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 \[d_{Earth Translation} = \theta_{Star Translated}*d_{star} \]

Blog #9 - Magnitude Scale

Magnitude Scale

In order to quantify the entire night sky, astronomers for ages have used the intensity and position of stars. Over all of the electromagnetic spectrum, intensity of light can be polarized and a spectrum can be created that lists the magnitude of a specific star. The first astronomer to catalog the "magnitude" of stars was the greek thinker Hipparchus who used a scale system from 1->6 to list the brightness of certain stars and their locations. 1 labeled the brightest stars and 6 the faintest. As technology improved and statistical methods were refined it became clear that the scale between "apparent" magnitudes not linear, but actually a log scale. With spectrometers, astronomers found that a difference of 5 magnitudes actually corresponded to a 100-fold increase in brightness giving us the now-standard quantity of a 2.5 log step between magnitudes. For example, one step in magnitude is equivalent to \( 2.5^2 \approx 6.25 \) times brighter.

Now astronomers can detect incredibly distinct magnitudes down to the .01 difference. They have also extended Hipparchus' scale (1->6) to a much larger scope (-26.83 (Sun) -> 30 (faintest stars)). This gives \(\approx 10^23 \) ratio between the brightness of our sun and the faintest stars.

To quantify the magnitude scale a bit more, we call the "brightness" of a star in terms of luminosity and radiant flux. Radiant flux is defined as the amount of energy (over all wavelengths) that a star gives off per second that is received on a square meter detector. This flux is based on the luminosity of star which is simply the energy exuded per second. Thus the relationship between flux and luminosity is

Where r is the distance from the observed star to the detector, L is the luminosity of the star, and F is the radiant flux. This relationship can be pulled back to the apparent magnitude of a star through the following relationship of radiant fluxes between two comparative stars. 

Where F is the flux of a respective star and m is its apparent magnitude. Once the relationship between the apparent magnitude and flux is determined, astronomers use what is called "absolute" magnitude, or the apparent magnitude a star would have exactly 10 parsecs away from the observer. Considering the fact that a magnitude change of 5 corresponds to a flux difference of 100. Expanding this equation outward to distance we can get the final relationship between apparent magnitude, absolute magnitude, and distance to a star as: 

With these quantities in mind, astronomers can calculate distance to a star solely based on the energy they receive across all wavelengths. Similarly, if the distance to a star is known based on its geometry, brightnesses can be compared and the lifecycle of a star can be approximated. 

Blog #8 - Solar Luminosity

Solar Luminosity 

Use a lamp fitted with a 100 Watt incandescent light bulb to estimate the luminosity of the Sun. Do this with the knowledge that:

• The Earth-Sun distance is 1 astronomical unit (AU), or \(a = 1.5 × 10^{13} cm \)

• An estimate of the distance from the bulb you need to hold your hand at for it to feels like a sunny Spring (low-humidity) day.

This problem is fairly straightforward as we simply need to compare the flux and luminosity of both the sun and the 100W lightbulb. We begin with the set up of the problem and the equivalence of solar and light bulb flux. \[F_B = F_{\odot}\]We can then use the conversion from flux to luminosity \[F_B = F_{\odot} \rightarrow \frac{L_B}{4* \pi * D_B^2} = \frac{L_{\odot}}{4* \pi *D_{\odot}^2} \]We know that the luminosity of the bulb is 100W and that the distance to the sun, \(D_{\odot} = 1.5*10^{13} cm\). We then need to make an assumption about the distance from the bulb to our hand on a spring day to simulate the heat transferred from the Sun. We will call that distance 10cm as an estimate. With that information we simply solve our equation to find the luminosity of the Sun. \[ \frac{100W}{(10cm)^2} = \frac{L_{\odot}}{(1.5*10^{13}cm)^2}\]Solving this equation gives us the Sun's approximate luminosity of \[L_{\odot} = 2.25*10^26 Watts \] Which is very close to the measured solar luminosity of \(3.828*10^{26} Watts \) 

Blog #7 - Y Dwarfs

Y Dwarfs

Recently, a new type of astrophysical object known as the Y dwarf was discovered. Y dwarfs are a subclass of brown dwarfs that we'll learn about later in the term. Consider a Y dwarf residing near a Sun-like star (where near refers to a small angular separation between the two objects on the sky). Y dwarfs have a temperature of about 350K (for reference, 0 degrees Celcius corresponds to 273K), and as a result they are extremely faint.

a) At what wavelength should observe to have the best change of detecting the Y dwarf? 

Using Wein's displacement law:

and having the information the Y dwarfs have a temperature around 350K, we find a wavelength of \[\lambda_{max} = 8285nm \] which puts the peak wavelength in the long-wavelength infrared range. 

b) Assume the star has a radius equal to that of the Sun, \(R_{\star} = R_{\odot} \), and that the Y dwarf has a radius equal to Jupiter's \(R_Y = R_{Jup} \). As measured at \(\lambda_{max} \), how many photons per second, per \(cm^2\) emitted from a Y dwarf at a distance of 30 light years would reach the Spitzer space telescope in a 1 \(\mu m\)-wide wavelength interval near \(\lambda_{max} \)? 

We begin by solving for the standard blackbody equation at our given Y dwarf temperature (350K) and \(\lambda_{max} \) (8285 nm). 

which, when plugging in constants: \[h = 6.626*10^{-34} \quad c = 3*10^8 \quad k = 1.38*10^{-23} \] gives us \[B_{\lambda}(T) = 7.14*10^6 \frac{W}{Sr * m^3} \] Converting this to new units we get \[B_{\lambda}(T) = .714 \frac{erg}{s*cm^2* A * sr} \] With this information we can now solve for the total number of photons that will strike at a 1 \(\mu m\)-wide wavelength on our space telescope. 

Now we must solve for the steradian size of the planet based on its radius and the distance from the telescope. Doing simple trigonometry with the distance and radius (similar to Jupiter's radius) at a small angle approximation we get \[tan(\theta) = \frac{2*R_{Jup}}{31 \space light \space years} \] where \[R_{Jup} = 69.911*10^6 m \quad and \quad 31 \space light \space years = 2.93*10^17m \]

This gives an angle of \[\theta = 4.77*10^{-10}\] which using the relationship of \(steradian = radian^2 \) we get \[d \Omega = 2.277*10^{-19} \] Now we can multiply our \(B_{\lambda}(T) \) term by both the steradian value and the value of our 1 \(\mu m\)-wide wavelength range to get a value in terms of \(\frac{erg}{s * cm^2}\)

Now our value becomes \[B_{\lambda}(T) * 2.277*10^{-19} \space steradians \space * 10,000 angstroms = 1.625*10^{-15} \frac {erg}{s*cm^2} \] But we want our value in terms of how many photons not how much energy in the form of ergs. So we find the average energy value of a photon at our \(\lambda_{max} = 8285nm \) which comes out to be \[E = \frac{h*c}{\lambda} = 2.4*10^{-20}\frac{joules}{photon} = 2.4*10^{-13}\frac{ergs}{photon}\] We can now finally do our the last calculation to find the number of photons that strike the Spitzer space telescope and find \[\frac{1.625*10^{-15} \frac{erg}{s*cm^2}}{2.4*10^{-13} \frac{ergs}{photon}} = .0067 \frac {photons}{s*cm^2} \]

c) How many photons arrive from the sun-like star (T = 5800K) in a 1 \(\mu m\)-wide wavelength interval near \(\lambda_{max}\) of the Y dwarf? 

Following the same process but for a sun-like star with temperature 5800K. From there we can compute our \(B_{\lambda}\) We find \[B_{\lambda} = .029 \frac{erg}{s*cm^2* A * sr}\] We continue to follow the same procedure as part B and find that the steradian value is \[d \Omega = 2.255*10^{-17}\] where \[R_{star} \approx R_{sun} = 695.7*10^6m\]Again multiplying through our \(B_{\lambda}\) value by the steradian value and the 1 \(\mu m\) wavelength range value we get \[B_{\lambda}(T) * 2.255*10^{-17} \space steradians \space * 10,000 angstroms = 6.57*10^{-15} \frac {erg}{s*cm^2} \]And again finding the energy at our peak wavelength of 8285nm, we get \[E = \frac{h*c}{\lambda} = 2.4*10^{-20}\frac{joules}{photon} = 2.4*10^{-13}\frac{ergs}{photon}\] and get a final result of \[\frac{6.57*10^{-15} \frac{erg}{s*cm^2}}{2.4*10^{-13} \frac{ergs}{photon}} = .027 \frac {photons}{s*cm^2} \]

d) What is the flux ratio of the Y dwarf to the star near \(\lambda_{max}\)? This should illustrate why it's so difficult to detect sub-stellar companions around sun-like stars. 

To find the flux ratio between the two bodies we simply divide out their fluxes found in parts b) and c). \[\frac{.027 \frac {photons}{s*cm^2}}{.0067 \frac {photons}{s*cm^2}} \approx 4\] This explains why its so hard to detect sub-stellar companions around stars given that the star typically emits almost 4 times as many photons at the Y dwarf's peak wavelength (even though it is nowhere near the star's peak wavelength). 

Blog #6 - Reading Summary - Blackbody Radiation

Blackbody Radiation

If you look up at the night sky, it becomes immediately evident that stars are not created equally. Some shimmer bright blue while others tend towards a darker red hue. In 1792, the first steps were taken to creating a link between the composition of stars and their color by Thomas Wedgewood. A porcelain maker at the time, Wedgewood began to notice that regardless of the size or shape of his kilns, the turned red at approximately the same temperature. Physicists began to believe that temperature was directly related to the emission of colored visible wavelengths. Eventually the term for a general mass that can perfect absorb and emit EM energy was dubbed a blackbody. To a rough approximation, planets and stars act as blackbodies allowing astronomers to make large leaps in the categorization of the universe based solely on temperature of stars and their surrounding planets.

As technology advanced, full spectrums were plotted and, as expected, the general shape of blackbody curves with respect to the emitted wavelength was constant and only the magnitude of the energy at said wavelengths changed based on the temperature of the body. Eventually laws were found to characterize the blackbody spectra. Wein's Displacement law relates the wavelength at which the blackbody has maximum energy to it's temperature with a relation of:

Josef Stefan then went on to bring a standardized equation to relate the energy of a blackbody to its temperature. The term, known as luminosity is simply related to the surface area, temperature and a constant (known as the Stefan-Boltzmann constant) as seen below. 

Where 

With these few equations astrophysicists are able to characterize a star with a number of techniques including spectrometry, or the separation of light into its spectrum. With just a wavelength or temperature as well as a relative size of a body (through distance analysis or comparison to nearby bodies) astronomers can get the relative age, temperature, and energy of a blackbody. 

Blog #5 - Worksheet 4.3 Interferometers

Interferometers

Interferometers are giant telescopes comprised of multiple smaller telescopes. Imagine two 1-meter telescopes in the CHARA array, separated by 330 meters (you should look CHARA up later). Use the double-slit experiment to think about how these two telescopes might be able to measure the angular diameter of a star with \(R = 5*R_\odot \) at a distance of 150 light years. What is the maximum wavelength \(\lambda\) at which this star can be resolved?

From the double slit experiment earlier in the worksheet, we discovered that angular resolution follows a simple relationship of \[\theta = \frac{\lambda}{D}\] From this we can extrapolate out to the CHARA example by considering the 330m separation to be the distance between our "slots", the 150 light years to be the distance L to get to the detecting plane, R to be the distance y between constructive light points. (See diagram below).

To begin we must convert all units to the same standard. The length to the star becomes \[150 \space light \space years = 1.42*10^18 m\] and \[5*R_\odot \approx 3.5*10^9 m \] Thus we can begin to solve for theta by finding the tangent. \[tan \theta = \frac{3.5*10^9}{1.42*10^18} \] which when using the small angle approximation of \(tan \theta = \theta \) we end up with \[ \theta = 2 *10^{-9} \space radians\] Now with \(\theta\) we can solve for the wavelength using the above equation, \(\theta = \frac{\lambda}{D}\). \[\theta = \frac{\lambda}{D} \Rightarrow 2*10^{-9} = \frac{\lambda}{330m} \] Which gives us the desired final quantity: \[\lambda_{max} \approx 700nm\] which means that we would be observing this star as a dark red visible light!

Blog #4 - Reading Summary - Seeing

Seeing 

Despite the advanced technologies of telescopes in their ability to improve resolution and focal length through improving optics, there are still some factors for telescopic observation that are out of humankind's hands. One of these important factors is known simply as seeing. Seeing is the ability to get a clear image through the natural distortion caused by changes in density and temperature in Earth's atmosphere. Changes from the centimeter to meter scale can cause significant decreases to an image's resolution. Seeing is actually what causes stars to "twinkle" when the light from a point source star is diffracted through the atmosphere.

I found it particularly interesting that seeing is the main reason that many observatories are located where they are. I have always heard that remote places like Hawaii and Chile have "better" night skies but did not realize that these locations have less dense atmospheres and less variable atmospheric temperatures giving the incoming light waves less distortion on their way to large optical telescopes.

In fact, the best location for seeing is out of the atmosphere all together! The telescope currently with the least diffraction is the Hubble Space Telescope located 380 miles above the surface of the earth. At this location there is nearly 0 diffraction limitations and only manmade errors in the grinding of telescope mirrors cause obscuration of the incoming light from galaxies, stars, planets, and other celestial bodies. According to calculations in the book, the resolution of the Hubble Space Telescope is .0127'' or enough to see a quarter from over 400km, a true mind bending thought.

Blog #3 - Local Sidereal Time

Local Sidereal Time 

The Local Sidereal Time (LST) is the right ascension that is at the meridian right now. LST = 0:00 is at noon on the Vernal Equinox (the time when the Sun is on the meridian is March 20th, 2017).

a) What is the LST at midnight after the Vernal Equinox?

In class we solved that there is a 1° change in earth's meridian every day due to its rotation around the sun and its own rotation period. Thus there is roughly a 4 minute difference between Universal Time (UT) and LST. Considering midnight is half a day past noon and the LST is 0:00 at noon on the Vernal Equinox, we know that midnight must actually occur at 12:02. the extra two minutes occur from half of the 4 minute LST difference.

b) What is the LST 24 hours later (after midnight of part a)?

Following the same logic as part 'a', 24 hours later would produce the expected 4 minute difference between UT and LST giving us an LST of 12:06 at midnight the next day.

c) What is the LST right now (to the nearest hour)?

At the time the worksheet was completed, the time was 3 o'clock on January 30th. This produces a 48 difference from the metric of March 20th for the Vernal Equinox. \(48 days = 192 minutes = 3 hours 12 minutes\) and we must move back in time from the future Vernal Equinox. Thus we move back 3 hours and 12 minutes to get a time of roughly 20:48. However this would be the LST for noon on January 30th. We want 3 o'clock and thus add 3 hours (and technically 30 seconds to correct for the 4 minute/day difference). This brings the LST to 23:48:30, which rounds up to 0:00 at the nearest hour.

d) What will the LST be tonight at midnight (to the nearest hour)?

Following the same process as part 'c' we know that at 3pm on January 30th the LST was 23:48:30 which rounded up to 0:00. Moving forward 9 hours to get to midnight and adjusting for the 1.5 minutes to correct for LST we get 8:50 which rounded to the nearest hour gives us 9:00.

e) What LST will it be at the Sunset on your birthday?

My birthday is November 18th which is 243 days after the Vernal Equinox on March 20th. 243 days converts to 16 hours and 12 minutes of LST variation from the UT given the 4 min/day rate found earlier. Moving up from the 0:00 noon time on March 20th, noon on November 18th would occur at 16:12 instead. If we assume the sun sets around 6pm we add 6 more hours to this time and the appropriate 1 minute for LST adjustment and end up with 22:13 for sunset on my birthday.