From b160c3b67851a74de8c31de8b5cc71168a99547b Mon Sep 17 00:00:00 2001 From: Billy Quarles <4674360+saturnaxis@users.noreply.github.com> Date: Mon, 11 Dec 2023 15:34:02 -0500 Subject: [PATCH] Updated Chapter 8 --- ...terstellar-medium-and-star-formation.ipynb | 277 ++++++++++++------ ...nterstellar-medium-and-star-formation.html | 265 +++++++++-------- .../interaction-of-light-and-matter.html | 68 ++--- docs/Chapter_9/stellar-atmospheres.html | 104 +++---- ...terstellar-medium-and-star-formation.ipynb | 277 ++++++++++++------ docs/objects.inv | Bin 7318 -> 7353 bytes docs/searchindex.js | 2 +- 7 files changed, 594 insertions(+), 399 deletions(-) diff --git a/Chapter_12/interstellar-medium-and-star-formation.ipynb b/Chapter_12/interstellar-medium-and-star-formation.ipynb index aabcb77..c343a63 100644 --- a/Chapter_12/interstellar-medium-and-star-formation.ipynb +++ b/Chapter_12/interstellar-medium-and-star-formation.ipynb @@ -20,7 +20,9 @@ "### Interstellar Extinction\n", "On a dark night (far from city lights), some of the dust clouds within the Milky Way Galaxy can be found because they obscure regions of sky that would otherwise be populated with stars. These dark regions are not devoid of stars, but the intervening dust blocks the background starlight. Dust that blocks starlight is due to the combined effet of scattering and absorption, which is called **interstellar extinction**. The apparent magnitude is a measure of how bright a star appears and if it is dimmed by intervening dust, the distance modulus equation must be modified appropriately. In a given wavelength band centered on $\\lambda$, we have\n", "\n", - "$$ m_\\lambda = M_\\lambda + 5\\log_{10}\\,d-5 + A_\\lambda, \\tag{1}$$\n", + "\\begin{align}\n", + "m_\\lambda = M_\\lambda + 5\\log_{10}\\,d-5 + A_\\lambda, \n", + "\\end{align}\n", "\n", "which is a modified version of the distance modulus equation with a correction factor $A_\\lambda>0$ that adjusts for the interstellar extinction along the line of sight. The extinction magnitude $A_\\lambda$ is positive because it always acts to dim objects along the magnitude scale. If $A_\\lambda$ is large enough, a star can fall behind the limiting magnitude of the background sky for the naked eye or a telescope. This is the cause for the dark bands running through the Milky Way.\n", "\n", @@ -54,15 +56,21 @@ "\n", "The change in apparent magnitude is simply the extinction magnitude $A_\\lambda$, or\n", "\n", - "$$ A_\\lambda = 1.086 \\tau_\\lambda, \\tag{2}$$\n", + "\\begin{align}\n", + " A_\\lambda = 1.086 \\tau_\\lambda, \n", + "\\end{align}\n", "\n", "which restates that *the change in magnitude due to extinction is approximately equal to the optical depth along the line of sight*. The optical depth through the cloud is \n", "\n", - "$$ \\tau_\\lambda = \\int_0^s n_d(s^\\prime)\\sigma_\\lambda\\,ds^\\prime, \\tag{3} $$\n", + "\\begin{align}\n", + "\\tau_\\lambda = \\int_0^s n_d(s^\\prime)\\sigma_\\lambda\\,ds^\\prime, \n", + "\\end{align}\n", "\n", "which depends on the number density of scattering dust grains $n_d(s^\\prime)$ and the scattering cross section $\\sigma_\\lambda$. If the scattering cross section is constant along the line of sight, then \n", "\n", - "$$ \\tau_\\lambda = \\sigma_\\lambda \\int_0^sn_d(s^\\prime),ds^\\prime = \\sigma_\\lambda N_d, \\tag{4}$$\n", + "\\begin{align}\n", + "\\tau_\\lambda = \\sigma_\\lambda \\int_0^sn_d(s^\\prime),ds^\\prime = \\sigma_\\lambda N_d, \n", + "\\end{align}\n", "\n", "and the dust grain *column density* $N_d$ represents the number of scattering dust particles in a thin cylinder with a cross section of 1 ${\\rm m^2}$ stretching from the observer to the star. The magnitude of extinction depends on the amount of interstellar dust that the light passes through." ] @@ -78,11 +86,15 @@ "\n", "where the extinction coefficient $Q_\\lambda$ depends on the composition of the dust grains. When the wavelength of the light is similar to the size of the dust grains, then $Q_\\lambda \\sim a/\\lambda$, implying that\n", "\n", - "$$ \\sigma_\\lambda \\propto \\frac{a^3}{\\lambda} \\qquad (\\lambda \\gtrsim a). \\tag{5}$$\n", + "\\begin{align}\n", + "\\sigma_\\lambda \\propto \\frac{a^3}{\\lambda} \\qquad (\\lambda \\gtrsim a).\n", + "\\end{align}\n", "\n", "The extinction coefficient $Q_\\lambda$ goes to zero when the wavelength $\\lambda$ is large compared to $a$. Conversely, the extinction coefficient $Q_\\lambda$ approaches a constant (independent of $\\lambda$) when the wavelength $\\lambda$ is very small compared to $a$ or\n", "\n", - "$$ \\sigma_\\lambda \\propto a^2 \\qquad (\\lambda \\ll a). \\tag{6} $$\n", + "\\begin{align}\n", + "\\sigma_\\lambda \\propto a^2 \\qquad (\\lambda \\ll a). \n", + "\\end{align}\n", "\n", "```{note}\n", "Gustav Mie made similar assumptions concerning dust particles in 1908 and showed the relationship between the scattering cross section $\\sigma_\\lambda$ and the radius of a dust grain.\n", @@ -95,6 +107,8 @@ "Strongly scattered blue light can leave the cloud in any direction, where a bright star behind a cloud casts a blue **reflection nebula** for observers that are not along the line of sight (e.g., the Pleiades). This process is analogous to Rayleigh scattering, which produces a blue sky on Earth. The difference between Mie scattering and Rayleigh scattering is that the scattering particles are much smaller than the wavelength of visible light in Rayleigh scattering and leads to $\\sigma_\\lambda \\propto \\lambda^{-4}$.\n", "\n", "```{exercise}\n", + ":class: orange\n", + "\n", "**A star is found to be dimmer than expected at 550 nm by 1.1 magnitudes in the visual (i.e., $A_V = 1.1$). If $Q_{550}=1.5$ and the radius of the dust grains are 0.2 ${\\rm \\mu m}$, estimate the average density $\\overline{n}$ of the material between the star and Earth. The star is 0.8 kpc from Earth.**\n", "\n", "The extinction magnitude $A_\\lambda$ is related to the optical depth $\\tau_\\lambda$ by a faction of 1.086, where $\\tau_{550} = 1.1/1.086 \\simeq 1$. Given the radius of the dust grains $(a = 0.2\\,{\\rm \\mu m})$ and the extinction coefficient $Q_{550} = 1.5$ we have,\n", @@ -232,12 +246,15 @@ "\n", "The rarity of 21 cm emission (or absorption) from individual atoms means that the central wavelength can remain optically thin over large interstellar distances. If we assume that the line profile is approximately Gaussian (like the shape of the Doppler profile), then the optical depth of the line is given by\n", "\n", - "$$ \\tau_H = 5.2 \\times 10^{-23} \\frac{N_H}{T\\,\\Delta v}, \\tag{7} $$\n", + "```{math}\n", + ":label: tau_hydro\n", + "\\tau_H = 5.2 \\times 10^{-23} \\frac{N_H}{T\\,\\Delta v}, \n", + "```\n", "\n", "which depends on the column density of neutral hydrogen $N_H$, the temperature $T$ (in K) of the gas, and the full width of the line at half maximum $\\Delta v$ (in ${\\rm km/s}$).\n", "\n", "```{note}\n", - "In the print version of the textbook (Carroll & Ostlie (2007)), the coefficient for Eqn. 7 is given as $5.2 \\times 10^{-15}$. The value above is corrected to an appropriate value.\n", + "In the print version of the textbook (Carroll & Ostlie (2007)), the coefficient for Eq. {eq}`tau_hydro` is given as $5.2 \\times 10^{-15}$. The value above is corrected to an appropriate value.\n", "```\n", "\n", "As long as the 21 cm hydrogen line is optically thin (i.e., on the linear part of the curve of growth), the optical depth is proportional to the neutral hydrogen column density. Studies of **diffuse ${\\rm H\\,I}$ clouds** show temperatures of $30-80\\,{\\rm K}$, number densities ranging from $1-8 \\times 10^8\\,{\\rm m^{-3}}$, and masses between $1-100\\,M_\\odot$. When $A_V<1$, the gas and dust are distributed together throughout the ISM. This correlation breaks down for $A_V > 1$, where the column density of ${\\rm H\\,I}$ no longer increases rapidly with the column density of the dust and other physical processes are involved as the dust becomes optically thick.\n", @@ -418,49 +435,72 @@ "\n", "By the virial theorem, the condition for collapse becomes\n", "\n", - "$$ \\frac{3M_c kT}{\\mu m_H} < \\frac{3}{5} \\frac{GM_c^2}{R_c}. \\tag{8}$$\n", + "```{math}\n", + ":label: virial_collapse\n", + "\\frac{3M_c kT}{\\mu m_H} < \\frac{3}{5} \\frac{GM_c^2}{R_c}. \n", + "```\n", "\n", "The cloud radius may be replaced by using the initial mass density of the cloud $\\rho_o$, which is assumed to be constant throughout the cloud to get\n", "\n", - "$$ R_c = \\left( \\frac{3 M_c}{4\\pi \\rho_o}\\right)^{1/3}. \\tag{9} $$\n", + "```{math}\n", + ":label: cloud_radius\n", + "R_c = \\left( \\frac{3 M_c}{4\\pi \\rho_o}\\right)^{1/3}. \n", + "```\n", "\n", - "Then Eqn. 8 can be solved to determine the minimum mass necessary to initiate the spontaneous cloud collapse, which is a condition known as the **Jeans criterion** $(M_c \\gtrsim M_J)$, or\n", + "Then Eq. {eq}`virial_collapse` can be solved to determine the minimum mass necessary to initiate the spontaneous cloud collapse, which is a condition known as the **Jeans criterion** $(M_c \\gtrsim M_J)$, or\n", "\n", - "$$ M_J \\simeq \\left(\\frac{5kT}{G\\mu m_H}\\right)^{3/2}\\left(\\frac{3}{4\\pi \\rho_o}\\right)^{1/2}. \\tag{10} $$\n", + "```{math}\n", + ":label: Jeans_crit\n", + "M_J \\simeq \\left(\\frac{5kT}{G\\mu m_H}\\right)^{3/2}\\left(\\frac{3}{4\\pi \\rho_o}\\right)^{1/2}.\n", + "```\n", "\n", - "The critical mass $M_J$ is called the **Jeans mass**. The Jeans criterion may also be expressed in terms of the minimum radius $(R_c > R_J)$ through the cloud density $\\rho_o$ (and solving Eq. 8 in terms of $M_c$) to get the **Jeans length**, or\n", + "The critical mass $M_J$ is called the **Jeans mass**. The Jeans criterion may also be expressed in terms of the minimum radius $(R_c > R_J)$ through the cloud density $\\rho_o$ (and solving Eq. {eq}`virial_collapse` in terms of $M_c$) to get the **Jeans length**, or\n", "\n", - "$$ R_J \\simeq \\sqrt{\\frac{15kT}{4\\pi G\\mu m_H \\rho_o}}. \\tag{11} $$\n", + "\\begin{align}\n", + "R_J \\simeq \\sqrt{\\frac{15kT}{4\\pi G\\mu m_H \\rho_o}}. \n", + "\\end{align}\n", "\n", "The Jeans mass derivation neglected the existence of an external pressure on the cloud due to the surrounding interstellar medium. The critical mass required for gravitational collapse in the presence of an external gas pressure $P_o$ is given by the **Bonnor-Ebert mass**,\n", "\n", - "$$ M_{\\rm BE} = \\frac{c_{\\rm BE}v_T^4}{P_o^{1/2}G^{3/2}}, \\tag{12} $$\n", + "\\begin{align}\n", + "M_{\\rm BE} = \\frac{c_{\\rm BE}v_T^4}{P_o^{1/2}G^{3/2}}, \n", + "\\end{align}\n", "\n", "where the *isothermal sound speed* $v_T$ $(\\gamma = 1)$ is\n", "\n", - "$$ v_T \\equiv \\sqrt{\\frac{kT}{\\mu m_H}}, \\tag{13} $$\n", + "\\begin{align}\n", + "v_T \\equiv \\sqrt{\\frac{kT}{\\mu m_H}}, \n", + "\\end{align}\n", "\n", - "and the dimensionless constant $c_{\\rm BE}$ is approximately $1.18$. A dimensionless constant $c_J$ can be derived from Eqn. 10 by substitution of $v_T$ and the ideal gas law, where $c_J \\simeq 5.46$. The smaller constant for the Bonnor-Ebert mass is expected since an external compression force (due to $P_o$) is being exerted on the cloud.\n", + "and the dimensionless constant $c_{\\rm BE}$ is approximately $1.18$. A dimensionless constant $c_J$ can be derived from Eq. {eq}`Jeans_crit` by substitution of $v_T$ and the ideal gas law, where $c_J \\simeq 5.46$. The smaller constant for the Bonnor-Ebert mass is expected since an external compression force (due to $P_o$) is being exerted on the cloud.\n", "\n", - "- **For a typical diffuse hydrogen cloud assumed to be entirely composed of ${\\rm H\\, I}$, $T = 50\\,{\\rm K}$ and $n = 5 \\times 10^8\\, {\\rm m^{-3}}$, what is the minimum mass necessary to cause the cloud to collapse spontaneously?**\n", + "```{exercise}\n", + ":class: orange\n", "\n", - ">Since the cloud is entirely neutral hydrogen, the initial density $\\rho_o$ is calculated by \n", - ">\n", - ">$$ \\rho_o = n_H m_H = 8.4 \\times 10^{-19}\\, {\\rm kg/m^3}.$$\n", - ">\n", - ">Taking $\\mu = 1$ and substituting into the Jeans mass equation (Eqn. 10), we get\n", - ">\n", - ">$$ M_J \\simeq \\left(\\frac{5kT}{G\\mu m_H}\\right)^{3/2}\\left(\\frac{3}{4\\pi \\rho_o}\\right)^{1/2} \\sim 1500\\,M_\\odot. $$\n", - "> \n", - ">This value significantly exceeds the estimated $1-100\\,M_\\odot$ believed to be contained in ${\\rm H\\,I}$ clouds. Hence diffuse hydrogen clouds are stable against gravitational collapse.\n", + "**For a typical diffuse hydrogen cloud assumed to be entirely composed of ${\\rm H\\, I}$, $T = 50\\,{\\rm K}$ and $n = 5 \\times 10^8\\, {\\rm m^{-3}}$, what is the minimum mass necessary to cause the cloud to collapse spontaneously?**\n", "\n", - "- **For a dense core of a giant molecular cloud (i.e., composed of ${\\rm H_2}$), the typical temperatures and number densities are $T = 10\\,{\\rm K}$ and $n = 10^{10}\\, {\\rm m^{-3}}$, what is the minimum mass necessary to cause the cloud to collapse?**\n", + "Since the cloud is entirely neutral hydrogen, the initial density $\\rho_o$ is calculated by \n", "\n", - ">For a cloud of molecular hydrogen, the initial density $\\rho_o$ is calculated by\n", - ">\n", - ">$$ \\rho_o = 2n_H m_H = 3 \\times 10^{-17}\\, {\\rm kg/m^3},$$\n", - ">\n", - ">and now $\\mu = 2$. In this case the Jeans mass is $M_J \\sim 7\\,M_\\odot$, where the characteristic mass of dense cores is $\\sim 10\\,M_\\odot$. Apparently, the dense cores of GMCs are unstable to gravitational collapse and consistent with being sites of star formation. If the Bonnor-Eber mass is used as the critical collapse conditions, then the required mass reduces to approximately $2\\,M_\\odot$." + "$$ \\rho_o = n_H m_H = 8.4 \\times 10^{-19}\\, {\\rm kg/m^3}.$$\n", + "\n", + "Taking $\\mu = 1$ and substituting into the Jeans mass equation (Eq. {eq}`Jeans_crit`), we get\n", + "\n", + "$$ M_J \\simeq \\left(\\frac{5kT}{G\\mu m_H}\\right)^{3/2}\\left(\\frac{3}{4\\pi \\rho_o}\\right)^{1/2} \\sim 1500\\,M_\\odot. $$\n", + " \n", + "This value significantly exceeds the estimated $1-100\\,M_\\odot$ believed to be contained in ${\\rm H\\,I}$ clouds. Hence diffuse hydrogen clouds are stable against gravitational collapse.\n", + "```\n", + "\n", + "```{exercise}\n", + ":class: orange\n", + "\n", + "**For a dense core of a giant molecular cloud (i.e., composed of ${\\rm H_2}$), the typical temperatures and number densities are $T = 10\\,{\\rm K}$ and $n = 10^{10}\\, {\\rm m^{-3}}$, what is the minimum mass necessary to cause the cloud to collapse?**\n", + "\n", + "For a cloud of molecular hydrogen, the initial density $\\rho_o$ is calculated by\n", + "\n", + "$$ \\rho_o = 2n_H m_H = 3 \\times 10^{-17}\\, {\\rm kg/m^3},$$\n", + "\n", + "and now $\\mu = 2$. In this case the Jeans mass is $M_J \\sim 7\\,M_\\odot$, where the characteristic mass of dense cores is $\\sim 10\\,M_\\odot$. Apparently, the dense cores of GMCs are unstable to gravitational collapse and consistent with being sites of star formation. If the Bonnor-Eber mass is used as the critical collapse conditions, then the required mass reduces to approximately $2\\,M_\\odot$.\n", + "```" ] }, { @@ -504,19 +544,22 @@ "metadata": {}, "source": [ "### Homologous Collapse\n", - "Following Jean's derivation, the criterion for gravitational collapse of a molecular cloud can be satisfied in the absence of rotation, turbulence, or magnetic fields. If we make another simplifying assumption that *any existing pressure gradients are small*, then the cloud is essentially in free-fall. Throughout the free-fall phase, the temperature of the gas remains nearly constant (or *isothermal*). This is true as long as the collapse remains optically thin and the gravitational potential energy released during collapse can be efficiently radiated away (i.e., the gas can remain cold). In this case the spherically symmetric hydrodynamic equation can be used to describe the contraction. From Eqn. 5 of [Sect. 7.1.2](https://saturnaxis.github.io/ModernAstro/Chapter_11/interiors-of-stars.html#the-derivation-of-the-hydrostatic-equilibrium-equation), we have\n", + "Following Jean's derivation, the criterion for gravitational collapse of a molecular cloud can be satisfied in the absence of rotation, turbulence, or magnetic fields. If we make another simplifying assumption that *any existing pressure gradients are small*, then the cloud is essentially in free-fall. Throughout the free-fall phase, the temperature of the gas remains nearly constant (or *isothermal*). This is true as long as the collapse remains optically thin and the gravitational potential energy released during collapse can be efficiently radiated away (i.e., the gas can remain cold). In this case the spherically symmetric hydrodynamic equation can be used to describe the contraction. From Eq. {eq}`diff_mass`, we have\n", "\n", - "$$ \\frac{d^2r}{dt^2} = -\\frac{GM_r}{r^2}. \\tag{14} $$\n", + "```{math}\n", + ":label: hydro_equil_gravity\n", + "\\frac{d^2r}{dt^2} = -\\frac{GM_r}{r^2}. \n", + "``````\n", "\n", - "The right-hand side (RHS) of Eqn. 14 is just the local acceleration due to gravity at a distance $r$ from the center of the spherical cloud and depends on the mass contained $M_r$ at that radius.\n", + "The right-hand side (RHS) of Eq. {eq}`hydro_equil_gravity` is just the local acceleration due to gravity at a distance $r$ from the center of the spherical cloud and depends on the mass contained $M_r$ at that radius.\n", "\n", - "To describe the behavior of the surface of a sphere within the collapsing cloud as function of time, Eqn. 14 must be integrated with respect to time. Since we are using the mass enclosed at $r$, $M_r$ remains a constant in a similar way as the charge in the integration of Gauss' law. As a result, we can replace the mass enclosed $M_r$ by the product of the initial density $\\rho_o$ and the spherical volume. Then we apply a \"trick\" used in many problems, where we multiply by the velocity $dr/dt$ on both sides so that the second derivative can be replaced by an anti-derivative through the chain rule. The resulting expression is\n", + "To describe the behavior of the surface of a sphere within the collapsing cloud as function of time, Eq. {eq}`hydro_equil_gravity` must be integrated with respect to time. Since we are using the mass enclosed at $r$, $M_r$ remains a constant in a similar way as the charge in the integration of Gauss' law. As a result, we can replace the mass enclosed $M_r$ by the product of the initial density $\\rho_o$ and the spherical volume. Then we apply a \"trick\" used in many problems, where we multiply by the velocity $dr/dt$ on both sides so that the second derivative can be replaced by an anti-derivative through the chain rule. The resulting expression is\n", "\n", - "$$ \\begin{align*}\n", + "\\begin{align*}\n", "\\frac{dr}{dt}\\frac{d^2 r}{dt^2} &= \\frac{1}{2}\\frac{d}{dt}\\left(\\frac{dr}{dt}\\right)^2 \\\\\n", "- \\left(\\frac{4}{3}\\pi G \\rho_o r_o^3 \\right)\\frac{1}{r^2}\\frac{dr}{dt} &= \\left(\\frac{4}{3}\\pi G \\rho_o r_o^3 \\right)\\frac{d}{dt}\\left(\\frac{1}{r}\\right) \\\\\n", "\\frac{1}{2}\\frac{d}{dt}\\left(\\frac{dr}{dt}\\right)^2 &= \\left(\\frac{4}{3}\\pi G \\rho_o r_o^3 \\right)\\frac{d}{dt}\\left(\\frac{1}{r}\\right),\n", - "\\end{align*} $$\n", + "\\end{align*}\n", "\n", "which can be integrated with respect to time to give\n", "\n", @@ -528,29 +571,44 @@ "\n", "Through substitution, we can solve for the velocity at the surface as\n", "\n", - "$$ \\frac{dr}{dt} = -\\left[ \\frac{8}{3}\\pi G \\rho_o r_o^2 \\left(\\frac{r_o}{r}-1 \\right)\\right]^{1/2}, \\tag{15} $$\n", + "```{math}\n", + ":label: vel_surf\n", + "\\frac{dr}{dt} = -\\left[ \\frac{8}{3}\\pi G \\rho_o r_o^2 \\left(\\frac{r_o}{r}-1 \\right)\\right]^{1/2},\n", + "```\n", "\n", - "where the negative root was chosen because the cloud is collapsing. To integrate Eqn. 15, we make the variable substitution $\\theta \\equiv r/r_o$ and $\\chi \\equiv \\sqrt{(8/3)\\pi G \\rho_o}$, which leads to a simpler differential equation,\n", + "where the negative root was chosen because the cloud is collapsing. To integrate Eq. {eq}`vel_surf`, we make the variable substitution $\\theta \\equiv r/r_o$ and $\\chi \\equiv \\sqrt{(8/3)\\pi G \\rho_o}$, which leads to a simpler differential equation,\n", "\n", - "$$ \\frac{d\\theta}{dt} = -\\chi\\left(\\frac{1}{\\theta} - 1 \\right)^{1/2} \\tag{16}.$$\n", + "```{math}\n", + ":label: vel_surf_angle\n", + "\\frac{d\\theta}{dt} = -\\chi\\left(\\frac{1}{\\theta} - 1 \\right)^{1/2} \n", + "```\n", "\n", - "The integral of Eqn. 16 is quite messy and we make yet another substitution,\n", + "The integral of Eq. {eq}`vel_surf_angle` is quite messy and we make yet another substitution,\n", "\n", - "$$ \\theta \\equiv \\cos^2 \\xi, \\tag{17} $$\n", + "\\begin{align}\n", + "\\theta \\equiv \\cos^2 \\xi, \n", + "\\end{align}\n", "\n", - "where $\\frac{d\\theta}{dt} = -2\\cos \\xi \\sin \\xi \\frac{d\\xi}{dt}$ and Eqn. 16 becomes\n", + "where $\\frac{d\\theta}{dt} = -2\\cos \\xi \\sin \\xi \\frac{d\\xi}{dt}$ and Eq. {eq}`vel_surf_angle` becomes\n", "\n", - "$$ \\cos^2 \\xi \\frac{d\\xi}{dt} = \\frac{\\chi}{2}. \\tag{18} $$\n", + "```{math}\n", + ":label: vel_surf_sub\n", + "\\cos^2 \\xi \\frac{d\\xi}{dt} = \\frac{\\chi}{2}. \n", + "```\n", "\n", - "Equation 18 can be integrated directly to yield\n", + "Equation {eq}`vel_surf_sub` can be integrated directly to yield\n", "\n", - "$$ \\frac{\\xi}{2} + \\frac{1}{4}\\sin 2\\xi = \\frac{\\chi}{2}t + C_2. \\tag{19} $$\n", + "\\begin{align}\n", + "\\frac{\\xi}{2} + \\frac{1}{4}\\sin 2\\xi = \\frac{\\chi}{2}t + C_2.\n", + "\\end{align}\n", "\n", "The integration constant $C_2$ can be evaluated using the same initial condition as before ($r=r_o$ at $t=0$), which implies that $d\\theta/dt = 0$ or $\\xi = 0$ at the beginning of the collapse. Therefore, $C_2=0$.\n", "\n", "The equation of motion for the gravitational collapse of the cloud (in parameterized form) is\n", "\n", - "$$ \\xi + \\frac{1}{2}\\sin 2\\xi = \\chi t. \\tag{20} $$\n", + "\\begin{align}\n", + "\\xi + \\frac{1}{2}\\sin 2\\xi = \\chi t.\n", + "\\end{align}\n", "\n", "The **free-fall timescale** $t_{\\rm ff}$ for a cloud is the time when the radius of the collapsing sphere reaches zero $(\\theta = 0\\,{\\rm or}\\, \\xi = \\pi/2)$ or at least becomes very small. Then \n", "\n", @@ -558,17 +616,24 @@ "\n", "Back-substituting our value for $\\chi$, we have\n", "\n", - "$$ t_{\\rm ff} = \\left( \\frac{3\\pi}{32}\\frac{1}{G\\rho_o} \\right)^{1/2}. \\tag{21} $$\n", + "```{math}\n", + ":label: tau_ff\n", + "t_{\\rm ff} = \\left( \\frac{3\\pi}{32}\\frac{1}{G\\rho_o} \\right)^{1/2}. \n", + "```\n", "\n", "The free-fall time is actually *independent* of the initial radius of the sphere. AS long as the original density of the spherical molecular cloud was uniform, all parts of the cloud will take the same amount of time to collapse and the density will increase at the same rate everywhere. This behavior is known as **homologous collapse**. However, if the cloud is somewhat centrally condensed when the collapse begins, the free-fall time will be shorter for material near the center compared to material farther out (i.e., $t_{\\rm ff} \\propto 1/\\sqrt{\\rho_o}$). As the collapse progresses, the density will increase more rapidly near the center than in the other regions and in this case, the collapse is known as **inside-out collapse**.\n", "\n", - "- **For a dense core of a giant molecular cloud, what is the time required for collapse?**\n", + "```{exercise}\n", + ":class: orange\n", "\n", - ">From the example in the Jeans Criterion, we have the initial density $\\rho_o$ and we can find the free-fall timescale by\n", - ">\n", - ">$$ t_{\\rm ff} = \\left( \\frac{3\\pi}{32}\\frac{1}{G\\rho_o} \\right)^{1/2} = 3.6 \\times 10^5\\,{\\rm yr}. $$\n", - ">\n", - ">Equation 20 is transcendental and cannot be solved explicitly. But, numerical root finding techniques can be used. The collapse is slow initially and accelerates as $t_{\\rm ff}$ is approached. At the same time, the density increases most rapidly during the final stages of collapse." + "**For a dense core of a giant molecular cloud, what is the time required for collapse?**\n", + "\n", + "From the example in the Jeans Criterion, we have the initial density $\\rho_o$ and we can find the free-fall timescale by\n", + "\n", + "$$ t_{\\rm ff} = \\left( \\frac{3\\pi}{32}\\frac{1}{G\\rho_o} \\right)^{1/2} = 3.6 \\times 10^5\\,{\\rm yr}. $$\n", + "\n", + "Equation 20 is transcendental and cannot be solved explicitly. But, numerical root finding techniques can be used. The collapse is slow initially and accelerates as $t_{\\rm ff}$ is approached. At the same time, the density increases most rapidly during the final stages of collapse.\n", + "```" ] }, { @@ -677,9 +742,11 @@ "\n", "For an adiabatic process the gas pressure is related to its density by $\\gamma$ (the ratio of specific heats). An adiabatic relation between density and temperature can be obtained (using the ideal gas law),\n", "\n", - "$$ T = K^{\\prime\\prime}\\rho^{\\gamma -1}, \\tag{22} $$\n", + "\\begin{align}\n", + "T = K^{\\prime\\prime}\\rho^{\\gamma -1}, \n", + "\\end{align}\n", "\n", - "where $K^{\\prime\\prime}$ is a constant. Substituting into the equation for the Jeans mass (Eqn. 10), we find that\n", + "where $K^{\\prime\\prime}$ is a constant. Substituting into the equation for the Jeans mass (Eq. {eq}`Jeans_crit`), we find that\n", "\n", "$$ M_J \\propto \\rho^{(3\\gamma-4)/2}, $$\n", "\n", @@ -701,9 +768,11 @@ "\n", "$$ M_J^{5/2} = \\frac{4\\pi}{G^{3/2}}R_J^{9/2}e\\sigma T^4. $$\n", "\n", - "Using the initial cloud radius at a constant density (Eqn. 9), and using Eqn. 10 to write the density in terms of the Jeans mass, we can determine the minimum obtainable Jeans mass:\n", + "Using the initial cloud radius at a constant density (Eq. {eq}`cloud_radius`), and using Eq. {eq}`Jeans_crit` to write the density in terms of the Jeans mass, we can determine the minimum obtainable Jeans mass:\n", "\n", - "$$ M_{J_{\\rm min}} = 0.03 \\left(\\frac{T^{1/4}}{e^{1/2} \\mu^{9/4}}\\right)\\, M_\\odot, \\tag{23} $$\n", + "\\begin{align} \n", + "M_{J_{\\rm min}} = 0.03 \\left(\\frac{T^{1/4}}{e^{1/2} \\mu^{9/4}}\\right)\\, M_\\odot, \n", + "\\end{align}\n", "\n", "which depends on the temperature $T$ (in K), mean molecular weight $\\mu$, and the efficiency factor $e$. If we take $\\mu \\sim 1$, $e\\sim 0.1$, and $T \\sim 1000\\,{\\rm K}$ to be representative when adiabatic effects become significant, then $M_J \\sim 0.5 M_\\odot$. Fragmentation ceases when the segments of the original cloud begin to reach the range of solar mass objects. The estimate is relatively insensitive to other reasonable choices, where if $e\\sim 1$, then $M_J \\sim 0.2 M_\\odot$.\n" ] @@ -729,21 +798,29 @@ "\n", "The virial theorem was invoked during the derivation of the Jean's criterion as a balance between the gravitational potential energy and the cloud's internal (thermal) kinetic energy. The magnetic field energy was ignored, where including the effects of the magnetic field produces a critical mass of\n", "\n", - "$$ M_B = c_B \\frac{\\pi R^2 B}{G^{1/2}}, \\tag{24}$$\n", + "\\begin{align}\n", + "M_B = c_B \\frac{\\pi R^2 B}{G^{1/2}}, \n", + "\\end{align}\n", "\n", "where $c_B = 380\\,{\\rm N^{1/2}/m/T}$ for a magnetic field permeating a spherical, uniform cloud. If the magnetic field strength $B$ is expressed in ${\\rm nT}$ and the cloud radius $R$ in units of ${\\rm pc}$, then $M_B$ can be written as\n", "\n", - "$$ M_B \\simeq 70\\,M_\\odot \\left(\\frac{B}{1\\,{\\rm nT}} \\right) \\left(\\frac{R}{1\\,{\\rm pc}} \\right)^2. \\tag{25} $$\n", + "\\begin{align}\n", + "M_B \\simeq 70\\,M_\\odot \\left(\\frac{B}{1\\,{\\rm nT}} \\right) \\left(\\frac{R}{1\\,{\\rm pc}} \\right)^2. \n", + "\\end{align}\n", "\n", "If the mass of the cloud $M_c$ is less than $M_B$, the cloud is said to be **magnetically subcritical** and stable against collapse. Otherwise, the cloud is **magnetically supercritical** and the force due to gravity will overwhelm the ability of the magnetic field to resist collapse.\n", "\n", - "- **What is the critical mass if the dense core has a magnetic field strength of $100\\,{\\rm nT}$ and a radius of $0.1\\,{\\rm pc}$?**\n", + "```{exercise}\n", + ":class:orange\n", "\n", - ">The critical mass after including the magnetic field can be calculated as,\n", - ">\n", - ">$$ M_B \\simeq 70\\,M_\\odot \\left(\\frac{100}{1\\,{\\rm nT}} \\right) \\left(\\frac{0.1}{1\\,{\\rm pc}} \\right)^2 = 70\\,M_\\odot, $$\n", - ">\n", - ">implying that a dense core of $10\\,M_\\odot$ would be stable against collapse $(M_c < M_B)$. However, if the magnetic field was weaker $(B=1\\,{\\rm nT})$, then $M_B \\simeq 0.7\\,M_\\odot$ and collapse would occur." + "**What is the critical mass if the dense core has a magnetic field strength of $100\\,{\\rm nT}$ and a radius of $0.1\\,{\\rm pc}$?**\n", + "\n", + "The critical mass after including the magnetic field can be calculated as,\n", + "\n", + "$$ M_B \\simeq 70\\,M_\\odot \\left(\\frac{100}{1\\,{\\rm nT}} \\right) \\left(\\frac{0.1}{1\\,{\\rm pc}} \\right)^2 = 70\\,M_\\odot, $$\n", + "\n", + "implying that a dense core of $10\\,M_\\odot$ would be stable against collapse $(M_c < M_B)$. However, if the magnetic field was weaker $(B=1\\,{\\rm nT})$, then $M_B \\simeq 0.7\\,M_\\odot$ and collapse would occur.\n", + "```" ] }, { @@ -763,15 +840,21 @@ "\n", "To determine the relative impact of ambipolar diffusion, we need to estimate the characteristic timescale. If we know the drift velocity of the neutrals and the distance across the molecular cloud, then it can be shown that\n", "\n", - "$$ t_{\\rm AD} \\simeq \\frac{2R}{v_{\\rm drift}} \\simeq 10\\,{\\rm Gyr} \\left( \\frac{n_{\\rm H_2}}{10^{10}\\,{\\rm m^{-3}}}\\right) \\left( \\frac{B}{1\\,{\\rm nT}}\\right)^{-2} \\left( \\frac{R}{1\\,{\\rm pc}}\\right)^{2}. \\tag{26} $$\n", + "\\begin{align}\n", + "t_{\\rm AD} \\simeq \\frac{2R}{v_{\\rm drift}} \\simeq 10\\ {\\rm Gyr} \\left( \\frac{n_{\\rm H_2}}{10^{10}\\,{\\rm m^{-3}}}\\right) \\left( \\frac{B}{1\\,{\\rm nT}}\\right)^{-2} \\left( \\frac{R}{1\\,{\\rm pc}}\\right)^{2}.\n", + "\\end{align}\n", "\n", - "- **What is the timescale for ambipolar diffusion within a dense core of a GMC, if $B = 1\\,{\\rm nT}$ and $R = 0.1\\,{\\rm pc}$?**\n", + "```{exercise}\n", + ":class: orange\n", "\n", - ">The characteristic number density for molecular hydrogen in a dense core is $n_{\\rm H_2} = 10^{10}\\,{\\rm m^{-3}}$. The timescale for ambipolar diffusion (AD) is then determined as\n", - ">\n", - ">$$ t_{\\rm AD} \\simeq 10\\,{\\rm Gyr} \\left( \\frac{0.1\\,{\\rm pc}}{1\\,{\\rm pc}}\\right)^{2} \\simeq 100\\,{\\rm Myr}. $$\n", + "**What is the timescale for ambipolar diffusion within a dense core of a GMC, if $B = 1\\,{\\rm nT}$ and $R = 0.1\\,{\\rm pc}$?**\n", + "\n", + "The characteristic number density for molecular hydrogen in a dense core is $n_{\\rm H_2} = 10^{10}\\,{\\rm m^{-3}}$. The timescale for ambipolar diffusion (AD) is then determined as\n", + "\n", + "$$ t_{\\rm AD} \\simeq 10\\,{\\rm Gyr} \\left( \\frac{0.1\\,{\\rm pc}}{1\\,{\\rm pc}}\\right)^{2} \\simeq 100\\,{\\rm Myr}. $$\n", ">\n", - ">This is ~$1000\\times$ longer than the free-fall timescale determined for [homologous collapse](https://saturnaxis.github.io/ModernAstro/Chapter_12/interstellar-medium-and-star-formation.html#homologous-collapse). Clearly the ambipolar diffusion process can control the evolution of a dense core for a long time before free-fall collapse begins." + "This is ~$1000\\times$ longer than the free-fall timescale determined for [homologous collapse](https://saturnaxis.github.io/ModernAstro/Chapter_12/interstellar-medium-and-star-formation.html#homologous-collapse). Clearly the ambipolar diffusion process can control the evolution of a dense core for a long time before free-fall collapse begins.\n", + "```" ] }, { @@ -920,29 +1003,35 @@ "\n", "which solving for the radius produces\n", "\n", - "$$ r \\simeq \\left( \\frac{3N}{4\\pi \\alpha n_H^2} \\right)^{1/3} \\tag{27} $$\n", + "\\begin{align}\n", + "r \\simeq \\left( \\frac{3N}{4\\pi \\alpha n_H^2} \\right)^{1/3} \n", + "\\end{align}\n", "\n", "and is called the **Strömgren radius** $r_S$.\n", "\n", - "- **What is the Strömgren radius due to an O6 star in a typical ${\\rm H\\,II}$ region?** (Assume that $T_e\\simeq 45,000\\,{\\rm K}$ and $L\\simeq 1.3\\times10^5\\,L_\\odot$)\n", + "```{exercise}\n", + ":class:orange\n", "\n", - ">From Wien's law, the peak wavelength is given as\n", - ">\n", - ">$$ \\lambda_{\\rm max} = \\frac{0.0029\\,{\\rm m\\,K}}{T_e} \\approx 64\\,{\\rm nm}.$$\n", - ">\n", - ">This is significantly shorter (or higher energy) than the 91.2-nm limit ([${\\rm Ly_{\\rm limit}}$ from the wavelength of hydrogen](https://saturnaxis.github.io/ModernAstro/Chapter_5/interaction-of-light-and-matter.html#the-wavelength-of-hydrogen)) and it can be assumed that most of the photons created by an O6 star are capable of causing ionization. The energy of one 64-nm photon can be calculated giving\n", - ">\n", - ">$$E_\\gamma = \\frac{hc}{\\lambda} = 19\\,{\\rm eV}.$$\n", - ">\n", - ">Assuming that all of the emitted photons have the same (peak wavelength), then the total number of photons produced *per second* is just\n", - ">\n", - ">$$ N \\simeq \\frac{L}{E_\\gamma} \\simeq 1.6 \\times 10^{49}\\,{\\rm photons/s}.$$\n", - ">\n", - ">Lastly, a typical the number density within a GMC is $n_H \\sim 10^8\\,{\\rm m^{-3}}$, we find\n", - ">\n", - ">$$ r_S \\simeq \\left( \\frac{3N}{4\\pi \\alpha n_H^2} \\right)^{1/3} \\simeq 3.48\\,{\\rm pc}$$\n", - ">\n", - ">Values of $r_S$ range from less than 0.1 pc to greater than 100 pc." + "**What is the Strömgren radius due to an O6 star in a typical ${\\rm H\\,II}$ region?** (Assume that $T_e\\simeq 45,000\\,{\\rm K}$ and $L\\simeq 1.3\\times10^5\\,L_\\odot$)\n", + "\n", + "From Wien's law, the peak wavelength is given as\n", + "\n", + "$$ \\lambda_{\\rm max} = \\frac{0.0029\\,{\\rm m\\,K}}{T_e} \\approx 64\\,{\\rm nm}.$$\n", + "\n", + "This is significantly shorter (or higher energy) than the 91.2-nm limit ([${\\rm Ly_{\\rm limit}}$ from the wavelength of hydrogen](https://saturnaxis.github.io/ModernAstro/Chapter_5/interaction-of-light-and-matter.html#the-wavelength-of-hydrogen)) and it can be assumed that most of the photons created by an O6 star are capable of causing ionization. The energy of one 64-nm photon can be calculated giving\n", + "\n", + "$$E_\\gamma = \\frac{hc}{\\lambda} = 19\\,{\\rm eV}.$$\n", + "\n", + "Assuming that all of the emitted photons have the same (peak wavelength), then the total number of photons produced *per second* is just\n", + "\n", + "$$ N \\simeq \\frac{L}{E_\\gamma} \\simeq 1.6 \\times 10^{49}\\,{\\rm photons/s}.$$\n", + "\n", + "Lastly, a typical the number density within a GMC is $n_H \\sim 10^8\\,{\\rm m^{-3}}$, we find\n", + "\n", + "$$ r_S \\simeq \\left( \\frac{3N}{4\\pi \\alpha n_H^2} \\right)^{1/3} \\simeq 3.48\\,{\\rm pc}$$\n", + "\n", + "Values of $r_S$ range from less than 0.1 pc to greater than 100 pc.\n", + "```" ] }, { @@ -1111,7 +1200,7 @@ "```\n", "\n", "```{admonition} Problem 5\n", - "Assuming that the free-fall acceleration of the surface of a collapsing cloud remains constant during the entire collapse, derive an expression for the free-fall time. Show that your answer differs from Eqn. 21 only by a term of order unity.\n", + "Assuming that the free-fall acceleration of the surface of a collapsing cloud remains constant during the entire collapse, derive an expression for the free-fall time. Show that your answer differs from Eq. {eq}`tau_ff` only by a term of order unity.\n", "``` \n", "\n", "```{admonition} Problem 6\n", diff --git a/docs/Chapter_12/interstellar-medium-and-star-formation.html b/docs/Chapter_12/interstellar-medium-and-star-formation.html index b142e08..dc03d21 100644 --- a/docs/Chapter_12/interstellar-medium-and-star-formation.html +++ b/docs/Chapter_12/interstellar-medium-and-star-formation.html @@ -508,8 +508,10 @@

8.1.1. The Interstellar Medium

8.1.2. Interstellar Extinction#

On a dark night (far from city lights), some of the dust clouds within the Milky Way Galaxy can be found because they obscure regions of sky that would otherwise be populated with stars. These dark regions are not devoid of stars, but the intervening dust blocks the background starlight. Dust that blocks starlight is due to the combined effet of scattering and absorption, which is called interstellar extinction. The apparent magnitude is a measure of how bright a star appears and if it is dimmed by intervening dust, the distance modulus equation must be modified appropriately. In a given wavelength band centered on \(\lambda\), we have

-
-\[ m_\lambda = M_\lambda + 5\log_{10}\,d-5 + A_\lambda, \tag{1}\]
+
+(8.1)#\[\begin{align} +m_\lambda = M_\lambda + 5\log_{10}\,d-5 + A_\lambda, +\end{align}\]

which is a modified version of the distance modulus equation with a correction factor \(A_\lambda>0\) that adjusts for the interstellar extinction along the line of sight. The extinction magnitude \(A_\lambda\) is positive because it always acts to dim objects along the magnitude scale. If \(A_\lambda\) is large enough, a star can fall behind the limiting magnitude of the background sky for the naked eye or a telescope. This is the cause for the dark bands running through the Milky Way.

https://github.com/saturnaxis/ModernAstro/blob/main/Chapter_12/Figure_1.jpg?raw=true? @@ -530,14 +532,20 @@

8.1.2. Interstellar Extinction\(A_\lambda\), or

-
-\[ A_\lambda = 1.086 \tau_\lambda, \tag{2}\]
+
+(8.2)#\[\begin{align} + A_\lambda = 1.086 \tau_\lambda, +\end{align}\]

which restates that the change in magnitude due to extinction is approximately equal to the optical depth along the line of sight. The optical depth through the cloud is

-
-\[ \tau_\lambda = \int_0^s n_d(s^\prime)\sigma_\lambda\,ds^\prime, \tag{3} \]
+
+(8.3)#\[\begin{align} +\tau_\lambda = \int_0^s n_d(s^\prime)\sigma_\lambda\,ds^\prime, +\end{align}\]

which depends on the number density of scattering dust grains \(n_d(s^\prime)\) and the scattering cross section \(\sigma_\lambda\). If the scattering cross section is constant along the line of sight, then

-
-\[ \tau_\lambda = \sigma_\lambda \int_0^sn_d(s^\prime),ds^\prime = \sigma_\lambda N_d, \tag{4}\]
+
+(8.4)#\[\begin{align} +\tau_\lambda = \sigma_\lambda \int_0^sn_d(s^\prime),ds^\prime = \sigma_\lambda N_d, +\end{align}\]

and the dust grain column density \(N_d\) represents the number of scattering dust particles in a thin cylinder with a cross section of 1 \({\rm m^2}\) stretching from the observer to the star. The magnitude of extinction depends on the amount of interstellar dust that the light passes through.

@@ -546,11 +554,15 @@

8.1.3. The Mie Theory \[ Q_\lambda \equiv \frac{\sigma_\lambda}{\sigma_g}, \]

where the extinction coefficient \(Q_\lambda\) depends on the composition of the dust grains. When the wavelength of the light is similar to the size of the dust grains, then \(Q_\lambda \sim a/\lambda\), implying that

-
-\[ \sigma_\lambda \propto \frac{a^3}{\lambda} \qquad (\lambda \gtrsim a). \tag{5}\]
+
+(8.5)#\[\begin{align} +\sigma_\lambda \propto \frac{a^3}{\lambda} \qquad (\lambda \gtrsim a). +\end{align}\]

The extinction coefficient \(Q_\lambda\) goes to zero when the wavelength \(\lambda\) is large compared to \(a\). Conversely, the extinction coefficient \(Q_\lambda\) approaches a constant (independent of \(\lambda\)) when the wavelength \(\lambda\) is very small compared to \(a\) or

-
-\[ \sigma_\lambda \propto a^2 \qquad (\lambda \ll a). \tag{6} \]
+
+(8.6)#\[\begin{align} +\sigma_\lambda \propto a^2 \qquad (\lambda \ll a). +\end{align}\]

Note

Gustav Mie made similar assumptions concerning dust particles in 1908 and showed the relationship between the scattering cross section \(\sigma_\lambda\) and the radius of a dust grain.

@@ -558,7 +570,7 @@

8.1.3. The Mie Theory\((\sigma_\lambda \sim 0)\). However, if the object (e.g., an island) is large compared to the waves, then they are simply blocked. The only waves that pass by the large object are those that miss it altogether. The light we detect from a dust cloud is the light that passes between the dust particles.

The extinction magnitude \(A_\lambda\) is clearly wavelength-dependent, where longer wavelengths (redder light) are not scattered as strongly as shorter wavelengths (bluer light). The starlight that passes through is preferentially red, or reddened, as the blue light is removed. This interstellar reddening causes stars to appear redder, where we might infer them to be cooler through Wien’s law. Fortunately, we can carefully analyze the star’s spectrum (i.e., absorption and emission lines) to detect the change.

Strongly scattered blue light can leave the cloud in any direction, where a bright star behind a cloud casts a blue reflection nebula for observers that are not along the line of sight (e.g., the Pleiades). This process is analogous to Rayleigh scattering, which produces a blue sky on Earth. The difference between Mie scattering and Rayleigh scattering is that the scattering particles are much smaller than the wavelength of visible light in Rayleigh scattering and leads to \(\sigma_\lambda \propto \lambda^{-4}\).

-
+

Exercise 8.1

@@ -653,12 +665,12 @@

8.1.6. 21-cm Radiation of Hydrogen

estimating magnetic fields using the Zeeman effect.

The rarity of 21 cm emission (or absorption) from individual atoms means that the central wavelength can remain optically thin over large interstellar distances. If we assume that the line profile is approximately Gaussian (like the shape of the Doppler profile), then the optical depth of the line is given by

-
-\[ \tau_H = 5.2 \times 10^{-23} \frac{N_H}{T\,\Delta v}, \tag{7} \]
+
+(8.7)#\[\tau_H = 5.2 \times 10^{-23} \frac{N_H}{T\,\Delta v}, \]

which depends on the column density of neutral hydrogen \(N_H\), the temperature \(T\) (in K) of the gas, and the full width of the line at half maximum \(\Delta v\) (in \({\rm km/s}\)).

Note

-

In the print version of the textbook (Carroll & Ostlie (2007)), the coefficient for Eqn. 7 is given as \(5.2 \times 10^{-15}\). The value above is corrected to an appropriate value.

+

In the print version of the textbook (Carroll & Ostlie (2007)), the coefficient for Eq. (8.7) is given as \(5.2 \times 10^{-15}\). The value above is corrected to an appropriate value.

As long as the 21 cm hydrogen line is optically thin (i.e., on the linear part of the curve of growth), the optical depth is proportional to the neutral hydrogen column density. Studies of diffuse \({\rm H\,I}\) clouds show temperatures of \(30-80\,{\rm K}\), number densities ranging from \(1-8 \times 10^8\,{\rm m^{-3}}\), and masses between \(1-100\,M_\odot\). When \(A_V<1\), the gas and dust are distributed together throughout the ISM. This correlation breaks down for \(A_V > 1\), where the column density of \({\rm H\,I}\) no longer increases rapidly with the column density of the dust and other physical processes are involved as the dust becomes optically thick.

Optically thick dust clouds shield hydrogen from sources of UV and a consequence is that molecular hydrogen can exist without the threat of dissociation by UV photons. Dust can also enhance \({\rm H_2}\) formation rate beyond what would be expected through the collision of hydrogen atoms. The enhancement occurs because:

@@ -780,45 +792,55 @@

8.2.1. The Jeans Criterion \[ N = \frac{M_c}{\mu m_H}. \]

By the virial theorem, the condition for collapse becomes

-
-\[ \frac{3M_c kT}{\mu m_H} < \frac{3}{5} \frac{GM_c^2}{R_c}. \tag{8}\]
+
+(8.8)#\[\frac{3M_c kT}{\mu m_H} < \frac{3}{5} \frac{GM_c^2}{R_c}. \]

The cloud radius may be replaced by using the initial mass density of the cloud \(\rho_o\), which is assumed to be constant throughout the cloud to get

-
-\[ R_c = \left( \frac{3 M_c}{4\pi \rho_o}\right)^{1/3}. \tag{9} \]
-

Then Eqn. 8 can be solved to determine the minimum mass necessary to initiate the spontaneous cloud collapse, which is a condition known as the Jeans criterion \((M_c \gtrsim M_J)\), or

-
-\[ M_J \simeq \left(\frac{5kT}{G\mu m_H}\right)^{3/2}\left(\frac{3}{4\pi \rho_o}\right)^{1/2}. \tag{10} \]
-

The critical mass \(M_J\) is called the Jeans mass. The Jeans criterion may also be expressed in terms of the minimum radius \((R_c > R_J)\) through the cloud density \(\rho_o\) (and solving Eq. 8 in terms of \(M_c\)) to get the Jeans length, or

-
-\[ R_J \simeq \sqrt{\frac{15kT}{4\pi G\mu m_H \rho_o}}. \tag{11} \]
+
+(8.9)#\[R_c = \left( \frac{3 M_c}{4\pi \rho_o}\right)^{1/3}. \]
+

Then Eq. (8.8) can be solved to determine the minimum mass necessary to initiate the spontaneous cloud collapse, which is a condition known as the Jeans criterion \((M_c \gtrsim M_J)\), or

+
+(8.10)#\[M_J \simeq \left(\frac{5kT}{G\mu m_H}\right)^{3/2}\left(\frac{3}{4\pi \rho_o}\right)^{1/2}.\]
+

The critical mass \(M_J\) is called the Jeans mass. The Jeans criterion may also be expressed in terms of the minimum radius \((R_c > R_J)\) through the cloud density \(\rho_o\) (and solving Eq. (8.8) in terms of \(M_c\)) to get the Jeans length, or

+
+(8.11)#\[\begin{align} +R_J \simeq \sqrt{\frac{15kT}{4\pi G\mu m_H \rho_o}}. +\end{align}\]

The Jeans mass derivation neglected the existence of an external pressure on the cloud due to the surrounding interstellar medium. The critical mass required for gravitational collapse in the presence of an external gas pressure \(P_o\) is given by the Bonnor-Ebert mass,

-
-\[ M_{\rm BE} = \frac{c_{\rm BE}v_T^4}{P_o^{1/2}G^{3/2}}, \tag{12} \]
+
+(8.12)#\[\begin{align} +M_{\rm BE} = \frac{c_{\rm BE}v_T^4}{P_o^{1/2}G^{3/2}}, +\end{align}\]

where the isothermal sound speed \(v_T\) \((\gamma = 1)\) is

-
-\[ v_T \equiv \sqrt{\frac{kT}{\mu m_H}}, \tag{13} \]
-

and the dimensionless constant \(c_{\rm BE}\) is approximately \(1.18\). A dimensionless constant \(c_J\) can be derived from Eqn. 10 by substitution of \(v_T\) and the ideal gas law, where \(c_J \simeq 5.46\). The smaller constant for the Bonnor-Ebert mass is expected since an external compression force (due to \(P_o\)) is being exerted on the cloud.

-
    -

  • For a typical diffuse hydrogen cloud assumed to be entirely composed of \({\rm H\, I}\), \(T = 50\,{\rm K}\) and \(n = 5 \times 10^8\, {\rm m^{-3}}\), what is the minimum mass necessary to cause the cloud to collapse spontaneously?

    • -
-
-

Since the cloud is entirely neutral hydrogen, the initial density \(\rho_o\) is calculated by

+
+(8.13)#\[\begin{align} +v_T \equiv \sqrt{\frac{kT}{\mu m_H}}, +\end{align}\]
+

and the dimensionless constant \(c_{\rm BE}\) is approximately \(1.18\). A dimensionless constant \(c_J\) can be derived from Eq. (8.10) by substitution of \(v_T\) and the ideal gas law, where \(c_J \simeq 5.46\). The smaller constant for the Bonnor-Ebert mass is expected since an external compression force (due to \(P_o\)) is being exerted on the cloud.

+
+ +

Exercise 8.2

+
+

For a typical diffuse hydrogen cloud assumed to be entirely composed of \({\rm H\, I}\), \(T = 50\,{\rm K}\) and \(n = 5 \times 10^8\, {\rm m^{-3}}\), what is the minimum mass necessary to cause the cloud to collapse spontaneously?

+

Since the cloud is entirely neutral hydrogen, the initial density \(\rho_o\) is calculated by

\[ \rho_o = n_H m_H = 8.4 \times 10^{-19}\, {\rm kg/m^3}.\]
-

Taking \(\mu = 1\) and substituting into the Jeans mass equation (Eqn. 10), we get

+

Taking \(\mu = 1\) and substituting into the Jeans mass equation (Eq. (8.10)), we get

\[ M_J \simeq \left(\frac{5kT}{G\mu m_H}\right)^{3/2}\left(\frac{3}{4\pi \rho_o}\right)^{1/2} \sim 1500\,M_\odot. \]

This value significantly exceeds the estimated \(1-100\,M_\odot\) believed to be contained in \({\rm H\,I}\) clouds. Hence diffuse hydrogen clouds are stable against gravitational collapse.

-
-
    -

  • For a dense core of a giant molecular cloud (i.e., composed of \({\rm H_2}\)), the typical temperatures and number densities are \(T = 10\,{\rm K}\) and \(n = 10^{10}\, {\rm m^{-3}}\), what is the minimum mass necessary to cause the cloud to collapse?

    • -
-
-

For a cloud of molecular hydrogen, the initial density \(\rho_o\) is calculated by

+ +
+
+ +

Exercise 8.3

+
+

For a dense core of a giant molecular cloud (i.e., composed of \({\rm H_2}\)), the typical temperatures and number densities are \(T = 10\,{\rm K}\) and \(n = 10^{10}\, {\rm m^{-3}}\), what is the minimum mass necessary to cause the cloud to collapse?

+

For a cloud of molecular hydrogen, the initial density \(\rho_o\) is calculated by

\[ \rho_o = 2n_H m_H = 3 \times 10^{-17}\, {\rm kg/m^3},\]

and now \(\mu = 2\). In this case the Jeans mass is \(M_J \sim 7\,M_\odot\), where the characteristic mass of dense cores is \(\sim 10\,M_\odot\). Apparently, the dense cores of GMCs are unstable to gravitational collapse and consistent with being sites of star formation. If the Bonnor-Eber mass is used as the critical collapse conditions, then the required mass reduces to approximately \(2\,M_\odot\).

-
+ +
from scipy.constants import G,k,pi
@@ -853,17 +875,17 @@ 
8.2.1. The Jeans Criterion
 
8.2.2. Homologous Collapse#

-
Following Jean’s derivation, the criterion for gravitational collapse of a molecular cloud can be satisfied in the absence of rotation, turbulence, or magnetic fields.  If we make another simplifying assumption that any existing pressure gradients are small, then the cloud is essentially in free-fall.  Throughout the free-fall phase, the temperature of the gas remains nearly constant (or isothermal).  This is true as long as the collapse remains optically thin and the gravitational potential energy released during collapse can be efficiently radiated away (i.e., the gas can remain cold).  In this case the spherically symmetric hydrodynamic equation can be used to describe the contraction.  From Eqn. 5 of Sect. 7.1.2, we have

-

-\[ \frac{d^2r}{dt^2} = -\frac{GM_r}{r^2}. \tag{14} \]

-
The right-hand side (RHS) of Eqn. 14 is just the local acceleration due to gravity at a distance \(r\) from the center of the spherical cloud and depends on the mass contained \(M_r\) at that radius.

-
To describe the behavior of the surface of a sphere within the collapsing cloud as function of time, Eqn. 14 must be integrated with respect to time.  Since we are using the mass enclosed at \(r\), \(M_r\) remains a constant in a similar way as the charge in the integration of Gauss’ law.  As a result, we can replace the mass enclosed \(M_r\) by the product of the initial density \(\rho_o\) and the spherical volume.  Then we apply a “trick” used in many problems, where we multiply by the velocity \(dr/dt\) on both sides so that the second derivative can be replaced by an anti-derivative through the chain rule.  The resulting expression is

-

-\[\begin{split} \begin{align*}
+
Following Jean’s derivation, the criterion for gravitational collapse of a molecular cloud can be satisfied in the absence of rotation, turbulence, or magnetic fields.  If we make another simplifying assumption that any existing pressure gradients are small, then the cloud is essentially in free-fall.  Throughout the free-fall phase, the temperature of the gas remains nearly constant (or isothermal).  This is true as long as the collapse remains optically thin and the gravitational potential energy released during collapse can be efficiently radiated away (i.e., the gas can remain cold).  In this case the spherically symmetric hydrodynamic equation can be used to describe the contraction.  From Eq. (7.5), we have

+

+(8.14)#\[\frac{d^2r}{dt^2} = -\frac{GM_r}{r^2}. \]

+
The right-hand side (RHS) of Eq. (8.14) is just the local acceleration due to gravity at a distance \(r\) from the center of the spherical cloud and depends on the mass contained \(M_r\) at that radius.

+
To describe the behavior of the surface of a sphere within the collapsing cloud as function of time, Eq. (8.14) must be integrated with respect to time.  Since we are using the mass enclosed at \(r\), \(M_r\) remains a constant in a similar way as the charge in the integration of Gauss’ law.  As a result, we can replace the mass enclosed \(M_r\) by the product of the initial density \(\rho_o\) and the spherical volume.  Then we apply a “trick” used in many problems, where we multiply by the velocity \(dr/dt\) on both sides so that the second derivative can be replaced by an anti-derivative through the chain rule.  The resulting expression is

+

+\[\begin{align*}
 \frac{dr}{dt}\frac{d^2 r}{dt^2} &= \frac{1}{2}\frac{d}{dt}\left(\frac{dr}{dt}\right)^2 \\
 - \left(\frac{4}{3}\pi G \rho_o r_o^3 \right)\frac{1}{r^2}\frac{dr}{dt} &=  \left(\frac{4}{3}\pi G \rho_o r_o^3 \right)\frac{d}{dt}\left(\frac{1}{r}\right) \\
 \frac{1}{2}\frac{d}{dt}\left(\frac{dr}{dt}\right)^2 &= \left(\frac{4}{3}\pi G \rho_o r_o^3 \right)\frac{d}{dt}\left(\frac{1}{r}\right),
-\end{align*} \end{split}\]

+\end{align*}\]

 
which can be integrated with respect to time to give

 

 \[\frac{1}{2}\left(\frac{dr}{dt}\right)^2 = \left(\frac{4}{3}\pi G \rho_o r_o^3 \right)\frac{1}{r} + C_1. \]

@@ -871,40 +893,48 @@ 
8.2.2. Homologous Collapse
 \[ C_1 = -\frac{4}{3}\pi G \rho_o r_o^2. \]

Through substitution, we can solve for the velocity at the surface as

-
-\[ \frac{dr}{dt} = -\left[ \frac{8}{3}\pi G \rho_o r_o^2 \left(\frac{r_o}{r}-1 \right)\right]^{1/2}, \tag{15} \]
-

where the negative root was chosen because the cloud is collapsing. To integrate Eqn. 15, we make the variable substitution \(\theta \equiv r/r_o\) and \(\chi \equiv \sqrt{(8/3)\pi G \rho_o}\), which leads to a simpler differential equation,

-
-\[ \frac{d\theta}{dt} = -\chi\left(\frac{1}{\theta} - 1 \right)^{1/2} \tag{16}.\]
-

The integral of Eqn. 16 is quite messy and we make yet another substitution,

-
-\[ \theta \equiv \cos^2 \xi, \tag{17} \]
-

where \(\frac{d\theta}{dt} = -2\cos \xi \sin \xi \frac{d\xi}{dt}\) and Eqn. 16 becomes

-
-\[ \cos^2 \xi \frac{d\xi}{dt} = \frac{\chi}{2}. \tag{18} \]
-

Equation 18 can be integrated directly to yield

-
-\[ \frac{\xi}{2} + \frac{1}{4}\sin 2\xi = \frac{\chi}{2}t + C_2. \tag{19} \]
+
+(8.15)#\[\frac{dr}{dt} = -\left[ \frac{8}{3}\pi G \rho_o r_o^2 \left(\frac{r_o}{r}-1 \right)\right]^{1/2},\]
+

where the negative root was chosen because the cloud is collapsing. To integrate Eq. (8.15), we make the variable substitution \(\theta \equiv r/r_o\) and \(\chi \equiv \sqrt{(8/3)\pi G \rho_o}\), which leads to a simpler differential equation,

+
+(8.16)#\[\frac{d\theta}{dt} = -\chi\left(\frac{1}{\theta} - 1 \right)^{1/2} \]
+

The integral of Eq. (8.16) is quite messy and we make yet another substitution,

+
+(8.17)#\[\begin{align} +\theta \equiv \cos^2 \xi, +\end{align}\]
+

where \(\frac{d\theta}{dt} = -2\cos \xi \sin \xi \frac{d\xi}{dt}\) and Eq. (8.16) becomes

+
+(8.18)#\[\cos^2 \xi \frac{d\xi}{dt} = \frac{\chi}{2}. \]
+

Equation (8.18) can be integrated directly to yield

+
+(8.19)#\[\begin{align} +\frac{\xi}{2} + \frac{1}{4}\sin 2\xi = \frac{\chi}{2}t + C_2. +\end{align}\]

The integration constant \(C_2\) can be evaluated using the same initial condition as before (\(r=r_o\) at \(t=0\)), which implies that \(d\theta/dt = 0\) or \(\xi = 0\) at the beginning of the collapse. Therefore, \(C_2=0\).

The equation of motion for the gravitational collapse of the cloud (in parameterized form) is

-
-\[ \xi + \frac{1}{2}\sin 2\xi = \chi t. \tag{20} \]
+
+(8.20)#\[\begin{align} +\xi + \frac{1}{2}\sin 2\xi = \chi t. +\end{align}\]

The free-fall timescale \(t_{\rm ff}\) for a cloud is the time when the radius of the collapsing sphere reaches zero \((\theta = 0\,{\rm or}\, \xi = \pi/2)\) or at least becomes very small. Then

\[ t_{\rm ff} = \frac{\pi}{2\chi}. \]

Back-substituting our value for \(\chi\), we have

-
-\[ t_{\rm ff} = \left( \frac{3\pi}{32}\frac{1}{G\rho_o} \right)^{1/2}. \tag{21} \]
+
+(8.21)#\[t_{\rm ff} = \left( \frac{3\pi}{32}\frac{1}{G\rho_o} \right)^{1/2}. \]

The free-fall time is actually independent of the initial radius of the sphere. AS long as the original density of the spherical molecular cloud was uniform, all parts of the cloud will take the same amount of time to collapse and the density will increase at the same rate everywhere. This behavior is known as homologous collapse. However, if the cloud is somewhat centrally condensed when the collapse begins, the free-fall time will be shorter for material near the center compared to material farther out (i.e., \(t_{\rm ff} \propto 1/\sqrt{\rho_o}\)). As the collapse progresses, the density will increase more rapidly near the center than in the other regions and in this case, the collapse is known as inside-out collapse.

-
    -

  • For a dense core of a giant molecular cloud, what is the time required for collapse?

    • -
-
-

From the example in the Jeans Criterion, we have the initial density \(\rho_o\) and we can find the free-fall timescale by

+
+ +

Exercise 8.4

+
+

For a dense core of a giant molecular cloud, what is the time required for collapse?

+

From the example in the Jeans Criterion, we have the initial density \(\rho_o\) and we can find the free-fall timescale by

\[ t_{\rm ff} = \left( \frac{3\pi}{32}\frac{1}{G\rho_o} \right)^{1/2} = 3.6 \times 10^5\,{\rm yr}. \]

Equation 20 is transcendental and cannot be solved explicitly. But, numerical root finding techniques can be used. The collapse is slow initially and accelerates as \(t_{\rm ff}\) is approached. At the same time, the density increases most rapidly during the final stages of collapse.

-
+ +
from scipy.constants import G
@@ -988,9 +1018,11 @@ 
8.2.3. The Fragmentation of Collapsing C
 
An important consequence of the collapse is that the density of the cloud increases by many orders of magnitude during free-fall.  Since the temperature \(T\) remains nearly constant throughout the collapse, it appears that the Jeans mass must decrease.  After the collapse begins, any initial inhomogeneities in density will cause individual sections of the cloud to satisfy the Jeans mass limit independently and begin to collapse locally.  This cascading collapse could lead to the formation of smaller objects.  However, it is likely that only about 1% of the cloud actually forms stars.

 
What stops the fragmentation process? The answer to this question lies in our implicit assumption that the collapse is isothermal, which in turn implies that the only that changes in the Jeans mass is the density.  Clearly this cannot be the case since stars have temperatures much larger than \(10-100\,{\rm K}\). The other extreme is an adiabatic collapse, where the temperature must rise.  The real situation lies somewhere between these two limits.

 
For an adiabatic process the gas pressure is related to its density by \(\gamma\) (the ratio of specific heats).  An adiabatic relation between density and temperature can be obtained (using the ideal gas law),

-

-\[ T = K^{\prime\prime}\rho^{\gamma -1}, \tag{22} \]

-
where \(K^{\prime\prime}\) is a constant.  Substituting into the equation for the Jeans mass (Eqn. 10), we find that

+

+(8.22)#\[\begin{align}
+T = K^{\prime\prime}\rho^{\gamma -1}, 
+\end{align}\]

+
where \(K^{\prime\prime}\) is a constant.  Substituting into the equation for the Jeans mass (Eq. (8.10)), we find that

 

 \[ M_J \propto \rho^{(3\gamma-4)/2}, \]

 
describes how the Jeans mass depends on density.  For atomic hydrogen (\(\gamma = 5/3\)), the dependence simplifies to \(M_J \propto \sqrt{\rho}\).  The Jeans mass increases with density for a perfectly adiabatic collapse of a cloud.  This behavior means the collapse results in a minimum fragment mass that can be produced.

@@ -1006,9 +1038,11 @@ 
8.2.3. The Fragmentation of Collapsing C
 
Equating the two expressions (free-fall and blackbody radiation) and solving in terms of the Jeans mass produces

 

 \[ M_J^{5/2} = \frac{4\pi}{G^{3/2}}R_J^{9/2}e\sigma T^4. \]

-
Using the initial cloud radius at a constant density (Eqn. 9), and using Eqn. 10 to write the density in terms of the Jeans mass, we can determine the minimum obtainable Jeans mass:

-

-\[ M_{J_{\rm min}} = 0.03 \left(\frac{T^{1/4}}{e^{1/2} \mu^{9/4}}\right)\, M_\odot, \tag{23} \]

+
Using the initial cloud radius at a constant density (Eq. (8.9)), and using Eq. (8.10) to write the density in terms of the Jeans mass, we can determine the minimum obtainable Jeans mass:

+

+(8.23)#\[\begin{align} 
+M_{J_{\rm min}} = 0.03 \left(\frac{T^{1/4}}{e^{1/2} \mu^{9/4}}\right)\, M_\odot, 
+\end{align}\]

 
which depends on the temperature \(T\) (in K), mean molecular weight \(\mu\), and the efficiency factor \(e\).  If we take \(\mu \sim 1\), \(e\sim 0.1\), and \(T \sim 1000\,{\rm K}\) to be representative when adiabatic effects become significant, then \(M_J \sim 0.5 M_\odot\).  Fragmentation ceases when the segments of the original cloud begin to reach the range of solar mass objects.  The estimate is relatively insensitive to other reasonable choices, where if \(e\sim 1\), then \(M_J \sim 0.2 M_\odot\).

 
 

@@ -1022,21 +1056,16 @@ 
8.2.4. Additional Physical Processes in
 
The term “frozen-in” refers to Alfvén’s theorem, which states that “in a fluid with infinite electric conductivity, the magnetic field is frozen into the fluid and has to move along with it.”

The virial theorem was invoked during the derivation of the Jean’s criterion as a balance between the gravitational potential energy and the cloud’s internal (thermal) kinetic energy. The magnetic field energy was ignored, where including the effects of the magnetic field produces a critical mass of

-
-\[ M_B = c_B \frac{\pi R^2 B}{G^{1/2}}, \tag{24}\]
+
+(8.24)#\[\begin{align} +M_B = c_B \frac{\pi R^2 B}{G^{1/2}}, +\end{align}\]

where \(c_B = 380\,{\rm N^{1/2}/m/T}\) for a magnetic field permeating a spherical, uniform cloud. If the magnetic field strength \(B\) is expressed in \({\rm nT}\) and the cloud radius \(R\) in units of \({\rm pc}\), then \(M_B\) can be written as

-
-\[ M_B \simeq 70\,M_\odot \left(\frac{B}{1\,{\rm nT}} \right) \left(\frac{R}{1\,{\rm pc}} \right)^2. \tag{25} \]
+
+(8.25)#\[\begin{align} +M_B \simeq 70\,M_\odot \left(\frac{B}{1\,{\rm nT}} \right) \left(\frac{R}{1\,{\rm pc}} \right)^2. +\end{align}\]

If the mass of the cloud \(M_c\) is less than \(M_B\), the cloud is said to be magnetically subcritical and stable against collapse. Otherwise, the cloud is magnetically supercritical and the force due to gravity will overwhelm the ability of the magnetic field to resist collapse.

-
    -

  • What is the critical mass if the dense core has a magnetic field strength of \(100\,{\rm nT}\) and a radius of \(0.1\,{\rm pc}\)?

    • -
-
-

The critical mass after including the magnetic field can be calculated as,

-
-\[ M_B \simeq 70\,M_\odot \left(\frac{100}{1\,{\rm nT}} \right) \left(\frac{0.1}{1\,{\rm pc}} \right)^2 = 70\,M_\odot, \]
-

implying that a dense core of \(10\,M_\odot\) would be stable against collapse \((M_c < M_B)\). However, if the magnetic field was weaker \((B=1\,{\rm nT})\), then \(M_B \simeq 0.7\,M_\odot\) and collapse would occur.

-

8.2.5. Ambipolar Diffusion#

@@ -1048,17 +1077,21 @@

8.2.5. Ambipolar Diffusion -\[ t_{\rm AD} \simeq \frac{2R}{v_{\rm drift}} \simeq 10\,{\rm Gyr} \left( \frac{n_{\rm H_2}}{10^{10}\,{\rm m^{-3}}}\right) \left( \frac{B}{1\,{\rm nT}}\right)^{-2} \left( \frac{R}{1\,{\rm pc}}\right)^{2}. \tag{26} \]

-
    -

  • What is the timescale for ambipolar diffusion within a dense core of a GMC, if \(B = 1\,{\rm nT}\) and \(R = 0.1\,{\rm pc}\)?

    • -
-
-

The characteristic number density for molecular hydrogen in a dense core is \(n_{\rm H_2} = 10^{10}\,{\rm m^{-3}}\). The timescale for ambipolar diffusion (AD) is then determined as

+
+(8.26)#\[\begin{align} +t_{\rm AD} \simeq \frac{2R}{v_{\rm drift}} \simeq 10\ {\rm Gyr} \left( \frac{n_{\rm H_2}}{10^{10}\,{\rm m^{-3}}}\right) \left( \frac{B}{1\,{\rm nT}}\right)^{-2} \left( \frac{R}{1\,{\rm pc}}\right)^{2}. +\end{align}\]
+
+ +

Exercise 8.5

+
+

What is the timescale for ambipolar diffusion within a dense core of a GMC, if \(B = 1\,{\rm nT}\) and \(R = 0.1\,{\rm pc}\)?

+

The characteristic number density for molecular hydrogen in a dense core is \(n_{\rm H_2} = 10^{10}\,{\rm m^{-3}}\). The timescale for ambipolar diffusion (AD) is then determined as

\[ t_{\rm AD} \simeq 10\,{\rm Gyr} \left( \frac{0.1\,{\rm pc}}{1\,{\rm pc}}\right)^{2} \simeq 100\,{\rm Myr}. \]

This is ~\(1000\times\) longer than the free-fall timescale determined for homologous collapse. Clearly the ambipolar diffusion process can control the evolution of a dense core for a long time before free-fall collapse begins.

-
+ +

8.2.6. Numerical Simulations of Protostellar Evolution#

@@ -1145,27 +1178,11 @@

8.3.8. \[ N \simeq \alpha n_H^2 \left(\frac{4}{3} \pi r^3 \right),\]

which solving for the radius produces

-
-\[ r \simeq \left( \frac{3N}{4\pi \alpha n_H^2} \right)^{1/3} \tag{27} \]
+
+(8.27)#\[\begin{align} +r \simeq \left( \frac{3N}{4\pi \alpha n_H^2} \right)^{1/3} +\end{align}\]

and is called the Strömgren radius \(r_S\).

-
    -

  • What is the Strömgren radius due to an O6 star in a typical \({\rm H\,II}\) region? (Assume that \(T_e\simeq 45,000\,{\rm K}\) and \(L\simeq 1.3\times10^5\,L_\odot\))

    • -
-
-

From Wien’s law, the peak wavelength is given as

-
-\[ \lambda_{\rm max} = \frac{0.0029\,{\rm m\,K}}{T_e} \approx 64\,{\rm nm}.\]
-

This is significantly shorter (or higher energy) than the 91.2-nm limit (\({\rm Ly_{\rm limit}}\) from the wavelength of hydrogen) and it can be assumed that most of the photons created by an O6 star are capable of causing ionization. The energy of one 64-nm photon can be calculated giving

-
-\[E_\gamma = \frac{hc}{\lambda} = 19\,{\rm eV}.\]
-

Assuming that all of the emitted photons have the same (peak wavelength), then the total number of photons produced per second is just

-
-\[ N \simeq \frac{L}{E_\gamma} \simeq 1.6 \times 10^{49}\,{\rm photons/s}.\]
-

Lastly, a typical the number density within a GMC is \(n_H \sim 10^8\,{\rm m^{-3}}\), we find

-
-\[ r_S \simeq \left( \frac{3N}{4\pi \alpha n_H^2} \right)^{1/3} \simeq 3.48\,{\rm pc}\]
-

Values of \(r_S\) range from less than 0.1 pc to greater than 100 pc.

-
from scipy.constants import h,c,Wien,pi,parsec
@@ -1298,7 +1315,7 @@ 
8.4. Homework
 
Problem 5

-
Assuming that the free-fall acceleration of the surface of a collapsing cloud remains constant during the entire collapse, derive an expression for the free-fall time.  Show that your answer differs from Eqn. 21 only by a term of order unity.

+
Assuming that the free-fall acceleration of the surface of a collapsing cloud remains constant during the entire collapse, derive an expression for the free-fall time.  Show that your answer differs from Eq. (8.21) only by a term of order unity.

Problem 6

diff --git a/docs/Chapter_5/interaction-of-light-and-matter.html b/docs/Chapter_5/interaction-of-light-and-matter.html index 57109cb..3c13275 100644 --- a/docs/Chapter_5/interaction-of-light-and-matter.html +++ b/docs/Chapter_5/interaction-of-light-and-matter.html @@ -603,8 +603,8 @@

2.1.1. Kirchoff’s Laws

2.1.2. Applications of Stellar Spectral Data#

Using the development of spectral fingerprints, a new element helium was discovered spectroscopically in the Sun (in 1868) and found on Earth in 1895. Another line of investigation was measuring the Doppler shifts of the spectral lines. For most cases at the time, the low-speed approximation (\(v_r \ll c\)) was adequate to determine the radial velocities \(v_r\) by the following equation,

-
-(2.1)#\[\begin{align} +
+(2.1)#\[\begin{align} \frac{\lambda_{\rm obs}-\lambda_{\rm rest}}{\lambda_{\rm rest}} = \frac{\Delta \lambda}{\lambda_{\rm rest}}=\frac{v_r}{c}. \end{align}\]

By 1887, the radial velocities of Sirius, Procyon, Rigel, and Arcturus were measured with and accuracy of a few km/s. The rest wavelength \(\lambda_{\rm rest}\) for the hydrogen spectral line (\({\rm H}_\alpha\)) is 656.281 nm (in air). Given the simplicity of hydrogen and its spectral lines, they are commonly used to measure the radial velocities of stars.

@@ -716,8 +716,8 @@

2.1.3. Spectrographs\(d\) is the slit spacing of the grating, \(n\) is the order of the spectrum, and \(\theta\) is measured relative the line perpendicular to the grating. The smallest measurable difference in wavelength \(\Delta \lambda\) depends on the order \(n\) and the total number of lines \(N\) through

-
-(2.2)#\[\begin{align} +
+(2.2)#\[\begin{align} \Delta \lambda = \frac{\lambda}{nN}, \end{align}\]

where \(\lambda\) is either of the closely spaced wavelengths. The ratio \(\lambda/\Delta \lambda\) is one way to express the resolving power of the grating.

@@ -736,8 +736,8 @@

2.2. Photons2.2.1. The Photoelectric Effect#

The photoelectric effect describes the light energy necessary to eject electrons from a metal surface. The electrons with the highest kinetic energy \(K_{max}\) originate from the surface of the metal and surprisingly, \(K_{max}\) does not depend on the light intensity. A higher intensity of light increases the number of electrons ejected, but not their maximum kinetic energy. The value of \(K_{max}\) varies with the frequency of the light and each metal has a characteristic cutoff frequency \(\nu_c\) and wavelength \((\lambda_c = c/\nu_c\)).

Einstein’s solution describes the light striking the metal surface as a stream of massless particles (i.e., photons). The energy of a single photon of frequency \(\nu\) and wavelength \(\lambda\) is just Planck’s quantum of energy

-
-(2.3)#\[\begin{align} +
+(2.3)#\[\begin{align} E_{\rm photon} = h\nu = \frac{hc}{\lambda}. \end{align}\]
@@ -778,8 +778,8 @@

2.2.1. The Photoelectric Effect

Einstein deduced that when a photon strikes the metal surface, its energy may be absorbed by a single electron. The photon supplies the energy required to free the electron from the metal (i.e., overcome the binding energy). The work function \(\phi\) defines the minimum binding energy of electrons in a metal and the maximum kinetic energy of the ejected electrons is

-
-(2.4)#\[\begin{align} +
+(2.4)#\[\begin{align} K_{\rm max} = E_{\rm photon} - \phi = h\nu - \phi, \end{align}\]

where the cutoff frequency and wavelength are \(\nu_c = \phi/h\) and \(\lambda_c = hc/\phi\), respectively. Albert Einstein was awarded the 1921 Nobel Prize for “his services to theoretical physics, and especially for his discovery of the law of the photoelectric effect”. Astronomers take advantage of the photoelectric effect through detectors that count photons by the work they do on a metal (e.g., charge-coupled devices or CCDs).

@@ -787,13 +787,13 @@

2.2.1. The Photoelectric Effect

2.2.2. The Compton Effect#

Another test of light’s particle-like nature was performed by Arthur Compton. He measured the change in the wavelength of X-ray photons as they were scattered by free electrons. Through special relativity, the energy of a photon is related to its momentum \(p\) by

-
-(2.5)#\[\begin{align} +
+(2.5)#\[\begin{align} E_{\rm photon} = h\nu = pc. \end{align}\]

compton considered the interaction (or “collision”) between a photon and a free electron (initially at rest). The electron is scattered by an angle \(\varphi\), while the photon is scattered by an angle \(\theta\). Due to the energy exchange during the interaction (i.e., electron gains kinetic energy) the photon’s energy is reduced and the wavelength has increased. The change in wavelength is the difference between the final and initial wavelength and can be expressed as

-
-(2.6)#\[\begin{align} +
+(2.6)#\[\begin{align} \Delta \lambda =\lambda_f - \lambda_i = \frac{h}{m_e c}(1-\cos \theta), \end{align}\]

where \(m_e\) is the mass of the electron. The prefactor \(h/m_e c\) is called the Compton wavelength \(\lambda_C\). Compton’s experiment provided evidence that photons carry momentum (although massless) and this is the physical basis for the force exerted by radiation upon matter.

@@ -817,8 +817,8 @@

2.3.2. The Wavelength of Hydrogen (2.7)#\[\lambda = \frac{1}{R_H}\left( \frac{4n^2}{n^2-4}\right), \]

where \(n\) is an integer greater than 2 (\(n=3,4,5,\ldots\)) and \(R_H = 1.09677583 \times 10^7 \pm 1.3 \;{\rm m}^{-1}\) is the Rydberg constant) for hydrogen. Equation (2.7) is accurate to a fraction of a percent, where values of \(n\) produce the wavelength for the spectral lines (\({\rm H}\alpha\), \({\rm H}\beta\), \({\rm H}\gamma\), etc. ). Balmer realized that the formula could be generalized (since \(2^2 = 4\)) producing the general form as

-
-(2.8)#\[\begin{align} +
+(2.8)#\[\begin{align} \lambda = \frac{1}{R_H}\left( \frac{m^2n^2}{n^2-m^2}\right), \end{align}\]

with \(m<n\) (both integers). There are only a few spectral lines that lie in the visible range for hydrogen, where the Lyman lines (\(m=1\)) are found in the ultraviolet and the Paschen lines (\(m=3\)) lie entirely in the infrared.

@@ -989,8 +989,8 @@

2.3.3. Bohr’s Semiclassical Atom

To analyze the interactions between a proton and electron, we start with the mathematical description given by Coulomb’s law. The electric force between two charges (\(q_1\) and \(q_2\)) separated by a distance \(r\) has the familiar form

-
-(2.9)#\[\begin{align} +
+(2.9)#\[\begin{align} {\bf F} = \frac{1}{4\pi\epsilon_o}\frac{q_1q_2}{r^2}\hat{{\bf r}}, \end{align}\]

where \(\epsilon_o = 8.854187817\ldots \times 10^{-12}\) F \({\rm m}^{-1}\) is the permittivity of free space and \(\hat{{\bf r}}\) is a unit vector that points from \(q_1\) to \(q_2\). In the hydrogen atom, the electron has a negative charge \(e^-\) and the proton a positive charge \(e^+\). The Coulomb force pulls opposite charges toward on another, which is definitely not a good description of the system. In the case of two masses orbiting each other, we set the gravitational force equal the centripetal force. Thus, it follows that we could try the same procedure for a proton of mass \(m_p\) and an electron of mass \(m_e\). Simplifying further, we can convert the two-body problem to an equivalent one-body problem using the reduced mass

@@ -1004,8 +1004,8 @@

2.3.3. Bohr’s Semiclassical Atom

which can be solved for the kinetic energy \(K\):

-
-(2.10)#\[\begin{align} +
+(2.10)#\[\begin{align} K = \frac{1}{2}\mu v^2 = \frac{1}{8\pi\epsilon_o}\frac{e^2}{r}. \end{align}\]

The electrical potential energy \(U\) of the Bohr atom is:

@@ -1014,14 +1014,14 @@

2.3.3. Bohr’s Semiclassical Atom

The total energy \(E = K + U\) of the atom is:

-
-(2.11)#\[\begin{align} +
+(2.11)#\[\begin{align} E = K + U = -K = -\frac{1}{8\pi\epsilon_o}\frac{e^2}{r}. \end{align}\]

The kinetic energy is a positive quantity, which implies that the total energy is negative and indicates that the electron and proton are bound. To ionize or free the electron, it must receive at least an amount of energy equal to \(|E|\).

Getting back to Bohr’s contribution, he suggested that the angular momentum \(L = \mu vr\) was quantized and could be expressed in terms of Planck’s constant \(h\) distributed over a unit circle as \(\hbar = h/2\pi\). The full condition is

-
-(2.12)#\[\begin{align} +
+(2.12)#\[\begin{align} L = \mu vr = n\hbar, \end{align}\]

which can be rewritten using the kinetic energy as

@@ -1031,19 +1031,19 @@

2.3.3. Bohr’s Semiclassical Atom

Solving the second line of the equation above for the radius \(r\) shows that the specific values are allowed by Bohr’s quantization condition as

-
-(2.13)#\[\begin{align} +
+(2.13)#\[\begin{align} r_n = \frac{4\pi\epsilon_o \hbar^2}{\mu e^2}n^2 = a_on^2, \end{align}\]

where \(a_o = 5.291772083 \times 10^{-11}\) m is known as the Bohr radius. Now that we know the scale of the atom (i.e., appropriate values of \(r\)), the energy in terms of fundamental constants are given as

-
-(2.14)#\[\begin{align} +
+(2.14)#\[\begin{align} E_n = -\frac{\mu e^4}{32\pi^2\epsilon_o^2 \hbar^2}\frac{1}{n^2} = \frac{-13.6\;{\rm eV}}{n^2}. \end{align}\]

The principal quantum number \(n\) completely determines the characteristics of the Bohr atom, where \(n\geq 1\).

The spectral lines are produced by electron transitions within the atom (i.e., moving between energy levels). Balmer tried to produce a formula that would predict the wavelength that resulted from a transition from \(n_{high}\rightarrow n_{low}\). Now consider the required energy to move between levels as the energy of a photon \(E_{photon} = hc/\lambda = E_{high} - E_{low}\), which gives

-
-(2.15)#\[\begin{align} +
+(2.15)#\[\begin{align} \lambda = \frac{h c}{13.6\;{\rm eV}}\frac{n_{high}^2n_{low}^2}{n_{high}^2-n_{low}^2}, \end{align}\]

where \(hc = 1240\) eV nm.

@@ -1224,8 +1224,8 @@

2.4.2. Heisenberg’s Uncertainty Princi

Werner Heisenberg developed a theoretical framework for this inherent “fuzziness” of the physical world. The product of the uncertainty in the position and momentum must be equal to or larger than \(\hbar/2\) and is known as Heisenberg’s uncertainty principle. A similar statement relates the uncertainty of energy and time measurements. Together they are described by

-
-(2.17)#\[\begin{align} +
+(2.17)#\[\begin{align} \Delta x \Delta p \approx \hbar \quad {\rm and} \quad \Delta E \Delta t \approx \hbar. \end{align}\]
@@ -1273,8 +1273,8 @@

2.4.2. Heisenberg’s Uncertainty Princi

2.4.3. Quantum Mechanical Tunneling#

Light changes its behavior at boundary transitions, where this is most easily seen for a light beam traveling from a prism (glass) into air. The light may undergo total internal reflection if its incident angle is greater than a critical angle \(\theta_c\) that is determined by the indices of refraction of the glass and air. The critical angle \(\theta_c\) can be derived from Snell’s law to get

-
-(2.18)#\[\begin{align} +
+(2.18)#\[\begin{align} \sin \theta_c = \frac{n_{\rm air}}{n_{\rm glass}}. \end{align}\]

The electromagnetic wave does enter the air, but it dies away exponentially (i.e., becomes evanescent). When another prism is placed next to the first, then the evanescent wave can begin propagating into the second prism without passing through the air gap between them. Photons have tunneled from one prism to another. The limitation to this effect is through Heisenberg’s uncertainty principle, where the location of a particle (i.e., photon) cannot be determined with an uncertainty less than its wavelength. For barriers that are only a few wavelengths wide, a particle can suddenly appear on the other side. Barrier penetration is important for radioactive decay and inside stars, where nuclear fusion rates depend upon tunneling.

@@ -1295,8 +1295,8 @@

2.4.4. Schrödinger’s Equation and the

The Schrödinger equation can be solved analytically for the hydrogen atom and it produces the same set of allowed energies obtained by Bohr. However, Schrödinger found that two additional quantum numbers \(\ell\) and \(m_\ell\) are required for a complete description of the electron orbitals. The principal quantum number \(n\) determines the energy level, where these quantum numbers describe the angular momentum vector L of the atom. The magnitude of the angular momentum \(L\) are

-
-(2.19)#\[\begin{align} +
+(2.19)#\[\begin{align} L = \sqrt{\ell(\ell+1)}\hbar, \end{align}\]

where \(\ell = \{0,\ldots,n-1 \}\). Early spectroscopists categorized some elements based on their alkali metal spectroscopic lines described as sharp, principal, diffuse, and fundamental and this historical nomenclature corresponded with \(\ell = 0,1,2,3\). The letters continues alphabetically where \(\ell = 4\rightarrow g\), \(\ell = 5\rightarrow h\), and so on (omitting \(j\)). Chemists now use a nomenclature that includes both the principal and angular momentum quantum numbers together (e.g., \(n\ell\)). For example, (\(n=2\), \(\ell=1\)) corresponds to \(2p\).

diff --git a/docs/Chapter_9/stellar-atmospheres.html b/docs/Chapter_9/stellar-atmospheres.html index 0b4eeb4..194eaf5 100644 --- a/docs/Chapter_9/stellar-atmospheres.html +++ b/docs/Chapter_9/stellar-atmospheres.html @@ -500,8 +500,8 @@

5.1.1. The Specific and Mean Intensities
\[ E_\lambda \equiv \frac{\partial E}{\partial \lambda}, \]

where \(E_\lambda d\lambda\) is the energy carried by the rays in the light cone in a time interval \(dt\). The specific intensity of the rays is defined as

-
-(5.1)#\[\begin{align} I_\lambda \equiv \frac{\partial I}{\partial \lambda} \equiv \frac{E_\lambda d\lambda}{d\lambda\:dt\:dA \cos\theta\:d\Omega}, \end{align}\]
+
+(5.1)#\[\begin{align} I_\lambda \equiv \frac{\partial I}{\partial \lambda} \equiv \frac{E_\lambda d\lambda}{d\lambda\:dt\:dA \cos\theta\:d\Omega}, \end{align}\]

which carries the units of energy per time per Area per steradian sr (solid angle) or more simply, energy per everything.

Intensity @@ -549,8 +549,8 @@

5.1.2. The Specific Energy Density \[ E_\lambda d\lambda = I_\lambda d\lambda\:dt\:dA \cos\theta\:d\Omega = I_\lambda d\lambda\:dA\:d\Omega\:\frac{dL}{c}, \]

using Eqn. (5.2) and the substitution of \(dt\). The quantity \(dA dL\) is the volume of the cylinder, where the specific energy density (\(u_\lambda d\lambda\)) is found by normalizing the energy \(E_\lambda d\lambda\) by the volume of the cylinder, integrating over all the solid angles, and using Eqn. (5.3):

-
-(5.4)#\[\begin{align} +
+(5.4)#\[\begin{align} u_\lambda d\lambda = \frac{1}{c} \int_{\phi = 0}^{2\pi} \int_{\theta=0}^{\pi} I_\lambda d\lambda\:\sin\theta d\theta\:d\phi = \frac{4\pi}{c} \langle I_\lambda \rangle d\lambda. \end{align}\]
@@ -613,8 +613,8 @@

5.1.4. Radiation Pressure \[ P_{\rm rad} = \int_0^\infty P_{\rm rad,\lambda} d\lambda, \]

where the substitution \(B_\lambda = I_\lambda\) in Eqn. (5.8) can be made blackbody radiation to get

-
-(5.9)#\[\begin{align} +
+(5.9)#\[\begin{align} P_{\rm rad} = \int_0^\infty B_\lambda(T)d\lambda = \frac{4\sigma T^4}{3c} = \frac{1}{3}u. \end{align}\]

The blackbody radiation pressure is one-third of the energy density, where an ideal monatomic gas is two-thirds of its energy density.

@@ -874,8 +874,8 @@

5.2.2. The Definition of Opacity

5.2.3. Optical Depth#

Scatter photons have a mean free path (given by Eqn. (5.10)), where \(\kappa_\lambda \rho\) and \(n\sigma_\lambda\) can be though of as the fraction of photons scatter per meter of distance traveled. Note that the mean free path is dependent on the photon wavelength \(\lambda\). However it is convenient to define an optical depth \(\tau_\lambda\) as

-
-(5.13)#\[\begin{align} d\tau_\lambda = -\kappa_\lambda \rho ds, \end{align}\]
+
+(5.13)#\[\begin{align} d\tau_\lambda = -\kappa_\lambda \rho ds, \end{align}\]

where \(s\) is the distance measured along the photon’s direction of motion (i.e., looking back along the path traveled by the photon). The difference in optical depth \(\Delta \tau_\lambda\) can be determined as the difference between the final (\(\tau_{\lambda,f}\)) and initial (\(\tau_{\lambda,o}\)) optical depths, or the integral along the photon’s path as

(5.14)#\[\Delta \tau_\lambda = \tau_{\lambda,f} - \tau_{\lambda,o} = -\int_0^s \kappa_\lambda \rho ds.\]
@@ -916,8 +916,8 @@

5.2.3. Optical Depth \[ \tau_\lambda = \int_0^s \kappa_\lambda \rho ds = \sec \theta \int_0^h \kappa_\lambda \rho dz = \tau_{\lambda,o} \sec \theta, \]

where \(\tau_{\lambda,o}\) refers to the optical depth of a vertically traveling photon ($\theta = 0). The intensity of the light at the telescope is

-
-(5.16)#\[\begin{align} +
+(5.16)#\[\begin{align} I_\lambda = I_{\lambda,o}e^{-\tau_{\lambda,o}\sec \theta}. \end{align}\]

For a single observation, both \(I_{\lambda,o}\) and \(\tau_{\lambda,o}\) are unknown. Over a night of observation, the Earth rotates on its axis and several measurements of the received intensity \(I_\lambda\) as a function of \(\sec \theta\) can be made. Then the best-fitting slope is \(-\tau_{\lambda,o}\) and \(I_{\lambda,o}\) can be determined through extrapolation

@@ -953,8 +953,8 @@

5.2.4. General Sources of Opacity

  • Electron scattering is as you would expect where a photon is scatter (not absorbed) by a free electron through the process of Thomson scattering. The electron “oscillates” (like a buoy) in the wake of a photon. Because the electron’s cross section is tiny, it is a poor target for an incident photon. The cross section for Thomson scattering \(\sigma_T\) has the same value (independent of wavelength),

    -
    -(5.17)#\[\begin{align} \sigma_T = \frac{1}{6\pi\epsilon_o^2}\left( \frac{e^2}{m_e c^2}\right) = 6.65 \times 10^{-29}\:{\rm m^2}. \end{align}\]
    +
    +(5.17)#\[\begin{align} \sigma_T = \frac{1}{6\pi\epsilon_o^2}\left( \frac{e^2}{m_e c^2}\right) = 6.65 \times 10^{-29}\:{\rm m^2}. \end{align}\]

    This is ~2 billion times smaller than the hydrogen cross section for photoionization from bound-free absorption, \(\sigma_{bf}\). Electron scattering is most effective when the electron density is very high (i.e., high temperatures for ionization) and other sources of opacity are eliminated due to the absence of necessary bound electrons. For high temperatures, the electron scattering opacity \(\kappa_{es}\) dominates the continuum opacity.

    A photon can also be scattered by loosely bound electrons. Compton scattering occurs when the photon’s wavelength is much smaller than the atom or Rayleigh scattering if the wavelength is much larger. Compton scatter is usually lumped together with Thomson scattering due to the small change in wavelength of the scattered photon. The cross section for Rayleigh scattering is smaller than the Thomson cross section and can be neglected in most stellar atmospheres. However, it is important in the UV for supergiant stars with their extended envelopes and in cool main-sequence stars. The opacity of stellar material suddenly increases at wavelengths \(\lambda \leq 364.7\:{\rm nm}\) and the measured radiative flux of the star suddenly decreases. This abrupt drop in the continuous spectrum is the Balmer jump and is evident in the Sun’s spectrum. The size of the Balmer jump in hot stars depends on the fraction of hydrogen atoms in the first excited state. The Balmer jump appears in the middle of the U band in the UBV system, which results in a decrease in the amount of light received in the bandwidth of the U filter and increase the measured values for both the U magnitude and color index (\(U-B\)).

    @@ -999,8 +999,8 @@

    5.2.4. General Sources of Opacity

    5.2.5. The Rosseland Mean Opacity#

    The opacity in Eqn. (5.15) depends on the wavelength of light, where a star is expected to produce photons covering a wide range of wavelengths. It is often useful to define an opacity that has been averaged over all wavelengths. This averaging produces an opacity that depends only on the composition, density, and temperature of the stellar atmosphere. The most commonly used scheme to compute a wavelength-independent opacity is the Rosseland mean opacity or the Rosseland mean. The Rosseland mean incorporates a weighting function that depends on how fast the blackbody spectrum varies with temperature, which is defined as

    -
    -(5.18)#\[\begin{align} \frac{1}{\bar{\kappa}} \equiv \frac{\int_0^\infty \frac{1}{\kappa_\lambda}\frac{\partial B_\lambda(T)}{\partial T} d\lambda}{\int_0^\infty \frac{\partial B_\lambda(T)}{\partial T} d\lambda}. +
    +(5.18)#\[\begin{align} \frac{1}{\bar{\kappa}} \equiv \frac{\int_0^\infty \frac{1}{\kappa_\lambda}\frac{\partial B_\lambda(T)}{\partial T} d\lambda}{\int_0^\infty \frac{\partial B_\lambda(T)}{\partial T} d\lambda}. \end{align}\]

    The absorption coefficient \(\kappa_\lambda\) varies with wavelength, but it also depends on the assumed source of opacity (e.g., bound-free, free-free, etc.) and there is not a simple equation to describe the average absorption coefficient for bound-bound transitions. Approximation formulae have been developed for the average bound-free and free-free opacities:

    The mass fractions are formally defined as

    -
    -(5.21)#\[\begin{align} X \equiv \frac{\rm total\: mass\:of\:hydrogen}{\rm total\: mass\:of\:gas} \end{align}\]
    -
    -(5.22)#\[\begin{align} Y \equiv \frac{\rm total\: mass\:of\:helium}{\rm total\: mass\:of\:gas} \end{align}\]
    -
    -(5.23)#\[\begin{align} Z \equiv \frac{\rm total\: mass\:of\:metals}{\rm total\: mass\:of\:gas} \end{align}\]
    +
    +(5.21)#\[\begin{align} X \equiv \frac{\rm total\: mass\:of\:hydrogen}{\rm total\: mass\:of\:gas} \end{align}\]
    +
    +(5.22)#\[\begin{align} Y \equiv \frac{\rm total\: mass\:of\:helium}{\rm total\: mass\:of\:gas} \end{align}\]
    +
    +(5.23)#\[\begin{align} Z \equiv \frac{\rm total\: mass\:of\:metals}{\rm total\: mass\:of\:gas} \end{align}\]

    such that \(X+Y+Z = 1\). The Gaunt factors (\(g_{bf}\) and \(g_{ff}\)) in Eqns. (5.19) and (5.20) are quantum-mechanical correction terms. These Gaunt factors are approximately 1 for the visible and UV wavelengths of interest. The guillotine factor \(t\) (Eddington 1932) describes a cutoff to the opacity of an atom after it has been ionized, where typical values lie between 1 and 100.

    Equations (5.19) and (5.20) have the functional form \(\bar{\kappa} \propto \kappa_o \rho/T^{3.5}\), where \(\kappa_o\) combines all the factors that depend on composition. H. A. Kramers derived the first forms (in 1923) using classical physics and the Rosseland mean. As a result, any opacity having a temperature and density dependence is referred to as Kramers opacity law.

    The electron scattering cross section is independent of wavelength and the Rosseland mean is simply

    -
    -(5.24)#\[\begin{align} \bar{\kappa}_{es} = 0.02(1+X) \:{\rm m^2/kg}. \end{align}\]
    +
    +(5.24)#\[\begin{align} \bar{\kappa}_{es} = 0.02(1+X) \:{\rm m^2/kg}. \end{align}\]

    The mean opacity is modified in cooler stars (later than F0) by the \(H^-\) ion over specific temperature range (3000 K \(\leq T \leq\) 6000 K), density range (\(10^{-7}\leq \rho \leq 10^{-2}\: {\rm kg/m^3}\)), and metal mass fraction (\(0.001 < Z < 0.03\)), when the hydrogen mass fraction \(X\sim0.7\) is lower. The average opacity is

    -
    -(5.25)#\[\begin{align} \bar{\kappa}_{H^-} \approx 7.9 \times 10^{-34} (50Z)\sqrt{\rho}T^9\:{\rm m^2/kg}. \end{align}\]
    +
    +(5.25)#\[\begin{align} \bar{\kappa}_{H^-} \approx 7.9 \times 10^{-34} (50Z)\sqrt{\rho}T^9\:{\rm m^2/kg}. \end{align}\]

    The total Rosseland mean opacity \(\bar{\kappa}\) is the average of the sum of the individual contributions or

    \[ \bar{\kappa} = \overline{\kappa_{bb}+\kappa_{bf}+\kappa_{ff}+\kappa_{es}+\kappa_{H^-}}. \]
    @@ -1155,8 +1155,8 @@

    5.3.3. Limb Darkening5.3.4. The Radiation Pressure Gradient#

    In addition to the photon’s meandering journey, there is a pressure that builds within the Sun’s atmosphere due to collisions. This is similar to the air pressure within a closed room, where an individual molecule moves ~500 m/s and collides with other air molecules several billion times per second. In the analogy of the closed room, a net flow of air (i.e., breeze) can occur after opening a window.

    The pressure outside is lower than the pressure inside the room, which can be expressed as a pressure differential using Bernoulli’s principle. In a star, the high pressure interior and vacuum of space also produces a pressure difference. The extent of the difference is not uniform, where the pressure decreases with stellar radius (i.e., as you get closer to the surface). But it is enough to create a light movement of photons toward the surface that described by

    -
    -(5.28)#\[\begin{align} \frac{dP_{\rm rad}}{dr} = -\frac{\bar{\kappa}\rho}{c}F_{\rm rad}. \end{align}\]
    +
    +(5.28)#\[\begin{align} \frac{dP_{\rm rad}}{dr} = -\frac{\bar{\kappa}\rho}{c}F_{\rm rad}. \end{align}\]

    The transfer of radiation as a flow with photons drifting toward the surface is a better description than a “beam” that propagate radially outward. The description of a beam is a convenient fiction used to define the direct of motion for a photon at a snapshot along its journey of continual absorption and scattering.

    • @@ -1210,8 +1210,8 @@

    5.4.3. The Special Case of Blackbody Rad
    \[ \ln \left(\frac{I_\lambda - S_\lambda}{I_{\lambda,o} - S_\lambda}\right) = -\kappa_\lambda \rho s, \]

    which simplifies to

    -
    -(5.32)#\[\begin{align} I_\lambda(s) = I_{\lambda,o} e^{-\kappa_\lambda \rho s} + S_\lambda(1-e^{-\kappa_\lambda \rho s}). \end{align}\]
    +
    +(5.32)#\[\begin{align} I_\lambda(s) = I_{\lambda,o} e^{-\kappa_\lambda \rho s} + S_\lambda(1-e^{-\kappa_\lambda \rho s}). \end{align}\]

    • Suppose the source function \(S_\lambda\) is twice the initial intensity \(I_{\lambda,o}\), then plot the transformation of the intensity in units of \(\kappa_\lambda \rho s\).

    @@ -1250,21 +1250,21 @@

    5.4.3. The Special Case of Blackbody Rad

    5.4.4. The Assumption of a Plane-Parallel Atmosphere#

    The intensity of light must depend on the path for the outward flow of energy. The absorption and emission coefficients are the same for light traveling in all directions, but these coefficients depend on the temperature and density in a complicated way. For astronomers to understand the stellar interior, they must know at what depth a spectral line is formed. Since we are interested in the depth, let’s rewrite the transfer equation in terms of the optical depth, which is simply

    -
    -(5.33)#\[\begin{align} \frac{dI_\lambda}{d\tau_\lambda} = I_\lambda - S_\lambda. \end{align}\]
    +
    +(5.33)#\[\begin{align} \frac{dI_\lambda}{d\tau_\lambda} = I_\lambda - S_\lambda. \end{align}\]

    Optical depth does not provide a unique geometric depth in the atmosphere and thus, must be replaced with a more meaningful measure of position. The atmospheres of main-sequence stars are physically thin compared with the size of the star. The atmosphere’s radius of curvature is much larger than its thickness, where we may consider the atmosphere as a plane-parallel slab. If we assume align the z-axis with the vertical direction, then \(z=0\) denotes the top of this plane-parallel atmosphere and a vertical optical depth \(\tau_{\lambda,v}(z)\) is defined as

    -
    -(5.34)#\[\begin{align} \tau_{\lambda,v}(z) \equiv \int_z^0 \kappa_\lambda \rho dz. \end{align}\]
    +
    +(5.34)#\[\begin{align} \tau_{\lambda,v}(z) \equiv \int_z^0 \kappa_\lambda \rho dz. \end{align}\]

    Note

    The \(v\) symbol is not to be confused with \(\nu\), where the former signifies the v in vertical.

    Recall the discussion for correcting the opacity of the Earth’s atmosphere, the depth was parameterized in a similar manner. A ray that travels upward at an angle \(\theta\) has farther to go through the atmosphere to reach the surface. The optical depth \(\tau_\lambda\) is greater than the vertical optical depth \(\tau_{\lambda,v}\). The two optical depths are related by

    -
    -(5.35)#\[\begin{align} \tau_\lambda = \frac{\tau_{\lambda,v}}{\cos \theta} = \tau_{\lambda,v} \sec \theta \end{align}\]
    +
    +(5.35)#\[\begin{align} \tau_\lambda = \frac{\tau_{\lambda,v}}{\cos \theta} = \tau_{\lambda,v} \sec \theta \end{align}\]

    and then the transfer equation becomes

    -
    -(5.36)#\[\begin{align} \cos \theta \frac{dI_\lambda}{d\tau_{\lambda,v}} = I_\lambda - S_\lambda \end{align}\]
    +
    +(5.36)#\[\begin{align} \cos \theta \frac{dI_\lambda}{d\tau_{\lambda,v}} = I_\lambda - S_\lambda \end{align}\]

    for the approximation of a plane-parallel atmosphere.

    plane parallel atmosphere @@ -1273,13 +1273,13 @@

    5.4.4. The Assumption of a Plane-Paralle

    The value of the vertical optical depth is wavelength dependent, but when considering the bulk flow of photons a gray atmosphere is assumed. Gray atmospheres are used when the measured flux does not show a strong wavelength dependence. Thus, the transfer equation for a plane-parallel gray atmosphere is

    -
    -(5.37)#\[\begin{align} \cos \theta \dfrac{dI}{d\tau_v} = I - S. \end{align}\]
    +
    +(5.37)#\[\begin{align} \cos \theta \dfrac{dI}{d\tau_v} = I - S. \end{align}\]

    Formulating the transfer equation in terms of an angle and optical depth leads to two useful relations describing the radiation field.

    1) Integrating over all solid angles

    The source function \(S\) depends only on the local conditions, so we can integrate over all solid angles to get

    -
    -(5.38)#\[\begin{align} \frac{d}{d\tau_v}\int I \cos \theta d\Omega = \int I d\Omega - S \int d\Omega. \end{align}\]
    +
    +(5.38)#\[\begin{align} \frac{d}{d\tau_v}\int I \cos \theta d\Omega = \int I d\Omega - S \int d\Omega. \end{align}\]

    This expression can be simplified using \(\int d\Omega = 4\pi\), definition of \(F_{rad}\) (Eqn. (5.7), integrated over \(\lambda\)) and the mean intensity (Eqn. (5.3), integrated over \(\lambda\)). Through substitution, we find

    \[ \frac{d F_{rad}}{d\tau_v} = 4\pi(\langle I \rangle - S). \]
    @@ -1295,8 +1295,8 @@

    5.4.4. The Assumption of a Plane-Paralle \[ \frac{d P_{rad}}{dr} = -\frac{\kappa \rho}{c}F_{rad}, \]

    in terms of a spherical coordinate system with its origin at the center fo the star. As mention in the radiation pressure gradient, this result describes the net radiative flux driven by differences in radiation pressure (i.e., “photon wind”).

    For a plane-parallel atmosphere to be in equilibrium, the radiative flux must have the same value at every level, including its surface, or

    -
    -(5.40)#\[\begin{align} F_{rad} = {\rm constant} = F_{surf} = \sigma T_e^4. \end{align}\]
    +
    +(5.40)#\[\begin{align} F_{rad} = {\rm constant} = F_{surf} = \sigma T_e^4. \end{align}\]

    This implies that the first of the above relations (\(dF_{rad}/d\tau_v\)) is zero and the mean intensity must equal the source function as

    \[ \langle I \rangle = S. \]
    @@ -1310,8 +1310,8 @@

    5.4.5. The Eddington ApproximationThe temperature as a function of the optical depth (i.e., temperature structure) can be determined if we knew how the radiation pressure varied with temperature in the general case. Equation (5.41) could be used if we also assume a description of the angular distribution of intensity. An approximation was made by Eddington, where the intensity of the radiation field is defined in the +\(z\)-direction \(I_{\rm out}\) and another in the -\(z\)-direction \(I_{\rm in}\). These intensities vary with depth in the atmosphere and when \(\tau_v = 0\), then \(I_{\rm in} = 0\) (i.e., radiation doesn’t flow inward at the top of the atmosphere). The Eddington approximation gives the mean intensity, radiative flux, and radiation pressure by:

    (5.42)#\[\langle I \rangle = \frac{1}{2}(I_{\rm out} + I_{\rm in}), \]
    -
    -(5.43)#\[\begin{align} F_{rad} = \pi(I_{\rm out} - I_{\rm in}), \end{align}\]
    +
    +(5.43)#\[\begin{align} F_{rad} = \pi(I_{\rm out} - I_{\rm in}), \end{align}\]

    and

    (5.44)#\[P_{rad}= \frac{2\pi}{3c}(I_{\rm out} + I_{\rm in}) = \frac{4\pi}{3c} \langle I \rangle .\]
    @@ -1322,8 +1322,8 @@

    5.4.5. The Eddington Approximation \[ \frac{2\pi}{3c} I_{\rm out} = \frac{2}{3c} F_{rad} = C, \]

    and substituting back into Eqn. (5.45), we find

    -
    -(5.46)#\[\begin{align} \frac{4\pi}{3c} \langle I \rangle = \frac{\tau_v}{c}F_{rad} + \frac{2}{3c} F_{rad} = \frac{F_{rad}}{c}\left(\tau_v + \frac{2}{3}\right). \end{align}\]
    +
    +(5.46)#\[\begin{align} \frac{4\pi}{3c} \langle I \rangle = \frac{\tau_v}{c}F_{rad} + \frac{2}{3c} F_{rad} = \frac{F_{rad}}{c}\left(\tau_v + \frac{2}{3}\right). \end{align}\]

    But we already know the radiative flux is a constant (\(F_{rad} = \sigma T_e^4\)) and thus, the mean intensity can be determined as a function of vertical optical depth by

    (5.47)#\[\langle I \rangle = \frac{3\sigma}{4\pi}T_e^4\left(\tau_v + \frac{2}{3}\right).\]
    @@ -1331,8 +1331,8 @@

    5.4.5. The Eddington Approximation (5.48)#\[S = B = \langle I \rangle = \frac{\sigma T^4}{\pi}.\]

    Substituting Eqn. (5.48) back into Eqn. (5.47) results in the variation of the temperature with vertical optical depth in a plane-parallel gray atmosphere in LTE, assuming the Eddington approximation:

    -
    -(5.49)#\[\begin{align} T^4 = \frac{3}{4}T_e^4\left(\tau_v + \frac{2}{3}\right). \end{align}\]
    +
    +(5.49)#\[\begin{align} T^4 = \frac{3}{4}T_e^4\left(\tau_v + \frac{2}{3}\right). \end{align}\]

    This relation reveals that \(T=T_e\) at \(\tau_v = 2/3\) and not at \(\tau_v = 0\). The “surface” of a star for which we measure an effective temperature \(T_e\) is not at the top of the atmosphere, but deeper down where \(\tau_v=2/3\). The average point of origin for the photons comes from within the stellar atmosphere. When looking at a star, we see down to a vertical optical depth of \(\tau_v \approx 2/3\), averaged over the disk of the star.

    @@ -1380,8 +1380,8 @@

    5.4.6. Limb Darkening Revisited \[ a_\lambda = \frac{\sigma T_e^4}{2\pi}\quad {\rm and} \quad b_\lambda = \frac{3\sigma}{4\pi} T_e^4. \]

    Using Eqn. (5.53), we have

    -
    -(5.54)#\[\begin{align} \frac{I(\theta)}{I(0)} = \frac{a_\lambda + b_\lambda \tau_v}{a_\lambda + b_\lambda} = \frac{2}{5} + \frac{3}{5}\cos \theta. \end{align}\]
    +
    +(5.54)#\[\begin{align} \frac{I(\theta)}{I(0)} = \frac{a_\lambda + b_\lambda \tau_v}{a_\lambda + b_\lambda} = \frac{2}{5} + \frac{3}{5}\cos \theta. \end{align}\]
    Eddington approximation
    @@ -1395,8 +1395,8 @@

    5.5. The Profiles of Spectral Lines

    5.5.1. Equivalent Widths#

    Absorption spectral lines are identified on graphs of the radiant flux \(F_\lambda\) as a function of the wavelength \(\lambda\). The radiant flux can be normalized relative to the flux of the continuous spectrum \(F_c\) outside the spectral line (i.e., \(F_\lambda/F_c\)). Near the middle of the absorption (i.e., core of the spectral line), there is the central wavelength \(\lambda_o\) and the wings extend outward, sweeping up to the continuum. The depth \(\delta\) of the line is simply the fractional decrease in radiant flux relative to the continuum (i.e., \(\delta F_\lambda = 1 - F_\lambda/F_c\)). The strength of absorption is measured by its equivalent width \(W\), which is a rectangle with height equal to unity and a specific width so that the area of the rectangle is equal to the total area removed from the continuum by the absorption. Mathematically, this is

    -
    -(5.55)#\[\begin{align} W = \int 1- \frac{F_\lambda}{F_c} d\lambda = \int \delta F d\lambda, \end{align}\]
    +
    +(5.55)#\[\begin{align} W = \int 1- \frac{F_\lambda}{F_c} d\lambda = \int \delta F d\lambda, \end{align}\]

    where the integral is taken from one side of \(\lambda_o\) to the other and is usually on the order of 0.01 nm. Another measure of spectral line width uses a point at half of the depth at the core wavelength (i.e., \(\delta F_{1/2} = 1/2 - F_{\lambda_o}/(2F_c)\)) and moves to the left/right a \(\Delta \lambda_{1/2} = \lambda_o \pm \lambda_{\delta F}\) until intersecting the absorption line so that \(\delta F = \delta F_{1/2}\), or

    \[ 1 - \frac{F_\lambda}{F_c} = \frac{1}{2}\left(1 - \frac{F_{\lambda_o}}{F_c} \right), \]
    diff --git a/docs/_sources/Chapter_12/interstellar-medium-and-star-formation.ipynb b/docs/_sources/Chapter_12/interstellar-medium-and-star-formation.ipynb index aabcb77..c343a63 100644 --- a/docs/_sources/Chapter_12/interstellar-medium-and-star-formation.ipynb +++ b/docs/_sources/Chapter_12/interstellar-medium-and-star-formation.ipynb @@ -20,7 +20,9 @@ "### Interstellar Extinction\n", "On a dark night (far from city lights), some of the dust clouds within the Milky Way Galaxy can be found because they obscure regions of sky that would otherwise be populated with stars. These dark regions are not devoid of stars, but the intervening dust blocks the background starlight. Dust that blocks starlight is due to the combined effet of scattering and absorption, which is called **interstellar extinction**. The apparent magnitude is a measure of how bright a star appears and if it is dimmed by intervening dust, the distance modulus equation must be modified appropriately. In a given wavelength band centered on $\\lambda$, we have\n", "\n", - "$$ m_\\lambda = M_\\lambda + 5\\log_{10}\\,d-5 + A_\\lambda, \\tag{1}$$\n", + "\\begin{align}\n", + "m_\\lambda = M_\\lambda + 5\\log_{10}\\,d-5 + A_\\lambda, \n", + "\\end{align}\n", "\n", "which is a modified version of the distance modulus equation with a correction factor $A_\\lambda>0$ that adjusts for the interstellar extinction along the line of sight. The extinction magnitude $A_\\lambda$ is positive because it always acts to dim objects along the magnitude scale. If $A_\\lambda$ is large enough, a star can fall behind the limiting magnitude of the background sky for the naked eye or a telescope. This is the cause for the dark bands running through the Milky Way.\n", "\n", @@ -54,15 +56,21 @@ "\n", "The change in apparent magnitude is simply the extinction magnitude $A_\\lambda$, or\n", "\n", - "$$ A_\\lambda = 1.086 \\tau_\\lambda, \\tag{2}$$\n", + "\\begin{align}\n", + " A_\\lambda = 1.086 \\tau_\\lambda, \n", + "\\end{align}\n", "\n", "which restates that *the change in magnitude due to extinction is approximately equal to the optical depth along the line of sight*. The optical depth through the cloud is \n", "\n", - "$$ \\tau_\\lambda = \\int_0^s n_d(s^\\prime)\\sigma_\\lambda\\,ds^\\prime, \\tag{3} $$\n", + "\\begin{align}\n", + "\\tau_\\lambda = \\int_0^s n_d(s^\\prime)\\sigma_\\lambda\\,ds^\\prime, \n", + "\\end{align}\n", "\n", "which depends on the number density of scattering dust grains $n_d(s^\\prime)$ and the scattering cross section $\\sigma_\\lambda$. If the scattering cross section is constant along the line of sight, then \n", "\n", - "$$ \\tau_\\lambda = \\sigma_\\lambda \\int_0^sn_d(s^\\prime),ds^\\prime = \\sigma_\\lambda N_d, \\tag{4}$$\n", + "\\begin{align}\n", + "\\tau_\\lambda = \\sigma_\\lambda \\int_0^sn_d(s^\\prime),ds^\\prime = \\sigma_\\lambda N_d, \n", + "\\end{align}\n", "\n", "and the dust grain *column density* $N_d$ represents the number of scattering dust particles in a thin cylinder with a cross section of 1 ${\\rm m^2}$ stretching from the observer to the star. The magnitude of extinction depends on the amount of interstellar dust that the light passes through." ] @@ -78,11 +86,15 @@ "\n", "where the extinction coefficient $Q_\\lambda$ depends on the composition of the dust grains. When the wavelength of the light is similar to the size of the dust grains, then $Q_\\lambda \\sim a/\\lambda$, implying that\n", "\n", - "$$ \\sigma_\\lambda \\propto \\frac{a^3}{\\lambda} \\qquad (\\lambda \\gtrsim a). \\tag{5}$$\n", + "\\begin{align}\n", + "\\sigma_\\lambda \\propto \\frac{a^3}{\\lambda} \\qquad (\\lambda \\gtrsim a).\n", + "\\end{align}\n", "\n", "The extinction coefficient $Q_\\lambda$ goes to zero when the wavelength $\\lambda$ is large compared to $a$. Conversely, the extinction coefficient $Q_\\lambda$ approaches a constant (independent of $\\lambda$) when the wavelength $\\lambda$ is very small compared to $a$ or\n", "\n", - "$$ \\sigma_\\lambda \\propto a^2 \\qquad (\\lambda \\ll a). \\tag{6} $$\n", + "\\begin{align}\n", + "\\sigma_\\lambda \\propto a^2 \\qquad (\\lambda \\ll a). \n", + "\\end{align}\n", "\n", "```{note}\n", "Gustav Mie made similar assumptions concerning dust particles in 1908 and showed the relationship between the scattering cross section $\\sigma_\\lambda$ and the radius of a dust grain.\n", @@ -95,6 +107,8 @@ "Strongly scattered blue light can leave the cloud in any direction, where a bright star behind a cloud casts a blue **reflection nebula** for observers that are not along the line of sight (e.g., the Pleiades). This process is analogous to Rayleigh scattering, which produces a blue sky on Earth. The difference between Mie scattering and Rayleigh scattering is that the scattering particles are much smaller than the wavelength of visible light in Rayleigh scattering and leads to $\\sigma_\\lambda \\propto \\lambda^{-4}$.\n", "\n", "```{exercise}\n", + ":class: orange\n", + "\n", "**A star is found to be dimmer than expected at 550 nm by 1.1 magnitudes in the visual (i.e., $A_V = 1.1$). If $Q_{550}=1.5$ and the radius of the dust grains are 0.2 ${\\rm \\mu m}$, estimate the average density $\\overline{n}$ of the material between the star and Earth. The star is 0.8 kpc from Earth.**\n", "\n", "The extinction magnitude $A_\\lambda$ is related to the optical depth $\\tau_\\lambda$ by a faction of 1.086, where $\\tau_{550} = 1.1/1.086 \\simeq 1$. Given the radius of the dust grains $(a = 0.2\\,{\\rm \\mu m})$ and the extinction coefficient $Q_{550} = 1.5$ we have,\n", @@ -232,12 +246,15 @@ "\n", "The rarity of 21 cm emission (or absorption) from individual atoms means that the central wavelength can remain optically thin over large interstellar distances. If we assume that the line profile is approximately Gaussian (like the shape of the Doppler profile), then the optical depth of the line is given by\n", "\n", - "$$ \\tau_H = 5.2 \\times 10^{-23} \\frac{N_H}{T\\,\\Delta v}, \\tag{7} $$\n", + "```{math}\n", + ":label: tau_hydro\n", + "\\tau_H = 5.2 \\times 10^{-23} \\frac{N_H}{T\\,\\Delta v}, \n", + "```\n", "\n", "which depends on the column density of neutral hydrogen $N_H$, the temperature $T$ (in K) of the gas, and the full width of the line at half maximum $\\Delta v$ (in ${\\rm km/s}$).\n", "\n", "```{note}\n", - "In the print version of the textbook (Carroll & Ostlie (2007)), the coefficient for Eqn. 7 is given as $5.2 \\times 10^{-15}$. The value above is corrected to an appropriate value.\n", + "In the print version of the textbook (Carroll & Ostlie (2007)), the coefficient for Eq. {eq}`tau_hydro` is given as $5.2 \\times 10^{-15}$. The value above is corrected to an appropriate value.\n", "```\n", "\n", "As long as the 21 cm hydrogen line is optically thin (i.e., on the linear part of the curve of growth), the optical depth is proportional to the neutral hydrogen column density. Studies of **diffuse ${\\rm H\\,I}$ clouds** show temperatures of $30-80\\,{\\rm K}$, number densities ranging from $1-8 \\times 10^8\\,{\\rm m^{-3}}$, and masses between $1-100\\,M_\\odot$. When $A_V<1$, the gas and dust are distributed together throughout the ISM. This correlation breaks down for $A_V > 1$, where the column density of ${\\rm H\\,I}$ no longer increases rapidly with the column density of the dust and other physical processes are involved as the dust becomes optically thick.\n", @@ -418,49 +435,72 @@ "\n", "By the virial theorem, the condition for collapse becomes\n", "\n", - "$$ \\frac{3M_c kT}{\\mu m_H} < \\frac{3}{5} \\frac{GM_c^2}{R_c}. \\tag{8}$$\n", + "```{math}\n", + ":label: virial_collapse\n", + "\\frac{3M_c kT}{\\mu m_H} < \\frac{3}{5} \\frac{GM_c^2}{R_c}. \n", + "```\n", "\n", "The cloud radius may be replaced by using the initial mass density of the cloud $\\rho_o$, which is assumed to be constant throughout the cloud to get\n", "\n", - "$$ R_c = \\left( \\frac{3 M_c}{4\\pi \\rho_o}\\right)^{1/3}. \\tag{9} $$\n", + "```{math}\n", + ":label: cloud_radius\n", + "R_c = \\left( \\frac{3 M_c}{4\\pi \\rho_o}\\right)^{1/3}. \n", + "```\n", "\n", - "Then Eqn. 8 can be solved to determine the minimum mass necessary to initiate the spontaneous cloud collapse, which is a condition known as the **Jeans criterion** $(M_c \\gtrsim M_J)$, or\n", + "Then Eq. {eq}`virial_collapse` can be solved to determine the minimum mass necessary to initiate the spontaneous cloud collapse, which is a condition known as the **Jeans criterion** $(M_c \\gtrsim M_J)$, or\n", "\n", - "$$ M_J \\simeq \\left(\\frac{5kT}{G\\mu m_H}\\right)^{3/2}\\left(\\frac{3}{4\\pi \\rho_o}\\right)^{1/2}. \\tag{10} $$\n", + "```{math}\n", + ":label: Jeans_crit\n", + "M_J \\simeq \\left(\\frac{5kT}{G\\mu m_H}\\right)^{3/2}\\left(\\frac{3}{4\\pi \\rho_o}\\right)^{1/2}.\n", + "```\n", "\n", - "The critical mass $M_J$ is called the **Jeans mass**. The Jeans criterion may also be expressed in terms of the minimum radius $(R_c > R_J)$ through the cloud density $\\rho_o$ (and solving Eq. 8 in terms of $M_c$) to get the **Jeans length**, or\n", + "The critical mass $M_J$ is called the **Jeans mass**. The Jeans criterion may also be expressed in terms of the minimum radius $(R_c > R_J)$ through the cloud density $\\rho_o$ (and solving Eq. {eq}`virial_collapse` in terms of $M_c$) to get the **Jeans length**, or\n", "\n", - "$$ R_J \\simeq \\sqrt{\\frac{15kT}{4\\pi G\\mu m_H \\rho_o}}. \\tag{11} $$\n", + "\\begin{align}\n", + "R_J \\simeq \\sqrt{\\frac{15kT}{4\\pi G\\mu m_H \\rho_o}}. \n", + "\\end{align}\n", "\n", "The Jeans mass derivation neglected the existence of an external pressure on the cloud due to the surrounding interstellar medium. The critical mass required for gravitational collapse in the presence of an external gas pressure $P_o$ is given by the **Bonnor-Ebert mass**,\n", "\n", - "$$ M_{\\rm BE} = \\frac{c_{\\rm BE}v_T^4}{P_o^{1/2}G^{3/2}}, \\tag{12} $$\n", + "\\begin{align}\n", + "M_{\\rm BE} = \\frac{c_{\\rm BE}v_T^4}{P_o^{1/2}G^{3/2}}, \n", + "\\end{align}\n", "\n", "where the *isothermal sound speed* $v_T$ $(\\gamma = 1)$ is\n", "\n", - "$$ v_T \\equiv \\sqrt{\\frac{kT}{\\mu m_H}}, \\tag{13} $$\n", + "\\begin{align}\n", + "v_T \\equiv \\sqrt{\\frac{kT}{\\mu m_H}}, \n", + "\\end{align}\n", "\n", - "and the dimensionless constant $c_{\\rm BE}$ is approximately $1.18$. A dimensionless constant $c_J$ can be derived from Eqn. 10 by substitution of $v_T$ and the ideal gas law, where $c_J \\simeq 5.46$. The smaller constant for the Bonnor-Ebert mass is expected since an external compression force (due to $P_o$) is being exerted on the cloud.\n", + "and the dimensionless constant $c_{\\rm BE}$ is approximately $1.18$. A dimensionless constant $c_J$ can be derived from Eq. {eq}`Jeans_crit` by substitution of $v_T$ and the ideal gas law, where $c_J \\simeq 5.46$. The smaller constant for the Bonnor-Ebert mass is expected since an external compression force (due to $P_o$) is being exerted on the cloud.\n", "\n", - "- **For a typical diffuse hydrogen cloud assumed to be entirely composed of ${\\rm H\\, I}$, $T = 50\\,{\\rm K}$ and $n = 5 \\times 10^8\\, {\\rm m^{-3}}$, what is the minimum mass necessary to cause the cloud to collapse spontaneously?**\n", + "```{exercise}\n", + ":class: orange\n", "\n", - ">Since the cloud is entirely neutral hydrogen, the initial density $\\rho_o$ is calculated by \n", - ">\n", - ">$$ \\rho_o = n_H m_H = 8.4 \\times 10^{-19}\\, {\\rm kg/m^3}.$$\n", - ">\n", - ">Taking $\\mu = 1$ and substituting into the Jeans mass equation (Eqn. 10), we get\n", - ">\n", - ">$$ M_J \\simeq \\left(\\frac{5kT}{G\\mu m_H}\\right)^{3/2}\\left(\\frac{3}{4\\pi \\rho_o}\\right)^{1/2} \\sim 1500\\,M_\\odot. $$\n", - "> \n", - ">This value significantly exceeds the estimated $1-100\\,M_\\odot$ believed to be contained in ${\\rm H\\,I}$ clouds. Hence diffuse hydrogen clouds are stable against gravitational collapse.\n", + "**For a typical diffuse hydrogen cloud assumed to be entirely composed of ${\\rm H\\, I}$, $T = 50\\,{\\rm K}$ and $n = 5 \\times 10^8\\, {\\rm m^{-3}}$, what is the minimum mass necessary to cause the cloud to collapse spontaneously?**\n", "\n", - "- **For a dense core of a giant molecular cloud (i.e., composed of ${\\rm H_2}$), the typical temperatures and number densities are $T = 10\\,{\\rm K}$ and $n = 10^{10}\\, {\\rm m^{-3}}$, what is the minimum mass necessary to cause the cloud to collapse?**\n", + "Since the cloud is entirely neutral hydrogen, the initial density $\\rho_o$ is calculated by \n", "\n", - ">For a cloud of molecular hydrogen, the initial density $\\rho_o$ is calculated by\n", - ">\n", - ">$$ \\rho_o = 2n_H m_H = 3 \\times 10^{-17}\\, {\\rm kg/m^3},$$\n", - ">\n", - ">and now $\\mu = 2$. In this case the Jeans mass is $M_J \\sim 7\\,M_\\odot$, where the characteristic mass of dense cores is $\\sim 10\\,M_\\odot$. Apparently, the dense cores of GMCs are unstable to gravitational collapse and consistent with being sites of star formation. If the Bonnor-Eber mass is used as the critical collapse conditions, then the required mass reduces to approximately $2\\,M_\\odot$." + "$$ \\rho_o = n_H m_H = 8.4 \\times 10^{-19}\\, {\\rm kg/m^3}.$$\n", + "\n", + "Taking $\\mu = 1$ and substituting into the Jeans mass equation (Eq. {eq}`Jeans_crit`), we get\n", + "\n", + "$$ M_J \\simeq \\left(\\frac{5kT}{G\\mu m_H}\\right)^{3/2}\\left(\\frac{3}{4\\pi \\rho_o}\\right)^{1/2} \\sim 1500\\,M_\\odot. $$\n", + " \n", + "This value significantly exceeds the estimated $1-100\\,M_\\odot$ believed to be contained in ${\\rm H\\,I}$ clouds. Hence diffuse hydrogen clouds are stable against gravitational collapse.\n", + "```\n", + "\n", + "```{exercise}\n", + ":class: orange\n", + "\n", + "**For a dense core of a giant molecular cloud (i.e., composed of ${\\rm H_2}$), the typical temperatures and number densities are $T = 10\\,{\\rm K}$ and $n = 10^{10}\\, {\\rm m^{-3}}$, what is the minimum mass necessary to cause the cloud to collapse?**\n", + "\n", + "For a cloud of molecular hydrogen, the initial density $\\rho_o$ is calculated by\n", + "\n", + "$$ \\rho_o = 2n_H m_H = 3 \\times 10^{-17}\\, {\\rm kg/m^3},$$\n", + "\n", + "and now $\\mu = 2$. In this case the Jeans mass is $M_J \\sim 7\\,M_\\odot$, where the characteristic mass of dense cores is $\\sim 10\\,M_\\odot$. Apparently, the dense cores of GMCs are unstable to gravitational collapse and consistent with being sites of star formation. If the Bonnor-Eber mass is used as the critical collapse conditions, then the required mass reduces to approximately $2\\,M_\\odot$.\n", + "```" ] }, { @@ -504,19 +544,22 @@ "metadata": {}, "source": [ "### Homologous Collapse\n", - "Following Jean's derivation, the criterion for gravitational collapse of a molecular cloud can be satisfied in the absence of rotation, turbulence, or magnetic fields. If we make another simplifying assumption that *any existing pressure gradients are small*, then the cloud is essentially in free-fall. Throughout the free-fall phase, the temperature of the gas remains nearly constant (or *isothermal*). This is true as long as the collapse remains optically thin and the gravitational potential energy released during collapse can be efficiently radiated away (i.e., the gas can remain cold). In this case the spherically symmetric hydrodynamic equation can be used to describe the contraction. From Eqn. 5 of [Sect. 7.1.2](https://saturnaxis.github.io/ModernAstro/Chapter_11/interiors-of-stars.html#the-derivation-of-the-hydrostatic-equilibrium-equation), we have\n", + "Following Jean's derivation, the criterion for gravitational collapse of a molecular cloud can be satisfied in the absence of rotation, turbulence, or magnetic fields. If we make another simplifying assumption that *any existing pressure gradients are small*, then the cloud is essentially in free-fall. Throughout the free-fall phase, the temperature of the gas remains nearly constant (or *isothermal*). This is true as long as the collapse remains optically thin and the gravitational potential energy released during collapse can be efficiently radiated away (i.e., the gas can remain cold). In this case the spherically symmetric hydrodynamic equation can be used to describe the contraction. From Eq. {eq}`diff_mass`, we have\n", "\n", - "$$ \\frac{d^2r}{dt^2} = -\\frac{GM_r}{r^2}. \\tag{14} $$\n", + "```{math}\n", + ":label: hydro_equil_gravity\n", + "\\frac{d^2r}{dt^2} = -\\frac{GM_r}{r^2}. \n", + "``````\n", "\n", - "The right-hand side (RHS) of Eqn. 14 is just the local acceleration due to gravity at a distance $r$ from the center of the spherical cloud and depends on the mass contained $M_r$ at that radius.\n", + "The right-hand side (RHS) of Eq. {eq}`hydro_equil_gravity` is just the local acceleration due to gravity at a distance $r$ from the center of the spherical cloud and depends on the mass contained $M_r$ at that radius.\n", "\n", - "To describe the behavior of the surface of a sphere within the collapsing cloud as function of time, Eqn. 14 must be integrated with respect to time. Since we are using the mass enclosed at $r$, $M_r$ remains a constant in a similar way as the charge in the integration of Gauss' law. As a result, we can replace the mass enclosed $M_r$ by the product of the initial density $\\rho_o$ and the spherical volume. Then we apply a \"trick\" used in many problems, where we multiply by the velocity $dr/dt$ on both sides so that the second derivative can be replaced by an anti-derivative through the chain rule. The resulting expression is\n", + "To describe the behavior of the surface of a sphere within the collapsing cloud as function of time, Eq. {eq}`hydro_equil_gravity` must be integrated with respect to time. Since we are using the mass enclosed at $r$, $M_r$ remains a constant in a similar way as the charge in the integration of Gauss' law. As a result, we can replace the mass enclosed $M_r$ by the product of the initial density $\\rho_o$ and the spherical volume. Then we apply a \"trick\" used in many problems, where we multiply by the velocity $dr/dt$ on both sides so that the second derivative can be replaced by an anti-derivative through the chain rule. The resulting expression is\n", "\n", - "$$ \\begin{align*}\n", + "\\begin{align*}\n", "\\frac{dr}{dt}\\frac{d^2 r}{dt^2} &= \\frac{1}{2}\\frac{d}{dt}\\left(\\frac{dr}{dt}\\right)^2 \\\\\n", "- \\left(\\frac{4}{3}\\pi G \\rho_o r_o^3 \\right)\\frac{1}{r^2}\\frac{dr}{dt} &= \\left(\\frac{4}{3}\\pi G \\rho_o r_o^3 \\right)\\frac{d}{dt}\\left(\\frac{1}{r}\\right) \\\\\n", "\\frac{1}{2}\\frac{d}{dt}\\left(\\frac{dr}{dt}\\right)^2 &= \\left(\\frac{4}{3}\\pi G \\rho_o r_o^3 \\right)\\frac{d}{dt}\\left(\\frac{1}{r}\\right),\n", - "\\end{align*} $$\n", + "\\end{align*}\n", "\n", "which can be integrated with respect to time to give\n", "\n", @@ -528,29 +571,44 @@ "\n", "Through substitution, we can solve for the velocity at the surface as\n", "\n", - "$$ \\frac{dr}{dt} = -\\left[ \\frac{8}{3}\\pi G \\rho_o r_o^2 \\left(\\frac{r_o}{r}-1 \\right)\\right]^{1/2}, \\tag{15} $$\n", + "```{math}\n", + ":label: vel_surf\n", + "\\frac{dr}{dt} = -\\left[ \\frac{8}{3}\\pi G \\rho_o r_o^2 \\left(\\frac{r_o}{r}-1 \\right)\\right]^{1/2},\n", + "```\n", "\n", - "where the negative root was chosen because the cloud is collapsing. To integrate Eqn. 15, we make the variable substitution $\\theta \\equiv r/r_o$ and $\\chi \\equiv \\sqrt{(8/3)\\pi G \\rho_o}$, which leads to a simpler differential equation,\n", + "where the negative root was chosen because the cloud is collapsing. To integrate Eq. {eq}`vel_surf`, we make the variable substitution $\\theta \\equiv r/r_o$ and $\\chi \\equiv \\sqrt{(8/3)\\pi G \\rho_o}$, which leads to a simpler differential equation,\n", "\n", - "$$ \\frac{d\\theta}{dt} = -\\chi\\left(\\frac{1}{\\theta} - 1 \\right)^{1/2} \\tag{16}.$$\n", + "```{math}\n", + ":label: vel_surf_angle\n", + "\\frac{d\\theta}{dt} = -\\chi\\left(\\frac{1}{\\theta} - 1 \\right)^{1/2} \n", + "```\n", "\n", - "The integral of Eqn. 16 is quite messy and we make yet another substitution,\n", + "The integral of Eq. {eq}`vel_surf_angle` is quite messy and we make yet another substitution,\n", "\n", - "$$ \\theta \\equiv \\cos^2 \\xi, \\tag{17} $$\n", + "\\begin{align}\n", + "\\theta \\equiv \\cos^2 \\xi, \n", + "\\end{align}\n", "\n", - "where $\\frac{d\\theta}{dt} = -2\\cos \\xi \\sin \\xi \\frac{d\\xi}{dt}$ and Eqn. 16 becomes\n", + "where $\\frac{d\\theta}{dt} = -2\\cos \\xi \\sin \\xi \\frac{d\\xi}{dt}$ and Eq. {eq}`vel_surf_angle` becomes\n", "\n", - "$$ \\cos^2 \\xi \\frac{d\\xi}{dt} = \\frac{\\chi}{2}. \\tag{18} $$\n", + "```{math}\n", + ":label: vel_surf_sub\n", + "\\cos^2 \\xi \\frac{d\\xi}{dt} = \\frac{\\chi}{2}. \n", + "```\n", "\n", - "Equation 18 can be integrated directly to yield\n", + "Equation {eq}`vel_surf_sub` can be integrated directly to yield\n", "\n", - "$$ \\frac{\\xi}{2} + \\frac{1}{4}\\sin 2\\xi = \\frac{\\chi}{2}t + C_2. \\tag{19} $$\n", + "\\begin{align}\n", + "\\frac{\\xi}{2} + \\frac{1}{4}\\sin 2\\xi = \\frac{\\chi}{2}t + C_2.\n", + "\\end{align}\n", "\n", "The integration constant $C_2$ can be evaluated using the same initial condition as before ($r=r_o$ at $t=0$), which implies that $d\\theta/dt = 0$ or $\\xi = 0$ at the beginning of the collapse. Therefore, $C_2=0$.\n", "\n", "The equation of motion for the gravitational collapse of the cloud (in parameterized form) is\n", "\n", - "$$ \\xi + \\frac{1}{2}\\sin 2\\xi = \\chi t. \\tag{20} $$\n", + "\\begin{align}\n", + "\\xi + \\frac{1}{2}\\sin 2\\xi = \\chi t.\n", + "\\end{align}\n", "\n", "The **free-fall timescale** $t_{\\rm ff}$ for a cloud is the time when the radius of the collapsing sphere reaches zero $(\\theta = 0\\,{\\rm or}\\, \\xi = \\pi/2)$ or at least becomes very small. Then \n", "\n", @@ -558,17 +616,24 @@ "\n", "Back-substituting our value for $\\chi$, we have\n", "\n", - "$$ t_{\\rm ff} = \\left( \\frac{3\\pi}{32}\\frac{1}{G\\rho_o} \\right)^{1/2}. \\tag{21} $$\n", + "```{math}\n", + ":label: tau_ff\n", + "t_{\\rm ff} = \\left( \\frac{3\\pi}{32}\\frac{1}{G\\rho_o} \\right)^{1/2}. \n", + "```\n", "\n", "The free-fall time is actually *independent* of the initial radius of the sphere. AS long as the original density of the spherical molecular cloud was uniform, all parts of the cloud will take the same amount of time to collapse and the density will increase at the same rate everywhere. This behavior is known as **homologous collapse**. However, if the cloud is somewhat centrally condensed when the collapse begins, the free-fall time will be shorter for material near the center compared to material farther out (i.e., $t_{\\rm ff} \\propto 1/\\sqrt{\\rho_o}$). As the collapse progresses, the density will increase more rapidly near the center than in the other regions and in this case, the collapse is known as **inside-out collapse**.\n", "\n", - "- **For a dense core of a giant molecular cloud, what is the time required for collapse?**\n", + "```{exercise}\n", + ":class: orange\n", "\n", - ">From the example in the Jeans Criterion, we have the initial density $\\rho_o$ and we can find the free-fall timescale by\n", - ">\n", - ">$$ t_{\\rm ff} = \\left( \\frac{3\\pi}{32}\\frac{1}{G\\rho_o} \\right)^{1/2} = 3.6 \\times 10^5\\,{\\rm yr}. $$\n", - ">\n", - ">Equation 20 is transcendental and cannot be solved explicitly. But, numerical root finding techniques can be used. The collapse is slow initially and accelerates as $t_{\\rm ff}$ is approached. At the same time, the density increases most rapidly during the final stages of collapse." + "**For a dense core of a giant molecular cloud, what is the time required for collapse?**\n", + "\n", + "From the example in the Jeans Criterion, we have the initial density $\\rho_o$ and we can find the free-fall timescale by\n", + "\n", + "$$ t_{\\rm ff} = \\left( \\frac{3\\pi}{32}\\frac{1}{G\\rho_o} \\right)^{1/2} = 3.6 \\times 10^5\\,{\\rm yr}. $$\n", + "\n", + "Equation 20 is transcendental and cannot be solved explicitly. But, numerical root finding techniques can be used. The collapse is slow initially and accelerates as $t_{\\rm ff}$ is approached. At the same time, the density increases most rapidly during the final stages of collapse.\n", + "```" ] }, { @@ -677,9 +742,11 @@ "\n", "For an adiabatic process the gas pressure is related to its density by $\\gamma$ (the ratio of specific heats). An adiabatic relation between density and temperature can be obtained (using the ideal gas law),\n", "\n", - "$$ T = K^{\\prime\\prime}\\rho^{\\gamma -1}, \\tag{22} $$\n", + "\\begin{align}\n", + "T = K^{\\prime\\prime}\\rho^{\\gamma -1}, \n", + "\\end{align}\n", "\n", - "where $K^{\\prime\\prime}$ is a constant. Substituting into the equation for the Jeans mass (Eqn. 10), we find that\n", + "where $K^{\\prime\\prime}$ is a constant. Substituting into the equation for the Jeans mass (Eq. {eq}`Jeans_crit`), we find that\n", "\n", "$$ M_J \\propto \\rho^{(3\\gamma-4)/2}, $$\n", "\n", @@ -701,9 +768,11 @@ "\n", "$$ M_J^{5/2} = \\frac{4\\pi}{G^{3/2}}R_J^{9/2}e\\sigma T^4. $$\n", "\n", - "Using the initial cloud radius at a constant density (Eqn. 9), and using Eqn. 10 to write the density in terms of the Jeans mass, we can determine the minimum obtainable Jeans mass:\n", + "Using the initial cloud radius at a constant density (Eq. {eq}`cloud_radius`), and using Eq. {eq}`Jeans_crit` to write the density in terms of the Jeans mass, we can determine the minimum obtainable Jeans mass:\n", "\n", - "$$ M_{J_{\\rm min}} = 0.03 \\left(\\frac{T^{1/4}}{e^{1/2} \\mu^{9/4}}\\right)\\, M_\\odot, \\tag{23} $$\n", + "\\begin{align} \n", + "M_{J_{\\rm min}} = 0.03 \\left(\\frac{T^{1/4}}{e^{1/2} \\mu^{9/4}}\\right)\\, M_\\odot, \n", + "\\end{align}\n", "\n", "which depends on the temperature $T$ (in K), mean molecular weight $\\mu$, and the efficiency factor $e$. If we take $\\mu \\sim 1$, $e\\sim 0.1$, and $T \\sim 1000\\,{\\rm K}$ to be representative when adiabatic effects become significant, then $M_J \\sim 0.5 M_\\odot$. Fragmentation ceases when the segments of the original cloud begin to reach the range of solar mass objects. The estimate is relatively insensitive to other reasonable choices, where if $e\\sim 1$, then $M_J \\sim 0.2 M_\\odot$.\n" ] @@ -729,21 +798,29 @@ "\n", "The virial theorem was invoked during the derivation of the Jean's criterion as a balance between the gravitational potential energy and the cloud's internal (thermal) kinetic energy. The magnetic field energy was ignored, where including the effects of the magnetic field produces a critical mass of\n", "\n", - "$$ M_B = c_B \\frac{\\pi R^2 B}{G^{1/2}}, \\tag{24}$$\n", + "\\begin{align}\n", + "M_B = c_B \\frac{\\pi R^2 B}{G^{1/2}}, \n", + "\\end{align}\n", "\n", "where $c_B = 380\\,{\\rm N^{1/2}/m/T}$ for a magnetic field permeating a spherical, uniform cloud. If the magnetic field strength $B$ is expressed in ${\\rm nT}$ and the cloud radius $R$ in units of ${\\rm pc}$, then $M_B$ can be written as\n", "\n", - "$$ M_B \\simeq 70\\,M_\\odot \\left(\\frac{B}{1\\,{\\rm nT}} \\right) \\left(\\frac{R}{1\\,{\\rm pc}} \\right)^2. \\tag{25} $$\n", + "\\begin{align}\n", + "M_B \\simeq 70\\,M_\\odot \\left(\\frac{B}{1\\,{\\rm nT}} \\right) \\left(\\frac{R}{1\\,{\\rm pc}} \\right)^2. \n", + "\\end{align}\n", "\n", "If the mass of the cloud $M_c$ is less than $M_B$, the cloud is said to be **magnetically subcritical** and stable against collapse. Otherwise, the cloud is **magnetically supercritical** and the force due to gravity will overwhelm the ability of the magnetic field to resist collapse.\n", "\n", - "- **What is the critical mass if the dense core has a magnetic field strength of $100\\,{\\rm nT}$ and a radius of $0.1\\,{\\rm pc}$?**\n", + "```{exercise}\n", + ":class:orange\n", "\n", - ">The critical mass after including the magnetic field can be calculated as,\n", - ">\n", - ">$$ M_B \\simeq 70\\,M_\\odot \\left(\\frac{100}{1\\,{\\rm nT}} \\right) \\left(\\frac{0.1}{1\\,{\\rm pc}} \\right)^2 = 70\\,M_\\odot, $$\n", - ">\n", - ">implying that a dense core of $10\\,M_\\odot$ would be stable against collapse $(M_c < M_B)$. However, if the magnetic field was weaker $(B=1\\,{\\rm nT})$, then $M_B \\simeq 0.7\\,M_\\odot$ and collapse would occur." + "**What is the critical mass if the dense core has a magnetic field strength of $100\\,{\\rm nT}$ and a radius of $0.1\\,{\\rm pc}$?**\n", + "\n", + "The critical mass after including the magnetic field can be calculated as,\n", + "\n", + "$$ M_B \\simeq 70\\,M_\\odot \\left(\\frac{100}{1\\,{\\rm nT}} \\right) \\left(\\frac{0.1}{1\\,{\\rm pc}} \\right)^2 = 70\\,M_\\odot, $$\n", + "\n", + "implying that a dense core of $10\\,M_\\odot$ would be stable against collapse $(M_c < M_B)$. However, if the magnetic field was weaker $(B=1\\,{\\rm nT})$, then $M_B \\simeq 0.7\\,M_\\odot$ and collapse would occur.\n", + "```" ] }, { @@ -763,15 +840,21 @@ "\n", "To determine the relative impact of ambipolar diffusion, we need to estimate the characteristic timescale. If we know the drift velocity of the neutrals and the distance across the molecular cloud, then it can be shown that\n", "\n", - "$$ t_{\\rm AD} \\simeq \\frac{2R}{v_{\\rm drift}} \\simeq 10\\,{\\rm Gyr} \\left( \\frac{n_{\\rm H_2}}{10^{10}\\,{\\rm m^{-3}}}\\right) \\left( \\frac{B}{1\\,{\\rm nT}}\\right)^{-2} \\left( \\frac{R}{1\\,{\\rm pc}}\\right)^{2}. \\tag{26} $$\n", + "\\begin{align}\n", + "t_{\\rm AD} \\simeq \\frac{2R}{v_{\\rm drift}} \\simeq 10\\ {\\rm Gyr} \\left( \\frac{n_{\\rm H_2}}{10^{10}\\,{\\rm m^{-3}}}\\right) \\left( \\frac{B}{1\\,{\\rm nT}}\\right)^{-2} \\left( \\frac{R}{1\\,{\\rm pc}}\\right)^{2}.\n", + "\\end{align}\n", "\n", - "- **What is the timescale for ambipolar diffusion within a dense core of a GMC, if $B = 1\\,{\\rm nT}$ and $R = 0.1\\,{\\rm pc}$?**\n", + "```{exercise}\n", + ":class: orange\n", "\n", - ">The characteristic number density for molecular hydrogen in a dense core is $n_{\\rm H_2} = 10^{10}\\,{\\rm m^{-3}}$. The timescale for ambipolar diffusion (AD) is then determined as\n", - ">\n", - ">$$ t_{\\rm AD} \\simeq 10\\,{\\rm Gyr} \\left( \\frac{0.1\\,{\\rm pc}}{1\\,{\\rm pc}}\\right)^{2} \\simeq 100\\,{\\rm Myr}. $$\n", + "**What is the timescale for ambipolar diffusion within a dense core of a GMC, if $B = 1\\,{\\rm nT}$ and $R = 0.1\\,{\\rm pc}$?**\n", + "\n", + "The characteristic number density for molecular hydrogen in a dense core is $n_{\\rm H_2} = 10^{10}\\,{\\rm m^{-3}}$. The timescale for ambipolar diffusion (AD) is then determined as\n", + "\n", + "$$ t_{\\rm AD} \\simeq 10\\,{\\rm Gyr} \\left( \\frac{0.1\\,{\\rm pc}}{1\\,{\\rm pc}}\\right)^{2} \\simeq 100\\,{\\rm Myr}. $$\n", ">\n", - ">This is ~$1000\\times$ longer than the free-fall timescale determined for [homologous collapse](https://saturnaxis.github.io/ModernAstro/Chapter_12/interstellar-medium-and-star-formation.html#homologous-collapse). Clearly the ambipolar diffusion process can control the evolution of a dense core for a long time before free-fall collapse begins." + "This is ~$1000\\times$ longer than the free-fall timescale determined for [homologous collapse](https://saturnaxis.github.io/ModernAstro/Chapter_12/interstellar-medium-and-star-formation.html#homologous-collapse). Clearly the ambipolar diffusion process can control the evolution of a dense core for a long time before free-fall collapse begins.\n", + "```" ] }, { @@ -920,29 +1003,35 @@ "\n", "which solving for the radius produces\n", "\n", - "$$ r \\simeq \\left( \\frac{3N}{4\\pi \\alpha n_H^2} \\right)^{1/3} \\tag{27} $$\n", + "\\begin{align}\n", + "r \\simeq \\left( \\frac{3N}{4\\pi \\alpha n_H^2} \\right)^{1/3} \n", + "\\end{align}\n", "\n", "and is called the **Strömgren radius** $r_S$.\n", "\n", - "- **What is the Strömgren radius due to an O6 star in a typical ${\\rm H\\,II}$ region?** (Assume that $T_e\\simeq 45,000\\,{\\rm K}$ and $L\\simeq 1.3\\times10^5\\,L_\\odot$)\n", + "```{exercise}\n", + ":class:orange\n", "\n", - ">From Wien's law, the peak wavelength is given as\n", - ">\n", - ">$$ \\lambda_{\\rm max} = \\frac{0.0029\\,{\\rm m\\,K}}{T_e} \\approx 64\\,{\\rm nm}.$$\n", - ">\n", - ">This is significantly shorter (or higher energy) than the 91.2-nm limit ([${\\rm Ly_{\\rm limit}}$ from the wavelength of hydrogen](https://saturnaxis.github.io/ModernAstro/Chapter_5/interaction-of-light-and-matter.html#the-wavelength-of-hydrogen)) and it can be assumed that most of the photons created by an O6 star are capable of causing ionization. The energy of one 64-nm photon can be calculated giving\n", - ">\n", - ">$$E_\\gamma = \\frac{hc}{\\lambda} = 19\\,{\\rm eV}.$$\n", - ">\n", - ">Assuming that all of the emitted photons have the same (peak wavelength), then the total number of photons produced *per second* is just\n", - ">\n", - ">$$ N \\simeq \\frac{L}{E_\\gamma} \\simeq 1.6 \\times 10^{49}\\,{\\rm photons/s}.$$\n", - ">\n", - ">Lastly, a typical the number density within a GMC is $n_H \\sim 10^8\\,{\\rm m^{-3}}$, we find\n", - ">\n", - ">$$ r_S \\simeq \\left( \\frac{3N}{4\\pi \\alpha n_H^2} \\right)^{1/3} \\simeq 3.48\\,{\\rm pc}$$\n", - ">\n", - ">Values of $r_S$ range from less than 0.1 pc to greater than 100 pc." + "**What is the Strömgren radius due to an O6 star in a typical ${\\rm H\\,II}$ region?** (Assume that $T_e\\simeq 45,000\\,{\\rm K}$ and $L\\simeq 1.3\\times10^5\\,L_\\odot$)\n", + "\n", + "From Wien's law, the peak wavelength is given as\n", + "\n", + "$$ \\lambda_{\\rm max} = \\frac{0.0029\\,{\\rm m\\,K}}{T_e} \\approx 64\\,{\\rm nm}.$$\n", + "\n", + "This is significantly shorter (or higher energy) than the 91.2-nm limit ([${\\rm Ly_{\\rm limit}}$ from the wavelength of hydrogen](https://saturnaxis.github.io/ModernAstro/Chapter_5/interaction-of-light-and-matter.html#the-wavelength-of-hydrogen)) and it can be assumed that most of the photons created by an O6 star are capable of causing ionization. The energy of one 64-nm photon can be calculated giving\n", + "\n", + "$$E_\\gamma = \\frac{hc}{\\lambda} = 19\\,{\\rm eV}.$$\n", + "\n", + "Assuming that all of the emitted photons have the same (peak wavelength), then the total number of photons produced *per second* is just\n", + "\n", + "$$ N \\simeq \\frac{L}{E_\\gamma} \\simeq 1.6 \\times 10^{49}\\,{\\rm photons/s}.$$\n", + "\n", + "Lastly, a typical the number density within a GMC is $n_H \\sim 10^8\\,{\\rm m^{-3}}$, we find\n", + "\n", + "$$ r_S \\simeq \\left( \\frac{3N}{4\\pi \\alpha n_H^2} \\right)^{1/3} \\simeq 3.48\\,{\\rm pc}$$\n", + "\n", + "Values of $r_S$ range from less than 0.1 pc to greater than 100 pc.\n", + "```" ] }, { @@ -1111,7 +1200,7 @@ "```\n", "\n", "```{admonition} Problem 5\n", - "Assuming that the free-fall acceleration of the surface of a collapsing cloud remains constant during the entire collapse, derive an expression for the free-fall time. Show that your answer differs from Eqn. 21 only by a term of order unity.\n", + "Assuming that the free-fall acceleration of the surface of a collapsing cloud remains constant during the entire collapse, derive an expression for the free-fall time. Show that your answer differs from Eq. {eq}`tau_ff` only by a term of order unity.\n", "``` \n", "\n", "```{admonition} Problem 6\n", diff --git a/docs/objects.inv b/docs/objects.inv index 451c8ede73d270d6858dabe15e01cd63310bf933..34f5d0a6fb81ce9af056c523d4afac50229dfbf0 100644 GIT binary patch delta 7047 zcmV;28+hcFIk`EIcYjhFNs|A0ii()M4cI_{1lVKGdM`8>kJ&Xg2i}>v4PuB=cS+q# zcUMnWHy}^=V&36?UBuj*+`s#O4>3=2`B7Ef5)!l|z!44u>guelto+Wb(m^EiQWY27 zoo!QP!j$35PzR#Z6$cYfwxTjk#K%a9vnpHj&$`=j2EW9*Fn=9A?3hv(W+sqX18bN^ z?wQa-afZ+8DSG^-DOHllqLZpHuF{UoLITnm>LQh8tY_oi_T?#5?09iZ;{}W-ti&5P zzD93%+W}WsI;rMjGGiQ1dMx9tbEbZ-R2C?j>O`BeGx_nEMd;AITO+XN0q0d>=96N^ zI7Yz1K5-w{uz&Zrhq6?-VW~7GzEXGW@SL$VW#7PK+}~qwB7AMXJwpKx-uP;tDr1^RRCBLkD_wc{CB@mPk-P?i8DJ2 zr++K-HT^`1Pbf`J*V?xPUuv@&=pz6d>-cM#^CbIO{yzrba%T${@QpN1G%^Wx> zB>noq+PkBJ!}n*0O-S}5e&^4b7G+!}>Mv_2g_7wYQQOC|xD54m))`0>2NPJ#8^ge0 z{|X&eynlhE7H&`t`ea;2`cA+TBW8o;AvMtJNTZgV3C=!aKtDuU$U=!SQbvhPmq228 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The Sun", "7. The Interiors of Stars", "8. The Interstellar Medium and Star Formation", "9. Main-Sequence and Post-Main-Sequence Stellar Evolution", "10. Stellar Pulsation", "11. The Fate of Massive Stars", "12. The Degenerate Remnants of Stars", "1. The Continuous Spectrum of Light", "2. The Interaction of Matter and Light", "3. Telescopes", "4. The Classification of Stellar Spectra", "5. 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"Chapter_3/continuous-spectrum-of-light.ipynb", "Chapter_5/interaction-of-light-and-matter.ipynb", "Chapter_6/telescopes.ipynb", "Chapter_8/classification-of-stellar-spectra.ipynb", "Chapter_9/stellar-atmospheres.ipynb", "LICENSE.md", "Preamble/Markdown-basics.ipynb", "Preamble/Python-basics.ipynb", "Preamble/who-for.ipynb", "README.md", "home.md"], "titles": ["6. The Sun", "7. The Interiors of Stars", "8. The Interstellar Medium and Star Formation", "9. Main-Sequence and Post-Main-Sequence Stellar Evolution", "10. Stellar Pulsation", "11. The Fate of Massive Stars", "12. The Degenerate Remnants of Stars", "1. The Continuous Spectrum of Light", "2. The Interaction of Matter and Light", "3. Telescopes", "4. The Classification of Stellar Spectra", "5. 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