From moving molecules to a line with a width

Count molecules that can make the transition

Begin with an imaginary molecule having just two energy levels. A downward transition releases one photon whose energy is the difference between the levels. Only molecules in the upper level are available for that particular spontaneous decay.

The upper-level fraction $p_u$ depends on temperature and on the degeneracies $g_u,g_l$ : the numbers of states sharing each energy. Local thermodynamic equilibrium, or LTE, assigns the Boltzmann ratio below. This is an excitation assumption. Establishing it in a real cloud would require a separate comparison of collisional and radiative rates.

The Lab uses a specified two-level teaching transition at $345\,\mathrm{GHz}$ , degeneracy ratio three and molecular mass 28 atomic mass units. Real CO has many rotational levels; the native-data tutorial keeps those spectroscopy inputs separate.

E_u-E_l=h\nu_0,\qquad p_u=\frac{g_u e^{-h\nu_0/(kT)}}{g_l+g_u e^{-h\nu_0/(kT)}}

The gap determines photon frequency. Temperature, in kelvin, and the state counts determine the emitting fraction.

Reading: Mangum & Shirley, How to Calculate Molecular Column Density, arXiv 1501.01703v4 (2017) · Leiden Atomic and Molecular Database, fixed CO data file

Let a distribution of velocities broaden the line

A cloud can be stationary while its molecules move thermally. The component of thermal velocity along one axis is Gaussian. Most molecules are near the mean; increasingly few occupy the wings.

Use positive $v$ for recession and $v_0$ for the parcel’s mean recession speed. Normalize the profile to unit area. The number of emitters can then set the total radiation while the profile decides how to distribute it. Consequently $\phi_v$ has units of inverse velocity.

The model also prescribes an unresolved, Gaussian nonthermal dispersion $\sigma_{\rm nt}$ . Convolving independent Gaussians adds their variances. This parameter describes a width; it does not generate a turbulent velocity field or identify a unique turbulent mechanism.

At 20 kelvin, the thermal dispersion of a 28-atomic-mass-unit molecule is $77.1\,\mathrm{m\,s^{-1}}$ . Adding a nonthermal dispersion of $150\,\mathrm{m\,s^{-1}}$ gives $\sigma=168.6\,\mathrm{m\,s^{-1}}$ . Its full width at half maximum is about $397\,\mathrm{m\,s^{-1}}$ . To obtain that width, set the Gaussian to half its peak, take logarithms and subtract the two symmetric roots.

\phi_v(v)=\frac{e^{-(v-v_0)^2/(2\sigma^2)}}{\sqrt{2\pi}\sigma},\quad\int\phi_v\,dv=1,\quad\sigma^2=\frac{kT}{m_{\rm mol}}+\sigma_{\rm nt}^2

Adding all velocities returns unity, so broadening does not create extra photons.

\mathrm{FWHM}=2\sqrt{2\ln2}\,\sigma

The full width between the half-height points is approximately 2.355 standard deviations.

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

Add the transparent sightline

For this step, neglect absorption and set the incident background to zero. Let $A_{ul}$ be the spontaneous transition rate per upper-state molecule and $n_u$ the upper-state number density. Isotropic emission spreads the power over a full sphere of directions.

Specific intensity is power per projected area, solid angle and frequency interval. A small path segment adds $j_\nu ds$ to the specific intensity. Integrating over depth and then over frequency removes the normalized profile, leaving the upper-state column $N_u$ : the number of emitting molecules per projected square metre.

This result explains why a transparent line area can constrain an emitting column. Inferring a total molecular column still requires the excitation fraction. Background absorption and opacity introduce further conditions, which the next chapter adds explicitly.

j_\nu=\frac{h\nu_0 A_{ul}n_u}{4\pi}\phi_\nu,\qquad \int I_\nu\,d\nu=\frac{h\nu_0 A_{ul}N_u}{4\pi}

The frequency-integrated intensity retains its per-solid-angle units. The result still refers to one viewing direction per unit solid angle.

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

Save a channel average, then recover an area

In the low-speed radio convention, $v=c(1-\nu/\nu_0)$ , so recession lowers frequency. The normalized profiles transform as $\phi_\nu=(c/\nu_0)\phi_v$ . The Jacobian matters: a per-hertz profile cannot simply be reused as a per-velocity profile.

A finite receiver channel averages the transferred brightness over its interval. The stored number is still brightness. Multiplying by channel width gives the contribution to integrated brightness. Combining two equal-width channels takes their mean and doubles the new width.

For a worked example, prescribe a transparent Gaussian brightness profile with peak one kelvin and dispersion $0.250\,\mathrm{km\,s^{-1}}$ . Its area is peak times $\sqrt{2\pi}\sigma$ , or $0.6267\,\mathrm{K\,km\,s^{-1}}$ . Double the dispersion at fixed emitting column and excitation: the peak halves while the area stays fixed. The one-kelvin peak is a supplied brightness, not a measurement of the gas temperature.

\overline T_i=\frac{1}{\Delta v_i}\int_{v_i^-}^{v_i^+}T(v)\,dv,\qquad W=\sum_i\overline T_i\Delta v_i

Channel widths convert the saved average brightnesses back into an integrated line.

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

From moving molecules to a line with a width

Count molecules that can make the transition

Let a distribution of velocities broaden the line

Add the transparent sightline

Save a channel average, then recover an area

Exercises and explanations

Try the prediction

Follow area and peak separately

Sources and calculation scope