research28 min studyLesson 912 in Physics

Disk luminosity follows energy transport and boundary conditions

Connect the one-face temperature profile to total luminosity while distinguishing Newtonian and relativistic efficiency.

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What you’ll understand

  • Keep one-face radiative flux distinct from two-face disk luminosity.
  • Derive a Newtonian radiative efficiency for a chosen inner radius.
  • Recognize when advection, irradiation or relativistic transfer invalidates a thin-disk inference.

A disk spectrum can reveal where orbital energy becomes radiation, but only after a transport model has been specified. Even a steady thin disk does not radiate each annulus’s local potential-energy loss in isolation: torques also move energy radially. We will connect the emitted flux to an integrated luminosity, keeping the number of radiating faces, the inner boundary, and the gravity approximation explicit.

Local radiation balances vertically integrated heating

Assume a steady, axisymmetric Newtonian disk with nearly circular Keplerian motion, negligible wind mass loss, and zero torque at Rᵢₙ. Suppose heat leaves vertically before it can be advected inward. Integrating the shear heating over height gives the energy emitted from both disk faces. Each face then has half that power per disk-plane area. Define an effective temperature by its frequency-integrated outward flux; a blackbody shape additionally requires suitable thermalization, rather than just a definition of temperature.

The flux is suppressed near the inner edge by the same boundary factor found from angular momentum conservation. Far outside that edge it scales as radius to the minus third power, so effective temperature scales as radius to the minus three quarters. The numerical factor is three times the naive annulus-only orbital-energy argument far from the boundary because the torque redistributes energy. This does not create extra total power: integrating the full boundary-dependent profile restores the global energy budget.

Fface(r)=σSBTeff4=3GMM˙8πr3(1−Rinr),r>RinF_{\rm face}(r)=\sigma_{\rm SB}T_{\rm eff}^4=\frac{3GM\dot M}{8\pi r^3}\left(1-\sqrt{\frac{R_{\rm in}}r}\right),\qquad r>R_{\rm in}
Fface is the outgoing flux of ONE face in W m⁻²; σSB is the Stefan–Boltzmann constant. Tₑff is in K, while the midplane and spectral color temperatures generally differ. The expression assumes local radiative cooling of internal dissipation.↔ Scroll wide equations horizontally.

Integrating both faces gives the efficiency

An annulus has area 2πrdr on each face, so its total radiating area is 4πrdr. Integrate the flux from the chosen inner radius to a large outer radius. For an effectively infinite outer disk the luminosity is half the gravitational potential-energy release per unit time at the inner edge. The other half is the orbital kinetic energy retained by material on that orbit. Additional surface boundary-layer emission from a slowly rotating star is a separate component, rather than another face of the disk.

Define gravitational radius r𝓰=GM/c², without the factor of two used for a Schwarzschild horizon radius. The Newtonian efficiency is r𝓰/(2Rᵢₙ). Substituting six gravitational radii into that Newtonian formula gives one twelfth, but that is not the Schwarzschild relativistic thin-disk result. In the ideal zero-torque Schwarzschild model, ignoring photon capture, the binding-energy efficiency is 1−√(8/9), about 0.0572. Strong gravity changes orbital energies and photon propagation; it cannot be repaired merely by choosing an inner radius in the Newtonian expression.

Ldisk=∫Rin∞4πrFface dr=GMM˙2Rin,ηN≡LdiskM˙c2=rg2Rin,rg=GMc2L_{\rm disk}=\int_{R_{\rm in}}^\infty4\pi rF_{\rm face}\,dr=\frac{GM\dot M}{2R_{\rm in}},\qquad \eta_N\equiv\frac{L_{\rm disk}}{\dot Mc^2}=\frac{r_g}{2R_{\rm in}},\quad r_g=\frac{GM}{c^2}
Ldisk is bolometric emitted power in W for the Newtonian model and ηN is dimensionless. No cosmological redshift, viewing projection, atmospheric color correction, or luminosity-distance conversion is included here.↔ Scroll wide equations horizontally.

Observed brightness does not determine efficiency by itself

Radiative efficiency means the fraction of supplied rest-mass power emerging as the specified radiation. At low densities, ions and electrons can exchange energy too slowly to radiate the available heat. At high optical depth and rapid inflow, photons can be trapped and advected. Magnetic extraction, winds and irradiation also modify the simple budget. A bright source’s isotropic-equivalent luminosity is not automatically the disk’s angle-integrated emitted luminosity. Inferring accretion rate or spin therefore requires a model for geometry, transfer, inner stress and spectral hardening.

The function lab isolates the fourth-power temperature sensitivity through a declared change of variable. Let its horizontal coordinate x represent (T/T₀)² and its vertical value represent normalized flux. Reset and set both controls to a=1,x=1, predicting one. Reset and set a=1,x=2, predicting four, corresponding to temperature √2 times larger. The instrument itself plots a quadratic; the fourth power arises only through our explicit squared-temperature coordinate. It contains no disk atmosphere or relativistic spectrum.

Compute the efficiency of a truncated Newtonian disk

Consider a hypothetical locally radiating, zero-torque disk truncated at Rᵢₙ=100r𝓰. It accretes Ṁ=10¹⁴ kg/s. Using c=3×10⁸ m/s, find its Newtonian efficiency, supplied rest-mass power and disk luminosity.

  1. The efficiency is ηN=r𝓰/(2Rᵢₙ)=1/200=0.005. The large truncation radius makes a Newtonian illustrative treatment more appropriate than an orbit near a horizon.

  2. The supplied rest-mass power is Ṁc²=10¹⁴×9×10¹⁶=9×10³⁰ W. It is an energy supply rate, not the radiated power.

  3. Multiply by the efficiency: Ldisk=0.005×9×10³⁰=4.5×10²⁸ W. This already includes radiation from both faces; doubling it again would overcount.

The result

The model radiates 0.5 percent of the supplied rest-mass power: Ldisk=4.5×10²⁸ W.

Explore the model

Interactive lab

Interactive lab · function

Read a temperature law through a squared coordinate

Interpret x=(T/T₀)² and y as normalized one-face flux. Reset and set BOTH controls a=1,x=1: predict y=1. Reset and set a=1,x=2: predict y=4, for T/T₀=√2. The actual instrument is y=ax² and computes no spectrum or disk structure.

Read a temperature law through a squared coordinateThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 rescaled coordinate can expose a scaling law only when its physical meaning is stated explicitly.

Check your understanding

Which statement correctly compares the inner radius and radiative efficiency?

Try it yourself

At a fixed radius and fixed Rᵢₙ in the same steady thin-disk model, increase Ṁ by a factor of sixteen. Find the changes in one-face flux, effective temperature and total disk luminosity. Keep M and the radiative assumptions fixed.

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 local flux and integrated luminosity are linear in Ṁ; effective temperature is the fourth root of flux.

Compare with the solution

The flux increases sixteenfold and Tₑff increases by 16¹ᐟ⁴=2. The total luminosity also increases sixteenfold, while ηN remains unchanged. These scalings hold only while local cooling and the same inner boundary remain valid; an observed color temperature need not follow without an unchanged atmosphere correction.

Take it with you

  • One-face flux must be integrated over both disk faces.
  • Newtonian and relativistic binding-energy efficiencies are different predictions.
  • Cooling, geometry and transfer assumptions are needed before interpreting observed brightness.

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