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advanced30 min studyLesson 890 in Physics

Eddington luminosity and radiation-supported flow

Derive an opacity-dependent radiation-force scale and separate hydrostatic balance from a closed outflow calculation.

Before you begin
On this page

The question for today

What you’ll understand

  • Use bolometric flux and flux-weighted momentum opacity with consistent SI units.
  • Derive the spherical Eddington ratio and its hydrostatic implication.
  • Identify the additional equations and boundaries needed to determine a steady outflow.

Radiation carries momentum as well as energy. A sufficiently strong outward flux can partly or wholly offset gravitational acceleration, changing the gas pressure gradient required for support. The Eddington luminosity expresses this competition for a declared geometry and opacity. It is a useful force scale, not a complete wind solution or an unconditional upper bound on every observed source. We will keep radiation momentum transfer explicit, use a spherical Newtonian toy flow, and distinguish local balance from the dynamics that selects an outflow.

The relevant opacity transfers momentum from a directed flux

Here F is the outward bolometric flux, the frequency integral of the earlier spectral F nu, in watts per square metre. For radial radiation force define the flux-weighted momentum opacity kappa F by weighting each frequency's momentum-transfer opacity with F nu. It has square metres per kilogram units. Isotropic elastic scattering transfers momentum despite exchanging little photon energy, whereas strongly forward scattering can have a smaller transport effect than its extinction opacity suggests. Energy absorption opacity and momentum opacity therefore need not be numerically identical.

For Thomson scattering by free electrons coupled to the ions, the opacity is the cross section per electron divided by the mass per free electron. Fully ionized pure hydrogen gives approximately 0.04 m2 kg−10.04\,\mathrm{m^2\,kg^{-1}}, equivalent to about 0.4 square centimetres per gram. Confusing those two numerical unit conventions produces a factor-ten error. Partial ionization, composition, spectral lines, dust and recoil can change the effective opacity. A fixed electron-scattering value is appropriate only when that interaction and the declared population dominate the momentum exchange.

F=∫0∞Fνdν,κF=∫0∞κνmomFνdνF,arad=κFFc,κes=σTμempF=\int_0^\infty F_\nu d\nu,\qquad \kappa_F=\frac{\int_0^\infty\kappa_\nu^{\mathrm{mom}}F_\nu d\nu}{F},\qquad a_{\mathrm{rad}}=\frac{\kappa_FF}{c},\qquad \kappa_{\mathrm{es}}=\frac{\sigma_T}{\mu_em_p}
The flux-weighted expression assumes a nonzero outward flux and a common radial direction for the one-dimensional moment equation. Mu-e is mass per free electron in proton-mass units. Radiation acceleration is in metres per second squared, with momentum opacity in m squared per kg.↔ Scroll wide equations horizontally.

Spherical force balance defines a luminosity scale

For a central luminosity L crossing spherical surfaces without net energy deposition, F equals L divided by four pi r squared. Newtonian gravity has magnitude G M divided by r squared. With constant kappa F their acceleration ratio is independent of radius, giving the Eddington ratio Gamma and its corresponding luminosity. This cancellation is geometrical; it does not imply that opacity, luminosity or enclosed gravitating mass are actually constant throughout every stellar envelope or accretion flow. State the radius range where those approximations are intended.

In a static gas, an outward radiation force reduces the gas pressure gradient needed to oppose gravity. Writing that force explicitly gives a gradient proportional to minus density times gravity times one minus Gamma. An alternative optically thick diffusion formulation can combine gas and radiation pressure in a total-pressure gradient. Adding both that radiation-pressure gradient and the same explicit radiation force would count its support twice. The chosen momentum equation must also be compatible with whatever radiation transfer approximation supplies its flux and opacity.

Γ=κFL4πGMc=LLEdd,LEdd=4πGMcκF,dPgasdr=−ρGMr2(1−Γ),M˙=4πr2ρv\Gamma=\frac{\kappa_FL}{4\pi GMc}=\frac{L}{L_{\mathrm{Edd}}},\qquad L_{\mathrm{Edd}}=\frac{4\pi GMc}{\kappa_F},\qquad \frac{dP_{\mathrm{gas}}}{dr}=-\rho\frac{GM}{r^2}(1-\Gamma),\qquad \dot M=4\pi r^2\rho v
The pressure equation is hydrostatic with the radiation force already included explicitly. The last expression is instead steady spherical mass continuity for radial speed v; it is not fixed by the hydrostatic equation. L and LEdd are in watts, density in kg m to the minus three, and mass flux in kg per second.↔ Scroll wide equations horizontally.

A force ratio alone cannot select a wind

For a moving steady spherical gas, the radial momentum equation includes v dv/dr=−(1/ρ)dPgas/dr−GM/r2+aradv\,dv/dr=-(1/\rho)dP_{\mathrm{gas}}/dr-GM/r^2+a_{\mathrm{rad}}. Mass continuity, an energy equation, a thermodynamic relation, opacity information and appropriate boundary or critical-point conditions are also needed. The Eddington ratio by itself supplies neither the base density nor the mass-loss rate. A gas below the scattering Eddington scale can still have a thermal or line-driven outflow, and a locally outward net radiative-gravity force does not specify the eventual speed or stable global structure.

Radiation-supported envelopes, disks and winds can involve anisotropic flux, convection, photon trapping, opacity changes and time dependence. Consequently a measured luminosity above a simple spherical scattering estimate is not by itself a violation of energy conservation. It may indicate that the reference assumptions are inappropriate. Our calculations isolate a transparent local momentum budget and a continuity estimate with supplied density and velocity. They do not solve coupled radiation hydrodynamics, determine an accretion efficiency, or establish which astrophysical flow will remain stationary under perturbations.

Work it through

A synthetic half-Eddington hydrostatic layer

Use GM=1020 m3 s−2GM=10^{20}\,\mathrm{m^3\,s^{-2}}, c = 3×1083\times10^8 m/s and κF=0.04 m2 kg−1\kappa_F=0.04\,\mathrm{m^2\,kg^{-1}}. Find the Eddington luminosity. At half that luminosity, radius 101010^{10} m and density 2×10−8 kg m−32\times10^{-8}\,\mathrm{kg\,m^{-3}}, find the radiation acceleration and required static gas-pressure gradient.

  1. Substitution gives LEdd=4π(1020)(3×108)/0.04=3π×1030 WL_{\mathrm{Edd}}=4\pi(10^{20})(3\times10^8)/0.04=3\pi\times10^{30}\,\mathrm W, approximately 9.42×1030 W9.42\times10^{30}\,\mathrm W.

  2. Gravity at the supplied radius is GM/r2=1 m s−2GM/r^2=1\,\mathrm{m\,s^{-2}}. Half the Eddington luminosity means Gamma = one half, so the outward radiation acceleration is 0.5 m s−20.5\,\mathrm{m\,s^{-2}}.

  3. The remaining inward effective acceleration is also 0.5. Hydrostatic balance therefore requires dPgas/dr=−(2×10−8)(0.5)=−10−8 Pa m−1dP_{\mathrm{gas}}/dr=-(2\times10^{-8})(0.5)=-10^{-8}\,\mathrm{Pa\,m^{-1}}.

The result

The Eddington scale is about 9.42 × 10^30 W. Radiation provides half the local gravitational acceleration, leaving a negative gas-pressure gradient of 10^−8 Pa per metre.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Compare inward gravity and outward radiation as separate vectors

Reset and set BOTH Angle = 180 degrees and Magnitude = 1; predict x projection −1 for a normalized inward contribution. For a fresh comparison, Reset and set Angle = 0 degrees and Magnitude = 0.50; predict x projection +0.50 for the separate outward contribution. Add those two readouts by hand to obtain −0.50.

Compare inward gravity and outward radiation as separate vectorsPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. 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 forEach run displays one vector, not two simultaneous forces. The comparison illustrates a signed force budget without opacity, spherical dilution, pressure gradients, mass continuity, or a radiation-hydrodynamics solver.

Pause & reason

Check your understanding

What does the spherical constant-opacity Eddington condition determine by itself?

Your turn

Try it yourself

Keep the worked example's mass and luminosity but double the effective opacity to 0.08 m2 kg−10.08\,\mathrm{m^2\,kg^{-1}}. Find Gamma and the hydrostatic gas-pressure gradient. Separately, if a steady spherical flow is supplied with r=1010r=10^{10} m, ρ=10−12 kg m−3\rho=10^{-12}\,\mathrm{kg\,m^{-3}} and v=104v=10^4 m/s, find its mass flux.

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

Give me a hint

Gamma is proportional to opacity at fixed mass and luminosity. The separate mass flux uses the supplied density and speed, not the force balance alone.

Compare with the solution

Doubling opacity raises Gamma from one half to one, giving zero effective gravity and hence zero static gas-pressure gradient locally in this restricted momentum model. That statement does not establish a global atmosphere or fix a wind density. For the separately supplied steady flow, M˙=4π(1010)2(10−12)(104)=4π×1012 kg s−1\dot M=4\pi(10^{10})^2(10^{-12})(10^4)=4\pi\times10^{12}\,\mathrm{kg\,s^{-1}}. This flux follows from the given kinematic and density data; it is not a mass-loss prediction deduced from Gamma = one.

Keep the idea

Take it with you

  • Radiation force uses a momentum opacity and bolometric directed flux with declared units.
  • The Eddington scale is an opacity-dependent spherical acceleration balance.
  • Hydrostatic support and steady outflow require different dynamical information; a force ratio alone does not determine mass loss.

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