Chapter 5. Energy Transport

How is energy generated in the core transported to the surface?

• Radiation: photons

• Convection: fluid motions

• Conduction: electrons, atoms

Radiation

The sun ($\bar\rho\odot\sim 1.4\text{ g/cm}^3, \bar n\odot\sim10^{24}\text{ cm}^{-3}$) is optically thick, that is, when we solely consider Thomson scattering (photons v.s. electrons), the mean free path for a photon is

$$l_\text{mfp, photons}\simeq\frac{1}{n\sigma_\text{T}}\sim 1\text{ cm}\ll R_{\odot}\sim 7\times10^{10}\text{ cm}$$

Therefore, photons must be scattered many times before reaching the surface, and energy is radiated by diffusion.

• Diffusion

  An example is that particles tend to escape from dense regions to the places with lower number density. The particle flux is approximately proportional to the density gradient

  $$\vec j=\rho\vec u\simeq-D\nabla n$$

  where $D$ is the diffusion coefficient

  $$D\sim\frac13\bar \mu l_\text{mfp},\text{ where } \bar\mu=\sqrt{\frac{kT}{\mu m_\text p}}$$

  Larger $l_\text{mfp}$ results in larger diffusion rate.

Radiation Energy

$$U_\text{rad}=aT_\text{rad}^4$$

where $a$ is the radiation coefficient derived from the Stefan–Boltzmann constant

$$\sigma_\text{SB}=\frac{ac}4$$

In a star, radiation and gas are thermalized due to numerous collisions, thus $T_\text{rad}\simeq T$.

We can similarly write down the radiation energy flux as

$$F_\text{rad}=-D\nabla U_\text{rad}$$

where

$$D\simeq\frac{1}{3}cl_\text{mfp}=\frac13\frac{c}{n\sigma_\text{T}}\equiv\frac{c}{3\rho\kappa}$$

Here we have defined the opacity $\kappa$ as

$$\kappa\equiv\frac{\sigma_\text{T}}{m}$$

For a spherical system (such as a star), the gradient is simply ${\partial}/{\partial r}$, thus

$$F_\text{rad}=-\frac{4ac}{3\rho\kappa}T^3\frac{\partial T}{\partial r}$$

The luminosity is

$$L=4\pi r^2 F_\text{rad}=-\frac{16ac\pi r^2ac}{3\rho\kappa}T^3\frac{\partial T}{\partial r}=-\frac{64ac\pi^2 r^4}{3\kappa}T^3\frac{\partial T}{\partial m}$$

But how can we know the opacity?

Rosseland mean opacity

In general, there are so many sources of opacity $\kappa_\nu(\rho,T)$ like scattering and absorption.

• Electron scattering

  $$\kappa=\kappa_e=0.35\text{ cm}^2\text{/g}$$
• Free-free transition (bremsstrahlung)

  $$k_\nu^\text{ff}\sim f(\nu)\rho T^{-7/2}$$

  

  From Fengwei Xu's notes

For each $\nu$, the radiation flux is

$$F_\nu=-\frac{c}{3\rho \kappa_\nu}\frac{\partial}{\partial r}U_{\text{rad},\nu}$$

where $U\nu$ is given by $4\pi B\nu(T)/c$,

$$B_\nu(T)=\frac{2h\nu^3}{c^2}\frac{1}{\exp{(h\nu/kT)}-1}$$

Thus

$$F_\nu=-\frac{4\pi}{3\rho \kappa_\nu}\frac{\partial B_\nu(T)}{\partial T}\frac{\partial T}{\partial r}$$

Obviously,

$$\frac1{\kappa_\nu}\frac{\partial B_\nu(T)}{\partial T}$$

is $\nu$-denpendent, thus we can define the Rosseland mean opacity as

$$\frac1{\kappa_R}\equiv\frac{\int\frac{1}{\kappa_\nu}\frac{\partial B_\nu(T)}{\partial T}\text d\nu}{\int\frac{\partial B_\nu(T)}{\partial T}\text d\nu}$$

Note that by integrating $B_\nu(T)$ over the frequency the integrated radiance $L$ is

$$L=\frac{2 \pi^{5}}{15} \frac{k^{4} T^{4}}{c^{2} h^{3}} \frac{1}{\pi}=\sigma_\text{SB} T^{4} \frac{1}{\pi}$$

Thus

$$\int\frac{\partial B_\nu(T)}{\partial T}\text d\nu=\frac{\partial}{\partial T}\left(\sigma_\text{SB} T^4\right)=\frac{ac}{\pi}T^3$$

Then

$${\int\frac{1}{\kappa_\nu}\frac{\partial B_\nu(T)}{\partial T}\text d\nu}=\frac1{\kappa_R}\frac{ac}{\pi}T^3$$

In this way,

$$F=-\frac{4ac}{3\rho \kappa_R}T^3\frac{\partial T}{\partial r}$$

Convection

Discussed in the next chapter.

Conduction

• Not important for normal stars

• Important for compact stars

Energy is transported via collision due to thermal motions of particles. Although the physics is different from radiation transport, the flux is simply given by

$$F_\text{cd}=-k_\text{cd}(T,\rho)\nabla T$$

So for a star without significant convection, the total energy flux is

$$F_\text{tot}=F_\text{rad}+F_\text{cd}=-\left(k_\text{rad}+k_\text{cd}\right)\nabla T$$