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!