weights. Combining layers preserves the solution only when the required shared-source conditions hold.

Try the prediction

Find where the peak loses column sensitivity

  1. Keep rotation and inclination fixed while changing from the transparent to the high-column case. Compare peak, area and wings separately.
  2. For the cold preset, predict the thick-line ceiling from the Planck expression. Then read a central spectrum and account for finite-channel and beam averaging.
  3. Run the offline two-layer, zero-mean-velocity example. Swap foreground and background and explain the changed central dip.

Sources and calculation scope

  1. Mangum & Shirley, How to Calculate Molecular Column Density, arXiv 1501.01703v4 (2017)

    Fixed v4: transfer and brightness equations 13–32, partition section 7, approximations 11–14 and appendices A–C. The printed Gaussian-width power in A4/A5 is checked against the normalized profile; no blanket non-LTE solution is inherited.

  2. Sandbox Physics independent molecular-line science producer and reader package

    Own synthetic input, independent frequency-space/continuous-ray references and retained resolution failures. The native reader obtains exact source files locally; no observed image, native table or app/Node Engine is bundled.

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