Protostars and Accretion Disks

A protostar is the accreting seed of a star: a hydrostatic core buried at the centre of a collapsing molecular-cloud fragment, still gaining almost all of its final mass from the envelope raining down around it. The collapse itself is easy — once a core exceeds its Jeans mass, gravity wins and the gas falls nearly in free-fall. The hard part, and the theme of this article, is angular momentum. Even the languid rotation of a cloud core, inherited from Galactic shear and turbulence, carries far more spin per gram than a star can hold. If it were conserved all the way down, the infalling gas would spin up so violently it could never reach the stellar surface. So it does not fall straight in: it falls until centrifugal force balances gravity — at the centrifugal radius, tens of astronomical units out — and there it piles into a rotationally supported accretion disk. The disk is not a bystander; it is the machine that makes a star. Neighbouring rings shear against one another through an effective viscosity that ferries mass inward while carrying angular momentum outward, so gas spirals down onto the star while its excess spin is dumped at large radius or flung out along magnetically launched jets. Meanwhile the energy budget is dominated not by fusion — which has not yet ignited — but by the gravitational energy liberated as gas slams onto the star, an accretion luminosity $L_{\mathrm{acc}}=GM\dot M/R$ that can outshine the nascent photosphere. Reprocessed by dust, that glow escapes only in the far-infrared and submillimetre — which is why the youngest protostars are invisible to the eye and are catalogued instead by the shape of their spectral energy distribution.

Key takeaways

A protostar grows by accretion; during the embedded phase its luminosity is gravitational infall, $L_{\mathrm{acc}}=GM\dot M/R$ , not nuclear burning — which has not started.
A rotating core cannot collapse straight to a point: conserved angular momentum halts infall at the centrifugal radius $r_{\mathrm c}=j^{2}/GM$ , tens of AU out, where the gas forms a Keplerian disk.
The disk carries mass inward and angular momentum outward through an effective (turbulent, magnetic) viscosity — the resolution of the classical angular-momentum problem.
A steady thin disk radiates as $T_{\mathrm{eff}}\propto r^{-3/4}$ , independent of the uncertain viscosity, giving the broad flat infrared spectrum that betrays a disk.
The embedded object is dust-shrouded and seen only in the infrared and submillimetre; the Class 0/I/II/III sequence tracks the envelope dispersing and the disk emerging.
MRI turbulence, gravitational torques, and bipolar jets are the mechanisms that let angular momentum leave so mass can arrive.
Key idea: a rotating cloud core cannot collapse directly to a star — angular-momentum conservation forces the infalling gas into a Keplerian disk, and the outward transport of angular momentum by viscous and magnetic stresses lets mass accrete while $L_{\mathrm{acc}}=GM\dot M/R$ powers the embedded object.

The core insight

A rotating cloud core cannot collapse directly to a star: angular-momentum conservation forces the infalling gas into a rotationally supported disk, and it is the outward transport of angular momentum by viscous and magnetic stresses that lets mass reach the protostar while the accretion luminosity $GM\dot M/R$ powers the embedded object.

Where this lives today

Protostellar disks are no longer inferred — they are imaged. ALMA resolves disks tens of AU across: the 2014 HL Tau image revealed concentric gaps and rings in a disk barely a million years old, and the DSHARP survey showed such substructure is ubiquitous, most likely carved by forming planets. Molecular-line maps recover the tell-tale $v_\phi\propto r^{-1/2}$ rotation, turning a disk into a scale that weighs the embedded star. Two puzzles drive current work. The angular-momentum transport the $\alpha$ -model hides is patchy: the MRI needs ionised gas, but the cold, dense midplane is nearly neutral, creating "dead zones" where non-ideal MHD and magnetically launched disk winds — not turbulence — may do the transporting. And the luminosity problem persists: embedded protostars are fainter than steady accretion predicts, pointing to episodic accretion punctuated by FU Orionis outbursts, in which a disk instability dumps mass and the luminosity leaps a hundredfold for decades. The James Webb Space Telescope now probes the inner-disk chemistry of the youngest Class 0 sources, testing where and how fast the first disks assemble.

The physical picture

Take a parcel of gas at the edge of a slowly rotating core, carrying fixed specific angular momentum $j=v_\phi r$ . As gravity pulls it inward and $r$ shrinks, the rotation speed $v_\phi=j/r$ must climb — like a spinning skater pulling in her arms — so the centrifugal acceleration $j^2/r^3$ rises far faster than gravity $GM/r^2$ and eventually overtakes it, halting the infall. That crossover is the centrifugal radius; interior to it the gas can no longer fall straight and instead orbits, accumulating into a thin, rotationally supported disk. The first figure fixes the scale: the barrier sits thousands of stellar radii out.

Figure

Infalling envelope gas conserves specific angular momentum j = v_φ r, so v_φ climbs as 1/r until centrifugal force balances gravity at the centrifugal radius r_c = j²/GM ≈ 30 AU — thousands of times the stellar radius R ≈ 3 R_⊙ ≈ 0.014 AU — where it piles into a Keplerian disk while bipolar jets carry angular momentum out along the axis

Now watch the disk work. It is a slow-motion conveyor belt for spin: rings orbit at different Keplerian speeds, so any friction between them — turbulence from the magnetorotational instability, magnetic tension, or self-gravity — brakes the inner gas (which sinks) and speeds the outer gas (which drifts out). Mass flows in, angular momentum flows out, and almost all the disk's mass reaches the star while a whisker at large radius carries the excess spin. As gas drains inward it releases gravitational energy locally, heating each annulus, so the disk glows — hottest at the centre, cooling outward. That temperature run is the disk's fingerprint, and the second figure is what an observer measures.

Figure

The steady thin-disk temperature runs as T ∝ r^−3/4 on a log–log plot: anchoring T ≈ 150 K at 1 AU for accretion rate Ṁ = 10⁻⁷ M_⊙ yr⁻¹ onto a solar-mass star gives ≈ 840 K at 0.1 AU and ≈ 27 K at 10 AU — the broad, flat infrared spectrum that reveals a disk

The core equations

The protostar's embedded-phase luminosity is set by infall, not fusion. A gram falling from rest at infinity to the stellar surface at radius $R$ releases $GM/R$ ; multiplying by the rate at which mass arrives gives the accretion luminosity.

L_{\mathrm{acc}}=\frac{GM\dot M}{R} \qquad (1)

where: $L_{\mathrm{acc}}$ — accretion luminosity; $G$ — gravitational constant; $M$ — protostellar mass; $\dot M$ — accretion rate; $R$ — protostellar radius. A heavier, more compact star and a heavier rain of gas all brighten it: a solar-mass protostar swallowing $10^{-5}\,M_\odot\,\mathrm{yr}^{-1}$ at $R\approx3\,R_\odot$ reaches $\sim\!100\,L_\odot$ , dwarfing its feeble pre-fusion output. Angular momentum sets where infall stops. A parcel with $j=v_\phi r$ orbits faster as it falls until the centrifugal acceleration $j^2/r^3$ equals gravity $GM/r^2$ ; that balance gives the centrifugal radius, the outer scale of the disk.

r_{\mathrm c}=\frac{j^{2}}{GM},\qquad j=v_\phi\, r \qquad (2)

where: $r_{\mathrm c}$ — centrifugal radius; $j$ — specific angular momentum of the infalling gas; $v_\phi$ — azimuthal velocity; $r$ — cylindrical radius; $M$ — central (enclosed) mass. Because $r_{\mathrm c}\propto j^2$ , even modest rotation places the barrier tens of AU out — thousands of stellar radii — so a disk is essentially unavoidable. Interior to $r_{\mathrm c}$ the gas is on near-circular orbits, and while the central mass dominates the disk's weight those orbits are Keplerian — the single fact ( $\Omega\propto r^{-3/2}$ ) behind all the disk's shear and transport.

v_{\mathrm K}(r)=\sqrt{\frac{GM}{r}},\qquad \Omega_{\mathrm K}(r)=\sqrt{\frac{GM}{r^{3}}} \qquad (3)

where: $v_{\mathrm K}$ — Keplerian orbital speed; $\Omega_{\mathrm K}$ — Keplerian angular velocity; $M$ — central mass; $r$ — disk radius. A resolved $v_\phi\propto r^{-1/2}$ rotation curve is the signature of a rotationally supported disk and a direct weigh-scale for the hidden star. Molecular viscosity is far too weak to drain a disk in the age of the Universe, so accretion needs a turbulent viscosity. The Shakura–Sunyaev prescription parameterises our ignorance in one number $\alpha\lesssim1$ , tying the viscosity to the largest eddies the disk can support.

\nu=\alpha\,c_s H,\qquad H=\frac{c_s}{\Omega_{\mathrm K}} \qquad (4)

where: $\nu$ — effective kinematic viscosity; $\alpha$ — dimensionless viscosity parameter ( $\lesssim1$ ); $c_s$ — sound speed; $H$ — disk scale height. The product $c_s H$ is (largest eddy speed) $\times$ (size), so $\alpha$ measures turbulence as a fraction of that maximum; the physical engine is usually the magnetorotational instability. In steady state the mass flux through every radius is the same constant $\dot M$ , and equating local viscous heating to blackbody re-radiation fixes the surface temperature — with the uncertain viscosity cancelling out.

\dot M=3\pi\nu\Sigma,\qquad T_{\mathrm{eff}}(r)=\left[\frac{3GM\dot M}{8\pi\sigma r^{3}}\left(1-\sqrt{\frac{R_\star}{r}}\right)\right]^{1/4} \qquad (5)

where: $\dot M$ — steady mass flux; $\nu$ — viscosity; $\Sigma$ — surface density; $\sigma$ — Stefan–Boltzmann constant; $R_\star$ — inner radius. Far from the inner edge the bracket $\to1$ and $T_{\mathrm{eff}}\propto r^{-3/4}$ , depending only on $M$ , $\dot M$ , $r$ — the profile in the second figure. Away from steady state, mass and angular-momentum conservation for a thin Keplerian disk collapse into one nonlinear diffusion equation for the surface density.

\frac{\partial\Sigma}{\partial t}=\frac{3}{r}\frac{\partial}{\partial r}\left[r^{1/2}\frac{\partial}{\partial r}\left(\nu\Sigma\,r^{1/2}\right)\right] \qquad (6)

where: $\Sigma$ — surface density; $\nu$ — viscosity; $r$ — radius; $t$ — time. An initial ring spreads both ways: most mass diffuses inward while a small tail carries the angular momentum outward — mass in, spin out, in one line.

Derivation AWhy a rotating core must build a disk

The goal is to turn the qualitative worry — that a spinning cloud cannot collapse to a point — into a quantitative criterion, by comparing the specific angular momentum a molecular-cloud core carries with the most a star can hold at its surface, and showing the mismatch is enormous. We estimate the core's $j$ , note it is conserved during free-fall, locate the centrifugal radius where infall stalls, and read off that a disk, not a bare star, is the generic outcome of collapse.

Step 1. Estimate the core's spin. A core of radius $R_0\sim0.05\,\mathrm{pc}$ rotating at the observed $\Omega_0\sim10^{-14}\,\mathrm{s}^{-1}$ has equatorial $j\sim\Omega_0 R_0^2\sim2\times10^{16}\,\mathrm{m^2\,s^{-1}}$ (a few $\times10^{20}\,\mathrm{cm^2\,s^{-1}}$ , matching core maps). Why: this is the reservoir the collapsing gas starts with, set by Galactic shear and turbulence.

Step 2. Compare with what a star can hold. The most a solar-mass star of radius $3\,R_\odot$ can carry without flying apart is $j_\star=\sqrt{GM R_\star}\approx5\times10^{14}\,\mathrm{m^2\,s^{-1}}$ — so the core has $j/j_\star\approx45$ times more. Why: keeping its spin, the gas would reach breakup long before the surface; the star cannot swallow it directly.

Step 3. Conserve $j$ and find where infall halts. Absent torques each element keeps $j$ and stops where centrifugal support equals gravity, $j^2/r^3=GM/r^2$ , i.e. $r_{\mathrm c}=j^2/GM\approx30\,\mathrm{AU}$ — some two thousand stellar radii out. Why: the abstract inequality becomes a length, and it is the observed size scale of young disks.

Step 4. Read off the resolution. Because $r_{\mathrm c}\gg R_\star$ , incoming gas accumulates in a disk; reaching the star then demands angular momentum be exported outward — the job of viscous stresses (MRI turbulence, gravitational torques) and magnetically launched jets. Why: the disk is the necessary intermediary that lets a spinning cloud become a slowly spinning star. *The three key numbers check out: $j\sim\Omega_0R_0^2\approx2\times10^{16}\,\mathrm{m^2\,s^{-1}}$ , $j_\star=\sqrt{GMR_\star}\approx5\times10^{14}\,\mathrm{m^2\,s^{-1}}$ , and $r_{\mathrm c}=j^2/GM\approx30\,\mathrm{AU}\approx2000\,R_\star$ — the quantitative statement of the classic angular-momentum problem.*

Go deeper · Derivation B · 1 minThe steady thin disk and its flat infrared spectrum

Derivation BThe steady thin disk and its flat infrared spectrum

The goal is to derive the temperature profile of a steady, thin, Keplerian disk and show that $T_{\mathrm{eff}}\propto r^{-3/4}$ is independent of the deeply uncertain viscosity — which is what makes disk spectra predictable at all. We fix the mass flux from steady state, compute the viscous heating from the shear, balance it against blackbody re-radiation, and eliminate the viscosity by substituting the mass flux, leaving a profile set only by central mass, accretion rate, and radius.

Step 1. Fix the flux. In steady state the diffusion equation integrates to a constant inward mass flux $\dot M=-2\pi r\Sigma v_r=3\pi\nu\Sigma$ , with the constant set by a zero-torque boundary at $R_\star$ . Why: the same $\dot M$ passes every ring, so $\nu\Sigma$ is pinned to a measurable rate rather than left free.

Step 2. Heat each ring. The viscous dissipation per area, set by the Keplerian shear, is $D(r)=\tfrac{9}{8}\nu\Sigma\,\Omega_{\mathrm K}^2$ ; inserting $\nu\Sigma=\dot M/3\pi$ gives $D(r)=\dfrac{3GM\dot M}{8\pi r^3}\left(1-\sqrt{R_\star/r}\right)$ . Why: substituting the flux removes $\nu$ entirely — the heating no longer depends on the unknown turbulence.

Step 3. Radiate it away. Each optically thick annulus re-emits as a blackbody from both faces, $2\sigma T_{\mathrm{eff}}^4=D(r)$ , giving $T_{\mathrm{eff}}=\left[\dfrac{3GM\dot M}{8\pi\sigma r^3}\left(1-\sqrt{R_\star/r}\right)\right]^{1/4}\to r^{-3/4}$ for $r\gg R_\star$ . Why: the outer disk is a nested set of ever-cooler blackbodies whose sum is the broad, flat infrared SED that observers use to flag a disk.

Step 4. Account the energy. Integrating, $L_{\mathrm{disk}}=2\!\int_{R_\star}^{\infty}D(r)\,2\pi r\,\mathrm{d}r=\dfrac{GM\dot M}{2R_\star}$ , exactly half the total accretion energy $GM\dot M/R_\star$ ; the rest is released in the boundary layer where disk meets star. Why: the disk radiates only half the budget — the hot inner boundary layer carries the rest, a clue when interpreting ultraviolet excesses. *The three results are mutually consistent: $\dot M=3\pi\nu\Sigma$ , $T_{\mathrm{eff}}\propto r^{-3/4}$ , and $L_{\mathrm{disk}}=GM\dot M/2R_\star$ ; notably, the $r^{-3/4}$ temperature law is viscosity-independent, as the derivation shows.*

Go deeper · Derivation C · 1 minThe timescale hierarchy that lets a disk sit still

Derivation CThe timescale hierarchy that lets a disk sit still

The goal is to justify treating the disk as a quasi-steady, near-Keplerian structure by ordering its three clocks — the orbital, thermal, and viscous times — and showing they are cleanly separated for a thin $\alpha$ -disk. We write each timescale in terms of $\Omega_{\mathrm K}$ , $\alpha$ , and the aspect ratio $H/r$ , and show that a thin ( $H\ll r$ ), weakly turbulent ( $\alpha<1$ ) disk relaxes dynamically and thermally long before it can drain — exactly the condition that legitimises the steady-state derivation.

Step 1. The orbital clock. $t_{\mathrm{dyn}}=\Omega_{\mathrm K}^{-1}=\sqrt{r^3/GM}$ ; vertical hydrostatic balance is set up within a fraction of an orbit. Why: the fastest clock, so vertical structure is established almost instantly.

Step 2. The thermal clock. The time to radiate the stored heat is $t_{\mathrm{th}}\sim1/(\alpha\Omega_{\mathrm K})$ , longer than dynamical by $1/\alpha$ . Why: heating and cooling balance on this clock, so Derivation 2's temperature profile is a valid snapshot.

Step 3. The viscous clock. The time for mass to spiral in is $t_{\mathrm{visc}}\sim r^2/\nu=\dfrac{1}{\alpha\Omega_{\mathrm K}}\left(\dfrac{r}{H}\right)^2$ , longer than thermal by $(r/H)^2$ . Why: with $H/r\ll1$ , accretion is glacially slow next to orbiting and cooling — the disk drains over millions of years.

Step 4. Order them. Together $t_{\mathrm{dyn}}\ll t_{\mathrm{th}}\ll t_{\mathrm{visc}}$ , so the disk stays near-Keplerian and near-thermal-equilibrium while it evolves quasi-statically. Why: precisely the licence to use the steady-state disk, and it predicts the few-Myr protoplanetary-disk lifetimes. *The timescale ordering $t_{\mathrm{dyn}}\ll t_{\mathrm{th}}\ll t_{\mathrm{visc}}$ holds with explicit ratios $1:\alpha^{-1}:\alpha^{-1}(r/H)^2$ , consistent with the observed $\sim$ Myr disk-dispersal times.*

Limiting cases

Case	Result	Meaning
Free-fall collapse	$t_{\mathrm{ff}}=\sqrt{\dfrac{3\pi}{32\,G\rho}}$	Gravity overwhelms pressure and rotation; collapse runs on the free-fall time set only by density $\rho$ , capping how fast the envelope rains down.
Rotationally supported disk	$v_\phi(r)=\sqrt{\dfrac{GM}{r}}$	Interior to $r_{\mathrm c}$ , radial infall is replaced by slow viscous inflow on near-Keplerian orbits.
Accretion-dominated luminosity	$L_{\mathrm{acc}}=\dfrac{GM\dot M}{R}\gg L_{\star,\mathrm{nuc}}$	In the Class 0/I phase infall powers the object because hydrogen fusion has not ignited ( $L_{\star,\mathrm{nuc}}\approx0$ ).
Optically thick first core	$\tau=\int\kappa\rho\,\mathrm{d}s\gtrsim1$	Dust makes the gas opaque, heat is trapped, collapse stalls, and a hydrostatic first core forms.
Far-field disk annulus	$T_{\mathrm{eff}}\propto r^{-3/4}$	For $r\gg R_\star$ the boundary term $\to1$ ; the temperature, hence the IR spectrum, depends only on $M$ , $\dot M$ , $r$ , not $\nu$ .

Common trapThe accretion luminosity is the star's own power, and gravity rivals fusion

Two related traps. First, $L_{\mathrm{acc}}=GM\dot M/R$ is not the star's photospheric or nuclear luminosity — it is gravitational energy liberated by infall, and it dominates the embedded phase only because deuterium and hydrogen burning have barely begun. Second, do not read "accretion powers the star" as "gravity releases as much energy as fusion will." Per gram, infall to a bloated protostar yields $GM/R\approx6\times10^{10}\,\mathrm{J\,kg^{-1}}$ , some ten thousand times less than hydrogen fusion's $\sim6\times10^{14}\,\mathrm{J\,kg^{-1}}$ — which is why gravitational (Kelvin–Helmholtz) contraction supports a star for only tens of millions of years while fusion lasts billions. Accretion dominates now because fusion is switched off, not because gravity out-supplies the nucleus. A third, geometric trap: real disks rotate differentially and nearly Keplerian, so "the disk rotates rigidly with the star" is wrong — the shear is the whole engine.

Worked examples

1 · The accretion luminosity of an embedded protostar (fully worked)

A solar-mass protostar of radius $R=3\,R_\odot$ accretes at $\dot M=10^{-5}\,M_\odot\,\mathrm{yr}^{-1}$ . Find its accretion luminosity and compare it with the Sun.

Step 1. Convert to SI. With $M_\odot=1.989\times10^{30}\,\mathrm{kg}$ and $1\,\mathrm{yr}=3.156\times10^{7}\,\mathrm{s}$ , $\dot M=10^{-5}M_\odot/\mathrm{yr}=\mathbf{6.30\times10^{17}\,kg\,s^{-1}}$ (about twenty Earth masses a year).

Step 2. Fix the radius. $R=3\,R_\odot=2.087\times10^{9}\,\mathrm{m}$ — protostars are bloated, several times their eventual main-sequence size.

Step 3. Apply (1): $L_{\mathrm{acc}}=GM\dot M/R=(6.674\times10^{-11})(1.989\times10^{30})(6.30\times10^{17})/(2.087\times10^{9})=\mathbf{4.0\times10^{28}\,W}$ .

Step 4. Compare. Dividing by $L_\odot=3.828\times10^{26}\,\mathrm{W}$ gives $\mathbf{L_{\mathrm{acc}}\approx105\,L_\odot}$ : the still-forming star outshines the present Sun a hundredfold, yet buried in dust this power emerges only in the far-infrared. Per gram the yield is $GM/R\approx6\times10^{10}\,\mathrm{J\,kg^{-1}}$ , far below fusion's $\sim6\times10^{14}$ — the object is luminous because $\dot M$ is huge, not because gravity beats the nucleus.

2 · The angular-momentum problem in numbers (fully worked)

A core of radius $R_0=0.05\,\mathrm{pc}$ rotates at $\Omega_0=10^{-14}\,\mathrm{s}^{-1}$ and collapses onto a solar-mass protostar of radius $3\,R_\odot$ . Where does the gas stall, and by how much does it overshoot what the star can hold?

Step 1. Specific angular momentum of the gas. With $R_0=0.05\,\mathrm{pc}=1.54\times10^{15}\,\mathrm{m}$ , $j\sim\Omega_0R_0^2=10^{-14}\times(1.54\times10^{15})^2=\mathbf{2.4\times10^{16}\,m^2\,s^{-1}}$ .

Step 2. What the star can hold. The surface-Keplerian value is $j_\star=\sqrt{GMR_\star}=\mathbf{5.3\times10^{14}\,m^2\,s^{-1}}$ , so the gas carries $j/j_\star\approx\mathbf{45}$ times too much spin to land directly.

Step 3. Centrifugal radius, from (2): $r_{\mathrm c}=j^2/GM=(2.4\times10^{16})^2/[(6.674\times10^{-11})(1.989\times10^{30})]=4.3\times10^{12}\,\mathrm{m}=\mathbf{29\,AU}$ .

Step 4. Interpret. The stellar radius is only $0.014\,\mathrm{AU}$ , so $\mathbf{r_{\mathrm c}/R_\star\approx2000}$ : infalling gas builds a disk tens of AU across and reaches the star only after shedding almost all its angular momentum — the reservoir the observed protoplanetary disks represent.

3 · Disk temperature and the snow line

For the same star accreting at the gentler $\dot M=10^{-7}\,M_\odot\,\mathrm{yr}^{-1}$ (a typical T Tauri rate), estimate the disk's effective temperature at 1 AU and locate where water ice can condense.

Step 1. In SI, $\dot M=10^{-7}M_\odot/\mathrm{yr}=6.30\times10^{15}\,\mathrm{kg\,s^{-1}}$ and $r=1\,\mathrm{AU}=1.496\times10^{11}\,\mathrm{m}$ , far outside $R_\star$ so the boundary term is negligible.

Step 2. Apply the far-field form of equation (5), $T_{\mathrm{eff}}=\left(3GM\dot M/8\pi\sigma r^3\right)^{1/4}$ with $\sigma=5.67\times10^{-8}$ : numerator $3GM\dot M=2.5\times10^{36}$ , denominator $8\pi\sigma r^3=4.8\times10^{27}$ , so $T_{\mathrm{eff}}^4=5.3\times10^{8}\,\mathrm{K^4}$ and $\mathbf{T_{\mathrm{eff}}\approx150\,K}$ .

Step 3. Scale with $r^{-3/4}$ : $T\approx840\,\mathrm{K}$ at $0.1\,\mathrm{AU}$ and $T\approx27\,\mathrm{K}$ at $10\,\mathrm{AU}$ — the anchor points of the second figure.

Step 4. Interpret. Water ice sublimates near $150{-}170\,\mathrm{K}$ , so the snow line sits around 1 AU for this rate: rocky planetesimals inside, icy ones outside — a boundary this simple viscous-heating estimate already places in the right region (stellar irradiation shifts it modestly outward).

Physics of the portrait

The accretion disk is where celestial mechanics, thermodynamics, and magnetohydrodynamics meet — an idea two centuries in the making, from a philosopher's rotating nebula to a computed, turbulent structure.

Origins · 1 minaccretion disk and how this was discovered

accretion disk

the thin viscous disk, ν = α c_s H, T ∝ r^−3/4 (Shakura and Sunyaev 1973) · thin-disk accretion theory and star formation

The disk began as a philosophical guess. In 1796 Pierre-Simon Laplace's nebular hypothesis pictured the Solar System condensing from a rotating, cooling cloud that shed rings as it contracted — the first recognition that rotation forces a collapsing cloud to flatten rather than fall to a point. For a century and a half the idea languished on the angular-momentum objection: the Sun holds almost none of the system's spin, and no one could say where it had gone. The modern theory arrived from an unexpected direction — X-ray binaries. In 1973 Nikolai Shakura and Rashid Sunyaev wrote down the thin accretion disk, parameterising the turbulent viscosity by a single number, $\nu=\alpha c_s H$ , and deriving the profile $T\propto r^{-3/4}$ any steady disk must radiate. A year later Donald Lynden-Bell and Jim Pringle showed such a disk spreads — mass in, angular momentum out — resolving Laplace's paradox: the lost spin went to the far edge of the disk and beyond. Frank Shu's 1977 inside-out collapse supplied the protostellar accretion rate and, with Fred Adams and Susana Lizano in 1987, assembled the standard picture of low-mass star formation with the disk as mediator. The last piece was the viscosity itself: in 1991 Steven Balbus and John Hawley showed that a weak magnetic field in a differentially rotating disk is violently unstable — the magnetorotational instability — supplying exactly the turbulence Shakura and Sunyaev had only named. What Laplace sketched as rings is now a computed, magnetised disk, and it is where planets are born.

History

1796

— Laplace's nebular hypothesis: a rotating contracting cloud sheds rings and flattens

1973

— Shakura and Sunyaev formulate the α-viscosity thin accretion disk

1974

— Lynden-Bell and Pringle show viscous disks spread, carrying mass in and angular momentum out

1977

— Shu's inside-out collapse fixes the protostellar accretion rate

1991

— Balbus and Hawley identify the magnetorotational instability as the disk's viscosity

Going further

Texts. Carroll and Ostlie, An Introduction to Modern Astrophysics, 2nd ed., Ch. 12 · Frank, King and Raine, Accretion Power in Astrophysics, 3rd ed. · Hartmann, Accretion Processes in Star Formation, 2nd ed. · Armitage, Astrophysics of Planet Formation, 2nd ed. · Stahler and Palla, The Formation of Stars. The key numerical results are: (1) $L_{\mathrm{acc}}=GM\dot M/R\approx105\,L_\odot$ for $M=1\,M_\odot$ , $R=3\,R_\odot$ , $\dot M=10^{-5}\,M_\odot\,\mathrm{yr}^{-1}$ ; (2) $r_{\mathrm c}=j^2/GM\approx30\,\mathrm{AU}$ with $j\sim\Omega_0R_0^2$ and $j/j_\star\approx45$ ; (3) the steady-disk relations $\dot M=3\pi\nu\Sigma$ and $T_{\mathrm{eff}}\propto r^{-3/4}$ , including $T\approx150\,\mathrm{K}$ at 1 AU for $\dot M=10^{-7}\,M_\odot\,\mathrm{yr}^{-1}$ ; (4) the surface-density diffusion equation (6); (5) the timescale ordering $t_{\mathrm{dyn}}\ll t_{\mathrm{th}}\ll t_{\mathrm{visc}}$ .

Prerequisites. Molecular Clouds · The Jeans Instability and Jeans Mass · Virial Theorem for Molecular Clouds · Thermal Dust Emission · Interstellar Dust Grains. Leads to. Protoplanetary Disks and Planet Formation · Pre-Main-Sequence Evolution and the Hayashi Track · Bipolar Outflows and Herbig–Haro Objects.

Ask the tutor: Ask me to weigh a real protostar from its disk's Keplerian rotation curve, or to trace a parcel of gas from the cloud core through the centrifugal barrier and disk onto the star.

Why can't the infalling gas just fall straight onto the star?

Because it is rotating, and it conserves that rotation. A parcel with $j=v_\phi r$ spins faster as it falls, so its centrifugal acceleration $j^2/r^3$ overtakes gravity $GM/r^2$ and infall halts at $r_{\mathrm c}=j^2/GM$ — tens of AU out, thousands of stellar radii. The gas cannot cross that barrier without shedding angular momentum, so it accumulates in a disk. Only the slow outward transport of that angular momentum, by viscous and magnetic stresses, lets it finally reach the star.

What actually provides the disk's viscosity?

Not ordinary molecular viscosity — that is far too feeble to drain a disk in the age of the Universe. The effective viscosity is turbulent, led by the magnetorotational instability (MRI): in a differentially rotating disk threaded by even a weak field, two fluid elements linked by a field line exchange angular momentum in a runaway that drives vigorous turbulence. The Shakura–Sunyaev prescription $\nu=\alpha c_s H$ hides all of this in one number $\alpha\lesssim1$ . In the cold, poorly ionised midplanes the MRI can be quenched ("dead zones"), and there magnetically launched winds and gravitational torques may take over.

Why is a protostar invisible in the optical?

Because it is buried. In the embedded phase the star and disk are wrapped in an infalling envelope of large optical depth, so visible and ultraviolet photons are absorbed within a short distance. The dust that soaks up that radiation re-emits it thermally at its own temperature — tens to a few hundred kelvin — peaking in the far-infrared and submillimetre. So the true luminosity, mostly $L_{\mathrm{acc}}=GM\dot M/R$ , still escapes, but reprocessed to long wavelengths. This is why the youngest protostars (Class 0) are found by submillimetre and infrared telescopes, not optical ones.

Does the star inherit the disk's spin?

Almost none of it, and that is the whole point. If the star kept the angular momentum of the gas it swallowed it would be spinning at breakup, whereas young stars are observed to rotate far more slowly. The disk ships angular momentum outward: as most of the mass spirals in, a little gas at large radius carries the excess spin away, and magnetically launched jets and disk winds fling still more out along the axis, while magnetic star–disk coupling brakes the star further. So the star gains mass while shedding spin, and the disk plus outflows are the sink for the angular momentum it cannot hold.

What do the Class 0/I/II/III labels mean?

They are stages in the dispersal of the envelope, read off the slope of a young stellar object's infrared SED. Class 0 sources are the youngest and most deeply embedded, detected only in the submillimetre with most mass still in the envelope. Class I show a rising infrared SED from a substantial envelope plus a growing disk. Class II — the classical T Tauri stars — have lost the envelope and show an infrared excess from an optically thick disk around a now-visible star. Class III retain only a weak disk signature. The sequence traces the progression from envelope-dominated infall, to disk-dominated accretion, to a revealed pre-main-sequence star.

Why doesn't the disk temperature depend on the viscosity?

It seems paradoxical, since viscosity heats the disk — but viscosity sets how fast mass drains, not how much heat each ring makes at a given accretion rate. In steady state $\dot M=3\pi\nu\Sigma$ is the same at every radius, so a larger $\nu$ just means a smaller $\Sigma$ passing the same $\dot M$ . Writing $D(r)=\tfrac98\nu\Sigma\,\Omega_{\mathrm K}^2$ and substituting $\nu\Sigma=\dot M/3\pi$ , the viscosity cancels, leaving $T_{\mathrm{eff}}\propto(GM\dot M/r^3)^{1/4}\propto r^{-3/4}$ . The temperature and the infrared spectrum depend only on $M$ , $\dot M$ , and $r$ — which is why disk SEDs are predictable despite our ignorance of the turbulence.