Today's Problem: Dim Stars at Great Distances

What is the power output of a star that is 100 light years away that you can barely see from a dark site at night?

This problem is, in part, informed by another one that isn't on my blog, so some further background info is called for. Part of the setup of that problem was the following: "The eye must receive ~ 10 photons in order to send a signal to the brain that says, 'Yep, I see that.'" Pretty straightforward, right? Maybe, except you could interpret that a few different ways. 10 photons every yoctosecond is going to be plenty more light than the brain would need (or, I imagine, want). 10 photons every year probably isn't going to cut it. We figured that if the brain can register frames of a movie running at around 30 fps, it needs somewhere on the order of 300 photons per second. Obviously if those photons aren't being emitted in visible light, that won't be too helpful either, so we figured a wavelength of 500 nanometers. The tools we have to work with, then, are as follows:

-The rate $R$ of photons received, 300 photons/s

-The wavelength of our photons, 500nm, or $5\times10^{-7}$ meters.

-The approximate area $A$ of a pupil (appropriately dilated for night observing), 1 cm$^2$, or $10^{-4}$ meters$^2$.

-The distance the the star, $D = 100ly$

-The speed of light $c$, $3\times10^8$ meters/s.

-Equation $(1)$: the energy of a photon, known to be $E = h\nu$, where $\nu$ is the photon's frequency and $h$ is Planck's constant $h = 6.6\times10^{-28} ergs/s$. 

-Equation $(2)$: the relationship between frequency and wavelength, $\lambda = \frac{c}{\nu}$.

-Equation $(3)$: flux, the amount of energy received in a given area over a given time, is given by $F = \frac{L}{4\pi D^2} ergs/s$ for a source of luminosity $L$.

That last equation comes from the fact that light propagates based on an inverse-square law. In other (more useful) words, if you imagine a star emitting light, you can imagine a spherical shell exactly 100 light years in radius consisting of all of the light that star emitted exactly 100 years ago. The amount of light that you observe is equal to the area of your detector (in this case, your pupil) divided by the total surface area of the spherical shell, $4\pi D^2$.

What we're ultimately looking for is the luminosity, $L$, of the star emitting the light we're observing. We can find this by using that last equation--we know $D$, so what we need first is $F$. For that, we'll need a way to translate photons received per second every second into energy per second--or, more precisely, photons into energy. We know that for a single photon, $E = h\nu$, and so our first step will be to deal with $\nu$, the frequency of our photons. Since we have the wavelength of the photons, we can plug it into $(2)$, which allows us to substitute $\frac{c}{\lambda}$ for $\nu$, giving us the following:\[E = \frac{hc}{\lambda}\]For 300 photons per second, we can simply multiply this by $R$ and and units of inverse seconds, giving us power. Dividing that by A gives us our flux as\[F = \frac{Rhc}{\lambda A}\]Now we can take this back to equation $(3)$ and do some algebra to solve for $L$. \[ \frac{Rhc}{\lambda A} = \frac{L}{4\pi D^2}\]\[L = {Rhc\times4\pi D^2}{\lambda A}\]This is worth unit checking before taking it as gospel. The units on the left side are power, or energy$*$time$^{-1}$. On the right we have $\frac{time^{-1}\times energy \times time^{-1}\times length \times time^{-1} \times time^2}{length^2 \times length}$, which also comes out to energy$\times $time$^{-1}$. So we're all set! At this point, all we need to do is plug in numerical values for all of the constants on the right side of the equation. That gives us the following:\[\frac{3\times 10^2 \times 7 \times 10^{-28} \times 4\times \pi \times 10^{36} \times 3\times 10^8}{10^{-4}5\times10^{-7}}\] Which evaluates to just about $2\times10^{31} ergs/s$. That's a big number, but it's actually pretty dim--this is a star with only about one percent of the luminosity of the sun. However, we're assuming otherwise uniformly dark skies, and operating at the bare minimum of your eye's sensitivity. It's a stretch, but in conditions as ideal as the ones we're describing, it might be possible.

We came into this problem with a fair amount of outside knowledge, particularly in the form of our equations (Planck's constant isn't something we'd be likely to estimate well in a casual experiment). However, with just a few relatively simple tools on hand, we were able to determine a pretty significant property of a hypothetical faraway star--pretty cool. As usual, I had help on this one, and once again credit goes to Anne Madoff, Jennifer Shi, and Louise Decoppet for working through this problem with me.

Estimations

In 240 BCE, Eratosthenes is thought to have measured the size of the Earth to within an error of either 1.6% or 16.3% (it depends on whether he was using an Egyptian or Attic Stadion). Even the latter isn't bad for someone whose tools were just well, a stick, and a pretty sketchy measurement of the distance between two cities. However, we're supposed to be smart, modern scientists, so let's see if we can do even better.

If we're told that the time difference between Los Angeles and Boston is 3.5 hours, and the two cities are 3,000 miles apart ($3\times10^{3}$ miles if we're going to be scientific about this), then we've functionally been given the same pieces of information Eratosthenes had--distance over ground between two points (let's assume it's along a great circle) and what fraction of the earth's circumference that distance is. Let's start by turning those hours and miles into slightly more useful units. The hours we can actually leave alone--we're just going to be using them in a ratio, so they'll cancel on their own. Miles, though, aren't too helpful, so let's get rid of those $3\times10^{3}$ miles and call them $5\times10^{8}$ cm. Now all we need to do is set up a pretty simple pair of ratios (since we're relating 3.5 hours to 24 hours, we're going to have to relate our distance over ground to the circumference--$2\pi R$ rather than just the radius itself):\[\frac{3.5}{24}=\frac{5\times10^{8}}{2\pi R}\] Now all we need to do is solve for $R$ in there and we come out with a radius of $6\times10^8$ cm. Not bad--that's an error of just under 6%. So in the interest of preserving our egos, let's say Eratosthenes' error was 16.3%.

This is subject to a fair few errors--the biggest of these is probably that we aren't relying on the altitude of the sun like Eratosthenes was, but on a time difference, which assumes that the distance between Boston and Los Angeles is purely longitudinal (that is to say, the two cities are the same distance from the equator and you could travel between the two along a line of latitude). If it doesn't seem like this could matter, imagine if we had traveled between Anchorage and Honolulu. The two are also about $3\times10^3$ miles apart, or $5\times10^8$ cm. But they're only one time zone apart. Replace the 3.5 hours in that equation from earlier with one hour, and you get a radius of $2\times10^9$ cm, and suddenly we're off by an order of magnitude. Of course, we could also have used Montreal and Bogota, $3\times10^3$ miles apart but in the same time zone, which would mean plugging zero into the equation and blowing the whole thing up. But Boston and Los Angeles are close enough to the same latitude that the estimation stays reasonable

And now with a couple extra pieces of equipment and information, this gives us a chance to estimate a whole bunch of other useful constants. Let's start by figuring out the mass of the Earth. Aside from the radius we figured before, we'll just need a scale, a measuring cup with some water in it, and a rock. The scale is to weigh the rock, the water is to measure its volume (by seeing how much it displaces), and the rock is to accurately represent the average density of the entire Earth (estimating, remember?). Our rock came in at 90 grams and 30 milliliters (or cubic centimeters), giving us a density of 90 grams/cm$^3$. Meanwhile, we can figure the volume of the Earth with the radius we calculated, since our planet is more or less spherical. The volume of a sphere is $\frac{4}{3}\pi r^3$, so plugging $6\times10^8$ in for $r$ gives us about $9\times10^{26}$ cm$^3$. Mass is just density times volume, so we just multiple three by nine, multiply that by $10^{26}$, and come out with just under $3\times10^{27}$ grams. We're off by about half here--not anywhere near as nice as our radius estimation--but we've got the right order of magnitude.

So let's throw another celestial body into the picture. Kepler's third law tells us that that if you take the time it takes the moon to orbit the Earth (a month or so--let's call it 30 days), and cube it, that's going to equal the distance from the Earth to the moon squared (along with a few other constants--in this case, the mass of the Earth, Newton's gravitational constant G, pi squared, and a four for good measure). In equation from, that looks like this:\[p^2 = \frac{4\pi^2 a^3}{GM_E}\] If we're not too sure about this shady Kepler guy, or (more likely) aren't too sure about our memory, we can check to make sure the units on either side match up (as long as we can remember the insane units of $G$, which are $\frac{distance^3}{mass*time^2}$. \[time^2 = \frac{distance^3}{mass*\frac{distance^3}{mass*time^2}}\] That simplifies to $time^2 = time^2$, so it looks like we're good! We can put 30 days in seconds (it's about $3\times10^6$), and if we remember that G is $7\times10^{-8}$ in cgs units, we've got everything we need to solve for $a$. It comes out to $2\times10^{10}$ cm. Again, we're off by a bit half--but given that we already knew that our mass was half off, that means everything else here is close to perfect!

With that distance in hand, along with one more little fact and some high school math, we can figure out how big the moon is, too. Aristarchus had a pretty cool way of doing this a few years earlier even than Eratosthenes figured the size of the Earth, using the size of the full moon relative to the Earth's shadow during a lunar eclipse (figure how large the diameter of the moon is relative to the diameter of the Earth's shadow, and you know the size of the moon relative to the size of the Earth). We'll do things a bit differently, using a frequently-seen (and incredibly simple) mathematical tool in astronomy: the small angle approximation. A full moon covers up about half a degree on the sky, which means you can draw a right triangle from an earthbound observer to two opposite edges of the moon. The vertex of that triangle poking into the Earth has an angle of half a degree. The tangent of that angle will be "opposite over adjacent"--in this case, the diameter of the moon over the distance from the Earth to the moon. But with a sufficiently small angle (and half a degree is sufficiently small), the tangent of an angle is about equal to the angle itself in radians. The trickiest part of the process is just the degree-radian conversion, which is easy enough with a bit of cross-multiplication. \[\frac{.5}{36}=\frac{\theta}{2\pi}\] Solving for $\theta$ tells us that half a degree is 1/120 radians. Now we can plug that into a simple equation using that small angle approximation. \[1/120 = \frac{2r_{moon}}{2\times10^{10}}\] This comes out to a radius of about $10^{8}$ cm, again just under half off but the right order of magnitude.

Finally, knowing that the moon is made up of more or less the same material as the Earth, we can figure its mass the same way we did the Earth's. Calculating its volume using the radius we just figured, we get \[\frac{4\pi(10^8)^3}{3} = 3\times10^{25} \: cm^3\]Multiply that by the density of 3 grams/cm$^3$ from earlier, and we end up with a mass of $10^{26}$ grams, an error of about 1/3.

I may have been able to work through this alone, but fortunately I didn't have to, so credit must go to my outstanding colleagues Anne Madoff, Louise Decoppet, and Jennifer Shi.

Introduction

My name is Tom Leith, and while I'm starting this blog at the outset of an introductory Astronomy course at Harvard, I could just as easily have started it five years ago--or even longer. I've loved outer space since I was a little kid, and instead of getting over that fascination like most sane people, I only became more and more interested as I grew older. The moment my house had a functional internet connection, my parents noticed that Space.com and NASA.gov had sprung up in their bookmarks, joined a short time later by NASA's Astronomy Picture of the Day (my background is an old Astronomy Picture of the Day, originally located here). I grew up in a house about a block away from the Harvard-Smithsonian Center for Astrophysics, and began religiously attending that institution's monthly Observatory Night public lectures--which were followed by stargazing through telescopes on the building's roof.

I'm embarrassed to admit that I don't remember what the first lecture I attended was about, but I was fortunate to be attending on a clear night, and I vividly remember the stargazing that followed. There was, as always, a big crowd, which meant that you were fortunate to get more than a few seconds on any of the few telescopes that were set up. Which was a shame, because Saturn was up, and I honestly think I could look at Saturn for hours. I felt an almost electric jolt seeing it for the first time--I'd seen thousands of pictures of it before, but I just couldn't believe that I could look through a telescope and just see it hanging there looking like that. Phil Plait, a science and astronomy blogger at Slate.com, wrote a great article not too long ago called "Opening the Sky to Everyone." I am, astonishingly, not a prostitute from the Bronx (the subjects of the article are), but what he had to say--about getting people to look up and the effect that having a chance to look at the sky through a telescope can have--resonated with me, and reminded me of that first look I had through a telescope at Saturn.

Five years ago my grandfather, Bill Leith, died. He was, among other things, an avid amateur astronomer, and I had always wanted to spend a night stargazing with him at my grandparents' house in rural Vermont. One of his most prized posessons was a beautiful Questar 3.5" reflector, which was as eager to share with me--the only one of his grandchildren to have inherited his interest--as I was to use it. Unfortunately, we never got the chance to. But after his death, the telescope found its way into my hands, and at 2:30 am on a brutally cold February night in high school I clambered through a skylight onto my parents' roof and pulled the telescope and its tripod after me. I had thus far done little more than flip through the manual and examine the telescope, and what followed was what seemed like an hour of trying to get the tripod set up, the telescope mounted on it, and figure out where all the knobs and levers to focus the telescope, switch back and forth from finder mode, and engage the Barlow lens belonged. Remarkably, with just an old red flashlight to examine the manual by, I figured it out. Even more remarkably, I didn't get frostbite. It was the first of many nights I spent using the telescope, both in the city and wherever I could find a dark sky.

This is definitely a longer introduction than I was planning to write, and probably longer than most folks will have the patience to read, but I think it informs in a very real way why I'm studying Astronomy now that I'm in college. I never had the chance to in high school, at least not in a formal way, and the idea that I could center an academic experience around what has always been very special to me is an exciting one. I'm not great at math, and I'm far worse still at physics. Those things worry me (a lot, considering that I have at least two more pure physics courses to go as an undergrad), but I've begun to see them as tools that I do have the persistence, if not the raw intelligence, to grasp in order to better understand a subject matter that has fascinated me my entire life. My hope for Astronomy 16 is not only that I will be a better and better-informed amateur astronomer--as much as I've emphasized my own observing here, it's really only one side of the coin. More than anything, I think I simply want to gain insight into what's out there, by accumulating raw information itself and even more so by assembling the toolkit to take that information, put it in context, and be able to turn it into real understanding.