Skip to main content
Sandbox Physics
Starspots, rotation and stellar ages

AC27 · L2

How a stellar wind slows rotation

Sandbox Physics ·

A magnetized wind can slow a star because the escaping gas carries angular momentum away. The braking rate depends on how the wind and magnetic field respond to rotation. A simple torque law makes that dependence calculable and shows where a square-root age clock comes from.

Gas carries more than mass

A parcel leaving the stellar surface starts with rotational motion. A large-scale magnetic field couples some of the outflow to the rotating star beyond the photosphere. The flow can therefore remove more angular momentum per unit mass than gas that loses all coupling at the surface.

For a star approximated as a rotating body, angular momentum is J=IΩJ=I\Omega, where II is its moment of inertia and Ω\Omega its angular velocity. The wind torque τw\tau_{\rm w} is negative when the wind removes angular momentum. Differentiating the product keeps structural evolution visible.

d(IΩ)dt=τw,dΩdt=τwI−ΩI˙I.\frac{d(I\Omega)}{dt}=\tau_{\rm w},\qquad \frac{d\Omega}{dt}=\frac{\tau_{\rm w}}{I}-\Omega\frac{\dot I}{I}.
A torque changes angular momentum; a changing moment of inertia also changes spin.

On a limited main-sequence interval, taking inertia as constant helps isolate the torque mechanism. During strong contraction or subgiant expansion that approximation can fail. The two terms should then be evolved together.

Reading: Matt et al. (2015), The Mass-Dependence of Angular Momentum Evolution in Sun-like Stars, arXiv1412.4786v1

Write a law with two rotation responses

The magnetic activity and wind do not keep strengthening with rotation at the same rate indefinitely. A teaching law lets the torque grow cubically with angular velocity for slow rotation, then linearly beyond a saturation threshold. Absorb the constant inertia and normalization into a positive coefficient κ\kappa.

Ω˙=−κΩmin⁡(Ω2,Ωsat2),P=2πΩ.\dot\Omega=-\kappa\Omega\min(\Omega^2,\Omega_{\rm sat}^2),\qquad P=\frac{2\pi}{\Omega}.
The two branches meet continuously at the saturation angular velocity.

Here angular velocity is in radians per day, evolutionary time is in gigayears, and radians are dimensionless. The coefficient has units of days squared per gigayear. In the interactive teaching curve, κ=0.7 d2/Gyr\kappa=0.7\,\mathrm d^2/\mathrm{Gyr} and Psat=3 dP_{\rm sat}=3\,\mathrm d. These chosen numbers illustrate the mechanism; they do not fit an observed star.

Matt and colleagues derive a related saturated/unsaturated structure from magnetic-field and wind scalings. Their mass, radius and convective-turnover factors matter for population evolution. Their equations 6–11 explain the two responses and constant-inertia limits. Our scalar example keeps those stellar properties fixed. A real mass sequence requires their values and evolutionary changes to be restored.

Reading: Matt et al. (2015), The Mass-Dependence of Angular Momentum Evolution in Sun-like Stars, arXiv1412.4786v1

Integrate the slow-rotation branch

Below saturation, divide the cubic differential equation by angular velocity cubed. Integrating from an initial spin Ω0\Omega_0 at zero elapsed time gives an inverse-square angular velocity that grows linearly with time.

∫Ω0ΩΩ−3 dΩ=−κt⟹Ω−2=Ω0−2+2κt.\int_{\Omega_0}^{\Omega}\Omega^{-3}\,d\Omega=-\kappa t\quad\Longrightarrow\quad \Omega^{-2}=\Omega_0^{-2}+2\kappa t.
The integration constant remembers the initial rotation.

Use the period definition to rewrite the solution in quantities we can compare with a light curve.

P2(t)=P02+8π2κt.P^2(t)=P_0^2+8\pi^2\kappa t.
At late times the accumulated torque term can dominate the initial-period term.

Only in that late-time limit does the period grow like the square root of elapsed time. Two stars with different initial periods need enough braking time before those histories become unimportant. A finite initial period, structural changes or a different torque response can all spoil a universal square-root age estimate.

Take an initial period of six days. After one gigayear under the displayed coefficient, the period is about 9.55 days. The advanced wind-clock panel shows this curve. Applying its green snapshot explicitly transfers that spin into a fresh sixty-day observation. This explicit handoff starts an observing clock measured in days.

0

Fast rotation first follows an exponential

Above saturation the squared threshold is constant. The differential equation is now linear in angular velocity, so the solution is exponential. The crossing time follows by setting the angular velocity equal to the threshold.

Ω(t)=Ω0e−κΩsat2t,ts=log⁡(Ω0/Ωsat)κΩsat2.\Omega(t)=\Omega_0e^{-\kappa\Omega_{\rm sat}^2t},\qquad t_s=\frac{\log(\Omega_0/\Omega_{\rm sat})}{\kappa\Omega_{\rm sat}^2}.
This crossing time applies only when the initial rotation is faster than the threshold.

At the crossing, start the slow-rotation solution from the saturation period and elapsed time after tst_s. Both spin and its first derivative agree at the join.

P2(t)=Psat2+8π2κ(t−ts),t≥ts.P^2(t)=P_{\rm sat}^2+8\pi^2\kappa(t-t_s),\qquad t\geq t_s.
The continuously joined branches avoid an artificial jump in the simulated star.

For a half-day initial period, a three-day saturation period and the same coefficient, the transition occurs after about 0.584 Gyr. At 1 Gyr the star reaches about 5.66 days. It has spent a substantial part of its history on the saturated branch. The independent scientific check integrates the differential equation directly and compares with both analytic branches.

0

Contraction can spin a star up

Turn the external torque off. Angular momentum is then conserved even when inertia changes. A shrinking star can spin faster simply because its mass moves closer to the rotation axis.

τw=0⟹I(t)Ω(t)=I0Ω0,P(t)=P0I(t)I0.\tau_{\rm w}=0\quad\Longrightarrow\quad I(t)\Omega(t)=I_0\Omega_0,\qquad P(t)=P_0\frac{I(t)}{I_0}.
If inertia falls by half, the period falls by half in this isolated example.

The package also checks an exponentially contracting inertia with zero torque. Angular momentum remains constant to numerical precision. Separating these terms is necessary to infer a wind torque from an observed period change.

0

Compare a mechanism with a population

The torque exercise explains why ageing and slower spin can be connected. Cluster data decide how well a particular connection works. In NGC 6811, cool stars stall relative to simple separable braking laws. Older-star magnetic-braking studies add seismic and magnetic information. Those observations ask which parts of the elementary torque picture must change.

The Lab keeps the wind curve and the empirical age density separate. One is a prescribed evolution mechanism; the other compares saved periods and temperature with observed cluster sequences. Their agreement would be an interesting comparison. It would not turn two calculations sharing assumptions into two independent age measurements.

Reading: Curtis et al. (2020), When Do Stalled Stars Resume Spinning Down? Advancing Gyrochronology with Ruprecht147, arXiv2010.02272v1 · Hall et al. (2021), Weakened magnetic braking supported by asteroseismic rotation rates of Kepler dwarfs, arXiv2104.10919v1

Exercises and explanations

  1. In the slow branch, a star starts at six days. If the torque term becomes four times larger than the initial-period-squared term, what is its period?

    Read the explanation

    The squared period is five times the initial squared period, so the period is six times the square root of five days, about 13.4 days. The initial term is still present.

  2. A contracting star has zero external torque and half its original inertia. Has angular momentum been lost when its period halves?

    Read the explanation

    No. Angular velocity doubles as inertia halves, so their product remains constant. Spin-up here records redistribution of a conserved angular momentum.

  3. Why does the saturated branch not give a square-root period law?

    Read the explanation

    Its torque is linear in angular velocity because the activity-dependent factor has reached a fixed threshold. Angular velocity decays exponentially until the star crosses into the unsaturated branch.

Try the prediction

Apply an evolutionary snapshot

  1. Open A separate wind-braking clock beneath the data. Keep the initial period at six days and move the evolutionary clock from zero to one gigayear.
  2. Apply the snapshot. Confirm that the star now starts a fresh day-scale observation at about 9.55 days equatorial spin.
  3. Change the age prior in the calibration controls. The wind curve stays the same; the empirical conditional age density can change.

Sources and calculation scope

  1. Curtis et al. (2020), When Do Stalled Stars Resume Spinning Down? Advancing Gyrochronology with Ruprecht147, arXiv2010.02272v1

    Sections2–6, AppendixA, Table4, Figures2and8–12; named Benchmark_Clusters.txt and Table_1-Ruprecht147_Catalog.txt. The independent temperature-selected fit uses the actual machine table, not the135YY selection described for another fit.

  2. Matt et al. (2015), The Mass-Dependence of Angular Momentum Evolution in Sun-like Stars, arXiv1412.4786v1

    Sections2.1–2.3, equations1–11 and Table1. Saturated/unsaturated torque structure and constant-inertia limits inform the teaching example; the full mass-dependent synthetic cluster is a separate model.

  3. Hall et al. (2021), Weakened magnetic braking supported by asteroseismic rotation rates of Kepler dwarfs, arXiv2104.10919v1

    Figures2–4 and Methods1–2. Read sample cuts, seismic/spot clocks and mixture inference together. The98.4%concerns posterior mass above a mixture weight, not a universal star fraction.

FeedbackSomething broken, confusing, or missing on this page?Tell us
What kind of feedback?

Sent with this page’s address, its Lab ID, the site language and your device type. Please leave out personal details. Feedback is kept for up to 12 months.Privacy policy