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 $dA$ and length $dr$. Radiation of intensity $I_\nu$ travels perpendicular to the bottom surface into a small solid angle $d\omega$. If the intensity changes by $dI_\nu$, the energy change over a time $dt$ is

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

Some of this energy is absorbed by the medium:

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

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

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

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

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

$$\tag{2.6.1} \implies \frac{1}{\alpha_\nu} \frac{dI_\nu}{dt} = -I_\nu + \frac{j_\nu}{\alpha_\nu}$$

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

$$\tag{2.6.2} S_\nu = \frac{j_\nu}{\alpha_\nu}$$

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

$$S_\nu = B_\nu (T)$$

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

$$\tag{2.6.3} \therefore \: \boxed{\frac{dI_\nu}{d \tau_\nu} = -I_\nu + S_\nu}$$

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

A general solution to this equation is:

$$\tag{2.6.4} I_\nu (\tau_\nu) = I_\nu (0) \, e^{-\tau_\nu} + \int_0^{\tau_\nu} S_\nu (t) \, e^{-(\tau_\nu - t)} \, dt$$

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

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

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

$$\tag{2.6.6} I_\nu (\tau_\nu) \approx S_\nu$$

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

$$\tag{2.6.7} I_\nu (\tau_\nu) \approx I_\nu (0) + S_\nu \tau_\nu$$

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 $R$ is at a distance $r$ from the Sun. The total energy flux hitting the planet is:

$$\tag{2.6.8} L_\text{in} = \pi R^2 \frac{L_\odot}{4 \pi r^2}$$

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

$$\tag{2.6.9} L_\text{out} = A L_\text{in}$$

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

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

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

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

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

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

$$\tag{2.6.13} p = \pi \Gamma$$

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

The flux density of reflected light is:

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

The solar flux at Earth is $F_\odot = L_\odot / 4\pi a^2$. Therefore,

$$\tag{2.6.15} m - m_\odot = -2.5 \log \frac{pR^2}{a^2} + 5 \log \frac{dr}{a^2} - 2.5 \log \Phi(\alpha)$$

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

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

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

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

The absolute magnitude at phase angle $\alpha$ is:

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

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

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

Thermal Equillibrium

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

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

If the planet rotates fast, the temperature is given by

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

where $T$ is the temperature of the planet, $A$ is the planet’s bond albedo, $T_s$ is the temperature of the star, $R$ is the radius of the planet, $R_s$ is the radius of the star and $r$ is the distance between them.