Graduate Seminar: Compact Objects in Dense Stellar Systems
Omega Centauri Society Problem Sets

Problem Set 1 — Stellar Kinematics & Mass Estimation

Estimated time: 3–4 hours Total marks: 40 Tools: velocity-dispersion · anisotropy-degeneracy-explorer omegacentauri.me/pset-1-kinematics.html
Learning objectives
  1. Apply the M–σ relation to predict a cluster's central velocity dispersion from an assumed IMBH mass
  2. Assess how velocity anisotropy (β) can mimic the kinematic signature of a central point mass
  3. Quantify the mass resolution needed to distinguish the Häberle (2024) and Baumgardt (2017) models
  4. Evaluate what follow-up observations would resolve the current kinematic tension

Background

Omega Centauri (NGC 5139) hosts the strongest current evidence for an intermediate-mass black hole (IMBH) in any globular cluster. Häberle et al. (2024) identified seven stars within 0.08 pc of the cluster center whose velocities require a central dark mass of M ≥ 8,200 M☉. That figure is a lower bound on the enclosed mass rather than a point estimate: the fast movers set a floor, and nothing in the measurement caps the mass from above. Baumgardt (2017) placed a 3σ upper limit of M < 3,000 M☉ from N-body modeling of the velocity dispersion profile. The two results do not overlap, and that tension is the central unresolved issue in the field.

The M–σ relation ties black hole mass to the host system's velocity dispersion. The calibration used throughout this set is Gültekin et al. (2009), Table 3, E sample, which is also what the OCS Velocity Dispersion tool implements:

M_BH = 10^8.12 M☉ × (σ / 200 km/s)^4.24 Inverted for σ: σ(M) = 200 km/s × (M / 10^8.12 M☉)^(1/4.24)

The relation was calibrated on galactic-nucleus black holes of 106–1010 M☉, three to six decades above a globular-cluster IMBH, with ~0.3 dex intrinsic scatter in MBH at fixed σ. Extrapolating it to cluster scales is an open question rather than an established result, and Problem 1 is partly about how far that extrapolation can be pushed.

Adopted values (used throughout PS 1–4)
  1. Distance to OC: d = 5.49 kpc (oMEGACat VI, arXiv:2503.04903; kinematic distance 5,494 ± 61 pc). 1 pc subtends 37.57″ at this distance.
  2. Central velocity dispersion: σ0 = 18.2 km/s (van de Ven et al. 2006, A&A 445:513). Two other figures circulate and are not merged into this one: Baumgardt & Hilker (2018) catalogue ~16.8 km/s for the global dispersion, and the OCS Jeans-model tools produce ~19–21 km/s in the central region depending on radius. Where an answer depends on σ0, the alternative is given alongside.
  3. Cluster mass and half-mass radius: Mcl = 4.0 × 106 M☉, rh = 7.0 pc (Harris/dynamical mass; Baumgardt & Hilker 2018 half-mass radius). The symbol rh is reserved for this radius; a black hole influence radius is written rinfl.
  4. Häberle mass: M ≥ 8,200 M☉, a lower bound. Baumgardt: M < 3,000 M☉, a 3σ upper limit.

The Keplerian velocity at projected radius R from a point mass M is:

v_K(R) = sqrt(G M / R) G = 6.674 × 10⁻¹¹ m³ kg⁻¹ s⁻² 1 M☉ = 1.989 × 10³⁰ kg 1 pc = 3.086 × 10¹⁶ m

The spherical Jeans equation (isotropic, power-law density profile) gives the projected velocity dispersion as a function of anisotropy parameter β (0 = isotropic, β > 0 = radially anisotropic, β < 0 = tangentially anisotropic). The OCS Anisotropy Degeneracy Explorer computes this numerically.

Problems

1. M–σ prediction at two mass hypotheses

The OCS Velocity Dispersion tool composes two published results: the virial theorem (σ² ≈ G Mcl / rh, from the cluster's total mass and half-mass radius) and the Gültekin et al. (2009) M–σ relation quoted in the Background. It takes a cluster mass as input and returns a predicted black hole mass. It does not accept an IMBH mass as input, so parts (b) and (c) run the relation backwards by hand and then use the tool's σ-override slider to confirm.

  1. Open the pre-filled link below (Mcl = 4.0 × 106 M☉, rh = 7.0 pc, σ override off). Record the virial σ and the M–σ predicted black hole mass. Compare the virial σ with the adopted σ0 = 18.2 km/s, and the predicted mass with the Häberle bound. Why does the virial route fail here? [5 marks]
  2. Invert the Gültekin relation by hand for M = 8,200 M☉ and for M = 3,000 M☉. What σ does each mass require, and by how many km/s do they differ? Then set the tool's σ-override slider (0.5 km/s steps) and find the half-km/s interval that brackets each mass. [5 marks]
  3. HST proper-motion surveys measure σ to roughly ±0.5 km/s. Is the σ difference between the two mass hypotheses distinguishable at that precision? Before answering yes, account for the 0.3 dex intrinsic scatter of the M–σ relation. [5 marks]
↗ Velocity Dispersion tool — pre-filled Mcl = 4.0 × 10⁶ M☉, rh = 7.0 pc, virial σ
Show solution
Part (a): σ_virial = sqrt(G M_cl / r_h) = sqrt(6.674×10⁻¹¹ × 4.0×10⁶ × 1.989×10³⁰ / (7.0 × 3.086×10¹⁶)) = sqrt(5.310×10²⁶ / 2.160×10¹⁷) = sqrt(2.458×10⁹) = 4.96×10⁴ m/s = 49.6 km/s M_BH = 10^8.12 × (49.6 / 200)^4.24 = 10^(8.12 + 4.24 × log₁₀ 0.248) = 10^(8.12 − 2.568) = 10^5.552 = 3.6×10⁵ M☉ The virial σ of 49.6 km/s is 2.7× the measured σ0 = 18.2 km/s, and the predicted 3.6 × 105 M☉ is 44× the Häberle bound and an order of magnitude above the highest claimed detection. The virial estimate is inflated because σ² ≈ G M / rh drops the structure-dependent virial coefficient (of order 0.4 for a Plummer sphere), and because it returns a mass-weighted global dispersion rather than the line-of-sight dispersion HST measures in the core. Feeding that σ into an exponent of 4.24 turns a factor of 2.7 into a factor of 2.74.24 ≈ 47.

Part (b): σ(M) = 200 × (M / 1.3183×10⁸)^(1/4.24) M = 8,200 M☉: 8200 / 1.3183×10⁸ = 6.220×10⁻⁵ log₁₀ = −4.2063; /4.24 = −0.9920 σ = 200 × 10^(−0.9920) = 200 × 0.10186 = 20.4 km/s M = 3,000 M☉: 3000 / 1.3183×10⁸ = 2.276×10⁻⁵ log₁₀ = −4.6428; /4.24 = −1.0950 σ = 200 × 10^(−1.0950) = 200 × 0.08034 = 16.1 km/s Δσ = 20.4 − 16.1 = 4.3 km/s Tool check with the σ-override slider: σ = 20.0 km/s returns 7,586 M☉ and σ = 20.5 km/s returns 8,423 M☉, so 8,200 M☉ falls in the 20.0–20.5 km/s interval. σ = 16.0 km/s returns 2,945 M☉ and σ = 16.5 km/s returns 3,356 M☉, bracketing 3,000 M☉. Both agree with the hand inversion.

Part (c): The 4.3 km/s separation is roughly 9× the ±0.5 km/s measurement precision, so the measurement is not the limiting factor. The relation is. A scatter of 0.3 dex in MBH at fixed σ is a factor of 2.0 in mass, which propagates to 0.3/4.24 = 0.071 dex in σ, or ±17 per cent, about ±3.4 km/s at 20 km/s. That scatter alone nearly spans the gap between the two hypotheses, and it is a floor derived from galactic-nucleus calibration data, before any question of whether the relation extends down to cluster masses at all. Baumgardt (2017) reached a low-mass conclusion from the full radial dispersion profile in N-body models rather than from a single global σ, which is the discriminating measurement M–σ cannot supply.
2. Keplerian velocities of the fast-moving stars

Häberle et al. (2024) report the seven fast-moving stars lie within a projected radius R ≈ 0.04–0.08 pc of the cluster center. Use the Keplerian velocity formula above (assuming circular orbits and M = 8,200 M☉).

  1. Compute v_K at R = 0.04 pc and at R = 0.08 pc. Express your answers in km/s. [5 marks]
  2. Häberle et al. report peak 3D velocities of ~50 km/s. Is this consistent with your Keplerian estimate for a bound orbit at these radii? [5 marks]
  3. If instead M = 3,000 M☉, what Keplerian velocity would you predict at R = 0.04 pc? Can this explain the observed ~50 km/s stars? [5 marks]
Show solution
Part (a): M = 8200 × 1.989×10³⁰ = 1.631×10³⁴ kg R₁ = 0.04 × 3.086×10¹⁶ = 1.234×10¹⁵ m R₂ = 0.08 × 3.086×10¹⁶ = 2.469×10¹⁵ m v_K(R₁) = sqrt(6.674e-11 × 1.631e34 / 1.234e15) = sqrt(8.823e8) ≈ 29,700 m/s ≈ 29.7 km/s v_K(R₂) = sqrt(6.674e-11 × 1.631e34 / 2.469e15) = sqrt(4.411e8) ≈ 21,000 m/s ≈ 21.0 km/s So: ~30 km/s at 0.04 pc, ~21 km/s at 0.08 pc.

Part (b): The Keplerian circular-orbit speed at 0.04 pc is ~30 km/s. The reported ~50 km/s is for the 3D space velocity; for a circular orbit seen at some inclination angle, the projected velocity can differ. A 50 km/s 3D speed at 0.04 pc requires M ~ (50/v_K)² × 8200 ≈ 2.8 × 8200 ≈ 23,000 M☉ for a purely circular orbit, above the Häberle lower bound (which the bound permits, since it caps nothing from above). However, for eccentric orbits, stars can reach higher velocities near periapsis than a circular orbit of the same semi-major axis, so 50 km/s 3D velocities are consistent with M ≈ 8,200 M☉ on eccentric orbits at small periapsis.

Part (c): For M = 3,000 M☉: v_K(0.04 pc) = 29.7 × sqrt(3000/8200) ≈ 29.7 × 0.605 ≈ 18.0 km/s This is far below 50 km/s. A 3,000 M☉ central mass cannot produce bound stars at 0.04 pc with 3D velocities of ~50 km/s; such stars would need to be unbound (escaping the cluster), which is dynamically implausible in large numbers.
3. The anisotropy degeneracy

A centrally-concentrated population of stellar-mass black holes, or radial velocity anisotropy (β > 0), can produce a central σ cusp that mimics an IMBH. Use the OCS Anisotropy Degeneracy Explorer.

  1. The tool always draws two curves: the current model (solid) and a no-IMBH model with the same β and Mrem (dashed). Load the link below and read the solid curve's σ at the innermost plotted radius, R = 0.05 pc, for MIMBH = 8,200 M☉ at β = 0. Then raise β in steps of 0.01 and read the dashed no-IMBH curve until it reaches that same σ. What β is required? [5 marks]
  2. The IMBH hypothesis predicts a Keplerian σ ∝ R−1/2 cusp inside the influence radius rinfl = G M / σ02. Compute rinfl for M = 8,200 M☉. Anisotropy produces a much shallower central profile. What observational signature would most cleanly distinguish the two cases? [5 marks]
↗ Anisotropy Degeneracy Explorer — pre-filled M = 8,200 M☉, Mrem = 0, β = 0, d = 5.49 kpc
Show solution
Part (a): With the tool's shipped cluster model (Plummer, Mcl = 4.0 × 106 M☉, scale radius 7.0 pc) the isotropic no-IMBH curve sits at σ = 19.07 km/s at R = 0.05 pc, and adding an 8,200 M☉ point mass lifts it to 20.12 km/s. Raising β on the no-IMBH curve gives 19.86 km/s at β = 0.04, 20.07 at β = 0.05 and 20.29 at β = 0.06, so β ≈ 0.05 reproduces the IMBH signal. Two things are worth noticing. The required β is small and entirely ordinary for a globular cluster, so the degeneracy does not live in an exotic corner of parameter space. And the whole IMBH signal is 1.05 km/s, roughly twice the ±0.5 km/s precision of the measurement it has to be extracted from.

Part (b): r_infl = G M / σ₀² = 6.674×10⁻¹¹ × 8200 × 1.989×10³⁰ / (1.82×10⁴)² = 1.0885×10²⁴ / 3.3124×10⁸ = 3.286×10¹⁵ m = 0.106 pc (4.0″ at 5.49 kpc) At the Baumgardt & Hilker σ0 = 16.8 km/s the same expression gives 0.125 pc. Neither value is anywhere near 1 pc, so the discriminating region is small.

The clearest distinguishing signature is the shape of the σ profile inside rinfl. An IMBH produces a Keplerian σ ∝ R−1/2 rise there and nothing outside it; radial anisotropy raises σ across a decade or more in radius with no characteristic scale at 0.1 pc. The tool's own curves show why the innermost plotted radius is not enough: at R = 0.05 pc the β = 0.05 and IMBH models agree to 0.05 km/s, and they still agree to within 0.3 km/s out to 0.3 pc. Discriminating requires resolved kinematics inside rinfl, at R < 0.05 pc (~2″), which is ELT/MICADO territory. There the two hypotheses separate a second way as well: the IMBH case predicts isotropic orbits with a Keplerian velocity distribution, while the anisotropy case predicts a measurably radially-biased tangential-to-radial velocity ratio.

References