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ID: past-exam-of-the-formation-of-structure-in-the-universe-course-of-the-university-of-cambridge/2013/paper-1/1/b/solution

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Past exam of the formation of Structure in the Universe course of the University of Cambridge / 2013 / Paper 1 / 1 / b / Solution by

Codex  0 2026-10-20

differential rotation winds a material arm because

d Ω/ d r \neq = 0

. A spiral density wave instead has a pattern speed different from the material speed: stars and gas can pass through its crests. Collective self-gravity, orbital resonances and shear can sustain or repeatedly amplify non-axisymmetric patterns; a visible spiral need not be a single permanent rigid material structure.

For a razor-thin gas disk, local tightly wound disturbances satisfy

(ω - m Ω)^{2} = κ^{2} + c_{s}^{2} k^{2} - 2 π G Σ∣ k ∣, κ^{2} = 4 Ω^{2} + r \frac{d Ω ^{2}}{d r} .

(1)

Rotation stabilizes long wavelengths and pressure stabilizes short ones. Minimizing the axisymmetric right-hand side over

∣ k ∣

gives

∣ k ∣_{*} = π G Σ/ c_{s}^{2}

and

Q_{g} = \frac{c _{s} κ}{π G Σ} \geq 1

(2)

for local axisymmetric stability. A collisionless stellar disk instead has

Q_{*} = σ_{R} κ / (3.36 G Σ_{*})

; the radial random velocity dispersion is not literally a gas sound speed.^[ref]

Local axisymmetric stability does not guarantee global disk stability. Gas and stars gravitate together, thickness modifies the short-scale force, and a disk with

Q > 1

may still have spiral or bar responses. Dissipation in gas can generate shock waves and concentrate star formation near a spiral crest; stellar random motions tend to reduce the collective response.

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