# Brightness - The Intro When it comes to light, we humans have two settings we use - "bright" and "dark". While applicable on human scales, scientists aren't very happy with this - meaning that they've had to go out and make their own scale, just to prove that they're quirky. With a table lamp, even you can find its 'brightness' in scientific terms - just take the power incident across its surface! Welcome to the land of light! Don't go blind, see the light, tread the path between street lamps and stellar furnaces. Onwards! ![[Pasted image 20240202115757.png]] # Luminosity Now, the scientific word for 'brightness' is luminosity. Just by applying the 'power per unit area' logic, we get a quick expression for the *flux*, which we'll go into later on. For our purposes, the *luminosity* is the **total power incident on the surface**. Some of the biggest 'lights' in the universe are [[Stars (Astro)|stars]] [^1] - big, stellar furnaces that churn out energy. Let's figure out how to find their luminosities - from our point of view!! First, let's take some constraints. Answers in the callout, but try finding some yourself first! >[!Success]- Deriving the Luminosity >Ask yourself some questions: how can we take the *power incident across an area?* > >1. A star is a sphere. When a body emits, it does so along its surface area. We can use the equation $A = 4\pi r^2$ for a sphere's surface area, so what now? > >2. If we're looking at the star point-blank, the power incidence is given by the equation $P = \sigma T^4$. [[Heat Transfers (Thermo)#Radiative Processes - The Stefan Boltzmann Constant|More about this here!]] > >3. The power incidence depreciates as a square relationship over time - an inverse-square law. After a bit of back-and-forth with the math, you should find that the luminosity is: $L = \frac{4\pi r^2\sigma T^4}{d^2}$ Where $d$ is the distance of the object relative to us. Notice the division by the distance squared at the bottom - it's another example of the *inverse-square law!* As the equation suggests, this means that we have to find the distance, either by the [[Celestial Measurements (Astro)#Understanding the Parsec|parallax method]] or using luminosity functions based on the predicted spectrum. >[!Abstract]- Example Question >A star has a radius of ## Bolometric Luminosity While luminosity's all well and complicated, why not add on the *rest of the spectrum* on top of it? Stars are [[Blackbody Radiation (Thermodynamics)|black-body radiators.]] Unfortunately, this means that the Sun isn't strictly yellow - it's actually white, given that it emits as a spectrum! To us, this 'whiteness' occurs due to the sun's black-body peak being at the visible light wavelengths, By emitting as a spectrum, we can fit the spectra of [[Intergalactic Dust (Astro)|heavily-reddened]] stars, ignoring the spectrum of higher wavelengths which are more easily scattered. ![[Pasted image 20240128144236.png]] >[!Example]- Example Question >A star has a temperature of 8000K and a # Flux Flux is the function of the energy per unit area second. This means it measures the amount of energy impacting a body or being emitted from a body. To test this, let's design an experiment! Think of an aluminium plate with easy dimensions - say, 1x1 meters, and place it on the ground, out in the sun. ![[Pasted image 20240128123514.png]] You don't want to be touching that plate. Trust me. Moving back to the experiment, you can calculate the incident solar flux over the area by matching it (approximately!) to the heat energy gained by the plate: $\frac{\Delta T}{\Delta t} = f \frac{A}{c \cdot m}$ **Where:** - $f$ is the flux in watts - $A$ is the surface area of the plate - $c$ is the specific heat capacity - $m$ is the mass of the plate - $T$ is the temperature Remember - the [[Kinetic Theory & Ideal Gases (Thermodynamics)#Internal Energy Electric Boogaloo - Molar Specific Heats|heat energy]] of the plate is given by $E = mc\Delta T$. Let's rearrange it to get another expression for the flux! $f = \frac{E}{At}$ This gives the solar incident on a certain surface area over a period of time. This value changes during the winter or summer (with varying degrees depending on where you live) and over the course of the day. >[!Abstract]- Linking Flux and Luminosity >Flux is proportional to the fourth power of temperature. This gets us an equation: >$f \propto T^4$ >The constant in the proportionality is the [[Heat Transfers (Thermo)#Radiative Processes - The Stefan Boltzmann Constant|Stefan-Boltzmann constant]], denoted by $\sigma$. This changes our equation, giving us: >$f = \sigma T^4$ >Since this is the solar flux over one unit area, we need to multiply this equation by the formula for the surface area of a sphere (which is approximately a stellar object), giving us a formula for luminosity $L$: >$L = 4\pi r^2\sigma T^4$ >Where $r$ is the radius of the body that is radiating. # Magnitude ## Intro - Science, Destroyer of Logic Scientists. They always have to be quirky, do they? From the philosophers of Ancient Greece to the engineers complaining about their homework, science involves diverse groups of people, all vying for the few titles that cement you as one of the greats. ![[Pasted image 20240206083352.png]] From life's greatest mysteries and a burning desire to answer questions comes forth the greatest theories. In 129 BC, Hipparchus, an ancient greek astronomer, noticed that all the stars in the night sky (light pollution wasn't as severe back then!) were of different brightnesses - and so he classified them into categories - the brightest being classified as "1" and the faintest "6". With the advent of the telescope, we're now able to peer far into the twenties and even thirties of magnitude, as well as into stars that shine in wavelengths too dim for us to see. So without further ado, let's go learn what all of that means! ## Apparent Magnitude Remember - the magnitude scale goes in increments of *5* for every 100-fold decrease in brightness. This tells us two things: 1. *The scale is inverse.* 2. **The scale is logarithmic.** Essentially, the equation is *empirical* - it swings on the definition of the magnitude scale itself, not on any fancy mathematical proofs! $m - m_{ref} = -\sqrt[5]{ 100 }\log_{10} \left(\frac{f}{f_{ref}}\right) $ This equation is also called **Pogson's Equation** Simple as that! What you see is what you get... ## Absolute Magnitude Mind you, this often isn't enough for certain surveys - even though we can get the flux of a star, we still want a baseline to go off of! To address this, we've developed the **absolute magnitude** - allowing us to adjust the apparent brightness of a star if viewed from 10 [[Celestial Measurements (Astro)#Understanding the Parsec|parsecs]]. Absolute magnitude is given by the equation: $M - M_{ref} = -\sqrt[5]{ 100 }\log_{10}\left( \frac{L}{L_{0}} \right)$ Where $L_{0}$ is the **zero-point** luminosity of the object, where the bolometric luminosity of a star 10 parsecs away would be equal to the total (across all wavelengths) electromagnetic flux of our sun. $L_0 = 3.1028 \times 10^{28}$ Mind you - it's the same thing as the absolute magnitude. There's not much else to say... >[!Tip]- Bolometric and Visible >Remember - the sun [[Blackbody Radiation (Thermodynamics)|emits as a spectrum.]] This means that the actual luminosity per-band is going to change! > >For example, the bolometric luminosity of the sun is the value above. The visible luminosity's different though - it's actually an order of magnitude or two less, at $3.846 \times 10^{26}$. Depending on the question, this is going to change the values you get - so be warned! >[!Example]- Distance Modulus Problem Taster >[[Cepheid Stars (Astro)|Cepheid Variables]] are variable stars used to extrapolate intergalactic distances. They were instrumental in ## The Distance Modulus To find the difference in apparent and absolute magnitudes of a star, we can use the **distance modulus**. $m-M = 5\log(D) - 5$ *Where:* - $m$ is the **apparent magnitude** - $M$ is the **absolute magnitude** - $D$ is the **distance to the star from the Earth, measured in parsecs.** There are two ways of rearranging this equation - one to find the apparent and another to find the absolute magnitude! Just get along with it - it's quite nice rearranging it! >[!Success]- Apparent Magnitude Equation To find the apparent magnitude $m$ from the distance modulus equation, we can rearrange the equation as follows: $m = -5\log_{10}(d) + 5 + M$ >[!Success]- Absolute Magnitude To find the absolute magnitude from the distance modulus equation, we can rearrange the equation once more: $M = m - 5\log_{10}(d) + 5$ >[!Abstract]- Proving the Distance Modulus >We can write the expression as a function of logarithms: >$m-M = -2.5\log_{10}(d^2) + 2.5\log_{10}(10^2)$ >*Where:* >- $m$ is the apparent magnitude >- $M$ is the absolute magnitude >- $d$ is the distance > >This gives us an expression after simplifying with log rules: >$m - M = -2.5\log_{10}\left( \frac{d^2}{10^2} \right)$ >More simplifying and cancelling-out then gives us our distance modulus! >$m - M = -5\log_{10}(d) + 5\log_{10}(10)$ >$m - M = -5\log_{10}(d) + 5$ ## The Curious Case of the Binary Star Binary stars are common. Heck, some scientists even believe them to be more common that our own sun, a single star. If that's not a call for action to you, then I don't know what is. ![[Pasted image 20240206091126.png]] *Three types of binary star system!* When we conduct all-sky surveys, such as ESA's [GAIA](https://gea.esac.esa.int/archive/), we take apparent magnitude data through the use of light incidents on certain filters. In our case, let's assume that we're taking data measurements in the G-Band - or the visible light spectrum. Amazing! $\frac{L}{L_{0}} = 10^{-0.4m}$ **Where**: - $m$ is the apparent magnitude - $L$ is the total flux (luminosity) of the star - $f_{ref}$ is the reference flux, which is the zero-point luminosity of the sun! $m_{tot} = -\sqrt[5]{ 100 }\log_{10}(10^{-0.4m_{1}} + 10^{-0.4m_{2}})$ >[!Abstract]- Flux Ratio Proof - Binaries >To get a better idea of this, let's first get an expression for the total power incident from the stars: >$\frac{f_{tot}}{f_{ref}} = \frac{f_{1}}{f_{ref}} + \frac{f_{2}}{f_{ref}}$ > >We can just substitute the exponential above - $10^{-0.4m}$ - to get the expression we've created for the total mass of the binary. >$10^{-0.4m_{tot}} = 10^{-0.4m_{1}} + 10^{-0.4m_{2}}$ >Since the absolute magnitude's directly related to the luminosity of the star, we can just make the addition equation - as seen in the initial ratio. We're done! # Extinction and Optical Depth in Magnitude *To be used in tandem with the article on* [[Intergalactic Dust (Astro)|interstellar dust!]] Space isn't empty. Dust litters the interstellar highways between us and far-flung stars, blotting out certain parts of their light by virtue of [[Rayleigh Scattering (Waves)|Rayleigh scattering.]] As a result, we can't rely on empirical values - we have to correct for the light that's blocked. Apparent magnitude runs under the principle that all the light from a star actually hits us, after all! Usually, we just add an additional term onto the distance modulus equation, as shown: $m-M = 5\log(D) + 5 + A$ Where $A$ represents the *extinction coefficient*. There's actually a reason for this - and that's because of something called the *optical depth* - which we'll go into more detail next! ## Optical Depth 'Next' in our case manifests in the section right under. Someone wasn't patient enough to make a new file for this! Anyways, when we look at light from a point source, we're looking at light that's passed through a lot of dust - one which we can approximate to be a cylindrical volume: ![[Pasted image 20240309175321.png]] We can approximate the volume to the star as being a cylinder with a cross-sectional area $A$. Each of the particles have a cross-sectional area of $\pi r^2$, which we call $\sigma$. The volume of the cylinder is therefore $LA$, and the fraction of the star covered by the dust particles **provided that there is negligible overlap between the dust particles** is: $f_{\sigma} = nLA\sigma$ Where $n$ is the (approximate) number of dust particles; it's impossible to ascertain the true number of 'particles' in real life. We can further simplify to get the fraction of the *light* blocked, so apply the concept that luminosity's based on a per-area basis! $f_{L} = \frac{f_{\sigma}}{A} = nL\sigma = \tau$ Which is our optical depth, $\tau$! To tie it to the extinction coefficient, we can just use the equation $A = 2.5\log e^{\tau}$ $A = 2.5\tau \log e$ Perfect! >[!Example]- Bombom's Problem Taster - POTD 23 >*Find dedicated astronomy POTDs on* [Bombom's dedicated Instagram Account!](https://www.instagram.com/astro.potd/) *Uploaded with permission.* > >A star with absolute magnitude $2.31$ is at a distance of 1.5 kpc. Due to absorption of interstellar matter, when seen from Earth this star has a magnitude of 14.59. If the interstellar matter (a collection of dust particles) between the star and the observer is evenly distributed and the size of the dust is about 0.3 $\mu m$, determine the average density of dust particles in that viewing direction. Ignore the scattering effect. > >**Solution:** >First use the distance modulus to calculate the extinction coefficient: >$m = M - 5\log(D) + 5 = 13.19$ >$A \approx 1.4$ >Ignoring the scattering effect allows us to take the extinction coefficient as equivalent to the optical depth calculation. Let's go ahead and set up our equation! >$1.4 = 2.5\tau \log e$ >$\tau \approx 1.29$ > >In our case, the number of particles $n$ is the number of particles per cubic meter. Let's substitute the equation $nL\sigma$ in first: >$\tau = nL\sigma$ >$n = \frac{1.29}{\frac{1}{4} \pi (0.3 * 10^{-6})^2 * 3.086 * 10^{10}}$ >$n = 3.93 \times 10^{-7}$ >Provided we convert parsecs to meters $n$ in our case will define the number of particles per meter - hence the density calculation. There's no need to overthink this question! *Note that this represents a "primer" for optical depth! Optical Depth's typically used as a means to measure the level of polarisation of some light - so laser physics, flying way above my high-school head again!* *Source:* http://burro.astr.cwru.edu/Academics/Astr221/StarPhys/opticaldepthprimer.html # Additional Information - The Solar Constant *Wave if you've come from the IB syllabus!* With the advent of solar panels came the need for a standardised way to measure the power incident on each part of our Earth from the sun. So, the idea of the **solar constant** was conceived. %%diagram of the astronomical sphere%% Roughly speaking, the solar constant refers to **the average power per unit area incident** **perpendicularly** on the Earth's surface. To calculate it, we can just use the simple formula: $k = \frac{L}{4\pi d^2}$ Where: - $L$ is the **luminosity of the sun**, with a value of $3.846 \times 10^{26} \ \mathrm{W}$ - $d$ is the distance between the Sun and the Earth, or 1 astronomical unit ($1.5 \times 10^{11} \ \mathrm{m}$) - $k$ is the **solar constant**, given in units of $\mathrm{W m^{-2}}$ Crunching the numbers gives us a rough value for the solar constant at $1360 \ \mathrm{W m^{-2}}$ for our Earth. With that much energy impacting our surface at any given moment, it's no wonder that solar energy's being touted as the future! Still, there's things to consider. Since the planet is very, VERY small with regards to the astronomical sphere it encloses, the area to which light is perpendicularly incident can be approximated to a perfect circle, meaning that the sun's light shines across an area of size $\pi r^2$, $r$ being the radius of the Earth. Still, heat must transfer across the entire surface of the Earth, which has a surface area of $4 \pi r^2$, so the **mean radiative intensity** is the incident intensity / 4, commonly written as: $S_{m} = \frac{S \pi r^2}{4\pi r^2} = \frac{S}{4}$ **So be careful when you do these questions!** If a question asks you for the **net heat gain across the entire surface of the Earth due to the sun shining on it**, remember to divide the incident power by 4 before putting the energy from the power into the [[Heat Transfers (Thermo)#Radiative Processes - The Stefan Boltzmann Constant|Stefan-Boltzmann Law!]] # Footers [1]: Note the "some". To learn about the true brightest lights in our night sky, take a quick look at [[Quasars(Astro)|Quasars!]] # What's Next? For the next instalment in the astronomy series, go here: [[Celestial Measurements (Astro)]] To get back to the original astronomy landing page, go here: [[Everything Astronomy]]