# Chapter 4. Energy Conservation

Let us consider the net energy in a unit time passing the sphere of a radius $r$, $L(r)$, or $L(m)$.

In the shell $\[r,r+\text dr]$, or $\[m,m+\text dm]$

1. If no energy generation/absorption
   * Eulerian picture: fixed fluid position

     $$
     \frac{\partial L}{\partial r}=0
     $$
   * Lagrange's picture: the coordinates move with mass

     $$
     \frac{\partial L}{\partial m}=0
     $$
   * In astrophysics, Lagrange's picture is usually adopted. One example is the perturbation theory.

     > In the perturbation theory, people usually consider a mass element with certain thermaldynamical quantities, such as $\rho$, $T$, $P$, etc. As the mass element moves, the change in each quantity can be easily characterized with its partial derivative with respect to $m$. Important applications include
     >
     > * Stellar pulsation
     > * Tidal perturbation of NS/WD and the gravitational radiation (GW)
2. If the energy is generated by nuclear burning

   $$
   \text dL=4\pi r^2\rho\text dr\cdot\varepsilon\_\text{nuc}
   $$

   where $\varepsilon\_\text{nuc}$ is the **energy generation rate per unit mass**. Then

   $$
   \frac{\partial L}{\partial m}=\varepsilon\_\text{nuc}
   $$

   and

   $$
   \Delta E=\Delta mc^2\sim\Delta t\int\_0^M\varepsilon\_\text{nuc}\text dm
   $$
3. If the energy is used to increase the internal energy and/or expand/contract a mass shell

   The first law in thermodynamics requires

   $$
   \text de+P\text{d}\left(\frac1\rho\right)=T\text ds\Rightarrow \frac{\partial e}{\partial t}+P\frac{\partial}{\partial t}\left(\frac1\rho\right)=T\frac{\partial s}{\partial t}
   $$

   we $s$ is the **specific entropy**. Thus

   $$
   \frac{\partial L}{\partial m}=\varepsilon\_\text{nuc}-T\frac{\partial s}{\partial t}\equiv \varepsilon\_\text{nuc}+\varepsilon\_\text{gas}
   $$

   For contraction, $\varepsilo&#x6E;*\text{gas}>0$, or expansion, $\varepsilon*\text{gas}<0$. Therefore, if there is no nuclear burning, $\partial L/\partial m=\varepsilon\_\text{gas}$, and the total luminosity is given by

   $$
   \begin{align\*}
   L&=\int\_0^M\frac{\partial L}{\partial m}\text dm=-\int\_0^M\text dm\left\[\frac{\partial e}{\partial t}+P\frac{\partial}{\partial t}\left(\frac1\rho\right)\right]\\
   &=-\frac{\text dE\_\text{int}}{\text dt}-\int\_0^MP\frac{\partial}{\partial t}\left(\frac1\rho\right)\text dm
   \end{align\*}
   $$

   From the perspective of energy conservation, the second term should correspond to the time derivative of $E\_\text{g}$. Now we give a proof.

   The virial theorem reveals that

   $$
   E\_g=-3\int\_0^M\frac{P}{\rho}\text dm
   $$

   The time derivative is thus

   $$
   \frac{\text dE\_\text{g}}{\text dt}=-3\int\_0^M\left\[\frac{\partial P}{\partial t}\frac{1}{\rho}+P\frac{\partial}{\partial t}\left(\frac1\rho\right)\right]\text dm
   $$

   We just need to prove

   $$
   -3\int\_0^M\left\[+P\frac{\partial}{\partial t}\left(\frac1\rho\right)\right]\text dm=\int\_0^MP\frac{\partial}{\partial t}\left(\frac1\rho\right)\text dm \Leftarrow 3P\frac{\partial}{\partial t}\left(\frac1\rho\right)+4\frac{\partial P}{\partial t}\frac{1}{\rho}=0
   $$

   $$
   \iff 3\frac{\partial \ln(1/\rho)}{\partial t}+4\frac{\partial \ln P}{\partial t}=4\frac{\partial \ln P}{\partial t}-3\frac{\partial \ln \rho}{\partial t}=0
   $$

   $$
   \iff\frac{\partial}{\partial t}\ln\left(\frac{P}{\rho^{4/3}}\right)=0
   $$

   This is true for the radiation-dominated case, where

   $$
   P\propto\rho^{4/3}
   $$

   Finally, we have derived

   $$
   L=-\frac{\text d}{\text dt}\left(E\_\text{int}+E\_\text g\right)
   $$

   simply assuming spherical symmetry, hydrostatic equilibrium in radiation-dominated fluid.
4. If neutrinos carry energy away, we have to somehow modify our equation

   $$
   \frac{\partial L}{\partial m}=\varepsilon\_\text{nuc}+\varepsilon\_\text{gas}-\varepsilon\_\nu
   $$

   Neutrinos hardly interact with matter. The typical cross section for neutrino scattering is $\sigm&#x61;*\nu\sim10^{-44}$ cm$^{2}$, about 20 orders of magnitude smaller than the cross section of Thomson scattering. Take the Sun as an example, the mean free path $l*\text{mfp}$ of neutrinos in the Sun is

   $$
   l\_{\text{mfp},\nu}=\frac1{n\sigma\_\nu}\sim10^{19-20}\text{ cm}>1\text{ AU}\gg R\_\odot
   $$

   As a result, neutrinos produced in the Sun escape without any scattering. They can be treated as an energy sink in stars.

## Total Energy Conservation of a Star

The energy conservation law of the whole star is

$$
L+L\_\nu=-\frac{\text d}{\text dt}\left(E\_\text{int}+E\_\text{g}+E\_\text{nuc}\right)
$$

where the **nuclear energy** $E\_\text{nuc}$ is given by

$$
E\_\text{nuc}=\int\text dt\int\_0^M\varepsilon\_\text{nuc}\text dm
$$

On the other hand,

$$
L=\int\_0^M\left(\varepsilon\_\text{nuc}+\varepsilon\_\text{gas}-\varepsilon\_\nu\right)\text dm=-\frac{\text dE\_\text{nuc}}{\text dt}-L\_\nu+\int\_0^M\varepsilon\_\text{gas}\text dm
$$

$$
\Rightarrow \frac{\text d}{\text dt}\left(E\_\text{int}+E\_\text{gas}\right)+\int\_0^M\varepsilon\_\text{gas}\text dm=0
$$

## Energy Source of the Sun

**Nuclear timescale**

$$
t\_\text{nuc}=\frac{E\_\text{nuc}}{L}
$$

is the timescale of continuous, stable nuclear burning in a star. Considering the hydrogen fusion

$$
\ce{4 ^1H->^4He}\Rightarrow Q=\frac{\Delta mc^2}{4m\_\text{p}}=6.3\times10^{18}\text{ erg/g}
$$

where $Q$ stands for the available energy per unit mass.

For the Sun,

$$
t\_\text{nuc}\sim\frac{M\_\odot Q}{L\_\odot}\sim10^2\text{ Gyr}
$$

Recall that

$$
t\_\text{sc}\approx t\_\text{ff}\sim 10^3\text{ s}\quad\ll\quad t\_\text{KH}\sim 10^{0-1}\text{ Myr}\quad \ll \quad t\_\text{nuc}\sim10^2\text{ Gyr}
$$

This is usually the case for a normal, stable star.

**What Powers Our Sun?**

The energy conversion efficiency of the Sun is approximately

$$
\sim\frac{L\_\odot}{M\_\odot}=1.5\text{ erg/s/g}
$$

According to radioactive elements in comets, the age of our system is at least $\sim 10$ Gyr.

1. Gravity

   $$
   Q\sim\frac{E\_\text{g}}{M\_\odot}\sim\frac{GM\_\odot}{R\_\odot}=2\times10^{15}\text{ erg/s}
   $$

   Considering the energy conversion efficiency, the Sun would be burnt out after

   $$
   t\_\text{g}\sim10^{15}\text{ s}\sim10^2\text{ Myr}
   $$

   Not enough!
2. Chemical Reaction

   For hydrogen atoms, $E\sim13.6$ eV, so

   $$
   Q\sim \frac{13.6\text{ eV}}{m\_\text{p}}=10^{13}\text{ erg}/g\Rightarrow t\_\text{chem}<1\text{ Myr}
   $$

   Not enough!
3. Nuclear Burning

   $$
   t\_\text{nuc}\sim10^2\text{ Gyr}
   $$

   Sufficient!
