Advanced Topics

Advanced Topics

Radiative Energy Transfer

Understanding how energy moves through space and matter is essential in astrophysics. Radiative transfer describes how light travels through a medium, such as a star’s atmosphere or interstellar gas.

Imagine a small cylinder with cross-sectional area dAdA and length drdr. Radiation of intensity IνI_\nu travels perpendicular to the bottom surface into a small solid angle dωd\omega. If the intensity changes by dIνdI_\nu, the energy change over a time dtdt is

dE=dIνdAdνdωdt dE = dI_\nu \, dA \, d\nu \, d\omega \, dt

Some of this energy is absorbed by the medium:

dEabs=ανIνdrdAdνdωdt dE_\text{abs} = \alpha_\nu I_\nu \, dr \, dA \, d\nu \, d\omega \, dt

Here, αν\alpha_\nu is the opacity (how much the medium absorbs) at frequency ν\nu. The medium can also emit energy, described by the emission coefficient jνj_\nu:

dEem=jνdrdAdνdωdt dE_\text{em} = j_\nu \, dr \, dA \, d\nu \, d\omega \, dt

The total change in energy is the absorbed minus the emitted energy:

dE=dEabsdEem dE = dE_\text{abs} - dE_\text{em}

    1ανdIνdt=Iν+jναν(2.6.1)\tag{2.6.1} \implies \frac{1}{\alpha_\nu} \frac{dI_\nu}{dt} = -I_\nu + \frac{j_\nu}{\alpha_\nu}

We define the source function, which describes how much energy is emitted compared to how much is absorbed:

Sν=jναν(2.6.2)\tag{2.6.2} S_\nu = \frac{j_\nu}{\alpha_\nu}

In local thermodynamic equilibrium (LTE), the source function equals the Planck function:

Sν=Bν(T) S_\nu = B_\nu (T)

We also define the optical thickness as dτν=ανdrd \tau_\nu = \alpha_\nu \, dr. It is a measure of how opaque the medium is to radiation at frequency ν\nu.

dIνdτν=Iν+Sν(2.6.3)\tag{2.6.3} \therefore \: \boxed{\frac{dI_\nu}{d \tau_\nu} = -I_\nu + S_\nu}

This is the fundamental equation of radiative transfer. If Sν>IνS_\nu > I_\nu, the intensity increases as radiation travels; if Sν<IνS_\nu < I_\nu, it decreases. At equilibrium, absorption equals emission: dEabs=dEemdE_\text{abs} = dE_\text{em} and Iν=Sν=jν/ανI_\nu = S_\nu = j_\nu / \alpha_\nu.

A general solution to this equation is:

Iν(τν)=Iν(0)eτν+0τνSν(t)e(τνt)dt(2.6.4)\tag{2.6.4} I_\nu (\tau_\nu) = I_\nu (0) \, e^{-\tau_\nu} + \int_0^{\tau_\nu} S_\nu (t) \, e^{-(\tau_\nu - t)} \, dt

where Iν(0)I_\nu (0) is the background intensity. If the source function is independent of τν\tau_\nu, the solution simplifies to:

Iν(τν)=Iν(0)eτν+Sν(1eτν)(2.6.5)\tag{2.6.5} I_\nu (\tau_\nu) = I_\nu (0) \, e^{-\tau_\nu} + S_\nu \left( 1 - e^{-\tau_\nu} \right)

If the medium is optically thick (τν1\tau_\nu \gg 1), the first term becomes negligible, and the intensity approaches the source function:

Iν(τν)Sν(2.6.6)\tag{2.6.6} I_\nu (\tau_\nu) \approx S_\nu

If the medium is optically thin (τν1\tau_\nu \ll 1), the intensity is approximately:

Iν(τν)Iν(0)+Sντν(2.6.7)\tag{2.6.7} I_\nu (\tau_\nu) \approx I_\nu (0) + S_\nu \tau_\nu

This equation is widely used to model the atmospheres of stars and planets, as well as the interstellar medium.

Albedo

Albedo measures how well a surface reflects light. Suppose a planet of radius RR is at a distance rr from the Sun. The total energy flux hitting the planet is:

Lin=πR2L4πr2(2.6.8)\tag{2.6.8} L_\text{in} = \pi R^2 \frac{L_\odot}{4 \pi r^2}

The Bond albedo AA is the fraction of incoming energy that is reflected:

Lout=ALin(2.6.9)\tag{2.6.9} L_\text{out} = A L_\text{in}

The reflected light is not always spread evenly in all directions; it depends on the phase angle α\alpha (the angle between the Sun, planet, and observer). The observed flux at Earth is:

F=CΦ(α)Lout4πd2=CA4πΦ(α)1d2Lin(2.6.10)\tag{2.6.10} F = C \, \Phi (\alpha) \, \frac{L_\text{out}}{4\pi d^2} = \frac{CA}{4\pi} \, \Phi(\alpha) \, \frac{1}{d^2} \,L_\text{in}

Here, dd is the distance from the planet to Earth, CC is a constant, and Φ(α)\Phi(\alpha) is the phase function (how brightness changes with phase angle). When α=0\alpha = 0^\circ, Φ(0)=1\Phi(0) = 1. The total reflected flux is LoutL_\text{out}, so:

C4πd2SΦ(α)dS=1(2.6.11)\tag{2.6.11} \frac{C}{4 \pi d^2} \int_S \Phi(\alpha) \, dS = 1

    C=4πd2SΦ(α)dS=20πΦ(α)sinαdα=4q(2.6.12)\tag{2.6.12} \implies C = \frac{4 \pi d^2}{\int_S \Phi(\alpha) \, dS} = \frac{2}{\int_0^\pi \Phi(\alpha) \sin \alpha \, d\alpha} = \frac{4}{q}

The phase integral is defined as q=20πΦ(α)sinαdαq = 2 \int_0^\pi \Phi(\alpha) \sin \alpha \, d \alpha. We define Γ=CA4π\Gamma = \frac{CA}{4\pi}. The geometric albedo of the planet is:

p=πΓ(2.6.13)\tag{2.6.13} p = \pi \Gamma

Bond albedo and geometric albedo are related by A=pqA = pq.

📍

A Lambertian surface is a perfect diffuser that reflects all incoming light equally in all directions, so A=1A = 1 and

Φ(α)={cosα0α900otherwise(2.6.14)\tag{2.6.14} \Phi(\alpha) = \begin{cases} \cos \alpha \qquad &0^\circ \le \alpha \le 90^\circ \\ 0 \qquad &\text{otherwise} \end{cases}

For a Lambertian surface, p=q=1p = q = 1 and C=4C = 4. The flux density at α=0\alpha = 0^\circ is:

FL=1π1d2Lin F_L = \frac{1}{\pi} \, \frac{1}{d^2} \, L_\text{in}

For a non-Lambertian surface, the flux density at α=0\alpha = 0^\circ is:

F=CA4π1d2Lin F = \frac{CA}{4\pi} \, \frac{1}{d^2} \, L_\text{in}

So,

FFL=CA4=p \frac{F}{F_L} = \frac{CA}{4} = p

Thus, geometric albedo pp is the ratio of the flux at α=0\alpha = 0^\circ reflected by the planet to that reflected by a Lambertian disk of the same size.

Some surfaces can have p>1p > 1 (for example, a mirror, where pp is infinite).

The flux density of reflected light is:

F=pπΦ(α)1d2LR24r2 F = \frac{p}{\pi} \, \Phi(\alpha) \, \frac{1}{d^2} \, \frac{L_\odot R^2}{4 r^2}

The solar flux at Earth is F=L/4πa2F_\odot = L_\odot / 4\pi a^2. Therefore,

mm=2.5logpR2a2+5logdra22.5logΦ(α)(2.6.15)\tag{2.6.15} m - m_\odot = -2.5 \log \frac{pR^2}{a^2} + 5 \log \frac{dr}{a^2} - 2.5 \log \Phi(\alpha)

The absolute magnitude V(1,0)V(1, 0) of a planet is the magnitude it would have if it were 1 AU from both the Earth and Sun, at phase angle α=0\alpha = 0^\circ:

V(1,0)=m2.5logpR2a2(2.6.16)\tag{2.6.16} V(1, 0) = m_\odot - 2.5 \log \frac{pR^2}{a^2}

Although this situation is not physically possible, it is a useful reference for comparing relative brightness of planets.

m=V(1,0)+5logdra22.5logΦ(α)(2.6.17)\tag{2.6.17} \therefore m = V(1, 0) + 5 \log \frac{dr}{a^2} - 2.5 \log \Phi(\alpha)

The absolute magnitude at phase angle α\alpha is:

V(1,α)=V(1,0)2.5logΦ(α)(2.6.18)\tag{2.6.18} V(1, \alpha) = V(1, 0) - 2.5 \log \Phi(\alpha)

    m=V(1,α)+5logdra2(2.6.19)\tag{2.6.19} \implies m = V(1, \alpha) + 5 \log \frac{dr}{a^2}

At α=0\alpha = 0^\circ, p=(draR)2100.4(mm0)p = \left( \frac{dr}{aR} \right)^2 10^{-0.4 (m - m_0)} where m0=m(α=0)m_0 = m(\alpha = 0^\circ) is the apparent magnitude.

Thermal Equillibrium

If a slowly rotating planet is in thermal equillibrium, the temperature of the planet is given by

2πR2σT4=4πRs2σTs44πr2πR2(1A) 2\pi R^2 \sigma T^4 = \frac{4\pi R_s^2 \sigma T_s^4}{4\pi r^2} \cdot \pi R^2 \cdot (1 - A)

If the planet rotates fast, the temperature is given by

4πR2σT4=4πRs2σTs44πr2πR2(1A) 4\pi R^2 \sigma T^4 = \frac{4\pi R_s^2 \sigma T_s^4}{4\pi r^2} \cdot \pi R^2 \cdot (1 - A)

where TT is the temperature of the planet, AA is the planet’s bond albedo, TsT_s is the temperature of the star, RR is the radius of the planet, RsR_s is the radius of the star and rr is the distance between them.