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Reading a stellar wind

Stellar winds · L1

How much momentum can light carry?

Sandbox Physics ·

Light can push stellar gas because photons carry momentum. Deciding whether that push is sufficient requires two calculations: the momentum transferred each second and the energy needed to lift and accelerate the gas. Their units differ, and they limit a wind in different ways.

Calculate the momentum of one photon

A photon has no rest mass, but it has momentum. In vacuum its energy is Planck constant times frequency, and its momentum magnitude is energy divided by light speed. Substituting zero mass into the low-speed formula for ordinary matter would give the wrong result.

Frequency and wavelength are connected by light speed. A photon at 280 nanometres has energy about 7.09×10−19 J7.09\times10^{-19}\,\mathrm J and momentum about 2.37×10−27 kg m s−12.37\times10^{-27}\,\mathrm{kg\,m\,s^{-1}}. Each impulse is tiny; a luminous star supplies a vast number of photons per second.

Momentum tells us how an encounter changes motion. Energy tells us how much work it can perform. Transferring the same impulse to slow and fast gas can involve different amounts of work, which is why the two calculations must remain separate.

Eγ=hν=hc/λ,pγ=Eγ/c=h/λE_\gamma=h\nu=hc/\lambda,\qquad p_\gamma=E_\gamma/c=h/\lambda
Use h=6.62607015×10−34 J sh=6.62607015\times10^{-34}\,\mathrm{J\,s} and c=299792458 m s−1c=299792458\,\mathrm{m\,s^{-1}}. Convert wavelength to metres before substitution.

Reading: Owocki2014, Theory of Winds from Hot, Luminous Massive Stars · Standalone stellar-wind scientific reproduction

Conservative scattering transfers momentum

Take an outward photon from the stellar surface. An atom receives its incoming momentum, then recoils when a photon leaves in another direction. The gas receives the difference. For many approximately isotropic outgoing directions, the outgoing recoil averages toward zero in the atomic rest frame, leaving the directionality of the incident radiation.

Conservative resonance scattering means that these photons are not locally created or thermally destroyed. Their energies need not remain identical in the stellar frame. Moving gas can receive mechanical work from the radiation, changing the outgoing frequency.

For scattering coherent in the gas frame with negligible atomic recoil energy, the frame transformation gives the relation below. The photon energy lost is the gas velocity dotted with the transferred momentum. The standalone exercise checks randomly directed encounters and reproduces that work relation to numerical roundoff.

The simplified line profile separately follows photon number under its stated Doppler approximation. The event calculation checks energy exchange at a specified velocity. A wind velocity and mass-flow rate still require a dynamical calculation.

Δp=pin−pout,Ein−Eout=v⋅Δp\Delta\mathbf p=\mathbf p_{\rm in}-\mathbf p_{\rm out},\qquad E_{\rm in}-E_{\rm out}=\mathbf v\cdot\Delta\mathbf p
v\mathbf v is the scatterer velocity. The energy difference is the energy transferred from the photon to the gas.

Reading: Lamers, Cerruti-Sola & Perinotto1987, the SEI line-profile method · Owocki2014, Theory of Winds from Hot, Luminous Massive Stars

Turn impulses into a wind momentum rate

Mass-flow rate M˙\dot M counts the kilograms crossing a distant spherical shell each second. If the outer speed approaches v∞v_\infty, the outward momentum carried per second is approximately their product. Momentum per time has force units, newtons.

Luminosity LL is radiated energy per second. Dividing it by light speed gives a useful radiation momentum-rate scale. The dimensionless wind performance number η\eta compares the terminal wind momentum rate with that scale. It does not yet specify how efficiently any particular transition intercepts the radiation.

For an illustrative star, take luminosity 3.0×1031 W3.0\times10^{31}\,\mathrm W, mass flow 6.0×1016 kg s−16.0\times10^{16}\,\mathrm{kg\,s^{-1}} and terminal speed 1.0×106 m s−11.0\times10^6\,\mathrm{m\,s^{-1}}. The wind momentum rate is 6.0×1022 N6.0\times10^{22}\,\mathrm N. Luminosity divided by light speed is about 1.00×1023 N1.00\times10^{23}\,\mathrm N, giving a performance number near 0.60.

These are example inputs, not measurements of P Cygni. Their ratio also does not establish that the required fraction of light actually couples to the gas. A single resonance line sees a narrow spectral band, so assigning the entire stellar luminosity to that line would overestimate its available momentum.

p˙wind≃M˙v∞,η=M˙v∞L/c\dot p_{\rm wind}\simeq\dot Mv_\infty,\qquad\eta=\frac{\dot Mv_\infty}{L/c}
This compares radial terminal-wind momentum flux with a radiation scale. Repeated interactions can produce wind performance above the single-interaction L/cL/c scale.

Reading: Owocki2014, Theory of Winds from Hot, Luminous Massive Stars

The gas must also pay for gravity

Lifting one kilogram from near radius R⋆R_\star to infinity requires gravitational work as well as its final kinetic energy. Neglecting initial speed, enthalpy and other energy sources gives the specific energy below. Multiplying by mass flow turns it into mechanical power.

Continue the example with stellar mass 5.0×1031 kg5.0\times10^{31}\,\mathrm{kg}, radius 1.5×1010 m1.5\times10^{10}\,\mathrm m and G=6.6743×10−11 m3 kg−1 s−2G=6.6743\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}. The required specific energy is about 7.22×1011 J kg−17.22\times10^{11}\,\mathrm{J\,kg^{-1}}. Mechanical power is then about 4.33×1028 W4.33\times10^{28}\,\mathrm W, approximately 0.144 percent of the example luminosity.

A momentum rate approaching L/cL/c can therefore use only a small fraction of the luminosity as mechanical work. Multiple interactions can even allow performance above one without immediately exhausting the energy supply. Momentum conservation must include radiation, gas and the star; the sum of outward radial momentum over a sphere is not simply one fixed-direction vector.

Larger mass flow eventually consumes the available radiative energy. This photon-tiring limit is a necessary energy ceiling. An actual wind must also obtain sufficient force and satisfy the appropriate dynamical conditions.

E˙mech≃M˙(GMR⋆+v∞22),M˙≲LGM/R⋆+v∞2/2\dot E_{\rm mech}\simeq\dot M\left(\frac{GM}{R_\star}+\frac{v_\infty^2}{2}\right),\qquad\dot M\lesssim\frac{L}{GM/R_\star+v_\infty^2/2}
The upper bound assumes radiation is the only available energy source. Computing the velocity and critical conditions remains a separate task.

Reading: Owocki2014, Theory of Winds from Hot, Luminous Massive Stars

Why a deep line can still supply too little force

Write local force per unit mass as an acceleration, with outward positive. Neglect gas pressure. A prescribed flow must overcome inward gravity and achieve its radial acceleration. Those two requirements give the acceleration that radiation would need to supply.

The line-profile calculation uses a separate illustrative star, with a 20000 kelvin continuum, a 75 solar-radius photosphere, gravity corresponding to 25 solar masses and a declared reference line opacity. The force obtained from the same angular transfer reaches only about 4.65×10−54.65\times10^{-5} to 2.42×10−42.42\times10^{-4} of the required value.

The line leaves a conspicuous profile but supplies far too little acceleration. Other transitions, ion populations, continuum transfer and additional forces can matter. The displayed tracers follow the prescribed velocity law; determining wind launch requires the radiation and fluid equations together.

grequired=vdvdr+GMr2g_{\rm required}=v\frac{dv}{dr}+\frac{GM}{r^2}
The first term accelerates matter along its trajectory; the second compensates gravity. Gas pressure and other radiation forces are omitted from this conditional diagnostic.

Reading: Owocki2014, Theory of Winds from Hot, Luminous Massive Stars · Standalone stellar-wind scientific reproduction

Keep three physical quantities separate when saving a spectrum

Local conservative scattering preserves photon number, but the observer does not collect every photon. Some return to the opaque photosphere, and others miss the aperture. The collected line can retain a net absorption area. Thermal re-emission after absorption by the photosphere is outside this singlet calculation.

Photon number, radiative energy and mechanical work have different units and integrals. A photon equivalent width in Doppler coordinates describes photon paths under a specified continuum normalization. Comparing it with actual energy flux requires the wavelength conversion and moving-gas work approximation to remain explicit.

Keeping those distinctions visible makes a deep line, a large force and a high mass-flow rate three questions that can be tested separately.

Reading: Lamers, Cerruti-Sola & Perinotto1987, the SEI line-profile method · Standalone stellar-wind scientific reproduction

Exercises and explanations

  1. A photon wavelength is halved. How do its energy and momentum change? If beam luminosity stays fixed, what happens to the momentum rate on complete absorption?

    Read the explanation

    Each photon has twice the energy and momentum, but the fixed-luminosity beam contains half as many photons per second. Its completely absorbed momentum rate remains luminosity divided by light speed.

  2. For performance number 3 and terminal speed 1000 kilometres per second, what fraction of luminosity becomes kinetic power if gravity is ignored? Does that prove the wind exists?

    Read the explanation

    Kinetic power divided by luminosity is performance number times terminal speed divided by twice light speed, about 0.005, or 0.5 percent. Energy is not exhausted, but actual momentum coupling and dynamical conditions are still required.

  3. Why can a conservative-scattering model have positive net absorption equivalent width?

    Read the explanation

    Conservative describes the local scattering process. A finite aperture collects only some paths, and the opaque star absorbs returning photons. Collected emission can therefore have less area than the removed stellar light.

Try the prediction

Change the flow and follow its scattered light

  1. Increase mass flow and ask whether absorption keeps deepening equally strongly. Open the force diagnostic and compare line depth with force per unit mass.
  2. Choose the saturated preset. Identify which quantities describe photon number and which describe momentum transfer, using the different axis units.
  3. Export the observation and recompute equivalent width from integer photons and continuum calibration alone. Compare it with model return and aperture loss.

Sources and calculation scope

  1. Owocki2014, Theory of Winds from Hot, Luminous Massive Stars

    Pinned v1; section2 for CAK/Sobolev steady winds and sections3.1–3.4 for line instability, clumping, porosity and velocity gaps. Linked theory source; original is not redistributed.

  2. Lamers, Cerruti-Sola & Perinotto1987, the SEI line-profile method

    ApJ314,726, sectionsII–V and equations4–10/14–19. Our local escape singlet and fixed-source finite-width test retain their separate approximations; no complete SEI/non-LTE solver is claimed.

  3. Standalone stellar-wind scientific reproduction

    Original small data, independent transfer/photon/force/finite-width references,197producer checks and a proper synthetic Poisson exercise. Fixed truths and prior-averaged coverage remain separate.

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