Metallicity, Winds, and Angular Momentum

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Metallicity influences a stellar-origin black hole chiefly by changing how the star loses mass and angular momentum before collapse. Metal lines provide opacity for hot-star winds; surface stripping changes the final mass, the helium-core growth, and the lever arm through which spin is removed. Yet metallicity is not a dial that sets a unique black-hole mass or spin. Wind prescriptions, surface composition, binary interaction, internal torques, and which layers fall into the remnant all intervene. This page develops that chain and includes a reproducible numerical check of its simplest scalings.

Helpful background. Black-Hole Geometry and Mechanics supplies the dimensionless spin convention; Relativistic Matter and Compact Objects supplies rotating-star structure and transport concepts; Black-Hole Astrophysics supplies the observable distinction between natal and later-acquired spin; and Black-Hole Dynamics and Collapse supplies the collapse stage that decides which retained angular momentum crosses the horizon.

Why metals matter to a hot-star wind

Radiative acceleration transfers photon momentum to ions through spectral lines. Iron-group elements provide many of the relevant lines in hot, hydrogen-rich stars, so changing the abundance mixture changes the mass flux even when luminosity and surface gravity are held fixed. Over a restricted regime it is useful to summarize a wind grid by

\dot M(Z)=\dot M(Z_\odot) \left(\frac{Z}{Z_\odot}\right)^m,

where $\dot M\gt0$ denotes the mass-loss rate. This is a local fit, not a law valid across every temperature, composition, or evolutionary phase. Vink, de Koter, and Lamers obtain $m\simeq0.85$ at fixed terminal-to-escape-speed ratio over approximately $1/30\leq Z/Z_\odot\leq3$ . Including their adopted metallicity dependence of terminal speed gives $m\simeq0.69$ for the hotter O-star branch and $m\simeq0.64$ for the cooler B-supergiant branch, separated near an effective temperature of $25{,}000$ K Vink et al. 2001, § 7, eqs. (21)–(22), and § 8, eqs. (24)–(25).

The conditions behind the exponent matter. The ionization balance changes across the bi-stability region; cool supergiants and stripped helium stars require different prescriptions; eruptive loss need not follow the same power law; and a binary can remove an envelope through Roche-lobe overflow or common-envelope evolution. Thus low $Z$ often promotes mass retention, but it does not guarantee it.

Mass loss is also angular-momentum loss

The angular momentum inside a Lagrangian mass coordinate is

J(m)=\int_0^m j(m')\,dm',

where $m$ is enclosed baryonic mass and $j$ is angular momentum per unit baryonic mass. This gives the stellar angular-momentum budget. It becomes a black-hole spin only after one determines the retained angular momentum $J_{\rm BH}$ and gravitational mass $M_{\rm BH}$ , giving $a_*=J_{\rm BH}/(GM_{\rm BH}^2)$ in the site convention $c=1$ . If a wind removes material at a positive rate $\dot M_{\rm loss}$ with effective specific angular momentum $j_{\rm w}$ , then

\dot J_{\rm wind}=-\dot M_{\rm loss}j_{\rm w}.

Surface layers can have a large lever arm, so they may carry much more specific angular momentum than the core average. Removing a modest mass can therefore remove a disproportionate fraction of $J$ . The response of $a_*$ is not determined by the mass loss alone because both its numerator and denominator evolve.

For a simple example, a spherical star rotating uniformly at angular speed $\Omega$ loses mass uniformly over a surface of radius $R$ . A parcel at colatitude $\vartheta$ carries axial specific angular momentum $j_z=\Omega R^2\sin^2\vartheta$ . The area average is $j_{\rm w}=2\Omega R^2/3$ . This estimate explains the large surface lever arm; a magnetized wind with a larger effective corotation radius or a latitude-dependent outflow requires a different value.

Internal coupling decides whether the core replenishes the lost surface angular momentum. Meridional circulation and shear can transport both composition and spin; magnetic stresses can enforce stronger coupling; tides and mass transfer can add or remove angular momentum in a binary. Heger and collaborators compare rotating and mass-losing fate channels and emphasize the distinction between final mass and core rotation in Heger et al. 2003, §§ 6.1 and 6.3. A rapidly rotating main-sequence star can still make a slowly rotating core if coupling and winds act efficiently.

From a retained core to a natal spin

Even a well-resolved pre-collapse $j(m)$ profile does not equal the black hole’s final spin. Low- $j$ material can fall nearly radially, while sufficiently high- $j$ material circularizes outside the horizon and may build a disk. Disk winds can then eject some of the mass and angular momentum. Magnetic fields, nonaxisymmetric instabilities, fallback, and later binary accretion alter the result further. A stellar model should therefore export $j(m)$ , not only one total-spin number.

Two elementary checks remain powerful. First, $a_*$ must be dimensionless; in ordinary units $a_*=cJ/(GM^2)$ . Second, a proposed black-hole assembly history must respect the Kerr bound after radiative and outflow losses are included. If direct integration of the assigned infall gives an impossible result, that is evidence that circularization and angular-momentum shedding cannot be neglected, not evidence for a super-extremal black hole.

Detailed line-transfer calculations, multidimensional magnetic transport, and collapse-disk evolution lie outside this page. The intended output is the causal interface: wind and binary history determine a pre-collapse mass and $j(m)$ profile; collapse and accretion determine how much of each becomes part of the hole.

A reproducible scaling and integration check

The calculation below is a teaching reconstruction, not a claimed stellar-evolution rerun. It evaluates the fixed-velocity wind ratio with $m=0.85$ and integrates the analytic angular-momentum profile $j/(j_0)=1+x^2$ , where $x=m/M$ . Its exact solution is $J/(j_0M)=4/3$ , providing an independent benchmark for the trapezoidal integration algorithm. The displayed results were produced with Python 3.9.6 and NumPy==2.0.2. Save this script as wind_spin_check.py.

import numpy as np

EXPONENT = 0.85
EXACT_J = 4.0 / 3.0

print("Z/Zsun  Mdot/Mdot_sun")
for metallicity in (1.0, 0.1, 0.01):
    ratio = metallicity**EXPONENT
    print(f"{metallicity:.2f}  {ratio:.8f}")

print("N  J/(j0 M)  absolute_error")
for n in (32, 64, 128):
    x = np.linspace(0.0, 1.0, n + 1)
    j_over_j0 = 1.0 + x**2
    estimate = np.trapezoid(j_over_j0, x)
    print(n, f"{estimate:.10f}", f"{abs(estimate-EXACT_J):.3e}")

The program must produce three metallicity ratios and three refinement rows containing grid intervals, integrated angular momentum, and absolute error. The exact save-and-run recipe is:

python3 -m venv .venv-wind
. .venv-wind/bin/activate
python3 -m pip install numpy==2.0.2
python3 wind_spin_check.py

Grid intervals $N$	Result $J/(j_0M)$	Absolute error
32	1.3334960938	$1.628\times10^{-4}$
64	1.3333740234	$4.069\times10^{-5}$
128	1.3333435059	$1.017\times10^{-5}$

The error falls by approximately four when $N$ doubles, as expected for the composite trapezoidal method on a smooth quadratic. The wind outputs are $1$ , $0.14125375$ , and $0.01995262$ at $Z/Z_\odot=1$ , $0.1$ , and $0.01$ . The last point deliberately evaluates an algebraic extrapolation below the quoted power-law range; it is not a validated wind prediction. The main numerical failure mode is under-resolving a sharp feature in a real $j(m)$ profile; the main physical sensitivity is treating a regime-dependent wind exponent as universal. Grid convergence cannot cure the latter.

Common pitfalls

Equating metallicity with iron abundance in every context. A stated $Z$ may scale all elements together, whereas real abundance patterns and self-enriched surfaces do not. The opacity-relevant mixture should be declared.

Equating core rotation with natal black-hole spin. The retained $j(m)$ profile must pass through collapse, circularization, outflows, and possible fallback before it becomes a horizon quantity.

Exercise

Using only the power-law approximation, compare the mass-loss ratio at $Z=0.1Z_\odot$ for $m=0.85$ and $m=0.69$ . Explain why their difference is not an uncertainty interval for all massive-star winds.

Solution

The two ratios are

0.1^{0.85}=0.141, \qquad 0.1^{0.69}=0.204.

Thus the fixed-terminal-speed scaling predicts about $31\%$ less mass loss than the effective $m=0.69$ scaling at this metallicity. These exponents arise under particular hot-star assumptions, including how terminal speed is treated. They do not bracket cool-star, eruptive, helium-star, or binary-stripping losses, and the normalization can be as consequential as the exponent.

References

Heger, A., Fryer, C. L., Woosley, S. E., Langer, N., and Hartmann, D. H. (2003). “How Massive Single Stars End Their Life.” The Astrophysical Journal 591, 288–300. DOI.
Vink, J. S., de Koter, A., and Lamers, H. J. G. L. M. (2001). “Mass-loss Predictions for O and B Stars as a Function of Metallicity.” Astronomy & Astrophysics 369, 574–588. DOI.