advanced31 min studyLesson 889 in Physics

Radiative cooling functions and thermal instability

Define cooling normalization and energy content, then derive distinct isochoric and isobaric stability criteria for an explicit toy heating law.

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What you’ll understand

  • Keep volumetric cooling rates separate from spectral per-solid-angle emissivity.
  • Estimate a cooling time with a declared particle count and thermodynamic constraint.
  • Derive local stability rather than infer it from a cooling curve's slope alone.

Radiating gas can lose thermal energy through many continua and atomic transitions. A cooling function packages those processes into a useful coefficient, but its density convention and ionization assumptions must travel with it. A gas can also be thermally unstable near a heating-cooling balance: a small temperature change may induce a net energy exchange that amplifies the change. We separate an energy-loss timescale from this stability question and derive a restricted criterion with an explicit heating law and pressure constraint.

A cooling function includes population and escape assumptions

Define a positive optically thin volumetric radiative loss as C=nenHΛ(T)C=n_en_H\Lambda(T), where n H counts hydrogen nuclei and Lambda has SI units watts cubic metres. C is watts per cubic metre, integrated over escaping photon frequencies and solid angle. It is not the earlier spectral emissivity j. Tabulations using electron times ion density, hydrogen density squared, or energy loss per mass have different normalizations. Always identify the density factors before using a numerical coefficient, especially when combining composition or ionization models.

A temperature-only table commonly assumes specified abundances and ionization equilibrium. Collisional ionization equilibrium is not the same as equilibrium under an external photoionizing field, and neither describes every rapidly cooling parcel. Ion fractions can lag the temperature when recombination is slow. Line photons may also be trapped instead of removing energy locally. Metal abundance, radiation, dust, density-dependent collisional deexcitation and optical depth can therefore alter the loss. A curve supplied for one environment is not a universal property of matter at that temperature alone.

C=nenHΛ(T),uth=32ntotkBT,tcool,isoch=uthCC=n_en_H\Lambda(T),\qquad u_{\mathrm{th}}=\frac32n_{\mathrm{tot}}k_BT,\qquad t_{\mathrm{cool,isoch}}=\frac{u_{\mathrm{th}}}{C}
Lambda is in W m cubed, C in W m to the minus three and thermal energy density u in J m to the minus three. The timescale is a local estimate for an ideal nonrelativistic monatomic gas without heating; ntot counts all freely moving particles, not only hydrogen nuclei.↔ Scroll wide equations horizontally.

A timescale depends on which energy reservoir is changing

At fixed density, the translational thermal energy uses three halves k T per free particle. Fully ionized pure hydrogen has one electron and one proton per nucleus, so its total particle density is twice n H. At fixed pressure an ideal monatomic parcel also does expansion or compression work; its enthalpy gives a five-halves factor per particle instead. Ionization energy, changing particle counts, molecular degrees of freedom and relativistic particles require an appropriate energy equation. A familiar numerical cooling-time formula can fail when those assumptions change.

With volumetric heating H, define the net loss as L equal to C minus H. A cooling-only timescale can remain finite at thermal equilibrium even though the net loss is zero. Conversely, dividing thermal energy by a nearly vanishing net loss does not determine the response to a small perturbation. Stability requires the derivative of the net energy exchange along the permitted perturbation, together with the parcel's heat capacity and mechanical constraint. Pressure equilibration by sound waves, conduction and the perturbation wavelength help decide which constraint is physically appropriate.

L=n2Λ(T)−nΓ,Λ(T)∝Ta,(∂L∂T)P,eq=n2ΛT(a−1),(∂L∂T)n,eq=n2ΛTa\mathcal L=n^2\Lambda(T)-n\Gamma,\qquad \Lambda(T)\propto T^a,\qquad \left(\frac{\partial\mathcal L}{\partial T}\right)_{P,\mathrm{eq}}=\frac{n^2\Lambda}{T}(a-1),\qquad \left(\frac{\partial\mathcal L}{\partial T}\right)_{n,\mathrm{eq}}=\frac{n^2\Lambda}{T}a
This separate toy gas uses one effective particle density n, pressure nkBT, constant positive heat capacity, heating Gamma watts per particle, and equilibrium nLambda = Gamma. It does not silently identify n with the preceding hydrogen/electron factors. Positive derivatives restore a small temperature perturbation when net loss is defined as cooling minus heating.↔ Scroll wide equations horizontally.

The perturbation constraint changes the stability boundary

For the toy power law, a constant-density perturbation is stable when a is positive: a hotter parcel cools more while its per-volume heating stays fixed. At constant pressure, however, raising temperature lowers density. Cooling carries two density factors whereas the chosen heating carries only one. Differentiating at equilibrium gives a factor a minus one, so isobaric stability requires a greater than one. A cooling function that rises with temperature can consequently still be unstable to pressure-balanced perturbations under this particular heating prescription.

This is a local linear result around an equilibrium with a declared heating law, not a claim that every cooling cloud fragments whenever a slope is small. Marginal exponents require additional terms or higher-order analysis. Thermal conduction can damp sufficiently short wavelength temperature variations, and magnetic geometry can make conduction anisotropic. Gravity, turbulent mixing, acoustic dynamics and a changing ionization state can couple additional modes. Applying a named instability criterion therefore requires identifying the energy equation and response times, rather than reading a slope from an unrelated equilibrium table.

A cooling-only time with the correct particle count

A fully ionized pure-hydrogen parcel has nH=ne=106 m−3n_H=n_e=10^6\,\mathrm{m^{-3}}, T = 10610^6 K, and a supplied optically thin coefficient Λ=10−35 W m3\Lambda=10^{-35}\,\mathrm{W\,m^3}. Neglect heating and ionization changes. Find its volumetric loss, thermal energy density and local isochoric cooling-time estimate, using kB=1.38×10−23 J K−1k_B=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}.

  1. The density product is 1012 m−610^{12}\,\mathrm{m^{-6}}, so C=10−23 W m−3C=10^{-23}\,\mathrm{W\,m^{-3}}. The supplied Lambda convention already specifies its density factors.

  2. There are two free particles per hydrogen nucleus, hence ntot=2×106 m−3n_{\mathrm{tot}}=2\times10^6\,\mathrm{m^{-3}} and uth=(3/2)(2×106)(1.38×10−23)(106)=4.14×10−11 J m−3u_{\mathrm{th}}=(3/2)(2\times10^6)(1.38\times10^{-23})(10^6)=4.14\times10^{-11}\,\mathrm{J\,m^{-3}}.

  3. Dividing gives tcool,isoch=4.14×1012 st_{\mathrm{cool,isoch}}=4.14\times10^{12}\,\mathrm s, about 1.31×1051.31\times10^5 years for a year of 3.15×1073.15\times10^7 s. This is an instantaneous estimate, since Lambda changes as cooling proceeds.

The result

The loss is 10^−23 W m^−3 and the local isochoric time is about 1.31 × 10^5 years. Counting only protons would underestimate the thermal energy by two.

Explore the model

Interactive lab

Interactive lab · heat

Diffusion damps short spatial scales faster

Reset and set BOTH Spatial mode n = 1 and Diffusivity κ = 0.50. Step twice to time 1 s; predict Current amplitude exp(−0.5), about 0.607. For a fresh comparison, Reset and set n = 2 and κ = 0.50, then Step twice to 1 s; predict exp(−2), about 0.135.

Diffusion damps short spatial scales fasterThe exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Current readings: Current amplitude: 1; Spatial L² norm: 1.772; Decay rate · κn²: 3.6 /s. All parameters are adjustable with the labeled controls below.Periodic heat equation mode 3 at time 0: the amplitude is 1.0π/2π3π/22π−1−0.500.51x · periodic intervalTemperature perturbation uExact mode · n = 3
Current temperature modeInitial sineCurrent amplitude
Current amplitude
1
Spatial L² norm
1.772
Decay rate · κn²
3.6 /s
t = 0 s

The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe exact periodic heat mode illustrates conductive smoothing's n-squared rate. It has no heating-cooling balance, density response, ionization evolution, unstable thermal mode, or radiation-hydrodynamics solver.

Check your understanding

For the stated toy heating proportional to n and cooling proportional to n²T^a, which local linear statement is correct?

Try it yourself

In the separate toy gas let Λ∝T1/2\Lambda\propto T^{1/2} and begin at heating-cooling equilibrium. Compare a trial state at four times the temperature, first at fixed pressure and then at fixed density. Express heating and cooling relative to their common initial value C0 and interpret the restoring direction.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

At fixed pressure density falls by four. Heating has one density factor, cooling two, and Lambda increases by two.

Compare with the solution

At fixed pressure, cooling becomes (1/4)2×2 C0=C0/8(1/4)^2\times2\,C_0=C_0/8 while heating becomes C0/4C_0/4. Net loss is −C0/8-C_0/8, so the hotter trial state gains energy, consistent with the local isobaric instability for a = 1/2. At fixed density cooling becomes 2C02C_0 and heating stays C0C_0, giving positive net loss that pushes the temperature down. This finite comparison illustrates the derivative signs; it does not predict a nonlinear endpoint, condensation size or evolution time.

Take it with you

  • Cooling-function units, density normalization, abundances and ionization assumptions are essential inputs.
  • Thermal energy, enthalpy and net heating describe different timescale questions.
  • Thermal stability depends on the heating law and perturbation constraint as well as the cooling slope.

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