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research24 min studyLesson 908 in Physics

Account for angular momentum during a pulsar glitch

Connect superfluid vortices, a rotating crust, and rapid angular-momentum transfer while retaining coupling and entrainment uncertainty.

Before you begin
On this page

The question for today

What you’ll understand

  • Use a two-component conservation law for a short glitch.
  • Relate superfluid rotation to quantized vortex density.
  • Distinguish reservoir bookkeeping from a complete trigger mechanism.

A pulsar glitch is a sudden increase in the rotation frequency traced by its electromagnetic pulse train, often followed by partial recovery. A leading interpretation transfers angular momentum from a more rapidly rotating internal superfluid to the component coupled to the observed crust. Conservation provides useful quantitative constraints, but the participating inertia, pinning forces, entrainment, and trigger remain separate physical questions. The measured frequency jump is not by itself a direct measurement of crustal mass.

Identify the solid and superfluid components

The neutron-star crust contains a lattice of neutron-rich nuclei and electrons, with unbound neutrons in its inner region. Below the relevant pairing transition some neutrons can form a superfluid. The crust and magnetically or collisionally coupled material need not have the same effective moment of inertia as the entire star on a rapid glitch time scale. Neutrons in the core can also participate depending on their coupling and the assumed glitch mechanism.

A rotating neutral superfluid supports angular momentum through quantized vortices rather than rigid-body rotation of an irrotational phase everywhere. For neutron pairs the circulation quantum is Planck’s constant divided by twice the neutron mass. Coarse-graining many approximately straight vortices gives an areal density proportional to rotation frequency. This macroscopic relation does not determine a pinning force, a crust-breaking strain, or the probability of an avalanche; those require microscopic and dynamical input.

κn=h2mn,nv=2Ωsκn\kappa_n=\frac{h}{2m_n},\qquad n_v=\frac{2\Omega_s}{\kappa_n}
κn is circulation in m² s⁻¹, nv is vortices per square meter, and Ωs is superfluid angular frequency. The relation assumes coarse-grained uniform rotation by singly quantized neutron-pair vortices; curved or entrained configurations need the appropriate momentum-circulation treatment.↔ Scroll wide equations horizontally.

Conserve angular momentum over the rapid transfer

In a minimal two-component model, let Ic describe the observed, rapidly coupled component and Is a superfluid reservoir. Over a sufficiently short event neglect external torque and changes in those inertias. An increase in the observed angular frequency must then be accompanied by a decrease in the reservoir’s angular frequency. The relevant lag is reservoir frequency minus crust frequency. Both changes reduce that lag, so accounting for only the reservoir’s own decrease underestimates the total lag release.

Entrainment makes momentum and transport velocity nontrivially related in a mixture or lattice. It can reduce the freely available reservoir compared with a naive fraction of crustal mass. Vortex pinning, creep, mutual friction, and magnetic interactions affect how much inertia participates and how recovery proceeds. A simplistic reservoir calculation therefore supplies a conservation requirement, not a proof that the crust alone can or cannot explain every observed event.

IcΔΩc+IsΔΩs=0,Δ(Ωs−Ωc)=ΔΩs−ΔΩcI_c\Delta\Omega_c+I_s\Delta\Omega_s=0,\qquad \Delta(\Omega_s-\Omega_c)=\Delta\Omega_s-\Delta\Omega_c
This is the ideal two-component bookkeeping law with fixed inertias and negligible integrated external torque during the event. Ic and Is are effective participating moments of inertia, not automatically the full crust and core inertias in an entrained relativistic star.↔ Scroll wide equations horizontally.

Read timing recovery as a constrained inverse problem

A frequency step, a change in spin-down rate, and several relaxation components can be inferred from timing data. Mapping those features into internal regions depends on coupling times and on how much material is effectively locked to the observed crust during each interval. A crustquake can also change the moment of inertia or trigger a vortex response. Evidence for angular-momentum exchange does not select one unique microscopic initiation process.

The primary review by Chamel and Haensel develops the crust composition, pairing, elasticity, entrainment, and vortex framework behind these interpretations. Its 2008 observational examples are historical benchmarks. Modern glitch activity and recovery studies still require uncertain pinning and coupling inputs. A convincing model should account for accumulated spin lag, available angular momentum, event sizes, and recovery without assigning all of those observations independently adjustable reservoirs that violate conservation.

Work it through

A small observed jump can require a larger reservoir change

Use the ideal two-component model with total inertia I=1038 kg m2I=10^{38}\,\mathrm{kg\,m^2}, Is=0.02II_s=0.02I, and Ic=0.98II_c=0.98I. Initially the observed frequency is 100 s−1100\,\mathrm{s^{-1}}. A glitch has ΔΩc/Ωc=10−6\Delta\Omega_c/\Omega_c=10^{-6}. Find both frequency changes and the change in lag.

  1. The observed increase is ΔΩc=100×10−6=10−4 s−1\Delta\Omega_c=100\times10^{-6}=10^{-4}\,\mathrm{s^{-1}}. Its angular-momentum gain is 0.98×1038×10−4=9.8×1033 kg m2 s−10.98\times10^{38}\times10^{-4}=9.8\times10^{33}\,\mathrm{kg\,m^2\,s^{-1}}.

  2. Conservation requires ΔΩs=−(0.98/0.02)10−4=−0.0049 s−1\Delta\Omega_s=-(0.98/0.02)10^{-4}=-0.0049\,\mathrm{s^{-1}}. The smaller reservoir changes its frequency much more than the coupled component.

  3. The lag change is −0.0049−0.0001=−0.0050 s−1-0.0049-0.0001=-0.0050\,\mathrm{s^{-1}}. A preexisting positive lag at least that large is needed if the reservoir is not to finish rotating more slowly than the crust.

The result

The observed component rises by 10⁻⁴ inverse seconds, the reservoir falls by 0.0049, and the lag falls by 0.0050. This is a conservation check for the stipulated participating inertias, not a trigger calculation.

The living notebook

Explore the model

Interactive lab

Interactive lab · oscillator

Inspect a recovery time scale in a restricted response

Reset and set ζ = 1 and ω₀ = 1. After one Step predict displacement 1.5 exp(−0.5), about 0.9098. Compare after a fresh Reset with ζ = 1 and ω₀ = 2, then one Step, predicting 2 exp(−1), about 0.7358. This classical critically damped displacement starts at one with zero velocity. It is not a vortex-unpinning event, two-reservoir angular-momentum transfer, or quantitative glitch recovery model.

Inspect a recovery time scale in a restricted responseUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. 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 forA glitch reservoir is constrained by angular-momentum conservation.

Pause & reason

Check your understanding

In the ideal short-event model, why does a small superfluid reservoir change its frequency more strongly?

Your turn

Try it yourself

Use κn=2×10−7 m2 s−1\kappa_n=2\times10^{-7}\,\mathrm{m^2\,s^{-1}} and Ωs=100 s−1\Omega_s=100\,\mathrm{s^{-1}} in a uniform unentrained vortex model. Find areal vortex density. At cylindrical radius one kilometer, what outward vortex speed would accompany smooth spin-down Ω˙s=−10−10 s−2\dot\Omega_s=-10^{-10}\,\mathrm{s^{-2}}, using vr=−rΩ˙s/(2Ωs)v_r=-r\dot\Omega_s/(2\Omega_s)?

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

Give me a hint

The given drift relation follows by conserving the vortex count inside a cylinder while its uniform areal density tracks rotation.

Compare with the solution

The areal density is 2(100)/(2×10−7)=109 m−22(100)/(2\times10^{-7})=10^9\,\mathrm{m^{-2}}. The outward speed is 1000×10−10/200=5×10−10 m s−11000\times10^{-10}/200=5\times10^{-10}\,\mathrm{m\,s^{-1}}. A positive outward drift reduces the number of vortices within a fixed cylinder and therefore reduces rotation. This continuous coarse-grained estimate does not predict sudden unpinning, a glitch size distribution, or an entrainment-corrected circulation law; it checks the scale and direction of smooth vortex transport.

Keep the idea

Take it with you

  • A glitch reservoir is constrained by angular-momentum conservation.
  • Quantized circulation connects superfluid rotation to vortex density.
  • Entrainment, coupling and triggers remain additional physical inputs.

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