Blog #16 - Nuclear Fusion in Stars

Nuclear Fusion in Stars

We have found equations for how gravitational energy change much be matched by a radiative energy exchange in a star. But how the radiation is internally created within the star is a subject yet to be covered. This energy is produced by what is known as nuclear fusion. In fusion, two atomic nuclei combine to form new elements and in the process create subatomic particles (neutrons or protons) and in the process release energy through the change in mass of the resulting atoms.

Depending on the mass and age of a star, there are different types of atomic fusions that are common. Most common is the proton-proton, or hydrogen-hydrogen fusion, in which 4 hydrogen particles come together to form 2 helium particles with the help of free electrons. Given that hydrogen is the most basic atom unit, all other elements were originally formed through some form of hydrogen fusion. Today we know that most stars, including our sun, have intense luminosities fueled by hydrogen fusion.

Other common types of fusion include using helium or even carbon isotopes to change mass of interior atomic units. Elements up until iron can be used for fusion as iron sits at the peak of the energy curve that determines whether it takes more energy to produce fusion than is produced from the reaction. The curve below shows the energy required for fusion. It is noted that some heavier elements can still produce energy through fission, a completely different energy process.

The constant fusion of atoms inside of stars requires energy to constantly escape the system and while there are different convective and radiative processes to get rid of the massive quantities of energy, each process is ultimately responsible for the light we see coming off of the star, also known as its luminosity.

Blog #15 - Radiative Energy Transport

Radiative Energy Transport 

Stars generate energy in their cores, where nuclear fusion is taking place. The energy generated is eventually radiated out at the star's surface. As a result, there is a gradient in energy density from the center (high) to the surface (low). However, thermodynamic systems tend towards "equilibrium". We therefore need to determine how energy flows through the star.

a) Inside the star, consider a mass shell of width \(\Delta r\) and radius r. This mass shell has an energy density \(u + \Delta u \), and the next mass shell out at radius \(r + \Delta r\) will have an energy density u. Both shells can be thought to behave as blackbodies. The net outwards flow of energy, L(r), must equal the total excess energy in the inner shell divided by the amount of time needed to cross the shell's width \(\Delta r\). Use this to derive an expression for L(r) in terms of the energy density profile \(\frac{du}{dr}\). This is the diffusion equation describing the outward flow of energy.

We know that the outward flow of energy L(r) has units of energy/time so we begin with equation \[L(r) = \frac{E}{\tau} \]We can then solve for E as the energy density over volume or \[E = du*V\] But the to find the change in energy density over the change in radius between the two shells \(\frac{du}{dr}\) comes out to an energy difference of \(- \Delta u\) and a radius difference of \(\Delta r\). This gives us an energy equation of \[E = du*V = - \Delta u * (4 \pi r^2) * \Delta r \] Using this E equation in our L(r) formula we get \[L(r) = \frac{E}{\tau} = \frac{- \Delta u * (4 \pi r^2) * \Delta r}{\frac{\Delta r^2 k \rho}{c}}\]The equation for \(\tau\) comes from the random walk equation, where c is the speed of light, \(\rho\) the density of the star, k an optical density. Simplifying this equation we get \[L(r) = \frac{- \Delta u}{\Delta r} * \frac{4 \pi r^2 c}{\rho c}\] Which gives us our final differential form \[L(r) = \frac{-4 \pi r^2 c}{\rho c} \frac{du}{dr}\]

b) From the diffusion equation, use the fact that the energy density of a blackbody is \(u[T(r)] = a T^4 \) to derive the equation for Radiative Energy Transport.

We begin by solving for the \(\frac{du}{dr}\) of the given energy density by taking a derivative of u with respect to r. \[\frac{du}{dr} = \frac{d}{dr}(u[T(r)]) = 4 a T(r)^3 \frac{dT}{dr} \] With an equation for \(\frac{du}{dr} \) We plug into the differential from part (a) and get \[L(r) = \frac{-4 \pi r^2 c}{\rho c} \frac{du}{dr} = \frac{-16 \pi r^2 c a T(r)^3}{\rho k} \frac{dT}{dr} \] Which rearranges to the expected equation for radiative energy transport \[\frac{dT}{dr} = \frac{-L \rho k}{16 \pi r^2 c a T(r)^3}\]

Blog #14 - Rudolf Clausius and the Virial Theorem

Rudolf Clausius and the Virial Theorem 

Rudolf Clausius was a German born physicist and mathematician who forever changed the rules of thermodynamics, mathematics, and eventually astronomy with his findings regarding the way heat moved through systems and how it affected an amorphous term known as energy. Before serving in the Franco-Prussian war in 1870 and injuring himself (leading to a decrease in his ability to research) Clausius became one of the world's foremost scientists in the development of heat transfer and thermodynamics. Early in his career he developed a more sound mathematical model for the carnot cycle. Then in 1850 he published his landmark work On the Moving Force of Heat which alongside what would eventually be known as the Second Law of Thermodynamics Clausius published his work on the virial theorem.

In it's most simple form, the virial theorem is listed as:

Which states that the kinetic energy of a system is directly related to the potential energy of a system which can be represented by the force on the kth particle which is at a position r from a central location. This equation takes on many forms including electromagnetic versions, special relativity versions, and quantum mechanical versions. Overall the virial theorem helps relate the average kinetic energy over time in a system to the average potential energy over time of a system. This equation can be expanded to systems with many bodies in a variety of positions, coordinate planes, and physical situations.

With Clausius' ground breaking work, later developments such as Kepler's 3 Laws of Motion, electron movement within particles, and the discovery of dark matter were all formed. The scale of the virial theorem seems to hold for most scenarios and the beautifully simple equation has an incredibly impactful footprint on the history of physics and astronomy.

Blog #13 - Kelvin-Helmholtz Timescale

Kelvin-Helmholtz Timescale

We know that the Sun started from the gravitational collapse of a giant cloud of gas. Let's hypothesize that the Sun is powered solely by this gravitational contraction, as was once posited by astronomers long ago. As it shrinks, its internal thermal energy increases, increasing its temperature due to the virial theorem and thereby causing it to radiate. How long would the Sun last if it was thermally radiating its current power output, \(L_{\odot} = 4*10^{33} erg \space s^{-1}\)? This is known as the Kelvin-Helmholtz timescale. How does this timescale compare to the age of the oldest Moon rocks (about 4.5 billion years old, also known as Gyr)?

We begin with the formula for internal energy derived earlier in the worksheet \[U = -\frac{3GM^2}{5R}\] For this problem we know that the M and R are the mass and radius of the Sun, respectively. We solve the problem with \[M_{Sun} = 2*10^{30} kg \space \space R_{Sun} = 7*10^8 m \]Which gives us \[U_{Sun} = 2.27*10^{41} Joules\] Using the conversion of joules to ergs as \[1 Joule = 10^7 ergs\]We get \[U_{Sun} = 2.27*10^{48} erg\]We know that luminosity is in a scale of \([\frac{erg}{s}] \) meaning we need to divide our \(U_{Sun}\) by its luminosity in order to get the Kelvin-Helmholtz Timescale. \[KHT = \frac{U}{L} = \frac{2.27*10^{48} erg}{4*10^{33} \frac{erg}{s}} = 5*10^{14} seconds\]Converting from seconds to years we get \[KHT \approx 16 \space million \space years\]We're told that the oldest moon rocks are 4.5 billion years old, much older than the expected age of the Sun, thus there is a flaw in the assumptions and astronomers were forced to reassess how the Sun's energy is created.

Blog #12 - Deriving Kepler's 3rd Law

Kepler's 3rd Law 

For a planet of mass m orbiting a star of mass \(M_{\star} \) at a distance of a in a circular orbit, start with the virial theorem and derive Kepler's Third Law of Motion. Assume that \(m << M_{\star} \) and remember that, since m is so small, the semimajor axis, which is formally \(a = a_p + a_{\star} \) reduces to \(a = a_p \). 

We begin with the virial theorem that states \[<t> = -\frac{1}{2}<U> \space \rightarrow \frac{mv^2}{2} = \frac{GM_{star}m}{2a} \]From this theorem we can divide out \(m\) and constants to get an expression for \(v^2\) as \[v^2 = \frac{GM_{star}}{a}\]This gives us a velocity in units \([\frac{m}{s}]\). We need to find the relationship between velocity and period with units [s]. We know that a period is the time that it takes a body to completely through its orbit, which is, in this case, circular. The distance traveled for a circular orbit is simply \[d = 2 \pi a \] and we know that time (period) is \[t = \frac{d}{v} \] So we divide our two terms to get the relationship between period and semimajor axis. \[P = \frac{d}{v} = \frac{2 \pi a}{\sqrt{\frac{GM_{\star}}{a}}} \]Squaring this equation we get the relationship\[P^2 = \frac{4 \pi^2 a^3}{GM_{\star}} \]Or more commonly, Kepler's 3rd Law of Motion \[P^2 \propto a^3 \]

Blog #11 - Dust and Spectral Lines

Dust and Spectral Lines 

The matter between stars is known as the interstellar medium and is composed of mostly dust and gas. This dust absorbs stellar light and therefore reduces the apparent brightness of stars as viewed from the Earth. This effect is known as extinction and worsens as we observe more distant stars.

(a) Using trigonometric parallax you measure that the distance to a particular star is 200pc.You know that the absolute magnitude of this star should be 2 and measure an apparent magnitude of 12. How much flux have you lost due to the intervening dust? Compare your answer in magnitudes.

We follow the the relationships that we previously established between absolute and relative magnitudes and the distance to an observed star. \[M-m = -5*(log(d) - 1)\]With this equation we have every constant given to us besides the observed apparent magnitude. Thus we solve for "m" and get \[2-m = -5*(log(200pc) -1) \rightarrow m = 8.5\] With the expected apparent magnitude as 12, this gives us a magnitude loss of 3.5. 

(b) Optical depth is a measure of the absorption of photons as they travel through a medium.The definition of optical depth is:

τ = ln(Iin/Iout)

where I denotes the specific intensity of the source. In this problem, we are looking at a single source at a fixed distance, so we can also express τ in terms of the flux:

τ = ln(Fin/Fout)

Use this latter definition to determine the relationship between τ and apparent magnitude. What is the optical depth along the line of sight to the star in the previous problem?

We are told that \[\tau = ln(\frac{F_{in}}{F_{out}})\]and know that know that \[\frac{F_{in}}{F_{out}} = 2.5^{m-M} \] Thus we can change the equation to \[\tau = ln(2.5^{m_{expected}-m_{observed}}) = (m_{expected}-m_{observed})*ln(2.5)\]With the equation updated, we can plug in with the values found in part A to solve for \(\tau\). \[\tau = (12-8.5)*ln(2.5) \rightarrow \tau \approx 3.5\]

(c) We can now compute the amount of dust required along our line of sight to produce the observed extinction. The optical depth along a line of sight can be calculated from the absorption cross section of any intervening material and the number density of the particles with that cross section. This can be written as:

τ = Nσ

where N is the total number of particles per \(cm^2\) in the line of sight and σ is the cross section of the individual particles.

Assume that each dust grain has the typical size of r = 0.1µm and is spherical. Calculate the geometric cross section of a dust grain in units of \(cm^2\). Assume that this geometric cross section is the absorption cross section, which is true for visible light. Determine how many particles per cm2 you would need to obtain the calculated optical depth. This value is referred to as a “column number density.

First we compute the circular cross section of the spherical particle as \[\sigma = \pi * r^2 = \pi * (1*10^{-5} cm)^2 \approx 3*10^{-10} cm^2\]With this cross sectional area we can find the particle density with the \(\tau\) we solved for earlier \[\tau = N* \sigma \rightarrow 3.5 = N*(3*10^{-10}) \rightarrow N \approx 10^{10} \frac{particles}{cm^2} \]

(d) We know that the mass in gas is about 100 times more than in dust. What is the column number density of gas along the same line of sight?

To solve for the column number density of the more massive gas, we have to start by solving for the mass column of the dust. To do this we first assume the density of the dust to be somewhat similar to rock on Earth \[\rho_{dust} \approx 1 \frac{g}{cm^3}\] We can then multiply by the volume of a dust particle to get the \(\frac{g}{particle}\) fraction. \[1 \frac{g}{cm^3} * 4*10^{-15} \frac {cm^3}{particle} \approx 4*10^{-15} \frac {g}{particle} \] Now we can multiply by the number density from part C to get the number of grams per cm^2 of the dust. \[4*10^{-15} \frac {g}{particle} * 10^{10} \frac{particles}{cm^2} = 4*10^{-5} \frac{g}{cm^2} \]Which is the mass column of our dust. Now working backwards to get back to to number density of the gas. We first multiply by the mass ratio for the gas to dust. \[4*10^{-5} \frac{g_{dust}}{cm^2} * 100 \frac{g_{gas}}{g_{dust}} = 4*10^{-3} \frac{g_{gas}}{cm^2} \]Finally we assume that the majority of the gas in space is made of hydrogen and thus can approximate the weight of one particle of gas to be the weight of one particle of hydrogen giving us the final number density. \[4*10^{-3} \frac{g_{hydrogen}}{cm^2} * \frac{1 \space particle}{1.7*10^{-24} g_{hydrogen}} \approx 2*10^{21} \frac{N_{gas}}{cm^2}\]

(e) What is the average gas density along the same line of sight? Express your answer as the number of particles per cm3.

To find the average gas density along the line of sight we simply must take our column number density and extend it the length of the column, or the viewing distance from the star, 200pc. \[200pc = 6*10^{20} cm\]Giving us an average density of \[2*10^{21} \frac{N_{gas}}{cm^2} * \frac{1}{6*10^{20} cm} \approx 3 \frac{gas \space particles}{cm^3} \]

Blog #10 - Trigonometric Parallax

Trigonometric Parallax 

One of the most common ways to measure the distance to a star is using the "trigonometric parallax." This works by measuring the angular distance a star appears to move, with respect to a background field of much more distant stars, as the Earth moves one quarter of an orbit (i.e. as the Earth translates "sideways" by a distance of b = 1 AU). 

There are 60 arcminutes in a degree and 60 arcseconds in an arcminute. What is the distance, measured in cm and light years, of a star that moves by 1 arcsecond when the Earth moves by 1 AU? Give a general formula relating the distance D to the angle, \(\theta\), moved by the star. 

We begin with a sketch of the situation described in the problem seen below:

In the drawing we see an exaggerated angle, \(\theta\) created when between when viewing a star from Earth's original position around the sun and then once the Earth makes a quarter orbit around the sun and is now directly vertical when original it was 1 AU away from the sun. 

First we must convert the distance of 1 arcsecond to degrees and then to radians in order to find a distance from the geometry of the situation. We convert with dimensional analysis \[\frac{1 \space degree}{60 \space arcminutes} * \frac{1 \space arcminute}{60 \space arcseconds} * 1 \space arcsecond = \frac{1}{3600} \space degrees \]Then converting from degrees to radians \[\frac{1}{3600} \space degrees * \frac{2 \pi \space radians}{360 \space degrees} \approx \frac{1}{200,000} \space radians \]With the radian found, we can use the geometry of the problem to find the distance to the star. \[tan(\theta) \approx \theta = \frac{1}{200,000} \space radians = \frac{1 AU}{d} \rightarrow d \approx 200,000 AU \]Converting from AU to cm and then to lightyears we get \[200,000 AU \approx 3*10^18 cm \approx 3 lightyear \approx 1 parsec \] This unit is known as a parsec because it is the trigonometric parallax (or geometric angle) passed through after a star moves 1 arcsecond when the Earth moves 1 AU. 

Moving this relationship to a general equation for trigonometric parallax, we get \[d_{Earth Translation} = \theta_{Star Translated}*d_{star} \]