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--- courses:ast301:1 [2024/02/06 20:22] – [5.2 CNO cycle] asad
+++ courses:ast301:1 [2024/03/07 01:12] (current) – [4.2 Dynamical timescale] asad
@@ Line 113: / Line 113: @@
 where we assume $N(N-1)\approx N^2$ because $N$ is large. The factor $N(N-1)/2$ comes from the **combination** of two items from a total of $N$ items: $_NC_2=N!/[2!(N-2)!]=N(N-1)/2$. If total mass $M=Nm$ then
-$$ M \langle v_i^2 \rangle_{av} - G\frac{M^2}{2} \langle r_{ij}^{-1} \rangle_{av} $$
+$$ M \langle v_i^2 \rangle_{av} - G\frac{M^2}{2} \langle r_{ij}^{-1} \rangle_{av} = 0$$
 and hence the **virial mass** of a cluster
@@ Line 153: / Line 153: @@
 The time needed to fall by a distance $R$ is found from the constant-force expression $s=at^2/2$, i. e. the dynamical timescale
-$$ \tau_{in} = \sqrt{\frac{R}{a}} = \sqrt{R^3}{GM} $$
+$$ \tau_{in} = \sqrt{\frac{R}{a}} = \sqrt{\frac{R^3}{GM}} $$
 where density $\rho=M/R^3$ giving rise to
@@ Line 258: / Line 258: @@
 Here is the cycle of six reactions. $^{12}$C fuses with $^1$H first producing $^{13}$N and a photon. The nitrogen decays producing a positron and an electron neutrino. This is known as **beta decay** because electrons were called //beta rays// as opposed to //alpha rays// ($^4$He) in the early days of radioactivity studies.
-Here $^{12}$ works only as a catalyst.
+Here $^{12}$C works only as a catalyst.
 ==== - Energy production ====
+The atomic masses of H and He are 1.00783 amu and 4.00260 amu, respectively, where 1 amu = $1.66053\times 10^{-27}$ kg. The energy released by a single set of interactions in pp or CNO process
+$$ E_r = -\Delta E_{rest} = (4M_H-M_{He})c^2 = 4.29\times 10^{-12} \text{ J} = 26.75 \text{ MeV} $$
+which is 0.71% of the rest-mass energy of four H atoms. The //per capita// energy production
+$$ \frac{E_r}{4M_H} = 0.0071 c^2 = 6.4\times 10^{14} \text{ J/kg} $$
+or almost 640 TJ/kg. If I had 640 terra-joules of energy available per kg, I would be able to survive 10 ky. I would need only 10 g of mass for my whole life. Only 2% of this energy in the sun is carried away by neutrinos.
+Nuclear burning is is 10 million times more efficient than chemical burning.
+Only 10% of H is located in the core where temperature high enough for H fusion. Taking into account the 2% loss due to $\nu_e$, the total energy produced by H burning
+$$ E_H \approx (0.98 \times 6.4\times 10^{14} \text{ J/kg}) (0.1\times 2\times 10^{30} \text{ kg}) = 1.3\times 10^{44} \text{ J}  $$
+or almost 1300 tredecillion joules. So the lifetime of the sun
+$$ \tau_\odot = \frac{E_H}{L_\odot} $$
+which is around 10 Gy. This is much larger than the thermal timescale we found before.
+The **energy generation rate** $\epsilon_{pp}$ is an important quantity that describes the amount of energy produced per kg (W/kg). We know that $\epsilon \propto X^2 (\rho^2/\rho) T^\beta$ where $X$ is the fraction of mass that is H, $\rho$ is the density of the star and $T$ temperature. The order 2 arises because the rate depends on the flux of both the 'projectile' and 'target' particles. We divide this by $\rho$ to get the energy rate per mass.
+$$ \epsilon = \epsilon_0 X^2 \left(\frac{\rho}{10^5 \text{ kg/m}^3}\right) \left(\frac{T}{10^7 \text{ K}}\right)^\beta $$
+^ Chain/cycle ^ Temperature where dominant (MK) ^ $\beta$ factor ^ Stars where dominant ^
+| pp | $5\sim 15$ | 4 | Sun and less massive |
+| CNO | $\ge 20$ | 15 | [[uv:stars|Type A]] and more massive |
+For the sun, $X=0.71$, but it has been reduced to $0.36$ at the center. Put the other values as $\rho=150$ Mg/m$^3$, $T=16$ MK and $\beta=4$ to get $\epsilon_{pp}=2.4$ mW/kg.
+At higher temperatures, CNO cycle dominates the pp chain because of its higher sensitivity to temperature: $\epsilon\propto T^{15}$.
 ===== - Limiting luminosity =====
+The luminosity of a star $L\propto M^{\sim 3}$, but the maximum possible luminosity, called the Eddington luminosity, $L_E\propto M$ which sets an upper limit to the mass of a star. If the mass is more than this, the radiation blows away the photosphere.
+{{:courses:ast301:eddington.png?nolink&600|}}
+Assuming the is made of ionized hydrogen, i. e. free protons and electrons, the gravitational force $F_G$ primarily works on the protons and the radiative force $F_{\text{rad}}$ primarily on the electrons. Eddington luminosity is found by equating the two.
+For finding the outward radiative force, note that the energy carried by a photon $E=h\nu=pc$ where $p$ is the momentum. The radiation pressure
+$$ P_{\text{rad}} = \dfrac{dp}{dt}\frac{1}{A} = \left(\frac{dE}{dt}\frac{1}{A}\right)\frac{1}{c} = \frac{\phi}{c} $$
+where $\phi$ is the energy flux in units of W m$^{-2}$. Then the outward radiative force on a single electron
+$$ F_{\text{rad,e}} = P_{\text{rad}} \sigma_T = \frac{\phi\sigma_T}{c} = \frac{L}{4\pi r^2} \frac{\sigma_T}{c} $$
+because flux at a radius $r$ is $L/(4\pi r^2)$. And the inward gravitational force on a single proton at the same radius
+$$ F_{\text{G,p}} = -\frac{GM\mu_em_p}{r^2} $$
+where //electron molecular weight// $\mu_e$ is the number of nucleons (protons and neutrons) per electron. Now condition for the **Eddington limit** says
+$$ F_{\text{G,p}} + F_{\text{rad,e}} = 0 $$
+which entails
+$$ L_E = \frac{4\pi G M_\odot \mu_e m_p c}{\sigma_T} \frac{M}{M_\odot} = 1.26\times 10^{31}\mu_e \frac{M}{M_\odot} \text{ W} = 3.27\times 10^4 \mu_e\frac{M}{M_\odot} \text{ L}_\odot. $$
+Note that the Eddington luminosity is independent of distance from the center and it is 33,000 times greater than solar luminosity for a 1-solar-mass star. So gravity can confine the solar plasma within the sun except for the million-degree hot gas in the corona which propagates throughout the solar system as //solar wind//.
+In order to find the **mass limit** for a hydrogen burning star note that $L\propto M^{3.2}$ if the pp process is dominant. That means $L=(M/M_\odot)^{3.2} L_\odot$ which we can equate with the above equation to get
+$$ M_{\text{max}} = (3.27\times 10^4 \mu_e)^{1/2.2} M_\odot = 113 M_\odot $$
+when $\mu_e=1$ or the gas is made of only hydrogen. And the luminosity of such a star will be
+$$ L = 113^{3.2} = 3.6\times 10^6 L_\odot $$
+or a million times the luminosity of the sun. We have not found a star more than 130 times massive than the sun.
+The stars close to their Eddington limit are erratic and called //luminous blue variables// (LBV). Their brightness changes with periods of months to years. For example, $\eta$ Car and P Cyg have shown outbursts in historical times. The ejections from LBVs are due, in part, to the radiation pressure.
+==== - Mass accretion ====
+{{https://upload.wikimedia.org/wikipedia/commons/thumb/2/2a/Accretion_disk.jpg/960px-Accretion_disk.jpg?nolink&400}}
+The Eddinton luminosity for a 1.4 M$_\odot$ neutron star is around $1.8\times 10^{31}$ W and the maximum luminosity of neutron stars has been observed to be around that. This proves that the emission mechanism of a neutron star is mass falling onto a neutron star from a companion star.
+If a star has an accretion rate of $\dot{m}=dm/dt$, then the accretion luminosity
+$$ L_{\text{acc}} = \frac{GM \dot{m}}{R} $$
+which is associated with the potential energy lost by a mass $dm$ as it infalls from infinity to a radius $R$. We can define an Eddington accretion rate $\dot{m}_E$ by equating the luminosity to $L_E$:
+$$ \dot{m}_E \approx 1.26\times 10^{31} \frac{R}{GM_\odot} $$
+which does not depend on the mass of the recipient star. For a 10-km neutron star $ \dot{m}_E \approx 10^{15} $ kg/s or $10^{-8}$ M$_\odot$/year.
+{{https://upload.wikimedia.org/wikipedia/commons/thumb/3/38/Artist%27s_rendering_ULAS_J1120%2B0641.jpg/1024px-Artist%27s_rendering_ULAS_J1120%2B0641.jpg?nolink&500}}
+On the other hand, we have found quasars with luminosity of $10^{39}$ W. Setting this value in the equation of Eddington luminosity we find the mass of the accreting object to be $10^8$ M$_\odot$. As the emission is often variable it must be coming from a small region of light-year size. And only black holes can have 100 million solar masses within a light year. This was the proof that the radiation of quasars is coming from matter falling into a black hole.
+The radius of a black hole is around $R=2GM/c^2$ which would be 2 astronomical unit for 100 million solar masses. Putting this value in the accretion rate equation we find that such a black hole gobbles up only half a solar mass per year which is pretty modest.
 ===== - Instability and pulsation =====
+Stars are not totally stable, all stars are subjected to some variability, they all physically oscillate, shrink and expand in a periodic manner. We study the oscillations using atmospheric velocity vectors and brightness in the field rightly called //asteroseismology// and, in case of the sun, //helioseismology//.
+If the oscillation amplitude is high, we call the stars //pulsating variables// that include quasi-periodic //cepheid variables// and RR Lyrae variables which are used to measure the distance of galaxies. Let us deal with the basic thermodynamics controlling these two types of variable stars.
+{{:courses:ast301:starcarnot.webp?nolink&750|}}
+The oscillation of these stars can be explained using the 4-step Carnot cycle used for describing the changes of state of a volume of gas. During the cycle, the gas absorbs heat and does some work on the surroundings. On the P-V plot above, work $W=\int P dV$ is the area under a curve with certain limits. Work is positive for movement to the right (upper path, isothermal expansion), negative for movement to the left (lower path, isothermal compression). The lower path yields lower negative work due to lower pressure compared to the the positive work for the upper path. So the net work is positive, which is always the case for clockwise cycle. If the cycle was counterclockwise, work would be done on the gas, i. e. work would be negative.
+In a star, the heat $Q$ is provided by radiant energy of the hot gas and the work results in a physical expansion and compression of the whole star, and the internal energy $U$ does not change.
+First law of thermodynamics states the law of energy conservation for a reversible process:
+$$ \delta Q = dU + \delta W $$
+where the $d$ and $\delta$ make it clear that $U$ is a //state variable// (internal property of a system) while $Q$ and $W$ are not. Pressure $P$, volume $V$, temperature $T$ and entropy $S$ are also state variables. $\delta Q$ is positive when gas absorbs heat, $\delta W$ is positive when gas works on the surroundings. Over a complete cycle, the loop integral
+$$ \oint dU = 0 \Rightarrow W = + \oint \delta Q $$
+where the positive sign makes it explicit that pulsations occur only if work is positive. Because entropy is a state variable its integral over the cycle must be zero as well:
+$$ \oint dS = \oint \frac{\delta Q}{T} = 0. $$
+Now, the variation of temperature can be approximated as a small fluctuation $\Delta T(t)$ in time over an average temperature $T_0$:
+$$ T(t) = T_0 + \Delta T(t) = T_0\left(1+\frac{\Delta T(t)}{T_0}\right). $$
+Assuming $\Delta T / T \ll 1$ we can approximate via an expansion that
+$$ \frac{1}{T} = \frac{1}{T_0} \left(1+\frac{\Delta T}{T_0}\right)^{-1} = \frac{1}{T_0} \left(1-\frac{\Delta T}{T_0}\right) $$
+which can be plugged in the equation of entropy to get
+$$ W = \oint \delta Q \approx \oint \frac{\Delta T(t)}{T_0} \delta Q $$
+which means $W$ and $\Delta T$ have the same sign, heat is absorbed (work on the surrounding) when temperature increases and discharged (work on the gas) when temperature decreases. This is similar to a **heat engine**, for example, an internal combustion engine. In both cases, heat is introduced during an expansion at high temperature ($T_2$) and dumped during a compression at low temperature ($T_1$).
+The ultimate **condition for pulsation** is achieved when we integrate the above equation over the whole mass of the star:
+$$ W \approx \int_M \oint_Q \frac{\Delta T(t,m)}{T_0(m)} \delta Q(m) \ dm > 0. $$
+In a heat engine, the operative mechanism is the burning of fuel. In a star, the operative mechanism is the **valving of heat** by the changing opacity in a star's ionization transition zone as shown below.
+{{:courses:ast301:pulsation.webp?nolink&700|}}
+Inward of the transition zone gas is hot and ionized, outward the gas is cold and neutral. For H and He, the transition zone is close to the surface. The transition zone itself is a mix of hot and cold gas.
+(a) When a star compresses (due to a breakdown of equilibrium), temperature in the transition zone increases leading to more ionization. The number of free electrons increases leading to a higher **opacity**, meaning obstruction to photons. Photons are trapped by the free electrons via scattering. The trapped outward radiation provides heat which is absorbed at a high temperature a required by a heat engine.
+(b) The trapped heat causes the star to expand and cool. This leads to recombination for many atoms, so ionization and number of free electrons decrease. The highway of photons is open again and the photons carry away the trapped heat. Heat is released at a low temperature.
+Due to higher volume and lower temperature, pressure decreases ($P \propto T/V$) and the outer layers of the star collapse due to self-gravity leading to a compression which takes us back to (a).
+RR Lyrae and Cepheid variables can be used to measure large distance because they are very luminous (300 to 30,000 times higher than the sun) and their luminosity is related to their period.