AST1440: Radiative Transfer Fundamentals and Problems 1.1–1.4

Scope: Rybicki & Lightman, Radiative Processes in Astrophysics, §§1.1–1.4 and Problems 1.1–1.4, plus an extension on M82 X-1.
Course website: https://www.astro.utoronto.ca/~mhvk/AST1440/
Textbook: Rybicki & Lightman, Radiative Processes in Astrophysics (2004).

The original Chinese notes first explain the physical concepts and derivations in detail, then provide assignment-ready English answers. This edition translates the explanatory material into English as well.

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0. The logical thread of this chapter

Starting with how to describe light, this chapter gradually builds the theory of radiative transfer:

1. §1.1: Describe electromagnetic radiation using wavelength, frequency, photon energy, and the corresponding temperature scale.
2. §1.2: Use flux to describe the net energy transported per unit area per unit time, and derive the inverse-square law.
3. §1.3: Retain directional information through specific intensity, then calculate mean intensity, energy density, flux, and radiation pressure.
4. §1.4: Study how emission and absorption by matter change specific intensity; introduce the absorption coefficient, optical depth, source function, and mean free path; finally examine how radiation exerts a force on matter.

The central relations throughout the chapter are

$$\lambda\nu=c, \qquad E_\gamma=h\nu, \qquad F=\frac{L}{4\pi r^2},$$ $$F_\nu=\int I_\nu\cos\theta\,d\Omega, \qquad u_\nu=\frac{1}{c}\int I_\nu\,d\Omega =\frac{4\pi J_\nu}{c},$$ $$\frac{dI_\nu}{ds}=-\alpha_\nu I_\nu+j_\nu.$$

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1. §1.1 The electromagnetic spectrum and basic properties of radiation

1.1 Wavelength and frequency

Electromagnetic waves in vacuum satisfy

$$\boxed{\lambda\nu=c},$$

where:

• $\lambda$: wavelength;
• $\nu$: frequency, in Hz;
• $c$: the speed of light in vacuum, approximately $3.00\times10^{10}\ \mathrm{cm\,s^{-1}}$.

Thus, a shorter wavelength means a higher frequency. From low frequency and long wavelength to high frequency and short wavelength, the electromagnetic spectrum runs as follows:

$$\text{radio}\rightarrow\text{infrared}\rightarrow\text{visible} \rightarrow\text{ultraviolet}\rightarrow\text{X-ray}\rightarrow\gamma\text{-ray}.$$

1.2 Photon energy

The energy of a single photon is

$$\boxed{E_\gamma=h\nu=\frac{hc}{\lambda}}.$$

A shorter-wavelength photon therefore has more energy. Note that this refers to the energy per photon; the total energy of a beam also depends on the number of photons.

The textbook also defines a corresponding temperature scale

$$\boxed{T_{\rm equivalent}=\frac{E}{k}},$$

to compare photon energy with the thermal energy scale $kT$. This does not imply that the emitting object actually has that temperature.

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2. §1.2 Radiative flux

2.1 Definition of flux

If energy $dE$ crosses area $dA$ during time $dt$, the flux is defined by

$$\boxed{dE=F\,dA\,dt}, \qquad \boxed{F=\frac{dE}{dA\,dt}}.$$

Its commonly used units are

$$[F]=\mathrm{erg\,cm^{-2}\,s^{-1}}.$$

Flux describes net energy transport per unit area per unit time. It depends on the orientation of the chosen area: light crossing obliquely contributes only through the component of its propagation direction along the surface normal.

2.2 The inverse-square law

Consider an isotropic source of luminosity $L$. At distance $r$, its energy is spread over a sphere of area $4\pi r^2$, so

$$\boxed{F(r)=\frac{L}{4\pi r^2}}.$$

Hence

$$F\propto r^{-2}.$$

Doubling the distance quadruples the spherical area, reducing the energy received per unit area to one quarter. Light does not spontaneously lose energy in free space; the same total power is distributed over a larger area.

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3. §1.3 Specific intensity and its moments

3.1 Why flux alone is insufficient

Two equally strong, oppositely directed beams can have zero net flux; the net flux is also zero when no light is present. Thus, $F_\nu$ alone cannot tell us whether there is any local radiation energy.

We need a quantity that retains directional information:

$$\boxed{I_\nu(\boldsymbol r,\hat{\boldsymbol n})}.$$

$I_\nu$ is a direction-dependent scalar, not a vector:

• $\hat{\boldsymbol n}$ is the input propagation direction, represented by a unit vector;
• $I_\nu(\hat{\boldsymbol n})$ is the radiation strength in that chosen direction, a nonnegative scalar.

This is analogous to a distance measured in different directions: the distance depends on direction, but is not itself a vector.

3.2 Definition of specific intensity

For a small area $dA$ perpendicular to a ray, the energy crossing during time $dt$, within frequency interval $d\nu$ and directional interval $d\Omega$, is

$$\boxed{dE=I_\nu\,dA\,dt\,d\Omega\,d\nu}.$$

If the angle between the surface normal and the ray is $\theta$, the effective projected area is $dA\cos\theta$, giving

$$dE=I_\nu\cos\theta\,dA\,dt\,d\Omega\,d\nu.$$

The units are

$$[I_\nu]=\mathrm{erg\,cm^{-2}\,s^{-1}\,sr^{-1}\,Hz^{-1}}.$$

3.3 Solid angle

An ordinary plane angle is an arc length divided by its radius:

$$\theta=\frac{\ell}{r}.$$

Solid angle is the three-dimensional generalization. Draw an imaginary sphere centered on the observer. If a bundle of directions covers area $A_{\rm sphere}$ on that sphere, then

$$\boxed{\Omega=\frac{A_{\rm sphere}}{r^2}}.$$

The solid angle of a full sphere is

$$\boxed{4\pi\ \mathrm{sr}},$$

A hemisphere subtends $2\pi\ \mathrm{sr}$. Solid angle describes a range of directions, not an object's physical area or the strength of the light.

In spherical coordinates,

$$\boxed{d\Omega=\sin\theta\,d\theta\,d\phi}.$$

because a small spherical surface element has area

$$dA_{\rm sphere}=(r\,d\theta)(r\sin\theta\,d\phi) =r^2\sin\theta\,d\theta\,d\phi.$$

For a cone of directions with half-opening angle $\alpha$,

$$\Omega=2\pi\int_0^\alpha\sin\theta\,d\theta =\boxed{2\pi(1-\cos\alpha)}.$$

In the small-angle limit, with $\alpha\ll1$ expressed in radians,

$$\boxed{\Omega\simeq\pi\alpha^2}.$$

For a physical small area $dA$ at distance $s$, whose normal makes angle $\beta$ with the line of sight, the solid angle is

$$\boxed{d\Omega=\frac{dA|\cos\beta|}{s^2}}.$$

The projection must be onto the plane perpendicular to the actual line of sight. Since the solid angle is centered on the observer, the denominator also uses the actual oblique distance $s^2$.

3.4 Mean specific intensity $J_\nu$

Mean specific intensity is defined as

$$\boxed{J_\nu=\frac{1}{4\pi}\int_{4\pi}I_\nu(\hat{\boldsymbol n})\,d\Omega}.$$

This does not assume that $I_\nu$ is the same in every direction. It is simply the sum over all directions divided by the total solid angle, just as a class average does not imply that every student earned the average score.

Only for isotropic radiation does

$$I_\nu(\hat{\boldsymbol n})=J_\nu \quad\text{hold in every direction}.$$

3.5 Energy density and the factor $c$

Define the directional energy density per unit solid angle as $w_\nu(\hat{\boldsymbol n})$. Within volume $dV$, the energy in a given range of directions is

$$dE=w_\nu\,dV\,d\Omega\,d\nu.$$

Light travels distance $c\,dt$ in time $dt$, so a cylinder with cross-sectional area $dA$ perpendicular to the ray has volume

$$dV=dA\,c\,dt.$$

Comparing

$$dE=w_\nu dA\,c\,dt\,d\Omega\,d\nu =I_\nu dA\,dt\,d\Omega\,d\nu,$$

gives

$$\boxed{w_\nu=\frac{I_\nu}{c}}.$$

Adding the energy densities from all directions:

$$\boxed{u_\nu=\int w_\nu\,d\Omega =\frac{1}{c}\int I_\nu\,d\Omega =\frac{4\pi J_\nu}{c}}.$$

In general, the correct expression is $u_\nu=4\pi J_\nu/c$. Only in the isotropic case, where $I_\nu=J_\nu$, can we write

$$u_\nu=\frac{4\pi I_\nu}{c}.$$

The solid-angle integral adds all directions; division by $c$ converts energy flow per unit area per unit time into energy per unit volume.

3.6 The relation between flux and specific intensity

The flux vector is

$$\boxed{\boldsymbol F_\nu=\int I_\nu(\hat{\boldsymbol n}) \hat{\boldsymbol n}\,d\Omega}.$$

For its component along a chosen surface normal $\hat{\boldsymbol m}$,

$$\boxed{F_\nu=\int I_\nu\cos\theta\,d\Omega}, \qquad \cos\theta=\hat{\boldsymbol n}\cdot\hat{\boldsymbol m}.$$

All directions enter the integral, but their contributions through the chosen surface are weighted by $\cos\theta$:

• $\theta=0^\circ$: full positive contribution;
• $\theta=60^\circ$: half the contribution;
• $\theta=90^\circ$: parallel to the surface, with no crossing;
• $\theta>90^\circ$: crossing in the reverse direction, giving a negative contribution.

For isotropic radiation, opposite directions cancel, so $F_\nu=0$, even though $u_\nu>0$.

3.7 Radiation pressure and the two cosines

Pressure is defined by

$$\boxed{P=\frac{\text{normal momentum delivered to the surface per unit time}}{\text{area}}}.$$

For particles arriving from direction $\theta$:

1. The number crossing the surface per second contains $\cos\theta$;
2. Each particle's normal momentum $p_\perp=p\cos\theta$ contains another factor $\cos\theta$.

Thus

$$\boxed{P_\nu=\frac{1}{c}\int I_\nu\cos^2\theta\,d\Omega}.$$

The two cosines are not double counting: one determines how many particles arrive per second, the other how much normal momentum each collision delivers. The factor $1/c$ follows from the photon momentum–energy relation

$$p=\frac{E}{c}.$$

Pressure is not energy multiplied by momentum. The actual calculation is

$$\text{energy flux}\times\frac{\text{normal momentum}}{\text{energy}} =\text{momentum flux}.$$

For isotropic radiation, choosing $z$ as the normal gives $\cos\theta=n_z$. The unit vector satisfies

$$n_x^2+n_y^2+n_z^2=1.$$

Isotropy makes the mean squares of all three components equal, so

$$\langle n_x^2\rangle=\langle n_y^2\rangle =\langle n_z^2\rangle=\frac13.$$

Therefore

$$\boxed{P_\nu=\frac{u_\nu}{3}}, \qquad \boxed{P=\frac{u}{3}}.$$

Similarly, a gas with no net flow still has pressure: energy fluxes in opposite directions cancel, but particles continue to strike different container walls.

3.8 Specific intensity is conserved in free space

Choose two small areas $dA_1,dA_2$ perpendicular to the same ray, separated by $R$. Track only the same bundle of light passing through both.

Seen from the first area, the second subtends solid angle

$$d\Omega_1=\frac{dA_2}{R^2};$$

Looking backward from the second, the first subtends the directional range

$$d\Omega_2=\frac{dA_1}{R^2}.$$

Thus

$$dA_1d\Omega_1=dA_2d\Omega_2 =\frac{dA_1dA_2}{R^2}.$$

The energies of this same beam at the two locations are

$$dE_1=I_{\nu,1}dA_1dt\,d\Omega_1d\nu,$$ $$dE_2=I_{\nu,2}dA_2dt\,d\Omega_2d\nu.$$

With no emission, absorption, or scattering in vacuum, energy conservation gives $dE_1=dE_2$. Canceling the common geometric factors yields

$$\boxed{I_{\nu,1}=I_{\nu,2}}, \qquad \boxed{\frac{dI_\nu}{ds}=0}.$$

For time-dependent radiation, compare the times at which the same light reaches the two locations:

$$I_\nu(P_2,t+R/c)=I_\nu(P_1,t).$$

3.9 Angular radius and a uniformly bright sphere

A sphere has physical radius $R$, and the observer is at distance $r$ from its center. The line of sight to the limb is tangent to the sphere. The angle between the center and limb directions is the angular radius $\theta_c$:

$$\boxed{\sin\theta_c=\frac{R}{r}}.$$

In the distant, small-angle limit, $\theta_c\simeq R/r$. The angular diameter is $2\theta_c$.

If the projected disk has uniform specific intensity $B_\nu$, with zero intensity outside, then

$$F_\nu =2\pi B_\nu\int_0^{\theta_c}\cos\theta\sin\theta\,d\theta =\pi B_\nu\sin^2\theta_c.$$

Substituting $\sin\theta_c=R/r$:

$$\boxed{F_\nu=\pi B_\nu\left(\frac{R}{r}\right)^2}.$$

This example shows that although specific intensity $B_\nu$ stays constant along a ray, the source's solid angle shrinks as $r^{-2}$, so the total flux still falls as $r^{-2}$.

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4. §1.4 Radiative transfer

4.1 Emission coefficient

The emission coefficient $j_\nu$ is defined by

$$dE=j_\nu\,dV\,d\Omega\,dt\,d\nu.$$

A beam traveling distance $ds$ passes through volume $dV=dA\,ds$, so emission increases its specific intensity by

$$\boxed{dI_\nu=j_\nu ds}.$$

For isotropic emission, the total emitted power per unit volume per unit frequency, $P_\nu$, is related to $j_\nu$ by

$$j_\nu=\frac{P_\nu}{4\pi}.$$

4.2 Why define an absorption coefficient?

The absorption coefficient $\alpha_\nu$ describes the local fraction of light lost per unit path length:

$$\boxed{dI_\nu=-\alpha_\nu I_\nu ds}, \qquad \boxed{-\frac{dI_\nu}{I_\nu}=\alpha_\nu ds}.$$

Its units are $\mathrm{cm^{-1}}$. It multiplies $I_\nu$ because the absolute number of absorbed photons is normally proportional to the incident photon number.

Microscopically, if absorbers have number density $n$ and individual effective cross section $\sigma_\nu$, the absorbed fraction in a thin layer is $n\sigma_\nu ds$, so

$$\boxed{\alpha_\nu=n\sigma_\nu}.$$

Alternatively, use the mass density $\rho$ and opacity per unit mass $\kappa_\nu$:

$$\boxed{\alpha_\nu=\rho\kappa_\nu}.$$

• $\alpha_\nu$: absorption coefficient, with units $\mathrm{cm^{-1}}$;
• $\kappa_\nu$: opacity or mass absorption coefficient, with units $\mathrm{cm^2\,g^{-1}}$.

4.3 The radiative transfer equation

Combining absorption and emission:

$$\boxed{\frac{dI_\nu}{ds}=-\alpha_\nu I_\nu+j_\nu}.$$

With emission only:

$$I_{\nu,\rm out}=I_{\nu,\rm in}+\int j_\nu ds.$$

With absorption only:

$$I_{\nu,\rm out}=I_{\nu,\rm in} \exp\left(-\int\alpha_\nu ds\right).$$

The attenuation is exponential because each layer absorbs a fixed fraction of the remaining light reaching it.

4.4 Optical depth

Define

$$\boxed{d\tau_\nu=\alpha_\nu ds}, \qquad \boxed{\tau_\nu=\int\alpha_\nu ds}.$$

Optical depth is dimensionless and measures cumulative attenuation along a path. For pure absorption,

$$\boxed{\frac{I_{\nu,\rm out}}{I_{\nu,\rm in}}=e^{-\tau_\nu}}.$$

• $\tau_\nu\ll1$: optically thin;
• $\tau_\nu\gg1$: optically thick.

Optical thickness is not geometric thickness: a geometrically thin but strongly absorbing layer can have large $\tau_\nu$. The same material can also have different optical depths at different frequencies.

4.5 The source function and solutions of the transfer equation

Define the source function

$$\boxed{S_\nu=\frac{j_\nu}{\alpha_\nu}}.$$

The transfer equation becomes

$$\boxed{\frac{dI_\nu}{d\tau_\nu}+I_\nu=S_\nu}.$$

For constant $S_\nu$, multiply both sides by the integrating factor $e^{\tau_\nu}$:

$$\frac{d}{d\tau_\nu}\left(I_\nu e^{\tau_\nu}\right) =S_\nu e^{\tau_\nu}.$$

Integrating from the entrance at $0$ to the exit at $\tau_\nu$ gives

$$\boxed{ I_{\nu,\rm out} =I_{\nu,\rm in}e^{-\tau_\nu} +S_\nu(1-e^{-\tau_\nu}) }.$$

Physically:

• The first term is the incident background light remaining after absorption;
• The second is light emitted along the path that survives subsequent absorption.

If $S_\nu$ varies with position, the formal solution is

$$\boxed{ I_\nu(\tau) =I_\nu(0)e^{-\tau} +\int_0^\tau S_\nu(t)e^{-(\tau-t)}\,dt }.$$

When $\tau\gg1$, the incident light disappears and $I_\nu\rightarrow S_\nu$. The source function can be understood as the local equilibrium value toward which the medium drives the specific intensity.

4.6 Mean free path

In a uniform medium, the photon mean free path is

$$\boxed{\ell_{\rm mfp}=\frac{1}{\alpha_\nu}}.$$

Therefore

$$\tau_\nu=\frac{\ell}{\ell_{\rm mfp}}.$$

The mean free path is a statistical average; at $\tau=1$, a fraction $e^{-1}\simeq37\%$ of the light is still directly transmitted.

4.7 Radiation force and the factor $c$

The photon momentum–energy relation is

$$p=\frac{E}{c}.$$

Dividing the energy absorbed per unit time by $c$ therefore gives momentum transferred per unit time, i.e. force. The radiation force per unit volume is

$$\boxed{\boldsymbol f_{\rm vol} =\frac{1}{c}\int\alpha_\nu\boldsymbol F_\nu\,d\nu}.$$

Dividing by mass density and using $\alpha_\nu=\rho\kappa_\nu$:

$$\boxed{\boldsymbol a_{\rm rad} =\frac{1}{c}\int\kappa_\nu\boldsymbol F_\nu\,d\nu}.$$

If $\kappa$ is frequency-independent,

$$\boxed{\boldsymbol a_{\rm rad}=\frac{\kappa\boldsymbol F}{c}}.$$

No further angular integration is needed here: $\boldsymbol F$ is already the net flux vector after integration over directions.

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5. Problems 1.1–1.4: detailed derivations

Problem 1.1: Pinhole camera

The pinhole diameter is $d$, the perpendicular separation between pinhole and film planes is $L$, and the focal ratio is $f=L/d$. The ray reaching a film point makes angle $\theta$ with the camera axis.

The actual oblique distance $s$ from the pinhole center to the film point satisfies

$$s=\frac{L}{\cos\theta}.$$

The pinhole area is

$$A_h=\frac{\pi d^2}{4}.$$

Seen from the film point, the pinhole is tilted relative to the actual line of sight. Its projected area is $A_h\cos\theta$, so its solid angle is

$$\Delta\Omega_h\simeq\frac{A_h\cos\theta}{s^2} =\frac{A_h}{L^2}\cos^3\theta.$$

The flux on the film requires one additional projection factor for the receiving surface:

$$F_\nu\simeq I_\nu(\theta,\phi)\cos\theta\,\Delta\Omega_h.$$

Therefore

$$\boxed{F_\nu =\frac{\pi\cos^4\theta}{4f^2}I_\nu(\theta,\phi)}.$$

The four cosines come from one film projection, one pinhole projection, and two factors from the squared oblique distance $s=L/\cos\theta$. All are geometric; none involves momentum or pressure.

Problem 1.2: Photoionization

The number density of ionizable atoms is $n_a$, the photoionization cross section is $\sigma_\nu$, and the threshold photon energy is $h\nu_0$. The photoionization absorption coefficient is

$$\boxed{\alpha_\nu=n_a\sigma_\nu}.$$

The absorbed energy rate within a frequency and directional interval is $\alpha_\nu I_\nu\,d\Omega\,d\nu$. Dividing by the single-photon energy $h\nu$ gives the ionization rate:

$$d\dot n_{\rm ion} =\frac{\alpha_\nu I_\nu}{h\nu}\,d\Omega\,d\nu.$$

Integrating over directions and above-threshold frequencies:

$$\dot n_{\rm ion} =\int_{\nu_0}^{\infty}\int_{4\pi} \frac{n_a\sigma_\nu I_\nu}{h\nu}\,d\Omega\,d\nu.$$

Using $\int I_\nu d\Omega=4\pi J_\nu=cu_\nu$:

$$\boxed{ \dot n_{\rm ion} =4\pi n_a\int_{\nu_0}^{\infty} \frac{\sigma_\nu J_\nu}{h\nu}\,d\nu =cn_a\int_{\nu_0}^{\infty} \frac{\sigma_\nu u_\nu}{h\nu}\,d\nu }.$$

The second form also follows from microscopic collisions: the photon number density in interval $d\nu$ is $u_\nu d\nu/(h\nu)$. One atom presents effective target area $\sigma_\nu$; during time $dt$, photons traveling at light speed that can hit it come from a cylinder of volume $\sigma_\nu cdt$. The per-atom ionization rate is therefore $c\sigma_\nu u_\nu d\nu/(h\nu)$. Multiplying by atomic number density $n_a$ and integrating gives the expression above. The factor $c$ is the photon speed.

Net flux cannot replace $4\pi J_\nu$ here: photons moving in opposite directions both ionize atoms, so the ionization rate can be nonzero even at zero net flux. The factor $4\pi$ follows from the definition of $J_\nu$ and does not require the actual radiation to be isotropic.

Problem 1.3: Resolved and unresolved X-ray clouds

A cloud of radius $R$ at distance $d$ uniformly produces $\Gamma$ photons per unit volume per unit time. Assume isotropic emission and negligible absorption. The photon-number emission coefficient per unit solid angle is

$$j_N=\frac{\Gamma}{4\pi}.$$

(a) Resolved central intensity

The central line of sight crosses length $2R$, so

$$\boxed{\mathcal I_{\rm center} =\int j_N ds =\frac{\Gamma R}{2\pi}}.$$

There is no distance dependence because specific intensity is conserved along rays in free space. For impact parameter $b$, the general brightness profile is

$$\mathcal I(b)=\frac{\Gamma}{2\pi}\sqrt{R^2-b^2}.$$

Uniform volume emissivity therefore produces a projected image that is bright at the center and faint at the edge.

A volume-based calculation also works, but it must use the narrow column corresponding to the central line of sight, not the entire cloud. If a pixel subtends $\delta\Omega$, the column volume is approximately $2R d^2\delta\Omega$. Counting the isotropically emitted photons from that volume that reach the detector gives the same result.

(b) Unresolved field-averaged intensity

The cloud's total photon-number luminosity is

$$\dot N_\gamma=\Gamma\frac{4\pi R^3}{3}.$$

The photon-number flux at Earth is

$$\Phi_N=\frac{\dot N_\gamma}{4\pi d^2} =\frac{\Gamma R^3}{3d^2}.$$

For detector acceptance half-angle $\Delta\theta$, the small-angle approximation gives

$$\Delta\Omega_{\rm det}\simeq\pi(\Delta\theta)^2.$$

The field-averaged intensity is therefore

$$\boxed{ \overline{\mathcal I}_{\rm beam} =\frac{\Phi_N}{\Delta\Omega_{\rm det}} =\frac{\Gamma R^3}{3d^2\Delta\Omega_{\rm det}} \simeq\frac{\Gamma R^3}{3\pi d^2(\Delta\theta)^2} }.$$

The effective area $\Delta A$ cancels when count rate is converted back into intensity per unit area. This average is over the detector field of view, not the all-sky $4\pi$ average defining $J_\nu$.

Using the whole cloud volume and dividing by the source's own solid angle $\Omega_{\rm src}\simeq\pi R^2/d^2$ instead gives the disk-averaged intensity

$$\overline{\mathcal I}_{\rm src}=\frac{\Gamma R}{3\pi},$$

not the central intensity $\Gamma R/(2\pi)$.

Problem 1.4: Radiation driving and the Eddington limit

A central source has mass $M$ and luminosity $L$. An optically thin cloud of mass $m$ lies at distance $r$ and has opacity per unit mass $\kappa$.

(a) Condition for outward acceleration

The local flux is

$$F(r)=\frac{L}{4\pi r^2}.$$

If the cloud's illuminated cross-sectional area is $A$, its mass per unit area is $m/A$ and its optical depth is

$$\tau=\kappa\frac{m}{A}\ll1.$$

The incident energy per second is $FA$. In the optically thin limit, the fraction absorbed or effectively transferring momentum is approximately $\tau$, so

$$\dot E_{\rm int}\simeq FA\tau=\kappa mF.$$

Using $p=E/c$ to convert energy transfer rate into momentum transfer rate:

$$\mathcal F_{\rm rad}=\frac{\kappa mF}{c}, \qquad a_{\rm rad}=\frac{\mathcal F_{\rm rad}}{m}=\frac{\kappa F}{c}.$$

The radiative and gravitational accelerations are, respectively,

$$a_{\rm rad}=\frac{\kappa F}{c} =\frac{\kappa L}{4\pi r^2c}, \qquad a_{\rm grav}=\frac{GM}{r^2}.$$

Requiring $a_{\rm rad}>a_{\rm grav}$ gives

$$\boxed{\frac{M}{L}<\frac{\kappa}{4\pi Gc}}.$$

Both accelerations scale as $r^{-2}$, so their ratio is independent of distance.

(b) Terminal speed when starting from rest at $R$

The net outward acceleration is

$$a(r)=\frac1{r^2} \left(\frac{\kappa L}{4\pi c}-GM\right).$$

Using the work–energy relation per unit mass and integrating from $R$ to infinity:

$$\frac12v_\infty^2 =\int_R^\infty a(r)dr =\frac1R\left(\frac{\kappa L}{4\pi c}-GM\right).$$

Hence

$$\boxed{ v_\infty^2 =\frac{2GM}{R} \left(\frac{\kappa L}{4\pi GMc}-1\right) }.$$

The bracket is positive precisely when the outward-driving condition in (a) holds. Here, terminal velocity means the asymptotic speed as $r\rightarrow\infty$.

(c) Eddington luminosity for fully ionized pure hydrogen

There is one free electron per hydrogen mass $m_H$, so the Thomson scattering opacity is

$$\kappa_{\rm es}=\frac{\sigma_T}{m_H} \simeq0.40\ \mathrm{cm^2\,g^{-1}}.$$

Setting radiative acceleration equal to gravitational acceleration:

$$\boxed{ L_{\rm Edd} =\frac{4\pi GMc}{\kappa_{\rm es}} =\frac{4\pi GMcm_H}{\sigma_T} \simeq1.25\times10^{38} \left(\frac{M}{M_\odot}\right) \mathrm{erg\,s^{-1}} }.$$

More generally, $L_{\rm Edd}=4\pi GMc/\kappa$. The Eddington limit is not a single fixed number: it depends on the effective opacity of the material being driven. Larger $\kappa$ gives stronger radiation driving and a lower critical luminosity.

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6. Extension: M82 X-1

6.1 The problem statement

Original statement:

The ULX (Ultra-Luminous X-ray source) M82 X-1 was observed to have an X-ray flux $f_X=4\times10^{-12}\ \mathrm{erg\,cm^{-2}\,s^{-1}}$. Estimate its luminosity, given that M82 is at $d\approx3.6\ \mathrm{Mpc}$. What could you say about its mass?

Restatement (translated from the Chinese version):

The observed X-ray flux of the ultraluminous X-ray source M82 X-1 is

$$f_X=4\times10^{-12}\ \mathrm{erg\,cm^{-2}\,s^{-1}}.$$

Given that M82 is approximately $d=3.6\ \mathrm{Mpc}$ away, estimate the source luminosity and discuss what can be inferred about its mass.

6.2 X-ray luminosity

$$d=3.6\times10^6\times3.086\times10^{18} \simeq1.11\times10^{25}\ \mathrm{cm}.$$

Assuming isotropic emission:

$$\boxed{ L_{X,\rm iso}=4\pi d^2f_X \simeq6.2\times10^{39}\ \mathrm{erg\,s^{-1}} }.$$

Strictly speaking, this is the isotropic-equivalent X-ray luminosity.

6.3 $L_X$, $L_{\rm bol}$, and the mass constraint

Bolometric luminosity is the total electromagnetic luminosity summed over all bands:

$$L_{\rm bol}=\int_0^\infty L_\nu d\nu.$$

X-rays occupy only one band, so

$$L_X\le L_{\rm bol}.$$

If we further assume isotropic emission and a total luminosity no greater than the classical Eddington limit for fully ionized pure hydrogen,

$$L_{\rm bol}\le L_{\rm Edd},$$

then

$$L_X\le1.25\times10^{38}\left(\frac{M}{M_\odot}\right).$$

therefore

$$\boxed{M\gtrsim50M_\odot}.$$

This is a conditional lower mass limit, not a direct mass measurement. The classical inference relies on:

• An isotropic conversion of the observed flux to luminosity;
• An approximately spherically symmetric local flux near the central source;
• Optically thin irradiated gas;
• Standard effective scattering cross sections and opacity;
• A total luminosity that does not exceed the classical Eddington limit.

Strong magnetic fields can modify effective scattering cross sections; radiation can also be anisotropic, or accretion can be super-Eddington. The actual mass therefore cannot be determined from this one number alone.

6.4 Eddington limits for different compositions

For fully ionized material with hydrogen mass fraction $X_H$, the electron-scattering opacity is approximately

$$\kappa_{\rm es}\simeq0.20(1+X_H)\ \mathrm{cm^2\,g^{-1}}.$$

These are not different fundamental laws, but results of the same formula $L_{\rm Edd}=4\pi GMc/\kappa$ with different opacities.

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7. English assignment answers

Problem 1.1

Let the pinhole area be $A_h=\pi d^2/4$. For a film point corresponding to an incident angle $\theta$, the distance from the pinhole is

$$s=\frac{L}{\cos\theta}.$$

Viewed from the film point, the projected pinhole area is $A_h\cos\theta$. In the small-pinhole approximation, the solid angle of the hole is

$$\Delta\Omega_h\simeq\frac{A_h\cos\theta}{s^2} =\frac{\pi d^2}{4L^2}\cos^3\theta.$$

Specific intensity is conserved along rays in free space. Hence,

$$F_\nu\simeq I_\nu(\theta,\phi)\cos\theta\,\Delta\Omega_h =\frac{\pi d^2}{4L^2}I_\nu(\theta,\phi)\cos^4\theta.$$

Since $f=L/d$,

$$\boxed{F_\nu=\frac{\pi\cos^4\theta}{4f^2}I_\nu(\theta,\phi)}.$$

Problem 1.2

In the interval $[\nu,\nu+d\nu]$, the photon number density is

$$dn_\gamma=\frac{u_\nu}{h\nu}\,d\nu.$$

An atom with photoionization cross section $\sigma_\nu$ has an interaction rate $c\sigma_\nu dn_\gamma$. Integrating above the ionization threshold and multiplying by the atomic number density gives

$$\boxed{ \dot n_{\rm ion} =cn_a\int_{\nu_0}^{\infty}\frac{\sigma_\nu u_\nu}{h\nu}\,d\nu }.$$

Using $u_\nu=4\pi J_\nu/c$,

$$\boxed{ \dot n_{\rm ion} =4\pi n_a\int_{\nu_0}^{\infty} \frac{\sigma_\nu J_\nu}{h\nu}\,d\nu }.$$

The units are $\mathrm{cm^{-3}\,s^{-1}}$. No assumption of an isotropic radiation field is required.

Problem 1.3

For isotropic emission, the photon number emission coefficient is $j_N=\Gamma/(4\pi)$.

For a resolved central line of sight, the path length is $2R$, so

$$\boxed{\mathcal I_{\rm center}=\frac{\Gamma R}{2\pi}}.$$

For an unresolved source, the total photon production rate and the photon flux at Earth are

$$\dot N_\gamma=\Gamma\frac{4\pi R^3}{3}, \qquad \Phi_N=\frac{\Gamma R^3}{3d^2}.$$

With $\Delta\Omega_{\rm det}\simeq\pi(\Delta\theta)^2$, the average beam intensity is

$$\boxed{ \overline{\mathcal I}_{\rm beam} =\frac{\Gamma R^3}{3d^2\Delta\Omega_{\rm det}} \simeq\frac{\Gamma R^3}{3\pi d^2(\Delta\theta)^2} }.$$

The detector effective area cancels when the count rate is converted into intensity.

Problem 1.4

At distance $r$,

$$F=\frac{L}{4\pi r^2}, \qquad a_{\rm rad}=\frac{\kappa L}{4\pi r^2c}, \qquad a_{\rm grav}=\frac{GM}{r^2}.$$

The cloud accelerates outward if

$$\boxed{\frac{M}{L}<\frac{\kappa}{4\pi Gc}}.$$

Using the work–energy theorem for a cloud starting from rest at $R$,

$$\boxed{ v_\infty^2 =\frac{2GM}{R} \left(\frac{\kappa L}{4\pi GMc}-1\right) }.$$

For fully ionized pure hydrogen, $\kappa_{\rm es}=\sigma_T/m_H$, so

$$\boxed{ L_{\rm Edd} =\frac{4\pi GMcm_H}{\sigma_T} \simeq1.25\times10^{38} \left(\frac{M}{M_\odot}\right)\mathrm{erg\,s^{-1}} }.$$

M82 X-1 extension

With $d=3.6\ \mathrm{Mpc}=1.11\times10^{25}\ \mathrm{cm}$,

$$\boxed{L_{X,\rm iso}=4\pi d^2f_X \simeq6.2\times10^{39}\ \mathrm{erg\,s^{-1}}}.$$

If the emission is isotropic and the bolometric luminosity does not exceed the pure-hydrogen Eddington limit, then

$$L_X\le L_{\rm bol}\le L_{\rm Edd},$$

which implies

$$\boxed{M\gtrsim50M_\odot}.$$

This is a conditional lower bound rather than a direct mass measurement. Anisotropic emission, super-Eddington accretion, or a different effective opacity could alter the inference.

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8. Key English terminology

Common instruction words in problems:

• show that: establish the stated result;
• derive: obtain a result from underlying relations;
• estimate: calculate an approximate value;
• assume: adopt a stated premise;
• neglect: omit an effect;
• infer: draw a conclusion from the results;
• starting from rest: with zero initial velocity.

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9. Final checklist

After studying this material, you should be able to answer:

1. Why does $I_\nu$ depend on direction without being a vector?
2. Why is $J_\nu$ a directional average without requiring isotropy?
3. Why is $u_\nu=4\pi J_\nu/c$ correct, rather than generally writing $4\pi I_\nu/c$?
4. Where do the single $\cos\theta$ in flux and the two factors $\cos\theta$ in pressure come from?
5. Why is specific intensity conserved in free space while total flux can decrease as $r^{-2}$?
6. What do $\alpha_\nu$, $\kappa_\nu$, $\tau_\nu$, and $S_\nu$ describe?
7. Why does the transfer-equation solution contain both transmitted background and medium-emission terms?
8. Why is the central intensity in Problem 1.3 independent of distance, while the unresolved field-averaged intensity contains $d^{-2}$?
9. Which physical assumptions underlie the Eddington limit, and why is the M82 X-1 mass conclusion only a conditional lower bound?